[book] Modern_control_theory [u.a Bakshi].pdf
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View [book] Modern_control_theory [u.a Bakshi].pdf as PDF for free.
More details
- Words: 69,788
- Pages: 386
Premieraz
(2·1!1D(2·72)
Cl>aofe< • 3
Matrix Algebra & Doriva!ion of Tronstor Funciul
(3 • lj lo (3. 46)
Cl>aote< • 4
Solutlcn of Sta:e EQualions
(4-llloj4-~)
Chapt., . 5
Pde Pl~.enlTechnque
(5· l)lo(5·00)
"' ,Features of 'Book
:-;-us-;~r~i;;;,-;1~i~-~d-1~-cid-l~;;;;;;;lli~-;h;~-ci;;s~ci~~-;;;;·~~~~-I I • Excellent theory well supported with the practical examples and
I
I
I
illustrations.
I
1 I • Important concepts arc highlighted using key points throughout tho book.
I
I
I • Large number of solved examples. I 1 • Approach of the book resembles classroom leaching. I • Use of informative, self explanatory diagrams, plots and graphs.
I
I : I
: • Book provides detailed insight into the subject. I •
1
Stepwise explanation to mathematical derivations for easier
:
understanding.
L------------·----··-·--·-··-·---------------·-------··--·--······--······ ' "' .~ ~ -~ ~~ '~
Best of Technical Publications
I
..,
1
As per Revised Syllabus of VTU - 2006 Course Semester - V [EEE) Modem Control Theory Signals and Systems Transmission and Distribution Linear !Cs and Applications D.C. Machines and Synchronous Machines
Bakshi, Godse' Bakshi' '
(4)
M
Modern Control Theory ISBN 9788184315066
,.,.rved
All righls wilh Technicol Publicottons. No pot! of this book should be reproduced in ony loon, Eledn>nic, Mochonlcol, Pholoc.opror or
Publishtt.f by : Thchnlcal l"Ublk:aUonsPune• #1,,......
~"""'•it.
Printer: Nc.OTPilotcn
s..... 1 (11),sw..,.d J\.w.41104l
Rood,
si-.-
PnJ.. "'-. •11 030, l.d..
Preface The importance of Control Theory is well known in various engineering fields. Overwhelming response to our books on various subjects inspired us lo write this book. The book is structured lo cover the key aspects of the subject Modem Control Theory. The book uses plain, lucid language to explain fundamentals of this subject. The book provides logical method of explaining various complicated concepts and stepwise methods to explain tho importonl topics. Each chapter is well supported with necessary illustrations, procticol examples and solved problems. All the chapters in the book ore arranged in o proper sequence that permits eoch topic to build upon earlier studies. All care has been token lo make students comfortoblc in understanding the basic concepts of the subject. The book not only covers the entire scope of the subject bul explains the philosophy of the subject. This makes the understanding of this subject more clear ond makes it more interesting. The book will be very useful nol only lo the students but also to the subject teachers. The students hove to omit nothing and possibly hove to cover nothing more. We wish to express our profound thanks to oil those who helped in making this book a realily. Much needed moral suppor1 and encouragement is provided on numerous occasions by our whole family. We wish to thank the Publisher and the entire teom of Technical Publications who hove taken immense pain to get this book in time with qualify printing. Any suggestion for the improvement of the book will be acknowledged and well appreciated.
Authors
(1
:ij Jo:(h 14)
1.1 Background
1.1.1 ~or 1.1.2 ~tages
1- 1
1 -1
Conventional Approach
of State Variable Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 • 2
1.2 Concept of State
1 •2
1.2.1 lmpajanJpelinitions
\ I'
:.................
•
1.3 Stal? MY1!el-pf Lin.ear Systems
:
1.3.1 Stale Model of Si)!lle Input Single OutputSystem 1.4 State Diagram Representation
.
: •...•........
.
\
1 -4 1 -6 1 •9 1 -9
1.4.1 State Diagramof Standard State Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 • 10 1.5 Non Uniqueness of State Model
1·. 11
1.6 Llnearization of State Equation
1 -12
B.eYlew. Questions............ ..
..
·2.1 State Variable Representation using Physical Variables 2.1.1 Advantages 2.2 State Space Representation using Phase Variables
1 - 14
2•1 2•4 2 -4
2.2.1 Stale Modetfrom Differential Equation ..••.•...•••...•..............•.••..••.
2•5
2 2 2 Slate Model from Transfer Fuiict;on
2-8
2.2.3 Advantages. . • . . . . . • . . . • . . . . . . . . • . . . • . . . . . • • • . • . • • . • . . • • . . • . • • . • • • . • • . 2 • 18 2 2 4 11rnaa11ons
2.3 State Space Representation using Canonical Variables
2.18
2 - 19
2.3.2 Advantages of Canonical Variables
2 • 26
2.3.3 Disadvantages of canonical Variables ••.••......•.......•.•..•..••........•
2 • 26
2.4 State Model by Cascade Programming
2 - 29
Examples with Solutions
2 - 31
3.1 Background 3.2 Definition of Matrix., . . 3.2.1 Types of Matrices 3.2.2 Important Terminologies
: .'.~:
3-1
..
• 3-1 3·2 3•3
. .. . . . . .. .'
3.3 Elementary Matrix Operations
3-6
3 4 Inverse of a Matrix
3-7
3.4.1 PropertiesoflnveiseofaMatrix 3 5 Deriyation of Transfer E11nclion from State Model
3-9
3.5.1 Characteristic Equation 3.5.2 MIMO System
3·8 3 • 10
3 -11
3.6 Eigen Values
3 • 12
3.7 Eigen Vectors
3 - 13 3. 13
3 8 Modal Matrix M 3 8 1 Yan!ler Mende Matrix
3-14
3.9 Diagonalisation
3 - 19
3.10 Generalised Eigen Vectors 3. 23 3.10.1 Vander Monde Matrix for Generalised Bgen Vectors . .. .. .. .. . . • . . . .. . .. .. . • • 3 • 24 Examples with Solulions Reyjew Questions
3 • 26
4.1 Background
4• 1
4.1.1 Homogeneous Equation
4.1.2 Nonhomogeneous Equation 4 2 Review of Classical Method of
4•1
Snh11ion
4.2.1 Zero Input Response 4.3 Solution of Nonhomogeneous Equation
4•2 4-2 4•4 4-4
4.4 Properties of State Transition Matrix
4-5
4.5 Solution of State Equation by Laplace Transform Method
4- 6
4.6 Computation of State Transition Matrix
4 -7
4.7 Laplace Transform Method
4-8 4-9
4 B Power Series Method
4.9 Cavley Hamilton Method 4 91 Prncedure lo Calct~ale ""1
4 - 10
4.10 Similarity Transformation Method
4 -16
4.11 Controllability and Observability
4 - 22
4.12 Controllability
4. 23
4-12
4.12.1 Kalman's Test for Controllability
4 • 23
4.12.2 Condition for State Controllability Ins-Plane .•..•.•.............•..••.•..•..
4 - 26
4.12.3 Gilbert's Test for Controllability
4 • 26
:
4.12.4 Output Controllability
4.13 Observability
4 • 29
4 • 29
4.13.1 Kalman's Testier Obse!Vllbllity
4 - 30
4.13.2 Conditioo for Comple:e Observability In s-Pl<Wle
4 • 33
4.13.3 Gilbert's Test ftlf Observabi!lty . .. . . . . . . .. . . . . . . . .. . . . . . . . . . . . • . . • . . . • • . .. 4 • 33 Examples with Solutions
4 • 35
5 1 Introduction
5-1
5.2 Pole Placement Design
5-2
5.3 Necessary and Sufficient Condition for Arbitrary Pole Placement 5 4 Evah1a!joo of Slate Feedback Gain Matrix K
5-3 5. 15
5 4 1 Use of Traosfonnatjo1J Matrix I
5-15
5 4 2 Direct S11bstit11tjon Method
5-15
5 4 3 Ackeanann's Foanu!a
5.5 Selection of Location of Desired Closed Loop Poles
5-16 5 • 19
5 6 State Observers
5-24
5.7 Concept of State Observer
5. 25
5.8 Dual Problem
5 - 27
5.9 Necessary and Sufficient Condition for State Observation
5- 28
5.10 Evaluation of State Observer Gain Matrix!<..
5 - 28
5.11 Selection of Suitable Value of Observer Gain Matrix K..
5 - 30
5 12 Observer Based Controller
5 - 35
Examples with Solutions
5 - 38 5 -57
6.2.2 Variable Range .•...••.•....••.......•.......................•••.•...•.• 6.2.3 Controller OutputRange 6.2.4 Control Lag .....•.........•.•....•...•...•.•..•..••.••.....•.•.....•.•.
6•3
6 2 5 Dead Zone
6-3
6 3 Classification of Controllers 6 4 Continuous Contmllm..Made..._
6.5 Proportional Control Mode 6.5.1 Characteristic of Proportional Mode 6520ffset
6 •3
6-3 6-4 -'"'-',_,.
6-5 6•6 6-6
6.5.3 Applications
6• 7
6.6 Integral Control Mode
6.6.1 Step Response oflnteqral Mode 6.6.2 Characteristics oflntegral Mode 6.6.3 Applications 6 7 Oerlvatiye Control Mode
6 7 1 Characterisljcs of !ledyatjye Con!rol Mode 6.7 2 Applications
6-7
6•9 6-9 6 • 10 6-10 6 • 11 6 • 12
6.8 Composite Control Modes
6 -12
6.9 Proportional + Integral Mode (Pl Control Mode)
6 • 1>!
6 9 1 Cbaractertsticsof Pl
Mode
6-14
M
6.9.2 Applications
6.10 Proportional+ Derivative Mode (PD Control Mode)
6 • 15
6. 15
6 1 O 1 Charac!eristtcs of PD Mode
6-16
6.10.2 Applications
6 -17
6.11 Three Mode Controller (PIO Control Mode)
6 - 17
6.12. Response of Controllers to Various lnputs
6- 21
6.13 Effect of Composite Controllers on 2"" Order System
6 - 22
6.14 PD Type of Controller
6- 23
6.15 Pl Type of Controller
6 - 24
6.16 PIO Type of Controller
6 - 26
6.17 Rate Feedback Controller (Output Derivative Controller)
6 - 26
6.18 Review of Second Order System Specifications
6 - 28
6.19 Steady State Error
6 - 30
Examples with Solutions
6 - 31
Review Questions
6 - 48
7.1 Introduction to Nonlinear Systems
7-1
7 .2 Properties of Nonlinear Systems
7- 1
7 3 Classification of Nonlinearities
7-2
7 4 Inherent Nonlinearities
7-2
741 SabaraOOo
7-3
742Deadmne
7.3
7.4.3 On-Off Nonfineatity •......•..•..•..•..•..•...•..•..•...••..•..•...•.....•
7 -4
7.4.4 On·Off Nonlinearity with Hysteresis ...........••.•........•...............•..
7•4
z 4 5 Nonfineac Friction
7 4 6 Bacldash
7.4.7 NonlinearSpring
7•5
. ". 7. 5 7 -6
7.5 Absolute Value Nonlinearity
7-7
7 6 Intentional Nonlinearities
7-8 7-8
M
.,~ 8.1 lnt!'Oductjon
8-1
8.2 limit Cycle
8•1
8.3 Jump Resonance
8•2
8 4 The Phase Plane Method
8-3
8.5 Basic Concept of Phase-Plane Method
8-5
8.6 Isocline Method for Construction of Phase Tralectory
8•7
8.7 Application of Phase-Plane Method to linear Control System
8 - 10
8.8 Second Order Nonlinear System on Phase Plane
8 -13
8.9 Different Types of Phase Portraits
8 • 14
8.9.1 Phase Portraits for Type O System ......•...•..•....•....••......•.•..••.•.
8. 14
8.9.1.1 S1able SyslemIMth Conl)lex Roots . . . • .
. . 8· 15
8.9.1.2 Unstable System,.;in Complex Roots • . . .
••
8.9.1.3 Marginally Slab:e Sys1em IMth Complex Roots
8-16
. 8· 16
8.9.1.4 Stable SystemIMtll Real Roots • . . • • .
•. 8-17
8.9.1.5 Unslable System-Mth Positive Real Roots . .
..
8.9.1.6 U11$lable System~ Cloe Positive and Ooe Negati\19 Real Root . . 8.9.2 Forced Second Older TypeO System
8.9.3 Phase Portraits for Type 1 Svslem 8.9.4 Phase Portraits for Type 2 S)'Slem 8.1 O Singular Points of a Nonlinear System
8-18
. . 8· 19
~-19 8 -19
8 • 22 8 - 26
8 11 Delta Method
8 -27
Examples with Solutlons
8 • 28
Review Questions.
.
. .
..
8- 39
chip&r; g uapunm stiiluiiY,\na1Yili'_.,--.-- .., 9 1 fnf"mductjon
9-1
9.2 Stability in the Sense of Liapunov
9•2
9.3 Asymptotic Stability
9- 3
9.4 Asymptotic Stability In the Large
9-3
9.5 Instability
9•3
9.6 Graphical Representation
9•3
9,7 Some Important Definitions
9 -4
.. 9 7 1 poSfttvflDpfinileness
9-4
9.7.2 Negative Definiteness
9 .4
9 7 3 Positive Semjdefiniteness
9-5
..
9.7.4 Negative Semidefinite ... 97
9.5 9.5
5 !odefinUeness
............. 9- 5
9 8 011adralic Form 9 9 Hermitian Forro 9.1 O Liapunov's Second Method
9- 5 '
9-6
9.11 Liapunov's Stability Theorem
9-6
9.12 Stability of Linear and Nonlinear Systems
9-7
9.13 Construction of Liapunov's Functions for Nonlinear Systems by Krasovskii's Method 9.14 The Direct Method of Liapunov and the Linear System
9-8 9 - 11
Examples with Solutions
9 - 12
Reyjew Questions
9 - 24
1m~::in~~~~w~~~
J·~~~~m~Ei.W'~Vti~{;.~'1.{M-t' ;f~M-.;~:.£~
State Variable Analysis and Design 1.1 Background The conventional approach used lo study the behaviour of linear time invariant control systems, uses the time domain or frequency domain methods. In all these methods, the systems are modelled using transfer function approach, which is the ratio of Laplace transform of output to input, neglecting all the initial conditions. Thus this conventional analysis faces all the limitations associated with the transfer function approach.
1.1.1 limitations of Conventional Approach Some> of its limitations can be stated as : 1) Naturally. significant initial conditions in obtaining precise solution of any system,
loose their importance in conventional approach. 2) 111e method is insufficient and troublesome lo give complete time domain solution
of higher order systems. 3) It is not very much convenient for the analysis of Multiple Input Multiple Output systems. 4) It gives analysis of system for some specific types of inputs like Step, Ramp etc. 5) II is only applicable to Linear Time Invariant Systems. 6) The classical methods like Root locus, Bode plot etc. arc basically trial and error procedures which foil lo give the optimal solution required. Hence it is absolutely necessary lo use a method of analysing systems which overcomes most of the above said difficulties. The modem method discussed in this chapter uses the concept of total internal slate of the system considering all initial conditions. This technique which uses the concept of state is called State Variable Analy»is or St•te Space Analysis. Koy Point: State variable analysis is c.<Scntially a time domain approach but it has 11m11ber of adva11/ages compared to co11Vt'nlio11al mrtliods of analysis.
(1 • 1)
Modem Control Theory
State Variable Analysis & Design
1·2
1.1.2 Advantages of State Variable Analysis The various advantages of state variable analysis arc, 1) The method takes into account tl\C effect of all initial conditions. 2) It can be applied to nonlinear as well as time varying systems.
3) It can be conveniently applied to Multiple Input Multiple Output systems. 4) The system can be designed
for the optimal conditions precisely by using this
modem method. S) Any type of the input can be considered for designing the system. 6) As the method involves matrix algebra, can be conveniently adopted for the digital
computers. 7) The state variables selected need not necessarily be the physical quantities of the
system. 8) The vector matrix notation greatly simplifies the mathematical representation of the system.
1.2 Concept of State Consider a football match. The score in the football match must be updated at every instant from the knowledge and Information of the total score before that instant. nus procedure of updating the score continues till the end of the match when we get exact and precise score of the entire match. This updating procedure has main importance in the understating of the concept of state. Consider the network as shown in the Fig. 1.1 To find V0111, the knowledge of the initial capacitor voltage must be known. Only information about V1n will not be suffkknt to obtain precisely the V out at any time t ~ O. Such systems in which the output is not only dependent on the input but also on the initial conditions arc called the systems with memory or dynamic systems.
Fig. 1.1
f-f--
V Velocity X Displacement
ForceF-0 /?7777Tj777777 Zero &icbon
Flg. 1.2
While if in the above network capadtor 'C' is replaced by another resistance 'R1' then output will be dependent only on the input applied Y'". Such systems in which the output of the system depends only on the input applied at t 0, arc called systems with zero mttnory or
=
static systems.
Modem Control Themy
State Variable Analysis & Design
1·3
Consider another example of simple mechanical system as shown in the Fig. 1.2 Now according to Newton's law of motion,
i.e.
F
Ma
a
Acceleration of mass M
F
M
v(l)
=
ard (v(t))
id
((t) dt
Now velocity at any time 't' Is the result of the force F applied to the particle in the entire past, I
'M
v(t)
f
f(t) cit
~
'!
f(t)dt
+kf f(t)dt l
'o where t0 = Initial time. Now
-b J f(t) dt
=
v(t0)
I
v(t)
v(t0)+
J f(t) dt 'o
From the above equation it is clear that for the same input f(t), we get the different values of the velocities v(t) depending upon our choice of parameter t0 and the value of v(t0). U v(t11) is known and the input vector from 't0' to 'I' is known then we can obtain a unique value of the output v(t) at any time t > t0• Key Point : Thus initial conditions i.e. memory affects the system c/1aracterisatim1 and subsequent behaviour. Thus initial conditions describe the status or state of the system at t = t 11• The state an be regarded as a compact a.nd concise representation of the past history of the system. So the state of the system in brief separates the future from the past so that U1e state contains all the information concerning the past history of the system necessarily required to determine the response of the system for any given type of input. The state of the system at any time 't' is actually the combined effect of the values of all the different elements of the system which arc associated with the initial conditions of the system. Thus the complete slate of the system can be considered to be a vector having components which are the variables of system which are:tlosely associated with
State Variable Analysis & Design
Modem Control Theory
initial conditions. So stale can be defined as vedor X(I) called slate vector, This X(t) i.e. stale al any lime ·r is 'n' dimensional vector i.e. column matrix n x 1 as indicated below. X1(t) X2(t)
X(t)
Now these variables X t (t) , X2 (t) called the state wriablts of the system.
Xn (t} which constitute the state vector X(t) arc
If slate at t s t1 is to be decided then we must know X(t0) and knowledge of the input applied between t0 ~It. This new state will be X(t 1) which will act as initial state to find out the state at any time t > t 1• This is called the updating of the state. The output of the system at t = t 1 will be the function of X(t 1) and the instantaneous value of the input at t t 1, if any. The number of the state variables for a system is generally equal to the order of the system. The number of independent stale variables is normally equal to the number of energy storing elements (e.g.: capacitor voltage, current in inductor) contained in the system.
=
1.2.1 Important Definitions 1) State : The state of a dynamk system is defined as a minimal set of variables such that the knowledge of these variables at t = 10 together with the knowledge of the inputs fort :·t0. 2) State Variables : The variables involved in determining the state of a dynamic system X(t), arc called the state variables. X1(t), X2(t) ...... X"(t) are nothing but the state variables. These are normally the energy storing elements contained in the system.
31 State Vector : The 'n' state variables necessary to describe the complete behaviour of the system can be considered as 'n' components of a vector X(t) called the state vector at time · t', The state vector X(t) is the vector sum of all the state variables. 4) State Space : The space whose co-ordinate axes are nothing but the "n' state variables with time as the implicit variable is called the slate space. 5) State Trajectory : It is the locus of the tips of the state vectors, with time as the
Implicit variable.
Modem Control Theory
State Variable Analysis & Design
1·5
Definitions can be explained by considering second order system : Order is 2 so number of state variables required is 2 say X1(t)
and X2(t).
State vector will be the matrix of order 2 x I. X(t)
->
--t X(t)
i.e,
->
X1(t)+ X2(t) (vector addition)
The stale space will be a plane in this case as the number of variables arc two. Thus state space can be shown as in the Fig. 1.3. Xi(l)
X1(t)
L~ . ,
x,c1,> ~(I)
(0 .0)
X2tt1)
Fig. 1.3
Now consider t =
Flg.U
t1
-> X(t1)
->
->
= X1(l1)+ X2(11)
x,c11
·~(ti
Fig. 1.5
Now consider
t
Fig. 1.6
= t2
->
X(t2)
->
"' X1(t2)+
->
X2(t2)
The state trajectory can be shown by joining the tips of the two state vectors i.e. X(t 1) and X(t2).
Modem Control Theory
State Variable Analysis & Design
1-6
1.3 State Model of Linear Systems Consider Multiple Input Multiple Output, nth order system as shown in the Fig. 1.7. Number .of inputs
m
Number of outputs
p
u,
----<"'!
U2-~-.i
MIMO
Inputs
SYSTEM
t----v, t-~--v2 OutpulS
.____. __ VP
.,....,.-T""'i~=:T" X1 X2 X3 Slate varlables
Fig. 1.7 X1(1) X2(t)
U2(t)
"•' ')
U(t)
;
X(t)
=
Um(t)
Y1(t) Y2(t)
, Y(t) Xn(t)
= Yp(t)
All arc column vectors having orders m x 1, n x l and p x 1 respectively. For sucti a system, the state variable representation can be arranged in the form of "n' first order differential equations. dX1(t)
di"'
•
X1(t)mf1(X1,X2
.
....... Xn,U1,U2
......... Um)
X2(t) = 12 (X1, X2 , ...... X0, U1• U2 ......... Uml
1. 7
Modern Control Theory
Where f =
State Variable Analysis & Design
is the functional operator.
r, Integrating the above equation, X1(l)
=
I
X;(lo) +
J f; (X1
• X2 ......
x,
U1 • U2 ..... Um) dt
'o where i = 1, 2, ........ n. Thus "n' state variables and hence state vector at any time "t' can be determined uniquely. Any "n' dimensional time invariant system has state equations in the functional form as,
.
X(t) = f (X, U) While outputs of such system ore dependent on the state of system and instantaneous inputs. :. Functional output equation can be written as, Y(t) = g(X, U) where · g' is the functional operator. For time variant system, the same ~'<Juations can be written as, X(l)
f(X, U, t)
State equation
Y(t)
g(X, U, t)
Output equation
Diagramatically this can be represented as in the Fig. 1.8. U(t) lnstarltancous
lnput Input U(t) ----~
Y(t) Oulput
X('ol lnlUal state
Fig. 1.8 Input-output state description of a systom
Modem
1·8
Control Theory
State Variable Analysis & Design
The functional equations can be expressed intcrms of linear combination of system states and the input as,
.
X 1 = a11 X1 + a12X2 + ... + a10X0 + b11U1 + b12U2 + ... + b1m Um X2
= a21 X1 + ~X2 ...
+ a20Xo + b21U1 + b22U2 + ... + b2m U'"
.
Xn = a,,1 X1 + "nl~ + ... + a.,,,Xo + b01U1 + bn2U2 + ... + bnm Um For the linear time invariant systems, the coefficients a11 and bii arc constants. Thus all the equations can be written in vector matrix form as,
I
X
A X(t) + B U(t)
X(t}
State vector matrix of order n x 1
U(t)
Input vector matrix of order m >< 1
where
A B
System matdx or Evoliition matrix of order n x n
=
lnput matrix or control matrix of order n >< m
Similarly the output variables at time t can be expressed as the linear combinations of the Input variables and state variables at time t as, v,(t)
=Cu
x,(t) + ... + C1n Xn(t) + d11 u,(t) + ... + dim Um(I)
Yp(t) = Cpt X1(t) + ... +cpl X0(t) +dpt U1(t} + ... + dpm Um(t) For the linear time invariant systems, the coefficients c;; and d;J arc constants. Thus all the output equations can be written in vector matrix form as, Y(t) = C X(t) + D U(t) where
= =
Y(t) C
Output vector matrix of order p .>< 1 Output matrix or observation matrix of order p x n
D = Direct transmission matrix of order p x m The two vector equations together is called the stale model of the linear system. A X(t) + B tl(t)
... State equation
Y(t) "' C X(t) + D U(t)
... Output equation
X(t)
s
This Is state model of a system.
1 .g
Modem Control Theory
State Variable Analysis
& Design
For llnear time-variant systems, the matrices A, B, C and D are also time dependent. Thus, X
= A(t) X(t) + B(t) U(t)} = C(t) X(t) + D(t) U(t)
For linear time variant system
1.3.1 State Model of Slngle Input Slngle Output System
=
=
Consider a single input single output system i.e. m 1 and p l. But its order is 'n' hence n state variables arc required to define state of the system. In such a case, the state model Is
.
X(t)
A X(t) + B U(t) C X(t) + d U(t)
A
n x n matrix, B
c
1 x n matrix, d
where
nnd
=
Y(t)
U(t)
= n x I matrix = constant
single scalar input variable
In general remember the orders of the various matrices. A
=
Evolution matrix
U =- Control matrix
=> n x n => n x m
C = Observation matrix => p x n D = Transmission matrix
=pxm
1.4 State Diagram Representation It is the pictorial representation of the state model derived for the given system. It forms a close relationship amongst the state model, differential equations of the system and its solution. It is basically a block diagram type approach which is designed from the view of programming of a computer. The basic advantage of state diagram Is when it Is impossible to select the state variables as physical variables. When transfer function of system is given then slate diagram may be obtained first. And then by assigning mathematical state variables there in, standard state model can be obtained. State diagram of a linear lime invariant continuous system is discussed here for the sake of simplicity. It is a proper interconnection of three basic units. i) Scalars
ii) Adders
iii) Integrators
Modem Control Theory
1·10
State Variable Analysis & Design
Scalars arc nothing but like amplifiers having required gain.
~<1>Tx,<1>=~<1>·x3<11 ><:i(t)
(a)Scalar
(b)Adder
Fig. 1.9 Adders arc nothing but summing points. Integrators arc the clements which actually integrate the differentiation of state variable to obtain required state variable.
Now integrators arc denoted as '1/s' in Laplace transform. So transfer function of any lntegralor is always l/s. With these three basic units we can draw the state diagrams of any order system. Fig. 1.9 (c) Key Point : Tire output
of each integrator i.~ the state variable.
1.4.1 State Diagram of Standard State Model X(ll
.
A X(t) + B U(t) and
Y(tl
C X(t) + D U(t)
Consider standard state model
So its state diagram will bl? as in the Fig. 1.10.
Y(I)
Fig. 1.10 State diagram of MIMO system The thick arrows indicate that there arc multiple number of input, output and state variables. There must be n parallel integrators for n state variables. The output of each integrator is a seperatc state variable. If such a state diagram for the system is obtained then the state model from the diagram can be easily obtained.
1 • 11
Modem Control Theory
State Variable Analysis & Dnlgn
Remarks 1. To obtain the stale model from stale diagram, always choose output
of each integrator as a stale variable. Number or integrators always equals the order of the system i.e. ·n· 2. Differentiators arc not used in the stale diagram as they amplify the inevitable noise. ,..
Example 1.1 : Obtain the state diagram uf 5150 system reprtstnttdby equations,
.
X1(1)
=
a1 X1(t)+b1
.
U(t), X2(1J =a2 X1(t)+a3
X2(tH b2 U(I)
and Y(t) = c1X1(t)+c2X2(1) Solution : A.• there arc 2 state variables X1(t) and X2(t), the two Integrators arc required.
Ytl)
U(t)
Fig. 1.11 1.5 Non Uniqueness of State Model Considera standard state model for a system X(I)
.
A X(t) + B U(t)
... (1)
y(I)
C X(t) + D U(t)
.•. (2)
Let X = M Z where M is a non singular constant matrix. Koy Point: A matrix whose daerminatu is nonuro is called nonsingular matrix. This means Z is new set of stale variables which Is obtained by linear combinations of the original state variables.
1·12
Modem Control Theory
. .
X = MZ hence
State Varlable Analyala & Oealgn
X" MZ
M Z(t)
.
AM Z(t) + B U(t)
... (3)
Y(t)
CM Z(t) + D U(t)
... (4)
Substituting in a slate model
:. Premultiplying equation (3) by M" 1
.
M-11\M Z(t) + M-1 B U(t)
Z(t) WMC
M-1 M Z(t)
where
I i.e. Identity matrix c
A
Where
Z(t) +
B U(t)
... (5)
M-1 AM
fi and
A
M-1 B
C Z{t) +
Y(t)
c
D U(t)
... (6)
CM
Equations (5) and (6) forms a new state model of the system. This shows that state model is not a unique property. Key Point: Any /i11~ar combinatio11s of th;. original set of state wriables result» into a valid 11ew set of state variables.
1.6 Llnearizatlon of State Equation Any general time invariant system is said lo be in equilibrium, at a poinl when.
Ix~
f (Xo- Uo)
u
0
(l
U0)
I
The derivatives of all the stale variables are zero at a point of equilibrium, The system has a tendency lo lie at the equilibrium point unless and unlill it is forcefully disturbed .
.
The state equation X(t) = f (X, U) can be linearized for small deviations about an equilibrium point (X0, U0). This is possible by expanding the state equation using Taylor series and considering only first order terms, neglecting second and higher order terms.
Modem Control Theory
State Variable Analysis & Design
1 • 13
So expanding kth state equation, n ilf,(X, U)
•
~ = fk (Xo- U0) +I~ /:I
Now
rk(Xo- U0)
=
I
I
(X1 -X10)+ X=Xo U=Uo
m i)lk(X, U)I L.-mj XeXo
(Ui - Up)
j=I
U=Uo
0 at the equilibrium point.
And let the variation about the operating point is,
. .
x1-Xio and
henceX=X,
U;-U;o
Hence the kth state equation can be linearized as,
.
Xk
.
.In this equation xk,
x, and U; are the vector matrices and the remaining
terms arc the
matrices or order n x n and n x m respectively. Hence the linearized equation can be expressed in the vector matrix form as,
x
=
AX+
nu
vx;ilr2 where
A
~
cir,
dx;;"
ilf2 ax2
arn
«:
~
ib<.2
ilr2
dx;;-
ar,
a11 ()Um
i!f2
()(2
au,
iil2 aum
er, ~
()In
m;-
is n x n matrix
illn ilx.,
iJU2
ilf1
au, B
()fl
dx;
()In
aum
Is n x m matrix
Modorn
1·14
Control Theory
State Variable Analysis & Design
All the partial derivatives of the matrices A and B are lo be obtained at an equilibrium slate (Xo- U0). Such matrices A and B defined above interms of partial derivatives are called the Jacobian matrices.
Review Questions 1. Exp/run tl1e concept of stole. 2. l)tfim •md explain tire followingterms, a Slat/ a.1riables I>. Stott 11«tor of. Stale c. Slate lraf"<10ry < St{lt( SfXIU
J. l:.xplain adt1tmti1grs of state t1t:1rit1ble method over conventional ant. 4. Writ' a slrort note on ad•w11tagt>and limitatums of st.tit ~'llriablt approotl1, 5. Wn'tr a note on lincariznliatr of Malt equation:;. 6. l'row tire nomtniqtttnt."SS
of lite state modtl.
ODO
State Space Representation 2.1 State Variable Representation using Physical Variables The state variables are minimum number of variables which arc associated with all the initial conditions of the system. As their sequence is not important, the state model of the system is not unique. But for all the state models it is necessary that the number of state variables is equal and minimal. This number 'n' indicates the order of the system. For second order system minimum two state variables arc necessary and so on. To obtain the state model for a given system, it is necessary to select the state variables. Many a times, the various physical quantities of system itself arc selected as the state variables. For the electrical systems, the currents through various inductors and the voltage across the various capacitors arc selected to be the state variables. Then by any method of network analysis, the equations must be written i'ntcrms of the selected state variables, their derivatives and the inputs. The equations must be rearranged in the standard form so as to obtain the required state model. Key Point: It is important ilia! Ilic equation for diffcre11tiatio11 of 011e state t•11riable s/1011/d 1101 i1100/vc tltt differmtiation of any otner slate variable. In the mechanical systems the displacements and velocities of energy storing elements such as spring and friction arc selected as the stale variables. In general, the physical variables associated with energy storing elements, which are responsible for initial conditions, arc selected as the stale variables of the given system. 1~
Example 2.1
Obtain tire state model of tire given electrical system.
~
v;(t)
l
IC
~
~t)
-
Fig. 2.1 (2 • 1)
v0(t)
-l
tvru : July/Aug.-2006)
2-2
Modem Control Theory
State Space Representation
Solution : There arc two cn!!l'gy storing elements L and C. So the two state variables arc current through inductor i(t) and voltage across capacitor i.e. v0(t). i(t) And
and
X2(t)
= v (t) 0
v1(t) = Input variable
Applying KVL to the loop. v1(t)
=
i(t) R + L
di (t)
-di
+ v0(t)
Arrange it for di(t}/dt, di (I)
-dt
I
( ) R '(
IV; t
-1·
R
1
I t)-Iv.,(l)
b 1
ut
i.e,
- ·r. X1(t)-I
While
Voltage across capacitor =
2..c i(t) i.e.
"2(t)
X2(tl+I
di (I)
dt - =
•
Xt(t) ... (1)
U(t)
bJ
i(t) di
~v o(l) = {._ (I) di "2
but
·c1 X1(t)
... (2)
The equations (1) and (2) give required state equation.
[~J(t)] X2(t)
i.e.
.
- ~L _ _!_] L [ 0
t
1
Xz(t)
0
X(t) = A X(t) + U U(t)
While the output variable Y(t) Y(t) i.e.
[X (I)) + [·I.~] U(t)
(0 1]
Y(t) = C X(t)
=
v0(t) = X2{t)
[;~]+(OJ
U(t)
and D = (01
This is the required state model. As n
=
2, it is second order system.
Note : The order of the state variables is not important. X1(t) can be v 0(1) and X2(t) can be i(t) due to which state model matrices get changed. Hence state model Is not the unique property of the system.
2·3
Modem Control Theory
11u+
Example 2.2 : form.
State Space Representation
Obtain the state model of the givm electrica! network i11 tit<' standard
Fig. 2.2
IDNrilln :
U(t) ~ input Y(t) "'
«
output
State variables: X1(t) = i1(t),
c,(t)
= c.,(t)
X2(t) = i2(t), X3(t) = vc(t)
Writing the equations : <',(I)
I · 1 ··- c, (I)- --- v c(I) I.I I.I
i.e.
... (1)
Then,
... (2)
and
dvc(l)
C-d-,dvc(t)
Cit
.
X3(t)
bi1(t)-bi2(t) 1
1
C X1(t)- C X2(l)
... (3)
2-4
Modem Control Theory
. .
0
X2
0
X3
1
X1
X(t)
and
e0(t) Y(t)
Y(t)
i.e.
Y(t)
R2
-i=;-
1 L2
1
c -c:
.
i.e.
-r;-
0
0
State Space Representation
[~~]+
'l
Li ~
U(t)
AX(t) + BU(t) i2(t) R2 c
X2(t) R2
(o ~
OJ[~:]
~ CX(t) + DU(I)
where D = 0. This is the required state model.
2.1.1 Advantages The advantages of using available physical variables as the state variables are, 1. The physical variables which are selected as the state variables are the physical quantities and can be measured. 2. As state variables can be physically measured, the feedback may consists the information about state variables in addition to the output variables. Thus design with state feedback is possible. 3. Once the state equations arc solved and solution is obtained, directly the behaviour of various physical variables with time is available. But the important limitation of this method is that obtaining solution of such state equation with state variables as physical variables is very difficult and lime consuming.
2.2 State Space Representation using Phase Variables Let us study how to obtain state space model using phase variables. The phase variables arc those state variables which are obtained by assuming one of the system variable as a state variable and other state variables are the derivatives of the selected system variable. Most of the time, the system variable used is the output variable which is used to select the state variable. Such set of phase variables is easily obtained if the differential equation of the system is known or the system transfer function is available.
2·5
Modem Control Theory
State Space Representation
2.2.1 State Model from Differential Equation Consider a linear continuous time system represented by n•h order differential equation yn +an-lyn-1 +an-2 yn-2+ ... +
= b0U+b1 U+ ...+bm_1um-t
•
a1 Y+a0Y(t)
+bmU°'
••• (1)
In the equation, Y"(t) = dY" (t) = nth derivative of Y(t). dt" For time invariant system, the coefficients a0 _ 1, an- 2, . • . ao- ho- b1, • .. bm are constants. For the system, Y(t)
Output variable
U(t)
Input variable
Y(O), Y(O), ... Y(O) n- l represent the initfal conditions of the system. Consider the simple case of the system in which derivatives of the control force U(t) arc absent. Thus
.
U(t)
.: Y" + an _ 1 Y" -
= um(t) = 0 Y + n0 Y(t) = b0U(l)
U(I) " ..•
1 + ... + a1
Choice of state variable Is generally output variable Y(t) itulf. vmablcs arc derivatives of the selected state variable Y(t).
. .
X2(t)
Y(t) = X1(t)
X1(I)
.
Xi(!)
X(t)
X3(t)
.
Xn(t)
Thus the various state equations arc,
X,. - I (t) X,.(t)
... (2) And other state
State Space Representation
Modem Control Theory
Note that only n variables arc to be defined to keep their number minimum. ThU! • x,. _ 1(t) gives
Important
n
th
•
state variable X0(t). But to complete state model X,.(t) is necessary,
..
.
: X,.(t) is to be obtained by substituting the selected state 1 vartsbles In the a X;i. ...
original differentialequation (2). We have Y(t) = X1, Y(t) " X,, Y(t) yn-l(t) =~(I), Y0(t)
. .
= X,.(t)
:. X,.(t) + an_ 1 ~(t) + an_ 2 ~ _ 1(t) + ... + a1 X2t + a0 X1(t) = b0 U(t) :. X,.(t)
=-
a0 X 1 - a1 X, ... - an _ 2 ~ _ 1 - an_ 1 X0 + b0 U(t)
-Hcnce all the equations now can be expressed in vector matrix fonn
xx2 _ 1
rL Le.
-
.
X
z
1
o
[ 0
0
0
~ '.J [J:
-~o
... (3) QS,
l { "'
AX(t) + BU(t)
Such set of slate variables is called set of phasr va.rlablc1. The matrix A is called matrix in phase variable fonn and it has following features, 1) Upper off diagonal i.e. upper parallel row to the main principle diagonal contains all elements as 1. 2) All other clements CXC1..'Pt last row are zeros. 3) Last row consists of the negatives of the coefficients contained by the original differential eouation. Such a form of matrix A is also called Bush fonn or Companion method Is also called companion form reallution. The output equation is,
'Y(t)
[ I 0 .... OJ X,,(t)
Le.
Y(t)
CX(t)
where D = 0
fonn. Hence the
State Spac:e Representation
2·7
Modem Control Theory
This model in the Bush fonn can be shown in the stale diagram as in the
Fig. 2.3. Output of each integrator is a slate variable.
U(t}
Fig. 2.3 State diagram for phase variable form Observe that the transfer function of the blocks in the various feedback paths arc the coefficients existing in the original dilfcrcnlial equation. If the differential equation consists of the derivatives of the input control force U(t) then this method L< not useful. Jn such a case, the state model is to be obtained from the transfer function. ,_.
Example 2.3 : Construct tht state model using phase mriobles if tht sy.•tt111 is dtscribtd by tire differentialequation,
d2Y(t) +4 't2Y(t~ +7 dY(I) + 2Y(t) dt3 dt2 di
= 5U(I)
Draw the state diagram. Solution : Choose output Y(l) as the stale variable X1(t) and successive derivatives of it give us remaining state variables. As order of the equation is 3, only 3 state variables are allewed. X1(t)
Y(t)
-
• • X 1(1) = Y(t)
Xii) and Thus
.
J\(t)
"2
dY(t)
=di
= Y(t) = d2Y(t) dt2
X2(t)
... (1)
2-8
Modem Control Theory
State Space Representation
... (2) To obtain Xj(t), substitute state variables obtained in the differential equation. d3Y(t} -dtJ
.
'" d " dX • = Y (t) = -(Y(t)l = -3= X3(t) di di
.
:. X3(t) + 4X:J(t) + 7 X2(t) + 2X1(t) = SU(t) "3(t) "' - 2X1(t) - 7 X2(t) - 4X3(t) + 5 U(t)
... (3)
The equations (I), (2) and (3) give us required state equation. X(t)
A
where
The output is, Y(t)
Y(t)
c
where
AX(t) + BU(t)
u
1 0
-7
j]
and
B=[n
X1(tl CX(t) + DU(t) (10OJ,D=0
This is the required state model using phase variables.
Fig. 2.4
2.2.2 State Model from Transfer Function Consider a system characterized by the differential equation containing derivatives of the input variable U(t) as,
Y"
+ an -
I
yn -
I
+ ... + "t
y + •o Y(t)
"' boU + b1
U + ...
+ bm -
l
Um -
I
+ bm um ... (I)
2·9
Modem Control Theory
State Space Representation
In such a case, it is advantageous to obtain the transfer function. assuming zero initial conditions. Taking Laplace transform of both sides of equation (1) and neglecting initial conditions we get, Y(s) [sn +
a,,_ 1
Sn -
1
+ ... + a1 s + aoJ
Y(s) U(s)
T(s) =
= Ibo + sb1 + ...
+ bm _ 1 Sm - 1 + b tu sml U(s)
bo +sb,+ .... +b n-lsm-1 +blllsm ' a0 TSa J+ ... +a11_1sn-l
... (2)
+Sn
Practically in most of the control systems m < n but for general case, let us assume m =n.
From such a transfer function, state model can be obtained and then zero initial conditions can be replaced by the practical initial conditions to get required result. There arc two methods of obtaining state model from the transfer function, I) Using signal now graph approach 2) Using direct decomposition of transfer function 1) Using Signal Flow Graph Approach
The Mason's gain formula for signal flow graph slates that, T(s)
LTkAk
-6-
Gain of k1h forward path
where A
System determinant
p: Gain x Gain product ol} I - {L i111 loop gains}+j
,111 combinations of two
l non • louchin~ i.\k
- ...
loops
Value of II clliminating those loop gains and products which are touching lo k1h forward path
According to this formula, construct the signal now graph from the transfer function. From the signal now graph, state model can be obtained. While obtaining sii;nal flow graph, try to get the gains of branches as 't/s' representing the integrators. This helps to obtain the required state model. To clear this idea, let us consider the system having transfer function,
Modem
2·10
Control Theory
State Space Representation
Divide both numerator and denominator by highest power of s,
T(s)
a a a i+-t+-1.+~ s 52 s 3
T161 +T262 + T363 + ·r4t14 = I -(l:Aliioop gains)+[l:Gain products of 2 n~n-touching loops]-···
Assuming that there arc no combinations of 2 and more non-touching loops. Loop gains arc, And let ll.1 = 62 =
I).3 =
l\~ = 1 i.e. all loops arc touching to au the forward paths.
Hence forward path gains arc, T1
=
b3' T2
b2
bl
= s' T3 = 2' s
Involving various branches having gains ~· the signal flow graph can be obtained as,
Fig. 2.5 Each brand' with gain ~ represents an integrator. Output of each integrator is a state variable.
According to signal flow graph, value of the variable at the node is an algebraic
2-11
Modern Control Theory
State Space Representation
sum of all the signals entering at that node. Outgoing brunches docs not affect the value of variable. Hence from signal flow graph. X1
b2U + Xz - "2 y
. ~
b1U + X3 - «1 Y b0U - a0 Y
y
b3 U + X1
Xz and
Substituting Y in all the equations, ... (la)
... (lb)
.
. .. (le)
X3 These equations give the required state model. X(t) = A X(t) + B U(t) A
where
f-a2
1
l-·•
0 I 13=!1>1-.i1b1 0 OJ lbo-aob3
-.10
and
Y(t}
c
where
1
Ol
ft'2 -a2b3
l
CX(t) + DU(t) [1 0 0),
D = b3
Now as Y(O), Y(O), Y(O), U(O), U(O) and U(O) arc the known initial conditions, using the derived state model, the initial conditions X1(0), X2(0) and X3(0) can be obtained.
>•
Examplo 2.4 : T(s) =
A feedbacksi;stem is clwacteriud by
t11e
closed loop trausferfunction,
s2 +3s+3
~3+2s2+3s+I Draw a suitable signal flow graph and obtain lite state model. Solution : Divide N and D by s3. 1 3 3 -+-+s g2 SJ 1+~+1-+..!... s ,2 53
1 3 3 -+-+s S2 SJ
1J-~_1__..!...] L
s
s2
s3
Modem Control Theory
2 • 12
State Space Representatlon
3
1'2=-2, s
And 61 = d2 = 63 = 1 with no combinations of non-touching loops. There arc mnny signal flow graphs which can be obtained to satisfy above transfer function. Method 1 : The signal flow graph is as shown in the Fig. 2.6.
-1
Fig. 2.6 From signal flow graph, y
.
X1
Xi
X3 + 3U - 3Y ~ - 3X1 + X3 + 3U
X3
3U - Y = - X1 + 3U
Hence the state model is,
i.e,
.
x
AX(t) + BU(t)
Modem Control Theory
and
Y(t)
K
(1 0 OJ
i.e.
Y(t)
=
X(t) with D
x231 [~
l
2-13
State Space Representation
=0
Method 2 : When m -J. n and m < n then signal flow graph can be constructed so as to obtain matrix A in phase variable form. This is shown in the Fig. 2.7.
-1
Fig. 2.7 From the signal flow graph,
. Xz .
X2
X3
- X1 - 3X2 - 2X3 + U
y
3X1 + 3X2 + X3
X1
and
~
Hence state model has, A
= [ ~ ~ ~} -1
D
=
6
= [~}
-3 -2
C "' (3 3 1 J and
I
0
Thus the matrix A is in phase variable form. Note : When m < n, then transmission matrix D = 0.
Modern Control Theory 2) Using
State Space Representation
Direct Decomposition
of Transfer
Function
Th.is is also called direct programming. In this method, denominator of transfer {unction is rearranged in a specific form. To understand the rearrangement, consider an element with transfer function _]_. From block diagram algebra. the transfer function of
s +~1
minor feedback loop is
1
+~H for negative feedback,
.!.
Lei
where
_s_=_G_ I+~ l+GH s
s+a
.!.s = integrator and H = a
G
The feedback is negative and the transfer function can be simulated as shown in the Fig. 2.8. with a minor feedback loop. Now if such n loop is added in the forward path of another such loop then we get the block diagram ns shown in the Fig. 2.9.
H
Fig. 2.8 1
S+i
r-----------------------1 I I I I
I I I I I I I
I I I I
!-----------------------~ b
Fig. 2.9 The transfer function now becomes, s (s+a)
·;:-b
S (S.j.
J)
Modem Control Theory
State Space Representation
2 • 15
X = (s ... a)
where
U now the entire block shown in the Fig. 2.9 is added in the forward path of another minor loop with an integrator and feedback gain 'c', we get the transfer function as, = -1- where Y = sX + b sY ... c Thus the denominator of transfer function becomes s (sX + b) = [s (s (s + a) + b)l = s3 + as2 ... bs Thus denominator of any order can be directly programmed as discussed above.
s2 + as + b
=> Is (s + a) + bl
s3 + as2 + bs + c => l[(s + a) s +b] s + c] s4 + as2 + bs2 +cs+ d=> I([ (s + a) s + bl s + c) s + d} and so on. Now if numerator is b1s ... b0 and denominator. Simulation is obtained directly then the block diagram is as shown in the Fig. 2.10. But s = ~~· which is differcntfotor and is not used to obtain state model. In such a case, take off point 'I' is shifted before the last · integrator block.
OireCl simulation cf denominator
Fig. 2.10 (a)
C> [>
TI~J U(s)
'
'I
~------------Fig. 2.10 (b)
........... 1--'-<Xl---Y(s)
Modern Control Theory
State Space Representation
2-16
·~'I I
I I
-------------Fig. 2.10 (c) According to block diagram reduction rule, while shifting take off point before the block. the take off signal must be multiplied by transfer function of block before which it is to be shifted. Thus we get block of b1 with take off from input of last integrator. Similarly if there is a term b2s2 in the numerator ihen shift take off point before one more integrator as shown in the Fig. 2.10 (d).
~--~--_ _[_ '
'
: ----------·
:
~,. $
~
•
:
$
•
:
:
;
:
:
Sin'u..Qtionof
numer~or bg • b,s • bit2
Fig. 2.10 (d) Thus for any order of numerator, complete simulation of the transfer function can be achieved. Then assigning output of each Integrator phase variable form can be obtained.
as the state variable, state model in the
2 .17
Modem Control Theory
1.+
State Space Representatlon
Obtain state model by direct decomposition method of a system whose transfer/1111ctiDn is Example 2.5 :
Y(s)
5s2 +6s + S
U(s)
s3 + 3s2 +7s+9
Solution : Decompose denominator as below, s3 + 3s2 +7s+9
= {(Is+
3)s+7)s+9}
Its simulation starts from (s + J) in denominator
Fig. 2.11 To simulate numerator, shift take-off point once for 6s and shill twice for 5s2. Therefore complete state diagram can be obtained as follows.
Y(l)
U(l)
Fig. 2.12 Assign output of each integrator as the state variable.
Modem Control Theory
While output,
2 -18
x~
U(t} - 9X1(t) - 7X2(t) -3X3(1)
Y(t)
SX1(1) + 6X2(t) + SX3(l)
State Space Representation
: . Sta te model L•,
and
X(t)
AX(t) + BU(!}
Y(t)
CX(t) + DU(t) 1
where
A
[_ ~
c = Ia
0
()]1 •
-7 -3 6
B=m
st .
The matrix 'A' obtained is in the Bush form or Phase Variable form.
2.2.3 Advantages The various advantages of phase variables i.e. direct programming method arc, 1. Easy to implement.
2. The phase variables need not be physical variables hence mathematically powerful to obtain stale model. 3. It is easy to establish the link between the transfer function design and lime domain design using phase variables. 4. In many simple cases, just by inspection, the matrices A, ll, C and D can be obtained.
2.2.4 Limitations The various limitations of phase variables arc, 1. The phase variables arc not the physical variables hence they loose the practical
significance. They have mathematical importance. 2. The phase variables arc mathematical variables hence not available for the measurement point of view. 3. Also these variables are not available from control point of view. 4. The phase variables arc the output and its derivatives, if derivatives of input arc absent. But it is difficult to obtain second and higher derivatives of output. 5. The phase variable form, though special, does not offer any advantage from the mathematical analysis point of view. Due to all these disadvantages. canonical variables arc very popularly used to obtain the state model.
Modem
State Space Representation
2·19
Control Theory
2.3 State Space Representation
using Canonical Variables
This method of obtaining the slate model using the canonical variables is also called parallel programming method and matrix A obtained using this method is said to have canonical form, normal form or Foster's form. The matrix A in such a case is a diagonal matrix and plays an important role in the state space analysis. The method is basically based on Partial Fraction Expansion of the given transfer function Tis). Consider the transfer function T(s) as, bos"' +b,sm·t +b2srn·2+ ... +bm (s+a 1) (s+a2) .... (s+ a 0)
T(s) =
Case 1 : If the degree 'rn' is less than 'n' (m < n), then T(s) can be expressed using partial fraction expansion as,
c Now each group s + ~
can be simulated using the minor loop in state diagram as 1
shown in the Fig. 2.13.
------------------, I I
I
~-----------------' '-......,.......
I I
The outputs of all such groups arc to be added to obtain the resultant output.
'
To add the outputs, all the groups must be connected in parallel with each other. The input U(s) to all of them is same. Hence the method is called parallel programming. The overall state diagram is shown in the Fig. 2.14.
'
0
....,.,,
1 s • a1
Fig. 2.13
Then assign output of each integrator as a state variable and write the state equations as,
While X
AX + BU
and Y = CX + DU
Modem Control
Theory
Stale Space Representation
2 ·20
U(s)
Y(s)
..
. ~---·
~--· I I
Fig. 2.14 Foster's fonn sim\Jlation
0 where
A
[1
-a2
0 0
0
0
C = (c1 C2 ••••• C,,)
j and
B
•[j]
0=0
Case 2 : If the degree m = n i.c. numerator and the denominator have same degree then first divide the numerator by denominator and then obtain partial fractions of remaining factor. N(s)
--=eo+
T(s) where
Co
ocs>
=
~
C;
k-i•I s+a;
Constant obtained by dividing N(s) by D(s).
In such a case, the state diagram for partial fraction -.terms remains same as before and in addition to all the outputs, Co U(t) gets added to obtain the resultant output as shown in the Fig. 2.15.
2 ·21
Modem Control Theory
State Space Repl'8Sentatlon
U(s)
•
Additional elemellt Fig. 2.15 State diagram for T(s) with m • n Thus the state model consists of the matrices as,
=
c
(c1
C2 •••
c,,J,
D
= Co
Key Point: 17111s for m • n, direct transmission matrix D exists in the state model. When the denominator D(s) of the transfer function T(s) has non-repeated roots then the matrix A obtained by parallel programming has following features, 1. Matrix A is diagonal i.e, in canonical form or normal form. 2. The diagonal consists of the clements which arc the gains of all the feedback paths associated with the integrators. 3. The diagonal elements are the poles of the transfer function T(s). 1!11$> Example 2.6 :
Obtain the state model by Foster's form of a system tu/lost T.F. is,
s2 +4 (s + l)(s + 2) (s + 3)
Modem Control Theory
2-22
Stato Spaco Reprosontatlon
Solution : Find out partial fraction expansion of it s2 + 4 - ------,:; (s -e- 1 l(s ... 2)(s + 3) -
2.5 s -t- 1
s
-·-----
s+2
6.5 s+ 3
+ --
:. Total state diagram is as shown in the Fig. 2.16.
U(I)
Y(I)
Fig. ., 2.16
Y(t)
2.5X1(t)-8X2(1)+
State model is, X
AX+ BU
and
y
ex+ ou
where
A
[-10 -20 0OJ 0 0 -3
c
(2.5 - 8 6.SJ
6.5X3(t)
B=m D=O
Modem Control Theory
State Space Representation
2 ·23
2.3.1 Jordan's Canonical Form In the Foster's form it is assumed that the roots or denominator of T(s) are non-repeated, simple and distinct. But if the roots arc repeated then the parallel programming results matrix A in a form called Jordan's Canonical form.
Let TM has pole at s " - a1 which is repeated for r times as, T(s) =
N(s) (s+a1)'(s+a2) •.. (s+~,.)
The method of obtaining partial fraction for such a case is, T(s) = __ c_•-+ (s+a
il'
Cz (s+a ,)r-1
+... +-c-'-+~+ (s+a 1) (s+a2)
If the degree of N(s) and D(s) is same i.e, m
e
....+~ (s+a0)
... m
n we get additional constant c0 as,
... m-=n This can be mathematically expressed as,
... m < n r
and
T(s) -Co+~
C
,,
('
~ --'z: ---'-r-J·J (s+a, ) - + P=U·l £... (s+e .) i>1:::l I
... m""n
Key Point: Note that in partial fraction expansion, a septrate coefficient is assumed for caclt power of rqJtattdfactor. In simulating such an equation by parallel programming, --1-(s+a1)'
is simulated by
connecting -( -1-) groups, r times in series first. While all other distinct factors arc s+a1 1 simulated by parallel programming as before. The components of each power of -( . ) to s+a1 be added lo get output is to be taken from output of each integrator which arc connected in series. This is shown in the Fig. 2.17. Now assign state variables at the output of each integrator. For series i.ntegrators, assign the state v.oriablcs from right to left. as shown in the Fig. 2.17.
Modern Control Theory
State Space Representation
.:
'
/'• t
.
\, \
''
'
'
'' ''
G
.'
'
1
': '' '' '
''
'
''
: '
J
:
:
'' '' '' ' ' . --, ' ' ... • v ; -' ... ' ': '' '' ' : :: '' : ''' '
,''
-:'r-·---------'
'' '
t: t'
-rl
'' ' ''
'' '' J'
•''
.
'
1-~~~~~~~~~~~~~~---------L
3
' ...... J'........
..f
Modem Control Theory
State S'pace Representation
2 -25
For series integrators, the state equations arc,
While for parallel integrators, the state equations arc,
X,. +
I
=
-<12 X, + I + U(t)
While
Key Point: Y(t) has additional c0 U(I) term
if 111
= 11.
Hence the state model has matrices in the form, . --a1forrtimes~
1-•,, 0 9.
0
A~ Jordan block
f 1
Element tforr-1 times O·····O
-a, 1 ····· <, o -a1······1
0
"''
0
O······
0
0 ···· ······--··O
o.
o ..... ..:a,
0
0 ····· 0
-, -a,
O·· ·· 0
o
1
········0
0
O····--·--···--·
o
0
o ··············
o
0
0
0
-1----------'
a2
O········--·O
: ·O
0 ····
: "-L
~
Fig. 2.18
0
:
~~:
only diagonal elements
an
Modem Control Theory
2 ·26
~1
: B
State Space Representation
r-1 times
?J
C
=
(c1 C2 ••• c, <:, • 1 ... c0J
The matrix D is zero if m < n and is '<:o' if m = n. The matrix A in such a case has a Jordan block for the repeated factor and many times A is denoted as J. X(t) = JX(t) + BU(t) The matrix A
... For repeated roots
= J has the following features,
1. The principle diagonal consists of all the poles of transfer function with repeated pole 'r' times and other nonrepeated poles. 2. Upper off diagonal consists of (r - 1) times unity element, as indicated in the Jordan block. 3. All the remaining elements arc zero. Note that the matrix B has 'r - 1' zeros and all other clements as unity. The matrix C has all partial fraction coefficients c1, c2 ... c;,.
2.3.2 Advantages of Canonical Variables The main advantages of canonical variables arc 1. The matrix A is diagonal 2. The diagonal clement is very important in the mathematical analysis. 3. Due to diagonal feature, the decoupling between the state variable is possible. This
.
means all 'n' differential equations are independent of each other. Thus X1 depends on X1 alone.Xi depends 011 "2 alone and so on. Such decoupling is important in system design from controlling point of view.
2.3.3 Disadvantages of Canonical Variables The main disadvantages of canonical variables is similar lo the phase variables. These arc not the physical variables hence practically difficult to measure and control. Hence such variables arc not practically advantageous, though mathematically arc very important
2. 27
Modern Control Theory
1•
State Space Representation
Example 2.7: Obtain the statt model in ]ord1111'; canonicalform of a systtm wlrose T.F. . I IS (s+2) 2 (s+t)
Solution : Finding partial fraction expansion, T(s) : _A_ + _B_ + C (s + 2) 2 (s + 2) (s + 1) Take LCM on right hand side and equate numerator with numerator of T(s), ;.A(s + 1) + B(s + 2) (s + 1) + C(s+2)2:
1
Equate coefficients of all powers of s on both sides,
a ... c A+ 38 + 4C A+28+4C
s2
0,
from power of
0
from power of s
r
... from constant term
Solving we get, A " - 1, B " - 1, C ~ 1 -1 1 1 T(s) = (s+ 2)2 - (ST 2) + (s+ 1) Simulate first term by series integrators while other nonrepeated integrators.
terms by parallel
Yi!)
Fig. 2.19
.
Total simulation is, Xt
.
X2 = U(t) - 2X2
.
X3 "U(t) - X3(t)
Modern Control Theory :. State model
2 ·28
X
AX+ BU
A
r-02
-2
0
0
is ,
l
c = I- 1
-~J
- 1 1j '
State Space Representation
and
B = D
Y =ex+
ou
[!]
=0
The matrix A consists of Jordan block. Important Noto : If some of the poles o{ T(s) arc complex approach can be used. The quadratic or higher order polynomial be simulated by direct decomposition while real distinct roots parallel programming using canonical variable, lhe example procedure.
l*
in nature, then a mixed having complex roots can can be simulated by the 2.8 below explains this
Example 2.8 : C"111bi11ntio11 of Direct Dtco111positio11 1111d Foster's Form. Obtain tire state modelfor tire given 1·.F.
tto ~
·--2---
(s -e- 3) (•2 +25+2)
Solution : Obtain partial fractions as, T(s)
2
s
(s+ 3l(s2 + 25+ 2)
=~ + lls + C s+ 3 s2 + 2s+ 2
Find LCM on right hand side and equate numerators or both sides, .'\(s 2 + 2s + 2) + (lls + C) (s + 3) = 2
As: +2As+2J\ + Bs~ +Cs+ 3Bs+ 3C = 2 Equating coefficients or all powers of s, A+B
0
2A + C-3A C-A
2A+3C=2
2A+3A=2
0
c Solving,
0
2A+C+3B=O
A
T(s)
A
2 5.
2
B=-5,C•5
2
2 2 2 2 2 2 = _L + -5s+5 =l + -5s+5 (s + 3) s2 + 2s + 2 s+ 3 {(s + 2) s+ 2}
The quadratic s2 + 2s + 2 having complex roots is decomposed directly.
Modem
Control
Theory
2 ·29
State Space Representation
:. Complete state diagram is as shown in the fig. 2.20
U(t)
Fig. 2.20
and
X1
U(t)-3X1
X2
x3
X3
U(t) - 2X2 - 2X3
Y(t)
~X1(l)+~X2(t)-
~X3(t)
ex
:. State model is,
x
AX +BU and Y
where
A
[- ~ _!],Bom
Q
0 0 -2
Note that in such a case matrix A does not have any specific form.
2.4 State Model by Cascade Programming This is also called Pole-Zero Form or Cullemln's Form. This can be effectively used when both numerator and denominator can be factorised. In such form, group a pole and zero together and arrange given transfer function as the product of all such groups. Then simulate each group separately and connect all such simulations in cascade to get complete simulation. Then assigning output of each integrator as a state variable obtain a state model in standard form.
State Space Representation
2. 30
Modem Control Theory
Simulation of group : Consider
s+ a
'S+b
First simulate denominator _l_b as in the Fig. 2.21 (a) and then as discussed earlier
s+
simulate the numerator (s + a) as shown in Ille Fig. 2.21 (b),
~------·------
..
I
, _ - - - - - - - - - - - - _, (a)
s
!b
(b)
~: ~
Fig. 2.21 Simulation of a group of pole-zero
>Ill+
Example 2.9 : Obtain Ille state model of a system by cascade programming w/Jose transfer
junction is T(s) ,. Y(sl ,. (s+ 2)(s+ 41 l/ls) s (s+ 1} (s+ 3) Solution: Arrange the given T.F. as below Y(s)
(s+2)
(s+4)
U(s)
(s+l)
(s+ 3)
! Group l
! Group 2
l Group3
Now simulate each group as discussed and connect all of them in series to obtain T(s). The complete simulation is shown in the Fig. 2.22.
Fig. 2.22 Now
.
.
X2 = - 3X2 +2X3 +X3
Modem
2. 31
Control Theory
State Space Representation
X.1 = U(tl-X3
and
Substituting X 3 into X2 equation,
Substituting
x2
into X1 equation,
Model becomes,
X3
[' ' r· ['I
and
Y(t)
X1(l)
i.e.
Y(t)
[1
So
A
X1 Xz
0
- 3 -
1
X2
0
0
1
X3J
0
[O0-3 1 0
c
=
0
[I 0 OJ
rn:l
OJ
111JB=1 -l
J
1
+
I
U(I}
1,
n 1
0=0
Feature of A : Feature of Matrix A is its principle diagonal contains gains of all feedback paths associated with all integrators i.c, 0, -3, -1 in above problem. All terms below principle diagonal are zero. Thus the features of A having all poles of T(s) in its principle diagonal still continues in this method of programming. This method is also known as Iterative programming.
Examples with Solutions na+
Example 2.10 ; Obtain the state model of the given network i11 tire sta11d11rd form. Assume R1 l Q C1 1F
=
=
n
R2
2
R3
3Q
c, = 1 F
Modem Control Theory
2 -32
State Space Representation
Fig. 2.23 Solution : Selecting state variables as voltages across capacitors C1 and C2
i.e. c1 and c2,
R,
X1(1)
c1
=
~l
X2(1)
U(I)
J
+
t
1,(ll
Applying Kirchhoff's laws, U(t)-i1
R1 -c1
-R2
U(t)
=
o-------'------'---o{<')
(i1-i2)=0
Fig. 2.23 (a)
i1 R 1 "'- c 1 + R2 (i 1 - i2)
... (1)
fur second loop, ~1-i2
R;i -c2 -(i2 -i1)R2
o ... (2)
Solving simultaneously equations (!) and (2) U(t) "1
;, (R, + Rz)- iz R3 + c, - i, Rz + i2 (Rz + R3) .. l'2
Substituting the values, U(t) i.e ..
i 1 (I + 2) - 2i z + c 1
U(I)
. .. (3) ... (4)
Multiply equation (3) by 2 and equation (4) by 3 2 U(ll
6i1-4i2+2e1 - oi 1 ~ 15i~ • 3c~ and adding
3 U(t) + :'lc1
2 .33
Modem Control Theory
State Space Representation ... (5)
Now from equation (3) U(t) = 3i1
-
2i2 + e1
Substituting i2 from equation (5) 6 2 6 U(t)+ IT U(t)+ IT e1 -e1 - IT e2 U(t)
3i I
-fr
U(t) - ~el
+fr <'z
+el
17 9 6 33 U(t) - 33 cl - 33 cz
... (6)
Now
... Capacitor current is C (dvc/dt)
as C1 =I,
dt
dct
17 9 6 3 I 3 33 U(t) - 33 e I - 33 c2 - IT U(t) - IT ct + IT ez
dc1
8 12 33 U(t) - 33
dt
.
e1
3 + 33 cz ... (7)
X1 and
as
dc2 C2(11
... Capacitor current is C (dvc/dt)
Cz dc2
dt
3 1 3 ITU(l)+ITel-Tfez ... (8)
X2 and
Y(t)
l [l
:. State model is, X
AX + OU and
where
12 [-~ 11
A
1
y = ex + DU
-~,B=~ 11
8
11
c
= [O ,tJ
D
= IoI
2. 34
Modem Control Theory ,..,
State Space Representation
Example 2.11 : Consider t/1t mechanic.a/ system slrou•n iu figure. For sllown displacements and velocities obtain tire state 1twdel i11 staudard [arm.
Ass11me t•elocity of M 2 as output.
v,
v,
Fig. 2.24 Solution : Select the state variables as energy sloring elements i.e. displacements and velocities related lo spring and friction. Y1(t), Y2(t),
F1(t),
x3
= V1(t)
X4 = V2(t)
Y(t)
=
V2(t)
Draw the equivalent mechanical system. Due to F1(t), M1 will displace by Y1• Due to spring K1 and friction 81 which arc between the two masses, the displacement change from Y1 to Y2• While mas' M2, spring K2 and friction 8z are under the Influence of Y2 alone as K2 and 8z are with reference to fixed support and not between two moving points. Represent each displacement by separate node. Connect the clements in parallel
Fig. 2.25
2. 35
Modern Control Theory
State Space Representation
which arc under the influence of same displacement, thus K1, B1 parallel between Y and Y 2, m2, K2 and Bi parallel between Y 2 and reference i.e. fixed support and so on. The spring force is proportional to net displacement in spring while frictional force is proportional to the velocities. At node Yi... (1)
At node Y2, 0 ... (2} Substituting all values in tenns of state variables we get, d2Y __ 1
• = XJ
dt2
:. Substituting in equation {l) and equation (2) U(t) 0
M2
... (3)
.
X4 + K2X2 + B2X4 + K1 [X2
-
Xi)+ B1 (X4
-
X3]
... (4)
From equation (3) and equation (4) we can write
. .
X3
~
{U(t)-K1(X1-X2)-B1(X3-X4))
... (5)
I
and
X4
and
X1
Mz
X3,
[-K2 X2 -B2 X4 -K1 (X2-X1)-B1 X2
=
X4, Y{t)
=
V2(t)
=
:. State model can be constructed in the standard fonn
.
X
AX+ BU
v
ex .. ou
X4 (t)
(X4 -X3))
... (6)
2. 36
Modem Control Theory 0 0
0 0 A
Where
K1
0
+~ M1 -(K1+Kz)
-Mi K1 M2
,...,
State Space Representation
M2
Example 2.12 : For tht given T.F. of a system obtain t/ie stole model by i) Direct decomposition T(s} =
ii} Gui/em in 's Form
iii} Foster's Form
(s+2)(s+3) s(s+l)(s2 +9s+20)
Solution : I) Direct decomposition T(s) =
s2 +5s+6 s(sJ + 10s2 + 29s+ 20)
s2+5s+6 {([s+ IO]s+29)s+20
}s
State diagram is as follows
Y(sl
Fig. 2.26
X4 y
U - 20X2 - 29X3 6X1+SX2+X3
-
lOX4
,
2 .37
Modern Control Theory
State Space Representation
:.State model is
where
x
AX + llU and Y ; CX
A
c
0 0 0 0 0 -20
["
-29
[6 5 I
oI
0 I
0
0 0 I
-10
+
DU
•. [!]
D= (0
I
Ii) Gullemln'sForm T(s)
(s+ 2)(s+ 3) (s+ 2) (s+ 3) 1 s(s+ ))(s+ 4)(s+5) =(s+l).(s+4) (s+S) s
State diagram is,
UC•)
Fig. 2.27
Substituting X4 back in X J
Substituting X 3 back in X 2
and
y
2. 38
Modem Control Theory
State Space Representation
:. State model becomes
x
=
0
I
where
A
and Y =ex
AX+ BU
[;
-1
-5
-4 0 0 0
-11
iii) For Foster's form :
C = [ 1 O O OJ
··m
Find out partial fraction expansion of T(s) A ll C D T(s) ;;;; s + 1 + s + 4 + s + 5 + S -1/ 6 -s+ I
.
1 / 6 3 I 10 3 / 10 -----. -s+4 s+5 s
U(IJ
Y(t)
U(t)-X.,
Fig. 2.28 1
1
3
Y=-6Xt+6X2-l(jXl+l(jX4
3
2 .39
Modem Control Theory
State Space Representation
State model is, X =AX+ BU
and
Y
where
111•~
=ex A=
EJtample 2.13
[~00 1 - 0 -~05 ~01OJ B = [~11
C=[-i
6
-to
to]
Derive tire stale model in /orda11's ca11011ical form for a system lunri11g T.F.
T(s)
Solution : T(s) =
l
0o4
=
l s3 +4s2 +5s+2
A B C 1 = I =--+--+-(s + 1} 2 (s+ 1) (s + 2) s 3 + 4s 2 + 5s + 2 (s ... I) 2 (s + 2)
i.e, A(s + 2) + B(s + 1) (s + 2) + qs + l) 2 = 1 As + 2A + Bs 2 + 3Bs + 2B + Cs 2 + 2Cs + C B+C
0 , A + 313 + 2C
=0
r
2A <· 28 + C Now C = 1 , B = -I, A = 1 T(s)
1 (s+1)2
1 (s+l}
1 (s+2)
-----+--
State diagram L~ , U[I)
Y(I)
Fig. 2.29
X1
- Xi+ X2
x~
U(t) - X2
X3
U(t)- 2X3
x1
Y(s) :.State model is,
where
State Space Representation
2 -40
Modem Control Theoiy
.
-X2 +
x_,
x
AX+ BU
A
[~1 ~1.
Y =CX
-1
B=m,
0 -2
1•
c
= (1 - 1
1)
Example 2.14 : Obtai11 the state model of system whose T.F. is
s 3 + 3s2
+ 2s
------- by Foster's form.
T(s)
s3 +12s2 +47s +60
Solution:
T(s)
=
s3 + 3s2 +2s s3 +12s2 +47s+60
As numerator and denominator are of same order we cannot directly find out partial fractions. For partial fraction, numerator d~'b'TCC must be less than denominator. divide N(s) by D(s} and find partial fractions of the remainder. s 3 + 12s2 + 47s + 60) s 3 + 3s2 + 2s ( 1 s3+12s2
- 9s2 T(s}
=
-
+ 47s+ 60 45s-60
Y(s)
U(s)
_ 1
-
[ 9s2 + 45s + 60 ] - s 3 + 12s2 + 47s + 60
9s2 +45s+60
l - [ (s+3)(s+4)(s+5)
]
A
= l -'S+'3 -
1 - _3_ + ~ - 2Q._ s+3 s+4 s+S
B
C
s+4- s+S
So directly
2-41
Modem Control Th~ry
State Space RepresentatlO<•
State diagram :
V(I)
V(t)
Fig. 2.30
X3
U(t) -5X3
Y(s)
- 3X1+2-IX2
- 30X3 + U(s)
.·. State model is,
x
AX + BU
and
y
=
ex
+ DU
where
C = [ - 3 24 - 30],
D = [11
When N(sl and D(s) are having same order, or degree of N(s) > D(s) then, there is always direct transmission matrix D present in the model.
••
Example 2.15 : Obtain a stat« space model of tile system will! transferfunction Y(s} U(s)
6 $
3+6:>2+11$
+6 (VTU:J•n.1Feb.-2007,July/Aug.-2007,Dec)Jan·2008)
2·42
Modem Control Theory
State Space Ropresantatlon
Solution: Y(s)
The T.F. is
6
U(s)
s3 +6s2 + lls+6
Using factorisation
of denominator,
Y(s)
U(s)
6
=
(s+l)(s+2)(s
... J)
Taking partial fractions, Y(s) = -1.._ _ _2_+_3_ U(s) s+I s+2 s+3 Hence the state diagram is as shown in the Fig. 2.31 U(s)
Yis)
Fig. 2.31 This is Foster's form of representation. From above Fig. 231 we get,
and
Xi
U(s)-X1
X2
U(s) - 2X2
X3
U(s) - 3X3
Y(s)
3X1 - 6X2 + 3X3
Hence the state space model is,
.=
X
AX.;. BU
2 .43
Modern Controltheory
and
y
where
A
=
State Space Representation
ex 0
n -~l -2 0
B
e ,..,.
m = (3
- 6
31
Obtain Ille state model for system represented by
Example 2.16 :
•J>y + 6-J2y + 11-dy +I011=3U(t) dt3 dt2 tit . Solution : System is 3rd order, n
(VTU:
=3
May·99, Dcc.2007/J•n.-2008)
3 integrators and variables are required. Select y = X 1 and then successive diffcrcntfation of y as next variable.
x.
X2 ~ dy/dt
. .. (1)
X2
d2v X3=-' dt2
... (2}
Now as 3 variables arc defined, X 3 .,. X,1 but X3 must be obtained by substituting all selected variables in original differential equation. as
x3 y •
... (3)
x,
which is output equation.
:. State model can be written as,
.
x
AX+ BU
and
y
ex+ ou
Where
A
L~
1
0 -11
!]·B=m,e:Jl
0 0 J, D =JOI
Medem Control Theory
2-44
State Space Representation
State diagram :
U(sl
Fig. 2.32 ,...
Example 2.17 :
OblRin the state model of tl1t differential equation
4 ''.l'i.!.!.+ 3 'l1dl) +de(/)+ 2cllJ d 1J
d 11
tit
= Sr(t) (VTU : Aug·95)
Solution :The equation can be written as,
.
4 c (t) • 3 c (t)+ c(tJ+ 2c(t) = Sr(t)
The order of the equation is 3. Selecting state variables as, c(t}
. .
X1(t)
X2
X1(t) = c(I)
... (I)
X3(t)
X2(t) = c(t)
••. (2)
X3(t)
c(t)
Substituting in the original equation, 4X l (t) + 3X3 (t)+ X2 (I)+
2x, (t)
= Sr(t) 2
-4X,(t)-4
Hence the state model is,
[ f:
l [_j _Jl _J
I
3
5
X2(t}-4X3(t)+4r(t)
... (3)
2·45
Modem Control Theory
St.ta Space Repreaentatlon
While the output equation is, c(t)
XI (1)
1 (I)]
X
i.e.
y(t) " [ 1 0 OJ
X2(t) [
••
X3H)
Example 2.18 : Obtain thL stat~ modLJ of t/U! syst~ whose cl~
loop trtmsfufunction
is
cot
2(:;+ 3)
R(s)
(s+l)(s+2)
(VTU : Feb-97, Orl-98)
Solution : Let us use the parallel progranuning Le. Foster's form. Finding partial fractions of C(s)/R(s).
Hence the state diagram is as shown in the Fig. 2.33
C(s)
Fig. 2.33 Hence the state equations from the Fig. 2.33 arc ... (!)
R(s) - 2X2 and
C(s)
Hence the state model is,
... (2)
Modem Control Theory
2 ·46
State Space Representation
and
1•
Example 2.19 : Write a set of stale tq11nlions for tltt 11tlwork shoum itt lht F(~. 2.34 Clroose i1, 12 find vc as $/alt variables. (VTU: April-98,July/Aug.·2007)
Fig. 2.34 Solution : Let
c(t)
Input
Voltage across R2 i,(t) i2(t)
vc(t) .. x,(t) Let i1(t) and i1(t) arc the loop currents. Applying KVL to the loops. c(t)
..• (1)
Then,
... (2)
and
c~~ di dvc (fl
... Capacitor current 1 (.
a
C •1-lz
. )
Modem Control Theory
2 • 47
State Spaco Representation ..• (3)
... (4)
While Hence the state model is,
x, X2
l x3
RI
-Li
0
0
-i:;-
-Li
R2
I
I
-c;
e
r x,
l L2
X2
0
X3
+-
. [?
l ""'
where "'''
"~·>
and
,,_.
Example 2.20 : Considering'Ve am/ 11 as state mriab/e:; and Is as tire outinu variables 111
tht circuit shown below, obtain the statt model. R1
L
IVVV'-v-,1-.-c-·
'-3)'• "
1
Fig. 2.35 (VTU : Aug. • 97)
Solution : Convert the CWTClll source to voltage source as shown.
R,
Fig. 2.35 (a)
Two inputs,
c{t) and Ii: (t) i.e. U1 = c(t) and U2 = lg(t)
One output, Vnriablcs,
State Space Representation
2 ·48
Modem Control Theory
i.e. Y(t) = J,.(t)
l,(t)
vc
=
X1(t) and Ii
= X2(t)
Let l,(t) and l;(t) be the loop currents. Applying KVL to the two loops, Loop l,
-I, R1
-
"c + c(t)
=o
c(t) - "c 1 1 - e(t)-vc(t) R1 RI Y(t)
1 1 - U1 - -- X1{t) R1 R1
di, Loop 2' - L dt - I' R2 - IS R2 + v c
=
... Output equation 0
... State equation and current through capacitor, dvc
CCit 1
ell, -I ;I Substituting I, from output equation, ... Stale equation
- ~~]l~;] y
[-R;"1 0][X1Xz(I)(t)] + [ R;"l 0][Uu, 1]
Modem Control Theory ••
Example
2.21 :
State Space Representation
2·49
Derite two state models for thL system described '1y the differential
eqllation D3y+4D2y+5Dy+2y i) ii)
= 2D2u+6D11+Su
where D = d/dt
One in phase variable form. Other in /ordan-Ca11011ical form.
(VTU: Mardt-2001)
Solution : Take the Laplace transform of both sides of the equation and neglect the initial conditions to obtain transfer function of the system as, s3Y(s)+4s2Y(s)+5sY(s)+2Y(s) Y(s) U(s)
= 2s2U(s)+6s
U(s)+SU(s)
2s2 +6s+5 s3 +4s2 +5s+2
I) Phase variable form
Use the direct decomposition. Y(s) 2s2 +65+5 U(s) = {((sH)s+5]s+21 The state diagram can be shown as in the Fig. 2.36
Fig. 2.36
So
X1
and
Y(s)
2. 50
Modem Control Theory
So state model is having matrices,
A
= [ ~2
~S ~
l
=
B
m
State Space Representation
and C = (5 6 2)
ii) Jordan canol\i."I form : fllclorise the denominator as, Y(s) U(s)
A(s+2)+B(~+l)
2s2 +6s+5 = (s+1)2 (s+2)
c
=
(s+2)+C(s+l)2
_A_+_D_+~ (s+1)2 (s+l) (s+2) 2s2 +6s+5
As+2A + D~2 + 3Ds+2D+Cs2 +2Cs+C = 2s2 +6s+5 B+C A
= 2,
= I,
A + 38 + 2C = 6, B = I,
2A + 2B + C " 5
C =l
Y(s) _1_ + _1_ + _1_ U(s) = (s+1)2 (s+l) (s+2) The a.talc model is shown in the Fig.2.37 (a) U(s)
!------<><>---
Fig. 2.37 So and
Y(s)
So state model is having matrices,
A = [~
~1
~J
B
=
m
and C = [1 1 1)
Y(s)
Modem Conttol Theory ,..
2. 51
State Spa~ Representation
For lht system s/1~1 in the Fig. 2.38, obtain the state 111od4 choosing v1(t} and 112(1) as tlte state wriab/es, (VTU: July/Aug.·2005, Jan./Feb.-2007) Example 2.22 :
Fig. 2.38 Solution : Select the two currents as shown in the Fig. 2.38(a). And wfife tho equations using KVL and KCL ..
1 Mil v1(t) U(t) 1 Mil vi(t) +9-':"V\NV~-..~.....,,,,'llJ\~~.-~~-<> U(t) ~
_l
T
.,
1µF~ '2
1
lµF
Fig. 2.38 (a) U(t)-v2(tJ
txl06 Yz(t) i.e.
~J
... (I)
(l t -J2) dt
C dv2(t) dt v2 -v1
... (2) ... (3)
lxl06
And
~Ji2
dt ... (4)
Elliminate i1 and i2 from above equations and C = l x 10- 6 F.
Modem Control Theory
2 -52
State Space Representation
Substituting (I) and (3) in (2) we get,
c ~.l. dt dv2
di
[U(t)-v2(t)
]-[ v2(t)-v
ixio>
1+~-+! C 1x106
dv2
dt
1(t)]
lxto6 Vl(t) _,!_[2V2(t)]
C txto6
C 1x106 ... (5)
v1(t) - 2v2(t) + U(t)
Using (3) in (4),
c~dt dv1 Select
Vz -Vt
Jxt06
di
- V1(l) + V2(t)
X1(t)
v1(t) and X2(t)
.•• (6)
= v2(t)
Using selected state variables,
and and
X1 Y(t)
.
2Xi +
U(t)
V1(t) = X1(l)
Hence the stale model is,
and
X(t)
AX(t) + BU(t)
Y(t)
CX(t) + DU(t)
where
>•
Example 2.23 :
A
[-: ~a
8
= [~} C = (!
01 D
=0
DmrN two state models far thL system wit/1 transfer function,
i) For first model, matrix A must bt in companion form. ii) For second model, matrix A must bt in diagonal form.
2·53
Modem Control Theory
State Space Representation
Solution : Arrange the transfer function as, 50x~x(s+5)
1000(s+5) s(s+2)(s+50J
T(s) =
i) The companion form means decomposition of denominator, T(s) =
phase
variable
form
for
lOOOs+SOOO {[(s+52) s+ lOO]sl
1000(s+5) [s2 +52s+lOO]s
The state dlagram is as shown in the Fig. 2.39.
From state diagram, ~----l 1000 1----~
. .
X1 and While
)'3 Y(t)
A
X2,
.
Fig. 2.39
X2 = X3
- 100X2 - 52X3 + U(t) SOOOX1 + lOOOX2
[o0
o}
1
0
1
B=[H
0 -100 -52 C = [5000 1000 OJ,
D =0
ii) For diagonal form use partial fractions,
A
T(s) A
5+ s+2+
0
C s+SO
T(s) s]
=
• •O
1000x5 2x50
= 50
which
use
direct
2. 54
Modem Control Theory
.
B
C
T(s)
1(s)(s+2)
2
1000(3)
I. __ 2
,. (-2) (.JS)= - 31.25
• =
T(s)(s+SO)I
State Space Representation
s=-50
.:!.~.t~. =(-50) Hb/
18.75
50 31.25 18.75 s+2 - s+SO
s-
The state diagram is,
., 50
t---"-<X>--Y{t)
Fig. 2.39 (a) Thus the stale equations are, X1 and Thus,
Y(t) = 50X1 A
C
.
U(t}, ~
[~
-
0 -2 0
=-
.
2X2 + U(t), ~
=-
31.2SX2 - 18.75X3
_J}
ISO - 31.25
B = [~} - 18.75), D = O
50X3 + U(t)
Modem Control Theory
>'*
Example 2.24 : Fig. 240.
2-55
State Space Representation
Derivr tire state model for two input two output systnn shown in the
't
Fig. 2.40
Select output of simple lags as state variables. Solution : As sueeested the output of -1- is X ee ' s·H I• I . X s+5 IS 2' The state diagrams for simple lags arc,
,, s+1
Fig. 2.40 (a)
Modem Control Theory
State Space Representation ... (!)
Fig. 2.40 (b) .. (2)
0.4
s+o:5
,,
~
Fig. 2.40 (c)
... (J)
Fig. 2.40 (d) ~
... (4)
= r2 - 2X4
From the given block diagram, r1 = 1<1 [U1 - Y1)
and
r2 " ~ [U2 - Y21
... (5)
2 .57
Modem Control Theory
State Space Representation ..• (6 a)
and
... (6 b) Substituting in (5) we get,
- K1 X1
-
51<1 X2 + K1 U1
... (7)
- 0.4K2 X3 - 4K2 X4 + K2 U2
... (8)
Using (7) and (8) in the equations (1) to (4), ... (9 a)
... (9b) ... (9 c) ... (9d) TI\c equations (9a) lo (9d) and (6a) to (6b) give the required state model as,
x
= AX+ BU
['-' ;K,J -5
A=
c ,._
-Ki 0 =
l
and Y = CX + OU where -5K1 -SK1 0
[1 5 o o} 0 0 0.4 4
0 --0.4K2 --05 --0.4K2
0
~K,0 Hl
r·
' B = ~I
0
Ki 0
Ki
D =[OJ
=
=
Example 2.25 : A series RLC circuit witlt R 1 !l. L = 1 H and C 1 F is excited l!y v = 10 V. Write the stale eq11atio11s iu Ille matrix form. !B•ngalorc Univ., Oec.·95(
Solution : The circuit is shown in the Fig. 2.41.
Fig. 2.41 Applyin_g KVL to the circuit, . v = 1(1) R.,.
dl(t)
Ldt
+ vc(I)
... (1)
2 ·58
Modem Control Theory
~f i(t) dt
and dvc(I)
]_ i(t)
_d_l_
Let
State Space Representation
... (2)
c
X1 = Current through inductor
i(l)
=
X2
vc(t)
Voltage across capacitor
U(t) = 10 V
V
Substituting all the values, 10
Xi+ X1 + X2
. ~
-X1 - X2 + 10
X1 and
••. (3) ... (4)
X1 Hence state equalions in matrix form are,
[xX2•
1]
n•
=
r-1 -11 [x [10] l
O
X1] + 0 U(t) 2
Example 2.26 : Obtain tltt state space representation in phase variable Jann for the system rqiresmtl'd by. D4y + 20D3y + 45D2y+ 18Dy+100 y =10D2u+S Du+ 100 u witlt y as output a11d u as input. (VTU: Ja.nJFeb.•2005)
Solution : Phase varalble form Taking Laplace transform of both sides and neglecting initial conditions, s~Y(s)+ 20s3Y(s)+ 45s2Y(s)+ 18s Y(s)+ 100 Y(s) Y(s) U(s) =
54
%
IOs2U(s) + 5s U(s)+ 100 U(s)
tos2 + 5s+ 100 + 20s3 + 45s2+18s+100
Using direct decomposition, Y(s) U(s)
l0s2+5s+IOO {[([s+ 20)s+ 45) s+ IS]s+ 100}
The state diagram is as shown in Fig. 2.4.2
Modem Control Theory
State Space Representation
2 .59
Fig. 2.42
.
.
.
The state equations arc,
.
x, =X2, X2 =X3, X3 =X4, x~ =-toox, -t8X2 -4X3 -20X,, + u and
>'*
Example 2.27 : Obtain the state mode! of
flit
system rrsponsed by the followi11g differential
equation : y + 6 y + 5 y + y = u.
{VTU: July/Aug.· 2005)
Solution : Take Laplace transform of both sides neglecting initial conditions, s3Y(s)+ 6s2Y(s)+ 5sY(s)+ Y(s)
=
U(s}
Y(s) U(s)
s3 + 6s2 + Ss+l
Y(s) U(s}
{[(s+ 6)s+ 5]s+
I
l}
The state diagram is shown in the Fig. 2.43
1S!ttJ Fig. 2.43
Y(s)
.
.
State Space Representation
2 -60
Modem Control Theory
The state equations are, xi =X2. X2 =X3.
.
X3 =-XI -5Xz -6X3 + u
and O
1 ..
I
0 0 [-1 -5
A
Example 2.28 : Obtain two 1/ifferent slatt models for a system represented by tire followmg transfer function. Write suitable state diagram in tach rase. (VfU: July/Aug.·2005)
Y(s)
8s2 + 17s+ S
U(s)
(s+ 1)
(s2+Ss+1s)
Solution : 1) Direct decomposition : Y(s)
8s2+17s+ 8
Ss2+17s+ 8
s3+9s2+23s+15
t..:(s)
{l(s+9)s+23js+15}
The state diagram is shown in the Fig.2.44
.
Fig. 2.44
X3 "'-15Xt -23X2 -9X3 + U Y
"'
A
=
8X1 +17X2 +8X3
l·
r~ ~ ~ "'r~J. -15 -23
-9
B
1
C"' (817 8]
Modem Control
2. 61
Theory
State Space Representation
2) Foster's term : Y(s) U(s) A = - 0.125,
=
B = -7.25,
c = 15.375
Y(s) -0.125 7.25 15.375 -= -----+-U(s) s+l s+3 s+5 The state diagram is shown in the fig. 2.45
y
Fig. 2.45 The state equations arc,
Y = - 0.125X1 -7.25X2 +15.375X3
A=
,_.
[-100 -301 -50OJ ,B= [']11 ,C=(-0.125-7.2515.375] .
Example 2.29 : Choosing appropriate physical variables as state variables, obtain the slate modelfor tho tlectric circ11it shown i11 Fig. 2.46 (VTU: JanJFeb.·2006)
M
2 ·62
Modem Control Theory
State Space Representation
rn
IH
y(t)
Fig. 2.46 Solution : The various currents arc shown in the Fig. 2.46(a). 1F
Fig. 2.46 (a) The volatagc across resistance and second inductor is y(t).
and
y~) = y(t)
.•.. (1)
di 1x ~ = y(t) di
.... (2)
The voltage across capacitor is, ic(l) and
di I '< _!:..!. dt
From (4), Using in (2),
y(I) dil2 di
u(t)-y(t)
dvc(I) _ dvc(I) C-d-1- -d-t -
.... (3)
u(t) - y(t) ~ vc (t)
..... (4)
u(t) -
dil I dt = u(t)
u(I) - vc(t)
- vc(t) .... (5)
2 ·63
State Space Representation
u(l)-vc(tl-iu +i1.2
... (6)
Modem Control Theory Applying KCL at node A,
dvc(t)
.
·-d-1 - +I LI dvc(t)
-d-1- =
. . X2 . X3 X1
and
y
A
JI*
i1.2 + y(t)
dill
. .. from (4)
dl;X3 di, 2 dl
=-X3+
dvc(I) _d_t_ = di,
di2
[~ -1
•
-X, +X2-X3 + U
= X2 ~
0
...from (5)
U
=-X3
. .. from (6) .. .from (2)
+U
-11].B=[~].C=fO -1 I
0 -l),D=[l}
. Y(s) s(s+ 2)(s + 3) Example 2.30 : For lite transferfunction R(s) ~ --..,,--(s+ 1)2(s+4)
Obtain lite state model in i) Phase variable canoP1icalfarm ii) fordan CanoP1ica/ form Solution : i) Phase variable fonn Y(s) s(s+2) (s+3) R(s) = (s + 1)2(s+4) Using direct decomposition, Y(s) R(s)
s3 +ss2 +6s {[(s+6}s+9)s+4}
s3 +5s2 +6s s3 +6s2 +9s+4
(VfU: Jan./Feb.·2006)
Modem Control Thoory
State Space Representation
The state diagram is shown in the Fig. 2.47.
R
.
x,=X2,
x,
.
Fig. 2.47
X2=X3.
X3=-4X,-9Xz-6X3+R
Y = 6X2 +5X3 +X3 Y
= 6X2 +5X3
=
-4X1 -9X2 -6X3 +R
-4X1-3X2-X3+R
A [~4 ~9 ~JB=m,C=(-4 m Jordan
-3-1),D=(l)
Canonical form :
As the degree of denominator and numerator is same, first divide
partial fractions s3+6s2+9s+-I}
s3-f-5s2+6s
(1
s3 +6s2 +9s+4
A(s+4)+
B(s+ l){s+4) +C(s+1)2
s2+3s+4
and then obtain
Modem Control
:. As+4A + Bs2 +5sB+4B+Cs2 + 2Cs+C :.B+C = 1,
State Space Representation
2-65
Theory
A+5B+2C=3,
4A+4B+C=4
A= 0.666, B = 0.111,
C = + 0.888
Y(s)
I { 0.666
R(s) ~
= s2 + 3s+4
0.111
0.888}
- (s+t)2 + (s+l) + (s+4)
I _ 0.666 _ O. lll _ 0.888 (s+t)2
s+ I
s+4
Fig. 2.48
-1 A=
11..
[
~
y =
- o.111x1
0 -1
_n' =[~]·
0
B
-0.666X2
- o.sssx 3 + R
C = [-0.111 -0.666 -0.888), D
= [1]
Example 2.31 : Fig. 2.49 shows tltt block diagram of a speed control systtm tuillt stute variable feedback. Tltt drive motor is an armalurt co11trolled d.c. motor willr · amiature resistance R., armature inductance L., motor torque constant Kp inertia referredlo consta11I K~ and tachometer K1. The applied armature voltage is ctmtrolled by a three phase full·converter. ec is control voltage, e4 is an11ature voltage, e, is the reference voltage corre:.-pondi11glo the desired spted. Taking X1 = w(speed) and X2 = i4 (armaluu current} as lht stale variables, u = e, as tltt i11put, and y = w as the 011tp11t, derive a state variable mode for the feedback system. (VTU: July/Aug.·2006)
2 -66
Modern Control Theory
State Space Representation
Fig. 2.49 Solution : For armature controlled d.c, motor, v.(t) and
di, . L•{it+•.R,
+cb(t
)
...... (1)
KbW(t)
...... (2)
...... (3) The torque produced is used to drive the load against inertia Tm
J~~+BIJl
J and
friction B. ..... (4)
dlJl ldt+BIJl d~) dt
...... (5)
The state variables arc X1 =
From the equation (1), va(t) = L,, d~; + i, R, + Ki.(J)(t)
..... (7)
Modem Controi Theory
2 -67
State Space Representation
But the armature voltage v. is controlled by three phase full converter.
KclK1 x-K2
i, J
K
dt
....(8)
X2
.... (9)
y
And
The equations (6), (9) and (JO) give us the required state model with,
l
Kr A
-(/Kc
].c~ll
R )J'B=[~
2+ •
L,
,,.... Example 2.32 : For a transfer function given by G(s) ;
2 s2 +3s+2
in the: i) Phase variable
form
ii) Diagonal
form
Solution : i) Phase variable form using direct decomposition G(s) ; __2_; s2 +3s+2
2 [(s+3)s+2]
o]
•
, write tire state model
.... Modem ControlTheory
State Space Representation
2-68
y
Fig. 2.51 X1=X2,
X2=-2X1-3X2+U, A =
Y=2X1
[~2 _ ~], B = [~], C = {2 O]
ii) Diagonal form using Foster's form G(s)
A G(s)
=
2
s2 +3s+2
2 (s+I) (s+2)
A G =-+s+ I s+2
2 2 (2-1) = 2' B = (-2+1) = -2
2 2 s+1-s+2
u X1=-2X2+U y
Y = 2X1 +2X2
:.A=[-~ -02}B=[~] :.C
Fig. 2.52
= (2
2)
Modem
••
2. 69
Control Theory
State Space Representation
Example 2.33 : For the network shown i11 Fig. 2.53, c/ioosi11g i1(t) = X1(1) and ifl) =Xz(t) as statt variables, obtain lite state equation and output equation in vector matrix form.
-1
Hl
1H
-
i,(t)-i2(l) 1H
•,(t)
u(t)
~------12-(l_)l~---~ 1 0 10
y(t)
Fig. 2.53 (VTU: o.c.-2007/Jan.-2008)
Solution : Applying KVL to the two loops, di,
-dt-i1
di ,
-dt-i2
+u(t)
0
...Loop I
0
... Loop 2
di2
... (1)
dt Using (1) in equation for loop l, di,
dt
= -i1-(i1-2i2)-i2+u(t)=-2i1+i2+u(t)
... (3)
.
...(4)
X2
and
and
... (2)
... (5)
y(t)
A
(1
-t][~~]
[;2 _aB=[~],c=(l
-1},D=(O)
2. 70
Modem Control Theory
State Space Representation
Review Questions I. Discuss the 'dtrilllltion of the state model 11Sing fallowing mtthodJ of programming.
BU5h form ii) Gulltmin's form iii) Foster's form iv) fordan's form i)
2. fl•bornlt upon tht basis of Mktling suitable stalt rJariablts for • systtm. 3. Obtain lht sl•I• tnodd for the block diagram shown.
U{t)
Y(t)
4. Vo/toge nrross r«p<1dlor is Ye- With "c and il as n Ml
of M•lt variables dniot tJie shllt
ltW
5. Qm.sWt·rth~ pernianent mngrtrl movingcoil in.strummt 4.f shoton, Here, t
= Vollagt to b< '"""5urtd,
Coll
R = A rmalure Rtsistantt, L ~ Armal·urr Inductance ,
constant relating i = Currtnl through 1trmahtrt toil ,
I=
M.I. of rotating part ,
Kr
=
K,
= Spring amstant •
Tonpu conshud rclnting i .ind T ,
l)eriVo' its state model.
Aru. : x = AX + llU y = ex+ OU
4 -~ -ti ·{!'Jc·""' n
whtttA •
[
2. 71
Modem Control Theory
State Space Representation
6. For " mt'chanirnlsystem slwwn below obtnin its st.all model in sttlndardform. Assum1 output as Y(IJ. Ans.:X=AX+BD,
Y•CX
T Y(l)
where A
=[ ~
~ ] B =[
O } C 1/M
-M -M
= (0 1)
7. Obtain the slate model in standard Busl• form of a system shown in figure. Ans.:
U(sl
X =AX+
BU, Y=CX
Y(s)
where A=
0
l
0
0
0 L
-288
0 0 -176 -98
8. Statt whether lhL ffJllowing stetemm! is True or Faist rvith reasons : 1·~ stale 1pace approachof llie system nnalysi$ is n frequency domain analysU.
9. Statt wlrether tht follmuing stalrmtnt is Trut ~r Fnlst with r~on. Tht slalt space approach of the system analysis is a time domain analysis.
C
= [72
96 24 OJ
Modem Control Theory
2. 72
State Space Representation
JO. For tht merha11iat/ •yslem •hawn ill ftg11rt input is uft) and output is y(t). Obtain stalt equntiqn
awl output eq11ation. Repre;e11t lllL sysl<1n by blodc diagram.
QQQ
Matrix Algebra and Derivation of Transfer Function 3.1 Background The methods of obtaining the state model in various forms, from the transfer function of a system are discussed in the last chapter. It is also seen that the state models can be different and not a unique property for the given system. But most important point about the system Is Its transfer function is always unique. Though number of state models are derived for the system, the transfer function of the system, obtained from all of them is always unique for single input single output system. In this chapter the method of obtaining transfer function of the system from its state model is discussed. The concepts of diagonalisation of system matrix A, eigcn values and elgcn vectors arc also included in this chapter. Number of methods from matrix algebra are used to understand the relation between stale model and transfer function hence revision of matrix algebra is included in the beguuung of this chapter.
3.2 Definition of Matrix A matrix is an ordered rectangular array of elements which may be rcal numbers, complex numbers. functions or mathematical operators. The order of matrix is always defined as m x n. where, m n
Number of rows
=
Number of columns
For example, consider matrix A of order 2 x 4 of real numbers.
A
=
4 3 [6 1
-z
1]
_... 2 Rows
0 8 ....
!. !. .!. J. 4 Columns
(3 • 1)
3-2
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
Any clement of a matrix is denoted as ail where i indicates position of row and j indicates position of column. Thus for matrix A above a11 = 4, a13 = - 2, a22 1 and so on.
=
Koy Point: Note that mal'rix does 1101 have a value but its detenninant has a value. A matrix with number of columns as 1 Le. order m x 1 is called column matrix while a matrix with number of rows as 1 i.c, order 1 x n is called row matrix, It is seen that vector matrices X. X, Y are the column matrices.
3.2.1 Types of Matrices The various types of matrices are. 1. Square Matrix : If number of rows (ml is equal to number of columns (n) of a matrix, it is called a square matrix. The system matrix A in a state model is always a square matrix of order n x n. 2. Diagonal Matrix : It is a square matrix with a.II a1;
=
0 for all i ~ j. Only main diagonal clements arc present which arc nonzero while all other clements are zero. For example,
3. Unit Matrix : This is a diagonal matrix, with all the main diagonal elements equal to unity. It is denoted as 1 and also called identity matrix. The unit matrix of order 3 x 3 is, I =
-l,0
o o]
l 0 0 0 1
4. Null Matrix : This is a matrix having all its elements as zero. It need not be a square matrix. 5. Transpose of a Matrix : If the rows of columns of m x n order matrix arc interchanged then resulting n x m order matrix is called trnnspose of the original matrix. If given matrix is A then its transpose is dl'TIOll'CI as AT.
A
o [2
21
1 5 4j. 2x3
Modem Control
Matrix Algebra & Derivation of Transfer Function
3.3
Theory
The proper11es of transpose arc, !. (A + B)T =AT+ BT 2.
(AB)T =BT AT
6. Symmetric Matrix : If transpose of a matrix is matrix itself, it is called symmetric matrix. It is a square matrix. AT A
A
[ I -4] -4
I -4]
AT= lr -4
1 •
I
7. Conjugate Matrix : The conjugate of a matrix is the matrix obtained in which each clement is the complex conjugate of the corresponding clement of the original matrix. It is denoted as A'. _ [I +j2 A - 2 -j4 8. Singula.r Matrix singular matrix.
l+jl] 3+j2 '
A'=
rll
-j2 2 +j4
1-jl] 3-j2
A square matrix having its determinant value zero is called
and
IAI
=
j 0.51
1=2 - 2 = 0
4 2
Thus A is singular matrix. 9. Nonsingular Matrix : A square matrix whose determinant is nonzero is called nonsingular matrix. A=[:!]
and
IAl=1:
~1=24-10=14"0
10. Skew Symmetric Matrix : A square matrix which is equal to its negative transpose is called a skew symmetric matrix. AT = -A
3.2.2 Important Terminologies Let us study the important terms and their meanings related to the matrix. 1. Minor of an element : Consider a square (n x n) matrix A and the clement aij. If now ith row and jth column are deleted then the remaining (n - 1) rows and columns form a determinant M;i. The value of this determinant is called the minor of an clement ll;j·
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
3-4
Consider a square matrix A as,
A ~ [~ ~ ~] Let us obtain minor of element alt determinant M21.
=
1. Delete second row and first column to get
Thus minor of 'a21' is 7. 2. Cofactor of an Clement : If M;i is the minor of an element aij then cofactor defined as,
I
cii =
<- ll; + i Mij
C;i is
I
Thus for the matrix A considered above, the cofactor of "it is, = (- 1)2 • t M21 = (- 1)3 >< 7 = - 7
C21
3. Adjoint of a matrix : The adjoint matrix is the transpose of a cofactor matrix. The cofactor matrix is the matrix obtained by replacing each clement of a matrix by the respective cofactor.
I
A~j A
n•~ Example 3.1 :
=
(Cofactor matrix of A]T
I
Find tlrt adjoint matrix of the matrix A,
'
A=
1 0 3] [ 2 4 1 3 2 2
Solution : Th" cofactors of all the clements are, C11
= (-
C13 = (-1)1
C22
11 = 6,
C12 = (- 1) 1 + 212 3
!) 1+·14 2 2
+31~
~I= -
8,
= (- 1)2 + 2 113 ~I= - 7,
~I= - 1
C2t
= (- 1)2 + 110 2 31 2 = +6
C23
= (- !)2 • 31 ~ ~I=
-2
Matrix Algebra & Derivation of Transfer Function
3.5
Modem Control Theory
Cofactor matrix
[-1~
-1-7 -8] -2 +5
-1-7
Adj A
[ _1~ +5
4
=:r =[~
6 -12] 5 -2 4 -7
4. Rank of a Matrix : To find the rank of a matrix means to search for a highest order determinant from a given matrix which is having nonzero value. Thus if r x r determinant has a nonzero value in a given matrix then the rank of a matrix is 'r' and any determinant having order r + 1 or more than that has a zero value. For example, consider matrix A as,
A
112
Now
1
I
2
= [~
31 =
4 1 1 1
! ~]
-5:t0
Thus 3 x 3 determinant is nonzero hence rank of matrix A is 3.
i•
Example 3.2 :
8olutlon
Find rank of A = [~
!]
: For the matrix A,
I~ !I=
0
hence rank is not 2.
Any other determinant of order 1 x 1 is nonzero. Hence rank of matrix A is 1. S. Equality of Matrices : The two matrices A and B are said lo be equal if they have same number of rows and columns and the elements of the corresponding orientations arc equal.
Matrix Algebra & Derivation of Transfer Function
3-6
Modern Control Theory
3.3 Elementary Matrix Operations The various elementary matrix operations arc. 1. Addition : The two matrices A and B can be added to get C B arc of same order and,
LS:o
=
A + B if both A and
I
+ b,j for all i and j
a;1
=
2. Subtraction : The two matrices A and B can be subtracted to get C = t. - B if both A and B arc of ~me order and,
I
C;i = a11 - b;J for all i and j
I
Associative law holds good for both addition and subtraction of the matrices. [A ± BJ
±c =
A ±
IB ±CJ
where all A, B and C matrices have same order. 3. Multiplication by scalar 'ex' : A matrix is multiplied by a scalar 'ex' if all the clements of a matrix are multiplied by that scalar 'ex'. cxa 11
exxA=
exalnl
Cltl 21
<XU nl
r.tan2
cxann
,
Ctn2n
[ 4. Multiplication of Matrices : The multiplication of two matrices is possible if number of columns of the first matrix is equal to the number of rows of the second matrix. Such matrices arc called conformal matrices. Thus if A is of order m x n and B is of order n x p. Then the multiplication C = A x B is of order m x p. The clements of matrix C arc obtained as, C;i
= I,"
a Ik bkj for i = 1, 2 ... n and j ~ 1, 2, ... P
l =I
Thus if A is of order 2 x 2 and B is of order 2 x 3 then C = A x B has a order 2 x 3. A
C
["11 a 21
•p] a~
B =[b11 b21
b12 b22
b13] b23
AxB •11b11+a12b2t [ 321b11 +a22b21
•11b12+a12b22 321b12 +a22b22
Matrix Algebra & Derivation of TransferFunction
3.7
Modem Control Theory Properties of Multiplication
J. Commutative law is not valid for the matrix multiplication. AB "' BA 2. The transpose of a product is the product of transposes of individual matrices in the reverse order.
,..,.
Example 3.3 : Obtain tire matrix multiplicationof. 3
.4
-Ji =[1 0 -1.]
= [~ ~
B
2 1
0
Solution : A has order 3 x 2 and ll has order 2 x 3 hence C = A x ll has order 3 x 3. 3xl-lx2
c
Ax B
=
Oxl+lx2 [ 2x I +Ox2
3x0-lxl OxO+lxl 2xO+Ox1
3x-1-lx0] Ox-l+lxO 2x-1 +OxO
[~ -: -~] 2
0 -2
3.4 Inverse of a Matrix In algebraic calculations we write ax
=y
i.e. Similarly in matrix algebra we can write, AX
B
i.e.
where
Inverse of matrix A
The inverse of a matrix exists only under the following conditions, I. The matrix is a
square matrix.
2. The matrix is a nonsingular matrix.
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
3-8
Mathematically inverse of a matrix can be calculated by the expression, [Adjoint of A)
_ 1
"-jAj
A
3.4.1 Properties of Inverse of a Matrix 1. When a matrix is multiplied by its inverse, the result is the identity matrix.
=
A-1A
AA-l =I
2. Inverse of inverse of a matrix is the matrix itself. (A-lf
l = A
3. Inverse of a product of two matrices is the' product of their individual inverses taken in the reverse order. (ABrt
.......
1....,..
= 5-l A-1
Example 3.4 : Find the inverse of A
= lf24 13}
Solution : Check that A is nonsingular.
IAI
I~
:1=2 - 12 = _ 10
[Adjoint o( A)
IAI Adj A
[-~1 --3]~r
[Cofactor of A)T
[~
Cross check:
AA-1=
[~
-10
= [_:
-:1
= [--0.1
:][:~::
+0.3] +0.4 -0.2 :~:~] ·[~
~]-1
Important Observation : For 2 x 2 matrix, the adjoint of the matrix can be directly obtained by changing the positions of main diagonal elements and by changing the signs of the remaining clements.
Modem
For example,
Matrix Algebra & Derivation of Transfer Function
3-9
Control Theory
A ~ [~ :] ,
Adj A= [ _:
-n
This result is directly used hereafter while solving the problems, when matrix has an order 2 x 2. Note that for any other order of matrix the adjoint must be obtained as transpose of cofactor matrix. With this study of matrix algebra, let us discuss the method of obtaining transfer function from the state model.
3.5 Derivation of Transfer Function from State Model Consider a standard state model derived for linear time invariant system as •
.
and
X(t)
A X(t) + B U(t)
... (la)
Y(t)
C X(t) + D U(t)
... (lb)
Taking Laplace transform of both sides, [s X(s) - X(O)I
A X(s) + B U(s}
... (2a)
Y(s)
C X(s} + D U(s}
... (2b)
and
Note that as the system is time invariant, the coefficient of matrices A, B, C and D arc constants. While the definition of transfer function l~ based on the assumption of zero initial conditions i.e. X(O) 0.
=
s X(s) s X(s) - A X(s)
A X(s) + B U(s)
B U(s)
Now s is an operator while A is matrix of order n x n hence to match the orders of two terms on left hand side, multiply 's' by identity matrix I of the order n x n, sl X(s) - A X(s) (sl -
Al
X(s)
B U(s) B U(s)
Premultiplying both stdes by [sl - Ar [sl - Af
1
1,
Isl - A) X(s)
[sl - Ar 1 [sl - AJ
Now
X(s)
[sr - Ar 1 BU(s) l [sl - Ar 1 BU(s)
Substituting in the equation (2b ), Ar 1 B U(s) + D U(s)
Y(s)
c Isl -
Y(s)
{C [sl - Ar I B + 0) U(s)
... (3)
Matrix Algebra & Derivation of Transfer Function
3 -10
Modern Control Theory
Hence the transfer function is, Tts)
Now
Isl
=
Y(s) =
U(s)
c (51 - Ar
l B+D
... (4a)
Adj(sl-AI Isl-Al
-Ar1 T (s)
=
c Adj(sl-AI n Isl-al
D
... (4b)
+
Key Point: Tire state model of a sys~11 is 11ot unique, but tire tra1L,fer f1111ctio11 of obtai1wtl from any stale model is unique. 11 is indrpcndent of tire met/rod used lo express tire system in slate model form.
3.5.1 Characteristic Equation It is seen from the expression of transfer function that the denominator is Isl - A [. Now the equation obtained by equating denominator of transfer function to zero is called characteristic equation. The roots of this equation arc the closed loop poles of the system. Thus the characteristic equation of the system is, ... Characteristic equation
= 0
Isl-A(
The stability of the system depends on the roots of the characteristic equation. In matrix algebra, the roots of the equation I sf - A I = 0 arc called cigcn values of matrix A and these arc generally denoted by A.
i•
Example 3.5 : Consider a system having stale model
f XI l = [-2 I• ! L X2 j
.1
-31J 2
(; ~] + [~] U anti Y = I 1 1
I [~~]
witlr D "' 0. 0!1tain its T.F.
(V'fU: )anJFcb.·2008)
Solution: T.F.
c Isl-Ar'
s
Adj[sl - A] Isl -Al (sf - A)
5
[1 o]-(-2 0 J
~
-312j = [s0
o]-[-2 -3] s
·t4
2
Modem Control Theory
[sl
Matrix Algebra & Derivation of Transfer Function
3·11
l"s+2
-Al
3]
-4
s-2
- 3]
Adj Isl -
s+ 2
Al
(s + 2) (s - 2) + 12 = s2 - 4 + 12 = s2 + 8
[s~2 s~2] s2 + 8
-3] f3]
[l I
[s~2 __ _,,___s_+_2-:-,_L 5_,,_ = l 1 1 s2 + 8
T.F.
[3s-21]
I
Ss+ 22
s2
+8
{Ss + 1] s2 + 8 3.5.2 MIMO System For multiple input multiple output systems, a single transfer function docs not exist. There exists a mathematical relationship between each output and all the inputs. Hence for such systems there exists a transfer matrix rather than the transfer function. Bul the method of obtaining transfer matrix remains same as before. 11111+
Example 3.6 : Determine the lrausfer matrix for MIMO system givtn by,
[~1]J L-2o 3][x'] x Xi
-5
2
=[11
1J[u'], [Y']=[2 o1J[xx u
1
1]
1
Y2
2
1
Solution : From the given state model, A
-2
B = [:
-5
C!sl-A]-l
T.M.
:}
C = [~ ~}
B+D
-3]
s [2 s+S
[sl-AI
Adj [sl - A) Isl-A
[ 0 3}
I
ell [C21
C1.2]T C22
s2 +5s+6=
= [s+5 3
-2JT s
(s + 2) (s + 3)
= [s+5 -2
.s3]
D =[OJ
Matrix Algebra & Derivation of Transfer Function
3-12
Modem Control Theory
Adj (sl-A) (sl-AI
[s_+: ~] (s..2)(s..3)
T.M.
s+S] [3s+14 3s+14] [2I 1]0 [s+S s-2 s-2 s+S s+S
,. ,. l '"" l
~2)(s+3-)-=
14
(s+2)(s+3)
3s+ (s+2) (s+3) (s+2) (s+3) s+S s+8 [ (s+2) (s+ 3) (s+2) (s+3) T.M. U(s)
Y(s)
i.e.
3s+ 14 (s+2) (s+ 3) s+S [ (s+2) (s+ 3)
The above relation indicates that each output depends on both the inputs.
3.6 Eigen Values Consider an equation AX = Y which indicates the transformation matrix X into 'n x 1' vector matrix Y by 'n x n' matrix operator A.
of 'n x 1' vector
If there exists such a vector X such that A transforms it to a vector JJ<. then X is called the solution of the equation,
0
i.e. (Al-
i.e.
A)X
0
..• (1)
The set of homogeneous condition,
I
p.. 1-AI
~o
equations
I
(1) have a nontrivial solution only under the
... (2)
The determinant I}.. I - A ( is called characteristic polynomial while the equation (2) is called the characteristic equation.
Modem Control
Matrix Algebra & Deriv1tlon of Transfer Function
3-13
Theory
After expanding, we get the characteristic equation as,
p..r - A I
= )." + a1 A" -
1
+ ... + a0 = 0
..• (3)
The 'n' roots of the equation (3) i.e. the values of ). satisfying the above equation (3) are called elgen values of the matrix A.
=
The equation (2) is similar to Isl - A I 0, which is the characteristic equation of the system. Hence values of ), satisfying characteristic equation are the dosed loop poles of the system. Thus eigen values are the closed loop poles of the system,
3.7 Eigen Vectors Any nonzero vector X; such that AX; = A; X; is said to be cigcn vector associated with eigenvalue ),;·Thus let ). = A; satisfies the equation, ().1 I-A)
X = 0
Then solution of this equation is called eigen vector of A associated with cigen value ).1 and is denoted as M1. If the rank of the matrix P·; I - A) is r, then there arc (n - r) independent eigcn vectors. Similarly another important point is that if the eigen values of matrix A are all distinct, then the rank r of matrix A is (n - 1) where n is order of the system. Mathematically, the cigen vector can be calculated by taking cofactors of matrix 0.1 I - A) along any row •
I
..--~~~~~~~~~~~~~~~~~~~ ckl
Thus
M; = Eigcn vector for A; = :~
where k = 1, 2, ... n
[
where Cid is cofactor of matrix (/,.; I - A) of k1h row. Key Point: If the cofactors along a particular row givts null solution i.e. nil elements of corresponding eigen vectors are zero then cofactors along any other row must be ob/nilled.
Otherwise inverse of modal matrix M cannot exist. 3.8 Modal Matrix M Let :>..1, 1.2, ..• /,.n are then eigenvalues of the matrix A while M1, M2, ... ~are the eigen vectors corresponding to the eigcn values ).1, ).2 ... An respectively. Each eigen vector is of the order n x 1 and placing all the cigcn vectors one after another as the columns of another matrix, n x n matrix can be obtained. Such a matrix obtained by placing all the eigen vectors together is called a modal matrix or diagonalising matrix of matrix A.
Modem ControlTheory
I
Matrix Algebra -& Derivation of Transfer Function
3 -14
M ; Modal Matrix = (M1 : !\,~ : ... :
M.,l
The important property of the modal matrix is that, AM
A (M1 : M2 : ... : M,,I (AM1 : AM2 : •.. AMnJ [i-i M1 : ~ M2 : .•• : Ar, M,,)
AM
.•. (1)
MA
r1..1 where
A
:o :
Diagonal matrix = j
I.co
0 0 1..2 0 0
Thus premultiplying equation (1) by M • 1•
= M- 1M A M - 1 AM = A (Diagonal
M-
1
AM
matrix)
It can be noted that, 1. Both A and A have same characteristic equation. 2. Both A and A have same eigen values. Hence due to transformation M • 1AM, eigcn values of matrix A remain unchanged and matrix A gets converted to diagonal form.
3.8.1 Vander Monde Matrix If the state model i.~ obtained using the phase variables then the matrix A is in Bush form or phase variable form as,
0 0
I
0
0
A
0 -<-ln
0 --an-I
0 -;) n-2
:I
-a
1J
And the characteristic equation i.e. denominator of T(s) is, F(s)
= s" + a1 s" - 1 + ... + an_ 1 s + a0
=0
Matrix Algebra & Derivation of Transfer Function
3 -15
Modern Control Theory
In such a case, modal matrix lakes a form of a special matrix as,
Ai
i..2
),.n
)..2
},~
)..~,
A)- i
l.!r2
;.n-1
v
I
'11
Such a modal matrix for matrix A which is in phase variable form, is called Vander Mondc matrix . ,....,
Example 3.7 :
Consider a state model witlt malrir A as, 2
0
A
-3~
OJ
\l
-9
Determine, a) Characteristic equation. b) Eigen values c) Eigen vectors and d) Modnl matrix. Also prove that the transformn/io11 M- 1AM results n diagonal matrix. Solution : a) The characteristic equation is p,1 - A I 1 0 ). 0 1 0 -
011
10 0 1
I
40
2 0
01
1
0
-48 -3~ -9
I -~ -~ -~ l 48
:.)..2
=0
... (!)
= 0
34 ).+9
(I.+ 9) + 2 x 48 + 0 + 0 - 8 (1.. .. 9) + 34 ). = 0
=
)..3+91?+26).+24
O
This is required characteristic equation. b) To find eigen values, test A= - 2 for its root. 9
26
24
- 2
- 14
- 24
7
12
-2
0
p.2 + n + 12)
o
(), + 2) (/..+ 3) (). + 4)
0
(), +
i.e.
Matrix Algebra & Derivation of Transfer Function
3 -16
Modorn Control Theory
2>
i.e,
- 2, 1..2 = - 3
and /,3 = - 4
These arc the cigcn values of matrix A. c) To find eigen vectors. obtain matrix (l.i I - Al for each cigen value by substituting value of I.. in equation (1).
r-2 =34~ -7~]
l~:
... The common factor can be taken oul. l
For i.2 = - 3,
p, i I -
A)
M2
For 1..3 = - 4,
[l..i I -A]
M3
=
[~:~l Hl = [~;4] =
o]
-4 = [-4
-2 -4 -1 48 34 5
=[~::] [-!] =h] =
M1, M2 and M3 are the eigenvectors corresponding to the eigenvalues :\.1, :\.2 and :\.3.
Matrix Algebra & Derivation of Transfer Function
3 -17
Modem Control Theory d) The modal matrix is,
M = IM1 : M2 : M3)
-! -~4jl
= (-~
-2
Let us prove M - 1 AM is a diagonal matrix. = AdjlMJ
M-1
!Mr=
Adj [M]
[
(CofoclorofMJT
IMI
-10 +8 -7]T r-10 - 7 -1
-Ii
-7 8 6 1 -7 -5 -1
=
6 -5 1 -1
I _\ -~ -~1=-l
IMI
-2
1
4
Adj [Ml
IMI= AM
L~s [10-8
M-1AM
7
u
[-810 7 1] 7
2 0
~
-; 2
ol1 [-11
-3
-34 -9 -2 7 -6 5
-11 l r-22 I
0 -3 0
4
-6 9
-3
-4]
-1!
4]
-6 9 8 -3 -16
OJ 0 = [i.01 -4
1 l r-2 -: = !
0
0 ).2
0
~l=
h
h3
Thus M - 1AM is a diagonal matrix.
l'*
Example 3.8 :
A=[l
Obtain tire Elgen values, Eigen vectors and Modal matrL~ for the matrix
1 OJ
0 l -11 -6
3 -18
Modem Control Theory
MatrixAlgebra& Derivation of Transfer Function
l
Solution: Eigenvalues are roots ofjt .. I-Al=O
!).. I-Al
=
>. -1
o >.
[
6
11
0 -1 =0 >.+6
1.. 3+61..2+11>.+6
i.e.
0
(>.+ !)(/.+ 2){).+ 3) /,1 =-1,
J..2 =-2,
0
J..3 =-3 are eigenvalues.
To find Eigen vector,
>.
Let
"-1
=-!
[~1 =~ ~1] 6
M1
M1
For
>-2
11
5
[en]
where C = Cofactor
C12 C 13 K•2
[~]
i.e,
M1=Hl
=-2 -1
(J..21-A)
M2
[~2 -2 [c11 C12 C 13
l
11
-:1]
for K~3
For J..3 =-3 -1
(1.31-A)
-3
[Y 11
~]
=[~] =[~2]
Modem Control Theory
3 -19
Matrix Algebl'll & Derivation of Transfer Function
Key Point: The cofactors about any ruw can bt obltlined. :. Modal matrix M
=
(M1 :M2 :M3)
M [~l ~2 ~] It can be easily checked that,
M-1 AM
=
A=[~
~2 ~3]
3.9 Diagonalisatlon The diagonal matrix plays an important role in the matrix algebra. The eigen values and inverse of a diagonal matrix can be very easily obtained just by an inspection. Thus in state variable analysis, if the matrix A is diagonalised then it is very easy to handle mathematically. Whc!l matrix A ls diagonalised, then the elements along its principle diagonal are the eigcn values. The eigen values are the closed loop poles of the system, from which stability of the system can be analysed. The diagonal matrix A indicates noninteraction of various state variables. Tile derivative of one state variable is dependent on the corresponding state variable alone and other state variables do not interact in it, when A is diagonal matrix. This is practically important from controlling point of view. Due to all these reasons, the diagonaUsation of matrix A is always motivated. The techniques used to transform the general state model into diagonal form i.e. canonical form are called dlagonallutlon techniques. Consider nlh order state model in which matrix A is not diagonal . X(t)
.
A X(t) + B U(t)
... (1)
Y(t)
C X(t) + D U(t)
... (2)
Let Z(t) is new state vector such that the ITansformalion is, X(I)
=
MZ(t)
..• (3)
Modern Control
Theory
M
Herc
Matrix Algebra & Derivation of Transfer Function
3 -20
Modal matrix of A
X(t)
MZ(t)
... (4)
Using equations (3) and (4) into equations (1) and (2),
and
M Z(t)
.
AM Z(t) + B U(t)
... (5)
Y(t)
CM Z(t) + DU(t)
... (6)
Prcmultiplying equation (5) by M - 1 on both sides, M"1M Z(I)
.
M" 1AMZ(t) + M"1 B U(t)
Z(t)
M - 1 AM Z(t) + M - 1B U(t)
... (7)
Y(l)
CM Z(t) + D U(t)
.•. (8)
and
The equations (7) and (8) gives the canonical state model in which M - 1 AM is a diagonal matrix denoted as II. The matrix M which transforms A into diagonal form is called diagonalising matrix or modal matrix. The new canonical state model is represented as, Z(t)
11 Z(t) + ii U(t)
Y(t}
C Z(t) +
where
... (10)
M - 1 AM " Diagonal matrix
II
ii
c
... (9)
D U(t)
M01B 3
CM
The method of obtaining modal matrix M is already discussed in the last section. The transformation M - 1AM is called diagonalisation of matrix A.
I,_.
Example 3.9 : matrix A,
&duce tire given state model into its canonical form by diagonalising
X
r~
-1:
-6 -11
and
Y(I)
-~i 5
[1 0 OJ X(I)
X(tJ + [~] U(tJ I
Matrix Algebra & Derivation of Transfer Function
3. 21
Modem Control Theory Solution : From the given state model,
B=[~l·
A=[~=~~-:]·
0 OJ
C=[l
Let us find eigen values, eigcn vectors and modal matrix of A.
=
1)..1-AI
H~
0 1 0
~]-[-~
-1 ~.+11
1:
-11 -6 -11
I
_:I =
-il I=
0
0
~--5
11
:. A. (A. + 11) (A. - 5) + 36 + 66 - 6 (A. + 11) + 6(1. - 5) + 66 ). t..3+6).2+111.+6
0
0-+
0
().+ 1)
2) ().+ 3)
Thus eigen values are, A. 1 = - 1, A.2 = - 2, A.3 = - 3 To find eigen vectors, For ).1 = - 1, li•1 1- A]
M1
n
-!
10 11
-~] -6
[~:~J=m=m
For ).2 = - 2, -1
[A.21-AJ
M2
r-:
9 11
-~] -7
[~:~l r~i = m =
=
O
MatrixAlgeb,. & Dertntlon
M = [M1 : M2:
M3) = [~
n
M-1s
Adj(MJ
[~
!MT""= [
ii and
! !]
=Modal matrix
n -! -~lcA
M-1AM While
ofT,.Mfer Function
3 ·22
Modem Control Theory
C
0.4285 --0.4285
0.H28
M-
18 5
j
-14
0.3571 -0.5714 0.2142
~r [1 ~~ ~] = -~
--0.2857] +0.4285 -0.1428
[=~·.~1 . r-~1 -0.1428
-1
= CM= (I 1 1)
Thus canonical state model is, Z(t) .. A Z(t) + and Y(t)
C Z(t)
B U(t) where X(t) ~ M Z(t)
... Multiplying by 7
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
3 ·23
3.10 Generalised Eigen Vectors The generalised eigen vectors are the eigen vectors corresponding to repeated eigcn values. Uptill now, it is assumed that tile eigenvalues are distinct but if an eigen value At is repeated for r times then the corresponding eigen vectors are called generalised eigen vectors.
U a particular cigen value A 1 is repeated r times then the rank of n x n matrix [A11- A) is n - 1 and there is only one independent cigen vector associated with A1 given by,
... Ckn are cofactors about kth row
Important Note : If cofactors along a particular row gives all values zero i.e. eigen vector as null matrix, then cofactors along other row must be obtained. For remaining (r - 1) eigenvectors for r - 1 times repeated value A1, generalised eigen vectors must be used as,
fj
a2ck,
ack, ~
2!
ilCk2
M2
n ar;ilC1;n
CJ2
1i ~ 1 (r-1)!
M,
ilA7
a2ck2 M3= 2! a At
c.,,
2! ~
a•-tcck1> ()}..~-·
a•-•cck2> (r-1)!
dif' ar-lcckn>
(r-1)!
CIA't-1
The modal matrix M then can be obtained by placing all the eigen vectors one aher the other. Note that in such a case, M- 1AM matrix is not diagonal but consists of a r x r Jordan block in it where r is the order of repetition of a particular eigen value.
Matrix Algebra & Dorivation of Transfer Function
3. 24
Modem Control Thoory
3.10.1 Vander Monde Matrix for Generalised Eigen Vectors If matrix A is in phase variable form and an eigen value 1..1 is repeated for r times, then n eigen values arc /,1, /,1 ... 1..1, I., .. 1 -,.. , hn. The corresponding modal matrix is Vander Mondc matrix in modified form as, 0
0
),,
M=
'·1
21 -r
I.~
31..~
in-I
,,..
()
••
0 0
1...
0 . ..
).,~+I
31..1
d l!t1
d2 '·'rl
...
2idT I
""
1
1..;,
'";~ 1
A.~.
1.n-l rt I
J..ll-1 I\
Fi11d the eigen values, eige11 uector« alld modal matrix JM,
Example 3.10 :
A
[~ ~ -~
l
Solutlon : Find eigcn values which arc roots of I i,J - A I
Ix
0
I
0
81..
0
-2 I..
-1
+S 2
A~S
/..2 (I. + 5) + 4 +
A.3+s1:+st..+4
=O
0
Try 1..=-1 -1
5
8
-4
-1
4
4
(I,+ 1) {1..2 + 41..+ 4)
0
{A.+ 1) {A.+ 2)2
0
4
-4
~
Matrix Algebra & Derivation of TransferFunction
3. 25
Modem Control Theory
A I = - 1, '·2 = - 2, A3 = - 2 Thus ·- 2' is repeated twice hence it is necessary lo use generalised cigen vectors. For A1 = -1, (1,11-
A)=[=~ -~ 8
2
-~i 4
Now A2 = - 2 is repealed twice.
P-21 - A) = [~~
~2
8 2
[c"] C13
L,
[ ~·'~
= 2),2 + 10
C12
M2
-4-SAz
l.z•-2
=[]Jan] dC11
and
M3
TI
TI
dC12 d1..2
TI
dC13 d1..z
=
[2'-~+5) -
=
l.z--2
Thus M1, M2 and M3 are the required eigen vectors. -1
:. Modal matrix M
=
fM1 : M2 : M3J = : [
-1 1
112
2 -8
Ul
Modorn Control Theory
Matrix Algebra & Dorivation of Transfer Function
3 -26
Examples with Solutions !...
Find lite T.f". of tl1t system /raving state modn,
Example 3.11 ;
X
= [-~ -~] X
+[~] U
and Y = (1 OJ X
Solution : From the given model,
A=[-~-~], T.F.
[sl -AJ
Adj (~I - A]
Y(s) U(s)
B=[~],C=[I
= c (sl -
OJ, D=O
Ar l B + D
o] [ o
l] [s
J
t -1 s [0 1 - -2 -3 = 2 s+ 3 ~+3 [ -2
+I]
... Change diagonal elements and
s
change signs of other clements. s (s + 3) + 2 = s2 + 3s + 2
. T.P.
C Adj (sl-AJ B Isl-Al
(1
=
=
(s + I) (s + 2}
OJ [s~23 ls][lo] (s+l)(s+2)
(s+3) (s+l) (s+2} ,_.
Fi114 the T.F. of tire system /raving stale model,
Example 3.12 :
eolutlon : Prom the given model,
A=[=~~]. a:[~], T.F.
[sl-A]
T(s) = U(s)
c (sl -
5(1 OJ-[-3 0 I
-2
01
C=[I
Ar l B + D l]=[s+3 0 2
-1] s
D=O
i Modem Control Theory
3 ·27
/Adj [sl-A) Isl-Al
(I 0) [-~ s~3][~]
CAdj[sl-A)B )sl-AI
T.F.
(s+l) (s+2)
1 (s+l) (s+2) ,,..
Obtain the transfer[unction matrix for 1iu, MIMO system having state
Example 3.13 : model,
.
[ ~ -11 ~jx+[-11
x
-1
amt
Y
=
0 I
0
G ~ ~] X
Solution : From the given state model, A
o]
2 -1 I 1 2 , [
-1
0 1
c
G ~ ~].
Transfer matrix
c [sl - Ar
[sl -A)
-1 OJ
=[~ ~ ,
0=0 I B+ 0
=
c Adj[sl -A) B Isl-Al
•[~ ~ ~]-[~ -: ~]=[s~2 0 0 1 -1 0 1 1 (s-1)2
Adj [sl-A)
B
~+1
[ -2
s-3 (s-2)(s-1) 2(s-2)
s-1
s2 -~s+3 ]
s-1 0 T
-~] s-1
Matrix Algebra & Derivation Modem Control Theory
3 -28
s2 -2s+1 s-3 [ s-1
Isl-Al
-s+l s2 -3s+2
-2 2s-4 s2 -3s+ 3
l
of Transfer Function
(s - 2) (s - 1 )2 - 2 + s - l s3 - 4s2 + 6s - 5 s2-2s+l s-3 [ s-1
Adj [sl- AIB
-s+l
s2 -3s+2 I
-s2 +5 s2 -4s+5 [ -s+2
C Adj [sI - Al B
[~ ~ ~]
-3s+5 [-s2 +2
Transfer matrix ~
A
[ ~ -12
and prove that M - 1 AM = Solution : For elgen values,
IA.I-Al
-~ -,~ ~21 12
7
-s2 +5 -4 s2 -4s+5 • 4s-8 [ -s+2 2s2 -6s+6
l
4s-12 ] 2s2 -6s+2
-4s2 +6s-5
Obtain tire eigen values, eige11 vectors and modal matrix for,
Example 3.14 :
I
l
-3s+5 4s-12 ] [-s2 +2 2s2 -6s+2 ~3
,_.
-4 4s-8 2s2 -6s+6
;\.+6
0 0
~ -7 A
~]
-6
= Diagonal matrix.
(VTU: July/Aug.-2007,Jan./Feb.-2008)
MatrixAlgebra & Derivation of TransferFunctlon
3. 29
Modern Control Theory
A.2('A. + 6) + 24 - 31..- 18 + 141..
0
)! + 61..2 + 111.. + 6
0
I
\
Test 'A.= - 1, -1
11
6
- 5
-6
6
~
6 0
- 1
5
i.e.
(l + 1) (l2 + 5A. + 6)
0
(A.+ 1) (A.+ 2) ('A.+ 3)
0
=-
2, A. 3
Hence eigen values arc, A 1 = - 1, },2
=-
3
For the eigcn vectors, A.1
[-1
-1 -3 -1
= -1, p,11-AJ
12
M1
[~:~ l c,J
A2
= - 2,
P-2 I - A)
M2
A.3 = - 3, [l3 I - A)
M3
M
7
-:]
= [-:)
-9
= [-~] -1
[-2 -1 OJ 12 7 4 -3 -2 -2
[~:~l =[-~]=HJ [-3 -1 -zo] -3 -3 12 7
3
[~: l =[-!:]=HJ Modal matrix
= [- ~ -1
2 -4
-i]
.
Moifani Conlroi Theory
3 ·30
Adj[M]
[
!Ml=
-9 -5
I
Matrix Algolira & Derivation of Ttnsfer Function
: ~iT [-: -: -~1
-2 2 -2 -2
- -2 -3 -2, -2
AM
Thus,
I ..
M- 1 AM = A= Diagonal matrix
Find I/it eigen valtttS, tigm 11ectors and
Example 3.15 :
A = [~ -~
"'f al matrix for,
I]
(VTU: Jan.!Ftb.·2.005)
Solution : For eigen values,
p.. I-AI
I
"-~4 -: -1
-~
O
I
0
A.-3
).. (1. - 4) (1. - 3) - 2 - 2 •· 2A. - 1. + 3 + V. - 8 = O
n2 + 1s1.. - 9
o
p.2 -
61. + 9)
0
(1.- 1)
p.. - 3)2
0
3 ).. -
(1.- 1)
:.
'-1
= 1, '-2
Now 1.2 = 3 is repeated twice hence for 1.2 used.
I
i3,
A.3
=
3
= 3, te generalised
eigen vectors must be
Modem Control Theory
Matrix Algebra & Derivallbn of Transfer !=unctioi\
3. 31
[=~
-1
-~]
-1
-2
[~:~J m =
As it is a null matrix, calculate cofactors about other row.
Now for A2
= 'A.3 = 3, use
'·2 -4 [
For
>.3 =
3,
-1 -1
-1
'·2 1
TI
dC11 dl.2
TI
dC12 d/,,2
i!
dC13 d/,,2
Hence the modal matrix M is,
M
[! ~ n
-22 1-i -3
l
= [2"'2:- 3J
_ ).2-J
•2~3
=
m
Modem Control Theory
n•
3-32
Matrix Algebra & Derivation of Transfer Function
The Fig. 3.1 slwws the block diagram of a control system using state wriab/e [eedbuck'and int1•gra/ co11trol. 111e state 1midel of plant is, Example 3.16 :
Y = [O
1]
[~:]
i) Derive the stale model of the en/ire system. ii)
Derive the trm1sfer [unction Y(s)/U(s).
U(t) --..•-O<J>---4
Y(t)
Fig. 3.1 Solution : i) The input of integrator shown is
. X:i
and
r(t}
x3.
- 1 x [U(t) - Y(t)] = Y(t} - U(t)
... (I)
- 3.5 X3
... (2)
-
2X1
-
1.5 X2
From the plant model given, - 3X1 + 2X2 + r(t)
... (3a) ... (3b)
and
... (3c)
Using equation (2) in equation (3a) we get,
.
X1
- 3X1 + 2X2 - 35 X3 - 2X1 - 1.5 X2 ... (4a) ... (4b)
Substituting equation (3c) in equation (I), ... (4c)
and
Matrix Algebra & Derivation of Transfer Function
3 .33
Modem Control Theory
... (4d)
Y(t) = ~
[ · i [-! ~:
Thus equations (4a) to (4d) give state model of the entire system.
~~ =
and
i.e.
XJ
0
Y(t)
(0 l
x
-~5]
1
0
[~~]+[ ~i U(t) X3
-1
l
OJ [~~
and Y = ex with D = o
= AX+ BU
ii) The transfer function Y(s)/U(s) is, Y(s) U(s)
[sl - A)
= s [~
~
001
Adj [st -A]
Al
Y(s)
~i-[-: ~·: -~· 1 5
0
[ •' .,.
=
(0
Isl-Al
=
s+5 -1
-3.5s-17 .5 -14 s2 +10s+23
sJ + 10s2 + 23s + 14
d D=O an
4s
[4
0~5s-3.5 s· +5s s+S
= 14 Y(s) 14 U(s) =Isl-Al=
-0.5
r
4
s+5 (s+5)2 -2
s2 +Ss l OJ
0
4s
= s (s + 5)2 + 14 - 2s CAdj[sl-A)B
[s_+:
s2 +5s -14
s2 +5s s+S
=
=
0
0.5s-3.5
4s 4
U(s)
C Adj [sl - A) B
B+D
[ •' +Ss = O.Ss-3.5 -3.S(s+S)
= Isl -
t
= c [sl - Ar
14 sJ +10s2 +23s+l4
-3.5s-17 -14 ~2
.51 [ o]0
+10s+23
-1
Modern Control Theory
Matrix Algebra & Derivation of Transfer Function
3. 34 14 (s+l)(s+2)(s+7)
~i
This is the required transfer function. ,,_.
Example 3.17 :
For the matrix A= [ ~ -2
~
-3
-1
Find the eigen values, eigen vectors and modal matrix M. Solution : For eigen values,
o
P..I-AI
I~ "
-1 4
ol
-1
- 0
l.+3
4J..
0
->< + 31.2 + 41. + 2
0
1.2 (I. + 3) + 2 +
Test
'-=
-1, 4
- 1 -.1
:v.. + 2)
0
(I.+ 1) (/.. + 1 + jl) (/.. + 1 - jl)
0
(I.+ 1) (1.2 +
"· Calculate eigcn vectors.
For >.1
=-
-2
1,
P-11-A)
M1
n
-1 -1 4
-rJ
[~:~l=H]=HJ
- 1,
-2
l..2 = - 1 - jl,
~-3
=-
1 + jl
Matrix Algebra & Derivation of Transfer Function
3. 35
Modern Control Theory
-1
[-1-jl For
},2
-1 -jl,
(/,21- A)
a
-1-jl
~
4
2~µ]
[~:] =[-;2-j]
M2
... Choosing 3rd row to calculate cofactors. -l+jl For A.3
= - 1 + jl, [A.3 I -
A)
=[ ~
-1-j
-1 0 2+jl
l
_;+jl
j2 -j2
M = [-:
ll*
-1 -l+jl 4
Example 3.18 : For a system wil/1 state model matrices.
[-1 0 1] [0] [l]T
A=~-:
~;B=~;C=~
Obtain tht system trans/trfunction.
(VTU: March·2001)
Solution : The T.F. is given by, T.P.
=
C(sI-Ar1s
(sI - A) =
rs~l ~l ] s~2
0
Adj[sl-A)
O
s-3
= [Cofactor sl-A)T
=
[(s+2)0(s-3)
(s-3) (s+l) (s-3)
0
-+{s+2)
+1
(s+l)O(s+2)
]T
Matrix Algebra & Derivation of Transfer Function
3. 36
Modem Control Theory
c
(s+2) (s-3) (s-3) [
0 (s+l)(s-3)
0
l
(s+l/(s+2)
0
(s + 1) (s + 2) (s - 3)
Isl-Al
Adjlsl-A) lsl-Aj
(s1-Ar1
0
s+l 1 (s+l) (s+2)
(s+2)
0
0
(s+l)(s-3) I (s+l) (s+2) (s-3) 1 s-3 0
s+f T.F.
(s+ 2)
[l
1
OJ
1 (s+l)(s+2)
(s+2)
0
0
(s+I) (s-3) (1
1
OJ
1 {s+l)±(s-3)
0 (s+l) (s-3) 1 (s+l) (s+2) (s-3) 1 s-3
0
l
(s-3) l + 1 (s+l)(s-3) (s+l)(s+2)(s-3) (s+3) (s+ I) (s+2) (s-3)
11••
Example 3.19 : Find the transformation matrix P which will co11vert the following matrix A lo diagonalform. Th« eigen tialutS are A. 1, A. 2 and ).. 3. A=
0 0
1 0
-01
-a2
[
]J
Modem Control Theory
3. 37
Solution : The matrix Al - A is,
p,1 - A) =
r "-
-1
0 at
).. a2
ll
To find P, means to find cigcn vectors and modal matrix,
-1 0 1., +a 3
l
To find eigen vector, find cofactors about 3rd row.
-1 0 A3 +a 3
[~:]=[~~]
l
Matrix Algebra & Derivation of Transfer Function
Matrix Algebra & Derivation Modern
Control Theory
of Transfer Function
3·38
Hence transformation matrix P is, I
p
A2
i,1
).,3
A.1 A.1 A.~ Thus if matrix A is in phase variable form,
A
U.,
0
I
r~
0
{)
()
0
()
-CX.n-l
...
-(l.n-2
-(XI
And the eigen values arc distinct as A.1, ),2, ••• ),n then it can be proved as above that the diagonalising matrix is,
i.,
..~
p
I
1n-l 'I
1•
"z
A..,
),~
A~,
.,
1.n-l
"ll-1
'·2
-u
Key Point: Note tha: while fi11di11g the eigen l·'<'Ctorsfind tire cofactorsabout last row. E.xample 3.20 :
Determine Ille trans/tr nwtrir for tlu: system.
=~
[ 1()J [~~ ]+ [~ ; ] [~~] : [~;] ~ [~ -11] [~: J [~I]= 2
(VTU: July/Aug.-2005}
Solution : Prom the given state model, A = [
=~ ~] ,
Adj[sl-AI Isl-Al Adj (sl-AJ
B
= [~S
60] r
-[s+3
_ '(sf
C = [~ - ~] , D
Al-
[ls -2s+3 ]T = [s-2 s+31 ]
2
-I
l
sJ
=[0J
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
3. 39
T.F.
[1 -1] 9s+ 18 . _ (s+1)(s+2) .. T.F. 27s-63 [ (s+1)(s+2)
l
[4s-5 6s 1 -Ss-23 -12J (s+l)(s+2)
8
6s+ 12 (s+l)(s+2) 48s-12 (s+l)(s+2)
l
9 s+l = 27s-63 [ (s+l)(s+2)
s+I 6 48s-12 (s+l)(s+2)
I
This is the necessary transfer matrix. ••
Example 3.21 : Convert the Jo/lawing state model into canonical [orm. (VTU: JulylAug.·2005)
Solution : From the given model.
= ('3 --64 } B = [O-1 ] • C =(I
A
0]
Let us find the cigen values, eigen vectors and model matrix of A.
-4JI=1i.-r3 1 11I .[1o oJ-[l 1 3 -6
l)J-AI (/..-l)(A.+6)+12 (/..+2)(A.+
0 i.e. A.2 +51.+6
=0
0 i.e. t..1 =-2. }..2 =-3
3)
To find eigen vectors,
=-i!J.
Fori.1
11-A)
= [=~ :]
i.e.M1
=[~:~]=[~]
~J
i. +
=0
Modom Control
For Az ;-3,
21-A
[l.
I= [~ :]
i.c.
Mr[~:~]=[~]=[:]
J
M = [M1:
M2 = [; ~]=Modal
matrix
r-20 - o]3 = "
M-11\M
M-1a
6
While
Matrix Algebra & Derivation of Transfer Function
3 ·40
Theory
[1 -1]
IMJ-n - _.,
M-1
ii
Adj M _ -3
4
.. [I
[~3 -~] [~1]
= [~]
C =CM= (1 Ol [43 I]= ('I 1
-l] 4
1]
1hc canonical state model is,
11.+
l(t)
A Z(t) -+
Bu
Y(t)
C Z(t)
where X(t)
= M Z(t)
Example 3.22 : Convert tire followi11g squart matrix A into [ordan canonical fonn using a suitable 11011-si11g11/ar transform11tio11 matrix P. (VTU: July/Aug.· 2005) A
=
o
1
0
0 -9
r l-.i
-~J
Solution : Find the cigen values or A.
1).1-AI
=
i,2(A+6)+4+9A=0 (AH)(/.+ l}(A+ 1) =0
I~
-1
A 9
~1 A+6
I= o
i.e. A3 +6i..2 +9i,+4 =0
Modem Control Theory
Matrix Algebra & Derivation of Transfer Func:tion
3-41
For A1= -4
cC1112 [C13
Mz
l -[).~ -
+6~.-4 +9]
2=-1
=
-[ 4]-[ Az=-1
Ul{~l
[2'-~ +6] -'1
1
- -4 - -1t 4 I
2
-4),2
I.
Az=-l
fl M
r -4 12r 2 1
Adj [M)
-17 15 -4 3
= [~24
2 -41] -17 15
3
2 1]
M-l =~~I~
=
[~24
-17 -4 15 3 9
M"1AM-f- 9
4 12
2 -4I] -17 15 3
[~4
O][l
I 0 l - 9 -6
1 -~-I 16 I
~oi)
l
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
3-42
..... Diagonal matrix with Jordan block
n•
Example 3.23 : Consider the matrix
A [: -2 31] I
3 -l
i) Find the eigen values and cige11 vectors .of A. ii) Write the modal matrix. iii) Shaw tlrat the modal matrix indud diagonalizes A. Solution ; For eigcn values I A.I -A
~;l
J
2 l.=1
-1
)..+:!
- 3
= 0
2) + 2-9- 3(/.-1)+2('A.+1 )- 3(>..- 2) =0
i.e, (J.-1)(!.+1)(1.i.e. 'A.3 -21! -5),+6=0 1..1 =+1,
I=0
i.e.
1..2 =-2,
p.- l)(A+2)('A.-3)=0
i.3 =3
For i.1 = l, [I.I-A)
For 1..2 =-2.
fl..1-AJ
As C11
=
C12
=
C13
=
0, calculate cofactors about other row.
rvruJan.lFeb.-2006)
Matrix Algebra & Derivation of Transfer Function
3 .43
Modem Control Theory
For i.3 = 3,
l
11 l -14
M
l 1
Adj M I Cofactor mat rix of M] 1 ~=!Ml
1 [15 -25
0 -2 10 -30 -15 - 3 -~2 15 - 25
l
1~
r~ - ~ ~
1 1 [~ -~ [l; -:5 ~~ l [~I \1 :1 -·lo
30
-}a
0 -2 [ -15 - 3 -12
- 45 - 9
-30 0 0
[0
60
- 36
3 -1
1
1
1
~1
1 \1 - '14 1
- 14 1
~1-[~ -~
0 - 90
.o
0
.
Thus modal matrix indeed diagonalizes A.
I,..
Examplo 3.24 : Cive11 the state model X =1\X-flu, y =CX
(VfU: JanJFeb.·2006)
1o wltert A
= [~
~] · B = [~] and C :
-1 - 2 -3 i) ii)
[I 0 OJ
1
Simulate and find the transfer function ~~~ using Mason's gain formula. Determine tile transferfunction from the state model formulation.
Matrix Algebra & Derivation of TransferFunction
3-44
Modem Control Theory
Solution : The state equations arc,
.
X1=X2•
.
.
X2i\X3,
X3,.-X1-2X2-3X3+u,y=X1
Hence the signal flow graph is, ,
1
1
u ~--~x\?27·"' -_s __ ~x~3--1s x9'2 s x?-1---·, 3 ---<>o Y
-3
-2 -1
Fig. 3.2 From Mason's gain formula, Y(s) U(s)
Herc K ~ 1,
T161 +T2a2+ 6
T1 =Forward
+TK6K
where K
= Number
of forward paths
path gain= !x!x.!. =...!_
s s s
53
The various loop gains are,
6 -3
Flg. 3.3 All loops are touching to each other. 6= 1-(L1 6K
a,
+L2+L3]:1+~+2.+...!_ s s2
53
Eliminating all loop gains from 6 wh.ich are touching to Kth forward path
Modem Control Thoory
MatrixAlgebra & Derivation of Transfer Function
3 -45
Y(s) U(s)
Now let us use the state model method of finding the transfer function. Y(s) U(s)
[sl-A]
... D ~o
C(st-Ar'n+o -1
[~
-I 0 s +3
2
s2 + 3s+2 Adj (sl-A( [
:+3
l
-1
-s
s(s+3)
-2s-1
]T [52+3s+2
s2
Y(s) U(s)
Isl-A
, :l
-2s-l
-s
I
-2s-1
s+3 s(s+ 3)
-1
Adj [sl-AJ
s(s+3)
-s
['' + ,,., [s1-Ar1
s+3
= -1
.i]
s3 +3s2 +2s+l
s2 +3s+2 (1 0 0) -1 [ -s
s+3 s(s+3)
.i][:]
-2s-1
s3 +3s2 +2s+I Y(s) U(s)
,...
... Same
Example 3.25 : Tlte vector [ ~] is a11 tigm vtetor of A -1 oo/11t of A corresponding. to lht vector given.
=
r-~
where
X,
obtained above
_:
x, X, = Elgen vector
= [ _~]
~ ~] Fmd tlte e1gm
-1 (V11J: July/Aug.·2006)
Solutlon : The eigen vector is that vector which is non-zero such that AX1
'15
s3 +3s2 +2s+t
Matrix Algebra & Derivation Modem Control Theory
A.I
of TransferFunction
3 -46
=
Elgen value = To be obtained
[-~-1 -2~ =!][~]=A.;[~] 0 -1 -1
[~~]A.;
[~ . l -A.1
= s
... Corresponding eigen value.
Review Questions l. Derise lllL traruferfunctionfrom statt rnDdtl.
Wh.1t is ch•racttristic equation of• system mDtrix A ? Dtfint eigm wluts and eigtn 11ttlors of a n1J1trix. Whtn lht gtncraliJtd tigen oeaor« •rt uStd ? How to use thtm ? What is rrwdal matrix ? St•lt its imporlana. 6. Stole the adwntagts of diAgonnlisationof a matrix. 2. J. 4. 5.
7. Now lo achinit dia,
X
= [~
~J
X + [:] U(t) and Y(t):
(1 OJ X(t)
9. What is Vandtr Monde matrix ? Whm it exists ? WIU1t is its imporiano: ? 10. Find tht T.F. of tht systttnJ with following state modtls, al
x [-~ -~Jx+[~]u.
bi
X •
[ ~ ~~] X + [~]
x
U-~
cJ
Y=/l
u
O/X
Y s 10 II X
~~)x·[~]u.
Y=ll
o 01x
11. Obtain the T.F. of the systrm having state modt/, X(t)
= [-;
Y(t)
=
=~]
X(t) + [~] U(t)
[1 2J X(t)
(
A
12s+ 59 )
ns,
<s + 2)(s + 4)
ODO
Solution of State Equations 4.1 Background Uptill now the methods of obtaining state model in various forms and obtaining transfer function from the state model arc discussed. It is seen that the output response depends on the state variables and their initial values. Hence it is necessary to obtain the state vector X(t) which satisfies the state equation X(t) " A X(t) + B U(t) at any time l. This is called solution of state equations, which helps to obtain the output response of a system. Consider the state equatio n of linear time invariant system as, X(t) = A X(t) + B U(t) The matrices A and B arc constant matrices. This state equation can be of two types, 1. Homogeneous and 2. Nonhomogcncous
4.1.1 Homogeneous Equation If A is a constant matrix and input control forces arc zero then the equation takes the form,
.
X(t) = A X(t)
...(1)
Such an equation is called homogeneous equation. The obvious equation is if input is zero,how output can exist ? In such systems, the driving force Is provided by the initial conditions of the system lo produce the output. For example, consider a series RC circuit in which capacitor is initially charged to V volts. The current is the output. Now there is no input control force i.e.. external voltage applied to the system. But the initial voltage on the capacitor drives the current through the system and capacitor starts discharging through the resistance R. Such a system which works on the initial conditions without any input applied lo it is called homogeneous system.
(4. 1)
Modem
4·2
Control Theory
Solution ·of State Equations
4.1.2 Nonhomogeneous Equation If A is a constant matrix and matrix U(t) is non zero vector I.e. the input control forces arc applied to the system then the equation takes normal form as,
.
X(t)
= A X(t) + B U(t)
... (2)
Such an equation is called nonhomogeneous equation. Most of the practical systems require inputs to drive them. Such systems are nonhomogcncous linear systems. The solution of the state equation is obtained by considering basic method of finding the solution of homogeneous equation.
4.2 Review of Classical Method of Solution Consider a scalar differential equation as, dx dt
=
ax
where
x(O)
= Xo
... (1)
This is a homogeneous equation without the input vector. Assume the solution of this equation as,
At t = 0,
x(t)
... (2)
x(O)
... (3)
The solution has to satisfy the original differential equation hence using (2) in (!), d di Ibo + b1 t + ...
+ bkt
k
I
a Ibo + b1 t + ... + bkt
k
I
For validity of this equation, the coefficients of various powers of 't' on both sides, must be equal. abo- 2b2 = ab1, ... kbk = abk. 1 a b0
.!. a b1 2
= .!. a2b0 2
= .!.2! a2 b0
.!. a bz = _1_ a1>o = ..!.. a3 bo 3
.!. a k b0 k!
3x2
and
3!
x(O)
= b0
Modem Control Theory
Solution of State Equations
Using all these values in the assumed solution, x(t) " bo + abo(t) +
..!.. a2 bl 2!
+ ... +
1.. ak ho tk k!
1 k t ~] b0 [ l +at +211 a 2 t 2 +.... +kia 2t2+ [l+at+.!..a 2!
But
I ... at +
.!_ a2 2!
.... +.!.aktk] x(O) k!
t2 + ... + _!_ ak tk: e11 k!
X{t) = e•1 x(O)
.•. (4)
This is the required solution of homogeneous equation in scalar form. Thus if the homogeneous state equation is considered, X(t)
=
A X(t)
then its solution can be written as, X(t) = eAt X(O) In this case, eAt is not a scalar but a matrix of order n x n as that of matrix A. Observation : It can be observed that without input, initial state X(O) drives the state X(t) at any time t. Thus there is transition of the initial state X(O) from initial time t : 0 to any time t through the matrix cA1. As it is an exponential term, the matrix cAl is called matrix exponential. It is also responsible for the transition of the state X(t) at any time t from initial time hence also called state transition matrix. It is denoted as o (t). $(1)
eAt = State transition matrix
And
eAt: 1 +At•..!_
21
A2 t2;. ... + _!_ Ak lk + ... k!
The each term of the above equation is a matrix of order n x n. If instead of initial time t
= 0, it Is selected
$(I) = eA {I-to)
as t
= to then
the state transition matrix is,
4.4
Modem Control Theory
Solution of State Equations
4.2.1 Zero Input Response The solution of the homogeneous state equation L~ under the condition of zero input. Such a response is called zero input response and for convenience denoted as ZIR. Thus the behaviour of X(t) under the initial conditions without any input U(t) is called the zero input response of the system. It is also called free, natural or unforced response of the system.
4.3 Solution of Nonhomogeneous Equation Consider a nonhomogeneous
.
X(t)
state equation as,
A X(t) + B U(t)
X(t) - A X(t)
BU(t)
Premultiplying both sides bye· At, e- At(Xct) - A X(t)I
=
e-At B U(t)
Bute· At X(t) - e-At A X(t)
= ~ [e-At X(t)) dt
Substituting in above equation, ~ [e-At X(t)) dt
=
e-At B U(t)
Assuming initial time as t = 0 and integrating both sides from t c-At X(t)
1: =
= 0 to t,
I
Jc-At B U(t) dr ,0
t
e-At X(t) - X(O)
= f c-At
B U(t) dt
0
Premultiplying both sides by eA•. t
J
:.eAt c-At X(t) - cAt X(O) = eAt e·At B U(t) dt 0 t
X(t)
eAt X(O) +
J eA(t-t)
B U('t) dt
0
This is the complete solution of the nonhomogeneous equation.
... (1)
4.5
Modem Control Theory
Solution of State Equations
Observations : 1. The solution is divided into two parts. The first part is eAt X(O) which is nothing but the homogeneous solution or zero input response (ZIR). t
J eA(t-t)
2. The other part
B U(t) dt is the part existing only due to application of
0
input U(t) from time 0 to t, It is called forced solution or zero state response (ZSR). Thus the solution is, t
X(l)
= eA• X(O)+
j eA(t-t) BU(t)dt 0
ZIR
ZSR
If the Initial time is considered as t = X{t)
=
to rather
than t = 0, the solution is,
t
cA <1-101x(t0)
J
+ eA Ct-•J B U(t)dt
'o ln general, the solution of nonhomogencous equation consists of both the parts, ZIR (zero input response) and ZSR (zero state response).
4.4 Properties of State Transition Matrix The various useful properties of the state transition matrix are, t)(t)
eAt = State transition matrix
1.
.;>(0)
cA •o
2.
Q(t)
CAI = (c-Atr I = ft) (- t)r 1
Le,
91(1)
= I ., Identity
matrix
$(-I) CA(t1d2) "'eAt1 ·eAt, •
3. t)(t1 + t2)
t) (t1l t) (l:z) ., t) (12) t) (11)
4.
eM• + •>
5. e(A + B) t 6.
7.
(9 (t)J"
eA• eAs ~ eAt eBt only if AB "
9(12 - 11) 't)(t1
[eA11" -
to)
=
= eAn• = Q (nt) D
t)(t2 -
to)
BA
4·6
Modem Control Theory
Solution of State Equations
This property states that the process of transition of state can be divided into number of sequential transition. Thus to to '2 can be divided as t0 to t1 and t1 lo t2, as stated in the property.
lntcrms of ~ (t), the solution is expressed as, I
X(t)
Q
(t - to) X(to) +
J Q(t-t) B U(t) dr 'o
where
and
to>
0 (t -
~(t - t)
8. o (t) is a nonsingular matrix for all futile values of t. 4.5 Solution of State Equation by Laplace Transform Method The Laplace transform method converts intcgro differential equations lo simple algebraic equations. Due this important property, it is very convenient to use Laplace transform method to obtain the solution of state equation. Consider the nonhomogcncous slate equation as, X(t) = A X(t} + B U(t)
... (1)
Taking Laplace transform of both sides, s X(s) - X(O} s X(s) - A X(s)
A X(s) + B U(s} X(O) + B U(s)
As s is operator, multiplying it by Identify matrix of order n x n, (sl - A) X(s) = X(O} + B U(s) Premultiplying both sides by (sl - Ar 1, :. [sl - A]" I [sl - A] X(s} = [sl - Ar I (X(O) + B U(s}I X(s)
[sl -
Ar
I X(O) + [sl -
Ar I B U(s)
... (2)
ZIR + ZSR Comparing zero input response obtained earlier, [sl - Ar and
1
= Hsl •.• (3)
=
The matrix Q (s) [sl - A]" 1 is called resolvant matrix of A. All the clements of this matrix are rational functions of s.
Modem
4.7
Control Theory
Solution of State Equations
Taking inverse Laplace transform of (2), L" 11X(s)) = L" 1 l(sI - Ar 11 X(O) + L" 1 l[sl - A)" I BU(s)I
X(l) Using
Isl - Ar
1
(s), L" 1(¢ (s)) X(O) + L" 1 (¢ (s) B U(s))
X(t)
... (4)
The term L" 1 [¢ (s) B U(s)] is called zero state response. From the convolution theorem in Laplace transform it is known that, f1(t) • f2(t) =convolution
L" 1 IF1(s) F2(s)l
I
Jf
1<1-t> r2c1)dt
0
Thus if F1(s) = $ (s) and F2(s) = BU(s) then, I
L"1{<>(s)BU(s)I
= Jo(t-t)BU(t)dt
... (5)
0
Thus equation (5) shows that L" 1 {Q (s) B U(s)I is the zero state response, as obtained earlier by classical approach. X(t) = L" 1
16 (s)) X(O)
Q(s) = [sl _A)" 1
=
+ L-
1[¢
(s) B U(s)]
Adj [sl-A]
lsi:At
Key Point : The advantage of tlris metlzod is witho11t carryi11g out actual i11tegration wliich is time consuming, the zero state response ca11 be easily obtained.
4.6 Computation of State Transition Matrix While obtaining the solution of stale equation, the computation of state transition matrix cAt plays an important role. There arc various methods of obtaining state transition matrix from the state model. These methods are, 1. Laplace transform method 2. Power series method 3. Caylcy Hamilton method 4. Similarity transformation method Let us discuss these methods of obtaining eAt in detail.
Solutionof State Equations
4·8
Modem Control Theory
4.7 Laplace Transform Method While studying the method of obtaining- the solution using Laplace transform method, it is seen that the Laplace transform of cAl i.e, Q (t) and given by, c>
()5
=
I5 I
Ari
- Adj[sl-A] - . I sI -A I
-
Thus knowing A, once [sl - A) is obtained then 6 (s) con be easily obtained. This c> (s) is again a matrix of order n x n i.c, same ns tha~ of i\. It is called rcsolvant matrix of A. The eA1 i.e. c> (t) then can be obtained by finding inverse Laplace transform of Q (s) i.e, Inverse Laplace transform of each clement of the resolvent matrix Q (s). This is most simple popularly used.
n'*
method of obtaining state transition
matrix cA• and hence
Find tl1e state lra11sitio11 matrix for,
Example 4.1
A =
o [+2
-1) -3
Solution : Use Laplace transform method, (st-A]=
Adj [sl -A]
Isl
-Al
1s1-Ar1
0 (s)
s[~
~H-~ -~]=[_;
s+ [ -!
3 2]T = 3 -1] s s
5:~]
[s +
2
!]
+3 = s2 + 3s + 2
s•.
Adj [st-AJ
= (s + 1)
(s + 2)
-1]
s+3 [ 2 s = (s+1) (s+2)
-·1-;1.:.-Al [sl -Ar
1
s+3 (s+l)(s+2} =
2 [ (s+l)(s+2)
(s+ l):s+2) -1 (s+1)(s+2)
l
4.9
Modem Control Theory
s+ 3 L-
I
[sl -
I
Ar = L-
l
Solutionof State Equations
I (s+l)(s+2}
[
2 (s+ l)(s+2)
(s+1):s+2) -1 (s+l)(s+2)
Using partial fraction expansion for all the elements.
cAt L-1 [s!1 ·· s!2 s~\ ,!2] +
2 2 ---s+1 s+2 e
At -- [ 22c·t -c·2t 2
e ·t - e ·21
-1 2 --+-s+I s+2
-c·t +c·2t ] - $ t -e ·I + 2e-21 - ()
This L~ the required state transition matrix.
4.8 Power Series Method The state transition matrix is matrix exponential and hence can be expressed in a power series as,
... (I)
cAt
Thus by calculating various powers of A and then few terms of power series, can be evaluated. As the power series is infinite, it is necessary to stop after calculating first few terms of the series. Calculating hlgh powers of A is practically time consuming and hence lhls method is not practically used to obtain stale transition matrix.
1•
Example 4.2 :
Find ~t of A = [~
=~]
using po~r serits me/hod.
Solution : Let us obtain various powers of A. A2
= [~ o [2
=~][~ =~]=[~ :] -1] [-2 3] [ 6 -7] -3
-6 7 = 14 -14
.... • Al :..10
Modem Control Theory cAt
Solutlon of State Equations
= I+ At+..!._ A2r + _!_ A3 21 31
+
13
[~ ~]+[~ =~]t+M=! ~}2+H -~;]13+ . . 1~
l-t 2 +I 3 +... 14 3 [ 21-31 . 2 +6l +...
Consider the first clement of
CAI
r + t3 + ...
which is l -
1 + [- t) + 2.[-tj2 +2.[-tj3 + ...
Now
21
2 - 2t + t2 -
3!
.!.
3
t3 + ...
+2- (-
I + (- 2t) + 2_(-2t)2 2!
and
. .. (I) 2t)3 + ...
3!
. .. (2)
1 - 2t + 2t2 - ~ ~ + ...
6
From (1) and (2) we can write, Zc" 1 - c- 21
=
l + 0- t2 +
t3 + ...
= 1 - t2 + t3 + ...
Hence the series present as the first clement is 2c- 1
-
c- 21•
But practically such a task is difficult to identify the combinations of series. Hence this method is rarely used.
4.9 Cayley Hamilton Method This method is based on the Cayley Hamilton theorem. The theorem states that, every square matrix A satisfies its own characteristics equation. Let f(IJ be the characteristic equation which is given by,
i.e,
f().)
II.I-Al
f().)
).n + al >.n - 1 + ... +
=0
a,, - 1
A.+
a,, = 0
... (!)
According to Caylcy Hamilton theorem, the matrix A has to satisfy the characteristic equation. Hence, ... (2)
4·11
Modem Control Theory where
Solution of State Equations
I = Identity matrix
The theorem is used for evaluating the function of a matrix. Consider a matrix polynomial as, p(A)
=
a01 + a1A + aiA2 + ... + anAn +an+ 1 A"•
1
+ ...
...(3)
This polynomial has degree more than the degree of A which is n. The above polynomial can be calculated considering the scalar polynomial, p(A)
= ao + a1A+ a2A.2 + ...
+ an A." +
a,. • I
An+ l + ...
... (4)
Divide this by the characteristic polynomial f(A),
where
~~~;
Q(/~+ ~:~
... (5)
R(A)
Remainder polynomial
p(A)
f().) Q(A) + R(A)
... (6)
Now the remainder polynomial R(A) has order less than the characteristic polynomial and can be expressed as, R(A) "' Clo + a,A. + Oz ),2 + ... + On - 1 An • I
Ai, ~' ... ).,, are the n distinct cigen values Ai, ~' ... >.,, are the roots of f(A) = 0. Evaluate
Let n as
But
p(/.i)
fP,) QO,) + RC),)
f(Ail
0 hence,
of matrix A and f(\) = 0 for i p(IJ al all the eigen values,
... (7)
= 1, 2, ...
for i = I, 2 .... n
.. , (8)
Substituting Ai· )2, •.. A,. in equation (8) we get the set of simultaneous equations which is to be solved for the values of Cl(). a1 ... a.,_ 1. Now replace A. in scalar polynomial by matrix A,
but
p(A)
f(A) Q(A) + R(A)
f(A)
0
p(A)
... By CayIcy Hamilton theorem
R(A) = OQI+a1A+CtzA2+
Thus knowing 0(), a1
...
a.,_
...
a.,.1A"-1
1, any matrix polynomial p(A) can be evaluated.
... (9) .
Solution of State Equations
4·12
Modem Control Theory
4.9.1 Procedure to Calculate
eA1
The procedure to calculate cAt using Cayley Hamilton theorem can be generalised as, 1. Find the eigen values of matrix A from
I A. I -
A I = 0.
2. Form the remainder polynomial R(~ of order (n - 1). R(~ = Clo + tt1 ). + ... + CX,.. - I An - I = p(~ 3. If all the eigen values are distinct then substitute At• 1'2 .. . ),,, in remainder polynomial R(~ to obtain simultaneous equations interms of ao, tt1 ... a,.._ 1. Important Note : If the eigen value \ is repeated r times, any one independent equation can be obtained by substituting 11< in R(~. Then remaining (r - 1) equations are to obtained by differentiating both sides of the equation,
diR(i..>j
-d).!
.
, J
4. Solve the simultaneous equations
1•
= 0, I, ... r -
1
... (10)
'-=l.k
ao, a1, ... a,.._ 1· a,.._1 An-l.
5. Then f(A) = R(A) =°<)I+ cx1A + ... +
Fi11df(A) = A
Example 4.3 :
12
for A = [-~ -~] using Cayley Hami/1011 theorem.
Solution : The function to be evaluated is, f(A) Step 1 :
I~ . ~31=
0
A.2 + 31.+ 2
Ai Step 2:
R(~
i.e.
O
i.e, (A.+ 1) (A.+ 2) = O
-1, ~ = - 2 Cf
i.. = p(),) ... as n
=
2.
Substitute /'1 and 1'2 in R(I,).
Clo + tt1 and
A 12 hence p().) = 1.12
Find eigcn values.
li..1-AI
Step 3 :
=
°<> +
Ai
tt1 ~
ao-
a1
0-i)12 (~)12 as p(J.} = (-1)12 = 1
,,u ... (1)
Modem Control Theory
and Step 4 :
Oo - 2a1
4 -13
Solution of State Equations
= (- 2)12 = 4036
... (2)
Subtracting (2) from (1 ), -4035
IX1
Step 5 :
Oo = -4034
and
A12 = R(A)
f(A)
aJ + U1A = - 4034 [~
~]- 4035 [-~ -~]
[-4034 -4035] 8070 8071 A12 =
[-403-1 -4035] 8070 8071
Similarly eAt which is a power series in\crms of A can be evaluated using Cayley Hamilton theorem.
>>*
Find llit state tra11silion matrix for
Example 4.4 :
-1] Using Cnyley Hamilton Theorem.
A = [o
2 -3
Solution : The function to be evaluated is, f(A) Step 1 :
).,2
O i.e,
+ 31.. + 2
O
At
-
I-~ ;,+ 1=0 3
i.e. I,
/"2.
(}., + 1) (}., + 2) = 0
=-
2
Construct R().) R(>.)
where
eAt hence p(A) = e>. 1
Find the elgen values, 11..t-AI
Step 2 :
=
ao + a11.. +
... +
ex,,_
1 1 /,n - but n = 2
R().)
... By Cayley Hamilton method
p(A)
... replacing A by /, in f(A).
Modem Control Theory
Solution of State Equations
4 • 14
Substituting values of >.1 and ~.
Step 3 :
. .. (1)
... (2)
and Step 4 :
Solving equations (I) and (2),
e-1 -e-21 2(,-1 _ e-21
·and Step 5 :
f(A)
c
eA•
R(A) = <Xol + u,A ... replace I.. by A in R(A)
« 21 [~
2e-c -
~] + (e- 1 _ e- 21) (~
=~]
This is the required stale transition matrix. 1111+
Fiml the state tra11sitio11 tnJ1trix of,
Example 4.5 :
A
=
o o
-21
1 0
3
0 1 OJ by Cayley Hamilto11 theorem. [
Solution : The function to be evaluated is, f(A) = eAt hence p(A) = ,/' Step f :
Find the eigcn values.
=1~
l'-1-AI
0 },-1 -1
0
=O
>.(A-1} (1..-3) + 2 (1..-1} (1..-1}
p,2 -31..+ A,
= 1,
~ I= 0
A-3
2)
= O i.e. {A-1)
~ = 1,
(1..-1} (1..-2)
~ =2
Thus A, ~ 1 is repeated 2 times Step 2 :
R(A)
<Xo + ail..+ fkz'-2
= p()J =
e'IJ.
... n = 3
=0
4-15
Modem Control Theory Step 3 :
a1
(Jo+
+CC:!
Ai = 1, as it
For
Solution of Sta~e Equations
For )1= 1,
=
d Al -e di.
\ ... (1)
is repeated use,
~P<' ·>I 11.=Ai di. i.e,
\
e1
I"'"2
te).tl ).=1·2
~Ro->J di. J.=A.i
2 II
d -[a0 +a1).+a2). di.
i).=).2
a I+ 2a2 /~l.a).2
\
te1
a1+2az
... (2)
Note that differentiation is with resped to I. and not t, and for
":i
= 2,
(Jo + 2 a1 + 4
Step 4 :
az =
.•. (3)
e21
Solve equations (1). (2) and (3). 2 Ciz
From (2),
CI1
t
From(!),
ao
et - a1
c1
-
-
az = ct -
2az -
tc1 +
«i
e1(1 - I) + f'2 Using in (3),
c1 (I -
t) +
Ciz + 21 e1 - 4 -12 + 4 Ciz = e21 c21 - c1 + t e1 - 2te1 " e21 - e1 - t e1
tc' - 2e21 + 2c' + 2te1 and Step 5 :
(Jo
f(A)
= 3te1
+ 2c1 - 2c21
e1 - te1 + e21 - e1 - te1 = - 2te1 + c21 R(A) =(Joi+
a1A
+
Clz
A2
r~ r -~H~ ~ -~J n r -il =
f(A) =
eA'=OQ[~ ~
Substituting values of (Jo. a1 and
az and
~l+a1[~ ! -~]+a{~! 1] rearranging,
Modem
Solution of State Equations
4 -16
Control Theory
~''"]
2e1 -c21
0
0
c'
,,.
-c1 +e21
0
2c21 -c'
eAt
'This is the required state transition matrix.
4.10 Similarity Transformation Method Consider a matrix A of order n x n having cigcn values At• >.i, ... ~· There exists a diagonal matrix I\ such that its diagonal clements are the eigen values of matrix A.
A., 0
0 ,_
;
;2
O
0
A
[
0 0
···
l
= Diagonal matrix
.•. (1)
A.0
Both A and A have same eigen values and hence arc called similar matrices. The diagonal matrix A can be obtained by defining a similarity transformation of state variables as, X(I)
= M Z{t)
In such a case, M is the modal matrix of A. And the diagonal matrix is obtained as,
Now
I\
M-1AM
c"t
I + /I.I +
i·I !
0 1 -
o-
0 0-
... (2)
r
_!. 11.2 t2 2!
0
0
'J I +
1
0
+ ... 0
o-o
"2
0-0
·- >.."
-2
t+.!. 2!
"•0. I!
0
0
'-~
0 -o-01 0
->..~,
o---·---·
o
0--·--
0
12 +
Modem Control Theory
4-17
c'\t
Solution of Stato Equations
... (3)
The transformation used is, X(t)
MZ(t)
X{O)
M Z(O) M-1M Z(O) = Z(O)
M-1 X(O)
. . .
and
X(t)
~t)
Thus
X(t)
AX(t) = AMZ(t)
... (4)
AMZ(t)
~t)
M-1Mi(t)
~C1AMZ(t)
Z(t)
... (5)
A Z(t)
This is transformed state model. The solution o( homogeneous state equation in X(t) is, X(t) = eAt X(O)
••. (6)
While solution of equation (5) is,
=
Z(t)
c"1 Z(O)
... (7)
Using equation (4) in (7), Z(t) = e"1 M - 1 X(O)
... (8)
Premultiplying by M, MZ(t) But
M
e"1 M - I X(O) ... Transformation used
MZ(t) = X(t) X(t)
••• (9)
Comparing equation (6) and (9) it can be written as, eAt = Mel\t M - 1 where
M : Modal matrix
and el\l is given by equation (3).
I
... (10)
4 -18
Modem Control Theory
,,..
Solution of state Equations
Fi11d stale l7a11s1tfon matrix of.
Example 4.6 :
A = (20 -l] by similarity transformation metl1od. -3 Solution : Find the eigen values.
1-~
p,J-AI (:>. + 1)
p, + 2)
)..::11=:>..2+3:>.+2=0
0 -1,
hi= -2
Calculate cigcn vector. For
Ai= -1,
[~.1 l - A)=[=~
M2
M
0At
M-1
~]
[~;~ J = [~] [M1: M2J
"l1rt ~] =
l c_q
[ 2 Adj [11.1] _ -1
-I] 1 _[ 2
[c~'
r c-•
O ] /-ii
0
JMT--1--
-1
-;] ... Required state transition matrix
Modem Control Theory ,,..
Example
4 -19
Obtain the complete time response of system given by,
4.7 :
. = [_0 1]
X(t)
and
Solution of State Equations
2
O
X(I)
where X(O) = [~]
Y(t) = (1 - 1) X(t)
(VTU:JanJFeb.·1008)
Solution : Given system is homogeneous whose solution is X{t) = c'" X(O) Now Isl -A) (sl-A)-1
Adj (sl -A)
Isl - Al (sl- A)-t
s [10 o]1 _ [-2o 01] = [s2 -1]s Adj(sl-A)
Isl-Al
[enc cc )r
= [1s
12 22
21
-2]r s 1] =[
s
-2 s
s2 + 2 s s2 + 2 -2
52
:+ 2]
s2 + 2
eAt
L-I
[s
2:
s/+ 21
2
s
-2
+2
+2
s2 L-1
L-1
L-1
{s2:i} {/+2} {s2-~ 2} CAI
L-1
lr __
L-1
{J_..fi. .
°s2
s__ }., cos .fit 52 + (..fi.)2
L-1 {-
..fi.
}
52 +(..fi.)2
../'i .
cos.fit [ -..fi. sin..fi. t
.fi
s2 + (.fi)2
2
J_J2
sin
} = - J2
~sin.fit] cosJ'i t
.n t sin
.fi t
Modem Conltol Theory X(t)
CAI
:. Output response Y(t) = (1 (1
X(O) "
CAI
[~] "
Jz
t]
Jz
[cos J2 t + s.in J2 cos./2 t-./2 sin J2 t
- 1) X(t) - 1)
cos
sin
J2 t
Y(t) = ~ sin
./2 t
t]
J2 t +
[ cos./2
,..
Solution of State Equations
'4·20
~ sin J2 v2 t-./2 sin J2 t
is the required response.
Example 4.8 : Find out the time response for u11it step input of a >1JSlem given by
Xct)
= [~2 ~3] X(t) + [~] U(t)
= [~2 -~] X(t)
and Y(t)
and X(O) •
Solution: (sl-AJ
Adj ( sl -Al Isl-Al eAt
[" o][o 1] = [" -1 ] [ell c!2 "[s+ c21 c22 0 s - -2 0
2 s+ 3
r
t
3 -2r = s
[·+-23
:]
s2 + 3s + 2 = (s +I) (s + 2) L.1 ((sl -
[ .. ,
A)-11
(s+ l)(s+ 2) -2 (s+ l)(s+ 2)
, I
(s+ l)(s+
2)
=
¢(s)
(s + 1)5(s + 2)
Finding partial fractions, 2 ---
1 s+2 (O(s)) = _2._ + _2_ [ s+t s+2 s+t
s+l s+2 I I -1 2 -+s+l s+2
l
m.
Modem Control Theory
Solution of State Equations
4 ·21 c-t - c - 2t ] -c-•
To ftnd
= e"'
+2c-21
[l] = [ 2·•c
ZIR
c111
ZSR
L-1 { ¢(s) 8 U(s)
U(t)
Unit step :. U(s) = 1 / s
ZSR
L-'
X(O)
0
l[(s+~~~:+2) (s+t):s+2)] [~]. /s]) [l
X(t)
ZIR +ZSR =
Y(t)
[-~ -~J o '] [-2 -3
(s + l)(s+ 2)
2.5- Je-1 + 1.5 c[
X(t)
[2.5- 3e3 C-
1
+1.5e-2']
I _
3c - 2 I
-e-11)
Y1(t)
3 (e-•
Y2(t)
- S + 3 (2c·21 -e-1)
Example 4.9 : For a system X
=
(_:~:]for
X(O)
=
2']
3c-' -3c-21
These arc the outputs for unit s~cp input applied.
X(t)
-2t]
J
(s+ l)(s + 2)
••
-c
-2c-1+2c-21
[-~J
Determine the systtm matrix A.
= AX,
X(I)
[
=
t _21 - 2t
-21
]
whm
X(O) • [-~]
and
4 ·22
Modem Control Theory Solution : System is homogeneous so ZSR
[Al A2] A3 A4
A
Now
and at t
I
X(t)
=
.
So
X(O).
Similarly
X(t)
So
X(O)
. .
-2• ]
, hence X(t)
L-e2e-2•
= [-2e-2') 4e - 21
[-42]
X(t)
0
=0
eAt X(O}
X(t) let
Solution of State Equations
=
Substituting in X(I) •
[-42] when X(O)
[_e:_1
1],
[:~]when
= [. ~]
• = [ e-e-•1
X(t)
I
X(O)
at t
=0
X(t) = [: ~]
= [:~]
A X(t)
and
A2J[l]-1 [-l]=[A1 1 A 3 A4
A1-2A2
-2
..• (1)
A 3 -2A4
+4
... (2)
-1
... (3)
-A2 A3 -A4
+A
I
... (4)
Solving these 4 equations simultaneously, A2
= 1,
A I = 0,
A4
=-
3,
A3
=-2
A "' [-02 _\] 4.11 Controllability and Observability In a control system analysis, it is necessary to find an optimal control solution for a given control problem. The existence of such a solution depends on the answers of two basic questions which are, 1. For a given system, is it possible to transfer any il).itial state to any other desired state in a finite time under the effect of suitable control input force ?
Modem Control Theory
Solution of State Equations
4-23
2. If the output is measured for finite time then with the knowledge of the input, is it possible to determine initial state of the system ? The answer to the first question gives the concept of the controllability while the answer to the second question gives the concept of observability of the system.
4.12 Controllability The answer to the first question means the concept of controllability of a system which is related to the transfer of any initial state of the system to any other desired state, in a finite length of time by application of proper inputs. Hence controllability can be defined as, A system is said to be completely state controllable if it L~ possible to transfer the system state from any initial state X(t0) to any other desired state X(t 1) in a specified finite time interval (t 1) by a control vector U(t). The concept of controllability and observability were originally introduced by Kalman hence Kalman's tests are used to find out whether the system is controllable and observable or not.
4.12.1 Kalman's Test for Controllablllty Consider n '" order multiple input Linear time invariant system represented by its state equation as, X = AX(t) + BU(t)
... (1)
where A has order n x n matrix and
U(t) is rn x 1
vector i.e, there arc m inputs.
X(t) is n x 1 state vector. The necessary and sufficient condition for the system to be completely state controllable is that the rank of the composite matrix Qc is 'n', The composite matrix Qc is given by, ... (2)
ln this composite matrix Qc, B, AB, A 2B ... arc the various columns. Proof : Let us see the proof for this condition. Assume that the final state is the origin of the state space while the initial time is zero i.e. t0
= 0.
As fmal state t = tr, is the origin of the state space then X(t1) = 0. . .. (3)
Solution of State Equations
4 ·24
Modem Control Theory
The solution of the state equation (1) is given by, X(t)
=
cAt X(O) +
J CA (HJ B U(t) dt
... (4)
0
Substituting t
= t1 for the final state, /t1
X(O)+J
CA(t1-t)
BU('t)dt
... (5)
0
Using (3) in (5),
J cA(t I -ti
I
0
At
e
I X(O) +
B U (t) dt
0
/11xco)
- J eAtl ·e-"'
BU(t)dt
0
'1
X(O)
= - J e-At
B U(t) dt
... (6)
0
This shows that any initial state X(O) 'nust satisfy the equation (6) for complete state controllability. Thus the controllability depends on matrices A and B and is Independent of matrix C. Now mathematically e-A' can be expressed as, ... (7) Using (7) in (6), ... (8)
X(O)
Let
J
ak(t) U(t)dt
0
X(O)
X(O)
- [B : AB : ... : An -
1
BJ [ ~;~
IJn-1
l
... (9)
Modem Control Theory
X(O)
= - 0., [
~;
Pn-1
l
4 -25
Solution of State Equations
... (10)
The vector matrix of Pk is of order n x 1 and if system has to be state controllable then any X(O) must satisfy equation (9) for which rank of 0., must be n which is a n x n matrix. For any rank of 0., less than 'n' ii indicates that some of the states arc not controllable. This proves that the rank of composite controllability of the system.
matrix
Oc
must be 'n' for complete state
The advantage of Kalman's test is that for any form of matrix A whether A is canonical or otherwise, the test can be applied. But its limitation is that it does not give a physical feel of the problem. If the answer to the Kalmans test is uncontrollable then it does not indicate which mode is uncontrollable. But due to mathematical simplicity of its application, Kalman's test is used to test the controllability. Key Point: The matrix has a rank r 111tans tire dt1em1i11a11t of ordtr r x r of the matrix has 1wnuro value and any detem1i111111t having order r + 1 or more than that has zero value. !Thus rank of 0., ••
= n for controllability
Example 4.10 Find the amtrollability of the system,
Solution:
Now
It
2
A
[~ -~1
AB
[_:]
c, =
(B AB]
I~ _; I = -
B = [~]
= [~ _;]
1 which is nonzero
Rank of Qc = 2 = n Hence the given system is completely controllable.
4-26
Modern Control Theory
i•
Cimlualt. tire co11tro/Jal>ilily of the system
Example 4.11:
.
willt,
x
AXT BU
and
A
[~
IiI
-) J
8
a
rq I J LO
n ,. 2
Solution:
Now
Solution of State Equations
QC
[B
All
[~ _:] [~] = [~]
QC
[~ ~]
, , 11 10 0
Hence rank of Qcr Rank of Qc
AB!
= Determinant of 2 x 2
=0
= 1 and "' n
:. The system is not state controllable.
4.12.2 Condition for State Controllability in s-Plane The system is represented by its transfer function in the s-plane. Such a system is completely state controllable only if the denominator and numerator polynomials of the transfer function do not have a common factor, except the constant. Such polynomials are called coprime. In other words, there should not be a pole-zero cancellation in the s-domain transfer function of the system for complete state controllability. If the pole-zero cancellation exists then a.JI the information about the dynamic behaviour of the system is not carried all along the system. In such a case, the system cannot be controlled in the direction of the cancelled mode.
4.12.3 Gilbert's Test for Controllability For the Gilbert's test it is necessary that the matrix A must be in canonical form. Hence the given state model is required to be transformed to the canonical form first, to apply the Gilbert's test.
.
Consider single input linear time invariant system represented by, X(t) = A X(t) + B U(t)
Modem Control Theory
Solutlon of State Equations
4 ·27
where A is not in the canonical form. Then it can be transformed to the canonical form by the transformation, X(t) • MZ(t) M
where
Modal matrix
The transformed state model, as derived earlier, takes the form, z.(t) where
c
A Z
B U(t)
A
Diagonal matrix
B
M -I B
It is assumed that all the eigcn values of A are distinct. In such a case the necessary and sufficient condition for the complete state controllability is that the vector matrix B should not have any zero elements. If it Ms zero elements then the corresponding state variables are not controllable.
If the eigen values are repeated then matrix A cannot be transformed to Jordan canonical form. If A has eigen values A,, A,, 1-z, 1'z, 1'z, ~ "4· ... A,, then the transformation results Jordan canonical form as, O················· O·················
0 0
J=
O··· ·············
0
Jordan block
O·················
0
O······~···· · ·
0
~················O 0
iJ ........••.•........................................
A,,
In such a case, the condition for the complete stale controllability is that the elements of any row of B that corresponds to the last row of each Jordan block are not all zero. ,_.,
Example 4.12:
Consider tlre system wit/i state equation.
~I
.
Xz [ XJ
Estimate tlte stale controllability by i) Ka/matt's test and ii) Gilbert's test
Modem Control Theory
4 ·28
Solution of State Equations
Solution : i) J<.lman's test
0c
(B : AB : A2BJ .•. n = 3
AB
~lj = [~
-6
I~ ~ ~I Hence the rank of
IOcl
= i. Thus
Oc is 3 which
36
0 -11 60
is nonsingular.
is 'n'.
Thus the system is completely state controllable. li) Gilbert's test
For this, it L~ necessary to express A ill the canonical form. Find eigen values of A.
II.I-Al
:. "K .. 61.2 +
=
)I ~.
-1,11 "" ~I
lo
0
}.+6
11). ... 6 = 0
:. (I.+ 1) (/.. + 2) (}, + 3) = 0 '·1 = -1,
~ = - 2,
A..!
=-3
l [ 1 1 1]
As matrix A is in phase variable form, the modal matrix M is Vander Monde Matrix.
1
i..2
i..3
).~
).~
=
-I
1
-2 -3 4
9
Modern Control
Theory
Solution of State Equations
4 ·29
[~
6
-l [j -5 -~i
-5 8 -3
---= =[-~ Adj(M]
M-1
IMI
li
W1B
As none of the clements of
i3 are zero,
8
-3 -2
2 -1
-1
-2
-1
= [-~
2.5 ~1 1.5
0.5] -1 0.5
2.5 0.5][°] = [°.5] -4 1.5
-1 0.5
0 l
-1 0.5
the system is completely stale controllable.
Key Point: As Gilbert's lest requires to transfonn matrix A into canonicalform, it is time
ccmsuming and hence Kal111a11's test is popularly used to ttSt controllability. 4.12.4 Output Controllablllty In the design of some practical systems, it is necessary to control the output variables rather than the stale variables. In such a case complete output controllability is defined. Consider the stale model of a linear lime invariant system as, X(I)
A X(t) + B U(l)
Y(I)
C X(t) + D U(t)
The system is said to be completely output controllable if it is possible to construct an unconstrained input vector U(t) which will transfer any given initial output Y(lo) to any final output Y(t1) in a finite time interval to St S t1. In such a case, construct the test matrix Q., as,
Q.,
=
[CB: CAB: CA2B: ... CAn-l
B: DJ
Thus if the order of matrix C is p x n i.e. there are p outputs then for the complete output controllability, the rank of test matrix Q< must be p.
4.13 Observability The observability is related to the problem of dcterrnlning the system ·state by measuring the output for finite length of time. Hence observability can be defined as, A system is said to be completely observable, if every stoic X(t0) can be completely identified by measurements of the outputs Y(t) over a finite time interval. If the system is not completely observable means that few of its state variables are not practically measurable and arc shielded from the observation. Similar to the controllability, the observabillty of the system can be obtained by using Kalman's test.
Solutlon of State Equations
4. 30
Modem Control Theory
4.13.1 Kalman's Test for Observability Consider nth order multiple input multiple output linear represented by its state equation as,
x and
Y(t)
where
Y(t)
and
time invariant system,
A X(t) + B U(t)
... (1)
CX(t)
... (2)
px l output vector
c
t x n matrix
The system is completely observable if and only if the rank of the composite matrix Q0 is 'n'.
TI1e composite matrix Q0 is given by,
c,
= (Cr,ATCT:
(A ,.)"-•cTJ
where
Transpose of matrix C
and
Transpose of matrix A
I
Thus if, rank of Q0 = n, then system is completely observable. Proof :
For the system described by the equation (1) and (2), the solution is, t
X(t) = cA• X(O) +
Y(t) =
J cA (t-t)
c CAI X(O) + c
I
B U(r) dr
CA (t-T)
B U(t) dr
... (3)
... (4)
0
Now as the matrices A, B and C are known and the input vector U(t) is known, the integral term in the equation (4) is known. This can be subtracted from observed value of Y(t), to modify the lefthand side. Hence the obserability can be investigated by considering the equation, Y(t) = C eAt X(O)
... (5)
as the left hand side is known due to measured output. Thus for observability, the unforced system can be considered i.e, homogeneous state equation can be considered. Consider the state model of unforced system as, X = A X(t)
and
Y(t)
= C X(t)
The output is given by, Y(t) = C eAt X(O)
... (6)
4. 31
Modem Control Theory
Solution of State Equations ... (7)
But •-1
Y(t)
I
Ccxk\t) A kX(O)
k=O
Y(t)
CJo(I) C X(O) + cx1(t) CA X(O) + ...... <Xn _ 1(t) C An - 1 X(O) ... (8)
For the system to be completely observable, from the given output Y(t) over the
interval 0 ~ t ~ t1, the X(O) must be uniquely determined. TIUs is possible if the rank of the matrix,
[J.] must be n. Taking transpose of the matrix, the condition can be stared as the rank of matrix Q., must be n for complete observability where, Qo;
>'*
[cT:ATCT:
... :(ATt-lCTJ
Example 4.13 : £va/11ate tire observability of tlrr sys/em
and
Y(I)
= (1 OJ [;:]
Solution : The order of the system is, A AT
[-~ -~J
and
c .. (1 OJ
[~ -21
and
CT .. [~]
I
Qo
n ~2
rc·r
-1
ATCT)
4 ·32
Modem Control Theory
Solution of State Equations
Consider the determinant
I 10 11I
1 =nonzero 2=n
:. Rank of
Hence the system is completely observable. ,...
Example 4.14 : £wlllate tire obstrVability of the system with
A
[~ -~
_r] •
B
= [~]
and C
= (3
4 IJ
Solution : The order of the system is n = 3 Qo
1cT
AT cT
AT
[! -~]
(A T)2CT)
0 0
-3
CT
[:] 0
ATCT
[! =~OJ [3j~ [~] 0
J
(A T)2CT
AT[ATCTJ=[!
0 Qo
[~
-~] -2
0 0
=~][~] = [=~]
Modem
4. 33
Control Theory
Solution of State Equations
Consider the determinant,
01
3 0 4 1 -2
11 I
-21
= - 6 + 0 +0 +0- 0+ 6
=0
Hence a nonzero determinant existing in Q., is having order less than 3. :. Rank of Q0
"'
3 "' n
Hence the system is not completely observable.
4.13.2 Condition for Complete Observability in s-Plane The condition for complete observability in s-plane also can be stated from the transfer functions or transfer matrices. The conditions remain same as before that the numerator and denominator polynomials of transfer function must be coprime. There should not be cancellation of pole and zero in the transfer function. If the cancellation occurs then the system is not observable and the cancelled mode cannot be observed in the output.
4.13.3 Gilbert's Test for Observability 11 is known that for Gilbert's test, the state model must be expressed in the canonical form. Consider the state model of linear time invariant system as,
and
X(t)
A X(ti + 13 U(t)
Y(t)
C X(t)
Use the transformation X(t) Y(t)
=
MZ(t) where M is the modal matrix.
CM Z{t) =
C Z(t)
CM
where
For a single input single output system,
Y(t)
[
~~
~:~]
Zn(t)
'11 Zi(t) + '12 Zz(t) + ··· + Cin Z,,(t) Due to the canonical form, all the states are decoupled and not linked to each other. Hence for the system to be observable, each term corresponding to each state must be observed in the output. Hence none of the coefficient of C must be zero.
Modem
4.34
Control Theory
Solution of State Equations
Thus the system is completely observable if all the coefficients of C arc nonzero, none of the coefficient is zero. If any clement is zero, the corresponding stale remains unobserved i.e. shielded from observation. n•
Example 4.15 : Evaluate lire obseroabilily of lhe system wil/1
A=[~
~1'
B = [~] a11d C = (3 4 1) l
~ 0 -2 -3
Using Gilbert's lest. Solution : For Gilbert's lest, find the cigcn values.
1~0 -~ -~ l=O
[),I-A)=
2 l.3+3'>..2+2i.
O
i.(i.2 + 3i.+ 2)
O
i.(),+1}(1.+2) :.)'! = 0,
i.+3
0
~ = - 1,
"3 = -
2
l
As the eigen values are distinct, the modal matrix M is a Vandcr Mondc matrix. 1 1 M = 1.1 1.2 [ ).~ ;.~
1 1.3 ;.~
[1
11
= 0 -II -2 0
I
4J
When the modal is transformed to canonical form,
C
=
CM = (3 4 1) [~ -~ -~] 0 I 4
= [3 0 - 1) As there is one zero element in
C,
the system is not completely ·observable.
4 .35
Modem Control Theory
Solutionof State Equations
Examples with Solutions
1•
Example 4.16 :
Find the statt transition Mitrix cf the state tquatio11.
Using lht inoerse transform method.
(Bang•lo"' Univy Aug.·96)
Solution:
Now
s [~ ~]-[~ ~] [s~ 1 s ~ 1]
sl-A
=
Adj (sl-A)
1s1-Ar1
Adjoint
[cofactor matrix) T
[sl - AJ Isl - A!
I
( sl-A
=
= [s-1 _1
0 ] s-1
(s - 1) (s - 1)
o] [
s-1 [ -I s- I (s-l)(s-1)
=
0]
s- 1 I (s - l)(s-1)
1 s-1
s-1
= [_1I (s-1)2
ol1 ;:I
Taking inverse Laplace transform. 4>(t) I~
Example 4.17 :
O]
e1 cl = [tel
.. . Required state transition matrix
For a certain system, wl!m
X(O)
while X(O) =
[_13] then X(I) = [
[:]Ihm
X(t>
~-:~Jt]
= [::]
Determine the system matrix A.
(llangolore Univ. Aug.·95, April-'J'J)
Solutionof State Equations
4·36
Modern Control Theory
Solution : The solution of the equation is, X(t}
Let
eA• X(O}
A
and the equation is,
.
X(t)
A X(t)
Now
X{t)
[
and
X{O)
[-13]
Now
X(O}
A X(O}
[-:] Similarly
• [ 3e-J1] hence X(t) = ~ e-Jt
-JI ] _; e-Jt
[Al
A3
A1-3A2
-3
A3-3A4
9
.
X(O}
[-3)9
[-~]
A2]
A4
... (1) ... (2)
.
X(t)
[::]
X(I) =
X(O)
[:]
X(O)
.
=
X(O)
[Al AJ
~:] X{O)
GJ
[A' AJ
~:] [~]
[::]
[~]
A1+A2
... (3)
A3 +A,
... (4)
Subtracting (3) from (1), -4 A2
-4
A2
1
Al
0
Modern Control Theory
4 -37
Solution
of State Equations
Subtracting (4) from (2),
8
n•
i.e. A4
= -
2
Example 4.18 : Obtai11 Ilic so/11tio11 of Ille liomoge11eo11s state equation X =AX ionere
A = rl1
-2]
and
X(O) =
fl0.51
1 -4
I
J
(Bangalore Univ. Aug.-97)
Solution: The state transition matrix is,
L-1 {cst-Al-1} [sl-A)
s[~ ~J-G =:] =[s~i' s::] Adj(sl-1\) tsl-AI s+4 = [ -2
Adj [sl-A) Isl-Al
(s-l)(s+4)+2
=
s2 + 3s- 2
[
L-1 [
s+ 4 ] (s-0.561) (s+ 3.561)
-2 ] (s- 0.561) (s+ '3.561)
]T
=[s+4 1
-2]
s- I
=s2-s+4s-H2 (s - 0.561) (s + 3.561)
(s-0.56;;(:+ 3.561) I (s-0.561) (s+ 3.561)
L-1 [
l.106 (s-0.561)
1.106 e + 0 561 L-1[
1
s- 1
(s-0.561;(!+
(s-0.561) (s+ 3.561)
0.106 ] (s+3.561) I -
1-1 [ -0.485 • (s-0.561)
0.106 e " J.56! I +
3.561)]
(s- I)
~1 (s+ 3.561)
-0.485 C~ 0.S61t+0.4850-
3.5611
Modem Control Theory i,-1 [
Solution of State Equations
4 -38 i,-1 [
1 ] (s-0.561) (s+ 3.561)
0.242 rs-0.561)
0.242 e " 0.56l
0
'"
:.
=
[1.106c0.S611-0.106c-J.5611 0.242 c115r.i' -0.242 e- 3·561'
-
0.2<12
t .,.
0.242 e" J.St.l t
-0. .i85 C0.~611 1 0.485 e" }.5611] -0.106co.>ol' ... 0.106c-3.561 •
0.068 CO.St.11 + 0.432 C. 3.561 [O.OlS e11•5611 ~ 0.9R5 e" J.S6l ,,_.
]
(s+ 3.561)
.
t] t
Example 4.19 : For a system represented l7y X = AX, the response is
X(I)
= [2e-1 e-4'1]
whm X(O)
• = [4e-c-2t21] = [2]1 and X(t)
whm X(O)
= [4] 1
Determine the system matrix A and the state transition matrix. (VTU:
Solution : Let the solution is, X(t)
cAt X(O)
A2] A4
Let
A = [At A3
Now
X(t)
[ 2c-4t ] e-.at
• X(t)
At t = 0,
X(O)
[~]
X(O)
And
X(t)
[4e-2 1] e-21
At t = 0
X(O)
[~]
= [ - 8 c ·41] . - 4 e -41
. = [-8) -4
• = [-se-21] -2c-21
X(t)
. [-8]
X(O) = -2
JanJFeb.-2005)
4 .39
Modem Control Theory X(t)
X(O)
= [~~ ~:] [~:
Solution of State Equations
X(t)
~:] X(O)
... (!) ••• (2)
and
-8
... (3)
-2
•.. (4)
Solving (1), (3) and (2), (4) simultaneously we get, A
=
[o -s] 1
-6
[sl-A)
Adj [sl-A) Isl-Al
[l l
s2 +6s+8
I
L-
= (s + 2) (s + 4)
-8]
s+6 1 s (s+2) (s+4)
s+6 L-t (s+2)1(s+4) [
(s+2)(s+4)
-8
l
(s+2)5(s+4) (s+2) (s+4)
Modem Control Theory
4 ·40
L-1[ ~2s~4 0.5 0.5
----s+2 (s-;-4)
1ui+
--+--4 4
s+ 2 s+4 1 2 (s+2) + (s+4)
l
Solutlon of State Equations
A linear timt invaritmt system is clwraderiud by the homoge11eous state
Example 4.20 :
equation:
Compute the solution of homogeneous equation, assume tire initial slate vector :
Solution : From the given model,
A [sl-A)
Adj [sI -A) Isl
-Al
=
[~ ~]
[s- I
[ 1 O] [l OJ
0]
s 0 I - l l " - I s- I
[s~
I ~~
r" [s~ s~ i] I
(s- l)l
s- I 0 ] Adjfsl-AJ - [ I s-1 jsl- Al (s- 1)1 I
•-I [
(s-\)2
L-11s1-Ar1
,; .] [
~
= L-1 ~~
(s- I)"
0]
_I_
s-
I
Modem Control Theory
4 -41
o]
c' [te' X(t)
cAt
SoluUon of State Equations
c'
X(O) ; zero input response
This is the required solution.
1..+
Example 4.21 : Obtain tire time response of the following system :
; [~ ~][~:] + [:] U (I)
[ ~:]
where U (I) is the unit step occurring at t = 0 and X T (0) = [1 OJ. Solution : From the given model, A
= [: ~]
B = [:)
X (0) = [~]
Now as matrix A is same as considered in Ex. 4.20 while X (0) is also same as consider in Ex. 4.20, the zero input n.'Sponsc is same as obtained in Ex. 4.20. · ZIR
=
X(t) = c' [t c'
J
Thi'.' zero state response which depends on U(t) and matrix B is given by, ZSR where
X(t) = L-1 { ets) B U(s)}
1 s-1
[ (s-11)2 B
(l(s) B U(s)
_l_]
[:] and U(s) =
1 si'I [(s-1)2
as obtained in Ex. ~.20.
(s-1)
0
as unit step
l G]m
_I (s-1)
m
l
4.42
Modem Control Theory
l
r_1 (s-1)I (•-1l' s-1 s 1 1 _1_ [
l
1_
(.!.] = r
(•-I)
5
ll
Solution of State Equations
l; 1
1..-1)'
s(s-1) 1
I
-(,-!)
. . . ... usmg partial fraction
(s-1)2
Total response = X(t) = ZTR+ ZSR
[e [e' -1] = [2e -1] 1
= ,,.
tc1
]
1
+ t c'
2tc1
Example 4.22 : Fitrd and sketch the response of the system witlr the followi11g transfer function, input and initial conditions. T(s) = -4 s+ 20 s+300 r(t) !1(0)
Solution :
r(t)
= input
10 u(t) ... (input) 0 and
y(t) = output
= Y(s) = -4s+20 T(s) R(s) s+JOO Dividing numerator by denominator as are of same order, Y(s}
_4 + 1220
R(s)
s+300
4 .43
Modem Control Theory
Solution of State Equations
The corresponding stale diagram is,
«o ,,..._,---y(t)
'---------i
4
1---------'
Fig. 4.1
and A=(-300],
X1
r(t)-300X1
y(t)
1220 X1 -4 r(t)
B=[I). (sl-A)
[s1-Ar1
C=[l220J.
D=(-4)
[s+ 300] [s+
~oo]
(e-3()01] Now
hence substituting in y(t)
y(Ol
0
y(O)
1220X1(0)-4 r{O),
0
r(O) =10 as given to be 10 u(t)
1220 X1(0)-40 40 1220
ZlR
CAT X"l(O)=~
.,-3001
1220
To find ZSR use, ZSR
L-1 [9(s) B U(s)) L-1 [-1-·l·!Q] as U(s)=R(s)=!Q (s+300) s s
L-1 [ ZSR
10 ]-t-1 s (s+ 300)
_..!___]_ 30 30
c·:lOOt
[1/3() s
1/30]
s+ 300
...... 4 .44 Solutlon of State Equations ~~~~~~~~~·~~~~~--~~~~~~~~~
Modem Control Theory X(t) X(t)
...!..-5.4x 10-4 c-JOO~
y(t)
1220 X(t)-! r(l)
30
1220 [ ~ -5.4x 10-1 c·· 3001
J -4
(10)
0.667 -0.667 e- JOO I The response y(t) is, y(t)
0.667 --------:;-:;·..----
Fig. 4.2 ,_..
Example 4.23 : A linear dynamic time int!Qriant system is rtprese11ted by
.
umere
X
A X(I) + B U(I)
A
[~ ~ ~1 B=[~ ~1 0 -2 -3
Find
if the
1 0
system is completely controllable.
Solutlon : For the system, n = 3
o,
[B: AB: A2BJ
AB
[~ ~ 0 -2 -3
~1 [~ ~i ~1 1 0
=[ ~ -3 0
Modem Control Theory
4 -45
1 A2B
= A [AB]=[~
OJ[ o1 o10 J = [-31 OJ0
0 I -2 -3
0 0 1 1 0 -3 0 -3 0 7 1
= [~
QC
Solution of State Equations
0
-3 0
~]
Consider the determinant,
[ ~ ~ ~1 I
= 1 = nonzero
0 -3
Hence rank of Qc = 3
=n
Thus the given system is completely controllable
Consider the syste~n represented by,
11111~ Example 4.24 :
x
[- ~~
Y(I)
~:~] X(t) +
G]
U(t)
(1 OJ X(I)
Find the complete obseroability of the system. Solution : For the system, n
=2
A [-0.0.12
0.4] -0.1
AT
[-0.2 0.1] 0.4 -0.1
Qo
(CT
ATCT
Qo
C = (1
CT
= [~]
A TCTJ
[-0.2 0.1][1]=[-0.2] 0.4 -0.l 0 0.4
[10
-0.2] 0.4
O]
7 0
Modem Control Theory
Solution of State Equations
Consider the determinant, 1 10
-
0·21
= Nonzero
0.4
0.4
Rank of Q.,
2=n
Hence the system is completely observable.
11*
Example 4.25 : Usi11g similJlrity tra11Sfonnationfind A=
I'' for,
[-4 3] -6 5
Solution : Find the eigen values.
IJ.:4 ;..3sl=o
p.I-AI
J. -A
-2
O
(A+l)(A-2}
O
2
,.,
-1,
~
=2
Find the eigcn vectors. For}.,
=-
1,
(}.,! - A)
=
[! ~]
[~:~]=[~]=[:~] a[: =~] :!] g ~] [ 2 -I] = .2.....2_
M1 = For ~ ~ - 2, (>'!! - A)
M2
M
M-1
[~:~]
2
[
= [~]
Adj (MJ
IMI
1
=[ 2 -1
-:1
Modem Control Theory
Solution of St.ate Equations
2c-1 -e21 = [ 2c-1 -2c21 ,..
-e-1 +e21 ] -c-1 +2c21
1'1 for
Example 4.26 : Using lAplace transform metltod, find
-3]
Solution : a) A = [oI _4 [sl -Al
=
Adj [sl-AI - [s~4
1s1-Ar1
Isl-Al
e ts) [
~1]"
- s(s+4)+3
s+4 (s+l\(s+3)
~ (s+l):s+3)
(s+1)(s+3)
(s+l)(s+3)
9 (t) = cAt =
l
[s~4
s+4 (s+ l) (s+ 3) (s+ I)-3 L- 1
:s+
(s+ I) (s+ 3)
1.5 0.5 L_ 1 s+I
s+3
0.5 0.5 [--s+I
s+3
-},5 s+l
3)
(s+l)(s+3)
T-1.:.?..l s+3
J
-0.5 1.5 --+-s+l
-:]
(s+l)(s+3)
s+3
I
4 ·48
Modern Control Theory
b) A=[-~
-n [sl -A)
[sI-Ar
s -1 ] [ 3 s+4 Adj[sl-A]
I
[
s+4 '] [ -3 s
l
s+4 (s+1~+3)
I (s+l):s+3)
(s+l)(s+3)
(s+l) (s+3) (s+I) I(s+3)
s+4 t_ 1 (s+l)(s+3) [
[s+4 -3
l
-3 s (s + l)(s+ 3) (s+ l)(s+ 3)
1.5 0.5 --s+l s+3
0.5 o.s ---
-1.5 1.5 [-+-
-0.5 1.5 --+-
s+l
s+3
s+l
s+l
s+3
a)A=[~ -~] b)A=[-~ ~] Solution : a) A = [-~ -~]
Step 1 :
hence
p(I.)
= c'"
Find cigcn values
p.. I-AI
s
I~
A~51=0
0
i.e,
(A + 2)
l
s+3
Example 4.27 : Using Caylty Hamilton method,find
f(A) = eA•
I]
= s2+4s+3 = (s+l)(s+3)
Isl-Al
6(s)
l ..
Solution of State Equations
p.. + 3)
~ O
it' for,
4 .49
Modem Control Theory
Step 2:
n=2 O'o + a1 A. = p(/.}
R(/.} i.e.
Step 3 :
Solution of State Equations
O'o +
a1 ).
e)J
Substitute values of A. ..• (1)
and
... (2)
Step 4 : Solving equations (1) and (2), <Xi = e- 21_e-31, O'o = 3e- 21 _ 2e-3•
Step 5:
f(A}
= R(A) = OQl
+ <X1 A
(3e- 21 _ 2c- 3t) [~ OJ+ (e-2•-e-31)[
eAI
l
[ 3e-2t -2e-Jt -6c-21 +6e-J1
CAI
bl
[-~ ~]
A
CAI
f(A)
Step 1 :
Step 2 :
hence p().)
Find eigen values.
IU-AI
I~
A.2+41.+4
0
(A.+ 2)2
·= =
"'n=2
-21
A.+4 = 0
0 -2,
'2 = - 2
R(I~ ~ O'o + a1 A.= p(~ i.e,
O'o + a,:i..
e -2t -e-31 ] -2c-21 + 3c-Jt
c>.:
= c)J
0
-6 -~J
Modem Control
Step 3 :
Theory
Solution
of State Equations
Substitute values of I.
Clo -
=
2(X1
c-21
.•. (1)
But H I. is repeated, second equation must be obtained as, dR(I.)
dp().) di.
di. (XI
t e'IJ at
o:,
t e-21
), = -
... (2)
2
Step 4:
... from (I)
Step 5:
f(A)
= R(A) = Cfol + <X1A
eAt
=
(e- 2t
+ 2t e-21) [~
c-21(1+21)
+ te-
2te-21
[ -2tc-2t ••
~]
21 [-~
~]
]
e-21(1-21)
Example 4.28 : Find the respans« ef the system,
X = [~2 and Y(I)
=G
]x •[~ ~]
~3
U(I), X(O)
= [~]
~] X lo tht following input,
U(I) = [~:~:~] = Solution : Find cAt for A
1s1-Ar1
$(s)
[e~;~::w] whtre = [~2
11(1)
=unit step function.
~3] thus [sl - A)
= [;
s+ 3 l] Adj(sl-A] _ [ -2 s _
Isl-Al
csr
-
-Ar 1 = [
~\] 5
[s+-23
I] s'
52 +3s+2 - (s+1)(s+2)
s+3(s+2) (s+I)
(s+l) 1(s+2)
-2 (s+l)(s+2)
s (s+l)(s+2)
l
4. 51
Modem Control Theory
2. s+l
1 s+2
s+I 1
-2 [ -+s+l
2. s+2
-1 2 -+s+l s+2
-1
Now
ZSR
U(•)
ZSR
s+2 1
l
Solution of State Equations
= [ --e2c-•-1 +-c-2 e21-21
L
lo(s)I
L-1
{<)(s) B U(s))
?
UJ L_1
![ .. ,
(s+1~:+3) (s+t) (s+2)
L-1
l [ "'
L-'l
(s+1)(s+2)
cs+3
(s+l) (s+2)
'"' :"''] r· : ,
s+3 (s+l)(s+3)
(s+l)s(s+2)
s(s1+3)
-2 (s+l)(s+2)
(s+l)(s+2)
s+3
(s+t~:+J)
l
'"'':'""] [: :] f : ] I (s+l) (s+2)
s+3
' ] ""'']l
l
4. 52
Modern Control Theory
Solution of State Equations
X(t)
(t)]
Y1 [Y2{l)
and ,,_.
[~ ~] [~~ ~: ]
Y1(t)
X1 (t) = 3 - 2.5 e-
Y2{t)
X1(t) + X2(t)
1
-
e-
21
= 1 + e-lt -
+ 0.5 e-
31
2 e-Jt
Example 4.29 : The Fig. 4.3 shows the block diagram of a process control systmi with state variablefeedback and feed forward control. Tire state model of process is,
. [-3 2] [l]
X=
and Y = (0
4 -5
X+
0
U
11 X
a) Derive state model for the eutire system. b) Find the response if input
r(t) is
unit step ass11mi11g
X(O) = (~}
r(I)
Fig. 4.3 Solution : a) From the block diagram, U(t) = + 7 r(t) - 3X1 - l.SX2
... (1)
From the given state model of the system, - 3X1 + 2X2 + U(t) and
... (2) ... (3)
4. 53
Modem Control Theory
Using (1) in (2),
.
Xi
Solution of State Equations
- 3 Xi + 2X2 + 7 r(t) - 3Xi - 1.5 X2 - 6 X1 + 0.5 X2 + 7 r(t)
... (4)
The equation (3) remain unchanged. Hence state model of the entire system is,
. [-6 0 5] [7]
X =
-~
4
X +
0
r(tl
The output equation remains unchanged. Y(t) = (0 1 I X(t)
Thiu, A=[-:
~:l 8 =fa],
bl
o(s)
Find
[sl
-Ar
[sI-Ar1 s+6 [ -4
(sl-A)
1
C = (0 11
-o·:] s+;,
Adj Isl-Al lsl-1\I s+S [ 4
s+5 [ 4
4
0.SJ' s+6
= (s+6) (s+S)-2 = (s2 +lls+28)
0.5] s+6
---(S+4)(s+7)
s+S 0.5 (s+4)4(s+7) (s+4) (s+7) c> (s} [
[s+S
0.5] s+6
(s+4) (s+7)
s+b (s+4)(s+7)
l J
No need to calculate eA• as X(O} = [~] the ZIR is null matrix i.e. zero. ZSR = L-1 ($ (s) B U(s)I but here input is r(t) and not U(t) to the entire system. ZSR = L-
1(9
(s) B R(s)I and R(s) =
!s
Modem Control Theory
ZSR
Solution of State Equations
[125-05S33c--11 -0.667 c-71] = + 1.333c-71
1-2.33c-4t
= 1-2.33 e-41+1.333
1•
[X (tl] 1
X2(t)
e-7t
This is the required response. Example 4.30 : Etialualt rontrol/JJbility and obseruabi/ity of tht. foll
-~l
a)A=r-~ b) A
= [-~ -~
c) A
= [~ ~
B=[~],
c -.u
-~l · [i ~l B=
~l · [~l · B
=
C
-1)
C = [~
= (0
0 1)
Solution: a) For controllability, ~ = [B : AB] as n
10
1
-21I
=2
1 hencerankof~=2=n
~]
4 .55
Modem Control Theory
Solutlon of State Equations
Hence system is completely controllable. For observability, Q.,
= (CT : ATCT) as n = 2
CT
(1- l)T
= [-~]
=1
:. Hence the rank of Q.,
Thus system is not completely observable. b) For controllability, Q.,
AB
=
A2B
Ii
-11
as n
= 3.
~i ~i ~i
r-~ -~ r: = r=~ 0
0 -3
A (AB]=
r-1~
2 I
0
0
-3
0 -1 0 -I 2 -2 --4 4 2 1 -6 -3 18
= - 7 - 0 hence
-6 -3
0 -2
[11
Q., 0 2 -2
= [B : AB : A2BJ
Ort-2 -4Ol = r-14 Ol8 J -6 -3 18 9
:]
rank of Q., = 3
=n
-6
Thus the system Is completely contrcllable. For observability, Q0 =(CT: AT CT =
AT CT
n
ATCT: A T2 CT) as n = 3 0 -2
0
0] [I1 3]1 = r-1-2 -3-2 ]
0 -3
AT (AT CT]=
2 5
r-1~
-6 -15
0 -2
0
ol r-1
0 -3
i [1
l
-3 = 4 4 3 -1 -2 -6 -15 18 45
4 -56
Modem Control Theory
1 3 -1 -3 1 1 -2 -2 [2 5 --6 -15
+7
'F
l
Solution of State Equations
3 4
4
18 .15
0 hence rank of Q0 = 3
=n
Thus the system is completely observable. 1For cl controllability, Oc = [B : AB : A]B] as n = 3
AB=[~ ~ =!][~}]=[;~] A2B A[ABJ=[~ ~ =~][~~]=[-~] Oc
=
40 0 [ 100 4010
OJ
-30
= - 12000 ~ 0
hence rank of
0
Thus the system is completely controllable. For observability, 0., =[CT: ATCT: A r2 CT] as n = 3
ATCT AT2cT
= [~ -~
_!][~]=[_:]
AT[ATCTJ=[~o
~
~][~]=[-:]
-3 -4
0.,
l
o0 1
o
[~
~ _:]
1
-4 13
11
1 -4 = 1 ~ 0 hence rank of Ou -4 13
Thus the system is completely observable.
-4
=3=n
13
Oc = 3 = n
4 .57
Mod•m Control Theory ,,...
Solution of State Equations
Example 4.31 : Consider the stale modtl,
X
= [~ ~] X{t),
Y(t) = (1
21
X(t)
Find the se! of initial conditions such that lite mode l' is suppressed in Y(t). Solution : f'md e At £or A =
lro
hen cc (s I - A] = [ _2 s s-l -I ]
l] 2 1
[s-1 l] [sl - Ar I = Adj [sl -Al = ~ Isl-A( s2 -s-2
<(s)
s-1 (s+ 1) (s-2)
(s+l) (s-2)
2 (s+l) (s-2)
(s+l)(s-2)
[21' 1/3 -1/3 1/3] (s+1) + (s-2)
--+-(s+l) (s-2)
-2/ 3 2/3 --+-(s+l) (s-2)
1 I 3 2/ 3 (s+l) + (s-2)
L- 1 [o (s)I
CAI
['-' 1 " 1 ., 1 -c +-c 3
CAI
3
-2 c-1 +!c21 3 3
-3c
+3c
The equation is homogcnoous hence the solution is, X(t)
2
Let X(O) = [ ~]
cAt X(O),
~t?-t
3
-2 -t [-c 3
+!e2• 3 2 2t +-c 3
"l
J -t 2 2t -c +-c 3 3
4.58
Modern Control Theory
.!.e-1(2a-b)+.!.c21(a+b)
[
And
Y(t)
~:-t
(1
l
Solution of Sblte Equations
(-2a +b)+~c21 (2a +2b)
2J [~~]
= X1
+ 2X2
.!. c- 1 (- 2a + b) + .!. c21 (Sa + Sb)
3
3
To suppress the mode c21 from Y(t), Sa+Sb i.e.
a
0
-b
Hence the initial state must be, X(O) 111.+
= [~:]
where m " 0
Example 4.32 : Write the state ~uatio11S of tire system sllow11 a11d determine its stale t:Q11trolillbility a11d obsutlQbi/ity. (VTU: }anJFeb.·2006)
Y(s)
Fig. 4.4 Solution : Consider the block ~. Its state diagram is, s(s+l)
X1
Fig. 4.5
4.59
Modem Control Theory From the state diagram.
x,
-
2
Solution of State EquaUons
.
= m--
x,
.•• (1)
2
Now s = .!!. which is a differentiator and input to it is X1. dt
Fig. 4.6 Thus, Xt
X3
z
... (2)
Now the block -2- has state diagram, s+3
. . "2
Fig. 4.7
X2 2
Now m = X2
-
U(t) - X1 = - 2X1
-
3X2
- -2-
3X2 + 2U(t)
... (3)
X3 hence using in (1), Xt 2
.=
but X1
Xt
2X2 - 2X3 -
X3 hence X t
. X:J
. =
=
.
x.
...(4)
X3 and using In (4),
2Xi - 2X3 -
X3 = 2X2 - 3X3
...(5)
Modem Control Theory
4·60
The equations (2), (3), (5) give us required state model. X{t) =
and
Y{t)
For conroUability,
=
0 0 1] [Ol [-20 -32 -3o X(t) + 2J0 U(t) [I 0 OJ X{t)
Oc = [B: AB: A BJ as n = 3] 2
AB=[-~-~ -~l[~]=[-:] A2B Q,,
o2 -6o [0 4
l
A{AB)=[-~ -~ -~l[-:]=[~:J [~ -~ 0
4
l
I~ -24
4 18 = + 32 " 0 hence rank of Q,, = 3 = n -24
Thus system is completely controtlable :
For observability, Q,, = (CT : AT CT : A T CT) as n = 3 ro -2 2
l A
I 0 :. 0 0 [ 0 I
~1
o
-3
I
0
T2 CT AT (AT CT] = [~ o, = [~ ~ -~] =-
2 7- 0 hence rank of Q0 = 3 = n
-3
Thus the system is completely observable.
Solution of State Equations
Modem Control Theory
u11+
4 ·61
Solution of State Equations
Example 4.33: Use co11trol/ability and obserobility matrices to determine whether t* system represmted lty the flow graph shown in Fig. 1 is completely co11trol/ablt and complttely observable. (VTU: JanJFeb.-2.005) -1
y
u
1
Fig. 4.8 Solution : The value of the variable at the node is an algebraic sum of all the signals entering at that node. X3
x1+x2+2u,
X1
-3X1 +X2,
.
A
For controllability,
AB
Xz= X1 -2X2 +X3 +2u
= X1 +2X2-X3 [;3 -:2 ~], B = m, C = r1 2 -JI o, = [B: All :A2B] 1 0~Jm=HJ Y
... n
[:3 -2 -5 [:3 -2 ~] [:3 1 0]0 [10-2 -1 l
A2
A2B
1 -2
[10-4 -2
-5 6 -1 2
Oc
-:2][~] = [-i]
[~ 2 -n -2
1
= -4
6
-:]
=3
Modem Control Theory
Solution of State Equations
4-62
= 3 "' n
I Qcl = -32 .,, 0 hence rank of~ The system is completely controllable.
Q,,=
For observability.
[cT
2c·i]
:ATCT:AT
r; 3 AT[ATcr]..l
u
QQ
IQ !
= 0 and
0
1
... n
=3
1 -2
l0
~2: _:] I~!- ~
= 0 thus 2 x 2 determinant is zero
Thus rank of Q0= I hence system Is not observable. n..
Example 4.34 :
Given the time invariant system :
a11d that 11(1) = e-1 and y(t) :: 2 - ate-1 , fi11d Xi(t) and Xi(t). Find also Xi(O) and Xz(O). Wlrat happens A
Solution :
[sl-A]
= ro
a}
LO -1
[~
if a
B = [0} C 1
=0 ?
= (I
-a] s+l , Adj [sl-A]
Adi (sl-A] Isl-A
I
(VTIJ : Ja.nJFeb.-2005)
0)
= [s+l 0
s(s~l)l s+l
Solution of State Equations
4 ·63
Modem Control Theory
L"1[s1-Ar1
ZLR
c'\t X(OI = [:
ZSR
el
a(t -e_e1-1 )]
= [Io
{¢(s}BU(s)}=L-'
1[: l
s(s~L][~] [e: ]j 1
c-1 ... U(s) = s
s+l
(.(C-1
L-1 s2(s+D
c-1
= e-1
s~ -~+ s:1 L-1
... Using partial fractions
[ 1
s(s+ 1)
1 s+I
X(t)
But given At t
X1~t)
y(t)
2-Ult'-l
= 0, y(t) = 2 = X1(t)J1;0 X
= X1(0)+C<X2(0)-e<X2(0)e-•
y(t)
=
X1(0)
.incl X2(0) = 0
2+ate-1 -te-1<:.<(e-t
1(1)
+c-1a.l-c-1a+e-1uc""'
-1)
and
«
l.f ,,_.
Example 4.35 :
X= [~~ y = (1
=
0, XI (I)
=2
Given Ilic state model of a system :
-~Jx+[~Ju
01 x
(VTU: July /Aug.- 2005)
Modem Control Theory
Solution of State Equations
with initial co11d1tio11s X(O) = [~] Determine: i)
Tire stat< trn11sitio11 matrix.
ii) The state transition equation X(t) a11d output Y(I) for an unit step i11p11t. iii) Inverse stale tra11Sillon matrix. Solullon: ii
A Isl-A)
[sr-Ar'
¢(s)
s
-1]
=
Adi [sl-A] _.;._ __ lsl-Aj
•1
(sl-A]
4/3 o(s)
:. Adj (s!-AJ = [ s+S 1] -4 s
s+S
[s+S 1 ] [s+S I ] .. -4 s "' -4 s -s2+5s+4 (s+l)(s+4}
=
s+5 (s+1Xs+4) -4 [ (s+1Xs+4)
- ~s+4
s+l
[ -4/3
+
s+l
1/3 s+l
:l.5R
= ""' X(O)
1/3 s+4
4/3
-1/3 + 4/3
s+4
s+l
5 -c
I iii
I (s+1Xs+4) s (s+1Xs+4)
= c"1 [
1
]
=
3
I
l
... Partial fractions
s+4
...
-5 [ --e 3
-t
--:-c23 8 +~-e 3
- ·ll
l
-·II
s+S ZfR
L-1 {¢{s) BU(s)}
= L-1
l[
(s+ I~ ~s+4)
{s+l):(s+4)]
(s+ I) (s+-1)
(s+I) (s+-1)
[~J m)
Modom
l
4 -65
Control Theory
L"' s(s+l) I (s+4) 1 [ (s+ l) (s+4) ---e I 1 4 3
ZIR
+-e I -4t 12
-t
l
Solutlon of State Equations
1/3 s+4 1/12 = L-1 [l/4 7-;::;:1+ 1/3 1/3 s+I - s+4
l
I -t --c I -41 [ -c
3
3
iii) The property of state transition matrix is, c!)-1(t)
,,_.
a
Cl>(-t)
Example 4.36 : Determine tire co11tro//ability attd oburvability of tire following stair model. 1
x
0 -ll
Y
=
(10
s l]x
Solution : From the given model.
A
= [~ -6
For controllability,
- ~I
(VTU : July/Aug.-2005)
-~J
o,
=
B
=
m,
[a:
C
=
~O 5 I]
AB:A2B]
..... n = 3
Modem
Control
[_~6 -~1 -~J m =L~l
AB
A (AB)= [ ~ -6
o, ~ [~ Now
I
Q< j =
-
Solution of State Equations
4. 66
Theory
0 ~ -11 -6
l
l [ ~1 [-l~l -12
61
-1°2 -1~ 61
84 "' 0 hence rank of Q< = n = 3.
The system is completely controllable. For observability, (4,;
ATCT
[CT :
T CT : A T2cT]
I\
=[~ ~ 0
1
-~ 1 [l~l =[=tl -6
1
-1
-6]
0 J\ T2CT
Q,, = Now
IQ j=96 0
[~o -6
[=t] = [~]
!l
-1 - 1
,eQ hence rank of
-11 -6
0
AT[ATcTJ=[~
J
Q0 = n = 3
The system is completely observable. 111•
A system represented by followi11g state model is controllable but 1101 observable. Slww that tire no11-01'urvability is due to a pole-zerocancelkuion in C(sl-Af1B. Example 4.37 :
X
r~
(VTU: July/Aug.-2005) 0
-6 -n
-6
JO] X+[~]
II
1
Y = (1 I OJ X Solution : From the given mod,cl,
A ;
~ ~].B=[~] 1 [~-6-11-6
,C
= (1
1 0)
Modem
Control
Theory
SoluUon of State Equations
4 ·67
0 ] -1
s2 +6s+11 s+6
Adj[sl-A] [
s+6
Adj fsl-AJ
fsr-Ar1
['' ''"" -6 -6s
s+6 s(s+6) -(lls+6)
-6s
+s(s+6)
-(1 ls+6) s2
l
,; I
Adj fsl-A]
Adi [sl-Aj
Isl-Al
s3 +6s2 +11s+6
[l l O]
-6
T
I
Adi (sl-AJ (s+l) (s+2) (s+3)
Adj [sl-A] [~] (s+1)(s+2)(s+3) 1
1 - (s+l)(s+2)(s+3)
l t[ l 1 1 0 1s [ ] s2
·-
s+ 1 (s+l)(s+2)(s+3)
Thus the factor (s+l) gets cancelled from the numerator and denominator. This shows that as then: is pole-zero cancellation in C(sl-Ar1 B which is the transfer function of the system, the system is not observable.
''*
.=
Examplo 4.38 : Consider the homogeneous equation X
Thefollcwing three solu[t~: folr -e-1
2~-1
AX where A is a .3 x 3 matrix.
[t:~:idiffilerf~e~tial conditions are
' -2e-:u
atailcbl«,
'l-6e-3• J
0
0
i) Identify the initial conditions ii) Find the state transition matrix iii) Hence or otherunse find the system matrix A Solution : i) For initial conditions, put t
= O.
(VTU: Jan.ffeb.·2006)
X(t)
X(O)
= [-~]
l ~- +:·]
X(O
[~~~'
~.. [~~]
ii)
Solution of Sblte Equ.tions
.. ·68
Modem Control Theory
~{~]
l ~·{:] ;,. [.;:l ~- f :1
~o
X(Ol
[·;:
=
1~~ :: 1 A7 A8
r-1] = r".I
:. . I -2
l
A4 A
L
AA36 X(O) A9
i.e
7
i.c.
A1-A2+2A3=-1
. .. (I)
A4 -A5 +2A6 =1
•.. (2)
A7 -A8 +2A9 =-2
... (3)
A1 -2A2 =-2
... (4)
A4 -2A5 =4
... (5)
A7-2A~
... (6)
=O
Solution of State Equations
4 ·69
Modem Control Theory
i.e
2/\ I - bi\ 2 "-6 2A4 -·6A, =IS
... (7)
=0
... (\1)
2A7 -6A8
... (8)
Solving the equations (1) to (9), the system matrix A is given by,
iii) State transition matrix
s [sl-AJ = ~ [
-1
o]
-1
s+S 0
(s+l) (s+S) Adj [st-A]=
Adj (sl-A) --Isl-Ai "
s3 +6s2 +11~+6
s2 +6s+5 -6(s+l)
(s+I) s(s+l)
0
0
[
l s
s2
s3+6s2+tts+6
+~.J
-----
s+S (s+2)(s+3)
(s +2)(s + 3)
-------(s+l)(s+2)(s·•3)
-6 (s+2)(s+3)
(s+2)(s+ 3)
(s+l)(s...2)(s+3)
..Y!:._ __
3
2
(s+2)-
(s+3)
(s+2) - (s+ 3j
-6 6 --+--
.2...._3_
(s+3)
1 s+l
0
0
0
s 2 +5s+6
(s+l) (s+2) (s+3)
(s1-Ar1
(s+2)
0 ]T 0
[
s+l
[si-Ar1
-6($+!) s(,+1)
(s~l)
(s+2)
(s+3) 0
(s+l)
1_.~_
(s+2)
..Y!:._ +-3(s+t)
(s+3)
- .2E...
(s+2) (s+3) 1 (s +1)
... Partial fractions
Modem Control Theory
Solution of State Equations
4-70 3c -21 _ Zc -Jt
0
0
Example 4.39: Obtain the lime response y(t)
of Ille system giMt bekn» by first transforming the state model into a 'Canonica/ model'. (VTU: [an, /F~b.·2006)
I,..
X = [~ ~ -6 -ll
l
Ol X + [~] u, y = [1 O OjX -6 2
u is a unit step [unction and Xr(O)
= fO 0 2]
Solution : First to transform given model into canonical model, find modal matrix M.
(i,1-AJ
= [~ -~. 6 1t
-~ ).+6
l
/
[)J-A]
t..3+6/..2+111..+6=0 (l..+1)(/.+2)(1..+3)
i.e.
1..1 = -1 , i,2 = -2, 1..3
For 1..1
For >..2
= -1, = -2,
= O
= -3
l~ _, '] [c"] ['] ['] f-1
-1-! ,M1= 11 :>
[Al-A]=
[>..I - A]
=
r-20
-1 -2
L6
11
[-3
For >..3 = -3, p,I-AJ = ~
-1 -3 11
M
[ ~1
C12 c13
= -6 6
= -1 1
~l· ~ ·[~:J[-:+Hl ~JMJ =[~::] =[~:]=[+]
~2 ~] = Vander monde matrix
Adj M
Solution of State Equations
4. 71
Modem Control Theoiy
~
-6 6 -2]T [-6 -5 -1] [ = 6 8 2 -2 -3 -1
- 5 8 -3 -1 2 -1
Ad.M Ad.M [3 -' -=-' -= -31 M -2
M-1
M-1AM
[~ 0
Z·S -4
O·S] -1
i.s o-s
~i
~2 =A 0 -3
C=CM=(l
11
I
The canonical state model is,
Z(t)
Y(t)
[~ 0
=
[1 1
-~ ~] Z( 0 -3
t)+[~z]u I
1] Z(t)
X(t) "' MZ(t) i.e, Z(t) "' M -1xct)
~~i [~] [~21
Z(O) = M-1 X(O) = [:13 :: 1 ·5 0·5
.~. l ZIR
Z.SR
... ,,., . r:· L-1 { o(s)BlJ( s)}
0
0
0 .,-3t
=
2
l[
t J
c-•
-2 1 ] = [ -2.,-21 1 c...J•
l
4-72
Modern Control Theory 0
s+I 0
1,-I
0
s+3
s(s+l) L-1
s s+l 1 1 L-1 _!+_I_
_:i_ s{~+2) I s(s+3)
::-21
1 -I -1 1 [ ---eI -Jt 3 3
ZSR
Z(I)
ZTR + ZSR
[;2] [l/•1
0
s+2 0
0
Solution of State Equations
1 l/~_\~~ l
l
=
J
s
-2t
-1 -e 1
s+
1
I 2 -Jt -+-e 3 3 Y(t) = [1 1 1) 1.(1) =
!. +~ c-Jt
1-1-e-21 +.!.+3.e-31 3
3
-c-11
3 3 II*
Example 4.40:
Git~I the system
X=[~'
~1]
x+[~
~] U. Find the input vtcior Ult!
to give the fo//oruing lime respons« :
x111J
6 (1-··-1)
X2(1)
3'--31 -2e-l1
(VTU : July/Aug.·
+6(t -e-1)
Solution : The given time response must satisfy the state equation.
Xt(ll
fi(~(1-c-1)]
= +6e"1
2006)
4 .73
Modem Control Theory
Solution of State Equations
k-JI ]+[~ ~] [6-6c:,1~;:~~-~2c- ]+[~ ~] U [-: ~ J
62~-~~-:~} -c-1
U
41
[33 0)2 u
[~ ~r
=
3
u [
U(t)
1•
O
-1 4 I
l
l
18-12e-• [-6+12 e •1 -6e"31 +6c-l1
[ _12+l2c.1
6-4c-1
-3c_31 + Jc_41
]
•••
J
Required input vector
Example 4.41 : Tiie foll11Wingis the state space representation of a linear systtm wl,_ eigtn wlitcs are -3, -2, -1. (VTU : July/Aug.·2006)
X =[ ~ -6 Givrn that u Solution:
~
~. lJ (X)
-11 -6
+[~]
II
2
= 0, X(O) = /0 0 1{. Find X(I).
:.
{sl-AI
~
[
s
-1
0 6
s 11
-1 0 s+6
l
s2 +6~+11 , Adj {sl-A)
s2 ... 6s+l1
r
Adj fsl-AI
fsl-A)
:. {sf-Ar'=
-b
l
-6s
s~6
n
-6s -6 s(s+6J -{11s+6)
[
s2
l
r
s+o s(s+6) -(11s+6)
= s3+os2+11s+6=(s+1)(s+2}(s+3)
Adilsl-AI
Jsl-AI
z
••
Solutionof State Equations
4. 74
Modem Controi Theory
j
s2 +6s+ II (s+ 1)(s+2)(s+ 3) -6 (s+1)(s+2)(s+3) -6s
s+6 (s+ l)(s+2){s+ 3) s(s+6) (s+l)(s+2)(s+3) -(I ls+6)
(s+l)(s+2)(s+3)
(S+l)(s+2)(s+3)
(s+ ll(s+2)(s+ 3) (s+ l)(s+2)(s+ 3) s~ (s+l)(s+2)(s+3)
2.5 4 1.5 0.5 1 0.5 3 3 ----+. -- 1 ----+------+-s+I s+2 s+3 s+I s+2 s+3 s+l s+2 s+ 3 2 1.5 -3 6 3 -2·5 8 4.5 -0.5 -+------+------+----s+I s+2 s+3 s+l s+2 s+'.\ s+l s+2 s+3 ~ 4,5 2.5 16 13·5 o.s 3 12 9 ----+------+------+-s+l s+2 s+'.\ s+l s+2 s+3 s+l s+2 s+3
2.se··• -4.e-21 +1.5e-31 -2.Sc-• +8e-21 -4.Se.J•
-0.Sc-t +2l'·2t -1.5'.>·JI
2.sc·1 -16e-2' + 13.se-31
o.se·•
O.se·• -e-21 +0.5c~11
l
-4c-21 +4.5e-3'
Example 4.42 : Find Ille lra11silio11 matrix <:(!) for a system wit~ system matrix is .~ivtn by A
=[-
35 -I
-1
1J
by tlze following tech11iqutS :
i)Laplact tra11sfom1 ii) lnfinit« series iii) Cayley-Hami11011
(VfU: July/Aug,·20061
Solution : A= . I)
[-5 3
Solutlon of Stato Equations
4.75
Modem Control Theory
-\] -1
Laplace transfonn method s+S I ] [s+ l -1 ] [ -3 s+ 1 Adi( sl -i\] = ~·' s+~•
(sl-A( (sl-/\
J
(s+;;)(s+1)+3=s2
s+ I = (s+2)(s+4)
Adj[sl-A)
Isl-Al
c'"
+6s+8=(H2)(s+4)
[
(s+2)(s+4)
-0.5 1.5 --+s+2 s+4
-0.5 0.5 --+s+2 s+4
1.5 1.5 [ --• s+2 s+4
1.5 0.5 --s+2 s+4
1.-1 [sl -Ar1=
I
(s+2)-1(s+4) (s+S) (s+2)(s+4)
l
3
-0.se-21+1.se-11 [ l.Sc-21 -l.5e~1
-0.s.c-21
1.Sc-21 -0.Sc-l•
II) Power series method A2
[-5 -I] [-5 -1] [ 22 6] r 22 6] r-s -1] = [-92 -28] 3
-I
.-18
-2
-t = -18 -2
3
3
A2t2
-\
84
A3t3
I+ At+--+--+ 2! 3!
[
I -51+1112 -¥13+ 3t-9t2 +14t3- ...
iii) Cayley • Hamilton meihod Stop 1 :
Find cigcn values.
j).. 1-A!;O
i.e,
l
i..+5 _3
I
I /..+I =Cl
-o.s ... -11]
20
...
... -t+3t 2 -61 28 3 + . 1-1-12+~13
6
.
l •
4. 76
Modem Control Theory i.e. 1.2 +6/..+8=0
Step 2:
i.eA, =-2,
Solution
of Stale Equations
'-2 a-4
Construct R().)
R(A.) = a0 +a1>. =p(I.) Where p(A.)=e'·1
Step 3 : Substituting values of >.1 and ).2
a0 -2a1
= c-21
= c~11
and a0-4a1
Step 4 : Solving above equations,
u = ..!.e-21 _.!.c-41 2
l
, <Xo = 2e-21 -e-41
2
f(A) = R(A)
Stop 5 :
:. cAt=
2c-21
= <Xol +
-c-'u[~
e"' ,,..
ii)
~]+Gc-21
-~c-~1 )[:: ::]
-0.5e-21 +l.5e_,1
-O.sc-21 +O.Sc-·11]
[ t.se-21 -t.Se__,1
t.Se-21 -0.se-4'
Obtain the stale transition matrix usi11g :
Examplo 4.43 : i)
lA
(l
Laplace transformation method n11d Cnyley-Hnmilto11 method.
For lite system describe by, X(t) = [_04 ~4] X(O)
Solution:
A = [~4 _\] X(O)
I) Laplace tansformmethod Isl-A)
[:
(sl-A]
s2 +4s+4
5:14].
Adj[sl-AJ =[s:~1
= (s+2)2
:]
()anJFcb.-2007)
Modem Control Thoory
l
4. 77
Solution of State Equations
s+4
l
Adj!sl-AI _ (s+2}. (s+\)2 ls! -Af - - _.:.'.!.__ (s+2)? (s+s2)2
[
I 2 s+2 + (s+2)2 -4
_!
(s+2)2 1 2_
s+2
(s+2)2
J
I
, . ... Partla! fractions
(s+2)2
Ii) Cayloy • Hamllton method
Step 1 : Find elgen values
ic.I:~
P.1-Al=O
-1 '
A+4j=O
().+2)2 «O i.e, A.1 =-2.
i.e, A.2 +41.. +4 =O
t.2
=-2
Step 2 : Construct R(>.) R(A.)= a0 +a1A. =p(A.) where p(l.)=c41
Step 3 : Substitute 1.1
For
"-z· as it is repeated use, ~ p(I) I 491.2 :~ T>f/.) I di.. • I d).. '" >-1.2 ~e'' l•=I "2 ~~!ao+r'11. .J! d). ,..) tc''
a1
= te-21
I
,),).2
).=•2 =ll1
and a0
= 2te-21 +e·21
4 -78
Modern Control Theory
Step 5; f(A)
= R{A) = Uol+ct1A
:. ~At= 2tc·2t +c-21
[I
I
0]+1c-21 lr 0 lJ 0 I -4 -4
l
tc-21
= [2tc-21 +e 21 -4tc-21 ,,_.
Solution of State Equations
c -21 - 21·c -21
J
Example 4.44 : Compute 1'1 for Ille gil1<'11 matrix :
Al=[~ ~]
CVTU : July/Aug.·2007)
=[~o ~lA =(_6w ~]
;A2
... (1)
This is the property of state transition matrix if A1A2 = AiA1. So verify this condition.
[~ ~) ( _0
(-~ro 6:) 0 6rol [ -6U> 0 J
00 ~)
A2A1 = [ _000
~]
[~
~]
As A1A2 = A2A1, the above property holds good. So calculate cA 11 and c"2' and then multiply to obtain e'". e"t'
(sl-A
1)
isl-A 11 (st-A
r
1
e"'' rsl-Azl
1
.r
L-1(s1-A [s-6 0
0 ] s-6 'Adj (Sl-A2J
(s-6)2
Adj[••-A,j jsl-A11
00
5
rs-6 0
+: . .'..l 0
C1(sl-AS1
[s -ro]
=
,
= [eO61
Adj(s1-A2]
e~'] -ro :]
= [s
0 ) s-6
4 .79
Modem Control Theory
e
,_..
A2t
-I[
~ L
sl-A 2
Solution of State Equations
[coscot sincot]
1_,
= -sin rot cos rot
Example 4.45 : Obtain tlie timt response of lht folluwing V«tor 11111trix differential
equation.
[~J
~s][~~)+[~]u y
oj[~~l
and = [1
= [:
wl~t 11(t) an wtit sttp
input and the initial conditions art X1(0) = Xz(O) = 0. Solution:
A
Thus
[_
06 ~5).B=[~lC=[l
[sl-Af
[s
(sl-A)
s2 +Ss+6=(s+2)(s+3)
[s•-Ar'
As X(O)
c
= [~] ,
6
-1 ] ,J\dj[sl-Af= s+S
Adj[sl-AJ Isl-Al
(VTU: July/Aug.·2007)
OJ [s+S _6
:]
[ "''
_ (s+2)(s+ 3) (.+2):.+3) ]-~·) -6 (s+2)(s+ 3) (s+2)(s+3)
cAt is not required.
ZIR
[~]
ZSR
1.-1 {<>(s)BU(s)}
L-'![(s+;~~+3) (s+2)(s+3)
... U(s) =!for unit step s
(s+2)~s+3)1 (s+2)(s+3)
(~]
m]
Solution of State Equations
4 -80
Modem Control Theory
X(t}
[1
Y(t)
Y(t)
,..,.
Example 4.46 :
T(s)
Y(s)
=
oj[Xi(t)] ~ X (t) Xz(I} I
.!__!_
<,.-21
6 2
+!.
c-31
3
Oblai11 lire observabl< phase variable stale model ~f 'l'T,
b ,s3 +b s1 +b,s +ll
=-
=---3.Drma tire :1ig11al flow graph of T(s).
= •.!.3--1~7-
ll(s}
s .,.,,Is- +n2s+nJ
(VTU: July/Aug.-2007)
Solution : The given transfer function Is Y(s} _
b0s3+b1s2+b2s+b3
U(s) -
s3 +.11s2 +d2s+a
:. (s3+d1s2
+a2s+a
3 3
l Y(s)=b s
0 3
+b1s2 +h2s+b3(U(s)I
Divide the equation by s3,
la Y(s}-b U(s))+_2_ fa Y(s)-b U(s)J:
Y(s)-b0U(s}+!f~1Y(s)-b1U(s)]+_2_2 S
Y(s} = b0U(s)+1(-a1
s
2
2
Y(s)+ b1 Ll(s)]
3
3
0
s3
S
+.}, 1-a 2Y(s)+b2U(s)) s·
+-\- f-.i 3 Y(s)+ b3 U(s)I s
... (!)
Now define the state variables as,
~lb1U(s)-d1Y(s)+X2(s)l
X3(s} X2(s)
:
Hb
2U(s)-a2
Y(s}+X1(s))
(2)
(3)
Modem Control Theory
>••
Solution
4·82
of State Equations
Example 4.47: Obtain tht contro/lablt p~ variabltformof ti"' transferfunction, Y(s) b0s3 +b1s2 +b2s+b3 T(s)
=-
U(s)
+a2s+n3
s3 +a1s2
(VTU: July/Aug.·20071
Solution: The given transfer function can be written as,
.::.<~
U(s)
Y(s)
Where
Y(s) Y(s)
U(s)
s3Q(s)+a 1s2Q(s)+a 2sQ(s)+a 3Q(s)
-a 1s2Q(s)-a2sQ(s)-a
s3Q(s)
3Q(s)+
U(s)
Define the state variables as, x.(s)
Q(s).
sX1(s)
X2(s)
sX2(s)
X3(s)
Taking inverse Laplace transform,
.
X1
=
.
X2, X2
=
X3,
Substituting selected state variables in the equation (1), s X 3(s) i.e,
.
X3
-a
1X 3(s)-a 2X2(s)-a
3X1(s)+
U(s)
... (1)
Modem Control Theory
Solution of State Equations
4. 83
From (2), Y(s)
= b0U(s)
+(b1-a1b0)X3(s)+(b2
-a2b0)X2(s)+(b3
-a 3b0)X1(s)
Taking inverse Laplace transform, Y
(b3 -a 3b0] X1 +(b2 +a 2b0] X2 +[b1-a1b0]
C
[(b3-a3b0){b2-a2b0){b1-a1b0)],
nus is controllable
X3 + b0U
D = (b0)
canonical form of the given transfer function.
Review Questions 1. WNzl is homogtn<1>us and nonhomognreous sliltt "'lualion 7 2. Otfi•t statt transition 1tmlrix usi•g classiail mtlhod of obtaining soluNon. J. What is zero input response and uro slate response ? 4. Obtain lht romp/tit solution of nonhomogtntous slalt equation using timt domain method. S. State tht imporlance of state transition matrix. 6. State tht various proptrlits of slalt transition matrix. 7. Obtain llrt solution of nonhomogtnt0us •lat< tquatian using Laplace lmnsform met/rod. 8. Whllt is resolrxtnt matrix ? 9. 10. 11. 12. 13.
i
Explain IAplare transformmtthod of obtaining 1. Expwn p
a[-~
-~1
b. [:
~] Ans.:
[ •' 0] le
I
e
I
I
Ans.:
f
cos./2 t Tzsin ../21 -.fi. sin ./21 cos./2 t
1
I
Modem
Control
4 .'a4
Theory
14. Oblai11 llU! ltomogenoous solutio1t of Iii• tqualiD1t ')({I} a. A= [ ~94
~1
Solution of State Equations
=A
X(I) wlttrc
nnd XIO) = [:]
l
-21
ti=['2 -ll-4J and X(O) = [ 0 1]
l I
I
[
b.
6 -71
-5e +5• Ans.: 7 ·ll --e
s
c
J •-e s 12 -71
c-3.$6!]
1.11 0·341 -0.11 Ans.: [ 0.48c0•56' -0.48c~'·56'
15. Find the output of tlrt systtm having state model.
X(t) = [ -12 0 1 ] X(t} + lPJ Ult) -7 I - l I Xltl
and Y(t} = [I
.
. . strp
Tirt inpu! U(I) ts unit
( = [10]0
mid X 01
(Ans.:Ylt)=~
3
... ~e-31 3
145e,1)
16. Show that tlzc following syst
=~ -:1
XII}=[-~ l and Y(I) =(I
Xfl) + [~] U(ll
0 -lJ
1
1
OJ X(tl
17. Find ti" solution is, X(I) = [~
-~J
X(t) with X(O) = [~]
18. Find the solution of,
xm =[~ -~] xw +[~]
Ult!
and Y(I) = (1 0/ Xltl if X(OI = [ ~] and syst
20. 21. 22. 23.
Explain Ka/man's lest for determining state controllability. Explain Cil!M!rt's lest for dtlennining stat« controllabilily. What is output controllaoi/ity? How tollunlc it for the system ? State lr.J: relation 0elwttn tram/er function in s domain and controllability, obesrwbility of a system.
Modem Control Theory
4. 85
Solution of State Equations
24. Dtfint a11d op/am concept of o/Js1rJ11bility. 25. Explain Ka/man's (PSI for tfetermi11i11g comp/tit 01Js1•rr.111bility. 26. Expl•n• Gilbert's test for detcm1i11111g complete obs(rt•abitrty. 27. F.1KJ/uatt tire ro11trt1llability of"" following sy5tems witlr,
a.A=[-2 -I01r ll=r31Cr b. A=
0} --121l5;
-1
fl 0
-2
r-1 c, A=
=fll]0
LI J
0
I
0 c 0
-1
H
~1
H =[~
~l
3 0J (Ans.: a. Controllable. b. Controllable c, Not controllable)
0 -2
28. £untuatt llu: obseroobility cf tire folfowi11g sy:
fl o0 0 2,J I· r 2
=[ ~
b. A=[-~ o}. < = _2
c,
A·[~
(O
I 0] c ~
0 -6
-11
I ,
-6
~
~]
l]
(4 5 1)
(Ans.: a. Observable b, and c, Not observable)
DOD
(4. 86)
Pole Placement Tuchnique 5.1 Introduction This chapter deals with the problem of pole placement. The pole placement technique or pole assignment technique is a method for design of control system which ls discussed in this chapter. If it is assumed that all the state variables for a given system arc measurable and available for the feedback then it can be proved that for a completely state controllable such system, its closed loop system poles can be placed at any desired location through state feedback by means of an appropriate state feedback gain matrix. The desired closed loop poles are firstly evaluated based on transient response and/or frequency response requirements such as say damping ratio, speed, bandwidth as well as the steady state requirements. Let us consider that the desired closed loop poles arc to be located at s=a1, s=u2 ......... s=a n- Then by properly selecting gain matrix for state feedback. il is possible to locate the closed loop poles al the desired locations provided that the original system is completely state controllable. In the following sections, we will consider the control signal to be scalar. It can then be proved that the necessary and sufficient condition that the closed loop poles can be placed at any arbitrary locations in the s-plane is that the system is completely state controllable Le. it is possible to transfer the system state from any initial state X(t0) to any other desired state X(t1) in a specified finite time interval (t1) by a control vector U(t). The required state feedback gain matrix can then be evaluated using various methods.
In this chapter, It is considered that the control signal is a scalar quantity and not a vector quantity. Under such case, the mathematical aspects of the pole placement technique is complicated as the state feedback gain matrix is not unique.
(5 - 1)
Modern Control Theory
5-2
Pole Placement Technique
5.2 Pole Placement Design The conventional method of design of single input single output control system consists of design of a suitable controller or compensator in such a way that the dominant dosed loop poles will have a desired damping ratio ~ and undamped natural frequency IDn. The order of the system in this case is increased by 1 or 2 if there arc no pole zero cancellation taking place. It is assumed in this method that the effects on the responses of non dominant closed loop poles to be negligible. Instead of specifying only the dominant dosed loop poles in the conventional method of design, the pole placement technique describes all the closed loop poles which requires measurements of alJ state variables or inclusion of a state observer in the system. The system dosed loop poles can be placed at arbitrarily chosen locations with the condition that the system is completely state controllable .. This condition can be proved and the proof is given below. Consider a control system described by following state equation ... (1)
x=Ax+Bu
Here x is a state vector, u is a control signal which is scalar, A is n x n state matrix. B is n x l constant matrix. The system defined by above equation represents open loop system. The state x is not fed back to the control signal u,
u
Let us select the control signal to be u Kx state. This indicates that the control signal is obtained from instantaneous state. This is called state feedback. The k is a matrix of order Ix n called state feedback gain matrix. Let us consider the control signal to be unconstrained. Substituting value of u in equation 1 . Fig. 5.1 Open loop control system
.
x = Ax+B(-Kx)=(A-13K)x
=-
. .. (2) The system defined by above equation is shown in the Fig. 5.2. It is a dosed loop control system as the system state x is fed back to the control system as the system slate x is fed back to control signal u. Thus this a system with state feedback.
Fig. 5.2
Pole Placement Technique
5-3
Modem Control Theory The solution of equation 2 is say x(t)
=
e,
x(O) is the initial state
... (3)
The stability and the transient response cha.racteristics arc determined by the eigcn values of matrix A - BK. Depending on the selection of state feedback gain matrix K, the matrix A - BK can be made asymptotically stable and it is possible to make x(t) approaching to zero 3$ time t approaches to infinity provided x(O) ,. 0. The eigen values of matrix A - BK are called regulator poles. These regulator poles when placed in left half of s plane then x(t) approaches zero as time t approaches infinity. The problem of placing the closed loop poles at the desired location is called a pole placement problem.
5.3 Necessary and Sufficient Condition for Arbitrary Pole Placement Consider a control system defined by following state equation. x = Ax+
Bu
Let the control signal selected be u x
=
Ax - BKx
. .. (1)
=-
Kx
= (A - BK) x
... (2)
The solution of the above equation is given as, x(t)
=
.,(A-6 K)I . x(O)
... (3)
The eigcn values of the matrix (A - BK) are nothing but desired closed loop poles. The necessary and sufficient condition for arbitrary pole placement is that the system is completely state controllable.
If suppose the system is not completely state controllable then there arc eigcn values of matrix A - BK that can not be controlled by state feedback. According to Kalrnan's test for state controllability, the system is said to be completely state controllable if the rank of the composite matrix Qc is n.
Q,, = [B: AB:A2B:
An-ls]
Let us suppose that the system is not completely state controllable. The rank of the controllabiUty matrix is less than n, rank(B:AB:A2B:
:A0-1B]
= s
Thus there are s linearly independent column vectors in the controllability matrix. Let these s linearly independent column vectors as c1, c2, c•. Also let us select n - s additional vectors t,.1, ts.2, ....... t,, such that
5.4
Modern Control Theory P
=
(c1:c2:
Pole Placement Technique
:c.:t1+1:ts+2:
..•..•...
:t0)
•.••.•••••.
Tile above matrix will have rank n By using matrix Pas the transfomation matrix, let us define Now we have
r·1AP=A.
p·1B=B
AP = PA
(Ac1: Ac2:
: Acs:Als+l:At.+2:
=
......•.... : At11)
[c1: c2: ..••...•. : c9:l_.1:t8•2:
.•....•....
: t0)
A
As there arc s linearly independent column vectors Ct- ~- •.••.••• c5 using Caylcy-Hamilton theorem vectors Ac1, Ac2.•..•.•..• Ac, can be expressed in terms of these s vectors. Ac1
au c1 + a21c2 + ·········· + as1 C.
Ac2
a12 c1 + a22c2 +
Ac,
1115
c1 + aisc2 .+
+ a52
c.
+ ass '•
Thus we can write
I
[Ac,: Ac2 :······:Ac,:
At,.,: Al,.2:
:At,,)
a82111 ····················821 ················ a1,
:·.::::::·.·.:·.·.·.:·.~.,., n
,:,a821',',', I I I I I
= [c, : Ci: . . .. -: c, : '•·• : t,.2: ..... t,,J
.
' ' . aan -----------------r----------------' ······················ : 8-•ts•1···· ···· 8.s•tn '' a~t
a,, : a~1•1
'' '' '
I
............. ........
I
; ;
:
:; : : ;:
: a"'., ?'
•
8m
5.5
Modem Control Theory
Polo Placement Technique
Let us define
~
u
a,,
a1,
a,1•1
a1n
321
a21
a2••1
azn
a;1
a5$
a~,.,
acn
]·
~
,~.. ..,,. zcromatra
•
~
,
···~~
b.,
~
We have,
(Ac,: A~ ....... : Ac,:
Ai..1 ; ....... : A1,,]e(c1: ~:
...... : C,,:
1,,,: 1>•2: ....... : •n]
[.:'!~-~-~'.L] 0
Hence we have,
'
Az2
Modern Control Theory Now we have,
Pole Placement Technique
5-6
B
PB
B
(c1: c2:
........•
:
c,:t •._1:t
2:
:
t.,]B
Now B can be eXJ>resscd in terms of s linearly independent column vectors
c;. Therefore we get,
c1, ei,
B = b11c1 +b21c-2+
b51c,
'o· I 0 b~l b11]
Let us define
bs1
We have
Now we define
u=
[-~n
K
We have
jsl-A+BKj
1r-1(sI-A+BK}Pj sKPj
jsr-r-1AP+r·1
But
"' [ ••••• An :1. •••• A12 ] P·I AP= A= 0 : ~
J
Modern
Control
Pole Placement Technique
5-7
Theory
Substituting
=
jsl-A+BKJ
js1-fi.+i3xkl
In the Identity matrix I, we have n rows and the diagonal clements as 1. nus matrix can be split as we have the matrix A.
o
1
0
0
0
0
1······· . "•.....
0 I•
I
::·.,
:
---------------i--------------0 0 1 ::::.:::: 0
0
......................
I
'
....................
0
.: ..,,
a , o
'' '' 1
':
0
:; : :
0
0
--------------+--------------...................... 0 ·: 1 :::·············· .. ···
.. I
0
:
0
'·, ••
· · ....
I
..........
~
0
"•,,
:·:::1
5-12
Modern Control Theory
Pole Placement Technique
The above equation can be rewritten as,
B
= Tm=QcWm [ B:i\B:i\ 2B]
But Q,,
r: ~ 1
[B:AB:A
T [~]
a11 0
s) [~] = 8
2
B
1T·m
r-18
r
r1B
= [ B: AB: A 2o)
m
=
The results which are proved for n = 3 can be extended and generalised so Iha! we have equations (2) and (3) as given below. 0 0
1 0
0··· 1
0 0
-1
T AT= 0 0 o.... -an -an-1 -an..2
1
-a,
5 ·13
Modem Control Theory
Pole P,lacement Technique
0
9
Now consider the followingequation
~ = r-1ATx+l1Bu The above equation can be transformed into the controllable canonical form if the system is completely state controllable. The state vector x can be transformed into state vector x by use of transformationmatrix T = MW. Now we will select a set of desired eigen values as a1,a2, a0• Then the desired characteristicequation becomes, ( s-a1
)( s-a2 )
(s-an) =sn +o1sn-l +
on-ts +on
.:o.
Let When u =
...(7) ... (8)
-f<x = -KTx
.
x
is used to control the system having the equation
(r1AT-11BKT) x
The characteristicequation is,
!sr-r-1AT+r1BKTI
c
o
This characteristic equation is the same as the characteristic equation for the system defined by ; =Ax+ Bu when u; = - Kx Is used as· a control signal. This ts proved below.
; = Ax+Bu=Ax+B(-Kx)=Ax-BKx=(A-BK}~.
Pole Placement Technique
5· 14
Modem Control Theory
The characteristic equation for this system is,
=I
!sl-A+BKI
Now we will simplify canonical form.
r1(sl-A+BK)Tl=ls1-r1AT+r1BKTI the characteristic equation of the system in the controllable
I
0 0
Iisl!
I sl-T-1AT-r1BKTI
0
0 0
0
+:
I
0
I
""""
I
9 [µ11µ11-1 ...... µ
0 -an-l
-1 0
iJ
I
I
"""1
0 0
l.L This is the characteristic equation for the system with state feedback. It is same as equation (7). 0I
d I+µ I
o.z
a2 +µ2
sn
an +µn
The above equations can be solved for µ; 's and substituting in equation (8) KT 1(
[Ii nl' n-1 ....... µ l) ... (9)
If the system is completely state controllable, all eigen values can be arbitrarily placed by choosing matrix K according to above equation no (9). nus is the sufficient condition.
Modem Control Theory
5-15
Pole Placement Technique
Thus the necessary and sufficient condition for arbitrary pole placement is proved and it requires the system to be completely state controllable. For a system which is not completely state controllable but is stabilizable, then the total system can be made stable by placement of closed loop poles at the desired locations for s controllable modes. The rest n - s uncontrollable modes are stable which shows that the total system can be made stable.
5.4 Evaluation of State Feedback Gain Matrix K Following are the different methods for evaluation of state feedback gain matrix K i) Using Transformation Matrix T ii) Using Direct Substitution Method
iii) Using Ackermann's Formula
5.4.1 Use of Transformation Matrix T Consider a system defined by a stale equation
.
x =Ax+
Bu, Let the control signal be
u = - Kx . = Ax-llKx=(A-BK)x
x=Ax+B(-Kx)
The state feedback gain matrix K forces the eigen values of (A - BK) to be a 1• a2 ..•••...••..... n". Then the desired characteristic equauon is ( s-tt.) ( s -n2 ) ..•...•..... (s-tt0) # s" + 61s0• 1 + i>2s0•2 + + 611_ 1s + 60 = 0 The desired cigen values 1x1,u2 Step I:
Step II:
The characteristic polynomial for matrix A is given by J
Step Ill:
n0 can be obtained from following steps.
The controllability condition is checked for the system. For the system which is completely state controllable following steps can be used.
sl-A
I
= s" +a 1s11-1 +a2s0-2+ From this values a 1,.12
+i111_1s+a0 a" can be obtained.
The transformation matrix T is determined which converts the system state equations into controllable canonical form. For system equations which arc already in controllable canonical form T = I. The system state equations are not required to be written in controllable canonical form but matrix T is found from. where Q., ts controllability matrix
Modem Control Theory
5 -16 0n-l 011:-2
Pole Placement Technique
0ra-2
a1
I
°n:-3
1
0
w 0 0 0 0
1
0 Step IV:
Using the desired eigen values which arc desired closed loop poles, the desired characteristic equation can be obtained ( s-«1 )( s-«2 ) (s-an) = s" +l>1sn-l +l>2s"-2+ l>.,_1s+l>n· The values of li1, 1i2 ,
Step V:
li0 can be obtained.
The state feedback gain matrix K can be obtained from following equation K = (l>n-an
lin-l-an-t·············li1-a1JT-I
5.4.2 Direct SubstitutionMethod For low order system with n less than or equal to 3 matrix I< can be directly substituted inlo the desired characteristic equation. If suppose n = 3 then the state feedback gain matrix K is I< = [1<1 1<2 K3) The above matrix is directly substituted in the desired characteristic equation Isl-A +BKI and equated to ( s-a1 )( s-a2 )( s-a3) i.e, lsl-A+BKI
=
(s-ai)
(s-a2)
(s-u3)
The both sides of the characteristic: equation are polynomials in s. Hence the co-efficient of similar powers in s on both sides can be equated so as to get the values of K1, K2 and K3. This method is restricted for n = 2 or 3. For higher orders of n the calculations become complex. It is not possible to determine matrix I< if the system is not completely controllable.
5.4.3 Ackermann's Formula Consider a system defined by state equation x =Ax+ Bu. let u = -Kx be state feedback control. Assuming' the system to be completely state controllable nnd the desired closed loop poles are assumed at s = a1, s = u2, ••••••••• s = a0• Using u = - Kx in state equation
5 -17
Modem Control Theory
Lei
Pole PlacementTechnique
A-BK
iix
x
The desired characteristic equation is, isl-A+
UKI
By cayley-Hamilton theorem, every matrix satisfies its own characteristic equation.
o (i\)
=
i\ n +o1A
n-1+
+011_1A +S.,J = 0
The above equation can be used for deriving Ackermann's formula. Let us consider the case of n .r 3 for simplification of derivation. We have following identities
A
.i\2
A-BK (A-BK)2=A2-2ABK+B2K2 A 2 -ABK-ABK+B2K2
=A
2
-ABK-BK(A-BK)
A2-ABK-BKA
j\3 I
(A-BK)3 =A 3 -A 2BK-ABKA-BKA 2 I
No~" consider
i\ 3 +o1.i\
2 +o2i\ +031
A 3 -A 2BK-ABKA -BKA
2
+o1( A 2
-
i\BK-BKI\)
+ o2(A -BK)+o31 A 3 -A 2BK-/\BKA-BKA
2 +o1A 2 -61ABK
- ll1BKA +l>iA -o213K+631 li3l+o2/\ +ll,A 2 +A 3 -ozBK-6,A13K - 61BKi\-A 2BK-ABKA-llKi\ We have,
and
= 63l+o2A+61A2
+A 3
2
= O (for n = 3)
Polo Placement Technique
5 -18
Modern Control Theory
substituting the above equations,
o(.i\.)
0 (A )-S2
BK-o, AllK-o,
UKA -A 213K-AllKA -llKA ~
0
B111
o21lK+1\113KA
6( A)
u(o2K+o1KA
+ 13KA 2 +61 ABK+;\13KA
+A 213K +KA 2)+An(o1 K+ KA)+,, 2nK
dl(A)
11, the system is completely state controllable, the inverse of the controllability matirx
Q., ~ [ 13: AB: A 2 B] exists. Multiplying bolhsides by inverse of controllability matrix.
Prcmultiplying both sides of above equation by [O 0 I) we get
10 0 l
I [13:Al3:A2Br'¢(A)
~ [O
0
1jro2K+ot10\ ~Ki\2] I StK+KA
IL
K
K K
to
0 1)[13:/\B:A213]o(A)
The above equation gives the required state feedback gain matrix K for arbitrary positive integer n, K
=
(0 0
[
,
0 I] 13:AB:A·ll:
:A0•1n
1·1 O(A)
This equation is called Ackermann's formula for the evaluation of state feedback gain matrix K.
Modem Control Theory
5·19
Pole .Placement Technique
5.5 Selection of Location of Desired Closed Loop Poles The locations of desired closed loop poles must first be chosen in the pole placement design technique. The commonly used method for such selection of poles is based on the experiences obtained from root locus design of placing a dominant pair of closed loop poles. The other poles are selected in such a way that they arc away from dominant dosed loop poles in left half of s plane. If the dominant closed loop poles are placed away from jlo or imaginary axis then the response obtained from the system is very quick but at the cost of system becoming nonlinear which should be avoided. The other method of selection of poles is Ute quadratic optimal control technique. In this method, the desired closed loop poles arc determined in such a way that it balances between the acceptable response and the amount of control energy required. The high speed response indicates that large amount of control eneri,'Y is required. In addition to this a larger and heavier actuator is required which increases the cost.
l '*
Example 5.1 : Consider the systtm defined by x = Ax+ll11
totier«
A
By using the state feedb
=-
Solution : Let us first check the controllability matrix of the system. The controllability matrix, Q,, =
[n:AB:A2n] B
[H AB=[~ ro
A2 =A.A.
A2B
I
-1
01[0
l~t
0 0 1 -5 -6J -1
[~
0 -5 +29 31
1 0
o] = ro 1
-1
-5
-6
c+6
~6] [~]=f ~] 1
l 31
0
~6]
-5 29 31
Modem Control Theory
l
o o ~
=
0
[1
Polo Placoment Tochnique
5-20
I -6 1 . -6 31
The rank of matrix ~ is J. ·111e system is controllable and hence it is possible to place the poles arbitrarily. Now the state. feedback gain matrix K can be obtained by using any of the three methods discussed previously.
ro~
Method 1 : Using tra:sf:rm[
mat']ix T
-1 -5 -6 The characteristic equation for the system is,
s ls(s+ 6) +5 J+ 1[1]
s [s2+6s+s]+1 s3 +6s2 +5>+1 =O Comparing this equation with s3+a1s2+a2s+,t3 :. a1
= 6.
a2
=
5, a3
=
=0
I
The desired characteristic equation is , (s+ 1-j2)(s+ 1+j2)(s+10) s3 +tos2
=
-t- 20s+2s2
[Cs+ 1)2 +4] (s+ IU) =[s2 +2s+5] +Ss·~SO
s3 +12s2 +25s+S(l
s,
=12, 02 =25. 03 =50 K
= fo3-a3
62-a2
o1-a11rl
As the given stale equation is given in controllable canonical from T = I.
11
r
K
Lo3-a3
K
[49 20 6)
1>2-.i2
-a.J
1>1
=[50-1
25-5
12-6]
Is+ 10(
Modem Control.Theory
5-22
•no
o -6 0 -5 =[~1 29 31_ -1 -
-5 29 -1~9
r-16 -31
[~l or I
0
1
1
[o
- 0
-5 -6
¢(A)
0 -5
-1
r-1
6 -31
Pole Placement Technique
:]
0 -5
~l -1571
:][~ -5
I
~] =[~1
0 -5
0
~ j+1{~-6
29 -149 -157
18
-5 29
0 -6 1 -5 29 31
~6]+25[ ~ 31
[ ·19
20 6 ] 19 -16 -128 74
115
[ B: All: A 2Bf1
0 1
K "'
l -6
i·lr 18
49
31 l-128
-6
20 6]
19 16 74 115
~ ~] [ ~: ~~ -~6} 0 0
-128
[1 0 0) [ ~: ~~ -~6]
-128
[49 20 6]
74
115
l
74 115
1
0 -1 -5
~]+so[~
0 1 0
~]
Modem Control Theory ,._.
Example 5.2 : It is desired to place tire closed loop poles of tire followi11g system al s = -3 and s = - 4 by a slate feedback controller witlr Ille control u = - Kx. Detennine tire state fttdbackgain matrix K a11d the control signs]. (VTU : Aug.-2005)
y
.
Solution : Equating x
e
Pole Placement Technique
5-23
=
[1
O]x
the state equations
with standard equations
with standard form
Ax +Bu and y=Cx we get
B
c = [1
= [~]
0)
Let us check the controllability matrix of the system (B :AB) AB
[~1 ~3][~] =[:6]
o, = [~ :] The rank of above matrix is 2. Hence the given system is completely controllable. The given equation is not in the controllable canonical form because B is other than[~] we use method of transformation matrix T to get state feedback gain matrix. T
Oc·W
W=["i 1]=[3 11 1
consider A
[-t0 -3I]
; jsl-AI=
I
s1
O
-1 s~ 3
1
I=
OJ
s(s + 3) + 1
= s2 + 3s+ 1
Equating with s2 + a1 s + a2 we gel
5 ·25
Modem Control Theory
Pole Placement Technique
5.7 Concept of State Observer In case of state observer, the state variables are estimated based on the measurements of the output and control variables. The concept of observability plays important part here in case of state observer. Consider a system defined by following state equations x ;
Ax+Bu
y
Cx
x
Let us consider as the observed state vectors. The observer is basically a subsystem which reconstructs the state vector of the system. The mathematical model of the observer is same as that of the plant except the inclusion of additional term consisting of estimation error to compensate for inaccuracies in matrices A and 'B and the lack of the initial error. The estimation error or the observation error is the difference between the measured output and the estimated output. The initial error is the difference between the initial state and the initial estimated state. Thus the mathematical model of the observer can be defined as, x
Ax+Bu+K.(y-Cx) Ax+Bu+K.y+KeCx (A-K.C)x+Bu+K.y
x
x
Here is the estimated state and C is the estimated output. The observer has inputs of output y and control input u. Matrix K0 is called the observer gain matrix. It is nothing but weighting matrix for the correction term which contains the difference between the measured output y and the estimated output CX. This additional term continuously corrects the model output and the performance of the observer is improved. Full order state observer The system equations are already defined as x ;
Ax+Bu
y
Cx
The mathematical model of the state observer is taken as,
Polo Placement Technique
5-26
Modem Control Theory
x = Ax+Bu +K.{y-Cx)
..
To determine the observer error equation, subtracting equation of x-x
i from
~ we gel
[Ax+Bu]-(Ax+Bu+K.(y-Cx)] Ax-Ax-K,(y-Cx)
A(x-x)-K0(Cx-CX)
..
A{x-x)-K.C(x-x)
x-x
(A-K.C) (x-x)
x-x e e : (A -K,C)e
Let
...
:. e=x-x
The block diagram of the system and full order state observer is shown in the Fig. 5.3.
>--------Y u
-----------------------·x·------------
Full order stateobsefver
Fig. 5.3 The dynamic behavior of the error vector is obtained from the cigen values of matrix A-K.C. If matrix A-K.C is a stable matrix then the error vector will converge to zero for any initial error vector e(O). Hence x(t) will converge to x(t) irrespective of values of x(O) and x(O). II the eigen values of matrix A-K.c are selected in such a manner that the dynamic behaviour of the error vector is asymptotically stable and is sufficiently fast then any of the error vector will tend to zero with sufficient speed. If the system is completely observable then it can be shown that it is possible to select matrix Ke such that A-K.C has arbitrarily desired eigen values. Le, observer gain matrix I<• can be obtained to get the desired matrix A-K.C.
5. 29
Modem Control Theory
Herc
T
Q.,.W = [
c• : NC·
Pole PlacementTechnique
: (N)2
c· :
: (N)"-1 c- I w
For the original system the observability matrix is,
I
c-.
: (A•}n-l c·
NC' T T'
w
As
rr-r'
I =
N
NW
w•N• = WN•
·: (AB)T =BT ·AT
w• (WN'r1
Taking conjugate transpose of both sides of equation I Bn -an On-1 ._-an-I
r Herc
w
N
i.l n-1
la'rz
a n-2 ''n-3
al
1
1
0
~
~
I ,
l
·~i
I
c- : NC' : (N)2 c- :
0
: (N)n-l c· I
The second method is based on direct substitution method to get the state observer gain matrix K,. This is applicable for system having low order. TI1e matrix Kc is directly substituted in the desired characteristic polynomial. Let us consider K~ :
r~:~Jl Ko3
Substitute this matrix Into desired characteristic polynomial.
By equaling the like power coefficients of s on both sides. the values of Ke1• K02 and be determined. This method is restricted for value of n at the most 3. The third method is use of Ackermann's formula. For the original system defined as
I<,,3 can
x=Ax+Bu and y = Cx we have already derived the formula known as Ackermann's formula given as,
5. 30
Modern Control Theory
K = [O 0
I]
Pole Placement Technique
or
[Il:Ail:
1\11•1
1'i)
(A)
Now let us consider the dual system z
A"z + c·v
y
n•z
The Ackermann's formuln for dual system is modified as K
= [O
0 .
1 ]I C' : A"C' :
: (A')n-l
c•
r' O (A ')
Both observer gain matrix, K0 = K'. -1
r :.Kc= K' = q>(A ')'
c~
l ;
()
c
0
C.A
0 0
= q>(A)
... (II)
CA n-2
Ci\ n-1
LCA"-l
The desired characteristic polynomial lor stale observer is
= ( s-u1
o(s)
)(
s-u2
) ......•......
(s-rtn)
Herc a 1, u2 •••..•••..•• n11 arc desired cigcn values. Equation (ll) is Ackermann's formula for determination of observer gain matrix K,.
5.11 Selection of Suitable Value of Observer Gain Matrix
K.
The feedback signal through the observer gain matrix K,, acts as a correction signal to the plant model to account for the unknowns in the plant. If the unknowns involved arc significant then the feedback signal through matrix Ke is comparatively large. But if the output signal is affected by disturbances and measurement noises then output y is not reliable and feedback signal through matrix K, should be small. The effects of disturbances and noises involved in output y should be neatly studied while evaluating matrix KeThe observer gain
is dependent on the desired characteristic equation 0. The selection of a1,u2 u., is many times not unique. But normally the observer poles should be two to five limes Caster than the controller poles. TIUs makes the observation or estimation error converging to zero quickly. The observer estimation error decays two to five times faster than does the stale vector x. Puc to this foster decay of observer error, the system response is dominated by controller poles.
(s-u1)(s-u2)
matrix
(s-o.,.)
=
Modem Control Theory
Pole Placement Technique
5 ·31
With the appreciable noise produced by sensor then the observer poles can be selected to be slower than two times the controller poles. The bandwidth of the system will then be less and noise wiU be smoothened. The system response is dependent on observer poles in such case. The observer poles arc located to the right of controller poles in left half of s-plane so that the system n.'Sponsc is influenced by observer poles. While designing thc state observer, it is required to evaluate various observer gain matrices Ke based on number of various desired characteristic equations. For each of these different matrices K. r the simulation tests can be run to check the overall system performance. Based on the performance, the best possible K. om be selected. In many systems, the best value of K0 is obtained after a compromise between fast response and sensitivity to disturbances and noise signals.
>'*
Example 5.3 Co11sider the system rrpresenled by
x =
r:
y " ll
Design -5. Solution:
A=
11
-~l
!J
x
+mll
0 O].r
full order observer s11cl1 t/1111 the observer eige11 oalues nre nl -2 ± j2./3 mrd (VTU: Aug.·2005)
r~
1
c = [1
0 -6 -11
Lei us check the rank of dual system of the original system I C : A'C· : {A')2 C" : : (A'ln-1 c· I Herc n
=3
:.( c-.
NC·: (A')2
c· {A')2
~ m;
c·
J
A'C•
A'. A'=
1[~
=[~ 0 0
()
0
~IJ m =m
~][~
0 0
-6]
[o
-6 -11
~· = ~
-6
~] 25
0 01
5 .34
Modem Control Theory
3
(A-)2
Pole Placement Technique
3 OJ =Ior1
0 7 -I
[~ ~][~ o L2 ~2] [~ ~][~]=[~] [~ 2 ~2]
=
-1
-1 2 1
0
(A"l2
c·
7 -1
0
I
c· : NC• : (A")2 C- I
2
0
The rank of above matrix 3. Hence the original system is observable.
Let
Ke
[K•l]
=
K.i
be observer gain matrix
K•3
The desired characteristic polynomJal is,
[s-(-3+j1)]
[s-(-3-jl)] [s-(-4)] = (Cs+3)2 +1] (s2 +6s+10) (s+4)
= sJ +6s2 +10s+4s2
SJ+ 10s2 +34s+40
I
Now consider sl-(A + K0
-C)j
=
0
2
s 0
-1 ~]-[~:~] (0 0 2
[~ :]-([: 2 [~ ~]-( [~ ~] -[~ 0
-1
0 0
2
0 K._,
0
0
=
OJ0 - [l3 -12 1-K.i -K.1 ]
I [~
0 s
I r-l-3
s+l
0
0
2
-2 -2
-I K.1 +K.2 ] s+K03
-Ke3
I
I
[s+4]
K,2 K
tJ)
+24s+40
5.37
Modem Control Theory
(A-K0C)
Pola Placement Technique
(x-x)
e = (A - K,, C ) e
... (Il)
Equations (I) and (JI) can be eombined and rewritten in matrix form as,
[~1 •
=
[x]
[A-BK BK J 0 A-K,C e
... (Ill)
eJ
Equation (Ill) shows the dynamics of observed state feedback control system. The characteristic equation for the system is
l
sl-A+BK
o
-BK sl-A+K.c
I
lsl-A+DKHsl-A+K0Cj
0 =
o
The closed loop poles of observed state feedback system consists of the poles due to pole placement design alone and poles due to observer design alone. Thus the pole placement design and observer design arc independent of each other. The design of each can be done independently and it can be combined to get observed state feedback control system. For n as the order of the plant, the order of the observer is also n whdc the resulting characteristic equation of the total dosed loop system is of order 2n. To derive the transfer function of observer based controller, let us consider the system defined as x
Ax+ Bu
y
Cx
Let the plant be completely observable and let the observed state feedback control is
u=-KX.
x
. :.x
Ax+Bu+K.(y-G) Ax+ B(-KX)+ K0(y-CX) (A-OK-K0C')x+
Key
Taking Laplace trans;... ::n of ~boh: equation sx(s)-x(O) = (A-OK-K.C)x(s)+K.Y(s)
Modem
Pole Placement Technique
5 -38
Control Theory
Assuming zero initial condition, [sl-(A-BK-K,.C)J
x(s)
K0 Y(s)
x(s}
(sl-A +BK+ Kccr1 K, Y(s)
Substituting this value of x(s) in equation for u, U(s)
-K >'.(s)
U(s}
- K (sl - A + BK + KcCf1 Kc Y(s)
The transfer Iunction is given by, U(s)
Y(s) U(s) -Y(s)
=
K(sl-J\+KCC+BKr1Kc
The block diagram representation of above system is shown in the Fig. 5.5
K{sl-A•K.c + BK)-1K•
H.
_.T
_P_1a_n_1
Fig. 5.5 The transfer function obtained above is called observer based controller transfer function. It can be seen that the observer controller matrix A - KcC - BK may or may not be stable even though A - BK and A - K.c are selected to be stable. In certain cases, the matrix A - K0C - BK may be poorly stable or even unstable also.
Examples with Solutions ,,..
Example 5.5 : A system represented by followi11g state model is controllable but 11ot obstrvablt. Show that lht 1101H1bsm;ability is due to a pole-zero cancellatio11 in Clsl - AJ-1. (VTU: July/Aug.·2ll05l
x
[ ~
~
-6 -11 Y
(1
1
o) x
~] x+[~]u
-6
1
Solution:
[-6~ -110 -6~ jx+[~]u l
x Y = (1
I 0) x l
Herc,
A
[~6
0 -11
q;
8 =
-6
[~ ~]-[~ 0
[sl-A]
0 s
jsl-AI
Pole Placement Technique
5-39
Modem Control Theory
m;
1
0
[~
lJ
-6 -II
sf(s+6)s+ll]+1(6)
c = -1 s 11
I 01
(1
0 ] -1 5*-6
s3 +6s2 +11s+6
(s+l)(s+2)(s+3) [s1-Ar1
Adj of fsi-A] jsl-A!
rs +6s+ll
T.F.
l
1 (s+l)(s+2)(s+
3)
I (s+l)(s+2)(s+
3) [
Y(s) D(s)
=
2
+s+6 1 s2 +6s+ll 6 . -6s
C(s1-Ar'n+o
[1 1 0 J
1 (s+l)(s+2)(s+3)
·,
-6 -6s ]T s2 +6s -lls-6 s2 s s+6 s(s+6) -(11s+6)
=
C[sl-AJ-1B
= (·:
O=O)
r s"+6s+11 '
l
(s+ l)(s+2)(s+ 3)
(s2 +6s+l1-6
1 (s+l)(s+2)(s+3)
s+l [. J
-6 -6s
s+6 S(H6)
IS
-(lls+b)
s2
(s+6)H(S+6)
l
[010 I
1+s] r~l]
Pole Placement Technique
5 -40
Modem Control Theory
s+ l
(s+i)(5:.:.2)(s+3) T.F.
Y(s) = -U(;)
s+l =-----(>+ l)(H-2)(~+
3)
Clearly it is seen that the factor {s.,.1) is cancelled from numerator and denominator. T.F. "
s+l s3+6s2+11s+6
s+I l~(s+6)+ 11 ]s+6
Let us draw the simulation of the transfer (unction.
y
Fig. 5.6 There is no coupling between x3 and y due to which state x3 can not be observed from measurement of output y. Thus the system is not completely observable due to pole-zero cancellation as matrix C contains one of the terms as zero.
1m+>
Example 5.6 : A regulator system has the plant
(VJ'U: J•n.!Feb.-2006)
i) Compute K so that the control law 11 = - Kx + ru), r(I) = reference input, places the closed loop poles at -2 ± jM,-5. (81 ii) Design an observer such that tne eigen values of the observer are located al -2±j./f2,-s. (61 iii) Draw a block diagram implementation of the control conjiguration. (JI iv) Obtain the stale model of the observer based stale feed back control system. (31
Modem Control Theory
Solution:
x
y A
Pole Placement Technique
5 ·41
0 0
[~ ~] [r ~]; (0 0 1) x
-.
0 0
B
C
= [~}
=
[O 0 1]
i) Let us first check the controllability matrix of the system the controllability matrix for n 3 is given by,
=
o, B
[B:AB:A2B]
0
m;
[~ ~][~] m [r ~][~ ~]= [~
AB=
0
0 0
A·A
0 0
-6 -11
-6
:i 25
-6 -11
[~ EJ[~l = m [~ ~] = = -6
0 1
IOcl
t(t)
l;t
o
0
Rank of Q, = 3 The system is completely state controllable as the rank of controllability matrix QC: n: 3. Now the desired characteristics equation is, [s-(-5)][s-(-2+jJ12)][
s-( -2-;./U))
(s+5)((s+2)-j"12) (s+5)[(s+2)2+12] (s+5)[s2
+4s+16]
[ (s+2)+
iJU)
5-46
Modem Control Theory
Pole Placement Technique
[~ ~s][~] [~sl [~ ;:] 0
A2B =
-6
=
30 19
1
19
J
=
1co-1i
0
QC Rank of
Oc
.I -5
=
IQd
;
= -1 ... o
3
The system is completely state controllable as rank of matrix Oc = n. Hence it is possible to place the poles arbitrarily. Let us now find the stale feedback gain matrix by use of transformation matrix T.
The characteristics equation for the system is,
s ls(s+5)+ 6 )-(-1 )(O] s(s2 +5s+6]
= s3 +5s2 +6s
Comparing the above equation with s3+a1s2 a1
=5
;
a2
=6
;
a3
+a 2s+a
3
=0
=0
The desired characteristic equation is,
[s-(-1 +ii)] [ s-(-1-jl)] [s-(-5)]
[(s+ 1 )-jl )[(s + 1) + jl J[s+5]
[Cs+1)2 +1J(s+SJ (s2 +2s+2)(s+5)
o3 we have o1 = 7; o2 = 12; o3 = 10 K = (1>3 -a 3 1>2 -a 2 61 -a iJ T-1 state equation is given in controllable form T = I T"1 = I
Comparing above equation with s3+ ots2 + o2s +
As the given
= s3 +7s2 +12s+!O
Modem Control Theory
K
5 -47
=
Pole Placement Technique
[!0-0 12-6 7-5]1 [10 6
21[~
01 OJ 0 = [10 6 2)
0 0 I
The state feedback gain matrix, K = ,_.
Example 5.8 :
Y(.-)
u'(,:)
Co11,;ider a
[to
6 2]
linear system described by
tlie
transfer function
lO Design a feedback controller roitlt a state fudback so I/tat closed loop = >(>+l)(s+ 2)
poles are placed at - Z, - 1 ± j.
(VTU : July/Aug.-2006)
Solution : The transfer furlction is given as ,
Y(s) U(s)
10 s(s+l)(s+2)
10
s(sz
+3s+2)
10 s2+3s2+2s
The denominator is decomposed as below, s3 +3s2 +2s = ((s+3)s+2]s Let us simulate this denominator
o(t)
Fig. 5.9 The complete slate diagram is as shown in the following Fig. 5.10 after simulating the numerator by shifting take off point.
y(t)
Fig. 5.10
Modem Control Theory
Pole Placement Technique
5-49
Now the desired characteristics equation is, ls-(-2)][ s-(-1 + jl)] [ s-(-1
-jl)]
(s+2} (s+l +jl) (s+l-jl) [s+2J[(s+1}2
+1]
s3 +2s2 +2s+2s2
= (s+2)(s2
+2s
•2)
+4s+4
s3 +4s2 +6s+4 Comparing above equation with s3+ o1s2 + o2s + &3 we have, &1 = 4;
o2
o3
= 6;
=4
Let us now find state feedback gain matrix by direct substitution desired state feedback gain matrix K be, K
Consider,
Isl
-A+ BK!
(K1
method. Let the
K3)
Ki
I
I[~ ~]-[~ -2 I3J+m[Ki K2 K3) I[~ ~]-[~ I3]+[11 JJI 0
0
1·
0
l[il
0
1
0
0
0 -2
Kz
2:~
0
s+f:KJ
2
s[ s( s+ 3+ K3 )-(-1)(2+K2)+1
[Ki]]
Equating coefficients of above equation with those from desired equation, 3 + K3
=4;
K3
=1
2 + Kz
., State feedback gain matrix,
=6 K
.'. Kz = 4
= [4
4 1]
K1
=4
Modem Control Theory
,,_.
Example
A
-1
"lr 1
2
J
Polo Placement
Consider tire 5ystem described l>y Ilic state model x = Ax. Y
5.9:
Jl
5-50
A
(~l
~l
Let us check the controllability
o,
[c'
C=[l
OJ
matrix of dual of given system for n
A'C°l
[~I ~][~]=[~I]
·A"C'
Rank of Qc = 2 = Number of state variables The dual system is completely state controllable :.Original system is observable.
=
[~cl]
be observer gain matrix
<2
The desired characteristics equation is, [s-(-5)Jfs-(-5)[
Cr where
(VTU:July/Aug.·2006)
Solution:
K0
=
0 I Design a full order stat( oltserver. The tlesi"·d eigL'>I talue» for the
C = [l
obseroer matrix arc 111 = -5, u2 = -5.
Let
Technique
= (s+5)(s+5)
= s2 +10s+25
Consider,
![s+ 1 + K,.1
-1
-l+Kc2
s+2
J
]I
=2
5. 51
Modom Control Theory
Polo Placoment Technique
( s+ 1+ Kc1 )(s+2)-(-I ){-1 + K,.2) ( s+I + Kcl )(s+2)+1
(-t
+ K02}
s2 +s+K01s+2s+2+2K01-l
+K02
s2 +(l+K,1 +2)s+(1+2Kc1+Kc2) s2 +(J+Kc1)s+(l
+2K01 +Kci)
Comparing the coefficients in above equation with those of desired characteristic equation, 3+Kd l + 2K,.1 + Kc2
10;
K01 =7
25;
K,,i=25-t-2K"'=24-2(7)"1(l
The observer gain matrix K0 is given as,
l
Example 5.10 : 0 A= 0 [ -]
l 0 -5
0 I -6
Consider I/re systtm drfined by
.=
.r
Ax + 811, toher«
Ol
J By 11si11g statt feedback control
/l = ~
[
=
lltt closed loop poft<s at s =-2±j4 ands "K" lJy any one method.
11
= -Kt, it ts desired to hat'I'
-10. Determin« lilt stale fe£dbackgoi11 matrix (VTU : J~n.fFeb.· 2008)
Solution : We have
Let us check the controllability matrix of the system. Herc n = 3. The controllability matrix • Q< =
B =
[0101:
All
(n:
All : A 21l]
=[~
~5
-1
lj
I A2 = A ·A
=
(~1
0
-s
:][~]=[:] 1
:J[~t q 0
-5 -6
n =[~I 6
-5
I ] -6
29
.}t
Modern
Control
5.52
Theory
0 l 1 -6 -6 31
l
Pole Placement Technique
:.IQcl
= 1[31-(36)]=
-5 ~
o
Rank of matrix Qc is 3. The system is completely state controllable as rank of
matrix
Oc = n.
Let us now find state feedback gain matrix by direct substitution method. The desired characteristic equation is,
f s-(-2 + j·l)]f s-(-2 -j4) J ls-(-10)]
= [(s + 2)-i4 J[(s + 2)+
[(s+2)2 +16] [s+IO}
i4]Ls + 10]
= (s2 +4s+20J[s+10)
s3 +14s2 +60s+200
s,
= 14,
02 = 60,
03= 200 K2 K3j
K = [ K1
let the desired feedback gain matrix K be Consider.
Jsl-A+BKj
I[~ ~]-[~ 0
0 s
I[~ ~]-[
:J+m lK1 0
1
0
0 s
I
0 -1 -5
~ ]+[K,~
0 -1 -5 -6
~
0
K2
-1
l[l+~KI 5+JS
s+6~KJI
s[s(
s+ 6+ K3) +(S+ JS)]+
s[s2
+(6+K3)s+(5+K2)]+(1+K1)
1 (1
+ K1)
s3 +(6+K3)s2 +(S+JS )s+( l+
K1)
Kz
K3)1
JJI
5 .57
Modem Control Theory
Ke
=
Pole Placement Technique
«•>[c~'r m [74 ~] [~ ~im [74 ~J m [~1l -18 -42
25 41 -95
0 0
25
K., = -18
-42
41
-95
This is the required observer feedback gain matrix by Ackermann's formula.
Review Questions J. Whal do you 17U'an be pole p/aremrnt probkm? 2. Prow the necessarya11d s11fflcinil rondilionfor arbitrary polt plactmtnl? 3 Write 5horl note of <1JOluation of stalef«dbiick gain matrix. 4. What art' the diffemtt tnrlhods of <1JOluating stolt fttdbackgain matrix? Explain any ont mtlhcd in dttnil. 5. lks
7. What do you mun by state obserwrs? 8. Write a note on full onler slate obstrwr? 9. What i• dual probltm? IO. What is nt«SS11ry and suffidtnl conditionforstate observation.
11. Wh41 art different mtlhods of <1JOluation of sl#lt
obstn,'l'f
goin rnalrix K, ?
12. Write• note on stltction of suiloble wlut of ~r g•in mmrix K., 13. Wrltt n 5hort nott 011 obstrotr bllstd controller. 14. Derite 11~ transfer [unction of observer hosed controlltr. 15. Con.sid" 11.. systtm defined hy x A
=
Ax+Bu 0
1
0
0
~l
[ -1 -5 -6
r
By using the statt fudbnck control u = -Kx, it is tksirtd to havt I/it dosed loop poles at s = -2 i j4 ands = -10. Dettrmine the state feedback gai11 matrix K. I Ans.: K = [ 199 SS 8 )J
5. 59
Modem Control Theory
Pole PlacementTechnique
22. Consider II.- systom r<pr'5ented by
x
y
r~ :l\Tm" -~l
= [1
0
o] .r
Dt.
000
Controllers 6.1 Background The concept of a control system is to sense deviation of the output from the desired value and correct it. till the desired output is achieved. The deviation of the actual output form its desired value is called an error. The measurement of error is possible because of feedback. The feedback allows us to compare the actual output with its desired value to generate the error. The error is denoted as e(t). The desired value of the output is also called reference input or a set point. The error obtained is required to be analysed to take the proper corrective action. The controller is an clement which accepts the error in some form and decides the proper corrective action. The output of the controller is then applied to the process or final control clement. This brings the output back to its desired set point value. The controller is the heart of a control system. The accuracy of the entire system depends on how sensitive is the controller to the error detected and how it is manipulating such an error. The controller has it' own logic lo handle the error. Now a days for better accuracy, the digital controllers such as microprocessors, microcontrollers, computers are used. Such controllers execute certain algorithm to calculate the manipulating signal. This chapter explains the basic discontinuous controllers such as on-off controller, continuous controllers such as proportional, integral etc. and composite controllers such "5 proportional plus integral, proportional plus derivative etc. Let us study first the general properties of the controller.
6.2 Properties of Controller Consider a control system shown in the Fig. 6.1 which includes a controller. The actual output is sensed by a sensor and converted to a proper feedback signal b(t) using a feedback clement. The set point value is the reference input r(t). For example the actual output variable may be temperature but using the thermocouple as the feedback element, the feedback signal b(t) is an electrical voltage. This is then compared with reference input which i~ also an electrical voltage. The thermocouple senses the output temperature and produces the corresponding electrical e.m.I as the feedback signal. Hence actual output variable sensed and the feedback signal may be having different forms. (6 • 1)
Modem Control Theory
8·2 Controller output (p)
•
l
ett)
r(I)
Controllers
Controller
Process lo be conlrotled
Controlledoutpul ~
Errot detector
c(t)
Sensing
'--/-~------! Feedback element 1-------' Feedback signal
Fig. 6.1 Basic control systam
6.2.1 Error The error detector compares the feedback signal b(t) with the reference input r(t) to generate an error. e{t)
=
r(t) - b(t)
This gives an absolute indication of an error.
For example if the set point for a range of 5 mV to 20 mV is 12 mV and the feedback signal is 11.8 mV then error is 0.2 mV. But actual vairable to be controlled may be different such as temperature, pressure etc. Hence to obtain correct information from the error, it is expressed in percentage forrn related to the controller operation. It is expressed as the percentage of the measured variable range. The range of the measured variable b(t) is also called span. span = b""" -bnun
Thus
Hence error can be expressed as percent of span as,
r-b er
cP = error as % of span
Where ••
b,ou. -bmin x 100
Example 6.1 : The rattgt of measured variable for a certoi11 co11trol system is 2 m V to 12 mV and a sttpoinl of 7 mV. Firrd the error as percent of span wlu"rl tire measured tariabl« is 6.5 m V.
Solution :
1.>11,,.,
= 12 mV, el' =
bmin
b
= 2 mV,
r-~ m.u: -
5%
xlOO min
b = 6.5 mV, 7-6.5 100 12-2 x
r = 1 mV
Modem Control Theory
6-3
Controllers
6.2.2 Variable Range In practical systems, the controlled variable has a range of values within which the control is required to be maintained. This range is specified as the maximum and minimum values allowed for the controlled variable. It can be specified as some nominal values and plus-minus tolerance allowed about this value. Sur.h range is important for the design of contollers.
6.2.3 Controller Output Range Similar to the controlled variable, a range is associated with a controller output variable. It is also specified intcrms of the maximum and minimum values. But oltcn the controller output is expressed as a percentage where minimum controller output is 0% and maximum controller output is 100 %. But 0 % controller output docs not mean, zero output. For example it is necessary requirement or the system that a steam flow corresponding to ,V.1h opening or the valves should be mininum. Thus 0% controller output in such case corresponds to the
}'.;th
opening of the valve.
The controller output as a percent of full scale when the output changes within the specified range is expressed as,
p = Where
u-umin
xlOO
um4x -umin
p
controller output as a percent of full scale
u
value of the output
u •"-'x
maximum value of controlling variable
u min
minimum value of controlling variable
6.2.4 Control Lag The control system can have a lag associated with it. The control lag is the time required by the process and controller loop to make the necessary changes to obtain the output at Its setpoinL The control lag must be compared with the process lag while designing the controllers. For example in a process a valve is required to be open or closed for controlling the output variable. Physically the action of opening or closing of the valve is very slow and is the part of the process lag. Jn such a case there is no point in designing a fast controller than the process lag.
6.2.5 Dead Zone Many a times a dead zone is associated with a process control loop. The time corresponding to dead zone is called dead time. The time elapsed between the instant when error occurs and the instant when first corrective action occurs is called dead time.
Modem Control Theory
Controllers
Nothing happens in the system, during this time though the error occurs. This part is also called dead band. The effect of such dead time must be considered while the design of the controllers.
6.3 Classification of Controllers The classification of the controllers is based on the response of the controller and mode of operation of the controller. For example in a simple temperature control of a room, the heater is to be controlled. It should be switched on or off by the controller when temperature crosses its setpoint. Such an operation of the controller is called discontinuous operation nnd the mode of operation is called disconti.nuous mode of controller. But in some process control systems, simple on/off decision is not sufficient. For example, controlling the steam flow by opening or closing the valve. In such case a smooth opening or closing of valve is necessary. The controller in such a case is said to be operating in a continuous mode. Thus the controllers are basically classified as discontinuous controllers and continuous controllers. The discontinuous mode controllers arc further classified as ON-OFF controllers and multiposition controllers. The continuous mode controllers arc further classified as proportional controllers, integral controllers and derivative controllers. Some continuous mode controllers can be combined to obtain composite controller mode. The examples of such composite controllers are Pl, PD and PIO controllers. The most of the controllers arc placed in the forward path of control system. But in some cases, input to the controller is controlled through a feedback path. The example of such a controller is rate feedback controller. In this chapter, only continuous and composite controllers are discussed.
6.4 Continuous Controller Modes In the discontinuous controller mode, the output of th" controller is discontinuous and not smoothly varying. But in the continuous controller mode, the controller output varies smoothly proportional to the error or proportional to some form of the error. Depending upon which form of the error is used as the input to the controller to produce the continuous controller output, these controllers arc classified as, L Proportional control mode 2. Integral control mode 3. Derivative control mode Let. us discuss these control modes in detail.
6-5
Modem Control Theory
Controllers
6.5 Proportional Control Mode In this control mode, the output of the controller is simple proportional to the error e(t). The relation between the error e(t) and the controller output p is determined by constant called proportional gain constant denoted as KP. The output of the controller is a linear function of the error e(t). Thus each value of the error has a unique value of the controller output. The range of the error which covers 0 % to 100 % controller output is called proportional band. Now though there exists linear relation between controller output and the error, for a zero error the controller output should not be zero, otherwise the process will come to halt. Hence there exists some controller output Po for the zero error. Hence mathematically the proportional control mode is expressed as, ... (!)
Proportional gain constant
Where Po
Controller output with zero error
The direct and reverse action is possible in the proportional control mode. The error may be positive or negative because error is r-b and b can be less or greater than reference setpoint r. If the controlled variable i.e. input to the controller increases, causing increase in the controller output, the action is called direct action. For example the output valve is to be controlled to maintain the liquid level in a tank. So if the level increases, the valve should be opened more to maintain the level. If the controlled variable decreases, causing increase in the controller output or increase in the controlled variable, causing decrease in the controller output, the action is called reverse action. For example simple heater control for maintaining temperature. If the temperature increases, the drive to the heater must be decreased.
So if e(t) is negative then Kpe(t) gets subtracted from p0 and c(t) is positive, then JS,e(t) gets added to Po- this Is reverse action. So equation (!) represents the reverse action. But using negative sign to the correction term Kpe(I), the direct action proportionnl controller can be achieved. The proportional controller depends on the proper design of the gain K.,. For fixed Peif gain KP is high then large output results for small error but narrow error band can be handled. Beyond these limit' of the error, output will be saturated. If the gain is small then the response is smaller but large error band can be handled. This is shown in the Fig. 6.2.
Conlrollcr
output
High gmn
100%t--------~~~~~~--· 50(14
0 ...
I
f
---
-· .
'
----~---..J----------
__ ..
-
Controllers
6-6
Modem Control Theory
I
'
1 I I I I
' 'I 0 '
Po= 50 %
--------r---1---- ·----,__
,
Q
'
Narrow
~
I,..
. !
t,.
Error%
error band
Wider error band
Fig. 6.2 The proportional band is mathematically defined by,
6.5.1 Characteristic of Proportional Mode The various characteristics of the proportional mode are, 1. When the error is zero, the controller output is constant equal to Po2. If the error occurs, then for every 1 % of error the correction of Kr % is achieved. If error is positive, KP% correction gets added to p0 and if error is negative, Kr% correction gets subtracted from Po· 3. The band of error exists for which the output of the controller is between 0 % to 100 % without saturation. 4. The gain KP and the error band PB are inversely proportional to each other.
6.5.2 Offset The major disadvantage of the proportional control mode is that it produces an offset error in the output. When the load changes, the output deviates from the setpoinl Such a deviation is called offset error or steady state error. Such an offset error is shown in the Fig. 6.3. The offset error depends upon the reaction rate of the controller. Slow reaction rate produces small offset error while fast reaction rate produces large offset error.
Modem Control Theory
6·7
Controllers
Oull)Ul
Fig. 6.3 Offset error in porportlonal mode The dead time or transfer lag present in the system further worsens the result. It produces not only the large offset at the output but the time required to achieve steady state is also large. The offset error can be minimized by the large proportional gain KP which reduces the proportional band. If KP is made very large, the proportional band becomes so small that it acts as an ON/OFF controller producing oscillations about the setpoint instead of an offset error.
6.5.3 Applications The proportional controller can be suitable where, 1. Manual reset of the operating point is possible. 2. Load changes arc small. 3. The dead time exists in the system is small.
6.6 Integral Control Mode In the proportional produces an offset error. another control mode is of the errors. This mode
control mode, error reduces but can not go to zero. It finally It can not adapt with the changing load conditions. To avoid this, oftenly used in the control systems which is based on the history is called integral mode or reset action controller.
In such a controller, the value of the controller output p(t) is changed at a rate which is proportional to the actuating error signal e(t). Mathematically it is expressed as, d p(t) dr Where
K;
=
K e(I) I
: Constant relating error and rate
The constant K, iq also called Integral constant. Integrating the above equation, actual controller output at any time t can be obtained as,
Modem
Control Theory
6•8
K, j c{t)dt.;. p(O)
p(t) Where
Controllers
... (2)
p(O) = Controller output when integral action starts i.e. at t = 0.
The output signal from the controller, at any instant is the area under the actuating error signal curve up to that instant. If the value of the error is doubled, the value of p(t) varies twice as fast i.c. rate of the controller output change also doubles. If the error is zero. the controller output is not changed. The control signal p{t) can have nonzero value when the error signal <'{t) is zero. This is because the output depends on the history of the error and not on the instantaneous value of the error. This is shown in the Fig. 6.4. e(l)
p(t)
(a)Error
(bl Controller output
Fig. 6.4 Integral mode The scale factor or constant K, expresses the scaling between error and the controller output. Thus a large value of K1 means that a small error produces a large rate of change of p(t) and vicevcrsa. This is shown in the Fig. 6.5. If there is positive error, the controller output begins to ramp up.
(+)
dp(t) di
0 (-) -'-------''-------Error 0
Fig. 6.5
Modem Control Thoory
6·9
Controllers
6.6.1 Step Response of Integral Mode e(t)
The step response of the Integral control mode is shown in the Fig. 6.6.
A
The integration lime constant is the time taken for the output to change by an amount equal to the input error step. This is shown in the Fig. 6.6. p(t)
A
I.___,' r,
ll can be seen that when error is positive, the outpul p(t) ramps up. For zero error, there is no change in the output. And when error is negative, the output p(t) ramps down.
Fig. 6.6 Step response
6.6.2 Characteristics of Integral Mode The integrating controller is relatively slow controller. It changes its output at a rate which is dependent on the integrating time constant, until the error signal is cancelled. Compared lo the proporlional control, the integral control requires lime lo build up an appreciable output. However it continues lo act till the error signal disappears. This corrects the problem of the offset error in the proportional controller. For example, Jct us assume that the integral controller is used to control the armature current of a d.c. motor and to keep its value constant ot 500 A. As long as the armature current is less U10n 500 A, the armature voltage, controlled by the controller, will increase. TI1us the output of the controller will increase and will continue to do so till the error becomes zero i.e. armature current becomes 500 A. Then the controller output will remain at that value reached. This is possible because the output of the controller can remain al any value within its r.mr,c. if the input is zero. 'The controller must not be overdriven as it will not then be effective. Thus for an iJllC!,'Tlll mode, I. If error is zero, the output remains at a fixed value equal to what is was, when the error became ·~ro. 2. If the error is not zero, then the output begins to increase or decrease, at a rate K; % per second for every ± l % of error. In some cases, the inverse of K, called integral time is specified, denoted as 1;.
Modem Control Theory
Controllers
6 -10
Ti : ..!....
Ki
:
Integral time
It is expressed in minutes instead of seconds.
6.6.3 Applications The comparison of proportional and integral mode behaviour at the time of occurrence of an error signal is tabulated below, Controller
Initial behaviour
p
Acts immediately. Action according to Kr.
Ofbet error always present. Larger the KP smaller lhe error.
I
Acts slowly. It Is the time integral of the error signal.
Error signal always
Steady state behaviour
becomes zero.
Table 6.1 It can be S(.'Cn that proportional mode is more favourable at the start while the integral is better for steady state response. In pure integral mode, error can oscillate about zero and can be cyclic. Hence in practice integral mode is never used alone but combined with the proportional mode, to enjoy the advantages of both the modes.
6.7 Derivative Control Mode In practice the error is Junction of time and at a particular instant it can be zero. But it may not remain zero forever after that instant. Hence some action is required corresponding to the rate at which the error is changing. Such a controller is called derivative controller. In this mode, the output of the controller depends on the time rate of change of the actual errors. Hence it is also called rate action mode or anticipatory action mode. The mathematical equation for the mode is. p(t) Where
K d c(t) J di Derivative gain constant.
The derivative gain constant indicates by how much % the controller output must change for every % per sec rate of change of the error. Generally Kd is expressed in minutes. The important feature of this type of control mode i.~ that for a given rate of change of error signal, there is a unique value of the controller output.
Modem Control Theory
Controllers
6·11
The advantage of the derivative control action is that ii responds to the rate of change of error and can produce the significant correction before the magnitude of the actuating error becomes too large. Derivative control thus anticipates the actuating error, initiates an early corrective action and tends to increase stability of the system improving the transient response.
6.7.1 Characteristics of Derivative Control Mode The Fig. 6.7 shows how derivative mode changes the controller output for the various rates of change of the error. e(t)
Error
0
Timet
p(%)
100% Conltoller output 50%
0%
Timet
Fig. 6.7 The controller output is SO% for the zero error. When error starts increasing, the controller output suddenly jumps to the higher value. It further jumps to a higher value for higher rate of increase of error. Then error becomes constant, the output returns to 50 %. When error is decreasing i.e. having negative slope, controller output decreases suddenly to a lower value.
Modem Control Theory
6-12
Controllers
The various characteristics of the derivative mode arc, I. For a given rate of change of error signal, there is a unique value of the controller output. 2. When the error is zero, the controller output is zero. 3. When the error is constant i.e. rate of change of error is zero, the controller output is zero. 4. When error is changing, the controller output changes by Kd % for even I % per second rate of change of error.
6.7.2 Applications When the error is zero or a constant, the derivative controller output is zero. Hence it is never used alone. Its gain should be small because faster rate of change of error can cause very large sudden change of controller output. This may lead to the instability of the system.
6.8 Composite Control Mo.des As mentioned earlier, due to offset error proportional mode is not used alone. Similarly integral and derivative modes are also not used individually in practice. Thus to tnke the advantages of various modes together, the composite control modes are used. TI1e various composite control modes are, I. Proportional + Integral mode (Pf) 2. Proportional + Derivative mode (PD) 3. Proportional + Integral + Derivative mode (PIO) let us sec U1e characteristics of these three modes.
6.9 Proportional + Integral Mode (Pl Control Mode) This is a composite control mode obtained by combining the proportional mode and the integral mode. The mathematical expression for such a composite control is, p(t)
=
I
Kr e(t)+KpK;J
e(t)dt+p(O) 0
Where
p(O)
Initial value of the output at t = 0
Modem Control Theory
Controllers
6 -13
The important advantage of this control is that one to one correspondence of proportional mode is available while the offset gets eliminated due to integral mode, the integral part of such a composite control provides a reset of the zero error output after a load change occurs. Consider the load change occurring at t.., t 1 and due to which error varies as shown in the Fig. 6.8. The controller output changes suddenly by amount VP due to the proportional action. After that the controller output changes linearly with respect to time at a rate KP I T1• The reset rate is defined as the reciprocal of T;.
e(I)
Error
p(I) Conlrollor output ------~--------' :
--- }------J: V1
} VP _ _.'--------
0
ti
1,
-----
--· .
Riso in output duo to intcg1et action between limes 11 and ti Immediate rise due to prol)0'1ionul action Tome
'---v--' Integral acton time
Fig. 6.8 Behaviour of Pl controller The response shown in the Fig. 6.8 is for the direct action of the controller. The response of composite Pl control mode for the reverse action is shown in the Fig. 6.9. In the reverse action, the proportional part is the image of the error. The sum of proportional plus Integral action finally leaves the error to zero.
Modem Control Theory
Controllers
6 -15
6.9.2 Applications The composite Pl mode completely removes the offset problems of proportional mode. Such a mode can be used in the systems with the frequent or large load changes. But the process must have relatively slow changes in the load, to prevent the oscillations.
6.10 Proportional + Derivative Mode (PD Control Mode) The series combination of proportional and derivative control modes gives proportional plus derivative control mode. The mathematical expression for the PD composite control is,
The behaviour of such a PD control to a romp type of the input is shown in the Fig. 6.10. Error e(I)
Ramp
lime
0 P(I) Controller output
:}v;-- --- -- ----. ------ fv;·--~--------------·
.
I
..
Rlso in signal due to proportional action between Um°' t1 :ind
tz
Immediate rtse due to derivative action
I
... 0t----..t.t1---, 2--------
Time
Fig. 6.10 Bohavlour of PD controllor The ramp function of error occurs at t = t 1. The derivative mode causes a step Vd at t 1 and proportional mode causes a rise of Vr equal lo Vd at t 2• This is for direct action PD control.
Modem Control Theory
Controllers
6 -19 I
0.02(3t+7j+0.02x0.04
J
(3t+7)dt+0.5
()
(0.06t]+0.14+8xl0-4[
3:2
+7{
+0.5
(0.06xl.5]+ (0.14}+ Bx 10-4[~+(1.5)2+7x1.5 0.09 + 0.14 + 0.0111 + 0.5 Le.
u•
=
]+ 0.5
0.7411
74.11 % after t = 1.5 minutes.
p(t)
Example 6.3: Suppose the error shown i11 the Fig. 6.14 is applitd lo a PD controller with KP = 6 and K,1 = 0.4 sec with p(O) = 25 %. Dr11w the graph of the controller output. e(t){%)
Fig. 6.14
Solutlon : The output of the PD controller is p(t)
=
d c(t) KP e(t)+K.,K,1 (i!+25%
It is necessary to obtain output response, dividing c(t) into Uu~'C sections, Section I)
0 e(t)
Cit p1(l)
s
t
s2
2t
... slope= 2
2
6x2t+6x0.4x2+25
= 12t + 29.8
Thus there is an instantaneous change of 29.8 - 25 ,. 4.8 % produced by this error in lhe output at t = 0.
Modem Control Theory
Controllers
6 -23
i
J\K
+
p
= 0 for the step
input
Lim s G(s)H(s) = ~;
• -o
2<;
e,.., = So there is a finite error for the ramp input of magnitude A, which depends on ~and For good time response, the system demands. 1. less settling time 2. Less overshoot 3. Less rise time 4. Smallest steady stale error. Increasing Kv, the steady state error for the ramp input can be reduced but it increases overshoot and settling time. This m:1y lead to instability of the system. So to keep steady state error and overshoot well within the acceptable limits, the various composite controllers arc used. Let us see the effect of PD, Pl, PID and rate feedback controller on the time rl'Sponsc of the second order system under consideration.
6.14 PD Type of Controller A controller in the forward path, which changes the controller output corresponding to proportional plus derivative ol error sit,'1lal is called PD controller.
i.e.
Output of controller Taking Laplace
=
= K e(t) + Td
K E(s) + ~TJ E(s)
=
de(l)
di
E(s) [ K + s1'0)
The T.F. of such controller is [K + sT,iJ. This can be realized as shown in the Fig. 6.17.
--"'-"-s(s + ~••nl
Ptanl
Fig. 6.17
Assuming K
Controllers
6·24
Modem Control Theory
= 1, we can write, (I .. sT,1)ro~ s(s+ 21',oo,,)
G(s)
C(s)
and
R(s)
Comparing denominator with standard form, oo., L~ same as in the previous P type controller. 2 ( Oln
and
=
21; oo,,
.. ~
Td
( = <, .. oon2Td Because of this controller, damping ratio increases by factor Lim
KP
= s _.... O
K,
= 5 -.a" s G(s)H(s) = 2~
G(s)H(s)
ro T -¥·
= ..
Lim
oo
As there is no change in coefficients, error also will remain same. Key Point: Hence PD co11troller has folluwing effects on system. i)
ii)
It increases damping ratio.
·oo,,'
for system remains unchanged.
ill) 'TYPE' of the system remains unchanged. iv) It reduces peak overshoot. v)
It reduces settling time.
vi) Steady state error remains unchanged. Key Point : In generRI P.O. controller improves transient part unihout affecting steady
state. 6.15 Pl Type of Controller A controller in the forward path, which changes the controller output corresponding to the proportional plus integral of the error signal is called Pl eontroller.
Modem Control Theory
I
Le,
6-25 K e(t) + K;
Output of controller
Taking Laplace
= K E(s) + ~
J
Controllers c(t) dt
~i]
E(s) = E(s) [ K+
:. The T.F. of such controller is [ K + ~i
)
and can be realised as shown in the Fig. 6.18.
__.,,,__ 2
s (s + 2~w.) Plant
Fig. 6.18 Assuming K = 1, we can write,
~~
G(s)
s(s + 2~0),,) (K, +sl%
s2 (s+ 2~ro0) i.e, system becomes TYPE 2 in nature. and
C(s) R(s)
(K, + s) ro~
=
s 3 + 2~ Cl),, s 2 + s
% + K; w~
i.e. it becomes third order. Now as order increases by one, system relatively becomes less stable as K; must be designed in such a way that system will remain in stable condition. Second order system is always stable. Key Point:
Hence lransitnl rtsponse gets a/Jtcled badly
if
controller is not dtsig11£d
properly. Lim
While
..... 5
K, =
0
G(s)H(s) = ...
c.,. = 0
Lim ..... s G(s)H(s) = - , c.,. = O 0
5
Key Point : Hence as type is increased by ont, error becomes zero for ram.~ t.VPe of inputs i.e. steady state of system gets improvtdand system bteomes more accurate in nature.
Modern
Control
6 -26
Theory
Controllers
Hence l'I controller has following effects : i) It increases order of the system. ii) It mcrcascs
TYPE of the system.
iii) Design of K, must be proper to maintain stability of system. So it makes system relatively less stable. iv) Steady state error reduces tremendously for same type of inputs. Koy Point: /11 gcr:ernl this controller improves steady state part ajfedi11g tlte transisnt part.
6.16 PIO Type of Controller As PD improves transien! and Pl improves steady state, combination of two may be used to improve overall time response of the system. This can be realized as shown in the Fig. 6.19(a).
Plant and
Contror.e<
Fig. 6.19(a) TI1e design of such controller is complicated in practice.
6.17 Rate Feedback Controller (Output Derivative Controller) This is achieved by fct.'Cling back the dertvativc of output signal internally using a tachogcnerator and comparing with slgnat proportional to error as shown. This is called minor loop feedback compcns.ition. Output of controller Output of the controller This can be realized as shown in the Fig. 6.19(b).
=
. dc(t) K l:(t) - K1 -dt
= K F.(s) -
sK1 C(s) E(sl
KE(s)
Assuming K = 1, lei us study its effect on same system which is considered earlier >
with G(s) = - --~~-- -·. >(> + 2.,
Fig. 6.19 (b)
Modern Control Theory
Controllers
6 ·30
Key Point: 1 Time co11sta11t 'T' is tire time required by tire system output to reach 63.2 of its final value d11ri11g the first attempt.
%
The equation for the actual response c(t) is,
... For unit step
Where
=
(l)d
(l)n J1 -1:,2 Damped frequency of oscillations tan- 1
a
and
{Ji
~1:,2 } radians.
Note : As ~ is Ute function of I:, alone, if ~ is known, the equation of ~ solved to obtain the corresponding value of damping ratio I:,.
can be
6.19 Steady State Error Consider a simple closed loop system using negative feedback as shown in the Fig. 6.21. C(s)
Fig. 6.21 Where
E(s) = Error signal,
and
Now,
E(s)
But
B(s)
C(s) · H(s)
R(s) - B(s)
E(s)
R(s) - C(s)H(s)
and
C(s)
E(s) · G(s)
E(s)
R(s) - E(s)G(s)H(s)
E(s) + E(s)G(s)H(s)
=
R(s)
B(s)
Feedback signal
6. 31
Modem Control Thoory R(s)
E(s)
E(s)
Controllers
1 + C(s)H(s) for nonunity feedback
=
R(s)
for unity feedback
I+ C(s)
This E(s) is the error in Laplace domain and is expression in "s'. We want to calculate the error value. In time domain, corresponding error will be e(t). Now steady state of the system is that state which remains as t ~ ... Steady state error, e..
. . ..
= Um L'(t)
Now we can relate this in Laplace domain by using final value theorem which states that,
,Lim ...
F(t)
Where F(s) = L I f(t) I
Lim s F(s)
s-+O
Lim e(t) = Lim s E(s)
Therefore,
I
-t•
Where E{s) is L I c(t) I
s-10
Substituting E(s) from the expression derived, we can write = Lim e,_
sR(s)
• _, o l + G{s)H(s)
For negative feedback systems use positive sign in denominator while use negative sign in denominator if system uses positive feedback. From the above expression it can be concluded that steady state error depends on, i) R{s) i.e. reference input, its type and magnitude. ii) C(s)H(s) i.e. open loop transfer furu:tion. iii) Dominant nonlinearities present if any.
Examples with Solutions ••
Example 6.4: Thefigure slurnrs PD controller used for the system. Determine the value of T4 so tltat system will be critically damped. Calculate its settling timt.
... , Solution:
G(s) = (I.- sTd)4 s(s+ 1.6) Cts) R(s)
Controllers
6. 32
Modern Control Theory
11( ) · s
I
=
(l.,. sTd)4 s (s + 1.6)
(1+s'fd)4
l+(l+sTd)4 s(s + 1.6)
s2+1.65+4'Tds+4
Comparing denominator with standard form, w~
4,
ron
=2
and
2!'; (!)11
=
1.6 + 4 Td
C. = 1.6+ 4Td • 4 Now system required is critically damped, i.e. E, = 1 1.6+ 4 Td --4-4
1.6+ 4T0 . . an d sett I mg time
0.6 T, ,._.
4
2X1=2
=
4
E,ro.,
sec.
Example 6.5: A 1111ity feedbacksystem is show11 in tirefollowingfigure. i) In tile abse11a of derivative feedback controller (K0 if E, is lo be modified lo 0.5 by use of controller.
= 0). Find E,
and ron' ii) Find K0,
C{s)
Solution:
G(s)
ior Ko = 0
50x-s(s+1- 3)
H(s)
C(s) R(s)
G(s) 1 + G(s)H(s)
50
:. w,.
_
50
- 82 + 3s + 50
= ./50 = 7.071 rad/sec.
Modern
,. ·Controne;s
6. 34
Control Theory
14
JM. = 3.7416
w,.
=
rad/sec.
14 K1.,. J.6 14K
+ l.o
~=0.5given 2x"H 0.1529
w,.
H
~
= 0.9695 sec.
= 3.7416 Ji -(0.5)2 =
3.2403 rad/sec
(l)d
e-•JP T,
11.+
4
i; w,,
=
x 100=16.3 'Yo
2.1381 sec.
Example 6.7: For tlu system shown determine % M1, a11d T,, when it is excited by 1111it ~tep i11p11t. If fbr the same system, I'D controller liaving constant T0 ~ 1/30 is used in fbr«Jard path, determine new values of damping ratio, MP andT,. Draw rtsptclive waveforms.
Solution : Without controller,
JOO
G(s) •
s(s+ 12)' H(s) =I
C(s)
100 s2 + 12s + 100
R(s) 0)2 n
100
:. Oln = 10 rad/sec.
~ = 0.6
12 OJ0
e
J1 - ~2
-·VJ·-~2 4
)
= 10x 0.8 = 8 rad/sec. = 9.47 'l'o
i;
1
I
•
6. 35
Modem Control Theory
Controllers
With controller 100
S('S'+i2i
G(s)
(1 +i )too
(s+ 30) x 3.33
s(s+ 12)
s(s+ 12)
H(s)
=1
(s+ 30) 3.33 C(s)
s(s + 12)
3.33(s ... 30)
3.33(s + 30)
R(s)
+ (s + 30) 3.33 1 s(s+ 12)
s2 + 12s + 3.33s + 100
s 2 + 15.33s + 100
100
:. ro.,= 10 rad/sec,
2sro.,
15.33
;
;:·3
:o
:. Sis improvcd,Old % MP
•ro,, =
= 0.7665
n
e-•VR
= 10J1 -(0.7665)2 xlOO
=
= 6.4224 rad/sec.
2.353 %
Overshoot decreased to 2.3 % from 9.47 %. T5 = ~
~
= 0.5218 sec.
Comparison : Following figure shows comparison between system with controller and system without controller. 0.47 %
t) 2.353%
Withoutcontroller
0.5218 sec With conttofler
tlmo
6-37
and
Controllers
2 ~
0.3162
This is the required damping ratio. For error calculation : C(s)H(s)
;
5
s (l + O.Ss)
For ramp input magnitude is unity i.e, A=I lim s G(s)H(s)3
•-o Cs~
lim -11 s.SO-:o-)= 5
.-us
+ ~•
=
Case bl Tbe derivative feedback is introduced in the system. The system becomes. Cts) =
K,, s [0.Ss + I + K1
and H(>)=l
I
1
~ 1+
C(sJ
$Ki
----
s ( 1•0.5s)
0.5s2 • s ( 1 + K1)
C(s) R(s)
KA s [0.5s + 1 + K1 I+
J
K,, slO.Ss+ I+ K1
K,..
J
0.5s2 +(1 + K1) HK,,
Modern
6 -40
Control Theory
=W
Ji -1;
2
Old
211
6.8676 rad I sec
Bui
w,, K,
and
0.4037
~ =
'"*
w:;
0
= 47.165
'.\+KP 2xro., 2.5449
Example 6.10 : An integral controller is used for tem,,....aturecontrol witlrin a range 40 - 60 •c. 17u: st/ point is 48 •c. Th« controller output is initially 12 % when error 15 uro. Tire intrgral constant K1 = - 0.2 % controlleroutput per ~nd per percent11geerror. If tire temperature increases to 54 •c, calcublte II~ controller output •fltr 2 s« for" constant error.
Solution : For integral controller,
... EP =error
p(I) p(O)
12 %,
K1
= - 0.2 % I
sec I "•error
Ep = Constnnt = b.,,.,.r-b 100 -bmm x Set point = 48 °C,
Now
60 EP JEl'dt
Al I= 2,
•c,
bmin = 40
48-54
--xlOO 60-40 ~ t
b
oe
= Actual
temperature
= 54 "C
= - 30%
as error is constant
p(t)
(- 0.2) (- 30) (t) + 12
p(t)
(0.2 x 30 x 2) + 12
= 24 %
This is the controller output after 2 sec.
111•
Example 6.11 ; For a proportional controller,tire amtrolltd variable is 11 tnnpemturc willr ra11gt of 50 to 130 'C with a st/ point of 73.5 •c. TM controlltr output is 50 % ftw uro error. The offset error is corresponding to a lead clumgt which causes 55 % c:mitrolkr 011tp11t. If tire proportional gain is 2, find tlit % control/tr output if thr ltmptral11rt is 11
61
-c.
Solutlun : For a proportional controller, p(t) Where
Kp 6r(t) + Po Kp
= 2,
Po
= 50 'l'o
Modem Control Theory
6-43
Controllers
After end of 2 sec, integral term gets accumulated to, P1(2) For 2 S t S 4, Ep2(t)
2xl.,f~t2]
1
+40=48.8% 1~2
-2.51+7
t
t
j Ep2(t)
dt
!(-25t+7)dt=[-2
2
+7{
- 1.25 (t2 - 4) + 7(t - 2) 2(-2.5t +7)+4.-l[-1.25(12
-4) +7(1-2)1+48.8
- St+ 14 - 5.5~ + 22 + 30.81 - 61.6 + 48.8 - 5.512 + 25.Bt + 23.2
for 2 S I S 4
... (2)
This is plotted for 2 - 4 sec. After end of 4 sec, the integral term gels accumulated to, P1(4) For 4 S t S 6, Ep3(1) !Ep:i(t)dt
-t.25(t2 -4)+7(t-2t%4 +48.8= 47.8% 1.51 - 9 = !(1.5t-9)dt
·;t
= [1
2
-911
0.75 (12 - 16) - 9 (t - 4) p3(I)
2[l .5t-9)+4.4[0.75(t2
-16)-9(1-4))+47.8
31 - 18 + 3.3t2 - 52.8 - 39.6t + 158.4 + 47.8 3.3t2 - 36.6 I + 135.4 This is plollcd for 4 - 6 sec. After 6 integral response of p3(t~1%6 34.6 %
=
S<.'C
for 4 S t S 6
... (3)
as error is zero hence output is accumulated
The graph of p(t) against time is shown in the Fig. 6.23. (Sec Fig. 6.23 on next Page) ,,_.
Example 6.14 : A proportional controlkr is employedfor the control of temperature i11 the range 50 •c - 130 •c •oith a set point of 73.5 •c. The zero error controller output is 50 %. Wlrat will be the offset mar re>uiting from a change in t/Je controller output to 55 % ? The proportionalgai11 is 2 % I %. Find tire offset in •c.
Solution : For proportional control mode
Modern Control Theory
6. 44
Controllers
Fig. 6.23
p Where
KpEP +Po
Po = Controller output with no error = 50 % 2%per%
p-po = 55-50 = l ..S % Kr 2 The offset error is 2.5 %.
= 130 "C and bmm = 50 "C.
The rani;c of 50 "C to 130 "C so b'"'" r-b
c = bnl,\\
2.5 b
xlOO
-blllHl
73.5-b 00 130_50x1 71•.s
·c
This is temperature corresponding to offset error.
Where r = set point
Modom Control
••
Example
Pi(OJ
6. 45
Thoory
Controllors
=
6.15 : A PID co11trnller lras K,. = 2.0. K1 2.2 scc"1• Ko= 2 sec and ~O %. Druw th« plot of controller 01111'111 for error of Fig. 6.24. ErmrE0k 4
2
0-'---+--_, _ _,.,,....
.,_
t Sec
-2 -4
Fig. 6.24 Solution : Kr= 2. K1 = 2.2
S«-1,
K0 = 2 sec, P1(0) = 40 %
from the i,.;vcn error plot, for O - 2 sec, Er= m1l
i;, For 2 - 4
St'C,
2-0
Where m1 =Slope=
2_0
=1
for 0 - 2 sec.
Ep
Two points on the line are (2, 2) and (4, -3). Yz-Y1
At (2, 2),
For 4 - 6 sec.
=-3-2,,_25 ·1-2
mi
x2 -XI
Ep
- 2.St + C2
2
,
- 2.5 x 2 + C2,
Ep
- 2.St + 7
EP
m3t + C3
for 2 - 4 sec.
Two points on the line arc (4. - 3) and (6, 0). 0-(-3) ~=+1.5 Ep At (4, -3),
-3
+l.5t+<; 1.5 x 4 + C3, ... 1.5 t -9
for 4 - 6 S« .
.: <; = 7
Modem Control Theory
Controllers
6 -46
Three mode equation for PID controller is,
dE
I
p = KrEp + KpKi} Epdl + KpKo -af + P1(0) 0
dE 2EP+4.4JEPdt+4-af+40 I
p
0
For 0 - 2 sec,
d
I
Pt
2t
+4.4J l di +4di(t)+40 0
... (!)
2t+2.2t2 +44
This is plotted for 0 - 2 sec. At the end of 2 sec, the integral term has accumulated to,
P1(2) = 4..J!tdl+40=4.4x[t;J:
For 2 - 4 sec,
+41l=48.8%
I d 2(-2.5t +7) +4.4J (-2.St +7) dt +4 di(-2.St+7)+
P2
48.8
2 &
-5t+l4+4.4[-1.25t2
+7t]~ -4x2.5+48.8
-5t + 14 +4.4{-t2S(t2
-4)+7(1-2)}-10
+48.8
-5.5t2+25.St+13.2
... (2)
This is plotted for 2 - 4 sec. At the end of 2 S<.'C, the integral term has accumulated to, P1(4) = 4.4{-1.25(12 -4)+7(t-2)}11".1+48.8
= 44.4 %
For 4 - 6 sec, I
p3
2(1.St-9)+.JAJ
I (1.St-9) dt +4Ft(l.5t-9)+44
..J
4
3t-18+4.4[075t2 3t-18+4.4{01s(12 3.3t2 -36.6t +138 This is plotted for 4 - 6 sec.
-9t)'.1 +.Jxl.5+44.4 -t6)-9(t-4)}+6+44.4 •.. (3)
Modern
Control Theory
6-47
Controllers
After 6 Sl'C, error is zero hence the output will simply be the accumulated integral response providing a constant output.
P1(6)
·l.4{075(t2
-16)-9(t-4)}1,.6
+~4.4
31.2 %
,,...
Example 6.16 : A temperature control system has tire block diagram given in Fig. 6.25. The input signal is a voltage and represents tire desired temperature 0, is a unit step
function and
i) D(s)
=1
ii) D{s)
= 1 + Pf
iii) D(s)
= 1 + 0.3 s. What is tire effect of
the integral term in the Pl controller a•rd the derivative term in PD controller on tire steady stale error ? (VTU: July/Aug.-2006, 8 Marks)
O(r)
~
200 Plant
Fig. 6.25
0
Controllers
6 ·48
Modem Control Theory
Solution : For the given system. G(sl =
200D(s) (s+ll(s ... 2)'
H() 002 s " · '
Lim sE(>)
Lim sl~s) ,.~o I +G(s)H(s)
1->.-,tl
i)
= 0.02 s
D(s} l.im ~-H 1 +
ii)
R(s)
D(s)
0.02 sx-s l()O x0.02 (s+l)(s+2)
0.02
;:20
&.66x10·3
2
I+ 0.1
s sx 0.02 s
Lim <-O
oo(i+¥Lom]
=0
2 1+
(s+l)(s+2) iii)
D(s)
1 + 0.3 s 0.02 sx--
l".,..
------"''----- = 6.66x10·3 :.-0 Lim
1+(200(1 +0.1s) x0.02) (>+l)(s~2)
Due to Pl controller, the steady state error reduces drastically while PD controller has no effect on the steady state error.
Review Questions l.
Whal is co11troller ? Explain its function i11 a systrrn.
2. Gill< /ht classification of controllers. 3. Stalt and txplain tlu: various J!TOJXrlks of controller. 4. State tlie various contin11ous controller mod.ts. 5. Explain tht prof}"'lional co11tro/ modr. State its d111racttristics. 6. Explain the integro!control mode, State its cbamctrristics. 7. Explain ti~ derivatiw control mode. State its tharocteristics. 8. Explain how constant K, •JJ
Modem Control Theory 11.
Controllers
6 -49
Explain the Pl control mod<, stating its characteristics.
(July·2005)
Explain the PD control mode, slatmg its chanwmslics. 13. Writ< a not« on tlzru moat control/er. 14. Discuss tht effrr:t of following controllers 011 th< second order con!rol systtm, al Pl controller bl PD <011trofler 12.
15. Explain 11" effrr:t of Pl controller on typical second order systtm. 16. E;rplain the ef!tct of PD controller on typical second order •yst<m.
(July·2005)
I 7. A fetdbnck systtm whid1 11$<$ a rate fetdlxic:k controller is slwrv11 in th< figure.
0.5s1 + s( I+
Kil
al For KA a in in absence of derioative fttdbtuk (Kh a 0) dtlmnint 11" damping ratio and naluralfrtqutncy of oscillations. Also find IN s s trror for unit ramp input. bl Determine tire constant K1, if th< damping foe/or nquired is 0.6, with KA = 10. With this oolue of Kh, dtter111inc ss error for writ ramp input. (Ans.: 0.316, 3.16 rad/sec, 0.2, 1.8, 0.38)
ODD
(6. 50)
Nonlinear Systems
7.1 Introductionto Nonlinear Systems It has been mentioned earlier that a control system is said lo be linear if it obeys law of superposition. Most of the control systems are nonlinear in nature and are treated to be linear, under certain approximations, from ease of analysis point of view. Let us discuss now the properties of nonlinear systems. ln practice nonlinearities may exist in the systems inherently or may be purposely introduced in the systems, to improve the performance. Hence knowledge of properties of nonlinear systems and various nonlinearities is important.
7.2 Properties of Nonlinear Systems The various characteristics of nonlinear systems arc, 1. The most important characteristics of a nonlinear system is that it docs not obey the law of superposition. Hence its behaviour with respect to standard test inputs can not be used as base to analyse its behaviour with respect to other inputs. Its response is different for different amplitudes of input signals. Hence while doing the analysis of nonlinear system, alongwith the mathematical model of the system, it is necessary to have information about amplitudes of the probable inputs, initial conditions etc. This makes the analysis of the nonlinear system difficult. 2. Unear system gives sinusoidal output for a sinusoidal input, may be introducing a phase shift. But nonlinear system produces higher harmonics and sometimes the subharmonics. Hence for sinusoidal input, the output of a nonlinear system is generally nonsinusoidal. The output consists of frequencies which are multiples of the input frequency i.e. harmonics. The subharmonics means the presence of frequencies which are lower than the input frequency. The input and output relations are not linear. 3. In linear systems, the sinusoidal oscillations depend on the input amplitude and the initial conditions. But in a nonlinear system, the periodic oscillations may exist
(7 • 1)
Nonlinear Systems
7-2
Modem Control Theory
which are not dependent on the applied input and other system parameter variations. ln nonlinear system, such periodic oscillations arc nonsinusoidal having fixed amplitude and frequency. Such oscillations are called limit cycles in case of nonlinear system. 4. Another important phenomenon which exists only in case of nonlinear system is jump resonance. nus can be explained by considering a frequency response. The Fig. 7.1 (a} shows the frequency response of a linear system which shows that output varies continuously as the frequency changes, Similarly though frequency is increased or decreased, the output travels along the same curve again and again. But in case of a nonlinear system, if frequency is increased, the output shows discontinuity l.e. it jumps at a certain frequency. And if frequency is decreased, it jumps back but at different frequency. nus is shown in the Fig. 7.1 (b).
0Utpu1
Peak resonant
Output
~ (a) Linear system
(b) Nonlinear system
Fig. 7.1 Jump reaonanc:e 5. There is no definite criterion for judging the stability of the nonlinear system. inc analysis and design techniques of linear systems cannot be applied to the nonlinear system.
7.3 Classification of Nonllnearitles Basically the nonlinearities are classified based on the fact that whether they are inherent or purposely introduced in the system. Depending upon this the two classes of nonlinearities are, 1. Inherent nonlinearities.
2. Intentional nonlinearities.
7.4 Inherent Nonlinearltles The nonlinearities which are unavoldeble in the control systems are called inherent nonlinearities. Let us discuss various types of inherent nonlinearities existing in practice.
7.5
Modem Control Theory
Nonlinear Systems
Ouiput
Output
(a) o,,_off with doadzone
(b) On·off with doadzone and hysteresis
Fig. 7.7
7.4.5 Nonlinear Friction In any practical system, where there is a relative motion between the two moving surfaces, there exists a friction. There are various Output types of such friction. All of them arc nonlinear Viscous except viscous friction. The nonlinear friction Is 0.()15 called Coulomb friction. The coulomb friction is a force acting in opposite direction of motion but it is constant in magnitude irrespective of velocity.
Coulomb friction
The viscous and coulomb friction are shown in the Fig. 7.8. It can be seen that viscous friction is linear while coulomb friction is constant irrespective of the input.
Fig. 7.8 Friction
The example of coulomb friction is the friction existing between the brushes resting against the commutator in an electrical motor.
7.4.6 Backlash This type of nonlinearity is generally found in the mechanical linkage where coupling is not perfect. A common example Is the existence of free play between teeth of drive gear and driven gear of a gear train. This is shown in the Fig. 7.9 (a). The driven ge..r docs not work until there is contact between drive and driven gear. So input member has to travel the distance B to have the output, as indicated by input-output characteristics shown in the Fig. 7.9 (b). Once the contact is made output follows the input. If Ule motion is to be reversed, then input member has lo travel distance of 2B in reverse direction, as shown in the Fig. 7.9 (b),
Nonlinear Systems
7-6
Modem Control Theory
(a}
(b}
Fig. 7.9 Backlash This is the most complex nonlinearity from modelling point of view. This is also memory dependent nonlinearity. It is a double valued nonlinearity like on-off with hysteresis. A process control valve with hysteresis is another example of backlash type of nonlinearity.
7.4.7 Nonlinear Spring A linear spring has a linear equation for its force and displacement. Linear spring force
... (1)
kx
where
k
positive spring constant
and
x
spring displacement
The nonlinear spring is not governed displacement are related by the equation. Nonlinear spring force
by the equation
(1)
but
kx+k'x3
where
k
Positive spring constant
and
k'
Nonlinear spring constant
its force and ... (2)
The k' can be positive or negative. If k' is positive, the spring is called hard spring while if k' is negative, the spring is called soft spring. If k' = 0, the spring is no longer a nonlinear but behaves in a linear fashion. The degree of nonlinearity in the spring is totally controlled by the magnitude of k'. If such a spring is used in the system consisting of mass and friction then it is experimentally observed that the frequency of the free oscillations increases or decreases as the amplitude decreases depending upon the value of k',
7.7
Modem Control Theory
Nonlinear Systems
Frequency k'>O Hard spring
k'
Amplitude Fig. 7.10 Frequency-amplitude curvesfor n7l1near spring The frequency-amplitude curves for a system with nonlinear spring arc shown in the Fig. 7.10. For k' > 0, as the amplitude decreases, frequency also decreases. This is hard spring. For k' < 0, as amplitude decreases, the frequency increases. This is soft spring. For a linear spring such characteristics is a straight line. These curves are often used to find out whether them exists a nonlinearity in the spring and also to find the degree of nonlinearity.
7.5 Absolute Value Nonlinearity In some systems, though the direction of the input is changed, the output direction remains same. The value of output is absolute irrespectiveof the direction of the input applied. The best example of such a nonlinearity is a relay operated by the solenoid. Irrespective of the direction of the current through the solenoid, relay gets operated. This is shown in the Fig. 7.ll (a) and (b). Output
· :]~~__..
Relay (a)
(b)
Fig. 7 .11 Absolute value nonlinearity Thus the output is not affected by the direction of the input applied.
Phase Plane Method
8.1 Introduction The phase plane method is an important method used for the analysis of nonlinear systems. The phase plane method is applied normally to first order and second order of linear and nonlinear systems. The system behaviour is qualitatively analysed ;ilong with design of system parameters so as to get the desired response from the system. The periodic oscillations in nonlinear systems called limit cycle can be identified with this method which helps in investigating the stability of the system.
8.2 Limit Cycle In case of Ilncar lime invariant systems a periodic oscillation is sinusoidal. The amplitude of such oscillations is a function of excitation applied and initial conditions. IJ the system parameters arc changed slightly i.e, if the system poles arc shifted from the imllginary axis of complex s plane, then the oscillations will no longer exist. In case of nonlinear systems, periodic oscillations are present but arc not dependent on amplitude of applied excitation. Also the oscillations arc not much sensitive to parameter variations. A periodic oscillation is case of a nonlinear system is called limit cycle. Normally the limit cycles are non-sinusoidal. The limit cycles which arc having fixed amplitude and time period can be observed over a finite range of system parameters. The limit cycle is also called self excited oscillation which is observed in certain nonlinear systems, Consider a system described by following equation, M;;-µ(l-x2);+kx=O Herc M, µ and k are positive quantities, This is a nonlinear equation. For small values of x, the damping will be negative which indicates the energy is actually put in the system. For large values ol -. thP dam;,1ng is positive which removes the energy from the system. Thus it can be expected that such system may exhibit a sustained oscillation. Now as it is not a forced system, these oscillations arc called self excited oscillations or limit cycle. (8. 1)
Modem Control Theory
8-2
Phase Plane Method
8.3 Jump Resonance The phenomenon of jump resonance is seen only in case of non linear systems. This can be explained by considering the frequency response plots. The Fig. 8.1 shows frequency n.'Sponse plot for a linear system while the Fig. 8.2 shows frequency response plot for nonlinear system.
t
Response
I\
.' [f\ : t:
~
.,_
Fig. 8.1 Llnoar system
Fig. 8.2 Nonlinear system
Consider a nonlinear system excited by sinusoidal input of constant amplitude. If the input frequency is increased then jump (or disrontinuity) occurs in the response amplitude. With decrease in frequency also, jump can be observed but at a different frequency. A nonlinear system is one which behaves qulte differently for various input functions. Consequently the logical design of a nonlinear system requires a complete description of the input signals. This dependence of the system behaviour on the actual input functions is demonstrated by the jump resonance phenomenon which is observed in certain closed loop systems with saturation.
Let us consider a nonlinear equation describing a certain system. M ~+ f=+ k1x+k2x3
= f cos rot
If the magnitude of the response for above system L~ plotted against frequency, then corresponding frequency response plot for k2 > 0 is shown in the Fig. 8.3. 8
F0tk,>O
Fig. 8.3 Jump resonance
8-3
Modem Control Theory
Phase Plane Method
Due to the presence of a nonlinear term k2 x 3, the resonant peak is bent towards the higher frequencies. As the input frequency is gradually increased from zero with input amplitude fixed, the measured response follows the curve through the points A, B and C. But at C, an increment in frequency results in a discontinuous jump down to the point D, after which with further increase in frequency, the rcsponsc curve follows through DE. If the frequency is decreased, the response follows the curve EDF with jump to B occurring at F and then move to A. For certain range of frequency the response function is double valued. For k, < 0, the resonant peak bends towards lower frequencies which is shown in the Fig. 8.4. similar Jump resonance takes place with the difference that there is upward jump when frequency is increased and downward jump when frequency is decreased. Response D F
Fork2< 0
w-
Fig. 8.4 Jump resonance Tile overall curve shows the familiar hysteresis or jump resonance phenomenon. It is well known in nonlinear mechanics and is frequently observed in high gain servo systems. In such cases, the response is multivalued and depends not only on the present value of the input but also on the past history.
8.4 The Phase Plane Method The equation of a second order nonlinear system having free motion has the following form,
.. . . . y-h(y,y)y+g(y,y)y
=0
At any particular moment,. the state of the system can be represented with the help of a point having coordinates (y, y) in rectangular coordinate system. This coordinate plane is called a 'phase plane'.
8-5
Modem Control Theory
Phase Plane Method
t,
Fig. 8.5 The starting point A with initial condilion but no initial velocity. the system comes to equilibrium i.e. to the origin with damped oscillatory behaviour.
8.5 Basic Concept of Phase-Plane Method Consider the diffcrentinl equation for a linear second order control system d2y2(t) +« dyd(t) +Jly(t) = u(t) dt t Herc u is input while y is outpuL The phase plane method is basicall y a graphical method of solving S<."COnd order linear as weU as non linear system equations. The co-ordinate plane whose axes
.
correspond to dependent variable y(t) and its first derivative y(t) is called phase plane. In phase plane plot y(t) and
y(t) arc plotted in a two axes plane. The trace of the phase
plane plot as time t increases is called trnjcctory. The starting point of the trajectory depends on the initial conditions imposed on the system. For different sets of initial conditions, various trajectories can be plotted. The complete dynamic behaviour of the system for various initial conditions can be obtained from the trajectories plotted in phase plane. The dynamic performance and analysis of given second order system differential equation can be made based on destination point of trajectories. The method has limitation of its application restricted to second order as for higher order system equations, three or higher dimensional plane is needed which is not much convenient to construct and visualise. But transient response of second order system excited by step input can be studied by this method. This method is thus applicable to linear time functions and can not be applied to systems excited by sinusoidal signals. The phase plane method can be considered as a special case of state space. With the use of slate variables known as phase variables, the state of second order system can be defined.
8-7
Modem Control Theory
Phase Plano Method
In phase plane analysis, the trajectories can be constructed analytically, graphically or experimentally. The analytical method is straight but not useful from practical point of view. The graphical methods are used when it is not possible to solve the given differential equation analytically. The graphical methods arc applicable to both linear and non linear equations.
8.6 Isocline Method for Construction of Phase Trajectory The isocline method i~ a graphical approach for determination of phase portrait. Consider the slate equation for a second order system.
111e slope m of trajectory at any point in the phase plane is given by, dx2
m = f2(xpx2)
~
f1("1·"2)
The point (x1,x2.' of the phase plane has with it associated slope of the trajectory except at singular point where the slope is indeterrniante. lsodines are the lines in the phase plane corresponding lo slopes of the phase portrait. In order to construct the phase trajectory, it is necessary to integrate the slope equation. This is very difficult except for few simple cases. We have, (~ ( x~.x2
)=
m1f1(x1.x2)
The above equation gives locus of all points in the phase plane where the slope of trajectory is m1. Such a locus is called isocline. TI1c direction of phase traiectroy at any point (x1,x2) can be obtained from the sign or ax1 and ax2 for small increment in at.
Consider once .1gain second order system equation. d2>•(1) --
dt2
dy(t) dt
-e- n-:-+(ly(t)
= u (t}
Subject to initial conditions y(t0)
= Yo
Modem Control Theory
Phase Plano Method
8-9
Based on various slopes of trajectories, isoclines arc drawn as shown in the Fig. 8.7
Fig. 8.7 conditions, the starting point is 0 which is the point on slope m1. The phase trajectory which is drown from point on m1. The phase trajectory which is drawn from point 0 on the to that having slope m2 is a straight line whose slope is the m +rn average of m1 and m2 i.e. -1-2--"This 2• line intersects isocline 2 at point P which has a From the given initial isocline 1 corresponding to the isocline whose slope is isodine whose slope is m1
slope of m2• Thus we get line segment OP. This procedure is continued at point P to get subsequent line segments PQ QR and so on as shown in the Fig. 8.7 Now consider undamped second order system such that ~ = 0 d2y(t) +2~(00 dy(t) +~ dt2 dt For simplicity let us select Uln d2v(t)
--·-+y(t) dt2 Let
\IS
=
y(t) =
=I
0
now select the state variables XI y(t) X2
The state equations arc,
.
y(t)
0
I
The above equation is
Modern Control Theory
Phase Plane Method
8 • 10
Integrating and rearranging above equation.
xi-x~
=
C
Here C is arbitrary constant obtained frorn initial conditions. The phase portrait in phase plane is as shown in the Fig. 8.8. The above equation represents a circle. The phase portrait consists of concentric circles depending on difforent initial conditions
(x1to x2lo x3to etc)
Initial oolldltion ~~~~--11---t-~--1--.~1----i~-t
Fig. 8.8
8. 7 Application of Phase-Plane Method to Linear Control System Before: studying the application of phase-plane method lo non-linear systems, let us see its application in linear systems. Let us consider second order linear control system as shown in the Fig.8.9 1------
Fig. 8.9
C(s)
8·11
Modem Control Theory
Phase Plane Method
The transfer function is given as,
C(s)
[~]
R(s) = [
1+---
(~)
K ] = s(T1s+l)+K
K
s(T,s+l)
Jn the forward path, the transfer function is C(s)
K
E(s)
!i(T1s+1) KE(s) KE(s)
Taking inverse Laplace transform, T1 c+c
Ke
e
r-c
I!
= r-<:
e
"
But
r-
..
c = r-c
c .. C=
r- f.'
Let the input be step input r (t)
I{
r(I)
r(I)
.
..
=0
The above equation becomes 0 Now At
e (OJ
fort>
0
r(O) - c (0)
I "' 0, c = 0 i.e. c (0) = 0
r (0) = R
Modem Control Theory
Phase Plane Mothod
8-13
When the value of K is adjusted in such a way as to get the system overdampcd then the phase plane trajectory is as shown in the Fig. 8.11
(a) Step response (Overdampod case)
(b) Corresponding phase portrait
Fig. 8.11 For the linear systems, where analytical solution can be computed for c and c then it is not required to draw phase portrait. It can be effectively applied in case of non-linear systems.
8.8 Second Order Nonlinear System on Phase Plane Let us consider the Van dcr Pol's differential equation,
..
.
.,
.
y-µ(1-y•)y+y
=0
.
This equation describes the behaviour in many nonlinear systems. Let us select the state variables as x 1 "' y and x2
= y.
TI\e stale equations are given as,
x2
)' .. µ(l-xi)x2-'1
The equilibrium point of this system is origin of the phase plane. Now there are two cases viz. µ > 0 and µ < 0. The phase portraits for both these cases is shown in the Fig. 8.12. When µ > 0 and initial value of x1 [i.e. x1 (0)1 is large then the system response is damped. The amplitude of x1(t)[ .. y1(1)] decreases till the system enters the limit cycle. This is shown by the outer trajectory. For small initial values of xi' the damping is negative.
Modem Control Theory
-
(a)~~>
Phase Plane Method
8·14
Stable llmilcycie
O
(b)~
Fig. 8.12 The amplitude of x1(t)[=y1(t)] increases till the system stale enters the limit cycle which is shown by inner trajectory. The limit cycle is stable as the paths in its neighbourhood arc converging. The limit cycle for µ < 0 is unstable. With increase in order of the system, the complexity is involved in applying phase plane method. The method of phase trajectories is restricted normally lo second order systems and ii is special case of phase space defined for nth order system. ln case of lime invariant systems, the total phase-plane contains trajectories with only one curve passing through every point of the plane except for some critical points. Through these critical points either none of the trajectories or infinite number of trajectories pass. For lime varying systems, two or more trajectories may pass through a single point. The phase portrait then becomes complex and not easy to interpret. Hence the method of phase-plane is mostly applied lo second order systems having constant parameters with no or constant input. Some simple lime varying systems can be analysed by this method.
8.9 Different Types of Phase Portraits In this section. the phase portraits of some commonly known linear systems is presented. Thl'SC portraits can be used for analysis of piecewise linear systems.
8.9.1 Phase Portraits for Type 0 System Consider a transfer function for a type 0 system. C(s)
ro;
R(s) = s2 +2sCJ\,s
+
Let us consider the input r(t) to be zero i.e. unforced system ls2 +2soons+ro~
JC(s) .. O
Modem Control Theory
c+2;col\ C+o>~C Let x1
a
C and x2
Phase Plane Method
8·15
a
;l = X2i ;2::
=
0
C. Therefore the state model is given by, -2~COnX2-(0~X]
Eliminating the time variable from the two equations by dividing the equations, dx2
-2~Ctl0Xz -C1)~x,
~
X2
dx
If we consider the point at which x1 = x2 = 0 then the slope al this point dx~ =
O
0
which is Indeterminate. This point is called singular point. We will study the behaviour of trajectories which are dosed to this singular point. To find out the singular points which arc nothing but the roots of characteristic equation s2+2;rons+oT,; (s-y1)(s-y2) = 0. Depending on the values of Yt and y2, six
=
different types of singular points arc obtained. These arc described as below. 8.9.1.1 Stable System with Complex Roots Let y1
= -a+jf1
y2 = -o.-jf1
a>O and ll>O
The output response in this case is c(t) = C1e-"1 sin
x2
=
C. The typical phase lrafcctory obtained is a logarithmic spiral into the
singular point which is shown in the Fig. 8.13. This type of singular point is called a stable focus.
jo>axis
~-----
ill
-a•
oaxis-
I I I I I
*----(a) Stable focus
-jjl
(b) Nature of roots Ins plane
Fig. 8.13
Modern Control Theory
Phase Plane Method
8 -17
t
jw jp
o-
-JP (a) Center
(b) Nature of roots Fig. 8.15
8.9.1.4 Stable System with Real Roots Let y1 and y2 be two real and distinct roots located in left half of s-plane. The response in this case is given by, C(t) = Ctcrt• +C2crz• The response obtained from the system is ovcrdamped. The singular point obtained after plotting the phase trajectory on (x 1 = y, x2 = y) plane for this case is called a stable node. The phase portrait contains two straight line trajectories having equations x2(t)=y1x1(t); x2(t) y2x1(t) which satisfies the differential equation of the given system.
=
Assuming that the term eT2' decays faster in the transient response than the term cT 11. As time t increases, x1 ~c,er11 ~a and x2 ~11C1cY1' ~a. All the trajectories are tangential at the origin to the straight line trajectory given for initial conditions x2(0) = /2x1(0).
2(t)=)'1x1(t).
The other trajectory is
All the trajectories arc asymptotic to this straight line trajectory. If the roots are repeated for stable systems then there is a single trajectory whose slope depends on value of root.
Modem Control
Theory
Phase Plane Method
8·18
t
jN
Fig. 8.17 Nature of roots
Fig. 8.16 Stable node
The phase trajectory and nature of roots is shown in the Fig. S.16 and Fig. S.17.
8.9.1.S Unstable System with Positive Real Roots Let y1 and r2 be two real and distinct roots localed in right half of s plane r1 is a smaller root. The singular point in this case is called unstable node. The phase portrait on
.
(x1 = y, x2 = y) plane is shown in the Fig. 8.18. Jh)
Fig. 8.18 Unstable node
Fig. 8.19 Nature of roots
The trajectories emerge from the singular point and go to the infinity. The trajectories arc tangential to the straight line trajectory x2(t) 11x1(t) al origin and asymptotic al infinity to other straight line trajectory. For the repeated roots, there is a single trajectory.
=
Modem Control Theory
Phase Plane Method
8 -19
8.9.1.6 Unstable System with One Positive and One Negative Real Root ln this case we have two straight line trajectories which have slopes dependent on values of root. The singular point obtained in this case is called a saddle. The stralghtli.ne obtained due to negative root gives trajectory entering into the singular point while due to positive part gives trajectory which leaves the singular point.
The other trajectories approach towards the singular point close to incoming straight line then takes a curve away and leaves the singular point in its vicinity. It approaches the second straight line asymptotically. The phase portrait on (x 1 plane is shown in the Fig. 8.20.
Fig. 8.20 Saddlo point
=
y, x2
=
.
y)
Fig. 8.21 Naturo of roots
8.9.2 Forced Second Order Type O System Let us consider a second order system excited by constant input with magnitude A. C(s)ls2 ·•2~o>ns+ro~
I = A~
The above differential equation can also be written in following form. d2 d • .z di2(C-A)+2~w0 dt(C-A)+w,,(C-A) = 0 The equation is similar lo that obtained for unforced system wilh the difference that the singular point is shifted from origin to point C A on the Y axis.
=
8.9.3 Phase Portraits for Type 1 System Consider the following transfer function which represents Type 1 system. This system is excited by constant input of magnitude A.
8-21
Modem Control Theory
Phase Plane Method
The total equation becomes,
l1 x 1(1) ~(x1
-
x1(0)J
=A - X2(t)-A/1~A - x2(t)]+
= -fx2
1 lx
1(0)-A
+x2(0)+A/1~A - x2(0)1
:;to)]
- x2(0)1-Al1{ AA_
The corresponding phase portrait for A > 0 is shown in the l'ig. 8.22.
A
Fig. 8.22 Phase portrait for A > 0 If initial conditions are zero i.e. x1(0) = x2(0) = 0 then ~XI
= -X7-A11{A~X2]
The corresponding phase trajectory is shown in the Fig. 8.22. The trajectory is asymptotic to the line x2 = A. If the initial point is different (xi (0), Xz(O)] then the trajectory will have same shape but shifted horizontally by k units so as to pass through the point (x1 (0), x2(0H The equation of trajectory in this case is given by, ~(x1-K)
=
-x2-Aln(A~xz)
Modem Control Theory
Phase Plane Method
8 ·22
The phase portrait for A < 0 is shown in the Fig. 8.23. The value of A = 0 is a special case which is unforced system and the phase portrait consists of number of straight lines ' 1 having the slope of - 'T. Fig. 8.23 Phase portrait for A < 0.
A
Fig. 8.23 Phase portraits for A < 0
8.9.4 Phase Portraits for Type 2 System Consider the transfer function of linear system representing type 2 system C(s) R(s)
= .l.. Js2
Let the input applied be of constant differential equation is given by, JC=
A
Let the state variable be x 1 = C, x2 = C The corresponding state model is given as, Xl Xz
c : Xz c =1A
magnitude, r(t)
A. The corresponding
8 ·23
Modem Control Theory
Phase Plane Method
By eliminating the time variable we get,
dx2
ii;z; tx2dx2
A
Jx2 A dx1
Integrating the above equation,
Where C is constant of Integration obtained from initial conditions.
J x~(t)
21\+C
J x~(O)
+C
2A C = X1(0)-
J x~(O) 2A
The nature of trajectory ls parabolic for an initial state point (x1(0), x2(0)1 passing through a point x 1 = C on the x-axis where C is given by above equation. This Is shown in the Fig. 8.24 for A < 0 and A > 0.
"1 =v
C
(a) For A>
o
(b) For A
Fig. 8.24
Phase Plane Method
8 -24
Modem Control Theory
With increase in time t, each trajectory is shown in clockwise direction. The direction of phase trajectory is obtained from the relationship;
1 ..
x2• As x1 increases with time in
upper half of phase plane, the state point hence moves from left to right in the lower half of phase plane. When x1 decreases with time, the state point moves from right to left
=
The time interval between the two points of a trajectory Is given by 6t 6x 1 I x2.tv. With the aid of this equation the time scale can be given to the trajectories. But the phase portrait is generally used to get an idea about transients in the system. A
The phase portrait or A < 0 is shown in the Fig. 8.24 (b). The unforced system with case. -,
= 0 is a special
dx2 dx';-
0
..·
dx 2- - 0
Integrating above equation
J dx
o
2
x2(t)
C
Where C is constant of integration obtained from initial conditions. C
=
x2(0)
:. We have equation of trajectory as x2(t) = x2(0). This equation represents the trajectories of straight lines parallel to x 1 axis. ,...
Example 8.1 : Determine the kind of singularity for eaclt of tltt following differt11tial
. ...
equations. i) y+3y+2y=O
= 34
ii) y-By+17y
(VTU: Jan/Feb.·2005)
Solution : i) y+ Jy+ly = 0 Let the state variables be x1 = y and x2 =
. .
.
The corresponding state model is
x1=y=x2;
x2=y=-3x2-2x1
Eliminating the time variable, dx2
-3x2 - 2x1
dXi'
X7
y
Modem Control Theory
8. 25
Phase Plane Method
The characteristic equation is given by, s2 +3s+2
0
(s+2)(s+t)
0
s = -2; s =-I We have, y1 = -1 and y2 = -2 The nature of phase trajectory is as shown in the Fig. 8.25 and the type of singular point obtained is stable node.
o--
(a)
(b)
Fig. 8.25 The response in this case is given by y(t) =
c1eYJ1
+C2cY21.
.
The transient term eY2•
decays at faster rate than eYt1• As time t-+~, x1 -.C1eY:1 -+0; x2
.
singular point node obtained in this case (0, OJ is stable in (y, y) plane . ii)
y-8y+17y
.. .
y-8y+t7y-34
34 0
The above equation can be rewritten as, d2y dy --8-+17y-34 dt2 dt d2(}'-2) d dt2-8dt(y-2)+17(y-2)
0 0
=
'Yt<'Y11 -+0. The
8 ·26
Modem Control Theory
Phase Plane Method
The characteristic equation is given by, 0
(s-2)2 -8(s-2)+17(s-2)
0
(s2 -4s+4)-8s+t6+17s-34
= 0
s2+5s-14 (s+7)(s-2) Y1
0 -7; y2
=2
There arc two roots one positive and one negative root. The singular point obtained In this case is called saddle point which is shifted from origin by 2 units. The slopes C1f the two straight line trajectories with slopes dependent on root values. The nature of phase trajectory is shown in the Fig. 8.26.
t
joo
Fig. 8.26
8.10 Singular Points of a Nonlinear System Singular point is a point on the phase plane where the slope of the trajectory i.' indeterminate. In the state model of a given system, when time variable is eliminated then we get,
By equating the numerator and denominator equations to zero i.e, solutions of the equations f(x1, x2) 0 and x2 0 gives the locations of the singular points along with Its number.
=
=
It is also Important lo know the behaviour of the trajectories in the vicinity of a singular point of a nonlinear system. The nonlinear equations are Ilnearized at the singular
8 -27
Modern Control Theory
Phase Plane Method
point using say Taylor series expansion and then the nature of phase trajectory around the singular point is determined by linear system analysis which is described earlier.
8.11 Delta Method In isocline method, entire phase plane must be filled with line segments of different possible slopes. If only a single solution curve is needed, only a few of these lines are put to use, thus wasting a lot of effort. A technique of construction known as the delta method leads more directly to the solution. This method applies to the solution of equations of the type,
.
x+f(x, x, t) ;; 0 Where f must be continuous dependent.
and single valued
.•• (1)
l>ut
may be nonlinear and time
The equation (1) is written as ... (2) The constant
ooij
may be determined from the equation (1) itself or m~y have to chosen
= °'o t
from other information. Once again it is convenient to introduce the definitions 1 dx . an d v = dt so as to give,
ooij vdx+ dv ooij x+(f(v,x,,t)-Cl>(jx) , dv dx
or where
li(v, x, t)
0
... (3)
_Ix+ ll(v, x, 1)1 v
1 x, t)-x 2f(v,
.•. (4)
... (5)
(l1j
·' Equation (4) is similar to that one used for the case of isocline method. Because of the . Jay, in which it is set up, the S method is most immediately applicable to the equations with oscillatory solutions, although it is not limited to this class of equation. Function o of equation (5) depends upon v, x and 1 but it is assumed constant for small changes in these variables. Under this assumption, the variables ol equation (4) can be separated and integrated to give, v2 +(x+5)2
= constant= R2 (say)
... (6)
Equation (6) represents a circle of radius R and centre at (-ll, 0). Thus for a suitable increment, the solution curve is the arc of a circle having these properties. The construction
Modem Control Theory
: I ·28
Phase Plana Method
is shown in Fig. 8.21 where [x(O), v(O)) represents a point on the solutiof\ curve at the time 1(0). These values used in equation (S) allow the ~kulatlon. of 6. This values of 6 determines the centre of the drcular arc located on the x-axis. The radius R Is 1utomatically fixed. A short drcular arc represents • portion of 10lulion curve. Actually it is more accurate to use the average values of v, x, 't for that Increment. Again, the allowable length of the arc is a rompromiae.
Fig. 8.27
Examples with Solutions ,,_.
Example 8.2 : A linear secottd order snw is described by tht! equation C+2l;w0 C+w~C = 0 where i; = 0.15, ~ = 1 rod/sec, CCO) = 1.5 and C ~ 0. Determine the singular point. Construct the phase trajectory, using the method of isoclines CVTU: July/Aug.·2006)
Solution : Consider the equation
C+21;w0 C+w~C Given !;= 0.15,
W0
=
0
= 1, C (0) = 1.5, C = 0 c+2(o.15H1>c+12c
o
C+0.3 C+C
0
Taking Laplace transform, s2Qs1+0.35 Qs)+C(s) (s2 +0.35+1] C(s) Consider
s2+0.35+1
=
0 0 0 -0.3±Jco.3)2 -4{t)(J) -0.3±i(t 2 = 2 -0.15±j0·9886
·9mJ
Modem
Control Theory
Phase Plane Method
8 ·29
The roots arc complex conjugate and located in left half of s plane. In time domain the response will be underdampcd consisting of oscillations and the singular point is thus sta bel focus. Consider again the differential equation d2qt) +0.3 dQt) +Qt) dt2 dt
. .
Let
:. XI = Qt) = X2,
X] = C(t),
0
.
X2 "' Qt)
X2 = Qt) = -
.
C(t) - 0.3 C(t) " -X1~.3X2 m
dx21 dx ,
=
dx2fdt dx1/dt
=
-x1 -0.3x2 x2
-X1~-3X2
-x,
-[m~o.3]x1
The above equation is an equation of isocline. At initial point (1.5,0) 0
--1-(1-5) m+0.3
m
-0.3--
= --
Due to this trajectory moves downwards Using the method of isocline, the phase portrait can be drawn as shown in the
Fig.8.28 Considering various values of m m
Isocline equation
1
•2 = -
5 10
-
0.7692 .,
xz. - 0.1886 •2
=-
9=tan-1m 45'
.,
78.69'
0.0970 •1
84.28°
•2.
0
-1
Xz =
1.4285 .,
-45'
-5
•2 • 0.2127 "'
-78.69'
-10
Xz
= 0.1030
- 84.28'
-
•2 c 0
90-
X1
-90'
.
PhasePlane Method
8 ·32
Modem Control Theory
The type of singular point obtained is stable focus . ii) Consider,
y + 3 y+ 2y
=0 y and
Let the state variables be x 1 =
. .
.
x2 = y
.
The corresponding state model is x1 =y=x2;
x2 = y=-3x2-2x1
Eliminating the time variable,
The characteristic equation is given by, s2+3s+2
0
(s+2)(s+l)
0 -2; s
= -1
We have, y1 =-I and y2 = - 2
t j(J)
x,
(a)
=y
(b)
Fig. 8.30 The nature of phase trajectory is as shown in the Fig. 8.30 and the type of singular point obtained is stable node. The response in this case is given by y(t)
= C1e111
+C2el21•
decays at faster rate than cl1'. As time t ~ -, x1 ~ C1cYJ1
~
The transient term eY2• O; x2 = y1c111
singular point node obtained in this case (0, 0) is stable in (y, y) plane.
~
0. The
8-33
Modem Control Theory Iii) Consider,
.
Phase Plane Method
=0 be x1 = y ; x2 = y
y + 3 y - 10
Let the state variables
The corresponding state model Is,
X2
.
Y =-3y+10=-3X2.,_JQ
Eliminating the time variable dx2
-3x2+10
di<";" :
X2
The .characteristic equation is given by, s2+3s-10
0
s2 +Ss-2s-10
0
s(s+5)-2(s+5)
0
(s + 5) (s - 2)
0
Y1 = -5; Y2 = 2 There are two roots one positive and one negative. The singular point obtained in this case is called saddle point. The slopes of the two straight line trajectories with slopes dependent on root values. The nature of phase trajectory Is shown In the Fig. 8.31.
v, =-5
x,
Fig. 8.31
=y
Modem Control Theory
8. 35
Phase Plane Method
Fig 8.32 C
I
M
Modem
m
lsoclino equation
1
Y2 • - 0.825y1
45•
2
Y2 • - 0.3846y1
63.43°
5
Y2 c
0.1785 Y1
78.69°
Y2 • - 0.0943y1
84.28°
10
-
O=tan-1 m
90"
_,
Y2 • 0
-2
Y2 • 0.7142Y1
-63.43"
-5
Y2 • 0.2272Y1
-78.69"
-10
Y2 • 0.1063y1
-84.28°
M
-45'
Y2 • 2.5y1
-
i•
Phase Plane Method
8 ·36
Control Theory
-90'
Y2 • 0
The phase-plane plot is shown in the Fig. 8.32.(See Fig. 8.32. on previous page)
Draw IM phase-plane trajtctoryfor lht following equation using isocline method : '.;+2!;(1);+w2x=O Given, 1;=·0.5, w2 l, Initial point (0, 6) Example 8.5 :
(VTU: JanJ Feb.-2007)
Solution: Consider the equation
'x"+21:,(1)11 ;+~x Given that I:,
= 0.5, Ci>n = 1 , x = 6, ~:
x+ x+x Let
= 0
Yt
=0
0
x
Y2 = x Yt
x = Y2 ; y 2 = x
= -x
m
~ = dyifdt = _y_2_
dy 2
-y l
dy ifdt
- x = -y2 -y I - )' 2
8 .37
Modem Control Theory
Phase Plane Method
(m+l)y2 Y2 The above equation is an equation of isocline. m+t
=
_l'..J. Y2
m
=
_l'..J._1 Y2
ll t initial point (6, 0).
:. The trajectory moves downwards Using the method of isocline, the phase portrait can be drawn considering various values of m. m
Isocline equation
1
Y2
0.5 Y1
45°
0.33 Y1
63.430
0.166 Y1
78.69"
Y2 • - 0.090 Y1
84.28'
2
Y2
5
Y2
10
~
Y2
-2
===-
O· tan-1m
=0
90'
Y2 = 1Y1
-45°
-5
Y2 • 0.25 Y1
-10
Y2 = 0.1111
-20
Y2
-
Y2 = 0
= 0.0526
-63,43' Y1
-78.69"
Y1
~.28'
The phase-plane plot is as shown in the Fig. 8.33.
-90"
Modem Control
Theory
8. 38
Fig. 8.33
Phase Plane Method
(8. 40)
Liapunov's Stability Analysis 9.1 Introduction
.
11'e state equation for a general time invariant system has the form x
=
f [x, u). If the
input u is constant then the equation will have form x = F(x). For this system, the points, at which derivetives of all state variables arc zero, arc the singular points. These singular point' are nothing but equilibrium points where the system stays if it is undisturbed when the system is placed at these points. The stability of such a system is defined in two different ways. If the input to the system is zero with arbitrary initial conditions, the resulting trajectory in phase-plane, discussed in earlier chapter, tends towards the equilibrium state. If U1e input to the system is provided then the stability is defined as for bounded input, the system output is also bounded. For linear systems with non-zero eigen values, there is only one equllibrium state and the behaviour of such systems about this equilibrium state totally determines the qualitative behaviour in the entire state space. In cruse of nonlinear systems, the behaviour for small deviations about the equilibrium point is different from that for large deviations. Hence local stability for such systems does not indicate the overall stability in the state space. Also the non-linear systems having multiple equilibrium states, the trajectories move from one equilibrium point and tend to other with time. Thus stability in case of non-linear system is always referred to equilibrium slate instead of global term stability which is the total stability of the system. Jn case of linear control systems, many of the stability criteria such as Routh's stability test. Nyquist stability criterion etc. arc available. But these cannot be applied for non-linear systems. The second method of Liapunov which is also called direct method of Liapunov is the most common method for obtaining the stability of non-linear systems. nus method is equally applicable to time varying systems. stability analysis of linear, time in variant systems and for solving quadratic optimal control problem.
(9 • 1)
Llapunov's Stability Analysis
9·2
Modem Control Theory
9.2 Stability in the Sense of Llapunov
.
Consider a system defined by the state equation x = f(x, t). Let us assume that this system has a unique solution starting at the given initial condition. Let us consider this solution as F(t : Xe- 10) where x = x0 at t = lo and t is the observed time. F(t0
:
xo- to)
=
Xo
=
If we consider a state x., for system with equation ; f (x, t) in such a way that ((•,~ ti e. 0 for all t then this x0 is called equilibrium state. For linear, time invariant systems having A non-singular, there is only one equilibrium state while there are one or more equilibrium states if A is singular. In case of non-linear systems as we have seen previously there are more than one equilibrium slates. The isolated equilibrium states that is isolated from each other can be shifted lo origin i.e, f(O, l) = 0 by properly shifting the coordinates. TilCSC equihbrium stales can be obtained from the solution oi equation f(x,,,. t) = 0. Now we will consider the stab!Uty analysis of equilibrium states at the origin. We "ill consider a spherical region of radius R about an equilibrium state x., such that
!Ix -
Xe
:J
S
R
U x - xc II is called Euclidean norm and is defined as
II
x-
•.II =
J2 + (x2 - x'k)1+ ...
[x1 - x1
(X,, -x,,il2
Let S(o) consists of all points such that 11 "o - x. II S Ii and let S(E) conststs of all points such that II F (t : >
=
Any equilibrium state x0 of the system ; f(x, t) is said lo be stable in the sense of Liapunov if corresponding to each S(t) there is S(~ such that trajectories starting in S(li) do not leave S(E) as time t Increases indefinitely. The real number I) depends on E and in general also depends on lo· If Ii does not depend on to- the equilibrium state is said to be uniformly stable. The region S(e) must be selected first and for each S(c), there must be a region S(ll) in such a way that the trajectories starting within S(o5) do not leave S(e) as time t progresses. There arc many types of stability definitions such as asymptotic stability, asymptotic stability in large. We will also see the definition of instability along with definitions of these types of stability.
Modem Control Theory
9.3
Liapunov'sStabilityAnalysis
9.3 Asymptotic Stability An equilibrium state xe is said to be asymptotically stable if it is stable in the sense of Liapunov and every solution starting within S(o) converges without leaving S(t) to x, as t increases indefinitely. The asymptotic stability is more important than mere stability. The asymptotic stability is a local stability. Hence establishing asymptotic stability does not indicate the proper operation of the system. The size of the largest region of the asymptotic stability is required. This region is called domain of attraction which is part of state space where asymptotically stable trajectories originate.
9.4 Asymptotic Stability in the Large I( the asymptotic stability holds for all states that is all points in state space, from which trajectories originate, the equilibrium state is said to be asymptotically stable in the large. Alternatively we can say that the system is asymptotically stable in the large if it is asymptotically stable for every initial state regardless of how near or how far it is from the origin.
The equilibrium state Xe of the system is said to be asymptotically stable in the large if it is stable and if every solution converges to x0 as t increases indefinitely. The necessary condition for asymptotic stability is to be determined which is normally different. Practically it stability in large is that there must be one equilibrium state in the whole state space. In control system problems, asymptotic stability in large is required in the absence of which, the largest region of asymptotic stability is to be determined which is normally difficult. Practically, it is sufficient to get a region of asymptotic stability large enough so that no disturbance will exceed it.
9.5 Instability An equilibrium state "• is said lo be unstable if for some real number E > 0 and any real number Ii > 0, irrespective of how small it is, there is always a state "o in S(o) such that the trajectory starting at this slates leaves S(e).
9.6 Graphlcal Representation The graphical representation of stability, asymptotic stability and instability is shown in the Fig. 9.1 (a), (b) and (c) respectively. The region S(o) is binding the initial state "o and the region S(E) corresponds to the boundary for the trajectory starting at "o· The correct region of allowable initial conditions are not specified for the definitions of stability that we have seen previously. Thus these
Modem Control Theory
9-4
Llapunov's Stability Analysls
definitions apply to the neighbourhood of the equilibrium slate only if S(I'.) is not corresponding to the entire state plane. S{t)
S(t)
S(6)
S(6)
(a) Stablllty
(b) Asymptotic stablllty
(c) tnstablllty
Flg. 9.1 In Fig. 9.1 (c), the trajectory is leaving S(E) and shows that the equilibrium state is unstable. It can not be stated that the trajectory will go to infinity as it may approach a limit cycle outside the region S(e). In case of linear, time invariant system if it is unstable trajectories starting near the unstable equilibrium state will go to infinity which is not necessarily true in case of non linear systems. The definitions of stability that we have seen so for are not the only ones defining the stability or equilibrium state. There arc other ways of defining st4bility also. In classical control theory, the systems which arc asymptotically stable are called stable systems and those which are stable in the sense or Liapunov but are not asymptotically stable are called unstable.
9.7 Some Important Definitions · In this section WC will consider some important definitions understanding Liapunov's stability criterion.
which are useful in
9.7.1 Positive Definiteness A scalar function F(x) is said to be positive definite in a particular region which includes the origin of state space if F(x) > 0 for all non-zero states x in that region and F(O)" 0. e.g. F(x) " xi +2x~
9.7.2 Negative Definiteness A scalar function F(x) is said to be negative definite if - F(x) is positive definite. e.g.
F(x)
= - xf-(3x
1
+2x2)2
Modem Control Theory
9.5
Liapunov's Stabllity Analysis
9.7.3 Positive Semidefiniteness A scalar function F(x) is said to be positive semidefinite if it is positive at all states in the particular region except al the origin and at certain other states where it is zero.
e.g. 9.7.4 Negative Semidefinite A scalar function F(x) is said to be negative semidefinite semidefinite.
if - F(x) is positive
9.7.5 Indefiniteness A scalar function F(x) is said to be indefinite in the particular region if it assumes both positive and negative values irrespective how small the region is e.g.
9.8 Quadratic Form A class of scalar functions which plays important role in the stability analysis based on Liapunov's second method is the quadratic form.
e.g.
P is real symmetric matrix and x is a real vector.
9.9 Hennltian Fonn If x is a complex n vector and P is a Hermitian matrix then the complex quadratic form is called Hermitian form.
e.g.
F(x)
The stability analysis in state space the Hermitian form is commonly used than the quadratic from as it is more general. The positive definiteness of the quadratic form or Hermitian form F(x) can be determined by Sylvester's criterion which states the necessary and sufficient conditions for quadratic or Hermitian form to be positive definite is
Modem Control Theory
9-6
Liapunov's Stability Analysis
=
F(x) x" Px is positive semidefinite ii P is singular and all the principle minors arc non-negative. F(x) is negative definite if - F(x) is positive definite. Similarly F(x) is negative semidefinite if - F(x) is positive semidefinite.
9.10 Liapunov's Second Method A system which is vibrating is stable lf Its total energy is continuously decreasing. This indicates that the time derivative o( the total cneri,>y must be negative. The energy is decreased till an equilibrium state is reached. The total energy ts a positive definite function.
This foct obtained from classical mechanics theory is generalised in Liapunov's second method. If the system has an asymptotically stable equilibrium state then the stored t'nergy decays with increase in time till it attains minimum value at the equilibrium stale. But there is no simple way for defining an energy function. For purely mathematical system. This difficulty was overcome as Uapunov introduced Liapunov function method which is fictitious energy function. Llapunov functions depend on x1, x1, ... xn and I. It is given as F(x1, x2, .•• Xn• t) or as F(x, t). In Liapunov's second method, the sign behaviour of F(x, t) and its time derivative F(x, t) " dF(x, t)/dt gives us information about stability, asymptotic stabil.ity or instability of an equilibrium state without requiring to solve the equations directly to get the solution.
9.11 Llapunov's Stability Theorem Consider a scalar (unction V(x), where x is n vector and is positive definite, then the states x that satisfy V(x) C, where C is a positive constant, lie on a closed hyper surface in n dimensional stale space at least in the neighbourhood of origin. This is shown in the Fig. 9.2.
=
If V(x) is a positive definite function obtained for a given system such that its time derivative taken along the trajectory is always negative then V(x) becomes smaller and smaller In terms o( C and finally reduced to zero as x reduces to zero. This indicates asymptotic stability of the origin. Liapunov's main stability theorem is based on this and gives a sufficient condition for asymptotic stability.
Modem Control
Liapunov's Stability Analysis
9-7
Theory
Llapunov's stability theorem is as given below.
.
Consider a system described by equation
=
x = f(x, t) where f(o, t) 0 for all t, If there exists a scalar function V(x, t) having continuous first partial derivatives and satisfying the conditions such as V(x, t) is positive definite and V(x, t) is negative definite then the equilibrium state at the origin is uniformly asymptotically stable. lf V(x, t) -+ oo as II x II -+ oo where II x II is norm of x, then the equillbrium state at the origin is uniformly asymptotically stable in the large.
Vincrease
Fig. 9.2
The above theorem is a basic theorem of
.=
second method. The theorem 2 is stated below . Consider the system described by x f(x, t) where f (0, t) = 0 for all t
.
.
positive definite, V (x, t) is negative semidefinite V (o (t : "& to), I) does not vanish Identically in t
.
definite In some region about the origin and U (x, t) is positive definite in the same region.
9.12 Stability of Linear and Nonlinear Systems 1f the equilibrium state in case of linear, time invariant system is asymptotically stable locally then it is asymptotically stable in the large. But in case of a nonlinear system. the equilibrium state has to be asymptotically stable in the large for the state to· be locally asymptotically stable. Hence the asymptotic stability of the equilibrium state of linear, time invariant systems and those of nonlinear systems Is different, lf it is required to tc~l thr. asymptotic stability of any equilibrium state for a nonlinear system then the stability anruysis of linearised models of non-linear systems is totally insufficient. The nonlinear systems arc to be tested without making them linearized. There are various method based on Llapunov's second method such as Krasovskii's method
9-8
Modem Control Theory
liapunov'sStabilityAnalysis
which can be used to test sufficient conditions for asymptotic stability of nonlinear systems. We shall study in detail Krasovskii's method for asymptotic stability test.
9.13 Construction of Liapunov's Functions for Nonlinear Systems by Krasovskli's Method Liapunov's direct method is a most powerful tool available for doing the stability analysis of nonlinear systems. The stability analysis by Liapunov's method includes determination of a positive definite function V(x) called Liapunov's function. But unfortunately there is no universal technique or method for selecting Liapunov's function which will be unique for a special problem. Some of the Liapunov's function will give better results than other functions, There are various techniques available for construction of Liapunov's functions. Similarly for a given function V(x), there is no general method which allows us to confirm that the function is positive definite. But if V(x) is in the quadratic form in x1's then Sylvester's theorem can be applied to test the positive definiteness of the function. If it is not possible to obtain the Liapunov's function of required type then it docs not indicate that the system is unstable. The theorems that we have seen so far just provides sufficient conditions for stability. The indication from the fact that Liapunov's function of required type is not obtained is that the attempt to establish the stability of the system is failed. Krasovskii's method gives a sufficient condition for the equilibrium state to be asymptotically stable. Consider the system described by following equation. x = f(x); f(O) = 0. The origin is a singular point.
Lei us consider Liapunov's function as, V = fT Pf Herc P is a symmetric positive definite matrix. Now,
v (If
ax_ JI
~·Ti-
9-9
Modem Control Theory
I
Llapunov's Stability Analysls
is called Jacobian Matrix given by
ar,
i)f,
ilf1
";ix1
~
ilx,.
i1f2
i)f2
i)f2
~
CJx;-
~
.
.
Substituting ( in expression for V we have, fT JT pf + fT PJf
V
= Let
(T (IT P +PI)
f
Q
As V is positive definite, for the system to be asymptotically stable, Q should be negative definite. If in addition V(x) is tending to infinity as norm of x tends to infinity (II x II~-) the system is asymptotically stable in the large. Alternatively if Liapunov function is taken as V(x)
a
fT(x) . f(x)
J(x) is the [acobian Matrix given earlier ilf1
rx;l(x)
ilf2
ilf2 ()~2
-n
f (x)
Now we have,
V (x)
ilf1
()7;
()~I i)(
We have
i)(l
~
ilf2
~:
i)fn
";ix2
() f(xl Clx
"'"Tx di
iJf,,
I
;rx;; -
• = l(x) x
ee
l(x) . f(x)
i
= f~ (x) · f(x) + fT(x) · (x) fT(x) ·
= [ J(x)
· f(x)IT f(x) + fT (x) · )(x) · f(x)
f (x) · f(x) + (T (x) · I(x) · f(x)
{f(x) ( JT(x) + J(x)] f(x)
Let j(x) : JT
Llapunov'sStabilityAnalysis
9·10
Modem Control Theory
(x) + J(x). lf j(x) is negative definite then
Vex)
will also be negative
definite. Therefore V(x) is Liapunov function and the origin is asymptotically stable. If V(x) is tending to infinity as stable in large.
I!
x
II
-1 -
then the equilibrium state is asymptotically
The Krasovskii's theorem is different from the normal lincari:zation technique. It is not limited to small departures from the equilibrium stale. The Liapunov function V(x) and V(x) arc expressed interms of f(x) or ~ rather than in terms of x, Krasovskii method gives sufficient condition for asymptotic stability of non-linear systems and necessary and sufficient condition in case of linear systems.
An equilibrium state of a non-linear system may be stable even if the conditions stated above are not satisfied. So comments on stability of non-linear systems cannot be vigoursly made by applying this method. Let us have an illustration of Krasovskii's method consider a second order non-linear system described by following equations
x1 + bx2
X:z
Let us consider that g1(0) We will also assume that
= g2(0) = 0 and g1
lg1 (x1) + g2 {x2)J2 + (x1 + bxiJ2 -> ec as
II x II
(x1) and g2("2) arc real and differentiable. -1 -
.
Now we will obtain the sufficient conditions for asymptotic stability of the equilibrium slate x = 0 f1 = x, = f (x,) • g, (x1) + 112 {x2) f2
=
"2 = f(xi) = x1 + bx2
Let J(x) be the Jacobian matrix given as
J(x)
* ~1=1
0~1
~
iJf2 UX2
ox,
[g'1~X1)
[:,(x,)+g7(X7)] a;;-!x1 +bx2) I
gl~2)]
Here
iJg,(x,) d '( .\ iJg2(X2) ax;an S2 xv = ~
Now we have, J(x)
jf {x)
+ J(x)
Modern Control Theory
9 -11
[:~::~~ ~Hg'1;x1)
Llapunov's
Stability
Analysis
s2~2>J
g] lx1l+s'1 (x1) 1+g'.zC•2 )] [ g'ztx2)+ I b+ b
By Krasovskii's theorem if i(x) is negative definite then the equilibrium state x "' O of the system considered is asymptotically stable in the large. Hence if g'1(~1) < 0 for all x1"/.0 and 4b g'1(x1)- [1 + s2(x2))2 > 0 for all Xt "/. 0, x2 7' 0 then the equilibrium state at x : 0 is asymptotically stable in the large. This two conditions arc sufficient for asymptotic stability.
9.14 The Direct Method of Liapunov and the Linear System For linear systems, Liapunov's direct method proves to be a simple method for stability analysis, Use of Liapunov's method for linear systems is helpful in extending the thinking towards nonlinear systems. Consider linear system described by state equation x = Ax The linear system described by above equation is asymptotica.lly stable in the large at the origin if and only if for any symmetric, positive definite matrix Q there exists a symmetric positive definite matrix r which is the unique solution A Tp + PA = - Q. The proof of above theorem can be given. For this we will assume the symmetric positive definite matrix P exists which is the unique solution of the equation V(x) xT Px.
=
Consider the scalar function, V(x) = x1 Px Herc V(x) > 0 for x We have,
7'
0 and V(O)
=0
V(x)
.
V(x)
A{x)1 Px + x1 PA x
x1 (ATP+ PA) x -XT
.
Qx
As Q is positive definite, V(x) is negative definite
Llapunov'sStabilityAnalysis
9 ·12
Modern Control Theory
Let norm of x define as
II x II
(xT Px)l/2
V(x)
II x
112
V(x) --> -
as II x II --> ee
The system is therefore asymptotically stable in the large at the origin. The result is also necessary. To prove this, assume that the system is asymptotically stable and P is negative definite. V(x)
xT Px
V(x)
-[~1Px+xrp;) -( - XT Qx)
> 0
XT Qx
=
This is the contradiction as V(x) xTPx satisfies instability theorem. Hence the conditions for the positive definiteness of P arc necessary and sufficient for asymptotic stability of the system. The Liapunov's direct method applied to linear time invariant systems is same as the Hurwitz stability criterion.
Examples with Solutions
n•
Example 9.1 : Show that tlze following q1111draticform is positive definite
V(x) = Sx~ +xi +4x~
+2X1X2 -4.l'1:t3 -2x2:t3
Solution : The above given V(s) can be written as
l
rs V(x) =
XT
p x
=
[x1 X2 X3J
1 -2] 1 -J [xXz
J 1 -2 -1
x3
4
l
Applying Sylvester's criterion we have, 8
> O;I~ ~I= 7>0;1~-2
1I -1
-21 -I = 20 > 0 4
As o.11 the successive principal minors of the matrix P arc positive, V(x) is positive definite.
9 -13
Modem Control Thoory
'"*
Example 9.2 :
X2
=-
Liapunov's Stability Analysis
Determine tire stability of a non-linear system governed by tire eouations
X2
Solution : Let us select the Llapunov's function to be V =xi +x~
We have,
xf +x~ i)V dx ,
Fx";.. dt i)V
Now,
i)V dx2
+ dXz".
dt
i)V
dXz" = 2x2
~
2x1;
dxt
• 2 Xt = - XI + 2Xl X2
dt
dx2
dt
X2=-X2
Substituting,
dV
dt
- 2xi (1 - 2 x1 Thus
~y
xv - 2x~
is negative definite if 1 - 2x1 x2 > 0. Thus Liapunov's criterion is satisfied
and hence the origin of the system is asymptotically stable.
11*
Determine the stab;Jity of tire system described by tire followingequation
Example 9.3 :
.
x=Ax;A=
[-1 1] -2 -4
Solution : To check the stability of the system described by equation ~
=
Ax we will solve the equation ATP+ PA= - Q where for any symmetric, positive definite matrix Q. there exists a symmetric positive definite matrix P. We can find out matrix P for any arbitrary choice of positive, definite real symmetric matrix Q. Let us select Q = I, where l is identity matrix
-Q
ATP+PA
Let
p
=
[Pn
P12
P21
P22
l
J
9 ·14
Modem Control Theory
Liapunov's StabilityAnalysis
o]
-1 [ 0 -1
But P is symmetric matrix ., P12 = P21 :.[
-2p11-4P12 P11 -5P12 -2P22
r11-Sr12-2Pn]=[-1 2P12 -Sp~2 0
-2 P11 - 4 P12 P11 - 5 P12 - 2 P22 2p12 - 5 P22
OJ -1
- 1 0
- I
Solving the three equations using Cramer's rule,
1-2
['. = 11
I
-I OI =-2(25+4)+~(-5) -5 -2 =-2(29)- 20 = -78 0 2 -5
-~ ~~ -1
1-2
I
I
01
=-1(25+4)+4(0-2) -2 =-29-8 2 -5 =-37
-1
01=-2(0-2)+1(-5)
1 0 -2 =4-5=-1 0 -1 -5
-~ ~ -~1 0
2
-1
rt-2(5)+4-1(2) =-10-4-2 =-16
Pu 6Pt2
P12
Pit
-1
1
= -ll- = -78 = 78
Modem Control Theory
9·15
Liapunov's Stability Analysis
Using Sylvester's criterion it can be seen U1J1l P is positive definite. Therefore origin of the sytcm under consideration is asymptotically stable in the large.
i•
Example 9. 4 : Use Krasouskii's theorem to show that the equilibrium state x = 0 of tllr system describedIn;
.
x1 = - 3
x1 = x1
-
X1
+ X2
x2 - xl
is asymptotically stable in the large.
(VTU: Jan./Feb.-2005)
Solution : We have, ;I : 1, (x) = - 3x1 + X2
~ =
f2(x)
= x1
-
x2
-
xi
The Jacobian matrix is given by
a12 ~
ar,
; ~=-1-3xi
1
,
2
(-3 1
1
-1-3x~
Let
Q
)TP ... PJ
Let
p
[Pn
P21
]
P12]~[1 ~]be identity matrix P22 0
Q
1 ][I OJ [I 0 ][-3 1 ] [~3 -1-3x~ 0 1 +0 l 1 -1-3xi
Q
I ] [-3 l ] [~ [~3 -l-3xi + 1 -l-3x~ = 2
2 ] -2-6x~
Modem
Llapunov's Stability Analysis
9·16
Control Theory
For system to be asymptotically stable Q must be negative definite i.e. - Q must be positive definite
-Q
= - [-{>2
2 ] [ 6 -2 ] -2-6"i = -2 2+6x~
Using Sylvester's criterion, 6 > 0 and 12 + 36 x~ - 4 > 0 i.e, 12 (1 + 3x~) > 4 then the equilibrium state x = 0 is asymptotically stable in the large.
of the follmving
Example 9.5 : lnwstigate th« stability method of Liaupnov.
,,...
• Xz
= - X1 - Xl2 X2
Solution : Given that ~1
no11-/i11erar
(VTU: July
sysl
using direct
/Aug.· 2005,J•nJFeb.-2005)
= "2• ~2 = - x1 - x~ x2
Let the Uapunov function be, V = x~ + xi
.
.
.
2 XI XI + 2 X2 "z
V
2 Xl
"2
+ 2 X2 (- Xl - X~ X2)
2x1 X2 - 2x1 X2 - 2 xf x~ - 2 xi xi It ran be seen that V < 0 for all non-zero values of x1 and x2. Hence the function is negative definite. Therefore the origin of the system is asymptotically stable in large.
"*
Example 9.6 : A smmd order systmi is reprtsnrttd by
·
x=AxwhereA=
[o 1] -1
-I
Assuming matrix Q to bt identity matrix, solo« for matrix P in the equation ATP+ PA= - Q. Use Liapunov theorem a11d detemrint the stability of tire origin of the system. Write th« Liapunou function V(x). (VTU: July/Aug.·2005, JanJFeb.-2007) Solution : ;
Let
/
=
Ax
p
P12} [P11 P21 P22
Q-['
- 0
CJ1
- 2 P12
P11 - P12 - P22 2 (P12 - Pn)
Uapunov's Stability Analysis
9 ·17
Modem Control Theory
= + 1/2
-1
:. P12
0
:. Pn - P22 = P12 = 2
-1
:. P12 - P22
I
I
2
1
P12 +
P11
P12 + Pzz = 2 + 1 = 2
=
2
l
P22
+
2
=-
I
2
= 1
1
3
' [! '.] Using Sylvester's criterion, P can be tested for positive definiteness
• P is positive definite. Hence the equilibrium state at origin is asymptotically stable in the large. The Liapunov's function is
V(•) • .'p
V(x)
>"
('> '>I
[f '.] [:;]
Modem Control Theory
''*
Llapunov's Stability Analysis
9·18
Example 9. 7 : A system is describtd by tM following equation : (VT\J: Jan/Feb.· 2008)
;=Ax Where A=[ ~I ~} Ass1m1i11g matrix Q to be the ide11tify matrix, sol!.lf! for matrix P and commmt 011 the stability of the system using the equation ATP + PA = -Q.
Solution : To check the stability of the system described by equation x =Ax. We will solve the equation, /\ T P+ PA =-Q where for any symmetric, positive definite matrix Q, there exists a symmetric, positive definite matrix Q, there exists a symmetric positive definite matrix P. We can find out matrix P for any arbitrary choice of positive, definite, real symmetric matrix Q. Let us select Q = I. where I is Identity matrix. A TP+PA =-Q. p
Let
=
[Pn
P12] P22
P21
A
=
[-1 -2] I
-I
,
AT=[=~
~]
But Pis symmetric matrix :.p12 =p21 -2pu +P12 +P21 [-2p11 -Sp12 + P22 -2p11+P12+P21
-1
-2p11 -SP12 +r22
0
-2P12 -2r21 -8P22
-2p11 -5p12 +P22 ]=[-I -2p12 -2P21 ··Sp22 0
-1
Solving these three equations using Cramer's rule
0] -1
Modem Control Theory
1
6 = [-2 -2 -5
is]
-2 -2
[-I 0
-5
-1
-2
[-2
-1
6p II =
"'Pu = -2 0 -2 -1
is]
~sl
~I]
[-2 6p22 = -2 -5 -2 -2 -1
9 -19
=-lliQ+fl-1(16+ =-2(42)-18-6
~+ 1(4-10)
=-84-18-6=-108 =-1(~0+ ~::!{.!)+ 1(-5) =-80-1-5 =-86
=-2(0+ 1) +!.{!§_+~+ =-2+11i+2 ill
!fn
=-?:.{5!-1(2)-1(4-10) =-10-2+6=-6
P11
-86
86
-6- = -108
108
6P12
P12
1.8
P21
tipll
Llapunov's Stability Analysis
-ti- = -108
6P22
-6
-ti- = -108 p -
86
6
108
-18]
108 106 [ -18 6 108 108
Using Sylvester's criterion, it can be seen that P is positive definite. Therefore origin of the system under consideration is asymptotically stable in the large. ..
>m+
•
,3
Example !Lii ; Consider lire system willr differential equalio11 c +kc+ k 1 c + ( = 0. Examine Lite stability by Uapu110t•'s met/rad, gi1Je11 tlrat k > 0 and k1 > 0. CVTU: July/Aug.-2007)
Solution: Consider the given equation, -
•
.J
c+ kc +k1 e +c=O,k > 0, k1 > 0
Let
e
">
M
.. Modom Control Thoory
9. 20
Liapunov's StabilityAnalysis
As there is no specific procedure for selecting Liapunov's function but it has to be selected from experience. Let us select f = x~ + x}ns Liapunov's function df
dt
The above equation is negative semidefinite and hence the system is stable.
I»*
Exami11e the stability of ti~ system d~cribed lly tlte foll1r.ui11g equation by Krasovaskii's theorem. Example 9.9 : XI= X2
-X1
= X1 -X2 -X~
(VTU: July/Aug.·2007!
Solution: We have,
The Jacobian matrix is given by,
r~l L
Let
Q
Let
p
Q
()
-1-3x~
]
)T P+P)
[P11 P12 ]=[1 P21 P22 0
[~l r-1 LO
~] be identity matrix
r
1 ][' OJ 0][-1 -1-3x~ 0 1 + 0 1 1
I
]
[-1
-1-3x~ + 1
0
-1-3x~
]
0 } -1-:Jx~
= [-21
l
-2-6x~
]
Modem Control Theory
Q
9 ·21
=
Llapunov's Stability Analysis
-2 1 ] [ 1 -2(1+3xn
For system to be asymptotically stable Q must be negative definite i.e. - Q must be positive definite. -2
- Q
= -[ 1
1 ] [2 -1 -2(1+3xl) = -1 2(1+3xn
]
Using Sylvester's criterion, 2>0 and 4+ 12x~ -1 >0 i.e. 4+12xi > 1 i.e. 4(1+3xD > l then equilibrium state x = 0 is asymptotically stable in the large. Example 9.10 : Deiereminewlitthu or not fall
,,_.
Solution : Consider the given equation Q(x,,x2)
= lOxi +4xi +x~ +2X1X2 -2X2X3 -4X1X3
Consider,
=
P11Xi +p22xi +p33x3 +2P12X1X2
+2p23x2X3
+2p13X1x3
Equating coefficients of above equation with those from given equation P11 =10 ;
P12 =4;
2P12 =2;
P12=1
2p23 =-2;
P23 =-1
2p13 =-4;
P13 =-2
P33=1
The given equation can be rewritten as,
4 -1
=~] [:~1 1
x3
9 ·22
Modern Control Thoory
Llapunov's StabllltyAnalysis
Applying Sylvester's criterion we have, 10 > 0;
1
10 1
~I
= 39 >
o;
10 1 -2] 1 4 -1 [-2 -1 I
10(-1-1)-1(1-2)-2(-1+8) 30+1-14 =17 >0 As all the successive principal minors of the matrix P are positive, Q(x1,x:z) is positive definite.
I"*
: Using Lyap1111ov's direct met/rod, find tire range of K lo guarani~ stability of the system slrown in the Fig. 9.3. (VfU: JanJFeb.· 2006) Example 9.11
Solution:
Fig. 9.3 The above diagram can be redrawn as,
~----------
9. 23
Modem Control Theory
Liapunov's Stability Analysis
In matrix from above equations arc written as,
.
XI X2 X3
Here
A
[!
0 -1
[~
0 -1
-:][:;l -1
X3
-:i
:.AT=[~
-K
-1
-1 0
IJ
To find the range of values of K for stability, we will solve the equation A r P+PA =-Q.
Let us select
Q
This selection of Q is valid as this docs not make xT Qx identically equal to zero except at the origin.
-l
0
l [::; ::: :~ l
~
-1
PtJ
P23
P33
-:1-[~ ~ ~1
P12
+ [::;
Pn
P1;
p23
:~] [~ ~1 p33 0 0 -1
Solving the equation we get 1 J«2-K) p
0 -1 2(2-K)
0 l 2(2-K) 1 2(2-K)
-1 2(2-K) 1 2(2-K) 1 (2-K)
0 0 -I
9 -24
Modem Control Theory
For
r
Liapunov'sStabilityAnalysis
to be positive definite it is necessary and sufficient that K(2-l<) >0 K>0;2-K>0,2>K
Thus for 0 < K < 2, the system is asymptotically stable.
Review Questions 1. Wrilt a note on stability in the sense of Liapunot1. 2. Define ii stability iii Asymptotic sta/lility iill Asymptotic stability in the large. J. Shotv the graphical rrpres
semirkfi11ittnt55vi inrkfinitmtss 5. Disc11ss Quadraticform mid Hermitian form. 6. Erplai11 Unpunov's second method and /Jap1111ov's stability theortm. 7. Write a nott 011 stability of lintJJr and nonlinear systems. 8. Write a short note on Kra5ovskli's mtlhod of constnicting Uapunov's functions for nonlinMr systems. 9. Examine th« stability of ti~ origin of the fo/lOtving systtm
- 6 XI - 5 X2
X2
JO. Write a Liapuno»function for the system
rl ;2;, Lr 1] [ x, J' J l 2 -3 -i
•i
Dtlermi"' the stabi/ity of ti.- origin of ll1t system. 11. Determine /ht stability of the equilibriumstale of the following syslmf. X1
- X1 - 2Xl:
.\'2
X1-4xz-]
+2
12. Statt and explainLiapunoo's thtoremson iJ Asymptotic stability
ii)
Global a.•ymptotic stability and iii) Instability. = 0 of system dtscribed by
13. Ust Krnsousk'ii's theorem to shaw that tht equilibrium slate x
x 1:::: - 3x1.,. x1 x2 ; >:z - x2 - x~
9 ·25
Modem Control Theory
Llapunov's StabilityAnalysis
14. Investigate tl"' stability of folluu1ing non-lineer system using direct mtthod Uap1111cro. Xl
= X2
X2=-x1-xl
X2
15. A 5'QJnd ordrr system is rrprtstnted by ~=Axrvl1LrtA=f0
L
-I
']·
-I
Assuming matrix Q lo be U!entity matrix, sol<>e for matrix p in the equation ATP + PA ~ - Q. Usr Liapuno» theorem and dtlem1i11e I/it stability of tht origin of systrm. Write tilt Uapunov func!iDn V(x).
DOD
(9.
26)
Modern Control Theory ·-=
~
•.
.........
Chapterwise University Questions with Answer Jon./Feb. • 2005
July/Aug. • 2005 Jon./Feb. • 2006
July/Aug.• 2006 Jon./Feb. • 2007
July/Aug.• 2007 Jon./Feb. • 2008
(P • 1)
CD
State Variable Analysis and Design
Q.1
Mention tlie diS11dlHl11tages of conr>enlionAJ amtTol lhtory and nplain how ~ art overco= in modern control lhtory with particulilr refermce lo (i) Non·lilWlr systems (ii) Time varying system (iii) Analysis (iv) DtSign and (11) Computerapplications.
Ans. :
Reier section 1.1.
Q.2
Explain the concepts of state NriAbk, state and state model of 11 lirmtr system.
Ans. :
Refer section 1.2.
Q.3
Define the concept of i) Stott ii) S14te variJlbles iii) Stolt space tu) State vector.
Ans. :
Refer section 1.2.
Q.4
Compare clsissical control theory against modern control theory.
(Ja.nJFeb.-2005,10 Marks)
(JanJFeb.·2006, 6 Marb)
(July/Aug.-2006,Jan./Feb.-2008,6 Marks; Jan.lFeb.-2007, 10 Marks)
(July/Aug.·2006,4 Marks) Ans. : · Refer section 1.1. Q.5
Ust advantages of modern control theory over conventional controt theory.
Ans. :
Refer section l.1.2.
Q.6
Drove Ille equationof the vector model differential state equation.
Ans. :
Refer section 1.3.
Q. 7
Mention tlie differences betwt't11 stalt space tedmiq11tSll1ld classicalapproach.
Ana. :
Refer section i.1.
(July/Aug.- 2007, 4 M•rla)
(July/Aug.- 2007, 1 Maries)
(Jan./Feb.-2008,6 Marks)
ODO
(P • 2)
p.4
Modem Control Theory
State Space Representation
Unmriu the fol/{1Wingequation in the neighbourhood of the origin.
Q.6
"29 =lnn-129-3sin9+211eul2 dt2
+113
Obtain tht approximate response O(t) for u equilibrium.
0.02, with the system initially at (JanJFcb.·2006, 6 Marlcsl
Ans.:
sin 0
.... in the neighbourhood of origin (20)3 (20)5 20+-3-+-5-+ 20-30+2ueu12 11
cu/2
(11/ 2)2
•29
+u3 (u/ 2)3
I+2+-2-!-+-3-!-+
d20
-0+2u
dtT
= - 0 + 2u = -
.
... neglecting higher order terms
d2a +9 = 2u dt2
X2(t)
in the neighbourhood of origin
... lineanzed equation
X1(t) + 2u
Thus the state model is,
[~::::] ~ [~1 ~][~~~:~]+[~]
u
p.5
Modem Control Theory
[sl-A)
Adj [sl-A[
[sI
s[l
=
OJ-[o-1
0 1
State Space Representation
']=(s -1]
0
1
s
[+1s -l]T s [s 1]s =
-I
:L[s/+1
-Ar'
Adj[sl-AJJ~l Isl-A I 52 +I
2_
s2 +I ZSR
L-1{q{s)BU(s)I where U(s)
L-1
![s~ l s :+ 5 1
s +1 s 2
2+1
l
l
52 s+I
l
= O.OZ s
[~)[0.~2] =L-1
[s(~s~~)l+1 l 2
L-1 {~+ Os+C} s +1
L-1 { s(~~~l)} As2 + A + Bs2 + sC
2
s2 +I I
52
0.04
i.e, A + 8 = 0, C = 0, A = 0.04, 8 = - 0.04.
0.04 - 0.04 cos t 0.04 - 0.04 cos I Q, 7
... u
= 0.02
Clioosi11g appropriate pliysica/ i.-oriables as slate variables, oblai11 tlu: slate model for I/re electric circuit sliown ill Fig. 2. (JanJFeb.·2006, 8 Marks) 1 F
1 H
10
y{I)
Fig. 2 Ans. :
Refer example 2.29.
M
P-6
Modern Control Theory Q.8
.
)'(.<)
State Space Representation
>(>+2)(.<+3)
For ll1e tmnsfcr [unction -- - = -··-·,--·-·-R(.<) (> + 1)· {; H) O/Jtai11 the stat« mode!
111
ii Phase oorwble canonical form iii lordun cm1011ical form
UanJFel>.-2006, 8 Marks)
Ans. :
Refer example 2.30.
Q.9
l'ig. 3 s/1ows tire />lock diagrcm of a speed co11lro/ system willi slate variable feedback. Th« driue motor ;5 mt annature controlled de motor with armnturt resistanr« R0, '1rmature inductance L,., motor torque constant Kr, inertia referred to motor shaft /, viscous friction coefficient referred to Ilic motor slrafl B, back emf constant Kb and tachometer K1. '111< applied armature voltage is controlled by a llrrce phase [ull-conoerter. c, is conlrol 110/tage, is armature voltage. e, is lite reference t'Oltage correspo11di11g lo the desired speeil. Taking X 1 = rn (speed) and Xz = ia (an11alure current) tis t/11• Mak variables. u = e, as Ifie input, and y = (O as tl1e output, derive 11 state vanable mode for the fe<>dbnck system. Uuly/Aug.·2006, 6 Marks)
•a
Fig. 3 Ans. :
Refer example 2.31.
Q.10
For a RLC network sho11111 i11 Fig. 4, write tire stale model in matrix 11ot11tio11 cl1oosing X1(t) = i(I) and X2(tl vcW where X1(1) and X2W are stale variabfrs, Vc(t) is 011tp11t, v(t) is input. Uuly/Aug.-2006, 8 Marks)
=
~T
~,, :u Tl Fig. 4 Ans.
Refer example 2.1..
M
p.7
Modem Control Theory Q.11
For a transfer ftmction gioen l1y G(s)
State Space Representation
= -,--2--write the stalt model i11 tltt ii Phase s·+JF+2
variable form ii) Diagonal form.
(Ju.ly/Aug.-2006, 8 Mari...)
Ans. :
Refer example 2.32.
Q.12
Stott
Ans. :
Refer section 2.3.1.
Q.13
Obtain the state space representation model for Giw11 R = I MO and C = l µI'.
tire
properties of [ordan matrix,
(July/Aug.·2006, 6 Marks)
tiff
follcnving electrical circuit i11 Fig. 5. (JanJFeb.·2007,10 Mades)
R
R
u(t)
y(t)
Fig. 5 Ans. : Q.14
Refer example 2.22.
... .. .
Obtain tit;, stale space representation of the folluwing system and draw its phase variable dillgram: Y +6 Y +II Y +6Y=6tL (JanJFeb.· 2007, 10 Marks; Ju.ly/Aug.·2007, 6 Marks)
Ans. : Taking Laplace transform of both sides and neglecting initial conditions we get, Y(s) 6 U(s)- s3+6s2 +lls+6· Then refer example 2.15. Q.15
Obtai11 the state modd of the electrical network shown in Fig. 6 selecting 'V', 'i1 ·and as state variables and voltage across R2 and cuntnt 12 through R2 are the output variables Y1 and ¥2. (July/Aug.·2007, 8 Marks)
·;2•
Fig. 6 Ans. : Refer example 2.19 for the procedure. The direction of i1 is opposite in this example. Thus the matrices A and B are,
.I
P-8
Modem Control Theory
RI -r; A
+-
0
-I-; i
1
Li
L1
R2
0
State Space Representation
+...!... L2
-c -c:
0
B
=
0 0
There are two outputs,
Eo(tl
Y2
i2 = X2
c Q.16
=
yI
[~
i2R2
=
R2X2
~2]
For tire network show11 in Fig. 7, dwosing i1(t) = X1(tl and iz(t) = X1(t} uariables. obtain tlie stale eq11atio11 and output equatio11 i11 vector matrix form.
as
state
I
(JanJFeb.·2008, 8 Marks)
1 H
Hl
IH l~t) u{t)
~------i i_t_)l~---~ 10
1n
y(t)
Fig. 7 Ans. :
Refer example 2.33.
Q.17
Ob1JZi11 the stat~ model in plrase variableform and write I~ block diagramfor t/1e system represented by,
(JanJFcb.·2008, 6 Marks)
,,Jy .i2y dy +6-+11-+10y=311(I). di 3 1112 dt Ans. :
Refer example 2.16.
Q.18
For tire following transferfunction obtain t~ state modtl in catwnicolform: (JanJFeb.·2008, 8 Marla)
G(sJ = Y(s) U(s) Ans. :
=
6 s3 + 6s2 + lls + 6
Refer example 2. I 5.
ODO
Matrix Algebra and Derivation of Transfer function
® Q.1
Determine lite transfer matrixfor lite system
QanJFeb.·2005, 10 Mules)
[x,x ] [-3 l][x' n 4 6]o ["'] :[YI ]=[J -1] [X1] 1
=
-2 0 X2
-5
u1
yz
B
1
Xi
Ans.:
Refer example 3.20.
Q.2
Find t11e transformation matrix P I/tat transforms lht matrix A into diago11al or Jordan formwhmA=
rl14 -1I -2] I
0
2. 3
QanJfeb.·2005, 10 Marks)
Ans . : Refer example 3.15. Q.3
What are ge11raliud eigen vectors? How are they determined? Quly/Aug.·2005, 5 Marks)
Ans. :
Refer section 3.10.
Q.4
Convert the followi11gstate model into canonicalform
QulyJAug.-2005,8 Marks)
x =[~ ~]x+[~ ]11 1
Y
= [1
OIX
Ans. :
Refer example 3.21.
Q.5
Convert the following square matrix A into Jordan canonical form using a suitable non-singular transformatio11 matrix P. QulyJAug.·2005, 7 Marks)
A=f~L-4 -9~ Ans. :
~i
-6
~for example 3.22.
(P • 9)
Modern Control Q.6
Matrix Algebra & Derivation of Transfer Function
p • 10
Theory
Consider the matrix
~i
A=[~1 -:3
-1 i) Find the eigen val11es and eigen vectors of A ii) Write tlte modal matrix iii) Show that lite modal matrix indeed diagonalizes A. Ans.:
Refer example 3.23.
Q.7
Given the stale model X " AX - Bu, y " CX toher« A
=[ ~
~
-1
-2
~i'
-3
B
= [~]
and C
=
(Jan.JFeb.-2006, 12 Marks)
[1 0 OJ
l
i) Sim11/nle and find tire transfer function ~~~ using Mason's gai11 formula. ii) Determine the transfer f1111ctio11 from tlie state model formulation. (JanJFeb.·2.006, 7 Marks) Ans. : Q.8
Refer example 3.24. ti« vector [_~] is an eigm wetor of A = [-~ ~ -I -2
~1
Find tire cigen value of A
0
Ans. :
corresponding to tlu vector given. (July/ Aug.·2006, 5 Marks) Refer example 3.25. Show that tlU' cliaracteristic equation and eigm Mluts of a sytem matrix art 111wariant under linear transformation. (July/Aug.·2006, 8 Marks) Refer section 3.9.
Q.10
Delennine tire transferJunction of the giwn state vector differentialequation below :
Ans. : Q.9
[~~]=[~ ~ X3
0 -4
~1[;:]+[~1]11
-5
X3
5 (July/Aug.·2007, 8 Marks)
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
p • 12 1 (;
s+5 s(s+-l)(s+l)
1 s+5 5 s - s(s+4)(s+1) + s(s+4)(s+1) 1 -s-5+5 s+ s (s+-1) (s+l)
=
s s (s+4) (s+l)
(s+4) (s+l) - s s (s+4) (s+l)
(s+.t)(s+l) T.F.
I
= ;;-
s2 +5s+4 -s s(s+4)(s+l)
s2 +4s+4 s (s+4) (s+l)
Obtai11 eigen values, eigm vectors a11d stale model in canonical form for a system described by the following stale model : (July/Aug.·2007, 10 Marks)
Q.11
~ ~ ~] [~;:]3 = [-12 -7 -6
[=:] .T3
+ [~] w11d y 2
= [1 o o J
[=:] X3
Ans. : Refer example 3.14 for obtaining M. Once M is obtained find M""1 and the new matrices B = M"1B and C =CM
[
Q.12
l
1 2 -3.1
M= -I -4 -1 1
[4.5 2.5-2 -1I } B= [6.-55}
M""1= -3
3
2.5
1
C=(I
-12 -7
Refer example 3.14.
~1
-6
2 1].
4.5
Obtain tht model matrix for tire matrix givt11 btlow :
A-[~~ Ans. :
1.5
(JanJFcb.·2008, 4 Marks)
Solution of State Equations
Q.1
For a o1Jslem represerrted by X = AX tire respc~ lo one set of i11itial co11diti1ms L~ X(I) 2''-41]
= [ c-1•
.r initia ... I cond.ition ' a11d anot I1er set oJ ts. X( I)
the state transition matrix OW.
=
c-it]
[4c_21
. matix . . D etermine
A and
(JanJFeb.-2005, 10 Marks)
Ans.:
Refer example 4.19.
Q.2
Mr11tio11 ilu: ccmditions for complete controllability and complete obsavability of conti11ous time systems. Using these, explain Ille principle of duality between controllability and observability.
Ans.:
Refer sections 4.12 and 4.13.
Q.3
Use controllability and observability matrices to determine wlretlier tlie system represented by tlie flow grapli show11 i11 Fig. 1 is completely co11trollable and completely obstroablt.
<J•nJfeb.-2005,JanJFeb,-2008 10 Morks,;July/Aug.-2007 6 Marks)
-1
y
u
Fig. 1 (JanJfeb.-2005, 10 Marks) Ans. :
Refer example 4.33.
Q.4
Given thr time i11varia11t system :
(P - 14)
M
Modern
p -15
Control Theory
=
Solution of State Equations
e 1 and
and 111111 11(tJ y(I) = 2-nk I, find X1(1) and X2(t). Find also XifOJ a11d X2(0). Wlmt li.1ppe11s if n = O? (JanJFcb.·2005,10 Marks! Ans. :
Refer example 4.34.
Q.S
What is a stat« transition matrix ? I.isl tire properties of state transitio11 matrix. (July/Aug.·2005, July/Aug.·2007,Jan.fcb.·2008,6 Marks; July/Aug.·2006, 5 Marks, Jan./Feb.·2007, 10 Mari<s)
Ans. :
Refer sections 4.3 ,111d 4.4.
Q.6
Gi11!'11
the stale model tif a sy~tem : X
[~
Y
:5] X
+[~]II
(1 OJ X
- [!.}
with initial condillo11s X(O) -
Determine : i)
Tile state transition matrix.
ii)
The stale trnnsitio11 equation X(t) and 011tp11t Y(t) for an 1111il step illput.
iii) lliverse state tra11sitio11 matrix.
Quly/Aug.·2005, 14 Mark51
Ans. :
Refer example 4.35.
Q.7
F.rp/ai11 tilt concep! of co11trollabi/ity and ob!:croabi/ity. Quly/Aug.·2005, )uly/Aug.·2006, 6 Marksl
Ana. :
Refer section 4.11.
Q.8
Determine tire controllability and observability of tire following stale model.
y
ftO 5 l)X
Quly/Aug.·2005, 8 Marksl
Modem
P-18
Control Theory
Solutionof State Equations
Ans. :
Refer example 4.36.
Q.9
A system represented by following statt model is controllable but not obsnvablt. Show t/ull tire no11-<Jbstruability is due to 11 pok-:uro ettnetllation in C/sl-Af18.
[! :Jx +mu
x
-~1
=
Y
[1 I O]X
(July/Aug.-2005, 6 M..001
Ans.:
Refer example 4.37.
Q.10
List 11tleast thrte imporl11nl propenie» of th« state transition matrix.
Ans.:
Refer section 4.4.
Q.11
Consider tlie lwmogenevus equation X = AX, wltere A is a 3 " 3 matrix. The following three solulio11 for three different initial comfitions art 111N1i111ble,
(JanJFeb.-2006, 3 Marks)
i:: lJ' [ .-1
[
i)
,-21 _;~-21
.
l[ l 2r31 ' -6~-31
Identify the initial conditio11s
ii) Find the stale transition matrix iii) Hence or otherwise find the system matrix A.
CJanJFeb.·2006, 10 Mulu)
Ans. :
Refer example 4.38.
Q.12
Obt11i11 the time response y(I) of lite system give.. btlow by first transforming the state model into a 'Canonical model'.
X=[ ~
~
-6
-11
~1
-6
X + [~] u, Y 2
u is a unit step function and Ans. :
Refer example 4.39.
= [1
xT (0) = [O
0 OJ X 0 2]
(JanJFeb.-2006, 12 Marb)
p ·18
Modem Control Theory
Solution of State Equations
Obtain the state transition matrix using :
Q.17
i)
l.llplace transformation method and Cayley-Ha111i/to11mtlhod. For tire system describe by,
ii)
X(I)= [~
~]X(O)
(JanJFeb.-2007, 10 Marks)
Ans. :
Refer example 4.43.
Q.18
State the conditions for C-Ompfetely controllability and complei« o/Jser.x1bility. Determin«
•;: :"[t~ [
c~nrro_;•]b[i;~l+[ai1]d[ulobseniability -11
4)
,J
6 X3
of
the
SIJSlem
described
by,
1
(JanJFeb.·2007, 10 M3rks) Ans. : Refer section 4.11 and example, 4.36 for the procedure. The system is completely controllable and completely observable. Q.19
Obtain the olJS<'rmble phas« variable slate model of T.F.
=
T(s)
=
)'(>) U(s)
= bo>3
+bis2 +bi> +b3. Draw tire signal .flow grap/1 of Tis). >3 +•1>2 ... azs +n3 (July/Aug.·2007, 8 Marks)
Ans. :
Refer example 4.46.
Q.20
Obtain the controllable phase t1ariable form of tlie transferfunction 'r.t:
a
Tts)
=
Y(sj
= bo> 3 +b1>2
0(>) Ans. :
Refer example 4.47.
Q.21
Compute ell• for the given matrix : A-
1 -
+b3.
... bzs s3 +n1s2 +a2s+a3
ro]
6 01 ·1h- [ o ·A[0 6} " - -C•) 0 '
Quly/Aug.·2007,6 Marks)
-l-(r 6l)
ooJ 6
(July/Aug.-2007,6 Marks)
Modem
Ans. :
Refer examples 4.27 and 4.28.
Q.26
Dd<•ri11111t
.X (I}=
t/w compfrtr tim1•
f[-2O
']s(t),
Solution of State Equations
p ·20
Control Theory
rt'S)JOllSC
of t/1r $ystrm i:1w11 /1y :
lnnd
uihrre X
(I
Ans. :
Refer example 4.7.
Q.27
Deter111i11r tile co11trol/ability dtSCl'ibed by.
IJ
Y(t)
=fl
-l I X(IJ. (JanJFel>.·2008,
12 Marks)
a11d obsernability 11si11K K11l111a11 's ltst for tire systn11
A (J;m./Fcb.-2008,
10 Marks)
Ans. : Refer example 4.36 for the procedure and the given system is completely controllable and observable.
DOD
Modern Control Theory
Q.11
P-23
Omsider the systcn defined by ;
Pole Placement Technique
= Ax + Bu, w/1ere A=[~ ~ ~] -1
B"'
-5 -6
[~.J1
IJy
l
usitrg slate feedback rontrol u a -Kx, ii is desired to /rave the c/OS
Refer example S.10
Q.12
Consider the systtm x =Ax+ Uu a11d Y = C><. w/rerr
A=[~ ~ -6
1111d C = [t 0 0 ]. Determine lite obsm.'fr gai11 matrix by the
US<'
-11
o
l J
I -6
rol B
=
loJ' L1
of:
The direct substitution me/hod. ii) Ackermann's formula. i)
Assume that µI =-2+j3.4641 Ans. :
the desired Eige11 values of tlrt and µ2 =-2-j3.4641, µ 3 =-S.
obserwr gai11 matrix arc <J•nJFeb.-2008, 8 M•rksl
Refer example 5.11.
ODD
'®
Controllers
I
Q.1
l:xplain effects of a Pl controller on the static and dynamic responseof a system.
Ans. :
Refer section 6.9.
Q.2
Consider a typical second order, type one system with u11ityfeedback, being controlled l1y a PD controller and stun» that i) Damping increase with PD control.
Ans. :
Refer section 6.14.
Q.3
A temperature control system has th« block diagram given i11 Fig.1. The i11p11t signal is a voltage a11d represents the desired temperature er Find the steady-state error of the
(July/Aug.·2005, 5 Marks)
(JulylAug.·2005, 4 Marks)
system whm
!+ is a unit step function a1ul i) D(s)
= 1 i1) D(s)= I+
0/
=
iii) Dis) 1 + 0.3 s. What is tl:e effect of the i11tegral lmn i11 lite Pl controllerand tl:t derivative term i11 PD controller on the steady state error ? (July/Aug.·2007, 8 Marks) O(r)
~
200
9
Plant
Fig. 1 Ans. :
Refer example 6.16.
Q.4
Wl1J1t is a controller ? £xplai11 P, l, Pl and PIO controllers.
Ans. :
Refer sections 6.1, 6.5, 6.6, 6.9 and 6.11
Q.5
Define controller. Explain properties of P, Pl and PIO controllers with llie help of block diagram. (JulylAug.-2007, 8 Marks)
Ans. :
Refer sections 6.1, 6.5, 6.9 and 6.11
Q.6
Deft11e co11tro/ler. Explain PD, Pl and PID coturollers. What are the advantages of PID controller ? (JanJfeb.2008, 6 Marks)
Ans. :
Refer sections 6.1, 6.10, 6.9 and 6.11 (P • 24)
(JanJfeb.-2007, 10 Marks)
ono M
jG)
Nonlinear Systems I
Q.1
£xplai11 the folluwingbehaviour of 11011-lincar systems : Iii Frequency amplitudedependence . (ii) Multivalued responses and jump resonances.
Ans. :
Refer section 7.2.
Q.2
Discuss tl1e basic featuresof the followingnon-linearities i)
Non-linear friction
ii)
On--0ff controllers
iii) Back lash
(Jan/Feb.-2005,10 Marks)
(Jan.lFcb.-2006, 9 Marks)
Ans. :
Refer sections 7.4.5 and 7.4.6.
Q.3
What are inherent nonlinearities ? Explai11 1my three of thmdJuly/Aug.-2006, 6 Marks)
Ans. :
Refer section 7.4.
Q.4
Sketch the following non-linearities : Ideal relay ii) Reilly with dead zone iii) Reilly with dead zone and lrystcrisis io) Relay with hysterisis v) Dead zone. (July/Aug.-2006,4 Marks) i)
Ans.: Q.5
Discuss tlte basic features of t/1t Back/as/1 non-linearities wit/1 suital>/efigures.
Ans. :
Refer section 7.4.6.
(July/Aug.-2007,4 Marks)
Q.6
Explain any three non-linearities in control systems.
Ans. :
Refer section 7.4.
QanJFeb.-2008,6 Marks)
000
(P - 25)
'® Q.1
Phase Plane Method
I
Determine the kind of si11gularity for tJJch of the fo/1111ui11gdiffermtioleq11atio11s. (i) y-3y+2y=O
(ii) y-8y+17y~34 Ans. : Q.2
(JanJfeJ>.·2005,10 Marks)
Refer example 8.1.
What is a phase 11/a11e 11/ot? Describe delta method or any other method of drawi11g phase plane trajectories.
Ans. : Q.3
(JanJFeb.·2005,10 Maries)
Refer sections 8.1 and 8.5.
Witlr rejerenc«to 11011-lineur system explain : (i) /1111111 r.•sona11ce (ii)
Limit cycles
(July/Aug.·2005,6 Maries)
Ans. :
Refer sections 8.3 and 8.2.
Q.4
Whal arc si11g11lar points? Explain tire classification of singular points based 011 the location of dgm values of tire system. (July/Aug.·2005; JanJfcb.·2008,8 Mark$)
Ans. :
Refer section 8.9.
Q.5
/'ig. 1 slunv« pl111se portraits for typt' • 0 systons. Classify tlrem i11to the categories. Stable focus. stable node. saddle point and w-on. (See Fig. 1 011 next page.)
Ans. :
Stable Iocus
(JanJfcb.·2006;July/Aug.-2007, 6 Marks)
Fig. 2 Stable focus (P • 26)
p ·29
Modem Control Theory
Phase Plane Method
Unstable node
x, ~y
Fig. 6 Unstable node Saddle point
Fig. 7 Saddle point Q.6
Explain l/1e concept of jump resonance with a suitable example. (JanJFcb.·2006, 5 Marks)
Ana. :
Refer section 8.3.
Q.7
Explain th« delta 1r1ttlrod of constructing phas» trajtctorie.s.
Ana. :
Refer section 8.11.
Q.8
Using isocline method, draw lite phas<e trajectory for tltt system. d2x+o.6'.!!.+x dt2 di with x = 1 a11d dx dt
Ans. :
Refer section 8.4.
OanJFcb.·2006, 7 Morks)
.. o
= 0 as initial amditia11.
(JanJFcb.-2006,8 Marks)
Modorn Control
Q.9
p ·30
Thoory
Phaso Plano Method
.
A linfar second order sen» is described by the equation C+2E,ll\,C+or,c
= 0 wliere
E, = 0.15. 10• = l rad/sec, CIOJ = l.5 and C = 0. Detmnirtt tile singular point. Constriid lite phase trajectory, using the method of isodine». (July/Aug.·2006,10 Marks) Ans. :
Refer example 8.2.
Q.10
Define singular point on a pnase plm1e. &plain diffoent types of sit1gular points · (July/Aug.-2006, 10 Marks)
Ans. :
Reier section 8.9.
Q.11 Ans. :
What are singular points? Explain differc1t singular points adopted in 1wn-li11ear control ;-ystems. (JanJFcb.-2007,8 Marks) Refer section 8.9.
Q.12
Find out singular points for tM followingsystems : i) x +0.5x+2x=O
. .
ii) .v+3y +2y=O iii) y +3y -10=0
(JanJFcb.·2007,12 Marks)
Ans. :
Refer example 8.3.
Q.13
Draw tlte phase-plane trajectoryfor thefollowittg equation using Isodine method: x+2E,ro;+w2x=U
Given, f,=0.5.
lll"'l,
Initial point
(0, 6).
QanJFcb.-2007,U Marks)
Ans. :
Refer example 8.5.
Q.14
Discuss lllC basic features of the jt••np rtSOna11ce 11011-linearities wit/1 suitable figures : /ump n~o111111cr. (July/Aug.·2007, 8 Marks)
Ans. :
Refer section 8.3.
Q, 15
&plain th« construction of a pllflSt trajectory by Delta method.
Ans. :
Refer section S.S.
QanJFeb.-2008, 6 Markli)
ODD
Liapunov's Stability Analysis
Q.1
State and explain Uapu11ov's theorems 011 Ii) Asymptotic stability (ii) Global asymptotic stability und (iii) l11stabilily. Qan./Feb.-2005;July/Aug.-2006;JanJFeb.·2008, 10 Marks)
Ans. :
Refer sections 9.3, 9.4 and 9.5.
Q.2
Use Krasovskii's theorem lo show that lite equilibrium slate x = 0 of lite system described by
.r
1
=-3x, +x2
.t2 =x1
-x2
-xi
is asymptotically stale i11 I/re large.
(Jan./Feb.·2005, 10 Marks)
Ans. :
Refer example 9.4.
Q.3
Define: Ii) Stability, Iii) Asymlolic stability (iii) Asymptotic stability in the large. Quly/Aug.·2005, 5 Marks, July/Aug.-2007, 6 Marks)
Ans. :
Refer sections 9.29.3, 9.4 and 9.5.
Q.4
Investigate I/re stability
of tlie following non-linear system
using direct method of
.
Liapunou. x1 =x2 -~2 =-.rl
-xfx2
(July/Aug.-2005,S Marks)
Ans. :
Refer example 9.5.
Q.5
A seco11d order system is represe11led by, • (l x=Ax where A= [ _1
1 ] -:
Assuming matrix Q to be identity matrix, solve for matrix P in tire eqw1lion ATP + PA Q. Use Liapunou theorem and determine I/re stability of tire origi11 of the system. Write tire Liapunot» [unction V(x). (July/Aug.-2005, 10 Marks)
=-
Ans. :
Refer example 9.6.
(P • 31)
- .. ,. . p -32
Modern Control Theory Q.6
Liapunov's Stability Analysis
Using Liaprmovs direct 111etl1od, find tlrr range of K lo guarantee sta/1ility of tire system show11 i11 Fig. I.
. _:~•~Ghl-f~~I
•O
s
Fig. 1 (JonJFeb.-2006,14 Marks) Ans. :
Refer example 9.11.
Q. 7
Choos« an appropriate Liapunov fimction and check tire stability of tire equilibrium state of tire system described by
X2 = - X7 - X~ X2
Ans. : Q,8
O•nJFeb.-2006, 6 Muks)
Refer example 9.5.
Determine whether or not folll1toing quadraticform is positive definite : Q (.q,x2) = 10xr +.ix~ +x~ + 2x1x2 -2xzx3 -4x1x3
(Jan.lFeb.-2007, 10 Marks)
Ans. :
Refer example 9.10.
Q.9
Explain with a11 example • i) Liapunov main stability thrown ii) Liapunop second method and iii) Krasooskii's tlrcorem. (JanJFcb.·2007,10 Marks)
Ans. :
Refer sections 9.10, 9.11 and 9.13.
Q.10
Fi11d the Liapwrovfunctionfor the system: X(I)
= [~ ~1]x.
Ans. :
Refer example 9.6.
Q.11
Explain Llap1111011's stability criterion.
Ans. :
Refer section 9.1 I.
(JanJFeb.·2007,8 Marks)
(July/Aug.-2007,6 Marks)
~
Modern Control Theory
Q.12
P-33
Liapunov's
Stability Analysis
Consider tlie system 1uitl1 diffmmt111/equation.
= U Examine tire stnbility
::+k;.+k1e.J +c
by Uap1111ov's method, gioen thot K > 0
and K 1 > 0.
(July/Aug.-2007,6 Marks]
Ans. :
Refer example 9.8.
Q.13
Exnmme ti~ stability of the system described by th« following equation by Krasovdskii's.
Ans. :
Refer example 9.9.
Q.14
A system is described by tire followingequation :
Ouly/Aug.-2007,8 Marks)
;=Ax Where A
=[-t 1
2].
-4
Q to be tire identify matrix, solve for matrix P and comment on the stability of tile system 11si11g the equation ATP+ PA -Q.
Assuming matrix
=
Qan/Fcb.- 2008, 10 Marks) Ans. :
Refer example 9.7.
ODO
(P • 34)
(2·1!1D(2·72)
Cl>aofe< • 3
Matrix Algebra & Doriva!ion of Tronstor Funciul
(3 • lj lo (3. 46)
Cl>aote< • 4
Solutlcn of Sta:e EQualions
(4-llloj4-~)
Chapt., . 5
Pde Pl~.enlTechnque
(5· l)lo(5·00)
"' ,Features of 'Book
:-;-us-;~r~i;;;,-;1~i~-~d-1~-cid-l~;;;;;;;lli~-;h;~-ci;;s~ci~~-;;;;·~~~~-I I • Excellent theory well supported with the practical examples and
I
I
I
illustrations.
I
1 I • Important concepts arc highlighted using key points throughout tho book.
I
I
I • Large number of solved examples. I 1 • Approach of the book resembles classroom leaching. I • Use of informative, self explanatory diagrams, plots and graphs.
I
I : I
: • Book provides detailed insight into the subject. I •
1
Stepwise explanation to mathematical derivations for easier
:
understanding.
L------------·----··-·--·-··-·---------------·-------··--·--······--······ ' "' .~ ~ -~ ~~ '~
Best of Technical Publications
I
..,
1
As per Revised Syllabus of VTU - 2006 Course Semester - V [EEE) Modem Control Theory Signals and Systems Transmission and Distribution Linear !Cs and Applications D.C. Machines and Synchronous Machines
Bakshi, Godse' Bakshi' '
(4)
M
Modern Control Theory ISBN 9788184315066
,.,.rved
All righls wilh Technicol Publicottons. No pot! of this book should be reproduced in ony loon, Eledn>nic, Mochonlcol, Pholoc.opror or
Publishtt.f by : Thchnlcal l"Ublk:aUonsPune• #1,,......
~"""'•it.
Printer: Nc.OTPilotcn
s..... 1 (11),sw..,.d J\.w.41104l
Rood,
si-.-
PnJ.. "'-. •11 030, l.d..
Preface The importance of Control Theory is well known in various engineering fields. Overwhelming response to our books on various subjects inspired us lo write this book. The book is structured lo cover the key aspects of the subject Modem Control Theory. The book uses plain, lucid language to explain fundamentals of this subject. The book provides logical method of explaining various complicated concepts and stepwise methods to explain tho importonl topics. Each chapter is well supported with necessary illustrations, procticol examples and solved problems. All the chapters in the book ore arranged in o proper sequence that permits eoch topic to build upon earlier studies. All care has been token lo make students comfortoblc in understanding the basic concepts of the subject. The book not only covers the entire scope of the subject bul explains the philosophy of the subject. This makes the understanding of this subject more clear ond makes it more interesting. The book will be very useful nol only lo the students but also to the subject teachers. The students hove to omit nothing and possibly hove to cover nothing more. We wish to express our profound thanks to oil those who helped in making this book a realily. Much needed moral suppor1 and encouragement is provided on numerous occasions by our whole family. We wish to thank the Publisher and the entire teom of Technical Publications who hove taken immense pain to get this book in time with qualify printing. Any suggestion for the improvement of the book will be acknowledged and well appreciated.
Authors
(1
:ij Jo:(h 14)
1.1 Background
1.1.1 ~or 1.1.2 ~tages
1- 1
1 -1
Conventional Approach
of State Variable Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 • 2
1.2 Concept of State
1 •2
1.2.1 lmpajanJpelinitions
\ I'
:.................
•
1.3 Stal? MY1!el-pf Lin.ear Systems
:
1.3.1 Stale Model of Si)!lle Input Single OutputSystem 1.4 State Diagram Representation
.
: •...•........
.
\
1 -4 1 -6 1 •9 1 -9
1.4.1 State Diagramof Standard State Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 • 10 1.5 Non Uniqueness of State Model
1·. 11
1.6 Llnearization of State Equation
1 -12
B.eYlew. Questions............ ..
..
·2.1 State Variable Representation using Physical Variables 2.1.1 Advantages 2.2 State Space Representation using Phase Variables
1 - 14
2•1 2•4 2 -4
2.2.1 Stale Modetfrom Differential Equation ..••.•...•••...•..............•.••..••.
2•5
2 2 2 Slate Model from Transfer Fuiict;on
2-8
2.2.3 Advantages. . • . . . . . • . . . • . . . . . . . . • . . . • . . . . . • • • . • . • • . • . . • • . . • . • • . • • • . • • . 2 • 18 2 2 4 11rnaa11ons
2.3 State Space Representation using Canonical Variables
2.18
2 - 19
2.3.2 Advantages of Canonical Variables
2 • 26
2.3.3 Disadvantages of canonical Variables ••.••......•.......•.•..•..••........•
2 • 26
2.4 State Model by Cascade Programming
2 - 29
Examples with Solutions
2 - 31
3.1 Background 3.2 Definition of Matrix., . . 3.2.1 Types of Matrices 3.2.2 Important Terminologies
: .'.~:
3-1
..
• 3-1 3·2 3•3
. .. . . . . .. .'
3.3 Elementary Matrix Operations
3-6
3 4 Inverse of a Matrix
3-7
3.4.1 PropertiesoflnveiseofaMatrix 3 5 Deriyation of Transfer E11nclion from State Model
3-9
3.5.1 Characteristic Equation 3.5.2 MIMO System
3·8 3 • 10
3 -11
3.6 Eigen Values
3 • 12
3.7 Eigen Vectors
3 - 13 3. 13
3 8 Modal Matrix M 3 8 1 Yan!ler Mende Matrix
3-14
3.9 Diagonalisation
3 - 19
3.10 Generalised Eigen Vectors 3. 23 3.10.1 Vander Monde Matrix for Generalised Bgen Vectors . .. .. .. .. . . • . . . .. . .. .. . • • 3 • 24 Examples with Solulions Reyjew Questions
3 • 26
4.1 Background
4• 1
4.1.1 Homogeneous Equation
4.1.2 Nonhomogeneous Equation 4 2 Review of Classical Method of
4•1
Snh11ion
4.2.1 Zero Input Response 4.3 Solution of Nonhomogeneous Equation
4•2 4-2 4•4 4-4
4.4 Properties of State Transition Matrix
4-5
4.5 Solution of State Equation by Laplace Transform Method
4- 6
4.6 Computation of State Transition Matrix
4 -7
4.7 Laplace Transform Method
4-8 4-9
4 B Power Series Method
4.9 Cavley Hamilton Method 4 91 Prncedure lo Calct~ale ""1
4 - 10
4.10 Similarity Transformation Method
4 -16
4.11 Controllability and Observability
4 - 22
4.12 Controllability
4. 23
4-12
4.12.1 Kalman's Test for Controllability
4 • 23
4.12.2 Condition for State Controllability Ins-Plane .•..•.•.............•..••.•..•..
4 - 26
4.12.3 Gilbert's Test for Controllability
4 • 26
:
4.12.4 Output Controllability
4.13 Observability
4 • 29
4 • 29
4.13.1 Kalman's Testier Obse!Vllbllity
4 - 30
4.13.2 Conditioo for Comple:e Observability In s-Pl<Wle
4 • 33
4.13.3 Gilbert's Test ftlf Observabi!lty . .. . . . . . . .. . . . . . . . .. . . . . . . . . . . . • . . • . . . • • . .. 4 • 33 Examples with Solutions
4 • 35
5 1 Introduction
5-1
5.2 Pole Placement Design
5-2
5.3 Necessary and Sufficient Condition for Arbitrary Pole Placement 5 4 Evah1a!joo of Slate Feedback Gain Matrix K
5-3 5. 15
5 4 1 Use of Traosfonnatjo1J Matrix I
5-15
5 4 2 Direct S11bstit11tjon Method
5-15
5 4 3 Ackeanann's Foanu!a
5.5 Selection of Location of Desired Closed Loop Poles
5-16 5 • 19
5 6 State Observers
5-24
5.7 Concept of State Observer
5. 25
5.8 Dual Problem
5 - 27
5.9 Necessary and Sufficient Condition for State Observation
5- 28
5.10 Evaluation of State Observer Gain Matrix!<..
5 - 28
5.11 Selection of Suitable Value of Observer Gain Matrix K..
5 - 30
5 12 Observer Based Controller
5 - 35
Examples with Solutions
5 - 38 5 -57
6.2.2 Variable Range .•...••.•....••.......•.......................•••.•...•.• 6.2.3 Controller OutputRange 6.2.4 Control Lag .....•.........•.•....•...•...•.•..•..••.••.....•.•.....•.•.
6•3
6 2 5 Dead Zone
6-3
6 3 Classification of Controllers 6 4 Continuous Contmllm..Made..._
6.5 Proportional Control Mode 6.5.1 Characteristic of Proportional Mode 6520ffset
6 •3
6-3 6-4 -'"'-',_,.
6-5 6•6 6-6
6.5.3 Applications
6• 7
6.6 Integral Control Mode
6.6.1 Step Response oflnteqral Mode 6.6.2 Characteristics oflntegral Mode 6.6.3 Applications 6 7 Oerlvatiye Control Mode
6 7 1 Characterisljcs of !ledyatjye Con!rol Mode 6.7 2 Applications
6-7
6•9 6-9 6 • 10 6-10 6 • 11 6 • 12
6.8 Composite Control Modes
6 -12
6.9 Proportional + Integral Mode (Pl Control Mode)
6 • 1>!
6 9 1 Cbaractertsticsof Pl
Mode
6-14
M
6.9.2 Applications
6.10 Proportional+ Derivative Mode (PD Control Mode)
6 • 15
6. 15
6 1 O 1 Charac!eristtcs of PD Mode
6-16
6.10.2 Applications
6 -17
6.11 Three Mode Controller (PIO Control Mode)
6 - 17
6.12. Response of Controllers to Various lnputs
6- 21
6.13 Effect of Composite Controllers on 2"" Order System
6 - 22
6.14 PD Type of Controller
6- 23
6.15 Pl Type of Controller
6 - 24
6.16 PIO Type of Controller
6 - 26
6.17 Rate Feedback Controller (Output Derivative Controller)
6 - 26
6.18 Review of Second Order System Specifications
6 - 28
6.19 Steady State Error
6 - 30
Examples with Solutions
6 - 31
Review Questions
6 - 48
7.1 Introduction to Nonlinear Systems
7-1
7 .2 Properties of Nonlinear Systems
7- 1
7 3 Classification of Nonlinearities
7-2
7 4 Inherent Nonlinearities
7-2
741 SabaraOOo
7-3
742Deadmne
7.3
7.4.3 On-Off Nonfineatity •......•..•..•..•..•..•...•..•..•...••..•..•...•.....•
7 -4
7.4.4 On·Off Nonlinearity with Hysteresis ...........••.•........•...............•..
7•4
z 4 5 Nonfineac Friction
7 4 6 Bacldash
7.4.7 NonlinearSpring
7•5
. ". 7. 5 7 -6
7.5 Absolute Value Nonlinearity
7-7
7 6 Intentional Nonlinearities
7-8 7-8
M
.,~ 8.1 lnt!'Oductjon
8-1
8.2 limit Cycle
8•1
8.3 Jump Resonance
8•2
8 4 The Phase Plane Method
8-3
8.5 Basic Concept of Phase-Plane Method
8-5
8.6 Isocline Method for Construction of Phase Tralectory
8•7
8.7 Application of Phase-Plane Method to linear Control System
8 - 10
8.8 Second Order Nonlinear System on Phase Plane
8 -13
8.9 Different Types of Phase Portraits
8 • 14
8.9.1 Phase Portraits for Type O System ......•...•..•....•....••......•.•..••.•.
8. 14
8.9.1.1 S1able SyslemIMth Conl)lex Roots . . . • .
. . 8· 15
8.9.1.2 Unstable System,.;in Complex Roots • . . .
••
8.9.1.3 Marginally Slab:e Sys1em IMth Complex Roots
8-16
. 8· 16
8.9.1.4 Stable SystemIMtll Real Roots • . . • • .
•. 8-17
8.9.1.5 Unslable System-Mth Positive Real Roots . .
..
8.9.1.6 U11$lable System~ Cloe Positive and Ooe Negati\19 Real Root . . 8.9.2 Forced Second Older TypeO System
8.9.3 Phase Portraits for Type 1 Svslem 8.9.4 Phase Portraits for Type 2 S)'Slem 8.1 O Singular Points of a Nonlinear System
8-18
. . 8· 19
~-19 8 -19
8 • 22 8 - 26
8 11 Delta Method
8 -27
Examples with Solutlons
8 • 28
Review Questions.
.
. .
..
8- 39
chip&r; g uapunm stiiluiiY,\na1Yili'_.,--.-- .., 9 1 fnf"mductjon
9-1
9.2 Stability in the Sense of Liapunov
9•2
9.3 Asymptotic Stability
9- 3
9.4 Asymptotic Stability In the Large
9-3
9.5 Instability
9•3
9.6 Graphical Representation
9•3
9,7 Some Important Definitions
9 -4
.. 9 7 1 poSfttvflDpfinileness
9-4
9.7.2 Negative Definiteness
9 .4
9 7 3 Positive Semjdefiniteness
9-5
..
9.7.4 Negative Semidefinite ... 97
9.5 9.5
5 !odefinUeness
............. 9- 5
9 8 011adralic Form 9 9 Hermitian Forro 9.1 O Liapunov's Second Method
9- 5 '
9-6
9.11 Liapunov's Stability Theorem
9-6
9.12 Stability of Linear and Nonlinear Systems
9-7
9.13 Construction of Liapunov's Functions for Nonlinear Systems by Krasovskii's Method 9.14 The Direct Method of Liapunov and the Linear System
9-8 9 - 11
Examples with Solutions
9 - 12
Reyjew Questions
9 - 24
1m~::in~~~~w~~~
J·~~~~m~Ei.W'~Vti~{;.~'1.{M-t' ;f~M-.;~:.£~
State Variable Analysis and Design 1.1 Background The conventional approach used lo study the behaviour of linear time invariant control systems, uses the time domain or frequency domain methods. In all these methods, the systems are modelled using transfer function approach, which is the ratio of Laplace transform of output to input, neglecting all the initial conditions. Thus this conventional analysis faces all the limitations associated with the transfer function approach.
1.1.1 limitations of Conventional Approach Some> of its limitations can be stated as : 1) Naturally. significant initial conditions in obtaining precise solution of any system,
loose their importance in conventional approach. 2) 111e method is insufficient and troublesome lo give complete time domain solution
of higher order systems. 3) It is not very much convenient for the analysis of Multiple Input Multiple Output systems. 4) It gives analysis of system for some specific types of inputs like Step, Ramp etc. 5) II is only applicable to Linear Time Invariant Systems. 6) The classical methods like Root locus, Bode plot etc. arc basically trial and error procedures which foil lo give the optimal solution required. Hence it is absolutely necessary lo use a method of analysing systems which overcomes most of the above said difficulties. The modem method discussed in this chapter uses the concept of total internal slate of the system considering all initial conditions. This technique which uses the concept of state is called State Variable Analy»is or St•te Space Analysis. Koy Point: State variable analysis is c.<Scntially a time domain approach but it has 11m11ber of adva11/ages compared to co11Vt'nlio11al mrtliods of analysis.
(1 • 1)
Modem Control Theory
State Variable Analysis & Design
1·2
1.1.2 Advantages of State Variable Analysis The various advantages of state variable analysis arc, 1) The method takes into account tl\C effect of all initial conditions. 2) It can be applied to nonlinear as well as time varying systems.
3) It can be conveniently applied to Multiple Input Multiple Output systems. 4) The system can be designed
for the optimal conditions precisely by using this
modem method. S) Any type of the input can be considered for designing the system. 6) As the method involves matrix algebra, can be conveniently adopted for the digital
computers. 7) The state variables selected need not necessarily be the physical quantities of the
system. 8) The vector matrix notation greatly simplifies the mathematical representation of the system.
1.2 Concept of State Consider a football match. The score in the football match must be updated at every instant from the knowledge and Information of the total score before that instant. nus procedure of updating the score continues till the end of the match when we get exact and precise score of the entire match. This updating procedure has main importance in the understating of the concept of state. Consider the network as shown in the Fig. 1.1 To find V0111, the knowledge of the initial capacitor voltage must be known. Only information about V1n will not be suffkknt to obtain precisely the V out at any time t ~ O. Such systems in which the output is not only dependent on the input but also on the initial conditions arc called the systems with memory or dynamic systems.
Fig. 1.1
f-f--
V Velocity X Displacement
ForceF-0 /?7777Tj777777 Zero &icbon
Flg. 1.2
While if in the above network capadtor 'C' is replaced by another resistance 'R1' then output will be dependent only on the input applied Y'". Such systems in which the output of the system depends only on the input applied at t 0, arc called systems with zero mttnory or
=
static systems.
Modem Control Themy
State Variable Analysis & Design
1·3
Consider another example of simple mechanical system as shown in the Fig. 1.2 Now according to Newton's law of motion,
i.e.
F
Ma
a
Acceleration of mass M
F
M
v(l)
=
ard (v(t))
id
((t) dt
Now velocity at any time 't' Is the result of the force F applied to the particle in the entire past, I
'M
v(t)
f
f(t) cit
~
'!
f(t)dt
+kf f(t)dt l
'o where t0 = Initial time. Now
-b J f(t) dt
=
v(t0)
I
v(t)
v(t0)+
J f(t) dt 'o
From the above equation it is clear that for the same input f(t), we get the different values of the velocities v(t) depending upon our choice of parameter t0 and the value of v(t0). U v(t11) is known and the input vector from 't0' to 'I' is known then we can obtain a unique value of the output v(t) at any time t > t0• Key Point : Thus initial conditions i.e. memory affects the system c/1aracterisatim1 and subsequent behaviour. Thus initial conditions describe the status or state of the system at t = t 11• The state an be regarded as a compact a.nd concise representation of the past history of the system. So the state of the system in brief separates the future from the past so that U1e state contains all the information concerning the past history of the system necessarily required to determine the response of the system for any given type of input. The state of the system at any time 't' is actually the combined effect of the values of all the different elements of the system which arc associated with the initial conditions of the system. Thus the complete slate of the system can be considered to be a vector having components which are the variables of system which are:tlosely associated with
State Variable Analysis & Design
Modem Control Theory
initial conditions. So stale can be defined as vedor X(I) called slate vector, This X(t) i.e. stale al any lime ·r is 'n' dimensional vector i.e. column matrix n x 1 as indicated below. X1(t) X2(t)
X(t)
Now these variables X t (t) , X2 (t) called the state wriablts of the system.
Xn (t} which constitute the state vector X(t) arc
If slate at t s t1 is to be decided then we must know X(t0) and knowledge of the input applied between t0 ~It. This new state will be X(t 1) which will act as initial state to find out the state at any time t > t 1• This is called the updating of the state. The output of the system at t = t 1 will be the function of X(t 1) and the instantaneous value of the input at t t 1, if any. The number of the state variables for a system is generally equal to the order of the system. The number of independent stale variables is normally equal to the number of energy storing elements (e.g.: capacitor voltage, current in inductor) contained in the system.
=
1.2.1 Important Definitions 1) State : The state of a dynamk system is defined as a minimal set of variables such that the knowledge of these variables at t = 10 together with the knowledge of the inputs fort :·t0. 2) State Variables : The variables involved in determining the state of a dynamic system X(t), arc called the state variables. X1(t), X2(t) ...... X"(t) are nothing but the state variables. These are normally the energy storing elements contained in the system.
31 State Vector : The 'n' state variables necessary to describe the complete behaviour of the system can be considered as 'n' components of a vector X(t) called the state vector at time · t', The state vector X(t) is the vector sum of all the state variables. 4) State Space : The space whose co-ordinate axes are nothing but the "n' state variables with time as the implicit variable is called the slate space. 5) State Trajectory : It is the locus of the tips of the state vectors, with time as the
Implicit variable.
Modem Control Theory
State Variable Analysis & Design
1·5
Definitions can be explained by considering second order system : Order is 2 so number of state variables required is 2 say X1(t)
and X2(t).
State vector will be the matrix of order 2 x I. X(t)
->
--t X(t)
i.e,
->
X1(t)+ X2(t) (vector addition)
The stale space will be a plane in this case as the number of variables arc two. Thus state space can be shown as in the Fig. 1.3. Xi(l)
X1(t)
L~ . ,
x,c1,> ~(I)
(0 .0)
X2tt1)
Fig. 1.3
Now consider t =
Flg.U
t1
-> X(t1)
->
->
= X1(l1)+ X2(11)
x,c11
·~(ti
Fig. 1.5
Now consider
t
Fig. 1.6
= t2
->
X(t2)
->
"' X1(t2)+
->
X2(t2)
The state trajectory can be shown by joining the tips of the two state vectors i.e. X(t 1) and X(t2).
Modem Control Theory
State Variable Analysis & Design
1-6
1.3 State Model of Linear Systems Consider Multiple Input Multiple Output, nth order system as shown in the Fig. 1.7. Number .of inputs
m
Number of outputs
p
u,
----<"'!
U2-~-.i
MIMO
Inputs
SYSTEM
t----v, t-~--v2 OutpulS
.____. __ VP
.,....,.-T""'i~=:T" X1 X2 X3 Slate varlables
Fig. 1.7 X1(1) X2(t)
U2(t)
"•' ')
U(t)
;
X(t)
=
Um(t)
Y1(t) Y2(t)
, Y(t) Xn(t)
= Yp(t)
All arc column vectors having orders m x 1, n x l and p x 1 respectively. For sucti a system, the state variable representation can be arranged in the form of "n' first order differential equations. dX1(t)
di"'
•
X1(t)mf1(X1,X2
.
....... Xn,U1,U2
......... Um)
X2(t) = 12 (X1, X2 , ...... X0, U1• U2 ......... Uml
1. 7
Modern Control Theory
Where f =
State Variable Analysis & Design
is the functional operator.
r, Integrating the above equation, X1(l)
=
I
X;(lo) +
J f; (X1
• X2 ......
x,
U1 • U2 ..... Um) dt
'o where i = 1, 2, ........ n. Thus "n' state variables and hence state vector at any time "t' can be determined uniquely. Any "n' dimensional time invariant system has state equations in the functional form as,
.
X(t) = f (X, U) While outputs of such system ore dependent on the state of system and instantaneous inputs. :. Functional output equation can be written as, Y(t) = g(X, U) where · g' is the functional operator. For time variant system, the same ~'<Juations can be written as, X(l)
f(X, U, t)
State equation
Y(t)
g(X, U, t)
Output equation
Diagramatically this can be represented as in the Fig. 1.8. U(t) lnstarltancous
lnput Input U(t) ----~
Y(t) Oulput
X('ol lnlUal state
Fig. 1.8 Input-output state description of a systom
Modem
1·8
Control Theory
State Variable Analysis & Design
The functional equations can be expressed intcrms of linear combination of system states and the input as,
.
X 1 = a11 X1 + a12X2 + ... + a10X0 + b11U1 + b12U2 + ... + b1m Um X2
= a21 X1 + ~X2 ...
+ a20Xo + b21U1 + b22U2 + ... + b2m U'"
.
Xn = a,,1 X1 + "nl~ + ... + a.,,,Xo + b01U1 + bn2U2 + ... + bnm Um For the linear time invariant systems, the coefficients a11 and bii arc constants. Thus all the equations can be written in vector matrix form as,
I
X
A X(t) + B U(t)
X(t}
State vector matrix of order n x 1
U(t)
Input vector matrix of order m >< 1
where
A B
System matdx or Evoliition matrix of order n x n
=
lnput matrix or control matrix of order n >< m
Similarly the output variables at time t can be expressed as the linear combinations of the Input variables and state variables at time t as, v,(t)
=Cu
x,(t) + ... + C1n Xn(t) + d11 u,(t) + ... + dim Um(I)
Yp(t) = Cpt X1(t) + ... +cpl X0(t) +dpt U1(t} + ... + dpm Um(t) For the linear time invariant systems, the coefficients c;; and d;J arc constants. Thus all the output equations can be written in vector matrix form as, Y(t) = C X(t) + D U(t) where
= =
Y(t) C
Output vector matrix of order p .>< 1 Output matrix or observation matrix of order p x n
D = Direct transmission matrix of order p x m The two vector equations together is called the stale model of the linear system. A X(t) + B tl(t)
... State equation
Y(t) "' C X(t) + D U(t)
... Output equation
X(t)
s
This Is state model of a system.
1 .g
Modem Control Theory
State Variable Analysis
& Design
For llnear time-variant systems, the matrices A, B, C and D are also time dependent. Thus, X
= A(t) X(t) + B(t) U(t)} = C(t) X(t) + D(t) U(t)
For linear time variant system
1.3.1 State Model of Slngle Input Slngle Output System
=
=
Consider a single input single output system i.e. m 1 and p l. But its order is 'n' hence n state variables arc required to define state of the system. In such a case, the state model Is
.
X(t)
A X(t) + B U(t) C X(t) + d U(t)
A
n x n matrix, B
c
1 x n matrix, d
where
nnd
=
Y(t)
U(t)
= n x I matrix = constant
single scalar input variable
In general remember the orders of the various matrices. A
=
Evolution matrix
U =- Control matrix
=> n x n => n x m
C = Observation matrix => p x n D = Transmission matrix
=pxm
1.4 State Diagram Representation It is the pictorial representation of the state model derived for the given system. It forms a close relationship amongst the state model, differential equations of the system and its solution. It is basically a block diagram type approach which is designed from the view of programming of a computer. The basic advantage of state diagram Is when it Is impossible to select the state variables as physical variables. When transfer function of system is given then slate diagram may be obtained first. And then by assigning mathematical state variables there in, standard state model can be obtained. State diagram of a linear lime invariant continuous system is discussed here for the sake of simplicity. It is a proper interconnection of three basic units. i) Scalars
ii) Adders
iii) Integrators
Modem Control Theory
1·10
State Variable Analysis & Design
Scalars arc nothing but like amplifiers having required gain.
~<1>Tx,<1>=~<1>·x3<11 ><:i(t)
(a)Scalar
(b)Adder
Fig. 1.9 Adders arc nothing but summing points. Integrators arc the clements which actually integrate the differentiation of state variable to obtain required state variable.
Now integrators arc denoted as '1/s' in Laplace transform. So transfer function of any lntegralor is always l/s. With these three basic units we can draw the state diagrams of any order system. Fig. 1.9 (c) Key Point : Tire output
of each integrator i.~ the state variable.
1.4.1 State Diagram of Standard State Model X(ll
.
A X(t) + B U(t) and
Y(tl
C X(t) + D U(t)
Consider standard state model
So its state diagram will bl? as in the Fig. 1.10.
Y(I)
Fig. 1.10 State diagram of MIMO system The thick arrows indicate that there arc multiple number of input, output and state variables. There must be n parallel integrators for n state variables. The output of each integrator is a seperatc state variable. If such a state diagram for the system is obtained then the state model from the diagram can be easily obtained.
1 • 11
Modem Control Theory
State Variable Analysis & Dnlgn
Remarks 1. To obtain the stale model from stale diagram, always choose output
of each integrator as a stale variable. Number or integrators always equals the order of the system i.e. ·n· 2. Differentiators arc not used in the stale diagram as they amplify the inevitable noise. ,..
Example 1.1 : Obtain the state diagram uf 5150 system reprtstnttdby equations,
.
X1(1)
=
a1 X1(t)+b1
.
U(t), X2(1J =a2 X1(t)+a3
X2(tH b2 U(I)
and Y(t) = c1X1(t)+c2X2(1) Solution : A.• there arc 2 state variables X1(t) and X2(t), the two Integrators arc required.
Ytl)
U(t)
Fig. 1.11 1.5 Non Uniqueness of State Model Considera standard state model for a system X(I)
.
A X(t) + B U(t)
... (1)
y(I)
C X(t) + D U(t)
.•. (2)
Let X = M Z where M is a non singular constant matrix. Koy Point: A matrix whose daerminatu is nonuro is called nonsingular matrix. This means Z is new set of stale variables which Is obtained by linear combinations of the original state variables.
1·12
Modem Control Theory
. .
X = MZ hence
State Varlable Analyala & Oealgn
X" MZ
M Z(t)
.
AM Z(t) + B U(t)
... (3)
Y(t)
CM Z(t) + D U(t)
... (4)
Substituting in a slate model
:. Premultiplying equation (3) by M" 1
.
M-11\M Z(t) + M-1 B U(t)
Z(t) WMC
M-1 M Z(t)
where
I i.e. Identity matrix c
A
Where
Z(t) +
B U(t)
... (5)
M-1 AM
fi and
A
M-1 B
C Z{t) +
Y(t)
c
D U(t)
... (6)
CM
Equations (5) and (6) forms a new state model of the system. This shows that state model is not a unique property. Key Point: Any /i11~ar combinatio11s of th;. original set of state wriables result» into a valid 11ew set of state variables.
1.6 Llnearizatlon of State Equation Any general time invariant system is said lo be in equilibrium, at a poinl when.
Ix~
f (Xo- Uo)
u
0
(l
U0)
I
The derivatives of all the stale variables are zero at a point of equilibrium, The system has a tendency lo lie at the equilibrium point unless and unlill it is forcefully disturbed .
.
The state equation X(t) = f (X, U) can be linearized for small deviations about an equilibrium point (X0, U0). This is possible by expanding the state equation using Taylor series and considering only first order terms, neglecting second and higher order terms.
Modem Control Theory
State Variable Analysis & Design
1 • 13
So expanding kth state equation, n ilf,(X, U)
•
~ = fk (Xo- U0) +I~ /:I
Now
rk(Xo- U0)
=
I
I
(X1 -X10)+ X=Xo U=Uo
m i)lk(X, U)I L.-mj XeXo
(Ui - Up)
j=I
U=Uo
0 at the equilibrium point.
And let the variation about the operating point is,
. .
x1-Xio and
henceX=X,
U;-U;o
Hence the kth state equation can be linearized as,
.
Xk
.
.In this equation xk,
x, and U; are the vector matrices and the remaining
terms arc the
matrices or order n x n and n x m respectively. Hence the linearized equation can be expressed in the vector matrix form as,
x
=
AX+
nu
vx;ilr2 where
A
~
cir,
dx;;"
ilf2 ax2
arn
«:
~
ib<.2
ilr2
dx;;-
ar,
a11 ()Um
i!f2
()(2
au,
iil2 aum
er, ~
()In
m;-
is n x n matrix
illn ilx.,
iJU2
ilf1
au, B
()fl
dx;
()In
aum
Is n x m matrix
Modorn
1·14
Control Theory
State Variable Analysis & Design
All the partial derivatives of the matrices A and B are lo be obtained at an equilibrium slate (Xo- U0). Such matrices A and B defined above interms of partial derivatives are called the Jacobian matrices.
Review Questions 1. Exp/run tl1e concept of stole. 2. l)tfim •md explain tire followingterms, a Slat/ a.1riables I>. Stott 11«tor of. Stale c. Slate lraf"<10ry < St{lt( SfXIU
J. l:.xplain adt1tmti1grs of state t1t:1rit1ble method over conventional ant. 4. Writ' a slrort note on ad•w11tagt>and limitatums of st.tit ~'llriablt approotl1, 5. Wn'tr a note on lincariznliatr of Malt equation:;. 6. l'row tire nomtniqtttnt."SS
of lite state modtl.
ODO
State Space Representation 2.1 State Variable Representation using Physical Variables The state variables are minimum number of variables which arc associated with all the initial conditions of the system. As their sequence is not important, the state model of the system is not unique. But for all the state models it is necessary that the number of state variables is equal and minimal. This number 'n' indicates the order of the system. For second order system minimum two state variables arc necessary and so on. To obtain the state model for a given system, it is necessary to select the state variables. Many a times, the various physical quantities of system itself arc selected as the state variables. For the electrical systems, the currents through various inductors and the voltage across the various capacitors arc selected to be the state variables. Then by any method of network analysis, the equations must be written i'ntcrms of the selected state variables, their derivatives and the inputs. The equations must be rearranged in the standard form so as to obtain the required state model. Key Point: It is important ilia! Ilic equation for diffcre11tiatio11 of 011e state t•11riable s/1011/d 1101 i1100/vc tltt differmtiation of any otner slate variable. In the mechanical systems the displacements and velocities of energy storing elements such as spring and friction arc selected as the stale variables. In general, the physical variables associated with energy storing elements, which are responsible for initial conditions, arc selected as the stale variables of the given system. 1~
Example 2.1
Obtain tire state model of tire given electrical system.
~
v;(t)
l
IC
~
~t)
-
Fig. 2.1 (2 • 1)
v0(t)
-l
tvru : July/Aug.-2006)
2-2
Modem Control Theory
State Space Representation
Solution : There arc two cn!!l'gy storing elements L and C. So the two state variables arc current through inductor i(t) and voltage across capacitor i.e. v0(t). i(t) And
and
X2(t)
= v (t) 0
v1(t) = Input variable
Applying KVL to the loop. v1(t)
=
i(t) R + L
di (t)
-di
+ v0(t)
Arrange it for di(t}/dt, di (I)
-dt
I
( ) R '(
IV; t
-1·
R
1
I t)-Iv.,(l)
b 1
ut
i.e,
- ·r. X1(t)-I
While
Voltage across capacitor =
2..c i(t) i.e.
"2(t)
X2(tl+I
di (I)
dt - =
•
Xt(t) ... (1)
U(t)
bJ
i(t) di
~v o(l) = {._ (I) di "2
but
·c1 X1(t)
... (2)
The equations (1) and (2) give required state equation.
[~J(t)] X2(t)
i.e.
.
- ~L _ _!_] L [ 0
t
1
Xz(t)
0
X(t) = A X(t) + U U(t)
While the output variable Y(t) Y(t) i.e.
[X (I)) + [·I.~] U(t)
(0 1]
Y(t) = C X(t)
=
v0(t) = X2{t)
[;~]+(OJ
U(t)
and D = (01
This is the required state model. As n
=
2, it is second order system.
Note : The order of the state variables is not important. X1(t) can be v 0(1) and X2(t) can be i(t) due to which state model matrices get changed. Hence state model Is not the unique property of the system.
2·3
Modem Control Theory
11u+
Example 2.2 : form.
State Space Representation
Obtain the state model of the givm electrica! network i11 tit<' standard
Fig. 2.2
IDNrilln :
U(t) ~ input Y(t) "'
«
output
State variables: X1(t) = i1(t),
c,(t)
= c.,(t)
X2(t) = i2(t), X3(t) = vc(t)
Writing the equations : <',(I)
I · 1 ··- c, (I)- --- v c(I) I.I I.I
i.e.
... (1)
Then,
... (2)
and
dvc(l)
C-d-,dvc(t)
Cit
.
X3(t)
bi1(t)-bi2(t) 1
1
C X1(t)- C X2(l)
... (3)
2-4
Modem Control Theory
. .
0
X2
0
X3
1
X1
X(t)
and
e0(t) Y(t)
Y(t)
i.e.
Y(t)
R2
-i=;-
1 L2
1
c -c:
.
i.e.
-r;-
0
0
State Space Representation
[~~]+
'l
Li ~
U(t)
AX(t) + BU(t) i2(t) R2 c
X2(t) R2
(o ~
OJ[~:]
~ CX(t) + DU(I)
where D = 0. This is the required state model.
2.1.1 Advantages The advantages of using available physical variables as the state variables are, 1. The physical variables which are selected as the state variables are the physical quantities and can be measured. 2. As state variables can be physically measured, the feedback may consists the information about state variables in addition to the output variables. Thus design with state feedback is possible. 3. Once the state equations arc solved and solution is obtained, directly the behaviour of various physical variables with time is available. But the important limitation of this method is that obtaining solution of such state equation with state variables as physical variables is very difficult and lime consuming.
2.2 State Space Representation using Phase Variables Let us study how to obtain state space model using phase variables. The phase variables arc those state variables which are obtained by assuming one of the system variable as a state variable and other state variables are the derivatives of the selected system variable. Most of the time, the system variable used is the output variable which is used to select the state variable. Such set of phase variables is easily obtained if the differential equation of the system is known or the system transfer function is available.
2·5
Modem Control Theory
State Space Representation
2.2.1 State Model from Differential Equation Consider a linear continuous time system represented by n•h order differential equation yn +an-lyn-1 +an-2 yn-2+ ... +
= b0U+b1 U+ ...+bm_1um-t
•
a1 Y+a0Y(t)
+bmU°'
••• (1)
In the equation, Y"(t) = dY" (t) = nth derivative of Y(t). dt" For time invariant system, the coefficients a0 _ 1, an- 2, . • . ao- ho- b1, • .. bm are constants. For the system, Y(t)
Output variable
U(t)
Input variable
Y(O), Y(O), ... Y(O) n- l represent the initfal conditions of the system. Consider the simple case of the system in which derivatives of the control force U(t) arc absent. Thus
.
U(t)
.: Y" + an _ 1 Y" -
= um(t) = 0 Y + n0 Y(t) = b0U(l)
U(I) " ..•
1 + ... + a1
Choice of state variable Is generally output variable Y(t) itulf. vmablcs arc derivatives of the selected state variable Y(t).
. .
X2(t)
Y(t) = X1(t)
X1(I)
.
Xi(!)
X(t)
X3(t)
.
Xn(t)
Thus the various state equations arc,
X,. - I (t) X,.(t)
... (2) And other state
State Space Representation
Modem Control Theory
Note that only n variables arc to be defined to keep their number minimum. ThU! • x,. _ 1(t) gives
Important
n
th
•
state variable X0(t). But to complete state model X,.(t) is necessary,
..
.
: X,.(t) is to be obtained by substituting the selected state 1 vartsbles In the a X;i. ...
original differentialequation (2). We have Y(t) = X1, Y(t) " X,, Y(t) yn-l(t) =~(I), Y0(t)
. .
= X,.(t)
:. X,.(t) + an_ 1 ~(t) + an_ 2 ~ _ 1(t) + ... + a1 X2t + a0 X1(t) = b0 U(t) :. X,.(t)
=-
a0 X 1 - a1 X, ... - an _ 2 ~ _ 1 - an_ 1 X0 + b0 U(t)
-Hcnce all the equations now can be expressed in vector matrix fonn
xx2 _ 1
rL Le.
-
.
X
z
1
o
[ 0
0
0
~ '.J [J:
-~o
... (3) QS,
l { "'
AX(t) + BU(t)
Such set of slate variables is called set of phasr va.rlablc1. The matrix A is called matrix in phase variable fonn and it has following features, 1) Upper off diagonal i.e. upper parallel row to the main principle diagonal contains all elements as 1. 2) All other clements CXC1..'Pt last row are zeros. 3) Last row consists of the negatives of the coefficients contained by the original differential eouation. Such a form of matrix A is also called Bush fonn or Companion method Is also called companion form reallution. The output equation is,
'Y(t)
[ I 0 .... OJ X,,(t)
Le.
Y(t)
CX(t)
where D = 0
fonn. Hence the
State Spac:e Representation
2·7
Modem Control Theory
This model in the Bush fonn can be shown in the stale diagram as in the
Fig. 2.3. Output of each integrator is a slate variable.
U(t}
Fig. 2.3 State diagram for phase variable form Observe that the transfer function of the blocks in the various feedback paths arc the coefficients existing in the original dilfcrcnlial equation. If the differential equation consists of the derivatives of the input control force U(t) then this method L< not useful. Jn such a case, the state model is to be obtained from the transfer function. ,_.
Example 2.3 : Construct tht state model using phase mriobles if tht sy.•tt111 is dtscribtd by tire differentialequation,
d2Y(t) +4 't2Y(t~ +7 dY(I) + 2Y(t) dt3 dt2 di
= 5U(I)
Draw the state diagram. Solution : Choose output Y(l) as the stale variable X1(t) and successive derivatives of it give us remaining state variables. As order of the equation is 3, only 3 state variables are allewed. X1(t)
Y(t)
-
• • X 1(1) = Y(t)
Xii) and Thus
.
J\(t)
"2
dY(t)
=di
= Y(t) = d2Y(t) dt2
X2(t)
... (1)
2-8
Modem Control Theory
State Space Representation
... (2) To obtain Xj(t), substitute state variables obtained in the differential equation. d3Y(t} -dtJ
.
'" d " dX • = Y (t) = -(Y(t)l = -3= X3(t) di di
.
:. X3(t) + 4X:J(t) + 7 X2(t) + 2X1(t) = SU(t) "3(t) "' - 2X1(t) - 7 X2(t) - 4X3(t) + 5 U(t)
... (3)
The equations (I), (2) and (3) give us required state equation. X(t)
A
where
The output is, Y(t)
Y(t)
c
where
AX(t) + BU(t)
u
1 0
-7
j]
and
B=[n
X1(tl CX(t) + DU(t) (10OJ,D=0
This is the required state model using phase variables.
Fig. 2.4
2.2.2 State Model from Transfer Function Consider a system characterized by the differential equation containing derivatives of the input variable U(t) as,
Y"
+ an -
I
yn -
I
+ ... + "t
y + •o Y(t)
"' boU + b1
U + ...
+ bm -
l
Um -
I
+ bm um ... (I)
2·9
Modem Control Theory
State Space Representation
In such a case, it is advantageous to obtain the transfer function. assuming zero initial conditions. Taking Laplace transform of both sides of equation (1) and neglecting initial conditions we get, Y(s) [sn +
a,,_ 1
Sn -
1
+ ... + a1 s + aoJ
Y(s) U(s)
T(s) =
= Ibo + sb1 + ...
+ bm _ 1 Sm - 1 + b tu sml U(s)
bo +sb,+ .... +b n-lsm-1 +blllsm ' a0 TSa J+ ... +a11_1sn-l
... (2)
+Sn
Practically in most of the control systems m < n but for general case, let us assume m =n.
From such a transfer function, state model can be obtained and then zero initial conditions can be replaced by the practical initial conditions to get required result. There arc two methods of obtaining state model from the transfer function, I) Using signal now graph approach 2) Using direct decomposition of transfer function 1) Using Signal Flow Graph Approach
The Mason's gain formula for signal flow graph slates that, T(s)
LTkAk
-6-
Gain of k1h forward path
where A
System determinant
p: Gain x Gain product ol} I - {L i111 loop gains}+j
,111 combinations of two
l non • louchin~ i.\k
- ...
loops
Value of II clliminating those loop gains and products which are touching lo k1h forward path
According to this formula, construct the signal now graph from the transfer function. From the signal now graph, state model can be obtained. While obtaining sii;nal flow graph, try to get the gains of branches as 't/s' representing the integrators. This helps to obtain the required state model. To clear this idea, let us consider the system having transfer function,
Modem
2·10
Control Theory
State Space Representation
Divide both numerator and denominator by highest power of s,
T(s)
a a a i+-t+-1.+~ s 52 s 3
T161 +T262 + T363 + ·r4t14 = I -(l:Aliioop gains)+[l:Gain products of 2 n~n-touching loops]-···
Assuming that there arc no combinations of 2 and more non-touching loops. Loop gains arc, And let ll.1 = 62 =
I).3 =
l\~ = 1 i.e. all loops arc touching to au the forward paths.
Hence forward path gains arc, T1
=
b3' T2
b2
bl
= s' T3 = 2' s
Involving various branches having gains ~· the signal flow graph can be obtained as,
Fig. 2.5 Each brand' with gain ~ represents an integrator. Output of each integrator is a state variable.
According to signal flow graph, value of the variable at the node is an algebraic
2-11
Modern Control Theory
State Space Representation
sum of all the signals entering at that node. Outgoing brunches docs not affect the value of variable. Hence from signal flow graph. X1
b2U + Xz - "2 y
. ~
b1U + X3 - «1 Y b0U - a0 Y
y
b3 U + X1
Xz and
Substituting Y in all the equations, ... (la)
... (lb)
.
. .. (le)
X3 These equations give the required state model. X(t) = A X(t) + B U(t) A
where
f-a2
1
l-·•
0 I 13=!1>1-.i1b1 0 OJ lbo-aob3
-.10
and
Y(t}
c
where
1
Ol
ft'2 -a2b3
l
CX(t) + DU(t) [1 0 0),
D = b3
Now as Y(O), Y(O), Y(O), U(O), U(O) and U(O) arc the known initial conditions, using the derived state model, the initial conditions X1(0), X2(0) and X3(0) can be obtained.
>•
Examplo 2.4 : T(s) =
A feedbacksi;stem is clwacteriud by
t11e
closed loop trausferfunction,
s2 +3s+3
~3+2s2+3s+I Draw a suitable signal flow graph and obtain lite state model. Solution : Divide N and D by s3. 1 3 3 -+-+s g2 SJ 1+~+1-+..!... s ,2 53
1 3 3 -+-+s S2 SJ
1J-~_1__..!...] L
s
s2
s3
Modem Control Theory
2 • 12
State Space Representatlon
3
1'2=-2, s
And 61 = d2 = 63 = 1 with no combinations of non-touching loops. There arc mnny signal flow graphs which can be obtained to satisfy above transfer function. Method 1 : The signal flow graph is as shown in the Fig. 2.6.
-1
Fig. 2.6 From signal flow graph, y
.
X1
Xi
X3 + 3U - 3Y ~ - 3X1 + X3 + 3U
X3
3U - Y = - X1 + 3U
Hence the state model is,
i.e,
.
x
AX(t) + BU(t)
Modem Control Theory
and
Y(t)
K
(1 0 OJ
i.e.
Y(t)
=
X(t) with D
x231 [~
l
2-13
State Space Representation
=0
Method 2 : When m -J. n and m < n then signal flow graph can be constructed so as to obtain matrix A in phase variable form. This is shown in the Fig. 2.7.
-1
Fig. 2.7 From the signal flow graph,
. Xz .
X2
X3
- X1 - 3X2 - 2X3 + U
y
3X1 + 3X2 + X3
X1
and
~
Hence state model has, A
= [ ~ ~ ~} -1
D
=
6
= [~}
-3 -2
C "' (3 3 1 J and
I
0
Thus the matrix A is in phase variable form. Note : When m < n, then transmission matrix D = 0.
Modern Control Theory 2) Using
State Space Representation
Direct Decomposition
of Transfer
Function
Th.is is also called direct programming. In this method, denominator of transfer {unction is rearranged in a specific form. To understand the rearrangement, consider an element with transfer function _]_. From block diagram algebra. the transfer function of
s +~1
minor feedback loop is
1
+~H for negative feedback,
.!.
Lei
where
_s_=_G_ I+~ l+GH s
s+a
.!.s = integrator and H = a
G
The feedback is negative and the transfer function can be simulated as shown in the Fig. 2.8. with a minor feedback loop. Now if such n loop is added in the forward path of another such loop then we get the block diagram ns shown in the Fig. 2.9.
H
Fig. 2.8 1
S+i
r-----------------------1 I I I I
I I I I I I I
I I I I
!-----------------------~ b
Fig. 2.9 The transfer function now becomes, s (s+a)
·;:-b
S (S.j.
J)
Modem Control Theory
State Space Representation
2 • 15
X = (s ... a)
where
U now the entire block shown in the Fig. 2.9 is added in the forward path of another minor loop with an integrator and feedback gain 'c', we get the transfer function as, = -1- where Y = sX + b sY ... c Thus the denominator of transfer function becomes s (sX + b) = [s (s (s + a) + b)l = s3 + as2 ... bs Thus denominator of any order can be directly programmed as discussed above.
s2 + as + b
=> Is (s + a) + bl
s3 + as2 + bs + c => l[(s + a) s +b] s + c] s4 + as2 + bs2 +cs+ d=> I([ (s + a) s + bl s + c) s + d} and so on. Now if numerator is b1s ... b0 and denominator. Simulation is obtained directly then the block diagram is as shown in the Fig. 2.10. But s = ~~· which is differcntfotor and is not used to obtain state model. In such a case, take off point 'I' is shifted before the last · integrator block.
OireCl simulation cf denominator
Fig. 2.10 (a)
C> [>
TI~J U(s)
'
'I
~------------Fig. 2.10 (b)
........... 1--'-<Xl---Y(s)
Modern Control Theory
State Space Representation
2-16
·~'I I
I I
-------------Fig. 2.10 (c) According to block diagram reduction rule, while shifting take off point before the block. the take off signal must be multiplied by transfer function of block before which it is to be shifted. Thus we get block of b1 with take off from input of last integrator. Similarly if there is a term b2s2 in the numerator ihen shift take off point before one more integrator as shown in the Fig. 2.10 (d).
~--~--_ _[_ '
'
: ----------·
:
~,. $
~
•
:
$
•
:
:
;
:
:
Sin'u..Qtionof
numer~or bg • b,s • bit2
Fig. 2.10 (d) Thus for any order of numerator, complete simulation of the transfer function can be achieved. Then assigning output of each Integrator phase variable form can be obtained.
as the state variable, state model in the
2 .17
Modem Control Theory
1.+
State Space Representatlon
Obtain state model by direct decomposition method of a system whose transfer/1111ctiDn is Example 2.5 :
Y(s)
5s2 +6s + S
U(s)
s3 + 3s2 +7s+9
Solution : Decompose denominator as below, s3 + 3s2 +7s+9
= {(Is+
3)s+7)s+9}
Its simulation starts from (s + J) in denominator
Fig. 2.11 To simulate numerator, shift take-off point once for 6s and shill twice for 5s2. Therefore complete state diagram can be obtained as follows.
Y(l)
U(l)
Fig. 2.12 Assign output of each integrator as the state variable.
Modem Control Theory
While output,
2 -18
x~
U(t} - 9X1(t) - 7X2(t) -3X3(1)
Y(t)
SX1(1) + 6X2(t) + SX3(l)
State Space Representation
: . Sta te model L•,
and
X(t)
AX(t) + BU(!}
Y(t)
CX(t) + DU(t) 1
where
A
[_ ~
c = Ia
0
()]1 •
-7 -3 6
B=m
st .
The matrix 'A' obtained is in the Bush form or Phase Variable form.
2.2.3 Advantages The various advantages of phase variables i.e. direct programming method arc, 1. Easy to implement.
2. The phase variables need not be physical variables hence mathematically powerful to obtain stale model. 3. It is easy to establish the link between the transfer function design and lime domain design using phase variables. 4. In many simple cases, just by inspection, the matrices A, ll, C and D can be obtained.
2.2.4 Limitations The various limitations of phase variables arc, 1. The phase variables arc not the physical variables hence they loose the practical
significance. They have mathematical importance. 2. The phase variables arc mathematical variables hence not available for the measurement point of view. 3. Also these variables are not available from control point of view. 4. The phase variables arc the output and its derivatives, if derivatives of input arc absent. But it is difficult to obtain second and higher derivatives of output. 5. The phase variable form, though special, does not offer any advantage from the mathematical analysis point of view. Due to all these disadvantages. canonical variables arc very popularly used to obtain the state model.
Modem
State Space Representation
2·19
Control Theory
2.3 State Space Representation
using Canonical Variables
This method of obtaining the slate model using the canonical variables is also called parallel programming method and matrix A obtained using this method is said to have canonical form, normal form or Foster's form. The matrix A in such a case is a diagonal matrix and plays an important role in the state space analysis. The method is basically based on Partial Fraction Expansion of the given transfer function Tis). Consider the transfer function T(s) as, bos"' +b,sm·t +b2srn·2+ ... +bm (s+a 1) (s+a2) .... (s+ a 0)
T(s) =
Case 1 : If the degree 'rn' is less than 'n' (m < n), then T(s) can be expressed using partial fraction expansion as,
c Now each group s + ~
can be simulated using the minor loop in state diagram as 1
shown in the Fig. 2.13.
------------------, I I
I
~-----------------' '-......,.......
I I
The outputs of all such groups arc to be added to obtain the resultant output.
'
To add the outputs, all the groups must be connected in parallel with each other. The input U(s) to all of them is same. Hence the method is called parallel programming. The overall state diagram is shown in the Fig. 2.14.
'
0
....,.,,
1 s • a1
Fig. 2.13
Then assign output of each integrator as a state variable and write the state equations as,
While X
AX + BU
and Y = CX + DU
Modem Control
Theory
Stale Space Representation
2 ·20
U(s)
Y(s)
..
. ~---·
~--· I I
Fig. 2.14 Foster's fonn sim\Jlation
0 where
A
[1
-a2
0 0
0
0
C = (c1 C2 ••••• C,,)
j and
B
•[j]
0=0
Case 2 : If the degree m = n i.c. numerator and the denominator have same degree then first divide the numerator by denominator and then obtain partial fractions of remaining factor. N(s)
--=eo+
T(s) where
Co
ocs>
=
~
C;
k-i•I s+a;
Constant obtained by dividing N(s) by D(s).
In such a case, the state diagram for partial fraction -.terms remains same as before and in addition to all the outputs, Co U(t) gets added to obtain the resultant output as shown in the Fig. 2.15.
2 ·21
Modem Control Theory
State Space Repl'8Sentatlon
U(s)
•
Additional elemellt Fig. 2.15 State diagram for T(s) with m • n Thus the state model consists of the matrices as,
=
c
(c1
C2 •••
c,,J,
D
= Co
Key Point: 17111s for m • n, direct transmission matrix D exists in the state model. When the denominator D(s) of the transfer function T(s) has non-repeated roots then the matrix A obtained by parallel programming has following features, 1. Matrix A is diagonal i.e, in canonical form or normal form. 2. The diagonal consists of the clements which arc the gains of all the feedback paths associated with the integrators. 3. The diagonal elements are the poles of the transfer function T(s). 1!11$> Example 2.6 :
Obtain the state model by Foster's form of a system tu/lost T.F. is,
s2 +4 (s + l)(s + 2) (s + 3)
Modem Control Theory
2-22
Stato Spaco Reprosontatlon
Solution : Find out partial fraction expansion of it s2 + 4 - ------,:; (s -e- 1 l(s ... 2)(s + 3) -
2.5 s -t- 1
s
-·-----
s+2
6.5 s+ 3
+ --
:. Total state diagram is as shown in the Fig. 2.16.
U(I)
Y(I)
Fig. ., 2.16
Y(t)
2.5X1(t)-8X2(1)+
State model is, X
AX+ BU
and
y
ex+ ou
where
A
[-10 -20 0OJ 0 0 -3
c
(2.5 - 8 6.SJ
6.5X3(t)
B=m D=O
Modem Control Theory
State Space Representation
2 ·23
2.3.1 Jordan's Canonical Form In the Foster's form it is assumed that the roots or denominator of T(s) are non-repeated, simple and distinct. But if the roots arc repeated then the parallel programming results matrix A in a form called Jordan's Canonical form.
Let TM has pole at s " - a1 which is repeated for r times as, T(s) =
N(s) (s+a1)'(s+a2) •.. (s+~,.)
The method of obtaining partial fraction for such a case is, T(s) = __ c_•-+ (s+a
il'
Cz (s+a ,)r-1
+... +-c-'-+~+ (s+a 1) (s+a2)
If the degree of N(s) and D(s) is same i.e, m
e
....+~ (s+a0)
... m
n we get additional constant c0 as,
... m-=n This can be mathematically expressed as,
... m < n r
and
T(s) -Co+~
C
,,
('
~ --'z: ---'-r-J·J (s+a, ) - + P=U·l £... (s+e .) i>1:::l I
... m""n
Key Point: Note that in partial fraction expansion, a septrate coefficient is assumed for caclt power of rqJtattdfactor. In simulating such an equation by parallel programming, --1-(s+a1)'
is simulated by
connecting -( -1-) groups, r times in series first. While all other distinct factors arc s+a1 1 simulated by parallel programming as before. The components of each power of -( . ) to s+a1 be added lo get output is to be taken from output of each integrator which arc connected in series. This is shown in the Fig. 2.17. Now assign state variables at the output of each integrator. For series i.ntegrators, assign the state v.oriablcs from right to left. as shown in the Fig. 2.17.
Modern Control Theory
State Space Representation
.:
'
/'• t
.
\, \
''
'
'
'' ''
G
.'
'
1
': '' '' '
''
'
''
: '
J
:
:
'' '' '' ' ' . --, ' ' ... • v ; -' ... ' ': '' '' ' : :: '' : ''' '
,''
-:'r-·---------'
'' '
t: t'
-rl
'' ' ''
'' '' J'
•''
.
'
1-~~~~~~~~~~~~~~---------L
3
' ...... J'........
..f
Modem Control Theory
State S'pace Representation
2 -25
For series integrators, the state equations arc,
While for parallel integrators, the state equations arc,
X,. +
I
=
-<12 X, + I + U(t)
While
Key Point: Y(t) has additional c0 U(I) term
if 111
= 11.
Hence the state model has matrices in the form, . --a1forrtimes~
1-•,, 0 9.
0
A~ Jordan block
f 1
Element tforr-1 times O·····O
-a, 1 ····· <, o -a1······1
0
"''
0
O······
0
0 ···· ······--··O
o.
o ..... ..:a,
0
0 ····· 0
-, -a,
O·· ·· 0
o
1
········0
0
O····--·--···--·
o
0
o ··············
o
0
0
0
-1----------'
a2
O········--·O
: ·O
0 ····
: "-L
~
Fig. 2.18
0
:
~~:
only diagonal elements
an
Modem Control Theory
2 ·26
~1
: B
State Space Representation
r-1 times
?J
C
=
(c1 C2 ••• c, <:, • 1 ... c0J
The matrix D is zero if m < n and is '<:o' if m = n. The matrix A in such a case has a Jordan block for the repeated factor and many times A is denoted as J. X(t) = JX(t) + BU(t) The matrix A
... For repeated roots
= J has the following features,
1. The principle diagonal consists of all the poles of transfer function with repeated pole 'r' times and other nonrepeated poles. 2. Upper off diagonal consists of (r - 1) times unity element, as indicated in the Jordan block. 3. All the remaining elements arc zero. Note that the matrix B has 'r - 1' zeros and all other clements as unity. The matrix C has all partial fraction coefficients c1, c2 ... c;,.
2.3.2 Advantages of Canonical Variables The main advantages of canonical variables arc 1. The matrix A is diagonal 2. The diagonal clement is very important in the mathematical analysis. 3. Due to diagonal feature, the decoupling between the state variable is possible. This
.
means all 'n' differential equations are independent of each other. Thus X1 depends on X1 alone.Xi depends 011 "2 alone and so on. Such decoupling is important in system design from controlling point of view.
2.3.3 Disadvantages of Canonical Variables The main disadvantages of canonical variables is similar lo the phase variables. These arc not the physical variables hence practically difficult to measure and control. Hence such variables arc not practically advantageous, though mathematically arc very important
2. 27
Modern Control Theory
1•
State Space Representation
Example 2.7: Obtain the statt model in ]ord1111'; canonicalform of a systtm wlrose T.F. . I IS (s+2) 2 (s+t)
Solution : Finding partial fraction expansion, T(s) : _A_ + _B_ + C (s + 2) 2 (s + 2) (s + 1) Take LCM on right hand side and equate numerator with numerator of T(s), ;.A(s + 1) + B(s + 2) (s + 1) + C(s+2)2:
1
Equate coefficients of all powers of s on both sides,
a ... c A+ 38 + 4C A+28+4C
s2
0,
from power of
0
from power of s
r
... from constant term
Solving we get, A " - 1, B " - 1, C ~ 1 -1 1 1 T(s) = (s+ 2)2 - (ST 2) + (s+ 1) Simulate first term by series integrators while other nonrepeated integrators.
terms by parallel
Yi!)
Fig. 2.19
.
Total simulation is, Xt
.
X2 = U(t) - 2X2
.
X3 "U(t) - X3(t)
Modern Control Theory :. State model
2 ·28
X
AX+ BU
A
r-02
-2
0
0
is ,
l
c = I- 1
-~J
- 1 1j '
State Space Representation
and
B = D
Y =ex+
ou
[!]
=0
The matrix A consists of Jordan block. Important Noto : If some of the poles o{ T(s) arc complex approach can be used. The quadratic or higher order polynomial be simulated by direct decomposition while real distinct roots parallel programming using canonical variable, lhe example procedure.
l*
in nature, then a mixed having complex roots can can be simulated by the 2.8 below explains this
Example 2.8 : C"111bi11ntio11 of Direct Dtco111positio11 1111d Foster's Form. Obtain tire state modelfor tire given 1·.F.
tto ~
·--2---
(s -e- 3) (•2 +25+2)
Solution : Obtain partial fractions as, T(s)
2
s
(s+ 3l(s2 + 25+ 2)
=~ + lls + C s+ 3 s2 + 2s+ 2
Find LCM on right hand side and equate numerators or both sides, .'\(s 2 + 2s + 2) + (lls + C) (s + 3) = 2
As: +2As+2J\ + Bs~ +Cs+ 3Bs+ 3C = 2 Equating coefficients or all powers of s, A+B
0
2A + C-3A C-A
2A+3C=2
2A+3A=2
0
c Solving,
0
2A+C+3B=O
A
T(s)
A
2 5.
2
B=-5,C•5
2
2 2 2 2 2 2 = _L + -5s+5 =l + -5s+5 (s + 3) s2 + 2s + 2 s+ 3 {(s + 2) s+ 2}
The quadratic s2 + 2s + 2 having complex roots is decomposed directly.
Modem
Control
Theory
2 ·29
State Space Representation
:. Complete state diagram is as shown in the fig. 2.20
U(t)
Fig. 2.20
and
X1
U(t)-3X1
X2
x3
X3
U(t) - 2X2 - 2X3
Y(t)
~X1(l)+~X2(t)-
~X3(t)
ex
:. State model is,
x
AX +BU and Y
where
A
[- ~ _!],Bom
Q
0 0 -2
Note that in such a case matrix A does not have any specific form.
2.4 State Model by Cascade Programming This is also called Pole-Zero Form or Cullemln's Form. This can be effectively used when both numerator and denominator can be factorised. In such form, group a pole and zero together and arrange given transfer function as the product of all such groups. Then simulate each group separately and connect all such simulations in cascade to get complete simulation. Then assigning output of each integrator as a state variable obtain a state model in standard form.
State Space Representation
2. 30
Modem Control Theory
Simulation of group : Consider
s+ a
'S+b
First simulate denominator _l_b as in the Fig. 2.21 (a) and then as discussed earlier
s+
simulate the numerator (s + a) as shown in Ille Fig. 2.21 (b),
~------·------
..
I
, _ - - - - - - - - - - - - _, (a)
s
!b
(b)
~: ~
Fig. 2.21 Simulation of a group of pole-zero
>Ill+
Example 2.9 : Obtain Ille state model of a system by cascade programming w/Jose transfer
junction is T(s) ,. Y(sl ,. (s+ 2)(s+ 41 l/ls) s (s+ 1} (s+ 3) Solution: Arrange the given T.F. as below Y(s)
(s+2)
(s+4)
U(s)
(s+l)
(s+ 3)
! Group l
! Group 2
l Group3
Now simulate each group as discussed and connect all of them in series to obtain T(s). The complete simulation is shown in the Fig. 2.22.
Fig. 2.22 Now
.
.
X2 = - 3X2 +2X3 +X3
Modem
2. 31
Control Theory
State Space Representation
X.1 = U(tl-X3
and
Substituting X 3 into X2 equation,
Substituting
x2
into X1 equation,
Model becomes,
X3
[' ' r· ['I
and
Y(t)
X1(l)
i.e.
Y(t)
[1
So
A
X1 Xz
0
- 3 -
1
X2
0
0
1
X3J
0
[O0-3 1 0
c
=
0
[I 0 OJ
rn:l
OJ
111JB=1 -l
J
1
+
I
U(I}
1,
n 1
0=0
Feature of A : Feature of Matrix A is its principle diagonal contains gains of all feedback paths associated with all integrators i.c, 0, -3, -1 in above problem. All terms below principle diagonal are zero. Thus the features of A having all poles of T(s) in its principle diagonal still continues in this method of programming. This method is also known as Iterative programming.
Examples with Solutions na+
Example 2.10 ; Obtain the state model of the given network i11 tire sta11d11rd form. Assume R1 l Q C1 1F
=
=
n
R2
2
R3
3Q
c, = 1 F
Modem Control Theory
2 -32
State Space Representation
Fig. 2.23 Solution : Selecting state variables as voltages across capacitors C1 and C2
i.e. c1 and c2,
R,
X1(1)
c1
=
~l
X2(1)
U(I)
J
+
t
1,(ll
Applying Kirchhoff's laws, U(t)-i1
R1 -c1
-R2
U(t)
=
o-------'------'---o{<')
(i1-i2)=0
Fig. 2.23 (a)
i1 R 1 "'- c 1 + R2 (i 1 - i2)
... (1)
fur second loop, ~1-i2
R;i -c2 -(i2 -i1)R2
o ... (2)
Solving simultaneously equations (!) and (2) U(t) "1
;, (R, + Rz)- iz R3 + c, - i, Rz + i2 (Rz + R3) .. l'2
Substituting the values, U(t) i.e ..
i 1 (I + 2) - 2i z + c 1
U(I)
. .. (3) ... (4)
Multiply equation (3) by 2 and equation (4) by 3 2 U(ll
6i1-4i2+2e1 - oi 1 ~ 15i~ • 3c~ and adding
3 U(t) + :'lc1
2 .33
Modem Control Theory
State Space Representation ... (5)
Now from equation (3) U(t) = 3i1
-
2i2 + e1
Substituting i2 from equation (5) 6 2 6 U(t)+ IT U(t)+ IT e1 -e1 - IT e2 U(t)
3i I
-fr
U(t) - ~el
+fr <'z
+el
17 9 6 33 U(t) - 33 cl - 33 cz
... (6)
Now
... Capacitor current is C (dvc/dt)
as C1 =I,
dt
dct
17 9 6 3 I 3 33 U(t) - 33 e I - 33 c2 - IT U(t) - IT ct + IT ez
dc1
8 12 33 U(t) - 33
dt
.
e1
3 + 33 cz ... (7)
X1 and
as
dc2 C2(11
... Capacitor current is C (dvc/dt)
Cz dc2
dt
3 1 3 ITU(l)+ITel-Tfez ... (8)
X2 and
Y(t)
l [l
:. State model is, X
AX + OU and
where
12 [-~ 11
A
1
y = ex + DU
-~,B=~ 11
8
11
c
= [O ,tJ
D
= IoI
2. 34
Modem Control Theory ,..,
State Space Representation
Example 2.11 : Consider t/1t mechanic.a/ system slrou•n iu figure. For sllown displacements and velocities obtain tire state 1twdel i11 staudard [arm.
Ass11me t•elocity of M 2 as output.
v,
v,
Fig. 2.24 Solution : Select the state variables as energy sloring elements i.e. displacements and velocities related lo spring and friction. Y1(t), Y2(t),
F1(t),
x3
= V1(t)
X4 = V2(t)
Y(t)
=
V2(t)
Draw the equivalent mechanical system. Due to F1(t), M1 will displace by Y1• Due to spring K1 and friction 81 which arc between the two masses, the displacement change from Y1 to Y2• While mas' M2, spring K2 and friction 8z are under the Influence of Y2 alone as K2 and 8z are with reference to fixed support and not between two moving points. Represent each displacement by separate node. Connect the clements in parallel
Fig. 2.25
2. 35
Modern Control Theory
State Space Representation
which arc under the influence of same displacement, thus K1, B1 parallel between Y and Y 2, m2, K2 and Bi parallel between Y 2 and reference i.e. fixed support and so on. The spring force is proportional to net displacement in spring while frictional force is proportional to the velocities. At node Yi... (1)
At node Y2, 0 ... (2} Substituting all values in tenns of state variables we get, d2Y __ 1
• = XJ
dt2
:. Substituting in equation {l) and equation (2) U(t) 0
M2
... (3)
.
X4 + K2X2 + B2X4 + K1 [X2
-
Xi)+ B1 (X4
-
X3]
... (4)
From equation (3) and equation (4) we can write
. .
X3
~
{U(t)-K1(X1-X2)-B1(X3-X4))
... (5)
I
and
X4
and
X1
Mz
X3,
[-K2 X2 -B2 X4 -K1 (X2-X1)-B1 X2
=
X4, Y{t)
=
V2(t)
=
:. State model can be constructed in the standard fonn
.
X
AX+ BU
v
ex .. ou
X4 (t)
(X4 -X3))
... (6)
2. 36
Modem Control Theory 0 0
0 0 A
Where
K1
0
+~ M1 -(K1+Kz)
-Mi K1 M2
,...,
State Space Representation
M2
Example 2.12 : For tht given T.F. of a system obtain t/ie stole model by i) Direct decomposition T(s} =
ii} Gui/em in 's Form
iii} Foster's Form
(s+2)(s+3) s(s+l)(s2 +9s+20)
Solution : I) Direct decomposition T(s) =
s2 +5s+6 s(sJ + 10s2 + 29s+ 20)
s2+5s+6 {([s+ IO]s+29)s+20
}s
State diagram is as follows
Y(sl
Fig. 2.26
X4 y
U - 20X2 - 29X3 6X1+SX2+X3
-
lOX4
,
2 .37
Modern Control Theory
State Space Representation
:.State model is
where
x
AX + llU and Y ; CX
A
c
0 0 0 0 0 -20
["
-29
[6 5 I
oI
0 I
0
0 0 I
-10
+
DU
•. [!]
D= (0
I
Ii) Gullemln'sForm T(s)
(s+ 2)(s+ 3) (s+ 2) (s+ 3) 1 s(s+ ))(s+ 4)(s+5) =(s+l).(s+4) (s+S) s
State diagram is,
UC•)
Fig. 2.27
Substituting X4 back in X J
Substituting X 3 back in X 2
and
y
2. 38
Modem Control Theory
State Space Representation
:. State model becomes
x
=
0
I
where
A
and Y =ex
AX+ BU
[;
-1
-5
-4 0 0 0
-11
iii) For Foster's form :
C = [ 1 O O OJ
··m
Find out partial fraction expansion of T(s) A ll C D T(s) ;;;; s + 1 + s + 4 + s + 5 + S -1/ 6 -s+ I
.
1 / 6 3 I 10 3 / 10 -----. -s+4 s+5 s
U(IJ
Y(t)
U(t)-X.,
Fig. 2.28 1
1
3
Y=-6Xt+6X2-l(jXl+l(jX4
3
2 .39
Modem Control Theory
State Space Representation
State model is, X =AX+ BU
and
Y
where
111•~
=ex A=
EJtample 2.13
[~00 1 - 0 -~05 ~01OJ B = [~11
C=[-i
6
-to
to]
Derive tire stale model in /orda11's ca11011ical form for a system lunri11g T.F.
T(s)
Solution : T(s) =
l
0o4
=
l s3 +4s2 +5s+2
A B C 1 = I =--+--+-(s + 1} 2 (s+ 1) (s + 2) s 3 + 4s 2 + 5s + 2 (s ... I) 2 (s + 2)
i.e, A(s + 2) + B(s + 1) (s + 2) + qs + l) 2 = 1 As + 2A + Bs 2 + 3Bs + 2B + Cs 2 + 2Cs + C B+C
0 , A + 313 + 2C
=0
r
2A <· 28 + C Now C = 1 , B = -I, A = 1 T(s)
1 (s+1)2
1 (s+l}
1 (s+2)
-----+--
State diagram L~ , U[I)
Y(I)
Fig. 2.29
X1
- Xi+ X2
x~
U(t) - X2
X3
U(t)- 2X3
x1
Y(s) :.State model is,
where
State Space Representation
2 -40
Modem Control Theoiy
.
-X2 +
x_,
x
AX+ BU
A
[~1 ~1.
Y =CX
-1
B=m,
0 -2
1•
c
= (1 - 1
1)
Example 2.14 : Obtai11 the state model of system whose T.F. is
s 3 + 3s2
+ 2s
------- by Foster's form.
T(s)
s3 +12s2 +47s +60
Solution:
T(s)
=
s3 + 3s2 +2s s3 +12s2 +47s+60
As numerator and denominator are of same order we cannot directly find out partial fractions. For partial fraction, numerator d~'b'TCC must be less than denominator. divide N(s) by D(s} and find partial fractions of the remainder. s 3 + 12s2 + 47s + 60) s 3 + 3s2 + 2s ( 1 s3+12s2
- 9s2 T(s}
=
-
+ 47s+ 60 45s-60
Y(s)
U(s)
_ 1
-
[ 9s2 + 45s + 60 ] - s 3 + 12s2 + 47s + 60
9s2 +45s+60
l - [ (s+3)(s+4)(s+5)
]
A
= l -'S+'3 -
1 - _3_ + ~ - 2Q._ s+3 s+4 s+S
B
C
s+4- s+S
So directly
2-41
Modem Control Th~ry
State Space RepresentatlO<•
State diagram :
V(I)
V(t)
Fig. 2.30
X3
U(t) -5X3
Y(s)
- 3X1+2-IX2
- 30X3 + U(s)
.·. State model is,
x
AX + BU
and
y
=
ex
+ DU
where
C = [ - 3 24 - 30],
D = [11
When N(sl and D(s) are having same order, or degree of N(s) > D(s) then, there is always direct transmission matrix D present in the model.
••
Example 2.15 : Obtain a stat« space model of tile system will! transferfunction Y(s} U(s)
6 $
3+6:>2+11$
+6 (VTU:J•n.1Feb.-2007,July/Aug.-2007,Dec)Jan·2008)
2·42
Modem Control Theory
State Space Ropresantatlon
Solution: Y(s)
The T.F. is
6
U(s)
s3 +6s2 + lls+6
Using factorisation
of denominator,
Y(s)
U(s)
6
=
(s+l)(s+2)(s
... J)
Taking partial fractions, Y(s) = -1.._ _ _2_+_3_ U(s) s+I s+2 s+3 Hence the state diagram is as shown in the Fig. 2.31 U(s)
Yis)
Fig. 2.31 This is Foster's form of representation. From above Fig. 231 we get,
and
Xi
U(s)-X1
X2
U(s) - 2X2
X3
U(s) - 3X3
Y(s)
3X1 - 6X2 + 3X3
Hence the state space model is,
.=
X
AX.;. BU
2 .43
Modern Controltheory
and
y
where
A
=
State Space Representation
ex 0
n -~l -2 0
B
e ,..,.
m = (3
- 6
31
Obtain Ille state model for system represented by
Example 2.16 :
•J>y + 6-J2y + 11-dy +I011=3U(t) dt3 dt2 tit . Solution : System is 3rd order, n
(VTU:
=3
May·99, Dcc.2007/J•n.-2008)
3 integrators and variables are required. Select y = X 1 and then successive diffcrcntfation of y as next variable.
x.
X2 ~ dy/dt
. .. (1)
X2
d2v X3=-' dt2
... (2}
Now as 3 variables arc defined, X 3 .,. X,1 but X3 must be obtained by substituting all selected variables in original differential equation. as
x3 y •
... (3)
x,
which is output equation.
:. State model can be written as,
.
x
AX+ BU
and
y
ex+ ou
Where
A
L~
1
0 -11
!]·B=m,e:Jl
0 0 J, D =JOI
Medem Control Theory
2-44
State Space Representation
State diagram :
U(sl
Fig. 2.32 ,...
Example 2.17 :
OblRin the state model of tl1t differential equation
4 ''.l'i.!.!.+ 3 'l1dl) +de(/)+ 2cllJ d 1J
d 11
tit
= Sr(t) (VTU : Aug·95)
Solution :The equation can be written as,
.
4 c (t) • 3 c (t)+ c(tJ+ 2c(t) = Sr(t)
The order of the equation is 3. Selecting state variables as, c(t}
. .
X1(t)
X2
X1(t) = c(I)
... (I)
X3(t)
X2(t) = c(t)
••. (2)
X3(t)
c(t)
Substituting in the original equation, 4X l (t) + 3X3 (t)+ X2 (I)+
2x, (t)
= Sr(t) 2
-4X,(t)-4
Hence the state model is,
[ f:
l [_j _Jl _J
I
3
5
X2(t}-4X3(t)+4r(t)
... (3)
2·45
Modem Control Theory
St.ta Space Repreaentatlon
While the output equation is, c(t)
XI (1)
1 (I)]
X
i.e.
y(t) " [ 1 0 OJ
X2(t) [
••
X3H)
Example 2.18 : Obtain thL stat~ modLJ of t/U! syst~ whose cl~
loop trtmsfufunction
is
cot
2(:;+ 3)
R(s)
(s+l)(s+2)
(VTU : Feb-97, Orl-98)
Solution : Let us use the parallel progranuning Le. Foster's form. Finding partial fractions of C(s)/R(s).
Hence the state diagram is as shown in the Fig. 2.33
C(s)
Fig. 2.33 Hence the state equations from the Fig. 2.33 arc ... (!)
R(s) - 2X2 and
C(s)
Hence the state model is,
... (2)
Modem Control Theory
2 ·46
State Space Representation
and
1•
Example 2.19 : Write a set of stale tq11nlions for tltt 11tlwork shoum itt lht F(~. 2.34 Clroose i1, 12 find vc as $/alt variables. (VTU: April-98,July/Aug.·2007)
Fig. 2.34 Solution : Let
c(t)
Input
Voltage across R2 i,(t) i2(t)
vc(t) .. x,(t) Let i1(t) and i1(t) arc the loop currents. Applying KVL to the loops. c(t)
..• (1)
Then,
... (2)
and
c~~ di dvc (fl
... Capacitor current 1 (.
a
C •1-lz
. )
Modem Control Theory
2 • 47
State Spaco Representation ..• (3)
... (4)
While Hence the state model is,
x, X2
l x3
RI
-Li
0
0
-i:;-
-Li
R2
I
I
-c;
e
r x,
l L2
X2
0
X3
+-
. [?
l ""'
where "'''
"~·>
and
,,_.
Example 2.20 : Considering'Ve am/ 11 as state mriab/e:; and Is as tire outinu variables 111
tht circuit shown below, obtain the statt model. R1
L
IVVV'-v-,1-.-c-·
'-3)'• "
1
Fig. 2.35 (VTU : Aug. • 97)
Solution : Convert the CWTClll source to voltage source as shown.
R,
Fig. 2.35 (a)
Two inputs,
c{t) and Ii: (t) i.e. U1 = c(t) and U2 = lg(t)
One output, Vnriablcs,
State Space Representation
2 ·48
Modem Control Theory
i.e. Y(t) = J,.(t)
l,(t)
vc
=
X1(t) and Ii
= X2(t)
Let l,(t) and l;(t) be the loop currents. Applying KVL to the two loops, Loop l,
-I, R1
-
"c + c(t)
=o
c(t) - "c 1 1 - e(t)-vc(t) R1 RI Y(t)
1 1 - U1 - -- X1{t) R1 R1
di, Loop 2' - L dt - I' R2 - IS R2 + v c
=
... Output equation 0
... State equation and current through capacitor, dvc
CCit 1
ell, -I ;I Substituting I, from output equation, ... Stale equation
- ~~]l~;] y
[-R;"1 0][X1Xz(I)(t)] + [ R;"l 0][Uu, 1]
Modem Control Theory ••
Example
2.21 :
State Space Representation
2·49
Derite two state models for thL system described '1y the differential
eqllation D3y+4D2y+5Dy+2y i) ii)
= 2D2u+6D11+Su
where D = d/dt
One in phase variable form. Other in /ordan-Ca11011ical form.
(VTU: Mardt-2001)
Solution : Take the Laplace transform of both sides of the equation and neglect the initial conditions to obtain transfer function of the system as, s3Y(s)+4s2Y(s)+5sY(s)+2Y(s) Y(s) U(s)
= 2s2U(s)+6s
U(s)+SU(s)
2s2 +6s+5 s3 +4s2 +5s+2
I) Phase variable form
Use the direct decomposition. Y(s) 2s2 +65+5 U(s) = {((sH)s+5]s+21 The state diagram can be shown as in the Fig. 2.36
Fig. 2.36
So
X1
and
Y(s)
2. 50
Modem Control Theory
So state model is having matrices,
A
= [ ~2
~S ~
l
=
B
m
State Space Representation
and C = (5 6 2)
ii) Jordan canol\i."I form : fllclorise the denominator as, Y(s) U(s)
A(s+2)+B(~+l)
2s2 +6s+5 = (s+1)2 (s+2)
c
=
(s+2)+C(s+l)2
_A_+_D_+~ (s+1)2 (s+l) (s+2) 2s2 +6s+5
As+2A + D~2 + 3Ds+2D+Cs2 +2Cs+C = 2s2 +6s+5 B+C A
= 2,
= I,
A + 38 + 2C = 6, B = I,
2A + 2B + C " 5
C =l
Y(s) _1_ + _1_ + _1_ U(s) = (s+1)2 (s+l) (s+2) The a.talc model is shown in the Fig.2.37 (a) U(s)
!------<><>---
Fig. 2.37 So and
Y(s)
So state model is having matrices,
A = [~
~1
~J
B
=
m
and C = [1 1 1)
Y(s)
Modem Conttol Theory ,..
2. 51
State Spa~ Representation
For lht system s/1~1 in the Fig. 2.38, obtain the state 111od4 choosing v1(t} and 112(1) as tlte state wriab/es, (VTU: July/Aug.·2005, Jan./Feb.-2007) Example 2.22 :
Fig. 2.38 Solution : Select the two currents as shown in the Fig. 2.38(a). And wfife tho equations using KVL and KCL ..
1 Mil v1(t) U(t) 1 Mil vi(t) +9-':"V\NV~-..~.....,,,,'llJ\~~.-~~-<> U(t) ~
_l
T
.,
1µF~ '2
1
lµF
Fig. 2.38 (a) U(t)-v2(tJ
txl06 Yz(t) i.e.
~J
... (I)
(l t -J2) dt
C dv2(t) dt v2 -v1
... (2) ... (3)
lxl06
And
~Ji2
dt ... (4)
Elliminate i1 and i2 from above equations and C = l x 10- 6 F.
Modem Control Theory
2 -52
State Space Representation
Substituting (I) and (3) in (2) we get,
c ~.l. dt dv2
di
[U(t)-v2(t)
]-[ v2(t)-v
ixio>
1+~-+! C 1x106
dv2
dt
1(t)]
lxto6 Vl(t) _,!_[2V2(t)]
C txto6
C 1x106 ... (5)
v1(t) - 2v2(t) + U(t)
Using (3) in (4),
c~dt dv1 Select
Vz -Vt
Jxt06
di
- V1(l) + V2(t)
X1(t)
v1(t) and X2(t)
.•• (6)
= v2(t)
Using selected state variables,
and and
X1 Y(t)
.
2Xi +
U(t)
V1(t) = X1(l)
Hence the stale model is,
and
X(t)
AX(t) + BU(t)
Y(t)
CX(t) + DU(t)
where
>•
Example 2.23 :
A
[-: ~a
8
= [~} C = (!
01 D
=0
DmrN two state models far thL system wit/1 transfer function,
i) For first model, matrix A must bt in companion form. ii) For second model, matrix A must bt in diagonal form.
2·53
Modem Control Theory
State Space Representation
Solution : Arrange the transfer function as, 50x~x(s+5)
1000(s+5) s(s+2)(s+50J
T(s) =
i) The companion form means decomposition of denominator, T(s) =
phase
variable
form
for
lOOOs+SOOO {[(s+52) s+ lOO]sl
1000(s+5) [s2 +52s+lOO]s
The state dlagram is as shown in the Fig. 2.39.
From state diagram, ~----l 1000 1----~
. .
X1 and While
)'3 Y(t)
A
X2,
.
Fig. 2.39
X2 = X3
- 100X2 - 52X3 + U(t) SOOOX1 + lOOOX2
[o0
o}
1
0
1
B=[H
0 -100 -52 C = [5000 1000 OJ,
D =0
ii) For diagonal form use partial fractions,
A
T(s) A
5+ s+2+
0
C s+SO
T(s) s]
=
• •O
1000x5 2x50
= 50
which
use
direct
2. 54
Modem Control Theory
.
B
C
T(s)
1(s)(s+2)
2
1000(3)
I. __ 2
,. (-2) (.JS)= - 31.25
• =
T(s)(s+SO)I
State Space Representation
s=-50
.:!.~.t~. =(-50) Hb/
18.75
50 31.25 18.75 s+2 - s+SO
s-
The state diagram is,
., 50
t---"-<X>--Y{t)
Fig. 2.39 (a) Thus the stale equations are, X1 and Thus,
Y(t) = 50X1 A
C
.
U(t}, ~
[~
-
0 -2 0
=-
.
2X2 + U(t), ~
=-
31.2SX2 - 18.75X3
_J}
ISO - 31.25
B = [~} - 18.75), D = O
50X3 + U(t)
Modem Control Theory
>'*
Example 2.24 : Fig. 240.
2-55
State Space Representation
Derivr tire state model for two input two output systnn shown in the
't
Fig. 2.40
Select output of simple lags as state variables. Solution : As sueeested the output of -1- is X ee ' s·H I• I . X s+5 IS 2' The state diagrams for simple lags arc,
,, s+1
Fig. 2.40 (a)
Modem Control Theory
State Space Representation ... (!)
Fig. 2.40 (b) .. (2)
0.4
s+o:5
,,
~
Fig. 2.40 (c)
... (J)
Fig. 2.40 (d) ~
... (4)
= r2 - 2X4
From the given block diagram, r1 = 1<1 [U1 - Y1)
and
r2 " ~ [U2 - Y21
... (5)
2 .57
Modem Control Theory
State Space Representation ..• (6 a)
and
... (6 b) Substituting in (5) we get,
- K1 X1
-
51<1 X2 + K1 U1
... (7)
- 0.4K2 X3 - 4K2 X4 + K2 U2
... (8)
Using (7) and (8) in the equations (1) to (4), ... (9 a)
... (9b) ... (9 c) ... (9d) TI\c equations (9a) lo (9d) and (6a) to (6b) give the required state model as,
x
= AX+ BU
['-' ;K,J -5
A=
c ,._
-Ki 0 =
l
and Y = CX + OU where -5K1 -SK1 0
[1 5 o o} 0 0 0.4 4
0 --0.4K2 --05 --0.4K2
0
~K,0 Hl
r·
' B = ~I
0
Ki 0
Ki
D =[OJ
=
=
Example 2.25 : A series RLC circuit witlt R 1 !l. L = 1 H and C 1 F is excited l!y v = 10 V. Write the stale eq11atio11s iu Ille matrix form. !B•ngalorc Univ., Oec.·95(
Solution : The circuit is shown in the Fig. 2.41.
Fig. 2.41 Applyin_g KVL to the circuit, . v = 1(1) R.,.
dl(t)
Ldt
+ vc(I)
... (1)
2 ·58
Modem Control Theory
~f i(t) dt
and dvc(I)
]_ i(t)
_d_l_
Let
State Space Representation
... (2)
c
X1 = Current through inductor
i(l)
=
X2
vc(t)
Voltage across capacitor
U(t) = 10 V
V
Substituting all the values, 10
Xi+ X1 + X2
. ~
-X1 - X2 + 10
X1 and
••. (3) ... (4)
X1 Hence state equalions in matrix form are,
[xX2•
1]
n•
=
r-1 -11 [x [10] l
O
X1] + 0 U(t) 2
Example 2.26 : Obtain tltt state space representation in phase variable Jann for the system rqiresmtl'd by. D4y + 20D3y + 45D2y+ 18Dy+100 y =10D2u+S Du+ 100 u witlt y as output a11d u as input. (VTU: Ja.nJFeb.•2005)
Solution : Phase varalble form Taking Laplace transform of both sides and neglecting initial conditions, s~Y(s)+ 20s3Y(s)+ 45s2Y(s)+ 18s Y(s)+ 100 Y(s) Y(s) U(s) =
54
%
IOs2U(s) + 5s U(s)+ 100 U(s)
tos2 + 5s+ 100 + 20s3 + 45s2+18s+100
Using direct decomposition, Y(s) U(s)
l0s2+5s+IOO {[([s+ 20)s+ 45) s+ IS]s+ 100}
The state diagram is as shown in Fig. 2.4.2
Modem Control Theory
State Space Representation
2 .59
Fig. 2.42
.
.
.
The state equations arc,
.
x, =X2, X2 =X3, X3 =X4, x~ =-toox, -t8X2 -4X3 -20X,, + u and
>'*
Example 2.27 : Obtain the state mode! of
flit
system rrsponsed by the followi11g differential
equation : y + 6 y + 5 y + y = u.
{VTU: July/Aug.· 2005)
Solution : Take Laplace transform of both sides neglecting initial conditions, s3Y(s)+ 6s2Y(s)+ 5sY(s)+ Y(s)
=
U(s}
Y(s) U(s)
s3 + 6s2 + Ss+l
Y(s) U(s}
{[(s+ 6)s+ 5]s+
I
l}
The state diagram is shown in the Fig. 2.43
1S!ttJ Fig. 2.43
Y(s)
.
.
State Space Representation
2 -60
Modem Control Theory
The state equations are, xi =X2. X2 =X3.
.
X3 =-XI -5Xz -6X3 + u
and O
1 ..
I
0 0 [-1 -5
A
Example 2.28 : Obtain two 1/ifferent slatt models for a system represented by tire followmg transfer function. Write suitable state diagram in tach rase. (VfU: July/Aug.·2005)
Y(s)
8s2 + 17s+ S
U(s)
(s+ 1)
(s2+Ss+1s)
Solution : 1) Direct decomposition : Y(s)
8s2+17s+ 8
Ss2+17s+ 8
s3+9s2+23s+15
t..:(s)
{l(s+9)s+23js+15}
The state diagram is shown in the Fig.2.44
.
Fig. 2.44
X3 "'-15Xt -23X2 -9X3 + U Y
"'
A
=
8X1 +17X2 +8X3
l·
r~ ~ ~ "'r~J. -15 -23
-9
B
1
C"' (817 8]
Modem Control
2. 61
Theory
State Space Representation
2) Foster's term : Y(s) U(s) A = - 0.125,
=
B = -7.25,
c = 15.375
Y(s) -0.125 7.25 15.375 -= -----+-U(s) s+l s+3 s+5 The state diagram is shown in the fig. 2.45
y
Fig. 2.45 The state equations arc,
Y = - 0.125X1 -7.25X2 +15.375X3
A=
,_.
[-100 -301 -50OJ ,B= [']11 ,C=(-0.125-7.2515.375] .
Example 2.29 : Choosing appropriate physical variables as state variables, obtain the slate modelfor tho tlectric circ11it shown i11 Fig. 2.46 (VTU: JanJFeb.·2006)
M
2 ·62
Modem Control Theory
State Space Representation
rn
IH
y(t)
Fig. 2.46 Solution : The various currents arc shown in the Fig. 2.46(a). 1F
Fig. 2.46 (a) The volatagc across resistance and second inductor is y(t).
and
y~) = y(t)
.•.. (1)
di 1x ~ = y(t) di
.... (2)
The voltage across capacitor is, ic(l) and
di I '< _!:..!. dt
From (4), Using in (2),
y(I) dil2 di
u(t)-y(t)
dvc(I) _ dvc(I) C-d-1- -d-t -
.... (3)
u(t) - y(t) ~ vc (t)
..... (4)
u(t) -
dil I dt = u(t)
u(I) - vc(t)
- vc(t) .... (5)
2 ·63
State Space Representation
u(l)-vc(tl-iu +i1.2
... (6)
Modem Control Theory Applying KCL at node A,
dvc(t)
.
·-d-1 - +I LI dvc(t)
-d-1- =
. . X2 . X3 X1
and
y
A
JI*
i1.2 + y(t)
dill
. .. from (4)
dl;X3 di, 2 dl
=-X3+
dvc(I) _d_t_ = di,
di2
[~ -1
•
-X, +X2-X3 + U
= X2 ~
0
...from (5)
U
=-X3
. .. from (6) .. .from (2)
+U
-11].B=[~].C=fO -1 I
0 -l),D=[l}
. Y(s) s(s+ 2)(s + 3) Example 2.30 : For lite transferfunction R(s) ~ --..,,--(s+ 1)2(s+4)
Obtain lite state model in i) Phase variable canoP1icalfarm ii) fordan CanoP1ica/ form Solution : i) Phase variable fonn Y(s) s(s+2) (s+3) R(s) = (s + 1)2(s+4) Using direct decomposition, Y(s) R(s)
s3 +ss2 +6s {[(s+6}s+9)s+4}
s3 +5s2 +6s s3 +6s2 +9s+4
(VfU: Jan./Feb.·2006)
Modem Control Thoory
State Space Representation
The state diagram is shown in the Fig. 2.47.
R
.
x,=X2,
x,
.
Fig. 2.47
X2=X3.
X3=-4X,-9Xz-6X3+R
Y = 6X2 +5X3 +X3 Y
= 6X2 +5X3
=
-4X1 -9X2 -6X3 +R
-4X1-3X2-X3+R
A [~4 ~9 ~JB=m,C=(-4 m Jordan
-3-1),D=(l)
Canonical form :
As the degree of denominator and numerator is same, first divide
partial fractions s3+6s2+9s+-I}
s3-f-5s2+6s
(1
s3 +6s2 +9s+4
A(s+4)+
B(s+ l){s+4) +C(s+1)2
s2+3s+4
and then obtain
Modem Control
:. As+4A + Bs2 +5sB+4B+Cs2 + 2Cs+C :.B+C = 1,
State Space Representation
2-65
Theory
A+5B+2C=3,
4A+4B+C=4
A= 0.666, B = 0.111,
C = + 0.888
Y(s)
I { 0.666
R(s) ~
= s2 + 3s+4
0.111
0.888}
- (s+t)2 + (s+l) + (s+4)
I _ 0.666 _ O. lll _ 0.888 (s+t)2
s+ I
s+4
Fig. 2.48
-1 A=
11..
[
~
y =
- o.111x1
0 -1
_n' =[~]·
0
B
-0.666X2
- o.sssx 3 + R
C = [-0.111 -0.666 -0.888), D
= [1]
Example 2.31 : Fig. 2.49 shows tltt block diagram of a speed control systtm tuillt stute variable feedback. Tltt drive motor is an armalurt co11trolled d.c. motor willr · amiature resistance R., armature inductance L., motor torque constant Kp inertia referredlo consta11I K~ and tachometer K1. The applied armature voltage is ctmtrolled by a three phase full·converter. ec is control voltage, e4 is an11ature voltage, e, is the reference voltage corre:.-pondi11glo the desired spted. Taking X1 = w(speed) and X2 = i4 (armaluu current} as lht stale variables, u = e, as tltt i11put, and y = w as the 011tp11t, derive a state variable mode for the feedback system. (VTU: July/Aug.·2006)
2 -66
Modern Control Theory
State Space Representation
Fig. 2.49 Solution : For armature controlled d.c, motor, v.(t) and
di, . L•{it+•.R,
+cb(t
)
...... (1)
KbW(t)
...... (2)
...... (3) The torque produced is used to drive the load against inertia Tm
J~~+BIJl
J and
friction B. ..... (4)
dlJl ldt+BIJl d~) dt
...... (5)
The state variables arc X1 =
From the equation (1), va(t) = L,, d~; + i, R, + Ki.(J)(t)
..... (7)
Modem Controi Theory
2 -67
State Space Representation
But the armature voltage v. is controlled by three phase full converter.
KclK1 x-K2
i, J
K
dt
....(8)
X2
.... (9)
y
And
The equations (6), (9) and (JO) give us the required state model with,
l
Kr A
-(/Kc
].c~ll
R )J'B=[~
2+ •
L,
,,.... Example 2.32 : For a transfer function given by G(s) ;
2 s2 +3s+2
in the: i) Phase variable
form
ii) Diagonal
form
Solution : i) Phase variable form using direct decomposition G(s) ; __2_; s2 +3s+2
2 [(s+3)s+2]
o]
•
, write tire state model
.... Modem ControlTheory
State Space Representation
2-68
y
Fig. 2.51 X1=X2,
X2=-2X1-3X2+U, A =
Y=2X1
[~2 _ ~], B = [~], C = {2 O]
ii) Diagonal form using Foster's form G(s)
A G(s)
=
2
s2 +3s+2
2 (s+I) (s+2)
A G =-+s+ I s+2
2 2 (2-1) = 2' B = (-2+1) = -2
2 2 s+1-s+2
u X1=-2X2+U y
Y = 2X1 +2X2
:.A=[-~ -02}B=[~] :.C
Fig. 2.52
= (2
2)
Modem
••
2. 69
Control Theory
State Space Representation
Example 2.33 : For the network shown i11 Fig. 2.53, c/ioosi11g i1(t) = X1(1) and ifl) =Xz(t) as statt variables, obtain lite state equation and output equation in vector matrix form.
-1
Hl
1H
-
i,(t)-i2(l) 1H
•,(t)
u(t)
~------12-(l_)l~---~ 1 0 10
y(t)
Fig. 2.53 (VTU: o.c.-2007/Jan.-2008)
Solution : Applying KVL to the two loops, di,
-dt-i1
di ,
-dt-i2
+u(t)
0
...Loop I
0
... Loop 2
di2
... (1)
dt Using (1) in equation for loop l, di,
dt
= -i1-(i1-2i2)-i2+u(t)=-2i1+i2+u(t)
... (3)
.
...(4)
X2
and
and
... (2)
... (5)
y(t)
A
(1
-t][~~]
[;2 _aB=[~],c=(l
-1},D=(O)
2. 70
Modem Control Theory
State Space Representation
Review Questions I. Discuss the 'dtrilllltion of the state model 11Sing fallowing mtthodJ of programming.
BU5h form ii) Gulltmin's form iii) Foster's form iv) fordan's form i)
2. fl•bornlt upon tht basis of Mktling suitable stalt rJariablts for • systtm. 3. Obtain lht sl•I• tnodd for the block diagram shown.
U{t)
Y(t)
4. Vo/toge nrross r«p<1dlor is Ye- With "c and il as n Ml
of M•lt variables dniot tJie shllt
ltW
5. Qm.sWt·rth~ pernianent mngrtrl movingcoil in.strummt 4.f shoton, Here, t
= Vollagt to b< '"""5urtd,
Coll
R = A rmalure Rtsistantt, L ~ Armal·urr Inductance ,
constant relating i = Currtnl through 1trmahtrt toil ,
I=
M.I. of rotating part ,
Kr
=
K,
= Spring amstant •
Tonpu conshud rclnting i .ind T ,
l)eriVo' its state model.
Aru. : x = AX + llU y = ex+ OU
4 -~ -ti ·{!'Jc·""' n
whtttA •
[
2. 71
Modem Control Theory
State Space Representation
6. For " mt'chanirnlsystem slwwn below obtnin its st.all model in sttlndardform. Assum1 output as Y(IJ. Ans.:X=AX+BD,
Y•CX
T Y(l)
where A
=[ ~
~ ] B =[
O } C 1/M
-M -M
= (0 1)
7. Obtain the slate model in standard Busl• form of a system shown in figure. Ans.:
U(sl
X =AX+
BU, Y=CX
Y(s)
where A=
0
l
0
0
0 L
-288
0 0 -176 -98
8. Statt whether lhL ffJllowing stetemm! is True or Faist rvith reasons : 1·~ stale 1pace approachof llie system nnalysi$ is n frequency domain analysU.
9. Statt wlrether tht follmuing stalrmtnt is Trut ~r Fnlst with r~on. Tht slalt space approach of the system analysis is a time domain analysis.
C
= [72
96 24 OJ
Modem Control Theory
2. 72
State Space Representation
JO. For tht merha11iat/ •yslem •hawn ill ftg11rt input is uft) and output is y(t). Obtain stalt equntiqn
awl output eq11ation. Repre;e11t lllL sysl<1n by blodc diagram.
QQQ
Matrix Algebra and Derivation of Transfer Function 3.1 Background The methods of obtaining the state model in various forms, from the transfer function of a system are discussed in the last chapter. It is also seen that the state models can be different and not a unique property for the given system. But most important point about the system Is Its transfer function is always unique. Though number of state models are derived for the system, the transfer function of the system, obtained from all of them is always unique for single input single output system. In this chapter the method of obtaining transfer function of the system from its state model is discussed. The concepts of diagonalisation of system matrix A, eigcn values and elgcn vectors arc also included in this chapter. Number of methods from matrix algebra are used to understand the relation between stale model and transfer function hence revision of matrix algebra is included in the beguuung of this chapter.
3.2 Definition of Matrix A matrix is an ordered rectangular array of elements which may be rcal numbers, complex numbers. functions or mathematical operators. The order of matrix is always defined as m x n. where, m n
Number of rows
=
Number of columns
For example, consider matrix A of order 2 x 4 of real numbers.
A
=
4 3 [6 1
-z
1]
_... 2 Rows
0 8 ....
!. !. .!. J. 4 Columns
(3 • 1)
3-2
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
Any clement of a matrix is denoted as ail where i indicates position of row and j indicates position of column. Thus for matrix A above a11 = 4, a13 = - 2, a22 1 and so on.
=
Koy Point: Note that mal'rix does 1101 have a value but its detenninant has a value. A matrix with number of columns as 1 Le. order m x 1 is called column matrix while a matrix with number of rows as 1 i.c, order 1 x n is called row matrix, It is seen that vector matrices X. X, Y are the column matrices.
3.2.1 Types of Matrices The various types of matrices are. 1. Square Matrix : If number of rows (ml is equal to number of columns (n) of a matrix, it is called a square matrix. The system matrix A in a state model is always a square matrix of order n x n. 2. Diagonal Matrix : It is a square matrix with a.II a1;
=
0 for all i ~ j. Only main diagonal clements arc present which arc nonzero while all other clements are zero. For example,
3. Unit Matrix : This is a diagonal matrix, with all the main diagonal elements equal to unity. It is denoted as 1 and also called identity matrix. The unit matrix of order 3 x 3 is, I =
-l,0
o o]
l 0 0 0 1
4. Null Matrix : This is a matrix having all its elements as zero. It need not be a square matrix. 5. Transpose of a Matrix : If the rows of columns of m x n order matrix arc interchanged then resulting n x m order matrix is called trnnspose of the original matrix. If given matrix is A then its transpose is dl'TIOll'CI as AT.
A
o [2
21
1 5 4j. 2x3
Modem Control
Matrix Algebra & Derivation of Transfer Function
3.3
Theory
The proper11es of transpose arc, !. (A + B)T =AT+ BT 2.
(AB)T =BT AT
6. Symmetric Matrix : If transpose of a matrix is matrix itself, it is called symmetric matrix. It is a square matrix. AT A
A
[ I -4] -4
I -4]
AT= lr -4
1 •
I
7. Conjugate Matrix : The conjugate of a matrix is the matrix obtained in which each clement is the complex conjugate of the corresponding clement of the original matrix. It is denoted as A'. _ [I +j2 A - 2 -j4 8. Singula.r Matrix singular matrix.
l+jl] 3+j2 '
A'=
rll
-j2 2 +j4
1-jl] 3-j2
A square matrix having its determinant value zero is called
and
IAI
=
j 0.51
1=2 - 2 = 0
4 2
Thus A is singular matrix. 9. Nonsingular Matrix : A square matrix whose determinant is nonzero is called nonsingular matrix. A=[:!]
and
IAl=1:
~1=24-10=14"0
10. Skew Symmetric Matrix : A square matrix which is equal to its negative transpose is called a skew symmetric matrix. AT = -A
3.2.2 Important Terminologies Let us study the important terms and their meanings related to the matrix. 1. Minor of an element : Consider a square (n x n) matrix A and the clement aij. If now ith row and jth column are deleted then the remaining (n - 1) rows and columns form a determinant M;i. The value of this determinant is called the minor of an clement ll;j·
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
3-4
Consider a square matrix A as,
A ~ [~ ~ ~] Let us obtain minor of element alt determinant M21.
=
1. Delete second row and first column to get
Thus minor of 'a21' is 7. 2. Cofactor of an Clement : If M;i is the minor of an element aij then cofactor defined as,
I
cii =
<- ll; + i Mij
C;i is
I
Thus for the matrix A considered above, the cofactor of "it is, = (- 1)2 • t M21 = (- 1)3 >< 7 = - 7
C21
3. Adjoint of a matrix : The adjoint matrix is the transpose of a cofactor matrix. The cofactor matrix is the matrix obtained by replacing each clement of a matrix by the respective cofactor.
I
A~j A
n•~ Example 3.1 :
=
(Cofactor matrix of A]T
I
Find tlrt adjoint matrix of the matrix A,
'
A=
1 0 3] [ 2 4 1 3 2 2
Solution : Th" cofactors of all the clements are, C11
= (-
C13 = (-1)1
C22
11 = 6,
C12 = (- 1) 1 + 212 3
!) 1+·14 2 2
+31~
~I= -
8,
= (- 1)2 + 2 113 ~I= - 7,
~I= - 1
C2t
= (- 1)2 + 110 2 31 2 = +6
C23
= (- !)2 • 31 ~ ~I=
-2
Matrix Algebra & Derivation of Transfer Function
3.5
Modem Control Theory
Cofactor matrix
[-1~
-1-7 -8] -2 +5
-1-7
Adj A
[ _1~ +5
4
=:r =[~
6 -12] 5 -2 4 -7
4. Rank of a Matrix : To find the rank of a matrix means to search for a highest order determinant from a given matrix which is having nonzero value. Thus if r x r determinant has a nonzero value in a given matrix then the rank of a matrix is 'r' and any determinant having order r + 1 or more than that has a zero value. For example, consider matrix A as,
A
112
Now
1
I
2
= [~
31 =
4 1 1 1
! ~]
-5:t0
Thus 3 x 3 determinant is nonzero hence rank of matrix A is 3.
i•
Example 3.2 :
8olutlon
Find rank of A = [~
!]
: For the matrix A,
I~ !I=
0
hence rank is not 2.
Any other determinant of order 1 x 1 is nonzero. Hence rank of matrix A is 1. S. Equality of Matrices : The two matrices A and B are said lo be equal if they have same number of rows and columns and the elements of the corresponding orientations arc equal.
Matrix Algebra & Derivation of Transfer Function
3-6
Modern Control Theory
3.3 Elementary Matrix Operations The various elementary matrix operations arc. 1. Addition : The two matrices A and B can be added to get C B arc of same order and,
LS:o
=
A + B if both A and
I
+ b,j for all i and j
a;1
=
2. Subtraction : The two matrices A and B can be subtracted to get C = t. - B if both A and B arc of ~me order and,
I
C;i = a11 - b;J for all i and j
I
Associative law holds good for both addition and subtraction of the matrices. [A ± BJ
±c =
A ±
IB ±CJ
where all A, B and C matrices have same order. 3. Multiplication by scalar 'ex' : A matrix is multiplied by a scalar 'ex' if all the clements of a matrix are multiplied by that scalar 'ex'. cxa 11
exxA=
exalnl
Cltl 21
<XU nl
r.tan2
cxann
,
Ctn2n
[ 4. Multiplication of Matrices : The multiplication of two matrices is possible if number of columns of the first matrix is equal to the number of rows of the second matrix. Such matrices arc called conformal matrices. Thus if A is of order m x n and B is of order n x p. Then the multiplication C = A x B is of order m x p. The clements of matrix C arc obtained as, C;i
= I,"
a Ik bkj for i = 1, 2 ... n and j ~ 1, 2, ... P
l =I
Thus if A is of order 2 x 2 and B is of order 2 x 3 then C = A x B has a order 2 x 3. A
C
["11 a 21
•p] a~
B =[b11 b21
b12 b22
b13] b23
AxB •11b11+a12b2t [ 321b11 +a22b21
•11b12+a12b22 321b12 +a22b22
Matrix Algebra & Derivation of TransferFunction
3.7
Modem Control Theory Properties of Multiplication
J. Commutative law is not valid for the matrix multiplication. AB "' BA 2. The transpose of a product is the product of transposes of individual matrices in the reverse order.
,..,.
Example 3.3 : Obtain tire matrix multiplicationof. 3
.4
-Ji =[1 0 -1.]
= [~ ~
B
2 1
0
Solution : A has order 3 x 2 and ll has order 2 x 3 hence C = A x ll has order 3 x 3. 3xl-lx2
c
Ax B
=
Oxl+lx2 [ 2x I +Ox2
3x0-lxl OxO+lxl 2xO+Ox1
3x-1-lx0] Ox-l+lxO 2x-1 +OxO
[~ -: -~] 2
0 -2
3.4 Inverse of a Matrix In algebraic calculations we write ax
=y
i.e. Similarly in matrix algebra we can write, AX
B
i.e.
where
Inverse of matrix A
The inverse of a matrix exists only under the following conditions, I. The matrix is a
square matrix.
2. The matrix is a nonsingular matrix.
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
3-8
Mathematically inverse of a matrix can be calculated by the expression, [Adjoint of A)
_ 1
"-jAj
A
3.4.1 Properties of Inverse of a Matrix 1. When a matrix is multiplied by its inverse, the result is the identity matrix.
=
A-1A
AA-l =I
2. Inverse of inverse of a matrix is the matrix itself. (A-lf
l = A
3. Inverse of a product of two matrices is the' product of their individual inverses taken in the reverse order. (ABrt
.......
1....,..
= 5-l A-1
Example 3.4 : Find the inverse of A
= lf24 13}
Solution : Check that A is nonsingular.
IAI
I~
:1=2 - 12 = _ 10
[Adjoint o( A)
IAI Adj A
[-~1 --3]~r
[Cofactor of A)T
[~
Cross check:
AA-1=
[~
-10
= [_:
-:1
= [--0.1
:][:~::
+0.3] +0.4 -0.2 :~:~] ·[~
~]-1
Important Observation : For 2 x 2 matrix, the adjoint of the matrix can be directly obtained by changing the positions of main diagonal elements and by changing the signs of the remaining clements.
Modem
For example,
Matrix Algebra & Derivation of Transfer Function
3-9
Control Theory
A ~ [~ :] ,
Adj A= [ _:
-n
This result is directly used hereafter while solving the problems, when matrix has an order 2 x 2. Note that for any other order of matrix the adjoint must be obtained as transpose of cofactor matrix. With this study of matrix algebra, let us discuss the method of obtaining transfer function from the state model.
3.5 Derivation of Transfer Function from State Model Consider a standard state model derived for linear time invariant system as •
.
and
X(t)
A X(t) + B U(t)
... (la)
Y(t)
C X(t) + D U(t)
... (lb)
Taking Laplace transform of both sides, [s X(s) - X(O)I
A X(s) + B U(s}
... (2a)
Y(s)
C X(s} + D U(s}
... (2b)
and
Note that as the system is time invariant, the coefficient of matrices A, B, C and D arc constants. While the definition of transfer function l~ based on the assumption of zero initial conditions i.e. X(O) 0.
=
s X(s) s X(s) - A X(s)
A X(s) + B U(s)
B U(s)
Now s is an operator while A is matrix of order n x n hence to match the orders of two terms on left hand side, multiply 's' by identity matrix I of the order n x n, sl X(s) - A X(s) (sl -
Al
X(s)
B U(s) B U(s)
Premultiplying both stdes by [sl - Ar [sl - Af
1
1,
Isl - A) X(s)
[sl - Ar 1 [sl - AJ
Now
X(s)
[sr - Ar 1 BU(s) l [sl - Ar 1 BU(s)
Substituting in the equation (2b ), Ar 1 B U(s) + D U(s)
Y(s)
c Isl -
Y(s)
{C [sl - Ar I B + 0) U(s)
... (3)
Matrix Algebra & Derivation of Transfer Function
3 -10
Modern Control Theory
Hence the transfer function is, Tts)
Now
Isl
=
Y(s) =
U(s)
c (51 - Ar
l B+D
... (4a)
Adj(sl-AI Isl-Al
-Ar1 T (s)
=
c Adj(sl-AI n Isl-al
D
... (4b)
+
Key Point: Tire state model of a sys~11 is 11ot unique, but tire tra1L,fer f1111ctio11 of obtai1wtl from any stale model is unique. 11 is indrpcndent of tire met/rod used lo express tire system in slate model form.
3.5.1 Characteristic Equation It is seen from the expression of transfer function that the denominator is Isl - A [. Now the equation obtained by equating denominator of transfer function to zero is called characteristic equation. The roots of this equation arc the closed loop poles of the system. Thus the characteristic equation of the system is, ... Characteristic equation
= 0
Isl-A(
The stability of the system depends on the roots of the characteristic equation. In matrix algebra, the roots of the equation I sf - A I = 0 arc called cigcn values of matrix A and these arc generally denoted by A.
i•
Example 3.5 : Consider a system having stale model
f XI l = [-2 I• ! L X2 j
.1
-31J 2
(; ~] + [~] U anti Y = I 1 1
I [~~]
witlr D "' 0. 0!1tain its T.F.
(V'fU: )anJFcb.·2008)
Solution: T.F.
c Isl-Ar'
s
Adj[sl - A] Isl -Al (sf - A)
5
[1 o]-(-2 0 J
~
-312j = [s0
o]-[-2 -3] s
·t4
2
Modem Control Theory
[sl
Matrix Algebra & Derivation of Transfer Function
3·11
l"s+2
-Al
3]
-4
s-2
- 3]
Adj Isl -
s+ 2
Al
(s + 2) (s - 2) + 12 = s2 - 4 + 12 = s2 + 8
[s~2 s~2] s2 + 8
-3] f3]
[l I
[s~2 __ _,,___s_+_2-:-,_L 5_,,_ = l 1 1 s2 + 8
T.F.
[3s-21]
I
Ss+ 22
s2
+8
{Ss + 1] s2 + 8 3.5.2 MIMO System For multiple input multiple output systems, a single transfer function docs not exist. There exists a mathematical relationship between each output and all the inputs. Hence for such systems there exists a transfer matrix rather than the transfer function. Bul the method of obtaining transfer matrix remains same as before. 11111+
Example 3.6 : Determine the lrausfer matrix for MIMO system givtn by,
[~1]J L-2o 3][x'] x Xi
-5
2
=[11
1J[u'], [Y']=[2 o1J[xx u
1
1]
1
Y2
2
1
Solution : From the given state model, A
-2
B = [:
-5
C!sl-A]-l
T.M.
:}
C = [~ ~}
B+D
-3]
s [2 s+S
[sl-AI
Adj [sl - A) Isl-A
[ 0 3}
I
ell [C21
C1.2]T C22
s2 +5s+6=
= [s+5 3
-2JT s
(s + 2) (s + 3)
= [s+5 -2
.s3]
D =[OJ
Matrix Algebra & Derivation of Transfer Function
3-12
Modem Control Theory
Adj (sl-A) (sl-AI
[s_+: ~] (s..2)(s..3)
T.M.
s+S] [3s+14 3s+14] [2I 1]0 [s+S s-2 s-2 s+S s+S
,. ,. l '"" l
~2)(s+3-)-=
14
(s+2)(s+3)
3s+ (s+2) (s+3) (s+2) (s+3) s+S s+8 [ (s+2) (s+ 3) (s+2) (s+3) T.M. U(s)
Y(s)
i.e.
3s+ 14 (s+2) (s+ 3) s+S [ (s+2) (s+ 3)
The above relation indicates that each output depends on both the inputs.
3.6 Eigen Values Consider an equation AX = Y which indicates the transformation matrix X into 'n x 1' vector matrix Y by 'n x n' matrix operator A.
of 'n x 1' vector
If there exists such a vector X such that A transforms it to a vector JJ<. then X is called the solution of the equation,
0
i.e. (Al-
i.e.
A)X
0
..• (1)
The set of homogeneous condition,
I
p.. 1-AI
~o
equations
I
(1) have a nontrivial solution only under the
... (2)
The determinant I}.. I - A ( is called characteristic polynomial while the equation (2) is called the characteristic equation.
Modem Control
Matrix Algebra & Deriv1tlon of Transfer Function
3-13
Theory
After expanding, we get the characteristic equation as,
p..r - A I
= )." + a1 A" -
1
+ ... + a0 = 0
..• (3)
The 'n' roots of the equation (3) i.e. the values of ). satisfying the above equation (3) are called elgen values of the matrix A.
=
The equation (2) is similar to Isl - A I 0, which is the characteristic equation of the system. Hence values of ), satisfying characteristic equation are the dosed loop poles of the system. Thus eigen values are the closed loop poles of the system,
3.7 Eigen Vectors Any nonzero vector X; such that AX; = A; X; is said to be cigcn vector associated with eigenvalue ),;·Thus let ). = A; satisfies the equation, ().1 I-A)
X = 0
Then solution of this equation is called eigen vector of A associated with cigen value ).1 and is denoted as M1. If the rank of the matrix P·; I - A) is r, then there arc (n - r) independent eigcn vectors. Similarly another important point is that if the eigen values of matrix A are all distinct, then the rank r of matrix A is (n - 1) where n is order of the system. Mathematically, the cigen vector can be calculated by taking cofactors of matrix 0.1 I - A) along any row •
I
..--~~~~~~~~~~~~~~~~~~~ ckl
Thus
M; = Eigcn vector for A; = :~
where k = 1, 2, ... n
[
where Cid is cofactor of matrix (/,.; I - A) of k1h row. Key Point: If the cofactors along a particular row givts null solution i.e. nil elements of corresponding eigen vectors are zero then cofactors along any other row must be ob/nilled.
Otherwise inverse of modal matrix M cannot exist. 3.8 Modal Matrix M Let :>..1, 1.2, ..• /,.n are then eigenvalues of the matrix A while M1, M2, ... ~are the eigen vectors corresponding to the eigcn values ).1, ).2 ... An respectively. Each eigen vector is of the order n x 1 and placing all the cigcn vectors one after another as the columns of another matrix, n x n matrix can be obtained. Such a matrix obtained by placing all the eigen vectors together is called a modal matrix or diagonalising matrix of matrix A.
Modem ControlTheory
I
Matrix Algebra -& Derivation of Transfer Function
3 -14
M ; Modal Matrix = (M1 : !\,~ : ... :
M.,l
The important property of the modal matrix is that, AM
A (M1 : M2 : ... : M,,I (AM1 : AM2 : •.. AMnJ [i-i M1 : ~ M2 : .•• : Ar, M,,)
AM
.•. (1)
MA
r1..1 where
A
:o :
Diagonal matrix = j
I.co
0 0 1..2 0 0
Thus premultiplying equation (1) by M • 1•
= M- 1M A M - 1 AM = A (Diagonal
M-
1
AM
matrix)
It can be noted that, 1. Both A and A have same characteristic equation. 2. Both A and A have same eigen values. Hence due to transformation M • 1AM, eigcn values of matrix A remain unchanged and matrix A gets converted to diagonal form.
3.8.1 Vander Monde Matrix If the state model i.~ obtained using the phase variables then the matrix A is in Bush form or phase variable form as,
0 0
I
0
0
A
0 -<-ln
0 --an-I
0 -;) n-2
:I
-a
1J
And the characteristic equation i.e. denominator of T(s) is, F(s)
= s" + a1 s" - 1 + ... + an_ 1 s + a0
=0
Matrix Algebra & Derivation of Transfer Function
3 -15
Modern Control Theory
In such a case, modal matrix lakes a form of a special matrix as,
Ai
i..2
),.n
)..2
},~
)..~,
A)- i
l.!r2
;.n-1
v
I
'11
Such a modal matrix for matrix A which is in phase variable form, is called Vander Mondc matrix . ,....,
Example 3.7 :
Consider a state model witlt malrir A as, 2
0
A
-3~
OJ
\l
-9
Determine, a) Characteristic equation. b) Eigen values c) Eigen vectors and d) Modnl matrix. Also prove that the transformn/io11 M- 1AM results n diagonal matrix. Solution : a) The characteristic equation is p,1 - A I 1 0 ). 0 1 0 -
011
10 0 1
I
40
2 0
01
1
0
-48 -3~ -9
I -~ -~ -~ l 48
:.)..2
=0
... (!)
= 0
34 ).+9
(I.+ 9) + 2 x 48 + 0 + 0 - 8 (1.. .. 9) + 34 ). = 0
=
)..3+91?+26).+24
O
This is required characteristic equation. b) To find eigen values, test A= - 2 for its root. 9
26
24
- 2
- 14
- 24
7
12
-2
0
p.2 + n + 12)
o
(), + 2) (/..+ 3) (). + 4)
0
(), +
i.e.
Matrix Algebra & Derivation of Transfer Function
3 -16
Modorn Control Theory
2>
i.e,
- 2, 1..2 = - 3
and /,3 = - 4
These arc the cigcn values of matrix A. c) To find eigen vectors. obtain matrix (l.i I - Al for each cigen value by substituting value of I.. in equation (1).
r-2 =34~ -7~]
l~:
... The common factor can be taken oul. l
For i.2 = - 3,
p, i I -
A)
M2
For 1..3 = - 4,
[l..i I -A]
M3
=
[~:~l Hl = [~;4] =
o]
-4 = [-4
-2 -4 -1 48 34 5
=[~::] [-!] =h] =
M1, M2 and M3 are the eigenvectors corresponding to the eigenvalues :\.1, :\.2 and :\.3.
Matrix Algebra & Derivation of Transfer Function
3 -17
Modem Control Theory d) The modal matrix is,
M = IM1 : M2 : M3)
-! -~4jl
= (-~
-2
Let us prove M - 1 AM is a diagonal matrix. = AdjlMJ
M-1
!Mr=
Adj [M]
[
(CofoclorofMJT
IMI
-10 +8 -7]T r-10 - 7 -1
-Ii
-7 8 6 1 -7 -5 -1
=
6 -5 1 -1
I _\ -~ -~1=-l
IMI
-2
1
4
Adj [Ml
IMI= AM
L~s [10-8
M-1AM
7
u
[-810 7 1] 7
2 0
~
-; 2
ol1 [-11
-3
-34 -9 -2 7 -6 5
-11 l r-22 I
0 -3 0
4
-6 9
-3
-4]
-1!
4]
-6 9 8 -3 -16
OJ 0 = [i.01 -4
1 l r-2 -: = !
0
0 ).2
0
~l=
h
h3
Thus M - 1AM is a diagonal matrix.
l'*
Example 3.8 :
A=[l
Obtain tire Elgen values, Eigen vectors and Modal matrL~ for the matrix
1 OJ
0 l -11 -6
3 -18
Modem Control Theory
MatrixAlgebra& Derivation of Transfer Function
l
Solution: Eigenvalues are roots ofjt .. I-Al=O
!).. I-Al
=
>. -1
o >.
[
6
11
0 -1 =0 >.+6
1.. 3+61..2+11>.+6
i.e.
0
(>.+ !)(/.+ 2){).+ 3) /,1 =-1,
J..2 =-2,
0
J..3 =-3 are eigenvalues.
To find Eigen vector,
>.
Let
"-1
=-!
[~1 =~ ~1] 6
M1
M1
For
>-2
11
5
[en]
where C = Cofactor
C12 C 13 K•2
[~]
i.e,
M1=Hl
=-2 -1
(J..21-A)
M2
[~2 -2 [c11 C12 C 13
l
11
-:1]
for K~3
For J..3 =-3 -1
(1.31-A)
-3
[Y 11
~]
=[~] =[~2]
Modem Control Theory
3 -19
Matrix Algebl'll & Derivation of Transfer Function
Key Point: The cofactors about any ruw can bt obltlined. :. Modal matrix M
=
(M1 :M2 :M3)
M [~l ~2 ~] It can be easily checked that,
M-1 AM
=
A=[~
~2 ~3]
3.9 Diagonalisatlon The diagonal matrix plays an important role in the matrix algebra. The eigen values and inverse of a diagonal matrix can be very easily obtained just by an inspection. Thus in state variable analysis, if the matrix A is diagonalised then it is very easy to handle mathematically. Whc!l matrix A ls diagonalised, then the elements along its principle diagonal are the eigcn values. The eigen values are the closed loop poles of the system, from which stability of the system can be analysed. The diagonal matrix A indicates noninteraction of various state variables. Tile derivative of one state variable is dependent on the corresponding state variable alone and other state variables do not interact in it, when A is diagonal matrix. This is practically important from controlling point of view. Due to all these reasons, the diagonaUsation of matrix A is always motivated. The techniques used to transform the general state model into diagonal form i.e. canonical form are called dlagonallutlon techniques. Consider nlh order state model in which matrix A is not diagonal . X(t)
.
A X(t) + B U(t)
... (1)
Y(t)
C X(t) + D U(t)
... (2)
Let Z(t) is new state vector such that the ITansformalion is, X(I)
=
MZ(t)
..• (3)
Modern Control
Theory
M
Herc
Matrix Algebra & Derivation of Transfer Function
3 -20
Modal matrix of A
X(t)
MZ(t)
... (4)
Using equations (3) and (4) into equations (1) and (2),
and
M Z(t)
.
AM Z(t) + B U(t)
... (5)
Y(t)
CM Z(t) + DU(t)
... (6)
Prcmultiplying equation (5) by M - 1 on both sides, M"1M Z(I)
.
M" 1AMZ(t) + M"1 B U(t)
Z(t)
M - 1 AM Z(t) + M - 1B U(t)
... (7)
Y(l)
CM Z(t) + D U(t)
.•. (8)
and
The equations (7) and (8) gives the canonical state model in which M - 1 AM is a diagonal matrix denoted as II. The matrix M which transforms A into diagonal form is called diagonalising matrix or modal matrix. The new canonical state model is represented as, Z(t)
11 Z(t) + ii U(t)
Y(t}
C Z(t) +
where
... (10)
M - 1 AM " Diagonal matrix
II
ii
c
... (9)
D U(t)
M01B 3
CM
The method of obtaining modal matrix M is already discussed in the last section. The transformation M - 1AM is called diagonalisation of matrix A.
I,_.
Example 3.9 : matrix A,
&duce tire given state model into its canonical form by diagonalising
X
r~
-1:
-6 -11
and
Y(I)
-~i 5
[1 0 OJ X(I)
X(tJ + [~] U(tJ I
Matrix Algebra & Derivation of Transfer Function
3. 21
Modem Control Theory Solution : From the given state model,
B=[~l·
A=[~=~~-:]·
0 OJ
C=[l
Let us find eigen values, eigcn vectors and modal matrix of A.
=
1)..1-AI
H~
0 1 0
~]-[-~
-1 ~.+11
1:
-11 -6 -11
I
_:I =
-il I=
0
0
~--5
11
:. A. (A. + 11) (A. - 5) + 36 + 66 - 6 (A. + 11) + 6(1. - 5) + 66 ). t..3+6).2+111.+6
0
0-+
0
().+ 1)
2) ().+ 3)
Thus eigen values are, A. 1 = - 1, A.2 = - 2, A.3 = - 3 To find eigen vectors, For ).1 = - 1, li•1 1- A]
M1
n
-!
10 11
-~] -6
[~:~J=m=m
For ).2 = - 2, -1
[A.21-AJ
M2
r-:
9 11
-~] -7
[~:~l r~i = m =
=
O
MatrixAlgeb,. & Dertntlon
M = [M1 : M2:
M3) = [~
n
M-1s
Adj(MJ
[~
!MT""= [
ii and
! !]
=Modal matrix
n -! -~lcA
M-1AM While
ofT,.Mfer Function
3 ·22
Modem Control Theory
C
0.4285 --0.4285
0.H28
M-
18 5
j
-14
0.3571 -0.5714 0.2142
~r [1 ~~ ~] = -~
--0.2857] +0.4285 -0.1428
[=~·.~1 . r-~1 -0.1428
-1
= CM= (I 1 1)
Thus canonical state model is, Z(t) .. A Z(t) + and Y(t)
C Z(t)
B U(t) where X(t) ~ M Z(t)
... Multiplying by 7
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
3 ·23
3.10 Generalised Eigen Vectors The generalised eigen vectors are the eigen vectors corresponding to repeated eigcn values. Uptill now, it is assumed that tile eigenvalues are distinct but if an eigen value At is repeated for r times then the corresponding eigen vectors are called generalised eigen vectors.
U a particular cigen value A 1 is repeated r times then the rank of n x n matrix [A11- A) is n - 1 and there is only one independent cigen vector associated with A1 given by,
... Ckn are cofactors about kth row
Important Note : If cofactors along a particular row gives all values zero i.e. eigen vector as null matrix, then cofactors along other row must be obtained. For remaining (r - 1) eigenvectors for r - 1 times repeated value A1, generalised eigen vectors must be used as,
fj
a2ck,
ack, ~
2!
ilCk2
M2
n ar;ilC1;n
CJ2
1i ~ 1 (r-1)!
M,
ilA7
a2ck2 M3= 2! a At
c.,,
2! ~
a•-tcck1> ()}..~-·
a•-•cck2> (r-1)!
dif' ar-lcckn>
(r-1)!
CIA't-1
The modal matrix M then can be obtained by placing all the eigen vectors one aher the other. Note that in such a case, M- 1AM matrix is not diagonal but consists of a r x r Jordan block in it where r is the order of repetition of a particular eigen value.
Matrix Algebra & Dorivation of Transfer Function
3. 24
Modem Control Thoory
3.10.1 Vander Monde Matrix for Generalised Eigen Vectors If matrix A is in phase variable form and an eigen value 1..1 is repeated for r times, then n eigen values arc /,1, /,1 ... 1..1, I., .. 1 -,.. , hn. The corresponding modal matrix is Vander Mondc matrix in modified form as, 0
0
),,
M=
'·1
21 -r
I.~
31..~
in-I
,,..
()
••
0 0
1...
0 . ..
).,~+I
31..1
d l!t1
d2 '·'rl
...
2idT I
""
1
1..;,
'";~ 1
A.~.
1.n-l rt I
J..ll-1 I\
Fi11d the eigen values, eige11 uector« alld modal matrix JM,
Example 3.10 :
A
[~ ~ -~
l
Solutlon : Find eigcn values which arc roots of I i,J - A I
Ix
0
I
0
81..
0
-2 I..
-1
+S 2
A~S
/..2 (I. + 5) + 4 +
A.3+s1:+st..+4
=O
0
Try 1..=-1 -1
5
8
-4
-1
4
4
(I,+ 1) {1..2 + 41..+ 4)
0
{A.+ 1) {A.+ 2)2
0
4
-4
~
Matrix Algebra & Derivation of TransferFunction
3. 25
Modem Control Theory
A I = - 1, '·2 = - 2, A3 = - 2 Thus ·- 2' is repeated twice hence it is necessary lo use generalised cigen vectors. For A1 = -1, (1,11-
A)=[=~ -~ 8
2
-~i 4
Now A2 = - 2 is repealed twice.
P-21 - A) = [~~
~2
8 2
[c"] C13
L,
[ ~·'~
= 2),2 + 10
C12
M2
-4-SAz
l.z•-2
=[]Jan] dC11
and
M3
TI
TI
dC12 d1..2
TI
dC13 d1..z
=
[2'-~+5) -
=
l.z--2
Thus M1, M2 and M3 are the required eigen vectors. -1
:. Modal matrix M
=
fM1 : M2 : M3J = : [
-1 1
112
2 -8
Ul
Modorn Control Theory
Matrix Algebra & Dorivation of Transfer Function
3 -26
Examples with Solutions !...
Find lite T.f". of tl1t system /raving state modn,
Example 3.11 ;
X
= [-~ -~] X
+[~] U
and Y = (1 OJ X
Solution : From the given model,
A=[-~-~], T.F.
[sl -AJ
Adj (~I - A]
Y(s) U(s)
B=[~],C=[I
= c (sl -
OJ, D=O
Ar l B + D
o] [ o
l] [s
J
t -1 s [0 1 - -2 -3 = 2 s+ 3 ~+3 [ -2
+I]
... Change diagonal elements and
s
change signs of other clements. s (s + 3) + 2 = s2 + 3s + 2
. T.P.
C Adj (sl-AJ B Isl-Al
(1
=
=
(s + I) (s + 2}
OJ [s~23 ls][lo] (s+l)(s+2)
(s+3) (s+l) (s+2} ,_.
Fi114 the T.F. of tire system /raving stale model,
Example 3.12 :
eolutlon : Prom the given model,
A=[=~~]. a:[~], T.F.
[sl-A]
T(s) = U(s)
c (sl -
5(1 OJ-[-3 0 I
-2
01
C=[I
Ar l B + D l]=[s+3 0 2
-1] s
D=O
i Modem Control Theory
3 ·27
/Adj [sl-A) Isl-Al
(I 0) [-~ s~3][~]
CAdj[sl-A)B )sl-AI
T.F.
(s+l) (s+2)
1 (s+l) (s+2) ,,..
Obtain the transfer[unction matrix for 1iu, MIMO system having state
Example 3.13 : model,
.
[ ~ -11 ~jx+[-11
x
-1
amt
Y
=
0 I
0
G ~ ~] X
Solution : From the given state model, A
o]
2 -1 I 1 2 , [
-1
0 1
c
G ~ ~].
Transfer matrix
c [sl - Ar
[sl -A)
-1 OJ
=[~ ~ ,
0=0 I B+ 0
=
c Adj[sl -A) B Isl-Al
•[~ ~ ~]-[~ -: ~]=[s~2 0 0 1 -1 0 1 1 (s-1)2
Adj [sl-A)
B
~+1
[ -2
s-3 (s-2)(s-1) 2(s-2)
s-1
s2 -~s+3 ]
s-1 0 T
-~] s-1
Matrix Algebra & Derivation Modem Control Theory
3 -28
s2 -2s+1 s-3 [ s-1
Isl-Al
-s+l s2 -3s+2
-2 2s-4 s2 -3s+ 3
l
of Transfer Function
(s - 2) (s - 1 )2 - 2 + s - l s3 - 4s2 + 6s - 5 s2-2s+l s-3 [ s-1
Adj [sl- AIB
-s+l
s2 -3s+2 I
-s2 +5 s2 -4s+5 [ -s+2
C Adj [sI - Al B
[~ ~ ~]
-3s+5 [-s2 +2
Transfer matrix ~
A
[ ~ -12
and prove that M - 1 AM = Solution : For elgen values,
IA.I-Al
-~ -,~ ~21 12
7
-s2 +5 -4 s2 -4s+5 • 4s-8 [ -s+2 2s2 -6s+6
l
4s-12 ] 2s2 -6s+2
-4s2 +6s-5
Obtain tire eigen values, eige11 vectors and modal matrix for,
Example 3.14 :
I
l
-3s+5 4s-12 ] [-s2 +2 2s2 -6s+2 ~3
,_.
-4 4s-8 2s2 -6s+6
;\.+6
0 0
~ -7 A
~]
-6
= Diagonal matrix.
(VTU: July/Aug.-2007,Jan./Feb.-2008)
MatrixAlgebra & Derivation of TransferFunctlon
3. 29
Modern Control Theory
A.2('A. + 6) + 24 - 31..- 18 + 141..
0
)! + 61..2 + 111.. + 6
0
I
\
Test 'A.= - 1, -1
11
6
- 5
-6
6
~
6 0
- 1
5
i.e.
(l + 1) (l2 + 5A. + 6)
0
(A.+ 1) (A.+ 2) ('A.+ 3)
0
=-
2, A. 3
Hence eigen values arc, A 1 = - 1, },2
=-
3
For the eigcn vectors, A.1
[-1
-1 -3 -1
= -1, p,11-AJ
12
M1
[~:~ l c,J
A2
= - 2,
P-2 I - A)
M2
A.3 = - 3, [l3 I - A)
M3
M
7
-:]
= [-:)
-9
= [-~] -1
[-2 -1 OJ 12 7 4 -3 -2 -2
[~:~l =[-~]=HJ [-3 -1 -zo] -3 -3 12 7
3
[~: l =[-!:]=HJ Modal matrix
= [- ~ -1
2 -4
-i]
.
Moifani Conlroi Theory
3 ·30
Adj[M]
[
!Ml=
-9 -5
I
Matrix Algolira & Derivation of Ttnsfer Function
: ~iT [-: -: -~1
-2 2 -2 -2
- -2 -3 -2, -2
AM
Thus,
I ..
M- 1 AM = A= Diagonal matrix
Find I/it eigen valtttS, tigm 11ectors and
Example 3.15 :
A = [~ -~
"'f al matrix for,
I]
(VTU: Jan.!Ftb.·2.005)
Solution : For eigen values,
p.. I-AI
I
"-~4 -: -1
-~
O
I
0
A.-3
).. (1. - 4) (1. - 3) - 2 - 2 •· 2A. - 1. + 3 + V. - 8 = O
n2 + 1s1.. - 9
o
p.2 -
61. + 9)
0
(1.- 1)
p.. - 3)2
0
3 ).. -
(1.- 1)
:.
'-1
= 1, '-2
Now 1.2 = 3 is repeated twice hence for 1.2 used.
I
i3,
A.3
=
3
= 3, te generalised
eigen vectors must be
Modem Control Theory
Matrix Algebra & Derivallbn of Transfer !=unctioi\
3. 31
[=~
-1
-~]
-1
-2
[~:~J m =
As it is a null matrix, calculate cofactors about other row.
Now for A2
= 'A.3 = 3, use
'·2 -4 [
For
>.3 =
3,
-1 -1
-1
'·2 1
TI
dC11 dl.2
TI
dC12 d/,,2
i!
dC13 d/,,2
Hence the modal matrix M is,
M
[! ~ n
-22 1-i -3
l
= [2"'2:- 3J
_ ).2-J
•2~3
=
m
Modem Control Theory
n•
3-32
Matrix Algebra & Derivation of Transfer Function
The Fig. 3.1 slwws the block diagram of a control system using state wriab/e [eedbuck'and int1•gra/ co11trol. 111e state 1midel of plant is, Example 3.16 :
Y = [O
1]
[~:]
i) Derive the stale model of the en/ire system. ii)
Derive the trm1sfer [unction Y(s)/U(s).
U(t) --..•-O<J>---4
Y(t)
Fig. 3.1 Solution : i) The input of integrator shown is
. X:i
and
r(t}
x3.
- 1 x [U(t) - Y(t)] = Y(t} - U(t)
... (I)
- 3.5 X3
... (2)
-
2X1
-
1.5 X2
From the plant model given, - 3X1 + 2X2 + r(t)
... (3a) ... (3b)
and
... (3c)
Using equation (2) in equation (3a) we get,
.
X1
- 3X1 + 2X2 - 35 X3 - 2X1 - 1.5 X2 ... (4a) ... (4b)
Substituting equation (3c) in equation (I), ... (4c)
and
Matrix Algebra & Derivation of Transfer Function
3 .33
Modem Control Theory
... (4d)
Y(t) = ~
[ · i [-! ~:
Thus equations (4a) to (4d) give state model of the entire system.
~~ =
and
i.e.
XJ
0
Y(t)
(0 l
x
-~5]
1
0
[~~]+[ ~i U(t) X3
-1
l
OJ [~~
and Y = ex with D = o
= AX+ BU
ii) The transfer function Y(s)/U(s) is, Y(s) U(s)
[sl - A)
= s [~
~
001
Adj [st -A]
Al
Y(s)
~i-[-: ~·: -~· 1 5
0
[ •' .,.
=
(0
Isl-Al
=
s+5 -1
-3.5s-17 .5 -14 s2 +10s+23
sJ + 10s2 + 23s + 14
d D=O an
4s
[4
0~5s-3.5 s· +5s s+S
= 14 Y(s) 14 U(s) =Isl-Al=
-0.5
r
4
s+5 (s+5)2 -2
s2 +Ss l OJ
0
4s
= s (s + 5)2 + 14 - 2s CAdj[sl-A)B
[s_+:
s2 +5s -14
s2 +5s s+S
=
=
0
0.5s-3.5
4s 4
U(s)
C Adj [sl - A) B
B+D
[ •' +Ss = O.Ss-3.5 -3.S(s+S)
= Isl -
t
= c [sl - Ar
14 sJ +10s2 +23s+l4
-3.5s-17 -14 ~2
.51 [ o]0
+10s+23
-1
Modern Control Theory
Matrix Algebra & Derivation of Transfer Function
3. 34 14 (s+l)(s+2)(s+7)
~i
This is the required transfer function. ,,_.
Example 3.17 :
For the matrix A= [ ~ -2
~
-3
-1
Find the eigen values, eigen vectors and modal matrix M. Solution : For eigen values,
o
P..I-AI
I~ "
-1 4
ol
-1
- 0
l.+3
4J..
0
->< + 31.2 + 41. + 2
0
1.2 (I. + 3) + 2 +
Test
'-=
-1, 4
- 1 -.1
:v.. + 2)
0
(I.+ 1) (/.. + 1 + jl) (/.. + 1 - jl)
0
(I.+ 1) (1.2 +
"· Calculate eigcn vectors.
For >.1
=-
-2
1,
P-11-A)
M1
n
-1 -1 4
-rJ
[~:~l=H]=HJ
- 1,
-2
l..2 = - 1 - jl,
~-3
=-
1 + jl
Matrix Algebra & Derivation of Transfer Function
3. 35
Modern Control Theory
-1
[-1-jl For
},2
-1 -jl,
(/,21- A)
a
-1-jl
~
4
2~µ]
[~:] =[-;2-j]
M2
... Choosing 3rd row to calculate cofactors. -l+jl For A.3
= - 1 + jl, [A.3 I -
A)
=[ ~
-1-j
-1 0 2+jl
l
_;+jl
j2 -j2
M = [-:
ll*
-1 -l+jl 4
Example 3.18 : For a system wil/1 state model matrices.
[-1 0 1] [0] [l]T
A=~-:
~;B=~;C=~
Obtain tht system trans/trfunction.
(VTU: March·2001)
Solution : The T.F. is given by, T.P.
=
C(sI-Ar1s
(sI - A) =
rs~l ~l ] s~2
0
Adj[sl-A)
O
s-3
= [Cofactor sl-A)T
=
[(s+2)0(s-3)
(s-3) (s+l) (s-3)
0
-+{s+2)
+1
(s+l)O(s+2)
]T
Matrix Algebra & Derivation of Transfer Function
3. 36
Modem Control Theory
c
(s+2) (s-3) (s-3) [
0 (s+l)(s-3)
0
l
(s+l/(s+2)
0
(s + 1) (s + 2) (s - 3)
Isl-Al
Adjlsl-A) lsl-Aj
(s1-Ar1
0
s+l 1 (s+l) (s+2)
(s+2)
0
0
(s+l)(s-3) I (s+l) (s+2) (s-3) 1 s-3 0
s+f T.F.
(s+ 2)
[l
1
OJ
1 (s+l)(s+2)
(s+2)
0
0
(s+I) (s-3) (1
1
OJ
1 {s+l)±(s-3)
0 (s+l) (s-3) 1 (s+l) (s+2) (s-3) 1 s-3
0
l
(s-3) l + 1 (s+l)(s-3) (s+l)(s+2)(s-3) (s+3) (s+ I) (s+2) (s-3)
11••
Example 3.19 : Find the transformation matrix P which will co11vert the following matrix A lo diagonalform. Th« eigen tialutS are A. 1, A. 2 and ).. 3. A=
0 0
1 0
-01
-a2
[
]J
Modem Control Theory
3. 37
Solution : The matrix Al - A is,
p,1 - A) =
r "-
-1
0 at
).. a2
ll
To find P, means to find cigcn vectors and modal matrix,
-1 0 1., +a 3
l
To find eigen vector, find cofactors about 3rd row.
-1 0 A3 +a 3
[~:]=[~~]
l
Matrix Algebra & Derivation of Transfer Function
Matrix Algebra & Derivation Modern
Control Theory
of Transfer Function
3·38
Hence transformation matrix P is, I
p
A2
i,1
).,3
A.1 A.1 A.~ Thus if matrix A is in phase variable form,
A
U.,
0
I
r~
0
{)
()
0
()
-CX.n-l
...
-(l.n-2
-(XI
And the eigen values arc distinct as A.1, ),2, ••• ),n then it can be proved as above that the diagonalising matrix is,
i.,
..~
p
I
1n-l 'I
1•
"z
A..,
),~
A~,
.,
1.n-l
"ll-1
'·2
-u
Key Point: Note tha: while fi11di11g the eigen l·'<'Ctorsfind tire cofactorsabout last row. E.xample 3.20 :
Determine Ille trans/tr nwtrir for tlu: system.
=~
[ 1()J [~~ ]+ [~ ; ] [~~] : [~;] ~ [~ -11] [~: J [~I]= 2
(VTU: July/Aug.-2005}
Solution : Prom the given state model, A = [
=~ ~] ,
Adj[sl-AI Isl-Al Adj (sl-AJ
B
= [~S
60] r
-[s+3
_ '(sf
C = [~ - ~] , D
Al-
[ls -2s+3 ]T = [s-2 s+31 ]
2
-I
l
sJ
=[0J
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
3. 39
T.F.
[1 -1] 9s+ 18 . _ (s+1)(s+2) .. T.F. 27s-63 [ (s+1)(s+2)
l
[4s-5 6s 1 -Ss-23 -12J (s+l)(s+2)
8
6s+ 12 (s+l)(s+2) 48s-12 (s+l)(s+2)
l
9 s+l = 27s-63 [ (s+l)(s+2)
s+I 6 48s-12 (s+l)(s+2)
I
This is the necessary transfer matrix. ••
Example 3.21 : Convert the Jo/lawing state model into canonical [orm. (VTU: JulylAug.·2005)
Solution : From the given model.
= ('3 --64 } B = [O-1 ] • C =(I
A
0]
Let us find the cigen values, eigen vectors and model matrix of A.
-4JI=1i.-r3 1 11I .[1o oJ-[l 1 3 -6
l)J-AI (/..-l)(A.+6)+12 (/..+2)(A.+
0 i.e. A.2 +51.+6
=0
0 i.e. t..1 =-2. }..2 =-3
3)
To find eigen vectors,
=-i!J.
Fori.1
11-A)
= [=~ :]
i.e.M1
=[~:~]=[~]
~J
i. +
=0
Modom Control
For Az ;-3,
21-A
[l.
I= [~ :]
i.c.
Mr[~:~]=[~]=[:]
J
M = [M1:
M2 = [; ~]=Modal
matrix
r-20 - o]3 = "
M-11\M
M-1a
6
While
Matrix Algebra & Derivation of Transfer Function
3 ·40
Theory
[1 -1]
IMJ-n - _.,
M-1
ii
Adj M _ -3
4
.. [I
[~3 -~] [~1]
= [~]
C =CM= (1 Ol [43 I]= ('I 1
-l] 4
1]
1hc canonical state model is,
11.+
l(t)
A Z(t) -+
Bu
Y(t)
C Z(t)
where X(t)
= M Z(t)
Example 3.22 : Convert tire followi11g squart matrix A into [ordan canonical fonn using a suitable 11011-si11g11/ar transform11tio11 matrix P. (VTU: July/Aug.· 2005) A
=
o
1
0
0 -9
r l-.i
-~J
Solution : Find the cigen values or A.
1).1-AI
=
i,2(A+6)+4+9A=0 (AH)(/.+ l}(A+ 1) =0
I~
-1
A 9
~1 A+6
I= o
i.e. A3 +6i..2 +9i,+4 =0
Modem Control Theory
Matrix Algebra & Derivation of Transfer Func:tion
3-41
For A1= -4
cC1112 [C13
Mz
l -[).~ -
+6~.-4 +9]
2=-1
=
-[ 4]-[ Az=-1
Ul{~l
[2'-~ +6] -'1
1
- -4 - -1t 4 I
2
-4),2
I.
Az=-l
fl M
r -4 12r 2 1
Adj [M)
-17 15 -4 3
= [~24
2 -41] -17 15
3
2 1]
M-l =~~I~
=
[~24
-17 -4 15 3 9
M"1AM-f- 9
4 12
2 -4I] -17 15 3
[~4
O][l
I 0 l - 9 -6
1 -~-I 16 I
~oi)
l
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
3-42
..... Diagonal matrix with Jordan block
n•
Example 3.23 : Consider the matrix
A [: -2 31] I
3 -l
i) Find the eigen values and cige11 vectors .of A. ii) Write the modal matrix. iii) Shaw tlrat the modal matrix indud diagonalizes A. Solution ; For eigcn values I A.I -A
~;l
J
2 l.=1
-1
)..+:!
- 3
= 0
2) + 2-9- 3(/.-1)+2('A.+1 )- 3(>..- 2) =0
i.e, (J.-1)(!.+1)(1.i.e. 'A.3 -21! -5),+6=0 1..1 =+1,
I=0
i.e.
1..2 =-2,
p.- l)(A+2)('A.-3)=0
i.3 =3
For i.1 = l, [I.I-A)
For 1..2 =-2.
fl..1-AJ
As C11
=
C12
=
C13
=
0, calculate cofactors about other row.
rvruJan.lFeb.-2006)
Matrix Algebra & Derivation of Transfer Function
3 .43
Modem Control Theory
For i.3 = 3,
l
11 l -14
M
l 1
Adj M I Cofactor mat rix of M] 1 ~=!Ml
1 [15 -25
0 -2 10 -30 -15 - 3 -~2 15 - 25
l
1~
r~ - ~ ~
1 1 [~ -~ [l; -:5 ~~ l [~I \1 :1 -·lo
30
-}a
0 -2 [ -15 - 3 -12
- 45 - 9
-30 0 0
[0
60
- 36
3 -1
1
1
1
~1
1 \1 - '14 1
- 14 1
~1-[~ -~
0 - 90
.o
0
.
Thus modal matrix indeed diagonalizes A.
I,..
Examplo 3.24 : Cive11 the state model X =1\X-flu, y =CX
(VfU: JanJFeb.·2006)
1o wltert A
= [~
~] · B = [~] and C :
-1 - 2 -3 i) ii)
[I 0 OJ
1
Simulate and find the transfer function ~~~ using Mason's gain formula. Determine tile transferfunction from the state model formulation.
Matrix Algebra & Derivation of TransferFunction
3-44
Modem Control Theory
Solution : The state equations arc,
.
X1=X2•
.
.
X2i\X3,
X3,.-X1-2X2-3X3+u,y=X1
Hence the signal flow graph is, ,
1
1
u ~--~x\?27·"' -_s __ ~x~3--1s x9'2 s x?-1---·, 3 ---<>o Y
-3
-2 -1
Fig. 3.2 From Mason's gain formula, Y(s) U(s)
Herc K ~ 1,
T161 +T2a2+ 6
T1 =Forward
+TK6K
where K
= Number
of forward paths
path gain= !x!x.!. =...!_
s s s
53
The various loop gains are,
6 -3
Flg. 3.3 All loops are touching to each other. 6= 1-(L1 6K
a,
+L2+L3]:1+~+2.+...!_ s s2
53
Eliminating all loop gains from 6 wh.ich are touching to Kth forward path
Modem Control Thoory
MatrixAlgebra & Derivation of Transfer Function
3 -45
Y(s) U(s)
Now let us use the state model method of finding the transfer function. Y(s) U(s)
[sl-A]
... D ~o
C(st-Ar'n+o -1
[~
-I 0 s +3
2
s2 + 3s+2 Adj (sl-A( [
:+3
l
-1
-s
s(s+3)
-2s-1
]T [52+3s+2
s2
Y(s) U(s)
Isl-A
, :l
-2s-l
-s
I
-2s-1
s+3 s(s+ 3)
-1
Adj [sl-AJ
s(s+3)
-s
['' + ,,., [s1-Ar1
s+3
= -1
.i]
s3 +3s2 +2s+l
s2 +3s+2 (1 0 0) -1 [ -s
s+3 s(s+3)
.i][:]
-2s-1
s3 +3s2 +2s+I Y(s) U(s)
,...
... Same
Example 3.25 : Tlte vector [ ~] is a11 tigm vtetor of A -1 oo/11t of A corresponding. to lht vector given.
=
r-~
where
X,
obtained above
_:
x, X, = Elgen vector
= [ _~]
~ ~] Fmd tlte e1gm
-1 (V11J: July/Aug.·2006)
Solutlon : The eigen vector is that vector which is non-zero such that AX1
'15
s3 +3s2 +2s+t
Matrix Algebra & Derivation Modem Control Theory
A.I
of TransferFunction
3 -46
=
Elgen value = To be obtained
[-~-1 -2~ =!][~]=A.;[~] 0 -1 -1
[~~]A.;
[~ . l -A.1
= s
... Corresponding eigen value.
Review Questions l. Derise lllL traruferfunctionfrom statt rnDdtl.
Wh.1t is ch•racttristic equation of• system mDtrix A ? Dtfint eigm wluts and eigtn 11ttlors of a n1J1trix. Whtn lht gtncraliJtd tigen oeaor« •rt uStd ? How to use thtm ? What is rrwdal matrix ? St•lt its imporlana. 6. Stole the adwntagts of diAgonnlisationof a matrix. 2. J. 4. 5.
7. Now lo achinit dia,
X
= [~
~J
X + [:] U(t) and Y(t):
(1 OJ X(t)
9. What is Vandtr Monde matrix ? Whm it exists ? WIU1t is its imporiano: ? 10. Find tht T.F. of tht systttnJ with following state modtls, al
x [-~ -~Jx+[~]u.
bi
X •
[ ~ ~~] X + [~]
x
U-~
cJ
Y=/l
u
O/X
Y s 10 II X
~~)x·[~]u.
Y=ll
o 01x
11. Obtain the T.F. of the systrm having state modt/, X(t)
= [-;
Y(t)
=
=~]
X(t) + [~] U(t)
[1 2J X(t)
(
A
12s+ 59 )
ns,
<s + 2)(s + 4)
ODO
Solution of State Equations 4.1 Background Uptill now the methods of obtaining state model in various forms and obtaining transfer function from the state model arc discussed. It is seen that the output response depends on the state variables and their initial values. Hence it is necessary to obtain the state vector X(t) which satisfies the state equation X(t) " A X(t) + B U(t) at any time l. This is called solution of state equations, which helps to obtain the output response of a system. Consider the state equatio n of linear time invariant system as, X(t) = A X(t) + B U(t) The matrices A and B arc constant matrices. This state equation can be of two types, 1. Homogeneous and 2. Nonhomogcncous
4.1.1 Homogeneous Equation If A is a constant matrix and input control forces arc zero then the equation takes the form,
.
X(t) = A X(t)
...(1)
Such an equation is called homogeneous equation. The obvious equation is if input is zero,how output can exist ? In such systems, the driving force Is provided by the initial conditions of the system lo produce the output. For example, consider a series RC circuit in which capacitor is initially charged to V volts. The current is the output. Now there is no input control force i.e.. external voltage applied to the system. But the initial voltage on the capacitor drives the current through the system and capacitor starts discharging through the resistance R. Such a system which works on the initial conditions without any input applied lo it is called homogeneous system.
(4. 1)
Modem
4·2
Control Theory
Solution ·of State Equations
4.1.2 Nonhomogeneous Equation If A is a constant matrix and matrix U(t) is non zero vector I.e. the input control forces arc applied to the system then the equation takes normal form as,
.
X(t)
= A X(t) + B U(t)
... (2)
Such an equation is called nonhomogeneous equation. Most of the practical systems require inputs to drive them. Such systems are nonhomogcncous linear systems. The solution of the state equation is obtained by considering basic method of finding the solution of homogeneous equation.
4.2 Review of Classical Method of Solution Consider a scalar differential equation as, dx dt
=
ax
where
x(O)
= Xo
... (1)
This is a homogeneous equation without the input vector. Assume the solution of this equation as,
At t = 0,
x(t)
... (2)
x(O)
... (3)
The solution has to satisfy the original differential equation hence using (2) in (!), d di Ibo + b1 t + ...
+ bkt
k
I
a Ibo + b1 t + ... + bkt
k
I
For validity of this equation, the coefficients of various powers of 't' on both sides, must be equal. abo- 2b2 = ab1, ... kbk = abk. 1 a b0
.!. a b1 2
= .!. a2b0 2
= .!.2! a2 b0
.!. a bz = _1_ a1>o = ..!.. a3 bo 3
.!. a k b0 k!
3x2
and
3!
x(O)
= b0
Modem Control Theory
Solution of State Equations
Using all these values in the assumed solution, x(t) " bo + abo(t) +
..!.. a2 bl 2!
+ ... +
1.. ak ho tk k!
1 k t ~] b0 [ l +at +211 a 2 t 2 +.... +kia 2t2+ [l+at+.!..a 2!
But
I ... at +
.!_ a2 2!
.... +.!.aktk] x(O) k!
t2 + ... + _!_ ak tk: e11 k!
X{t) = e•1 x(O)
.•. (4)
This is the required solution of homogeneous equation in scalar form. Thus if the homogeneous state equation is considered, X(t)
=
A X(t)
then its solution can be written as, X(t) = eAt X(O) In this case, eAt is not a scalar but a matrix of order n x n as that of matrix A. Observation : It can be observed that without input, initial state X(O) drives the state X(t) at any time t. Thus there is transition of the initial state X(O) from initial time t : 0 to any time t through the matrix cA1. As it is an exponential term, the matrix cAl is called matrix exponential. It is also responsible for the transition of the state X(t) at any time t from initial time hence also called state transition matrix. It is denoted as o (t). $(1)
eAt = State transition matrix
And
eAt: 1 +At•..!_
21
A2 t2;. ... + _!_ Ak lk + ... k!
The each term of the above equation is a matrix of order n x n. If instead of initial time t
= 0, it Is selected
$(I) = eA {I-to)
as t
= to then
the state transition matrix is,
4.4
Modem Control Theory
Solution of State Equations
4.2.1 Zero Input Response The solution of the homogeneous state equation L~ under the condition of zero input. Such a response is called zero input response and for convenience denoted as ZIR. Thus the behaviour of X(t) under the initial conditions without any input U(t) is called the zero input response of the system. It is also called free, natural or unforced response of the system.
4.3 Solution of Nonhomogeneous Equation Consider a nonhomogeneous
.
X(t)
state equation as,
A X(t) + B U(t)
X(t) - A X(t)
BU(t)
Premultiplying both sides bye· At, e- At(Xct) - A X(t)I
=
e-At B U(t)
Bute· At X(t) - e-At A X(t)
= ~ [e-At X(t)) dt
Substituting in above equation, ~ [e-At X(t)) dt
=
e-At B U(t)
Assuming initial time as t = 0 and integrating both sides from t c-At X(t)
1: =
= 0 to t,
I
Jc-At B U(t) dr ,0
t
e-At X(t) - X(O)
= f c-At
B U(t) dt
0
Premultiplying both sides by eA•. t
J
:.eAt c-At X(t) - cAt X(O) = eAt e·At B U(t) dt 0 t
X(t)
eAt X(O) +
J eA(t-t)
B U('t) dt
0
This is the complete solution of the nonhomogeneous equation.
... (1)
4.5
Modem Control Theory
Solution of State Equations
Observations : 1. The solution is divided into two parts. The first part is eAt X(O) which is nothing but the homogeneous solution or zero input response (ZIR). t
J eA(t-t)
2. The other part
B U(t) dt is the part existing only due to application of
0
input U(t) from time 0 to t, It is called forced solution or zero state response (ZSR). Thus the solution is, t
X(l)
= eA• X(O)+
j eA(t-t) BU(t)dt 0
ZIR
ZSR
If the Initial time is considered as t = X{t)
=
to rather
than t = 0, the solution is,
t
cA <1-101x(t0)
J
+ eA Ct-•J B U(t)dt
'o ln general, the solution of nonhomogencous equation consists of both the parts, ZIR (zero input response) and ZSR (zero state response).
4.4 Properties of State Transition Matrix The various useful properties of the state transition matrix are, t)(t)
eAt = State transition matrix
1.
.;>(0)
cA •o
2.
Q(t)
CAI = (c-Atr I = ft) (- t)r 1
Le,
91(1)
= I ., Identity
matrix
$(-I) CA(t1d2) "'eAt1 ·eAt, •
3. t)(t1 + t2)
t) (t1l t) (l:z) ., t) (12) t) (11)
4.
eM• + •>
5. e(A + B) t 6.
7.
(9 (t)J"
eA• eAs ~ eAt eBt only if AB "
9(12 - 11) 't)(t1
[eA11" -
to)
=
= eAn• = Q (nt) D
t)(t2 -
to)
BA
4·6
Modem Control Theory
Solution of State Equations
This property states that the process of transition of state can be divided into number of sequential transition. Thus to to '2 can be divided as t0 to t1 and t1 lo t2, as stated in the property.
lntcrms of ~ (t), the solution is expressed as, I
X(t)
Q
(t - to) X(to) +
J Q(t-t) B U(t) dr 'o
where
and
to>
0 (t -
~(t - t)
8. o (t) is a nonsingular matrix for all futile values of t. 4.5 Solution of State Equation by Laplace Transform Method The Laplace transform method converts intcgro differential equations lo simple algebraic equations. Due this important property, it is very convenient to use Laplace transform method to obtain the solution of state equation. Consider the nonhomogcncous slate equation as, X(t) = A X(t} + B U(t)
... (1)
Taking Laplace transform of both sides, s X(s) - X(O} s X(s) - A X(s)
A X(s) + B U(s} X(O) + B U(s)
As s is operator, multiplying it by Identify matrix of order n x n, (sl - A) X(s) = X(O} + B U(s) Premultiplying both sides by (sl - Ar 1, :. [sl - A]" I [sl - A] X(s} = [sl - Ar I (X(O) + B U(s}I X(s)
[sl -
Ar
I X(O) + [sl -
Ar I B U(s)
... (2)
ZIR + ZSR Comparing zero input response obtained earlier, [sl - Ar and
1
= Hsl •.• (3)
=
The matrix Q (s) [sl - A]" 1 is called resolvant matrix of A. All the clements of this matrix are rational functions of s.
Modem
4.7
Control Theory
Solution of State Equations
Taking inverse Laplace transform of (2), L" 11X(s)) = L" 1 l(sI - Ar 11 X(O) + L" 1 l[sl - A)" I BU(s)I
X(l) Using
Isl - Ar
1
X(t)
... (4)
The term L" 1 [¢ (s) B U(s)] is called zero state response. From the convolution theorem in Laplace transform it is known that, f1(t) • f2(t) =convolution
L" 1 IF1(s) F2(s)l
I
Jf
1<1-t> r2c1)dt
0
Thus if F1(s) = $ (s) and F2(s) = BU(s) then, I
L"1{<>(s)BU(s)I
= Jo(t-t)BU(t)dt
... (5)
0
Thus equation (5) shows that L" 1 {Q (s) B U(s)I is the zero state response, as obtained earlier by classical approach. X(t) = L" 1
16 (s)) X(O)
Q(s) = [sl _A)" 1
=
+ L-
1[¢
(s) B U(s)]
Adj [sl-A]
lsi:At
Key Point : The advantage of tlris metlzod is witho11t carryi11g out actual i11tegration wliich is time consuming, the zero state response ca11 be easily obtained.
4.6 Computation of State Transition Matrix While obtaining the solution of stale equation, the computation of state transition matrix cAt plays an important role. There arc various methods of obtaining state transition matrix from the state model. These methods are, 1. Laplace transform method 2. Power series method 3. Caylcy Hamilton method 4. Similarity transformation method Let us discuss these methods of obtaining eAt in detail.
Solutionof State Equations
4·8
Modem Control Theory
4.7 Laplace Transform Method While studying the method of obtaining- the solution using Laplace transform method, it is seen that the Laplace transform of cAl i.e, Q (t) and given by, c>
()5
=
I5 I
Ari
- Adj[sl-A] - . I sI -A I
-
Thus knowing A, once [sl - A) is obtained then 6 (s) con be easily obtained. This c> (s) is again a matrix of order n x n i.c, same ns tha~ of i\. It is called rcsolvant matrix of A. The eA1 i.e. c> (t) then can be obtained by finding inverse Laplace transform of Q (s) i.e, Inverse Laplace transform of each clement of the resolvent matrix Q (s). This is most simple popularly used.
n'*
method of obtaining state transition
matrix cA• and hence
Find tl1e state lra11sitio11 matrix for,
Example 4.1
A =
o [+2
-1) -3
Solution : Use Laplace transform method, (st-A]=
Adj [sl -A]
Isl
-Al
1s1-Ar1
0 (s)
s[~
~H-~ -~]=[_;
s+ [ -!
3 2]T = 3 -1] s s
5:~]
[s +
2
!]
+3 = s2 + 3s + 2
s•.
Adj [st-AJ
= (s + 1)
(s + 2)
-1]
s+3 [ 2 s = (s+1) (s+2)
-·1-;1.:.-Al [sl -Ar
1
s+3 (s+l)(s+2} =
2 [ (s+l)(s+2)
(s+ l):s+2) -1 (s+1)(s+2)
l
4.9
Modem Control Theory
s+ 3 L-
I
[sl -
I
Ar = L-
l
Solutionof State Equations
I (s+l)(s+2}
[
2 (s+ l)(s+2)
(s+1):s+2) -1 (s+l)(s+2)
Using partial fraction expansion for all the elements.
cAt L-1 [s!1 ·· s!2 s~\ ,!2] +
2 2 ---s+1 s+2 e
At -- [ 22c·t -c·2t 2
e ·t - e ·21
-1 2 --+-s+I s+2
-c·t +c·2t ] - $ t -e ·I + 2e-21 - ()
This L~ the required state transition matrix.
4.8 Power Series Method The state transition matrix is matrix exponential and hence can be expressed in a power series as,
... (I)
cAt
Thus by calculating various powers of A and then few terms of power series, can be evaluated. As the power series is infinite, it is necessary to stop after calculating first few terms of the series. Calculating hlgh powers of A is practically time consuming and hence lhls method is not practically used to obtain stale transition matrix.
1•
Example 4.2 :
Find ~t of A = [~
=~]
using po~r serits me/hod.
Solution : Let us obtain various powers of A. A2
= [~ o [2
=~][~ =~]=[~ :] -1] [-2 3] [ 6 -7] -3
-6 7 = 14 -14
.... • Al :..10
Modem Control Theory cAt
Solutlon of State Equations
= I+ At+..!._ A2r + _!_ A3 21 31
+
13
[~ ~]+[~ =~]t+M=! ~}2+H -~;]13+ . . 1~
l-t 2 +I 3 +... 14 3 [ 21-31 . 2 +6l +...
Consider the first clement of
CAI
r + t3 + ...
which is l -
1 + [- t) + 2.[-tj2 +2.[-tj3 + ...
Now
21
2 - 2t + t2 -
3!
.!.
3
t3 + ...
+2- (-
I + (- 2t) + 2_(-2t)2 2!
and
. .. (I) 2t)3 + ...
3!
. .. (2)
1 - 2t + 2t2 - ~ ~ + ...
6
From (1) and (2) we can write, Zc" 1 - c- 21
=
l + 0- t2 +
t3 + ...
= 1 - t2 + t3 + ...
Hence the series present as the first clement is 2c- 1
-
c- 21•
But practically such a task is difficult to identify the combinations of series. Hence this method is rarely used.
4.9 Cayley Hamilton Method This method is based on the Cayley Hamilton theorem. The theorem states that, every square matrix A satisfies its own characteristics equation. Let f(IJ be the characteristic equation which is given by,
i.e,
f().)
II.I-Al
f().)
).n + al >.n - 1 + ... +
=0
a,, - 1
A.+
a,, = 0
... (!)
According to Caylcy Hamilton theorem, the matrix A has to satisfy the characteristic equation. Hence, ... (2)
4·11
Modem Control Theory where
Solution of State Equations
I = Identity matrix
The theorem is used for evaluating the function of a matrix. Consider a matrix polynomial as, p(A)
=
a01 + a1A + aiA2 + ... + anAn +an+ 1 A"•
1
+ ...
...(3)
This polynomial has degree more than the degree of A which is n. The above polynomial can be calculated considering the scalar polynomial, p(A)
= ao + a1A+ a2A.2 + ...
+ an A." +
a,. • I
An+ l + ...
... (4)
Divide this by the characteristic polynomial f(A),
where
~~~;
Q(/~+ ~:~
... (5)
R(A)
Remainder polynomial
p(A)
f().) Q(A) + R(A)
... (6)
Now the remainder polynomial R(A) has order less than the characteristic polynomial and can be expressed as, R(A) "' Clo + a,A. + Oz ),2 + ... + On - 1 An • I
Ai, ~' ... ).,, are the n distinct cigen values Ai, ~' ... >.,, are the roots of f(A) = 0. Evaluate
Let n as
But
p(/.i)
fP,) QO,) + RC),)
f(Ail
0 hence,
of matrix A and f(\) = 0 for i p(IJ al all the eigen values,
... (7)
= 1, 2, ...
for i = I, 2 .... n
.. , (8)
Substituting Ai· )2, •.. A,. in equation (8) we get the set of simultaneous equations which is to be solved for the values of Cl(). a1 ... a.,_ 1. Now replace A. in scalar polynomial by matrix A,
but
p(A)
f(A) Q(A) + R(A)
f(A)
0
p(A)
... By CayIcy Hamilton theorem
R(A) = OQI+a1A+CtzA2+
Thus knowing 0(), a1
...
a.,_
...
a.,.1A"-1
1, any matrix polynomial p(A) can be evaluated.
... (9) .
Solution of State Equations
4·12
Modem Control Theory
4.9.1 Procedure to Calculate
eA1
The procedure to calculate cAt using Cayley Hamilton theorem can be generalised as, 1. Find the eigen values of matrix A from
I A. I -
A I = 0.
2. Form the remainder polynomial R(~ of order (n - 1). R(~ = Clo + tt1 ). + ... + CX,.. - I An - I = p(~ 3. If all the eigen values are distinct then substitute At• 1'2 .. . ),,, in remainder polynomial R(~ to obtain simultaneous equations interms of ao, tt1 ... a,.._ 1. Important Note : If the eigen value \ is repeated r times, any one independent equation can be obtained by substituting 11< in R(~. Then remaining (r - 1) equations are to obtained by differentiating both sides of the equation,
diR(i..>j
-d).!
.
, J
4. Solve the simultaneous equations
1•
= 0, I, ... r -
1
... (10)
'-=l.k
ao, a1, ... a,.._ 1· a,.._1 An-l.
5. Then f(A) = R(A) =°<)I+ cx1A + ... +
Fi11df(A) = A
Example 4.3 :
12
for A = [-~ -~] using Cayley Hami/1011 theorem.
Solution : The function to be evaluated is, f(A) Step 1 :
I~ . ~31=
0
A.2 + 31.+ 2
Ai Step 2:
R(~
i.e.
O
i.e, (A.+ 1) (A.+ 2) = O
-1, ~ = - 2 Cf
i.. = p(),) ... as n
=
2.
Substitute /'1 and 1'2 in R(I,).
Clo + tt1 and
A 12 hence p().) = 1.12
Find eigcn values.
li..1-AI
Step 3 :
=
°<> +
Ai
tt1 ~
ao-
a1
0-i)12 (~)12 as p(J.} = (-1)12 = 1
,,u ... (1)
Modem Control Theory
and Step 4 :
Oo - 2a1
4 -13
Solution of State Equations
= (- 2)12 = 4036
... (2)
Subtracting (2) from (1 ), -4035
IX1
Step 5 :
Oo = -4034
and
A12 = R(A)
f(A)
aJ + U1A = - 4034 [~
~]- 4035 [-~ -~]
[-4034 -4035] 8070 8071 A12 =
[-403-1 -4035] 8070 8071
Similarly eAt which is a power series in\crms of A can be evaluated using Cayley Hamilton theorem.
>>*
Find llit state tra11silion matrix for
Example 4.4 :
-1] Using Cnyley Hamilton Theorem.
A = [o
2 -3
Solution : The function to be evaluated is, f(A) Step 1 :
).,2
O i.e,
+ 31.. + 2
O
At
-
I-~ ;,+ 1=0 3
i.e. I,
/"2.
(}., + 1) (}., + 2) = 0
=-
2
Construct R().) R(>.)
where
eAt hence p(A) = e>. 1
Find the elgen values, 11..t-AI
Step 2 :
=
ao + a11.. +
... +
ex,,_
1 1 /,n - but n = 2
R().)
... By Cayley Hamilton method
p(A)
... replacing A by /, in f(A).
Modem Control Theory
Solution of State Equations
4 • 14
Substituting values of >.1 and ~.
Step 3 :
. .. (1)
... (2)
and Step 4 :
Solving equations (I) and (2),
e-1 -e-21 2(,-1 _ e-21
·and Step 5 :
f(A)
c
eA•
R(A) = <Xol + u,A ... replace I.. by A in R(A)
« 21 [~
2e-c -
~] + (e- 1 _ e- 21) (~
=~]
This is the required stale transition matrix. 1111+
Fiml the state tra11sitio11 tnJ1trix of,
Example 4.5 :
A
=
o o
-21
1 0
3
0 1 OJ by Cayley Hamilto11 theorem. [
Solution : The function to be evaluated is, f(A) = eAt hence p(A) = ,/' Step f :
Find the eigcn values.
=1~
l'-1-AI
0 },-1 -1
0
=O
>.(A-1} (1..-3) + 2 (1..-1} (1..-1}
p,2 -31..+ A,
= 1,
~ I= 0
A-3
2)
= O i.e. {A-1)
~ = 1,
(1..-1} (1..-2)
~ =2
Thus A, ~ 1 is repeated 2 times Step 2 :
R(A)
<Xo + ail..+ fkz'-2
= p()J =
e'IJ.
... n = 3
=0
4-15
Modem Control Theory Step 3 :
a1
(Jo+
+CC:!
Ai = 1, as it
For
Solution of Sta~e Equations
For )1= 1,
=
d Al -e di.
\ ... (1)
is repeated use,
~P<' ·>I 11.=Ai di. i.e,
\
e1
I"'"2
te).tl ).=1·2
~Ro->J di. J.=A.i
2 II
d -[a0 +a1).+a2). di.
i).=).2
a I+ 2a2 /~l.a).2
\
te1
a1+2az
... (2)
Note that differentiation is with resped to I. and not t, and for
":i
= 2,
(Jo + 2 a1 + 4
Step 4 :
az =
.•. (3)
e21
Solve equations (1). (2) and (3). 2 Ciz
From (2),
CI1
t
From(!),
ao
et - a1
c1
-
-
az = ct -
2az -
tc1 +
«i
e1(1 - I) + f'2 Using in (3),
c1 (I -
t) +
Ciz + 21 e1 - 4 -12 + 4 Ciz = e21 c21 - c1 + t e1 - 2te1 " e21 - e1 - t e1
tc' - 2e21 + 2c' + 2te1 and Step 5 :
(Jo
f(A)
= 3te1
+ 2c1 - 2c21
e1 - te1 + e21 - e1 - te1 = - 2te1 + c21 R(A) =(Joi+
a1A
+
Clz
A2
r~ r -~H~ ~ -~J n r -il =
f(A) =
eA'=OQ[~ ~
Substituting values of (Jo. a1 and
az and
~l+a1[~ ! -~]+a{~! 1] rearranging,
Modem
Solution of State Equations
4 -16
Control Theory
~''"]
2e1 -c21
0
0
c'
,,.
-c1 +e21
0
2c21 -c'
eAt
'This is the required state transition matrix.
4.10 Similarity Transformation Method Consider a matrix A of order n x n having cigcn values At• >.i, ... ~· There exists a diagonal matrix I\ such that its diagonal clements are the eigen values of matrix A.
A., 0
0 ,_
;
;2
O
0
A
[
0 0
···
l
= Diagonal matrix
.•. (1)
A.0
Both A and A have same eigen values and hence arc called similar matrices. The diagonal matrix A can be obtained by defining a similarity transformation of state variables as, X(I)
= M Z{t)
In such a case, M is the modal matrix of A. And the diagonal matrix is obtained as,
Now
I\
M-1AM
c"t
I + /I.I +
i·I !
0 1 -
o-
0 0-
... (2)
r
_!. 11.2 t2 2!
0
0
'J I +
1
0
+ ... 0
o-o
"2
0-0
·- >.."
-2
t+.!. 2!
"•0. I!
0
0
'-~
0 -o-01 0
->..~,
o---·---·
o
0--·--
0
12 +
Modem Control Theory
4-17
c'\t
Solution of Stato Equations
... (3)
The transformation used is, X(t)
MZ(t)
X{O)
M Z(O) M-1M Z(O) = Z(O)
M-1 X(O)
. . .
and
X(t)
~t)
Thus
X(t)
AX(t) = AMZ(t)
... (4)
AMZ(t)
~t)
M-1Mi(t)
~C1AMZ(t)
Z(t)
... (5)
A Z(t)
This is transformed state model. The solution o( homogeneous state equation in X(t) is, X(t) = eAt X(O)
••. (6)
While solution of equation (5) is,
=
Z(t)
c"1 Z(O)
... (7)
Using equation (4) in (7), Z(t) = e"1 M - 1 X(O)
... (8)
Premultiplying by M, MZ(t) But
M
e"1 M - I X(O) ... Transformation used
MZ(t) = X(t) X(t)
••• (9)
Comparing equation (6) and (9) it can be written as, eAt = Mel\t M - 1 where
M : Modal matrix
and el\l is given by equation (3).
I
... (10)
4 -18
Modem Control Theory
,,..
Solution of state Equations
Fi11d stale l7a11s1tfon matrix of.
Example 4.6 :
A = (20 -l] by similarity transformation metl1od. -3 Solution : Find the eigen values.
1-~
p,J-AI (:>. + 1)
p, + 2)
)..::11=:>..2+3:>.+2=0
0 -1,
hi= -2
Calculate cigcn vector. For
Ai= -1,
[~.1 l - A)=[=~
M2
M
0At
M-1
~]
[~;~ J = [~] [M1: M2J
"l1rt ~] =
l c_q
[ 2 Adj [11.1] _ -1
-I] 1 _[ 2
[c~'
r c-•
O ] /-ii
0
JMT--1--
-1
-;] ... Required state transition matrix
Modem Control Theory ,,..
Example
4 -19
Obtain the complete time response of system given by,
4.7 :
. = [_0 1]
X(t)
and
Solution of State Equations
2
O
X(I)
where X(O) = [~]
Y(t) = (1 - 1) X(t)
(VTU:JanJFeb.·1008)
Solution : Given system is homogeneous whose solution is X{t) = c'" X(O) Now Isl -A) (sl-A)-1
Adj (sl -A)
Isl - Al (sl- A)-t
s [10 o]1 _ [-2o 01] = [s2 -1]s Adj(sl-A)
Isl-Al
[enc cc )r
= [1s
12 22
21
-2]r s 1] =[
s
-2 s
s2 + 2 s s2 + 2 -2
52
:+ 2]
s2 + 2
eAt
L-I
[s
2:
s/+ 21
2
s
-2
+2
+2
s2 L-1
L-1
L-1
{s2:i} {/+2} {s2-~ 2} CAI
L-1
lr __
L-1
{J_..fi. .
°s2
s__ }., cos .fit 52 + (..fi.)2
L-1 {-
..fi.
}
52 +(..fi.)2
../'i .
cos.fit [ -..fi. sin..fi. t
.fi
s2 + (.fi)2
2
J_J2
sin
} = - J2
~sin.fit] cosJ'i t
.n t sin
.fi t
Modem Conltol Theory X(t)
CAI
:. Output response Y(t) = (1 (1
X(O) "
CAI
[~] "
Jz
t]
Jz
[cos J2 t + s.in J2 cos./2 t-./2 sin J2 t
- 1) X(t) - 1)
cos
sin
J2 t
Y(t) = ~ sin
./2 t
t]
J2 t +
[ cos./2
,..
Solution of State Equations
'4·20
~ sin J2 v2 t-./2 sin J2 t
is the required response.
Example 4.8 : Find out the time response for u11it step input of a >1JSlem given by
Xct)
= [~2 ~3] X(t) + [~] U(t)
= [~2 -~] X(t)
and Y(t)
and X(O) •
Solution: (sl-AJ
Adj ( sl -Al Isl-Al eAt
[" o][o 1] = [" -1 ] [ell c!2 "[s+ c21 c22 0 s - -2 0
2 s+ 3
r
t
3 -2r = s
[·+-23
:]
s2 + 3s + 2 = (s +I) (s + 2) L.1 ((sl -
[ .. ,
A)-11
(s+ l)(s+ 2) -2 (s+ l)(s+ 2)
, I
(s+ l)(s+
2)
=
¢(s)
(s + 1)5(s + 2)
Finding partial fractions, 2 ---
1 s+2 (O(s)) = _2._ + _2_ [ s+t s+2 s+t
s+l s+2 I I -1 2 -+s+l s+2
l
m.
Modem Control Theory
Solution of State Equations
4 ·21 c-t - c - 2t ] -c-•
To ftnd
= e"'
+2c-21
[l] = [ 2·•c
ZIR
c111
ZSR
L-1 { ¢(s) 8 U(s)
U(t)
Unit step :. U(s) = 1 / s
ZSR
L-'
X(O)
0
l[(s+~~~:+2) (s+t):s+2)] [~]. /s]) [l
X(t)
ZIR +ZSR =
Y(t)
[-~ -~J o '] [-2 -3
(s + l)(s+ 2)
2.5- Je-1 + 1.5 c[
X(t)
[2.5- 3e3 C-
1
+1.5e-2']
I _
3c - 2 I
-e-11)
Y1(t)
3 (e-•
Y2(t)
- S + 3 (2c·21 -e-1)
Example 4.9 : For a system X
=
(_:~:]for
X(O)
=
2']
3c-' -3c-21
These arc the outputs for unit s~cp input applied.
X(t)
-2t]
J
(s+ l)(s + 2)
••
-c
-2c-1+2c-21
[-~J
Determine the systtm matrix A.
= AX,
X(I)
[
=
t _21 - 2t
-21
]
whm
X(O) • [-~]
and
4 ·22
Modem Control Theory Solution : System is homogeneous so ZSR
[Al A2] A3 A4
A
Now
and at t
I
X(t)
=
.
So
X(O).
Similarly
X(t)
So
X(O)
. .
-2• ]
, hence X(t)
L-e2e-2•
= [-2e-2') 4e - 21
[-42]
X(t)
0
=0
eAt X(O}
X(t) let
Solution of State Equations
=
Substituting in X(I) •
[-42] when X(O)
[_e:_1
1],
[:~]when
= [. ~]
• = [ e-e-•1
X(t)
I
X(O)
at t
=0
X(t) = [: ~]
= [:~]
A X(t)
and
A2J[l]-1 [-l]=[A1 1 A 3 A4
A1-2A2
-2
..• (1)
A 3 -2A4
+4
... (2)
-1
... (3)
-A2 A3 -A4
+A
I
... (4)
Solving these 4 equations simultaneously, A2
= 1,
A I = 0,
A4
=-
3,
A3
=-2
A "' [-02 _\] 4.11 Controllability and Observability In a control system analysis, it is necessary to find an optimal control solution for a given control problem. The existence of such a solution depends on the answers of two basic questions which are, 1. For a given system, is it possible to transfer any il).itial state to any other desired state in a finite time under the effect of suitable control input force ?
Modem Control Theory
Solution of State Equations
4-23
2. If the output is measured for finite time then with the knowledge of the input, is it possible to determine initial state of the system ? The answer to the first question gives the concept of the controllability while the answer to the second question gives the concept of observability of the system.
4.12 Controllability The answer to the first question means the concept of controllability of a system which is related to the transfer of any initial state of the system to any other desired state, in a finite length of time by application of proper inputs. Hence controllability can be defined as, A system is said to be completely state controllable if it L~ possible to transfer the system state from any initial state X(t0) to any other desired state X(t 1) in a specified finite time interval (t 1) by a control vector U(t). The concept of controllability and observability were originally introduced by Kalman hence Kalman's tests are used to find out whether the system is controllable and observable or not.
4.12.1 Kalman's Test for Controllablllty Consider n '" order multiple input Linear time invariant system represented by its state equation as, X = AX(t) + BU(t)
... (1)
where A has order n x n matrix and
U(t) is rn x 1
vector i.e, there arc m inputs.
X(t) is n x 1 state vector. The necessary and sufficient condition for the system to be completely state controllable is that the rank of the composite matrix Qc is 'n', The composite matrix Qc is given by, ... (2)
ln this composite matrix Qc, B, AB, A 2B ... arc the various columns. Proof : Let us see the proof for this condition. Assume that the final state is the origin of the state space while the initial time is zero i.e. t0
= 0.
As fmal state t = tr, is the origin of the state space then X(t1) = 0. . .. (3)
Solution of State Equations
4 ·24
Modem Control Theory
The solution of the state equation (1) is given by, X(t)
=
cAt X(O) +
J CA (HJ B U(t) dt
... (4)
0
Substituting t
= t1 for the final state, /t1
X(O)+J
CA(t1-t)
BU('t)dt
... (5)
0
Using (3) in (5),
J cA(t I -ti
I
0
At
e
I X(O) +
B U (t) dt
0
/11xco)
- J eAtl ·e-"'
BU(t)dt
0
'1
X(O)
= - J e-At
B U(t) dt
... (6)
0
This shows that any initial state X(O) 'nust satisfy the equation (6) for complete state controllability. Thus the controllability depends on matrices A and B and is Independent of matrix C. Now mathematically e-A' can be expressed as, ... (7) Using (7) in (6), ... (8)
X(O)
Let
J
ak(t) U(t)dt
0
X(O)
X(O)
- [B : AB : ... : An -
1
BJ [ ~;~
IJn-1
l
... (9)
Modem Control Theory
X(O)
= - 0., [
~;
Pn-1
l
4 -25
Solution of State Equations
... (10)
The vector matrix of Pk is of order n x 1 and if system has to be state controllable then any X(O) must satisfy equation (9) for which rank of 0., must be n which is a n x n matrix. For any rank of 0., less than 'n' ii indicates that some of the states arc not controllable. This proves that the rank of composite controllability of the system.
matrix
Oc
must be 'n' for complete state
The advantage of Kalman's test is that for any form of matrix A whether A is canonical or otherwise, the test can be applied. But its limitation is that it does not give a physical feel of the problem. If the answer to the Kalmans test is uncontrollable then it does not indicate which mode is uncontrollable. But due to mathematical simplicity of its application, Kalman's test is used to test the controllability. Key Point: The matrix has a rank r 111tans tire dt1em1i11a11t of ordtr r x r of the matrix has 1wnuro value and any detem1i111111t having order r + 1 or more than that has zero value. !Thus rank of 0., ••
= n for controllability
Example 4.10 Find the amtrollability of the system,
Solution:
Now
It
2
A
[~ -~1
AB
[_:]
c, =
(B AB]
I~ _; I = -
B = [~]
= [~ _;]
1 which is nonzero
Rank of Qc = 2 = n Hence the given system is completely controllable.
4-26
Modern Control Theory
i•
Cimlualt. tire co11tro/Jal>ilily of the system
Example 4.11:
.
willt,
x
AXT BU
and
A
[~
IiI
-) J
8
a
rq I J LO
n ,. 2
Solution:
Now
Solution of State Equations
QC
[B
All
[~ _:] [~] = [~]
QC
[~ ~]
, , 11 10 0
Hence rank of Qcr Rank of Qc
AB!
= Determinant of 2 x 2
=0
= 1 and "' n
:. The system is not state controllable.
4.12.2 Condition for State Controllability in s-Plane The system is represented by its transfer function in the s-plane. Such a system is completely state controllable only if the denominator and numerator polynomials of the transfer function do not have a common factor, except the constant. Such polynomials are called coprime. In other words, there should not be a pole-zero cancellation in the s-domain transfer function of the system for complete state controllability. If the pole-zero cancellation exists then a.JI the information about the dynamic behaviour of the system is not carried all along the system. In such a case, the system cannot be controlled in the direction of the cancelled mode.
4.12.3 Gilbert's Test for Controllability For the Gilbert's test it is necessary that the matrix A must be in canonical form. Hence the given state model is required to be transformed to the canonical form first, to apply the Gilbert's test.
.
Consider single input linear time invariant system represented by, X(t) = A X(t) + B U(t)
Modem Control Theory
Solutlon of State Equations
4 ·27
where A is not in the canonical form. Then it can be transformed to the canonical form by the transformation, X(t) • MZ(t) M
where
Modal matrix
The transformed state model, as derived earlier, takes the form, z.(t) where
c
A Z
B U(t)
A
Diagonal matrix
B
M -I B
It is assumed that all the eigcn values of A are distinct. In such a case the necessary and sufficient condition for the complete state controllability is that the vector matrix B should not have any zero elements. If it Ms zero elements then the corresponding state variables are not controllable.
If the eigen values are repeated then matrix A cannot be transformed to Jordan canonical form. If A has eigen values A,, A,, 1-z, 1'z, 1'z, ~ "4· ... A,, then the transformation results Jordan canonical form as, O················· O·················
0 0
J=
O··· ·············
0
Jordan block
O·················
0
O······~···· · ·
0
~················O 0
iJ ........••.•........................................
A,,
In such a case, the condition for the complete stale controllability is that the elements of any row of B that corresponds to the last row of each Jordan block are not all zero. ,_.,
Example 4.12:
Consider tlre system wit/i state equation.
~I
.
Xz [ XJ
Estimate tlte stale controllability by i) Ka/matt's test and ii) Gilbert's test
Modem Control Theory
4 ·28
Solution of State Equations
Solution : i) J<.lman's test
0c
(B : AB : A2BJ .•. n = 3
AB
~lj = [~
-6
I~ ~ ~I Hence the rank of
IOcl
= i. Thus
Oc is 3 which
36
0 -11 60
is nonsingular.
is 'n'.
Thus the system is completely state controllable. li) Gilbert's test
For this, it L~ necessary to express A ill the canonical form. Find eigen values of A.
II.I-Al
:. "K .. 61.2 +
=
)I ~.
-1,11 "" ~I
lo
0
}.+6
11). ... 6 = 0
:. (I.+ 1) (/.. + 2) (}, + 3) = 0 '·1 = -1,
~ = - 2,
A..!
=-3
l [ 1 1 1]
As matrix A is in phase variable form, the modal matrix M is Vander Monde Matrix.
1
i..2
i..3
).~
).~
=
-I
1
-2 -3 4
9
Modern Control
Theory
Solution of State Equations
4 ·29
[~
6
-l [j -5 -~i
-5 8 -3
---= =[-~ Adj(M]
M-1
IMI
li
W1B
As none of the clements of
i3 are zero,
8
-3 -2
2 -1
-1
-2
-1
= [-~
2.5 ~1 1.5
0.5] -1 0.5
2.5 0.5][°] = [°.5] -4 1.5
-1 0.5
0 l
-1 0.5
the system is completely stale controllable.
Key Point: As Gilbert's lest requires to transfonn matrix A into canonicalform, it is time
ccmsuming and hence Kal111a11's test is popularly used to ttSt controllability. 4.12.4 Output Controllablllty In the design of some practical systems, it is necessary to control the output variables rather than the stale variables. In such a case complete output controllability is defined. Consider the stale model of a linear lime invariant system as, X(I)
A X(t) + B U(l)
Y(I)
C X(t) + D U(t)
The system is said to be completely output controllable if it is possible to construct an unconstrained input vector U(t) which will transfer any given initial output Y(lo) to any final output Y(t1) in a finite time interval to St S t1. In such a case, construct the test matrix Q., as,
Q.,
=
[CB: CAB: CA2B: ... CAn-l
B: DJ
Thus if the order of matrix C is p x n i.e. there are p outputs then for the complete output controllability, the rank of test matrix Q< must be p.
4.13 Observability The observability is related to the problem of dcterrnlning the system ·state by measuring the output for finite length of time. Hence observability can be defined as, A system is said to be completely observable, if every stoic X(t0) can be completely identified by measurements of the outputs Y(t) over a finite time interval. If the system is not completely observable means that few of its state variables are not practically measurable and arc shielded from the observation. Similar to the controllability, the observabillty of the system can be obtained by using Kalman's test.
Solutlon of State Equations
4. 30
Modem Control Theory
4.13.1 Kalman's Test for Observability Consider nth order multiple input multiple output linear represented by its state equation as,
x and
Y(t)
where
Y(t)
and
time invariant system,
A X(t) + B U(t)
... (1)
CX(t)
... (2)
px l output vector
c
t x n matrix
The system is completely observable if and only if the rank of the composite matrix Q0 is 'n'.
TI1e composite matrix Q0 is given by,
c,
= (Cr,ATCT:
(A ,.)"-•cTJ
where
Transpose of matrix C
and
Transpose of matrix A
I
Thus if, rank of Q0 = n, then system is completely observable. Proof :
For the system described by the equation (1) and (2), the solution is, t
X(t) = cA• X(O) +
Y(t) =
J cA (t-t)
c CAI X(O) + c
I
B U(r) dr
CA (t-T)
B U(t) dr
... (3)
... (4)
0
Now as the matrices A, B and C are known and the input vector U(t) is known, the integral term in the equation (4) is known. This can be subtracted from observed value of Y(t), to modify the lefthand side. Hence the obserability can be investigated by considering the equation, Y(t) = C eAt X(O)
... (5)
as the left hand side is known due to measured output. Thus for observability, the unforced system can be considered i.e, homogeneous state equation can be considered. Consider the state model of unforced system as, X = A X(t)
and
Y(t)
= C X(t)
The output is given by, Y(t) = C eAt X(O)
... (6)
4. 31
Modem Control Theory
Solution of State Equations ... (7)
But •-1
Y(t)
I
Ccxk\t) A kX(O)
k=O
Y(t)
CJo(I) C X(O) + cx1(t) CA X(O) + ...... <Xn _ 1(t) C An - 1 X(O) ... (8)
For the system to be completely observable, from the given output Y(t) over the
interval 0 ~ t ~ t1, the X(O) must be uniquely determined. TIUs is possible if the rank of the matrix,
[J.] must be n. Taking transpose of the matrix, the condition can be stared as the rank of matrix Q., must be n for complete observability where, Qo;
>'*
[cT:ATCT:
... :(ATt-lCTJ
Example 4.13 : £va/11ate tire observability of tlrr sys/em
and
Y(I)
= (1 OJ [;:]
Solution : The order of the system is, A AT
[-~ -~J
and
c .. (1 OJ
[~ -21
and
CT .. [~]
I
Qo
n ~2
rc·r
-1
ATCT)
4 ·32
Modem Control Theory
Solution of State Equations
Consider the determinant
I 10 11I
1 =nonzero 2=n
:. Rank of
Hence the system is completely observable. ,...
Example 4.14 : £wlllate tire obstrVability of the system with
A
[~ -~
_r] •
B
= [~]
and C
= (3
4 IJ
Solution : The order of the system is n = 3 Qo
1cT
AT cT
AT
[! -~]
(A T)2CT)
0 0
-3
CT
[:] 0
ATCT
[! =~OJ [3j~ [~] 0
J
(A T)2CT
AT[ATCTJ=[!
0 Qo
[~
-~] -2
0 0
=~][~] = [=~]
Modem
4. 33
Control Theory
Solution of State Equations
Consider the determinant,
01
3 0 4 1 -2
11 I
-21
= - 6 + 0 +0 +0- 0+ 6
=0
Hence a nonzero determinant existing in Q., is having order less than 3. :. Rank of Q0
"'
3 "' n
Hence the system is not completely observable.
4.13.2 Condition for Complete Observability in s-Plane The condition for complete observability in s-plane also can be stated from the transfer functions or transfer matrices. The conditions remain same as before that the numerator and denominator polynomials of transfer function must be coprime. There should not be cancellation of pole and zero in the transfer function. If the cancellation occurs then the system is not observable and the cancelled mode cannot be observed in the output.
4.13.3 Gilbert's Test for Observability 11 is known that for Gilbert's test, the state model must be expressed in the canonical form. Consider the state model of linear time invariant system as,
and
X(t)
A X(ti + 13 U(t)
Y(t)
C X(t)
Use the transformation X(t) Y(t)
=
MZ(t) where M is the modal matrix.
CM Z{t) =
C Z(t)
CM
where
For a single input single output system,
Y(t)
[
~~
~:~]
Zn(t)
'11 Zi(t) + '12 Zz(t) + ··· + Cin Z,,(t) Due to the canonical form, all the states are decoupled and not linked to each other. Hence for the system to be observable, each term corresponding to each state must be observed in the output. Hence none of the coefficient of C must be zero.
Modem
4.34
Control Theory
Solution of State Equations
Thus the system is completely observable if all the coefficients of C arc nonzero, none of the coefficient is zero. If any clement is zero, the corresponding stale remains unobserved i.e. shielded from observation. n•
Example 4.15 : Evaluate lire obseroabilily of lhe system wil/1
A=[~
~1'
B = [~] a11d C = (3 4 1) l
~ 0 -2 -3
Using Gilbert's lest. Solution : For Gilbert's lest, find the cigcn values.
1~0 -~ -~ l=O
[),I-A)=
2 l.3+3'>..2+2i.
O
i.(i.2 + 3i.+ 2)
O
i.(),+1}(1.+2) :.)'! = 0,
i.+3
0
~ = - 1,
"3 = -
2
l
As the eigen values are distinct, the modal matrix M is a Vandcr Mondc matrix. 1 1 M = 1.1 1.2 [ ).~ ;.~
1 1.3 ;.~
[1
11
= 0 -II -2 0
I
4J
When the modal is transformed to canonical form,
C
=
CM = (3 4 1) [~ -~ -~] 0 I 4
= [3 0 - 1) As there is one zero element in
C,
the system is not completely ·observable.
4 .35
Modem Control Theory
Solutionof State Equations
Examples with Solutions
1•
Example 4.16 :
Find the statt transition Mitrix cf the state tquatio11.
Using lht inoerse transform method.
(Bang•lo"' Univy Aug.·96)
Solution:
Now
s [~ ~]-[~ ~] [s~ 1 s ~ 1]
sl-A
=
Adj (sl-A)
1s1-Ar1
Adjoint
[cofactor matrix) T
[sl - AJ Isl - A!
I
( sl-A
=
= [s-1 _1
0 ] s-1
(s - 1) (s - 1)
o] [
s-1 [ -I s- I (s-l)(s-1)
=
0]
s- 1 I (s - l)(s-1)
1 s-1
s-1
= [_1I (s-1)2
ol1 ;:I
Taking inverse Laplace transform. 4>(t) I~
Example 4.17 :
O]
e1 cl = [tel
.. . Required state transition matrix
For a certain system, wl!m
X(O)
while X(O) =
[_13] then X(I) = [
[:]Ihm
X(t>
~-:~Jt]
= [::]
Determine the system matrix A.
(llangolore Univ. Aug.·95, April-'J'J)
Solutionof State Equations
4·36
Modern Control Theory
Solution : The solution of the equation is, X(t}
Let
eA• X(O}
A
and the equation is,
.
X(t)
A X(t)
Now
X{t)
[
and
X{O)
[-13]
Now
X(O}
A X(O}
[-:] Similarly
• [ 3e-J1] hence X(t) = ~ e-Jt
-JI ] _; e-Jt
[Al
A3
A1-3A2
-3
A3-3A4
9
.
X(O}
[-3)9
[-~]
A2]
A4
... (1) ... (2)
.
X(t)
[::]
X(I) =
X(O)
[:]
X(O)
.
=
X(O)
[Al AJ
~:] X{O)
GJ
[A' AJ
~:] [~]
[::]
[~]
A1+A2
... (3)
A3 +A,
... (4)
Subtracting (3) from (1), -4 A2
-4
A2
1
Al
0
Modern Control Theory
4 -37
Solution
of State Equations
Subtracting (4) from (2),
8
n•
i.e. A4
= -
2
Example 4.18 : Obtai11 Ilic so/11tio11 of Ille liomoge11eo11s state equation X =AX ionere
A = rl1
-2]
and
X(O) =
fl0.51
1 -4
I
J
(Bangalore Univ. Aug.-97)
Solution: The state transition matrix is,
L-1 {cst-Al-1} [sl-A)
s[~ ~J-G =:] =[s~i' s::] Adj(sl-1\) tsl-AI s+4 = [ -2
Adj [sl-A) Isl-Al
(s-l)(s+4)+2
=
s2 + 3s- 2
[
L-1 [
s+ 4 ] (s-0.561) (s+ 3.561)
-2 ] (s- 0.561) (s+ '3.561)
]T
=[s+4 1
-2]
s- I
=s2-s+4s-H2 (s - 0.561) (s + 3.561)
(s-0.56;;(:+ 3.561) I (s-0.561) (s+ 3.561)
L-1 [
l.106 (s-0.561)
1.106 e + 0 561 L-1[
1
s- 1
(s-0.561;(!+
(s-0.561) (s+ 3.561)
0.106 ] (s+3.561) I -
1-1 [ -0.485 • (s-0.561)
0.106 e " J.56! I +
3.561)]
(s- I)
~1 (s+ 3.561)
-0.485 C~ 0.S61t+0.4850-
3.5611
Modem Control Theory i,-1 [
Solution of State Equations
4 -38 i,-1 [
1 ] (s-0.561) (s+ 3.561)
0.242 rs-0.561)
0.242 e " 0.56l
0
'"
:.
=
[1.106c0.S611-0.106c-J.5611 0.242 c115r.i' -0.242 e- 3·561'
-
0.2<12
t .,.
0.242 e" J.St.l t
-0. .i85 C0.~611 1 0.485 e" }.5611] -0.106co.>ol' ... 0.106c-3.561 •
0.068 CO.St.11 + 0.432 C. 3.561 [O.OlS e11•5611 ~ 0.9R5 e" J.S6l ,,_.
]
(s+ 3.561)
.
t] t
Example 4.19 : For a system represented l7y X = AX, the response is
X(I)
= [2e-1 e-4'1]
whm X(O)
• = [4e-c-2t21] = [2]1 and X(t)
whm X(O)
= [4] 1
Determine the system matrix A and the state transition matrix. (VTU:
Solution : Let the solution is, X(t)
cAt X(O)
A2] A4
Let
A = [At A3
Now
X(t)
[ 2c-4t ] e-.at
• X(t)
At t = 0,
X(O)
[~]
X(O)
And
X(t)
[4e-2 1] e-21
At t = 0
X(O)
[~]
= [ - 8 c ·41] . - 4 e -41
. = [-8) -4
• = [-se-21] -2c-21
X(t)
. [-8]
X(O) = -2
JanJFeb.-2005)
4 .39
Modem Control Theory X(t)
X(O)
= [~~ ~:] [~:
Solution of State Equations
X(t)
~:] X(O)
... (!) ••• (2)
and
-8
... (3)
-2
•.. (4)
Solving (1), (3) and (2), (4) simultaneously we get, A
=
[o -s] 1
-6
[sl-A)
Adj [sl-A) Isl-Al
[l l
s2 +6s+8
I
L-
= (s + 2) (s + 4)
-8]
s+6 1 s (s+2) (s+4)
s+6 L-t (s+2)1(s+4) [
(s+2)(s+4)
-8
l
(s+2)5(s+4) (s+2) (s+4)
Modem Control Theory
4 ·40
L-1[ ~2s~4 0.5 0.5
----s+2 (s-;-4)
1ui+
--+--4 4
s+ 2 s+4 1 2 (s+2) + (s+4)
l
Solutlon of State Equations
A linear timt invaritmt system is clwraderiud by the homoge11eous state
Example 4.20 :
equation:
Compute the solution of homogeneous equation, assume tire initial slate vector :
Solution : From the given model,
A [sl-A)
Adj [sI -A) Isl
-Al
=
[~ ~]
[s- I
[ 1 O] [l OJ
0]
s 0 I - l l " - I s- I
[s~
I ~~
r" [s~ s~ i] I
(s- l)l
s- I 0 ] Adjfsl-AJ - [ I s-1 jsl- Al (s- 1)1 I
•-I [
(s-\)2
L-11s1-Ar1
,; .] [
~
= L-1 ~~
(s- I)"
0]
_I_
s-
I
Modem Control Theory
4 -41
o]
c' [te' X(t)
cAt
SoluUon of State Equations
c'
X(O) ; zero input response
This is the required solution.
1..+
Example 4.21 : Obtain tire time response of the following system :
; [~ ~][~:] + [:] U (I)
[ ~:]
where U (I) is the unit step occurring at t = 0 and X T (0) = [1 OJ. Solution : From the given model, A
= [: ~]
B = [:)
X (0) = [~]
Now as matrix A is same as considered in Ex. 4.20 while X (0) is also same as consider in Ex. 4.20, the zero input n.'Sponsc is same as obtained in Ex. 4.20. · ZIR
=
X(t) = c' [t c'
J
Thi'.' zero state response which depends on U(t) and matrix B is given by, ZSR where
X(t) = L-1 { ets) B U(s)}
1 s-1
[ (s-11)2 B
(l(s) B U(s)
_l_]
[:] and U(s) =
1 si'I [(s-1)2
as obtained in Ex. ~.20.
(s-1)
0
as unit step
l G]m
_I (s-1)
m
l
4.42
Modem Control Theory
l
r_1 (s-1)I (•-1l' s-1 s 1 1 _1_ [
l
1_
(.!.] = r
(•-I)
5
ll
Solution of State Equations
l; 1
1..-1)'
s(s-1) 1
I
-(,-!)
. . . ... usmg partial fraction
(s-1)2
Total response = X(t) = ZTR+ ZSR
[e [e' -1] = [2e -1] 1
= ,,.
tc1
]
1
+ t c'
2tc1
Example 4.22 : Fitrd and sketch the response of the system witlr the followi11g transfer function, input and initial conditions. T(s) = -4 s+ 20 s+300 r(t) !1(0)
Solution :
r(t)
= input
10 u(t) ... (input) 0 and
y(t) = output
= Y(s) = -4s+20 T(s) R(s) s+JOO Dividing numerator by denominator as are of same order, Y(s}
_4 + 1220
R(s)
s+300
4 .43
Modem Control Theory
Solution of State Equations
The corresponding stale diagram is,
«o ,,..._,---y(t)
'---------i
4
1---------'
Fig. 4.1
and A=(-300],
X1
r(t)-300X1
y(t)
1220 X1 -4 r(t)
B=[I). (sl-A)
[s1-Ar1
C=[l220J.
D=(-4)
[s+ 300] [s+
~oo]
(e-3()01] Now
hence substituting in y(t)
y(Ol
0
y(O)
1220X1(0)-4 r{O),
0
r(O) =10 as given to be 10 u(t)
1220 X1(0)-40 40 1220
ZlR
CAT X"l(O)=~
.,-3001
1220
To find ZSR use, ZSR
L-1 [9(s) B U(s)) L-1 [-1-·l·!Q] as U(s)=R(s)=!Q (s+300) s s
L-1 [ ZSR
10 ]-t-1 s (s+ 300)
_..!___]_ 30 30
c·:lOOt
[1/3() s
1/30]
s+ 300
...... 4 .44 Solutlon of State Equations ~~~~~~~~~·~~~~~--~~~~~~~~~
Modem Control Theory X(t) X(t)
...!..-5.4x 10-4 c-JOO~
y(t)
1220 X(t)-! r(l)
30
1220 [ ~ -5.4x 10-1 c·· 3001
J -4
(10)
0.667 -0.667 e- JOO I The response y(t) is, y(t)
0.667 --------:;-:;·..----
Fig. 4.2 ,_..
Example 4.23 : A linear dynamic time int!Qriant system is rtprese11ted by
.
umere
X
A X(I) + B U(I)
A
[~ ~ ~1 B=[~ ~1 0 -2 -3
Find
if the
1 0
system is completely controllable.
Solutlon : For the system, n = 3
o,
[B: AB: A2BJ
AB
[~ ~ 0 -2 -3
~1 [~ ~i ~1 1 0
=[ ~ -3 0
Modem Control Theory
4 -45
1 A2B
= A [AB]=[~
OJ[ o1 o10 J = [-31 OJ0
0 I -2 -3
0 0 1 1 0 -3 0 -3 0 7 1
= [~
QC
Solution of State Equations
0
-3 0
~]
Consider the determinant,
[ ~ ~ ~1 I
= 1 = nonzero
0 -3
Hence rank of Qc = 3
=n
Thus the given system is completely controllable
Consider the syste~n represented by,
11111~ Example 4.24 :
x
[- ~~
Y(I)
~:~] X(t) +
G]
U(t)
(1 OJ X(I)
Find the complete obseroability of the system. Solution : For the system, n
=2
A [-0.0.12
0.4] -0.1
AT
[-0.2 0.1] 0.4 -0.1
Qo
(CT
ATCT
Qo
C = (1
CT
= [~]
A TCTJ
[-0.2 0.1][1]=[-0.2] 0.4 -0.l 0 0.4
[10
-0.2] 0.4
O]
7 0
Modem Control Theory
Solution of State Equations
Consider the determinant, 1 10
-
0·21
= Nonzero
0.4
0.4
Rank of Q.,
2=n
Hence the system is completely observable.
11*
Example 4.25 : Usi11g similJlrity tra11Sfonnationfind A=
I'' for,
[-4 3] -6 5
Solution : Find the eigen values.
IJ.:4 ;..3sl=o
p.I-AI
J. -A
-2
O
(A+l)(A-2}
O
2
,.,
-1,
~
=2
Find the eigcn vectors. For}.,
=-
1,
(}.,! - A)
=
[! ~]
[~:~]=[~]=[:~] a[: =~] :!] g ~] [ 2 -I] = .2.....2_
M1 = For ~ ~ - 2, (>'!! - A)
M2
M
M-1
[~:~]
2
[
= [~]
Adj (MJ
IMI
1
=[ 2 -1
-:1
Modem Control Theory
Solution of St.ate Equations
2c-1 -e21 = [ 2c-1 -2c21 ,..
-e-1 +e21 ] -c-1 +2c21
1'1 for
Example 4.26 : Using lAplace transform metltod, find
-3]
Solution : a) A = [oI _4 [sl -Al
=
Adj [sl-AI - [s~4
1s1-Ar1
Isl-Al
e ts) [
~1]"
- s(s+4)+3
s+4 (s+l\(s+3)
~ (s+l):s+3)
(s+1)(s+3)
(s+l)(s+3)
9 (t) = cAt =
l
[s~4
s+4 (s+ l) (s+ 3) (s+ I)-3 L- 1
:s+
(s+ I) (s+ 3)
1.5 0.5 L_ 1 s+I
s+3
0.5 0.5 [--s+I
s+3
-},5 s+l
3)
(s+l)(s+3)
T-1.:.?..l s+3
J
-0.5 1.5 --+-s+l
-:]
(s+l)(s+3)
s+3
I
4 ·48
Modern Control Theory
b) A=[-~
-n [sl -A)
[sI-Ar
s -1 ] [ 3 s+4 Adj[sl-A]
I
[
s+4 '] [ -3 s
l
s+4 (s+1~+3)
I (s+l):s+3)
(s+l)(s+3)
(s+l) (s+3) (s+I) I(s+3)
s+4 t_ 1 (s+l)(s+3) [
[s+4 -3
l
-3 s (s + l)(s+ 3) (s+ l)(s+ 3)
1.5 0.5 --s+l s+3
0.5 o.s ---
-1.5 1.5 [-+-
-0.5 1.5 --+-
s+l
s+3
s+l
s+l
s+3
a)A=[~ -~] b)A=[-~ ~] Solution : a) A = [-~ -~]
Step 1 :
hence
p(I.)
= c'"
Find cigcn values
p.. I-AI
s
I~
A~51=0
0
i.e,
(A + 2)
l
s+3
Example 4.27 : Using Caylty Hamilton method,find
f(A) = eA•
I]
= s2+4s+3 = (s+l)(s+3)
Isl-Al
6(s)
l ..
Solution of State Equations
p.. + 3)
~ O
it' for,
4 .49
Modem Control Theory
Step 2:
n=2 O'o + a1 A. = p(/.}
R(/.} i.e.
Step 3 :
Solution of State Equations
O'o +
a1 ).
e)J
Substitute values of A. ..• (1)
and
... (2)
Step 4 : Solving equations (1) and (2), <Xi = e- 21_e-31, O'o = 3e- 21 _ 2e-3•
Step 5:
f(A}
= R(A) = OQl
+ <X1 A
(3e- 21 _ 2c- 3t) [~ OJ+ (e-2•-e-31)[
eAI
l
[ 3e-2t -2e-Jt -6c-21 +6e-J1
CAI
bl
[-~ ~]
A
CAI
f(A)
Step 1 :
Step 2 :
hence p().)
Find eigen values.
IU-AI
I~
A.2+41.+4
0
(A.+ 2)2
·= =
"'n=2
-21
A.+4 = 0
0 -2,
'2 = - 2
R(I~ ~ O'o + a1 A.= p(~ i.e,
O'o + a,:i..
e -2t -e-31 ] -2c-21 + 3c-Jt
c>.:
= c)J
0
-6 -~J
Modem Control
Step 3 :
Theory
Solution
of State Equations
Substitute values of I.
Clo -
=
2(X1
c-21
.•. (1)
But H I. is repeated, second equation must be obtained as, dR(I.)
dp().) di.
di. (XI
t e'IJ at
o:,
t e-21
), = -
... (2)
2
Step 4:
... from (I)
Step 5:
f(A)
= R(A) = Cfol + <X1A
eAt
=
(e- 2t
+ 2t e-21) [~
c-21(1+21)
+ te-
2te-21
[ -2tc-2t ••
~]
21 [-~
~]
]
e-21(1-21)
Example 4.28 : Find the respans« ef the system,
X = [~2 and Y(I)
=G
]x •[~ ~]
~3
U(I), X(O)
= [~]
~] X lo tht following input,
U(I) = [~:~:~] = Solution : Find cAt for A
1s1-Ar1
$(s)
[e~;~::w] whtre = [~2
11(1)
=unit step function.
~3] thus [sl - A)
= [;
s+ 3 l] Adj(sl-A] _ [ -2 s _
Isl-Al
csr
-
-Ar 1 = [
~\] 5
[s+-23
I] s'
52 +3s+2 - (s+1)(s+2)
s+3(s+2) (s+I)
(s+l) 1(s+2)
-2 (s+l)(s+2)
s (s+l)(s+2)
l
4. 51
Modem Control Theory
2. s+l
1 s+2
s+I 1
-2 [ -+s+l
2. s+2
-1 2 -+s+l s+2
-1
Now
ZSR
U(•)
ZSR
s+2 1
l
Solution of State Equations
= [ --e2c-•-1 +-c-2 e21-21
L
lo(s)I
L-1
{<)(s) B U(s))
?
UJ L_1
![ .. ,
(s+1~:+3) (s+t) (s+2)
L-1
l [ "'
L-'l
(s+1)(s+2)
cs+3
(s+l) (s+2)
'"' :"''] r· : ,
s+3 (s+l)(s+3)
(s+l)s(s+2)
s(s1+3)
-2 (s+l)(s+2)
(s+l)(s+2)
s+3
(s+t~:+J)
l
'"'':'""] [: :] f : ] I (s+l) (s+2)
s+3
' ] ""'']l
l
4. 52
Modern Control Theory
Solution of State Equations
X(t)
(t)]
Y1 [Y2{l)
and ,,_.
[~ ~] [~~ ~: ]
Y1(t)
X1 (t) = 3 - 2.5 e-
Y2{t)
X1(t) + X2(t)
1
-
e-
21
= 1 + e-lt -
+ 0.5 e-
31
2 e-Jt
Example 4.29 : The Fig. 4.3 shows the block diagram of a process control systmi with state variablefeedback and feed forward control. Tire state model of process is,
. [-3 2] [l]
X=
and Y = (0
4 -5
X+
0
U
11 X
a) Derive state model for the eutire system. b) Find the response if input
r(t) is
unit step ass11mi11g
X(O) = (~}
r(I)
Fig. 4.3 Solution : a) From the block diagram, U(t) = + 7 r(t) - 3X1 - l.SX2
... (1)
From the given state model of the system, - 3X1 + 2X2 + U(t) and
... (2) ... (3)
4. 53
Modem Control Theory
Using (1) in (2),
.
Xi
Solution of State Equations
- 3 Xi + 2X2 + 7 r(t) - 3Xi - 1.5 X2 - 6 X1 + 0.5 X2 + 7 r(t)
... (4)
The equation (3) remain unchanged. Hence state model of the entire system is,
. [-6 0 5] [7]
X =
-~
4
X +
0
r(tl
The output equation remains unchanged. Y(t) = (0 1 I X(t)
Thiu, A=[-:
~:l 8 =fa],
bl
o(s)
Find
[sl
-Ar
[sI-Ar1 s+6 [ -4
(sl-A)
1
C = (0 11
-o·:] s+;,
Adj Isl-Al lsl-1\I s+S [ 4
s+5 [ 4
4
0.SJ' s+6
= (s+6) (s+S)-2 = (s2 +lls+28)
0.5] s+6
---(S+4)(s+7)
s+S 0.5 (s+4)4(s+7) (s+4) (s+7) c> (s} [
[s+S
0.5] s+6
(s+4) (s+7)
s+b (s+4)(s+7)
l J
No need to calculate eA• as X(O} = [~] the ZIR is null matrix i.e. zero. ZSR = L-1 ($ (s) B U(s)I but here input is r(t) and not U(t) to the entire system. ZSR = L-
1(9
(s) B R(s)I and R(s) =
!s
Modem Control Theory
ZSR
Solution of State Equations
[125-05S33c--11 -0.667 c-71] = + 1.333c-71
1-2.33c-4t
= 1-2.33 e-41+1.333
1•
[X (tl] 1
X2(t)
e-7t
This is the required response. Example 4.30 : Etialualt rontrol/JJbility and obseruabi/ity of tht. foll
-~l
a)A=r-~ b) A
= [-~ -~
c) A
= [~ ~
B=[~],
c -.u
-~l · [i ~l B=
~l · [~l · B
=
C
-1)
C = [~
= (0
0 1)
Solution: a) For controllability, ~ = [B : AB] as n
10
1
-21I
=2
1 hencerankof~=2=n
~]
4 .55
Modem Control Theory
Solutlon of State Equations
Hence system is completely controllable. For observability, Q.,
= (CT : ATCT) as n = 2
CT
(1- l)T
= [-~]
=1
:. Hence the rank of Q.,
Thus system is not completely observable. b) For controllability, Q.,
AB
=
A2B
Ii
-11
as n
= 3.
~i ~i ~i
r-~ -~ r: = r=~ 0
0 -3
A (AB]=
r-1~
2 I
0
0
-3
0 -1 0 -I 2 -2 --4 4 2 1 -6 -3 18
= - 7 - 0 hence
-6 -3
0 -2
[11
Q., 0 2 -2
= [B : AB : A2BJ
Ort-2 -4Ol = r-14 Ol8 J -6 -3 18 9
:]
rank of Q., = 3
=n
-6
Thus the system Is completely contrcllable. For observability, Q0 =(CT: AT CT =
AT CT
n
ATCT: A T2 CT) as n = 3 0 -2
0
0] [I1 3]1 = r-1-2 -3-2 ]
0 -3
AT (AT CT]=
2 5
r-1~
-6 -15
0 -2
0
ol r-1
0 -3
i [1
l
-3 = 4 4 3 -1 -2 -6 -15 18 45
4 -56
Modem Control Theory
1 3 -1 -3 1 1 -2 -2 [2 5 --6 -15
+7
'F
l
Solution of State Equations
3 4
4
18 .15
0 hence rank of Q0 = 3
=n
Thus the system is completely observable. 1For cl controllability, Oc = [B : AB : A]B] as n = 3
AB=[~ ~ =!][~}]=[;~] A2B A[ABJ=[~ ~ =~][~~]=[-~] Oc
=
40 0 [ 100 4010
OJ
-30
= - 12000 ~ 0
hence rank of
0
Thus the system is completely controllable. For observability, 0., =[CT: ATCT: A r2 CT] as n = 3
ATCT AT2cT
= [~ -~
_!][~]=[_:]
AT[ATCTJ=[~o
~
~][~]=[-:]
-3 -4
0.,
l
o0 1
o
[~
~ _:]
1
-4 13
11
1 -4 = 1 ~ 0 hence rank of Ou -4 13
Thus the system is completely observable.
-4
=3=n
13
Oc = 3 = n
4 .57
Mod•m Control Theory ,,...
Solution of State Equations
Example 4.31 : Consider the stale modtl,
X
= [~ ~] X{t),
Y(t) = (1
21
X(t)
Find the se! of initial conditions such that lite mode l' is suppressed in Y(t). Solution : f'md e At £or A =
lro
hen cc (s I - A] = [ _2 s s-l -I ]
l] 2 1
[s-1 l] [sl - Ar I = Adj [sl -Al = ~ Isl-A( s2 -s-2
<(s)
s-1 (s+ 1) (s-2)
(s+l) (s-2)
2 (s+l) (s-2)
(s+l)(s-2)
[21' 1/3 -1/3 1/3] (s+1) + (s-2)
--+-(s+l) (s-2)
-2/ 3 2/3 --+-(s+l) (s-2)
1 I 3 2/ 3 (s+l) + (s-2)
L- 1 [o (s)I
CAI
['-' 1 " 1 ., 1 -c +-c 3
CAI
3
-2 c-1 +!c21 3 3
-3c
+3c
The equation is homogcnoous hence the solution is, X(t)
2
Let X(O) = [ ~]
cAt X(O),
~t?-t
3
-2 -t [-c 3
+!e2• 3 2 2t +-c 3
"l
J -t 2 2t -c +-c 3 3
4.58
Modern Control Theory
.!.e-1(2a-b)+.!.c21(a+b)
[
And
Y(t)
~:-t
(1
l
Solution of Sblte Equations
(-2a +b)+~c21 (2a +2b)
2J [~~]
= X1
+ 2X2
.!. c- 1 (- 2a + b) + .!. c21 (Sa + Sb)
3
3
To suppress the mode c21 from Y(t), Sa+Sb i.e.
a
0
-b
Hence the initial state must be, X(O) 111.+
= [~:]
where m " 0
Example 4.32 : Write the state ~uatio11S of tire system sllow11 a11d determine its stale t:Q11trolillbility a11d obsutlQbi/ity. (VTU: }anJFeb.·2006)
Y(s)
Fig. 4.4 Solution : Consider the block ~. Its state diagram is, s(s+l)
X1
Fig. 4.5
4.59
Modem Control Theory From the state diagram.
x,
-
2
Solution of State EquaUons
.
= m--
x,
.•• (1)
2
Now s = .!!. which is a differentiator and input to it is X1. dt
Fig. 4.6 Thus, Xt
X3
z
... (2)
Now the block -2- has state diagram, s+3
. . "2
Fig. 4.7
X2 2
Now m = X2
-
U(t) - X1 = - 2X1
-
3X2
- -2-
3X2 + 2U(t)
... (3)
X3 hence using in (1), Xt 2
.=
but X1
Xt
2X2 - 2X3 -
X3 hence X t
. X:J
. =
=
.
x.
...(4)
X3 and using In (4),
2Xi - 2X3 -
X3 = 2X2 - 3X3
...(5)
Modem Control Theory
4·60
The equations (2), (3), (5) give us required state model. X{t) =
and
Y{t)
For conroUability,
=
0 0 1] [Ol [-20 -32 -3o X(t) + 2J0 U(t) [I 0 OJ X{t)
Oc = [B: AB: A BJ as n = 3] 2
AB=[-~-~ -~l[~]=[-:] A2B Q,,
o2 -6o [0 4
l
A{AB)=[-~ -~ -~l[-:]=[~:J [~ -~ 0
4
l
I~ -24
4 18 = + 32 " 0 hence rank of Q,, = 3 = n -24
Thus system is completely controtlable :
For observability, Q,, = (CT : AT CT : A T CT) as n = 3 ro -2 2
l A
I 0 :. 0 0 [ 0 I
~1
o
-3
I
0
T2 CT AT (AT CT] = [~ o, = [~ ~ -~] =-
2 7- 0 hence rank of Q0 = 3 = n
-3
Thus the system is completely observable.
Solution of State Equations
Modem Control Theory
u11+
4 ·61
Solution of State Equations
Example 4.33: Use co11trol/ability and obserobility matrices to determine whether t* system represmted lty the flow graph shown in Fig. 1 is completely co11trol/ablt and complttely observable. (VTU: JanJFeb.-2.005) -1
y
u
1
Fig. 4.8 Solution : The value of the variable at the node is an algebraic sum of all the signals entering at that node. X3
x1+x2+2u,
X1
-3X1 +X2,
.
A
For controllability,
AB
Xz= X1 -2X2 +X3 +2u
= X1 +2X2-X3 [;3 -:2 ~], B = m, C = r1 2 -JI o, = [B: All :A2B] 1 0~Jm=HJ Y
... n
[:3 -2 -5 [:3 -2 ~] [:3 1 0]0 [10-2 -1 l
A2
A2B
1 -2
[10-4 -2
-5 6 -1 2
Oc
-:2][~] = [-i]
[~ 2 -n -2
1
= -4
6
-:]
=3
Modem Control Theory
Solution of State Equations
4-62
= 3 "' n
I Qcl = -32 .,, 0 hence rank of~ The system is completely controllable.
Q,,=
For observability.
[cT
2c·i]
:ATCT:AT
r; 3 AT[ATcr]..l
u
IQ !
= 0 and
0
1
... n
=3
1 -2
l0
~2: _:] I~!- ~
= 0 thus 2 x 2 determinant is zero
Thus rank of Q0= I hence system Is not observable. n..
Example 4.34 :
Given the time invariant system :
a11d that 11(1) = e-1 and y(t) :: 2 - ate-1 , fi11d Xi(t) and Xi(t). Find also Xi(O) and Xz(O). Wlrat happens A
Solution :
[sl-A]
= ro
a}
LO -1
[~
if a
B = [0} C 1
=0 ?
= (I
-a] s+l , Adj [sl-A]
Adi (sl-A] Isl-A
I
(VTIJ : Ja.nJFeb.-2005)
0)
= [s+l 0
s(s~l)l s+l
Solution of State Equations
4 ·63
Modem Control Theory
L"1[s1-Ar1
ZLR
c'\t X(OI = [:
ZSR
el
a(t -e_e1-1 )]
= [Io
{¢(s}BU(s)}=L-'
1[: l
s(s~L][~] [e: ]j 1
c-1 ... U(s) = s
s+l
(.(C-1
L-1 s2(s+D
c-1
= e-1
s~ -~+ s:1 L-1
... Using partial fractions
[ 1
s(s+ 1)
1 s+I
X(t)
But given At t
X1~t)
y(t)
2-Ult'-l
= 0, y(t) = 2 = X1(t)J1;0 X
= X1(0)+C<X2(0)-e<X2(0)e-•
y(t)
=
X1(0)
.incl X2(0) = 0
2+ate-1 -te-1<:.<(e-t
1(1)
+c-1a.l-c-1a+e-1uc""'
-1)
and
«
l.f ,,_.
Example 4.35 :
X= [~~ y = (1
=
0, XI (I)
=2
Given Ilic state model of a system :
-~Jx+[~Ju
01 x
(VTU: July /Aug.- 2005)
Modem Control Theory
Solution of State Equations
with initial co11d1tio11s X(O) = [~] Determine: i)
Tire stat< trn11sitio11 matrix.
ii) The state transition equation X(t) a11d output Y(I) for an unit step i11p11t. iii) Inverse stale tra11Sillon matrix. Solullon: ii
A Isl-A)
[sr-Ar'
¢(s)
s
-1]
=
Adi [sl-A] _.;._ __ lsl-Aj
•1
(sl-A]
4/3 o(s)
:. Adj (s!-AJ = [ s+S 1] -4 s
s+S
[s+S 1 ] [s+S I ] .. -4 s "' -4 s -s2+5s+4 (s+l)(s+4}
=
s+5 (s+1Xs+4) -4 [ (s+1Xs+4)
- ~s+4
s+l
[ -4/3
+
s+l
1/3 s+l
:l.5R
= ""' X(O)
1/3 s+4
4/3
-1/3 + 4/3
s+4
s+l
5 -c
I iii
I (s+1Xs+4) s (s+1Xs+4)
= c"1 [
1
]
=
3
I
l
... Partial fractions
s+4
...
-5 [ --e 3
-t
--:-c23 8 +~-e 3
- ·ll
l
-·II
s+S ZfR
L-1 {¢{s) BU(s)}
= L-1
l[
(s+ I~ ~s+4)
{s+l):(s+4)]
(s+ I) (s+-1)
(s+I) (s+-1)
[~J m)
Modom
l
4 -65
Control Theory
L"' s(s+l) I (s+4) 1 [ (s+ l) (s+4) ---e I 1 4 3
ZIR
+-e I -4t 12
-t
l
Solutlon of State Equations
1/3 s+4 1/12 = L-1 [l/4 7-;::;:1+ 1/3 1/3 s+I - s+4
l
I -t --c I -41 [ -c
3
3
iii) The property of state transition matrix is, c!)-1(t)
,,_.
a
Cl>(-t)
Example 4.36 : Determine tire co11tro//ability attd oburvability of tire following stair model. 1
x
0 -ll
Y
=
(10
s l]x
Solution : From the given model.
A
= [~ -6
For controllability,
- ~I
(VTU : July/Aug.-2005)
-~J
o,
=
B
=
m,
[a:
C
=
~O 5 I]
AB:A2B]
..... n = 3
Modem
Control
[_~6 -~1 -~J m =L~l
AB
A (AB)= [ ~ -6
o, ~ [~ Now
I
Q< j =
-
Solution of State Equations
4. 66
Theory
0 ~ -11 -6
l
l [ ~1 [-l~l -12
61
-1°2 -1~ 61
84 "' 0 hence rank of Q< = n = 3.
The system is completely controllable. For observability, (4,;
ATCT
[CT :
T CT : A T2cT]
I\
=[~ ~ 0
1
-~ 1 [l~l =[=tl -6
1
-1
-6]
0 J\ T2CT
Q,, = Now
IQ j=96 0
[~o -6
[=t] = [~]
!l
-1 - 1
,eQ hence rank of
-11 -6
0
AT[ATcTJ=[~
J
Q0 = n = 3
The system is completely observable. 111•
A system represented by followi11g state model is controllable but 1101 observable. Slww that tire no11-01'urvability is due to a pole-zerocancelkuion in C(sl-Af1B. Example 4.37 :
X
r~
(VTU: July/Aug.-2005) 0
-6 -n
-6
JO] X+[~]
II
1
Y = (1 I OJ X Solution : From the given mod,cl,
A ;
~ ~].B=[~] 1 [~-6-11-6
,C
= (1
1 0)
Modem
Control
Theory
SoluUon of State Equations
4 ·67
0 ] -1
s2 +6s+11 s+6
Adj[sl-A] [
s+6
Adj fsl-AJ
fsr-Ar1
['' ''"" -6 -6s
s+6 s(s+6) -(lls+6)
-6s
+s(s+6)
-(1 ls+6) s2
l
,; I
Adj fsl-A]
Adi [sl-Aj
Isl-Al
s3 +6s2 +11s+6
[l l O]
-6
T
I
Adi (sl-AJ (s+l) (s+2) (s+3)
Adj [sl-A] [~] (s+1)(s+2)(s+3) 1
1 - (s+l)(s+2)(s+3)
l t[ l 1 1 0 1s [ ] s2
·-
s+ 1 (s+l)(s+2)(s+3)
Thus the factor (s+l) gets cancelled from the numerator and denominator. This shows that as then: is pole-zero cancellation in C(sl-Ar1 B which is the transfer function of the system, the system is not observable.
''*
.=
Examplo 4.38 : Consider the homogeneous equation X
Thefollcwing three solu[t~: folr -e-1
2~-1
AX where A is a .3 x 3 matrix.
[t:~:idiffilerf~e~tial conditions are
' -2e-:u
atailcbl«,
'l-6e-3• J
0
0
i) Identify the initial conditions ii) Find the state transition matrix iii) Hence or otherunse find the system matrix A Solution : i) For initial conditions, put t
= O.
(VTU: Jan.ffeb.·2006)
X(t)
X(O)
= [-~]
l ~- +:·]
X(O
[~~~'
~.. [~~]
ii)
Solution of Sblte Equ.tions
.. ·68
Modem Control Theory
~{~]
l ~·{:] ;,. [.;:l ~- f :1
~o
X(Ol
[·;:
=
1~~ :: 1 A7 A8
r-1] = r".I
:. . I -2
l
A4 A
L
AA36 X(O) A9
i.e
7
i.c.
A1-A2+2A3=-1
. .. (I)
A4 -A5 +2A6 =1
•.. (2)
A7 -A8 +2A9 =-2
... (3)
A1 -2A2 =-2
... (4)
A4 -2A5 =4
... (5)
A7-2A~
... (6)
=O
Solution of State Equations
4 ·69
Modem Control Theory
i.e
2/\ I - bi\ 2 "-6 2A4 -·6A, =IS
... (7)
=0
... (\1)
2A7 -6A8
... (8)
Solving the equations (1) to (9), the system matrix A is given by,
iii) State transition matrix
s [sl-AJ = ~ [
-1
o]
-1
s+S 0
(s+l) (s+S) Adj [st-A]=
Adj (sl-A) --Isl-Ai "
s3 +6s2 +11~+6
s2 +6s+5 -6(s+l)
(s+I) s(s+l)
0
0
[
l s
s2
s3+6s2+tts+6
+~.J
-----
s+S (s+2)(s+3)
(s +2)(s + 3)
-------(s+l)(s+2)(s·•3)
-6 (s+2)(s+3)
(s+2)(s+ 3)
(s+l)(s...2)(s+3)
..Y!:._ __
3
2
(s+2)-
(s+3)
(s+2) - (s+ 3j
-6 6 --+--
.2...._3_
(s+3)
1 s+l
0
0
0
s 2 +5s+6
(s+l) (s+2) (s+3)
(s1-Ar1
(s+2)
0 ]T 0
[
s+l
[si-Ar1
-6($+!) s(,+1)
(s~l)
(s+2)
(s+3) 0
(s+l)
1_.~_
(s+2)
..Y!:._ +-3(s+t)
(s+3)
- .2E...
(s+2) (s+3) 1 (s +1)
... Partial fractions
Modem Control Theory
Solution of State Equations
4-70 3c -21 _ Zc -Jt
0
0
Example 4.39: Obtain the lime response y(t)
of Ille system giMt bekn» by first transforming the state model into a 'Canonica/ model'. (VTU: [an, /F~b.·2006)
I,..
X = [~ ~ -6 -ll
l
Ol X + [~] u, y = [1 O OjX -6 2
u is a unit step [unction and Xr(O)
= fO 0 2]
Solution : First to transform given model into canonical model, find modal matrix M.
(i,1-AJ
= [~ -~. 6 1t
-~ ).+6
l
/
[)J-A]
t..3+6/..2+111..+6=0 (l..+1)(/.+2)(1..+3)
i.e.
1..1 = -1 , i,2 = -2, 1..3
For 1..1
For >..2
= -1, = -2,
= O
= -3
l~ _, '] [c"] ['] ['] f-1
-1-! ,M1= 11 :>
[Al-A]=
[>..I - A]
=
r-20
-1 -2
L6
11
[-3
For >..3 = -3, p,I-AJ = ~
-1 -3 11
M
[ ~1
C12 c13
= -6 6
= -1 1
~l· ~ ·[~:J[-:+Hl ~JMJ =[~::] =[~:]=[+]
~2 ~] = Vander monde matrix
Adj M
Solution of State Equations
4. 71
Modem Control Theoiy
~
-6 6 -2]T [-6 -5 -1] [ = 6 8 2 -2 -3 -1
- 5 8 -3 -1 2 -1
Ad.M Ad.M [3 -' -=-' -= -31 M -2
M-1
M-1AM
[~ 0
Z·S -4
O·S] -1
i.s o-s
~i
~2 =A 0 -3
C=CM=(l
11
I
The canonical state model is,
Z(t)
Y(t)
[~ 0
=
[1 1
-~ ~] Z( 0 -3
t)+[~z]u I
1] Z(t)
X(t) "' MZ(t) i.e, Z(t) "' M -1xct)
~~i [~] [~21
Z(O) = M-1 X(O) = [:13 :: 1 ·5 0·5
.~. l ZIR
Z.SR
... ,,., . r:· L-1 { o(s)BlJ( s)}
0
0
0 .,-3t
=
2
l[
t J
c-•
-2 1 ] = [ -2.,-21 1 c...J•
l
4-72
Modern Control Theory 0
s+I 0
1,-I
0
s+3
s(s+l) L-1
s s+l 1 1 L-1 _!+_I_
_:i_ s{~+2) I s(s+3)
::-21
1 -I -1 1 [ ---eI -Jt 3 3
ZSR
Z(I)
ZTR + ZSR
[;2] [l/•1
0
s+2 0
0
Solution of State Equations
1 l/~_\~~ l
l
=
J
s
-2t
-1 -e 1
s+
1
I 2 -Jt -+-e 3 3 Y(t) = [1 1 1) 1.(1) =
!. +~ c-Jt
1-1-e-21 +.!.+3.e-31 3
3
-c-11
3 3 II*
Example 4.40:
Git~I the system
X=[~'
~1]
x+[~
~] U. Find the input vtcior Ult!
to give the fo//oruing lime respons« :
x111J
6 (1-··-1)
X2(1)
3'--31 -2e-l1
(VTU : July/Aug.·
+6(t -e-1)
Solution : The given time response must satisfy the state equation.
Xt(ll
fi(~(1-c-1)]
= +6e"1
2006)
4 .73
Modem Control Theory
Solution of State Equations
k-JI ]+[~ ~] [6-6c:,1~;:~~-~2c- ]+[~ ~] U [-: ~ J
62~-~~-:~} -c-1
U
41
[33 0)2 u
[~ ~r
=
3
u [
U(t)
1•
O
-1 4 I
l
l
18-12e-• [-6+12 e •1 -6e"31 +6c-l1
[ _12+l2c.1
6-4c-1
-3c_31 + Jc_41
]
•••
J
Required input vector
Example 4.41 : Tiie foll11Wingis the state space representation of a linear systtm wl,_ eigtn wlitcs are -3, -2, -1. (VTU : July/Aug.·2006)
X =[ ~ -6 Givrn that u Solution:
~
~. lJ (X)
-11 -6
+[~]
II
2
= 0, X(O) = /0 0 1{. Find X(I).
:.
{sl-AI
~
[
s
-1
0 6
s 11
-1 0 s+6
l
s2 +6~+11 , Adj {sl-A)
s2 ... 6s+l1
r
Adj fsl-AI
fsl-A)
:. {sf-Ar'=
-b
l
-6s
s~6
n
-6s -6 s(s+6J -{11s+6)
[
s2
l
r
s+o s(s+6) -(11s+6)
= s3+os2+11s+6=(s+1)(s+2}(s+3)
Adilsl-AI
Jsl-AI
z
••
Solutionof State Equations
4. 74
Modem Controi Theory
j
s2 +6s+ II (s+ 1)(s+2)(s+ 3) -6 (s+1)(s+2)(s+3) -6s
s+6 (s+ l)(s+2){s+ 3) s(s+6) (s+l)(s+2)(s+3) -(I ls+6)
(s+l)(s+2)(s+3)
(S+l)(s+2)(s+3)
(s+ ll(s+2)(s+ 3) (s+ l)(s+2)(s+ 3) s~ (s+l)(s+2)(s+3)
2.5 4 1.5 0.5 1 0.5 3 3 ----+. -- 1 ----+------+-s+I s+2 s+3 s+I s+2 s+3 s+l s+2 s+ 3 2 1.5 -3 6 3 -2·5 8 4.5 -0.5 -+------+------+----s+I s+2 s+3 s+l s+2 s+'.\ s+l s+2 s+3 ~ 4,5 2.5 16 13·5 o.s 3 12 9 ----+------+------+-s+l s+2 s+'.\ s+l s+2 s+3 s+l s+2 s+3
2.se··• -4.e-21 +1.5e-31 -2.Sc-• +8e-21 -4.Se.J•
-0.Sc-t +2l'·2t -1.5'.>·JI
2.sc·1 -16e-2' + 13.se-31
o.se·•
O.se·• -e-21 +0.5c~11
l
-4c-21 +4.5e-3'
Example 4.42 : Find Ille lra11silio11 matrix <:(!) for a system wit~ system matrix is .~ivtn by A
=[-
35 -I
-1
1J
by tlze following tech11iqutS :
i)Laplact tra11sfom1 ii) lnfinit« series iii) Cayley-Hami11011
(VfU: July/Aug,·20061
Solution : A= . I)
[-5 3
Solutlon of Stato Equations
4.75
Modem Control Theory
-\] -1
Laplace transfonn method s+S I ] [s+ l -1 ] [ -3 s+ 1 Adi( sl -i\] = ~·' s+~•
(sl-A( (sl-/\
J
(s+;;)(s+1)+3=s2
s+ I = (s+2)(s+4)
Adj[sl-A)
Isl-Al
c'"
+6s+8=(H2)(s+4)
[
(s+2)(s+4)
-0.5 1.5 --+s+2 s+4
-0.5 0.5 --+s+2 s+4
1.5 1.5 [ --• s+2 s+4
1.5 0.5 --s+2 s+4
1.-1 [sl -Ar1=
I
(s+2)-1(s+4) (s+S) (s+2)(s+4)
l
3
-0.se-21+1.se-11 [ l.Sc-21 -l.5e~1
-0.s.c-21
1.Sc-21 -0.Sc-l•
II) Power series method A2
[-5 -I] [-5 -1] [ 22 6] r 22 6] r-s -1] = [-92 -28] 3
-I
.-18
-2
-t = -18 -2
3
3
A2t2
-\
84
A3t3
I+ At+--+--+ 2! 3!
[
I -51+1112 -¥13+ 3t-9t2 +14t3- ...
iii) Cayley • Hamilton meihod Stop 1 :
Find cigcn values.
j).. 1-A!;O
i.e,
l
i..+5 _3
I
I /..+I =Cl
-o.s ... -11]
20
...
... -t+3t 2 -61 28 3 + . 1-1-12+~13
6
.
l •
4. 76
Modem Control Theory i.e. 1.2 +6/..+8=0
Step 2:
i.eA, =-2,
Solution
of Stale Equations
'-2 a-4
Construct R().)
R(A.) = a0 +a1>. =p(I.) Where p(A.)=e'·1
Step 3 : Substituting values of >.1 and ).2
a0 -2a1
= c-21
= c~11
and a0-4a1
Step 4 : Solving above equations,
u = ..!.e-21 _.!.c-41 2
l
, <Xo = 2e-21 -e-41
2
f(A) = R(A)
Stop 5 :
:. cAt=
2c-21
= <Xol +
-c-'u[~
e"' ,,..
ii)
~]+Gc-21
-~c-~1 )[:: ::]
-0.5e-21 +l.5e_,1
-O.sc-21 +O.Sc-·11]
[ t.se-21 -t.Se__,1
t.Se-21 -0.se-4'
Obtain the stale transition matrix usi11g :
Examplo 4.43 : i)
lA
(l
Laplace transformation method n11d Cnyley-Hnmilto11 method.
For lite system describe by, X(t) = [_04 ~4] X(O)
Solution:
A = [~4 _\] X(O)
I) Laplace tansformmethod Isl-A)
[:
(sl-A]
s2 +4s+4
5:14].
Adj[sl-AJ =[s:~1
= (s+2)2
:]
()anJFcb.-2007)
Modem Control Thoory
l
4. 77
Solution of State Equations
s+4
l
Adj!sl-AI _ (s+2}. (s+\)2 ls! -Af - - _.:.'.!.__ (s+2)? (s+s2)2
[
I 2 s+2 + (s+2)2 -4
_!
(s+2)2 1 2_
s+2
(s+2)2
J
I
, . ... Partla! fractions
(s+2)2
Ii) Cayloy • Hamllton method
Step 1 : Find elgen values
ic.I:~
P.1-Al=O
-1 '
A+4j=O
().+2)2 «O i.e, A.1 =-2.
i.e, A.2 +41.. +4 =O
t.2
=-2
Step 2 : Construct R(>.) R(A.)= a0 +a1A. =p(A.) where p(l.)=c41
Step 3 : Substitute 1.1
For
"-z· as it is repeated use, ~ p(I) I 491.2 :~ T>f/.) I di.. • I d).. '" >-1.2 ~e'' l•=I "2 ~~!ao+r'11. .J! d). ,..) tc''
a1
= te-21
I
,),).2
).=•2 =ll1
and a0
= 2te-21 +e·21
4 -78
Modern Control Theory
Step 5; f(A)
= R{A) = Uol+ct1A
:. ~At= 2tc·2t +c-21
[I
I
0]+1c-21 lr 0 lJ 0 I -4 -4
l
tc-21
= [2tc-21 +e 21 -4tc-21 ,,_.
Solution of State Equations
c -21 - 21·c -21
J
Example 4.44 : Compute 1'1 for Ille gil1<'11 matrix :
Al=[~ ~]
CVTU : July/Aug.·2007)
=[~o ~lA =(_6w ~]
;A2
... (1)
This is the property of state transition matrix if A1A2 = AiA1. So verify this condition.
[~ ~) ( _0
(-~ro 6:) 0 6rol [ -6U> 0 J
00 ~)
A2A1 = [ _000
~]
[~
~]
As A1A2 = A2A1, the above property holds good. So calculate cA 11 and c"2' and then multiply to obtain e'". e"t'
(sl-A
1)
isl-A 11 (st-A
r
1
e"'' rsl-Azl
1
.r
L-1(s1-A [s-6 0
0 ] s-6 'Adj (Sl-A2J
(s-6)2
Adj[••-A,j jsl-A11
00
5
rs-6 0
+: . .'..l 0
C1(sl-AS1
[s -ro]
=
,
= [eO61
Adj(s1-A2]
e~'] -ro :]
= [s
0 ) s-6
4 .79
Modem Control Theory
e
,_..
A2t
-I[
~ L
sl-A 2
Solution of State Equations
[coscot sincot]
1_,
= -sin rot cos rot
Example 4.45 : Obtain tlie timt response of lht folluwing V«tor 11111trix differential
equation.
[~J
~s][~~)+[~]u y
oj[~~l
and = [1
= [:
wl~t 11(t) an wtit sttp
input and the initial conditions art X1(0) = Xz(O) = 0. Solution:
A
Thus
[_
06 ~5).B=[~lC=[l
[sl-Af
[s
(sl-A)
s2 +Ss+6=(s+2)(s+3)
[s•-Ar'
As X(O)
c
= [~] ,
6
-1 ] ,J\dj[sl-Af= s+S
Adj[sl-AJ Isl-Al
(VTU: July/Aug.·2007)
OJ [s+S _6
:]
[ "''
_ (s+2)(s+ 3) (.+2):.+3) ]-~·) -6 (s+2)(s+ 3) (s+2)(s+3)
cAt is not required.
ZIR
[~]
ZSR
1.-1 {<>(s)BU(s)}
L-'![(s+;~~+3) (s+2)(s+3)
... U(s) =!for unit step s
(s+2)~s+3)1 (s+2)(s+3)
(~]
m]
Solution of State Equations
4 -80
Modem Control Theory
X(t}
[1
Y(t)
Y(t)
,..,.
Example 4.46 :
T(s)
Y(s)
=
oj[Xi(t)] ~ X (t) Xz(I} I
.!__!_
<,.-21
6 2
+!.
c-31
3
Oblai11 lire observabl< phase variable stale model ~f 'l'T,
b ,s3 +b s1 +b,s +ll
=-
=---3.Drma tire :1ig11al flow graph of T(s).
= •.!.3--1~7-
ll(s}
s .,.,,Is- +n2s+nJ
(VTU: July/Aug.-2007)
Solution : The given transfer function Is Y(s} _
b0s3+b1s2+b2s+b3
U(s) -
s3 +.11s2 +d2s+a
:. (s3+d1s2
+a2s+a
3 3
l Y(s)=b s
0 3
+b1s2 +h2s+b3(U(s)I
Divide the equation by s3,
la Y(s}-b U(s))+_2_ fa Y(s)-b U(s)J:
Y(s)-b0U(s}+!f~1Y(s)-b1U(s)]+_2_2 S
Y(s} = b0U(s)+1(-a1
s
2
2
Y(s)+ b1 Ll(s)]
3
3
0
s3
S
+.}, 1-a 2Y(s)+b2U(s)) s·
+-\- f-.i 3 Y(s)+ b3 U(s)I s
... (!)
Now define the state variables as,
~lb1U(s)-d1Y(s)+X2(s)l
X3(s} X2(s)
:
Hb
2U(s)-a2
Y(s}+X1(s))
(2)
(3)
Modem Control Theory
>••
Solution
4·82
of State Equations
Example 4.47: Obtain tht contro/lablt p~ variabltformof ti"' transferfunction, Y(s) b0s3 +b1s2 +b2s+b3 T(s)
=-
U(s)
+a2s+n3
s3 +a1s2
(VTU: July/Aug.·20071
Solution: The given transfer function can be written as,
.::.<~
U(s)
Y(s)
Where
Y(s) Y(s)
U(s)
s3Q(s)+a 1s2Q(s)+a 2sQ(s)+a 3Q(s)
-a 1s2Q(s)-a2sQ(s)-a
s3Q(s)
3Q(s)+
U(s)
Define the state variables as, x.(s)
Q(s).
sX1(s)
X2(s)
sX2(s)
X3(s)
Taking inverse Laplace transform,
.
X1
=
.
X2, X2
=
X3,
Substituting selected state variables in the equation (1), s X 3(s) i.e,
.
X3
-a
1X 3(s)-a 2X2(s)-a
3X1(s)+
U(s)
... (1)
Modem Control Theory
Solution of State Equations
4. 83
From (2), Y(s)
= b0U(s)
+(b1-a1b0)X3(s)+(b2
-a2b0)X2(s)+(b3
-a 3b0)X1(s)
Taking inverse Laplace transform, Y
(b3 -a 3b0] X1 +(b2 +a 2b0] X2 +[b1-a1b0]
C
[(b3-a3b0){b2-a2b0){b1-a1b0)],
nus is controllable
X3 + b0U
D = (b0)
canonical form of the given transfer function.
Review Questions 1. WNzl is homogtn<1>us and nonhomognreous sliltt "'lualion 7 2. Otfi•t statt transition 1tmlrix usi•g classiail mtlhod of obtaining soluNon. J. What is zero input response and uro slate response ? 4. Obtain lht romp/tit solution of nonhomogtntous slalt equation using timt domain method. S. State tht imporlance of state transition matrix. 6. State tht various proptrlits of slalt transition matrix. 7. Obtain llrt solution of nonhomogtnt0us •lat< tquatian using Laplace lmnsform met/rod. 8. Whllt is resolrxtnt matrix ? 9. 10. 11. 12. 13.
i
Explain IAplare transformmtthod of obtaining 1. Expwn p
a[-~
-~1
b. [:
~] Ans.:
[ •' 0] le
I
e
I
I
Ans.:
f
cos./2 t Tzsin ../21 -.fi. sin ./21 cos./2 t
1
I
Modem
Control
4 .'a4
Theory
14. Oblai11 llU! ltomogenoous solutio1t of Iii• tqualiD1t ')({I} a. A= [ ~94
~1
Solution of State Equations
=A
X(I) wlttrc
nnd XIO) = [:]
l
-21
ti=['2 -ll-4J and X(O) = [ 0 1]
l I
I
[
b.
6 -71
-5e +5• Ans.: 7 ·ll --e
s
c
J •-e s 12 -71
c-3.$6!]
1.11 0·341 -0.11 Ans.: [ 0.48c0•56' -0.48c~'·56'
15. Find the output of tlrt systtm having state model.
X(t) = [ -12 0 1 ] X(t} + lPJ Ult) -7 I - l I Xltl
and Y(t} = [I
.
. . strp
Tirt inpu! U(I) ts unit
( = [10]0
mid X 01
(Ans.:Ylt)=~
3
... ~e-31 3
145e,1)
16. Show that tlzc following syst
=~ -:1
XII}=[-~ l and Y(I) =(I
Xfl) + [~] U(ll
0 -lJ
1
1
OJ X(tl
17. Find ti" solution is, X(I) = [~
-~J
X(t) with X(O) = [~]
18. Find the solution of,
xm =[~ -~] xw +[~]
Ult!
and Y(I) = (1 0/ Xltl if X(OI = [ ~] and syst
20. 21. 22. 23.
Explain Ka/man's lest for determining state controllability. Explain Cil!M!rt's lest for dtlennining stat« controllabilily. What is output controllaoi/ity? How to
Modem Control Theory
4. 85
Solution of State Equations
24. Dtfint a11d op/am concept of o/Js1rJ11bility. 25. Explain Ka/man's (PSI for tfetermi11i11g comp/tit 01Js1•rr.111bility. 26. Expl•n• Gilbert's test for detcm1i11111g complete obs(rt•abitrty. 27. F.1KJ/uatt tire ro11trt1llability of"" following sy5tems witlr,
a.A=[-2 -I01r ll=r31Cr b. A=
0} --121l5;
-1
fl 0
-2
r-1 c, A=
=fll]0
LI J
0
I
0 c 0
-1
H
~1
H =[~
~l
3 0J (Ans.: a. Controllable. b. Controllable c, Not controllable)
0 -2
28. £untuatt llu: obseroobility cf tire folfowi11g sy:
fl o0 0 2,J I· r 2
=[ ~
b. A=[-~ o}. < = _2
c,
A·[~
(O
I 0] c ~
0 -6
-11
I ,
-6
~
~]
l]
(4 5 1)
(Ans.: a. Observable b, and c, Not observable)
DOD
(4. 86)
Pole Placement Tuchnique 5.1 Introduction This chapter deals with the problem of pole placement. The pole placement technique or pole assignment technique is a method for design of control system which ls discussed in this chapter. If it is assumed that all the state variables for a given system arc measurable and available for the feedback then it can be proved that for a completely state controllable such system, its closed loop system poles can be placed at any desired location through state feedback by means of an appropriate state feedback gain matrix. The desired closed loop poles are firstly evaluated based on transient response and/or frequency response requirements such as say damping ratio, speed, bandwidth as well as the steady state requirements. Let us consider that the desired closed loop poles arc to be located at s=a1, s=u2 ......... s=a n- Then by properly selecting gain matrix for state feedback. il is possible to locate the closed loop poles al the desired locations provided that the original system is completely state controllable. In the following sections, we will consider the control signal to be scalar. It can then be proved that the necessary and sufficient condition that the closed loop poles can be placed at any arbitrary locations in the s-plane is that the system is completely state controllable Le. it is possible to transfer the system state from any initial state X(t0) to any other desired state X(t1) in a specified finite time interval (t1) by a control vector U(t). The required state feedback gain matrix can then be evaluated using various methods.
In this chapter, It is considered that the control signal is a scalar quantity and not a vector quantity. Under such case, the mathematical aspects of the pole placement technique is complicated as the state feedback gain matrix is not unique.
(5 - 1)
Modern Control Theory
5-2
Pole Placement Technique
5.2 Pole Placement Design The conventional method of design of single input single output control system consists of design of a suitable controller or compensator in such a way that the dominant dosed loop poles will have a desired damping ratio ~ and undamped natural frequency IDn. The order of the system in this case is increased by 1 or 2 if there arc no pole zero cancellation taking place. It is assumed in this method that the effects on the responses of non dominant closed loop poles to be negligible. Instead of specifying only the dominant dosed loop poles in the conventional method of design, the pole placement technique describes all the closed loop poles which requires measurements of alJ state variables or inclusion of a state observer in the system. The system dosed loop poles can be placed at arbitrarily chosen locations with the condition that the system is completely state controllable .. This condition can be proved and the proof is given below. Consider a control system described by following state equation ... (1)
x=Ax+Bu
Here x is a state vector, u is a control signal which is scalar, A is n x n state matrix. B is n x l constant matrix. The system defined by above equation represents open loop system. The state x is not fed back to the control signal u,
u
Let us select the control signal to be u Kx state. This indicates that the control signal is obtained from instantaneous state. This is called state feedback. The k is a matrix of order Ix n called state feedback gain matrix. Let us consider the control signal to be unconstrained. Substituting value of u in equation 1 . Fig. 5.1 Open loop control system
.
x = Ax+B(-Kx)=(A-13K)x
=-
. .. (2) The system defined by above equation is shown in the Fig. 5.2. It is a dosed loop control system as the system state x is fed back to the control system as the system slate x is fed back to control signal u. Thus this a system with state feedback.
Fig. 5.2
Pole Placement Technique
5-3
Modem Control Theory The solution of equation 2 is say x(t)
=
e,
x(O) is the initial state
... (3)
The stability and the transient response cha.racteristics arc determined by the eigcn values of matrix A - BK. Depending on the selection of state feedback gain matrix K, the matrix A - BK can be made asymptotically stable and it is possible to make x(t) approaching to zero 3$ time t approaches to infinity provided x(O) ,. 0. The eigen values of matrix A - BK are called regulator poles. These regulator poles when placed in left half of s plane then x(t) approaches zero as time t approaches infinity. The problem of placing the closed loop poles at the desired location is called a pole placement problem.
5.3 Necessary and Sufficient Condition for Arbitrary Pole Placement Consider a control system defined by following state equation. x = Ax+
Bu
Let the control signal selected be u x
=
Ax - BKx
. .. (1)
=-
Kx
= (A - BK) x
... (2)
The solution of the above equation is given as, x(t)
=
.,(A-6 K)I . x(O)
... (3)
The eigcn values of the matrix (A - BK) are nothing but desired closed loop poles. The necessary and sufficient condition for arbitrary pole placement is that the system is completely state controllable.
If suppose the system is not completely state controllable then there arc eigcn values of matrix A - BK that can not be controlled by state feedback. According to Kalrnan's test for state controllability, the system is said to be completely state controllable if the rank of the composite matrix Qc is n.
Q,, = [B: AB:A2B:
An-ls]
Let us suppose that the system is not completely state controllable. The rank of the controllabiUty matrix is less than n, rank(B:AB:A2B:
:A0-1B]
= s
Thus there are s linearly independent column vectors in the controllability matrix. Let these s linearly independent column vectors as c1, c2, c•. Also let us select n - s additional vectors t,.1, ts.2, ....... t,, such that
5.4
Modern Control Theory P
=
(c1:c2:
Pole Placement Technique
:c.:t1+1:ts+2:
..•..•...
:t0)
•.••.•••••.
Tile above matrix will have rank n By using matrix Pas the transfomation matrix, let us define Now we have
r·1AP=A.
p·1B=B
AP = PA
(Ac1: Ac2:
: Acs:Als+l:At.+2:
=
......•.... : At11)
[c1: c2: ..••...•. : c9:l_.1:t8•2:
.•....•....
: t0)
A
As there arc s linearly independent column vectors Ct- ~- •.••.••• c5 using Caylcy-Hamilton theorem vectors Ac1, Ac2.•..•.•..• Ac, can be expressed in terms of these s vectors. Ac1
au c1 + a21c2 + ·········· + as1 C.
Ac2
a12 c1 + a22c2 +
Ac,
1115
c1 + aisc2 .+
+ a52
c.
+ ass '•
Thus we can write
I
[Ac,: Ac2 :······:Ac,:
At,.,: Al,.2:
:At,,)
a82111 ····················821 ················ a1,
:·.::::::·.·.:·.·.·.:·.~.,., n
,:,a821',',', I I I I I
= [c, : Ci: . . .. -: c, : '•·• : t,.2: ..... t,,J
.
' ' . aan -----------------r----------------' ······················ : 8-•ts•1···· ···· 8.s•tn '' a~t
a,, : a~1•1
'' '' '
I
............. ........
I
; ;
:
:; : : ;:
: a"'., ?'
•
8m
5.5
Modem Control Theory
Polo Placement Technique
Let us define
~
u
a,,
a1,
a,1•1
a1n
321
a21
a2••1
azn
a;1
a5$
a~,.,
acn
]·
~
,~.. ..,,. zcromatra
•
~
,
···~~
b.,
~
We have,
(Ac,: A~ ....... : Ac,:
Ai..1 ; ....... : A1,,]e(c1: ~:
...... : C,,:
1,,,: 1>•2: ....... : •n]
[.:'!~-~-~'.L] 0
Hence we have,
'
Az2
Modern Control Theory Now we have,
Pole Placement Technique
5-6
B
PB
B
(c1: c2:
........•
:
c,:t •._1:t
2:
:
t.,]B
Now B can be eXJ>resscd in terms of s linearly independent column vectors
c;. Therefore we get,
c1, ei,
B = b11c1 +b21c-2+
b51c,
'o· I 0 b~l b11]
Let us define
bs1
We have
Now we define
u=
[-~n
K
We have
jsl-A+BKj
1r-1(sI-A+BK}Pj sKPj
jsr-r-1AP+r·1
But
"' [ ••••• An :1. •••• A12 ] P·I AP= A= 0 : ~
J
Modern
Control
Pole Placement Technique
5-7
Theory
Substituting
=
jsl-A+BKJ
js1-fi.+i3xkl
In the Identity matrix I, we have n rows and the diagonal clements as 1. nus matrix can be split as we have the matrix A.
o
1
0
0
0
0
1······· . "•.....
0 I•
I
::·.,
:
---------------i--------------0 0 1 ::::.:::: 0
0
......................
I
'
....................
0
.: ..,,
a , o
'' '' 1
':
0
:; : :
0
0
--------------+--------------...................... 0 ·: 1 :::·············· .. ···
.. I
0
:
0
'·, ••
· · ....
I
..........
~
0
"•,,
:·:::1
5-12
Modern Control Theory
Pole Placement Technique
The above equation can be rewritten as,
B
= Tm=QcWm [ B:i\B:i\ 2B]
But Q,,
r: ~ 1
[B:AB:A
T [~]
a11 0
s) [~] = 8
2
B
1T·m
r-18
r
r1B
= [ B: AB: A 2o)
m
=
The results which are proved for n = 3 can be extended and generalised so Iha! we have equations (2) and (3) as given below. 0 0
1 0
0··· 1
0 0
-1
T AT= 0 0 o.... -an -an-1 -an..2
1
-a,
5 ·13
Modem Control Theory
Pole P,lacement Technique
0
9
Now consider the followingequation
~ = r-1ATx+l1Bu The above equation can be transformed into the controllable canonical form if the system is completely state controllable. The state vector x can be transformed into state vector x by use of transformationmatrix T = MW. Now we will select a set of desired eigen values as a1,a2, a0• Then the desired characteristicequation becomes, ( s-a1
)( s-a2 )
(s-an) =sn +o1sn-l +
on-ts +on
.:o.
Let When u =
...(7) ... (8)
-f<x = -KTx
.
x
is used to control the system having the equation
(r1AT-11BKT) x
The characteristicequation is,
!sr-r-1AT+r1BKTI
c
o
This characteristic equation is the same as the characteristic equation for the system defined by ; =Ax+ Bu when u; = - Kx Is used as· a control signal. This ts proved below.
; = Ax+Bu=Ax+B(-Kx)=Ax-BKx=(A-BK}~.
Pole Placement Technique
5· 14
Modem Control Theory
The characteristic equation for this system is,
=I
!sl-A+BKI
Now we will simplify canonical form.
r1(sl-A+BK)Tl=ls1-r1AT+r1BKTI the characteristic equation of the system in the controllable
I
0 0
Iisl!
I sl-T-1AT-r1BKTI
0
0 0
0
+:
I
0
I
""""
I
9 [µ11µ11-1 ...... µ
0 -an-l
-1 0
iJ
I
I
"""1
0 0
l.L This is the characteristic equation for the system with state feedback. It is same as equation (7). 0I
d I+µ I
o.z
a2 +µ2
sn
an +µn
The above equations can be solved for µ; 's and substituting in equation (8) KT 1(
[Ii nl' n-1 ....... µ l) ... (9)
If the system is completely state controllable, all eigen values can be arbitrarily placed by choosing matrix K according to above equation no (9). nus is the sufficient condition.
Modem Control Theory
5-15
Pole Placement Technique
Thus the necessary and sufficient condition for arbitrary pole placement is proved and it requires the system to be completely state controllable. For a system which is not completely state controllable but is stabilizable, then the total system can be made stable by placement of closed loop poles at the desired locations for s controllable modes. The rest n - s uncontrollable modes are stable which shows that the total system can be made stable.
5.4 Evaluation of State Feedback Gain Matrix K Following are the different methods for evaluation of state feedback gain matrix K i) Using Transformation Matrix T ii) Using Direct Substitution Method
iii) Using Ackermann's Formula
5.4.1 Use of Transformation Matrix T Consider a system defined by a stale equation
.
x =Ax+
Bu, Let the control signal be
u = - Kx . = Ax-llKx=(A-BK)x
x=Ax+B(-Kx)
The state feedback gain matrix K forces the eigen values of (A - BK) to be a 1• a2 ..•••...••..... n". Then the desired characteristic equauon is ( s-tt.) ( s -n2 ) ..•...•..... (s-tt0) # s" + 61s0• 1 + i>2s0•2 + + 611_ 1s + 60 = 0 The desired cigen values 1x1,u2 Step I:
Step II:
The characteristic polynomial for matrix A is given by J
Step Ill:
n0 can be obtained from following steps.
The controllability condition is checked for the system. For the system which is completely state controllable following steps can be used.
sl-A
I
= s" +a 1s11-1 +a2s0-2+ From this values a 1,.12
+i111_1s+a0 a" can be obtained.
The transformation matrix T is determined which converts the system state equations into controllable canonical form. For system equations which arc already in controllable canonical form T = I. The system state equations are not required to be written in controllable canonical form but matrix T is found from. where Q., ts controllability matrix
Modem Control Theory
5 -16 0n-l 011:-2
Pole Placement Technique
0ra-2
a1
I
°n:-3
1
0
w 0 0 0 0
1
0 Step IV:
Using the desired eigen values which arc desired closed loop poles, the desired characteristic equation can be obtained ( s-«1 )( s-«2 ) (s-an) = s" +l>1sn-l +l>2s"-2+ l>.,_1s+l>n· The values of li1, 1i2 ,
Step V:
li0 can be obtained.
The state feedback gain matrix K can be obtained from following equation K = (l>n-an
lin-l-an-t·············li1-a1JT-I
5.4.2 Direct SubstitutionMethod For low order system with n less than or equal to 3 matrix I< can be directly substituted inlo the desired characteristic equation. If suppose n = 3 then the state feedback gain matrix K is I< = [1<1 1<2 K3) The above matrix is directly substituted in the desired characteristic equation Isl-A +BKI and equated to ( s-a1 )( s-a2 )( s-a3) i.e, lsl-A+BKI
=
(s-ai)
(s-a2)
(s-u3)
The both sides of the characteristic: equation are polynomials in s. Hence the co-efficient of similar powers in s on both sides can be equated so as to get the values of K1, K2 and K3. This method is restricted for n = 2 or 3. For higher orders of n the calculations become complex. It is not possible to determine matrix I< if the system is not completely controllable.
5.4.3 Ackermann's Formula Consider a system defined by state equation x =Ax+ Bu. let u = -Kx be state feedback control. Assuming' the system to be completely state controllable nnd the desired closed loop poles are assumed at s = a1, s = u2, ••••••••• s = a0• Using u = - Kx in state equation
5 -17
Modem Control Theory
Lei
Pole PlacementTechnique
A-BK
iix
x
The desired characteristic equation is, isl-A+
UKI
By cayley-Hamilton theorem, every matrix satisfies its own characteristic equation.
o (i\)
=
i\ n +o1A
n-1+
+011_1A +S.,J = 0
The above equation can be used for deriving Ackermann's formula. Let us consider the case of n .r 3 for simplification of derivation. We have following identities
A
.i\2
A-BK (A-BK)2=A2-2ABK+B2K2 A 2 -ABK-ABK+B2K2
=A
2
-ABK-BK(A-BK)
A2-ABK-BKA
j\3 I
(A-BK)3 =A 3 -A 2BK-ABKA-BKA 2 I
No~" consider
i\ 3 +o1.i\
2 +o2i\ +031
A 3 -A 2BK-ABKA -BKA
2
+o1( A 2
-
i\BK-BKI\)
+ o2(A -BK)+o31 A 3 -A 2BK-/\BKA-BKA
2 +o1A 2 -61ABK
- ll1BKA +l>iA -o213K+631 li3l+o2/\ +ll,A 2 +A 3 -ozBK-6,A13K - 61BKi\-A 2BK-ABKA-llKi\ We have,
and
= 63l+o2A+61A2
+A 3
2
= O (for n = 3)
Polo Placement Technique
5 -18
Modern Control Theory
substituting the above equations,
o(.i\.)
0 (A )-S2
BK-o, AllK-o,
UKA -A 213K-AllKA -llKA ~
0
B111
o21lK+1\113KA
6( A)
u(o2K+o1KA
+ 13KA 2 +61 ABK+;\13KA
+A 213K +KA 2)+An(o1 K+ KA)+,, 2nK
dl(A)
11, the system is completely state controllable, the inverse of the controllability matirx
Q., ~ [ 13: AB: A 2 B] exists. Multiplying bolhsides by inverse of controllability matrix.
Prcmultiplying both sides of above equation by [O 0 I) we get
10 0 l
I [13:Al3:A2Br'¢(A)
~ [O
0
1jro2K+ot10\ ~Ki\2] I StK+KA
IL
K
K K
to
0 1)[13:/\B:A213]o(A)
The above equation gives the required state feedback gain matrix K for arbitrary positive integer n, K
=
(0 0
[
,
0 I] 13:AB:A·ll:
:A0•1n
1·1 O(A)
This equation is called Ackermann's formula for the evaluation of state feedback gain matrix K.
Modem Control Theory
5·19
Pole .Placement Technique
5.5 Selection of Location of Desired Closed Loop Poles The locations of desired closed loop poles must first be chosen in the pole placement design technique. The commonly used method for such selection of poles is based on the experiences obtained from root locus design of placing a dominant pair of closed loop poles. The other poles are selected in such a way that they arc away from dominant dosed loop poles in left half of s plane. If the dominant closed loop poles are placed away from jlo or imaginary axis then the response obtained from the system is very quick but at the cost of system becoming nonlinear which should be avoided. The other method of selection of poles is Ute quadratic optimal control technique. In this method, the desired closed loop poles arc determined in such a way that it balances between the acceptable response and the amount of control energy required. The high speed response indicates that large amount of control eneri,'Y is required. In addition to this a larger and heavier actuator is required which increases the cost.
l '*
Example 5.1 : Consider the systtm defined by x = Ax+ll11
totier«
A
By using the state feedb
=-
Solution : Let us first check the controllability matrix of the system. The controllability matrix, Q,, =
[n:AB:A2n] B
[H AB=[~ ro
A2 =A.A.
A2B
I
-1
01[0
l~t
0 0 1 -5 -6J -1
[~
0 -5 +29 31
1 0
o] = ro 1
-1
-5
-6
c+6
~6] [~]=f ~] 1
l 31
0
~6]
-5 29 31
Modem Control Theory
l
o o ~
=
0
[1
Polo Placoment Tochnique
5-20
I -6 1 . -6 31
The rank of matrix ~ is J. ·111e system is controllable and hence it is possible to place the poles arbitrarily. Now the state. feedback gain matrix K can be obtained by using any of the three methods discussed previously.
ro~
Method 1 : Using tra:sf:rm[
mat']ix T
-1 -5 -6 The characteristic equation for the system is,
s ls(s+ 6) +5 J+ 1[1]
s [s2+6s+s]+1 s3 +6s2 +5>+1 =O Comparing this equation with s3+a1s2+a2s+,t3 :. a1
= 6.
a2
=
5, a3
=
=0
I
The desired characteristic equation is , (s+ 1-j2)(s+ 1+j2)(s+10) s3 +tos2
=
-t- 20s+2s2
[Cs+ 1)2 +4] (s+ IU) =[s2 +2s+5] +Ss·~SO
s3 +12s2 +25s+S(l
s,
=12, 02 =25. 03 =50 K
= fo3-a3
62-a2
o1-a11rl
As the given stale equation is given in controllable canonical from T = I.
11
r
K
Lo3-a3
K
[49 20 6)
1>2-.i2
-a.J
1>1
=[50-1
25-5
12-6]
Is+ 10(
Modem Control.Theory
5-22
•no
o -6 0 -5 =[~1 29 31_ -1 -
-5 29 -1~9
r-16 -31
[~l or I
0
1
1
[o
- 0
-5 -6
¢(A)
0 -5
-1
r-1
6 -31
Pole Placement Technique
:]
0 -5
~l -1571
:][~ -5
I
~] =[~1
0 -5
0
~ j+1{~-6
29 -149 -157
18
-5 29
0 -6 1 -5 29 31
~6]+25[ ~ 31
[ ·19
20 6 ] 19 -16 -128 74
115
[ B: All: A 2Bf1
0 1
K "'
l -6
i·lr 18
49
31 l-128
-6
20 6]
19 16 74 115
~ ~] [ ~: ~~ -~6} 0 0
-128
[1 0 0) [ ~: ~~ -~6]
-128
[49 20 6]
74
115
l
74 115
1
0 -1 -5
~]+so[~
0 1 0
~]
Modem Control Theory ,._.
Example 5.2 : It is desired to place tire closed loop poles of tire followi11g system al s = -3 and s = - 4 by a slate feedback controller witlr Ille control u = - Kx. Detennine tire state fttdbackgain matrix K a11d the control signs]. (VTU : Aug.-2005)
y
.
Solution : Equating x
e
Pole Placement Technique
5-23
=
[1
O]x
the state equations
with standard equations
with standard form
Ax +Bu and y=Cx we get
B
c = [1
= [~]
0)
Let us check the controllability matrix of the system (B :AB) AB
[~1 ~3][~] =[:6]
o, = [~ :] The rank of above matrix is 2. Hence the given system is completely controllable. The given equation is not in the controllable canonical form because B is other than[~] we use method of transformation matrix T to get state feedback gain matrix. T
Oc·W
W=["i 1]=[3 11 1
consider A
[-t0 -3I]
; jsl-AI=
I
s1
O
-1 s~ 3
1
I=
OJ
s(s + 3) + 1
= s2 + 3s+ 1
Equating with s2 + a1 s + a2 we gel
5 ·25
Modem Control Theory
Pole Placement Technique
5.7 Concept of State Observer In case of state observer, the state variables are estimated based on the measurements of the output and control variables. The concept of observability plays important part here in case of state observer. Consider a system defined by following state equations x ;
Ax+Bu
y
Cx
x
Let us consider as the observed state vectors. The observer is basically a subsystem which reconstructs the state vector of the system. The mathematical model of the observer is same as that of the plant except the inclusion of additional term consisting of estimation error to compensate for inaccuracies in matrices A and 'B and the lack of the initial error. The estimation error or the observation error is the difference between the measured output and the estimated output. The initial error is the difference between the initial state and the initial estimated state. Thus the mathematical model of the observer can be defined as, x
Ax+Bu+K.(y-Cx) Ax+Bu+K.y+KeCx (A-K.C)x+Bu+K.y
x
x
Here is the estimated state and C is the estimated output. The observer has inputs of output y and control input u. Matrix K0 is called the observer gain matrix. It is nothing but weighting matrix for the correction term which contains the difference between the measured output y and the estimated output CX. This additional term continuously corrects the model output and the performance of the observer is improved. Full order state observer The system equations are already defined as x ;
Ax+Bu
y
Cx
The mathematical model of the state observer is taken as,
Polo Placement Technique
5-26
Modem Control Theory
x = Ax+Bu +K.{y-Cx)
..
To determine the observer error equation, subtracting equation of x-x
i from
~ we gel
[Ax+Bu]-(Ax+Bu+K.(y-Cx)] Ax-Ax-K,(y-Cx)
A(x-x)-K0(Cx-CX)
..
A{x-x)-K.C(x-x)
x-x
(A-K.C) (x-x)
x-x e e : (A -K,C)e
Let
...
:. e=x-x
The block diagram of the system and full order state observer is shown in the Fig. 5.3.
>--------Y u
-----------------------·x·------------
Full order stateobsefver
Fig. 5.3 The dynamic behavior of the error vector is obtained from the cigen values of matrix A-K.C. If matrix A-K.C is a stable matrix then the error vector will converge to zero for any initial error vector e(O). Hence x(t) will converge to x(t) irrespective of values of x(O) and x(O). II the eigen values of matrix A-K.c are selected in such a manner that the dynamic behaviour of the error vector is asymptotically stable and is sufficiently fast then any of the error vector will tend to zero with sufficient speed. If the system is completely observable then it can be shown that it is possible to select matrix Ke such that A-K.C has arbitrarily desired eigen values. Le, observer gain matrix I<• can be obtained to get the desired matrix A-K.C.
5. 29
Modem Control Theory
Herc
T
Q.,.W = [
c• : NC·
Pole PlacementTechnique
: (N)2
c· :
: (N)"-1 c- I w
For the original system the observability matrix is,
I
c-.
: (A•}n-l c·
NC' T T'
w
As
rr-r'
I =
N
NW
w•N• = WN•
·: (AB)T =BT ·AT
w• (WN'r1
Taking conjugate transpose of both sides of equation I Bn -an On-1 ._-an-I
r Herc
w
N
i.l n-1
la'rz
a n-2 ''n-3
al
1
1
0
~
~
I ,
l
·~i
I
c- : NC' : (N)2 c- :
0
: (N)n-l c· I
The second method is based on direct substitution method to get the state observer gain matrix K,. This is applicable for system having low order. TI1e matrix Kc is directly substituted in the desired characteristic polynomial. Let us consider K~ :
r~:~Jl Ko3
Substitute this matrix Into desired characteristic polynomial.
By equaling the like power coefficients of s on both sides. the values of Ke1• K02 and be determined. This method is restricted for value of n at the most 3. The third method is use of Ackermann's formula. For the original system defined as
I<,,3 can
x=Ax+Bu and y = Cx we have already derived the formula known as Ackermann's formula given as,
5. 30
Modern Control Theory
K = [O 0
I]
Pole Placement Technique
or
[Il:Ail:
1\11•1
1'i)
(A)
Now let us consider the dual system z
A"z + c·v
y
n•z
The Ackermann's formuln for dual system is modified as K
= [O
0 .
1 ]I C' : A"C' :
: (A')n-l
c•
r' O (A ')
Both observer gain matrix, K0 = K'. -1
r :.Kc= K' = q>(A ')'
c~
l ;
()
c
0
C.A
0 0
= q>(A)
... (II)
CA n-2
Ci\ n-1
LCA"-l
The desired characteristic polynomial lor stale observer is
= ( s-u1
o(s)
)(
s-u2
) ......•......
(s-rtn)
Herc a 1, u2 •••..•••..•• n11 arc desired cigcn values. Equation (ll) is Ackermann's formula for determination of observer gain matrix K,.
5.11 Selection of Suitable Value of Observer Gain Matrix
K.
The feedback signal through the observer gain matrix K,, acts as a correction signal to the plant model to account for the unknowns in the plant. If the unknowns involved arc significant then the feedback signal through matrix Ke is comparatively large. But if the output signal is affected by disturbances and measurement noises then output y is not reliable and feedback signal through matrix K, should be small. The effects of disturbances and noises involved in output y should be neatly studied while evaluating matrix KeThe observer gain
is dependent on the desired characteristic equation 0. The selection of a1,u2 u., is many times not unique. But normally the observer poles should be two to five limes Caster than the controller poles. TIUs makes the observation or estimation error converging to zero quickly. The observer estimation error decays two to five times faster than does the stale vector x. Puc to this foster decay of observer error, the system response is dominated by controller poles.
(s-u1)(s-u2)
matrix
(s-o.,.)
=
Modem Control Theory
Pole Placement Technique
5 ·31
With the appreciable noise produced by sensor then the observer poles can be selected to be slower than two times the controller poles. The bandwidth of the system will then be less and noise wiU be smoothened. The system response is dependent on observer poles in such case. The observer poles arc located to the right of controller poles in left half of s-plane so that the system n.'Sponsc is influenced by observer poles. While designing thc state observer, it is required to evaluate various observer gain matrices Ke based on number of various desired characteristic equations. For each of these different matrices K. r the simulation tests can be run to check the overall system performance. Based on the performance, the best possible K. om be selected. In many systems, the best value of K0 is obtained after a compromise between fast response and sensitivity to disturbances and noise signals.
>'*
Example 5.3 Co11sider the system rrpresenled by
x =
r:
y " ll
Design -5. Solution:
A=
11
-~l
!J
x
+mll
0 O].r
full order observer s11cl1 t/1111 the observer eige11 oalues nre nl -2 ± j2./3 mrd (VTU: Aug.·2005)
r~
1
c = [1
0 -6 -11
Lei us check the rank of dual system of the original system I C : A'C· : {A')2 C" : : (A'ln-1 c· I Herc n
=3
:.( c-.
NC·: (A')2
c· {A')2
~ m;
c·
J
A'C•
A'. A'=
1[~
=[~ 0 0
()
0
~IJ m =m
~][~
0 0
-6]
[o
-6 -11
~· = ~
-6
~] 25
0 01
5 .34
Modem Control Theory
3
(A-)2
Pole Placement Technique
3 OJ =Ior1
0 7 -I
[~ ~][~ o L2 ~2] [~ ~][~]=[~] [~ 2 ~2]
=
-1
-1 2 1
0
(A"l2
c·
7 -1
0
I
c· : NC• : (A")2 C- I
2
0
The rank of above matrix 3. Hence the original system is observable.
Let
Ke
[K•l]
=
K.i
be observer gain matrix
K•3
The desired characteristic polynomJal is,
[s-(-3+j1)]
[s-(-3-jl)] [s-(-4)] = (Cs+3)2 +1] (s2 +6s+10) (s+4)
= sJ +6s2 +10s+4s2
SJ+ 10s2 +34s+40
I
Now consider sl-(A + K0
-C)j
=
0
2
s 0
-1 ~]-[~:~] (0 0 2
[~ :]-([: 2 [~ ~]-( [~ ~] -[~ 0
-1
0 0
2
0 K._,
0
0
=
OJ0 - [l3 -12 1-K.i -K.1 ]
I [~
0 s
I r-l-3
s+l
0
0
2
-2 -2
-I K.1 +K.2 ] s+K03
-Ke3
I
I
[s+4]
K,2 K
tJ)
+24s+40
5.37
Modem Control Theory
(A-K0C)
Pola Placement Technique
(x-x)
e = (A - K,, C ) e
... (Il)
Equations (I) and (JI) can be eombined and rewritten in matrix form as,
[~1 •
=
[x]
[A-BK BK J 0 A-K,C e
... (Ill)
eJ
Equation (Ill) shows the dynamics of observed state feedback control system. The characteristic equation for the system is
l
sl-A+BK
o
-BK sl-A+K.c
I
lsl-A+DKHsl-A+K0Cj
0 =
o
The closed loop poles of observed state feedback system consists of the poles due to pole placement design alone and poles due to observer design alone. Thus the pole placement design and observer design arc independent of each other. The design of each can be done independently and it can be combined to get observed state feedback control system. For n as the order of the plant, the order of the observer is also n whdc the resulting characteristic equation of the total dosed loop system is of order 2n. To derive the transfer function of observer based controller, let us consider the system defined as x
Ax+ Bu
y
Cx
Let the plant be completely observable and let the observed state feedback control is
u=-KX.
x
. :.x
Ax+Bu+K.(y-G) Ax+ B(-KX)+ K0(y-CX) (A-OK-K0C')x+
Key
Taking Laplace trans;... ::n of ~boh: equation sx(s)-x(O) = (A-OK-K.C)x(s)+K.Y(s)
Modem
Pole Placement Technique
5 -38
Control Theory
Assuming zero initial condition, [sl-(A-BK-K,.C)J
x(s)
K0 Y(s)
x(s}
(sl-A +BK+ Kccr1 K, Y(s)
Substituting this value of x(s) in equation for u, U(s)
-K >'.(s)
U(s}
- K (sl - A + BK + KcCf1 Kc Y(s)
The transfer Iunction is given by, U(s)
Y(s) U(s) -Y(s)
=
K(sl-J\+KCC+BKr1Kc
The block diagram representation of above system is shown in the Fig. 5.5
K{sl-A•K.c + BK)-1K•
H.
_.T
_P_1a_n_1
Fig. 5.5 The transfer function obtained above is called observer based controller transfer function. It can be seen that the observer controller matrix A - KcC - BK may or may not be stable even though A - BK and A - K.c are selected to be stable. In certain cases, the matrix A - K0C - BK may be poorly stable or even unstable also.
Examples with Solutions ,,..
Example 5.5 : A system represented by followi11g state model is controllable but 11ot obstrvablt. Show that lht 1101H1bsm;ability is due to a pole-zero cancellatio11 in Clsl - AJ-1. (VTU: July/Aug.·2ll05l
x
[ ~
~
-6 -11 Y
(1
1
o) x
~] x+[~]u
-6
1
Solution:
[-6~ -110 -6~ jx+[~]u l
x Y = (1
I 0) x l
Herc,
A
[~6
0 -11
q;
8 =
-6
[~ ~]-[~ 0
[sl-A]
0 s
jsl-AI
Pole Placement Technique
5-39
Modem Control Theory
m;
1
0
[~
lJ
-6 -II
sf(s+6)s+ll]+1(6)
c = -1 s 11
I 01
(1
0 ] -1 5*-6
s3 +6s2 +11s+6
(s+l)(s+2)(s+3) [s1-Ar1
Adj of fsi-A] jsl-A!
rs +6s+ll
T.F.
l
1 (s+l)(s+2)(s+
3)
I (s+l)(s+2)(s+
3) [
Y(s) D(s)
=
2
+s+6 1 s2 +6s+ll 6 . -6s
C(s1-Ar'n+o
[1 1 0 J
1 (s+l)(s+2)(s+3)
·,
-6 -6s ]T s2 +6s -lls-6 s2 s s+6 s(s+6) -(11s+6)
=
C[sl-AJ-1B
= (·:
O=O)
r s"+6s+11 '
l
(s+ l)(s+2)(s+ 3)
(s2 +6s+l1-6
1 (s+l)(s+2)(s+3)
s+l [. J
-6 -6s
s+6 S(H6)
IS
-(lls+b)
s2
(s+6)H(S+6)
l
[010 I
1+s] r~l]
Pole Placement Technique
5 -40
Modem Control Theory
s+ l
(s+i)(5:.:.2)(s+3) T.F.
Y(s) = -U(;)
s+l =-----(>+ l)(H-2)(~+
3)
Clearly it is seen that the factor {s.,.1) is cancelled from numerator and denominator. T.F. "
s+l s3+6s2+11s+6
s+I l~(s+6)+ 11 ]s+6
Let us draw the simulation of the transfer (unction.
y
Fig. 5.6 There is no coupling between x3 and y due to which state x3 can not be observed from measurement of output y. Thus the system is not completely observable due to pole-zero cancellation as matrix C contains one of the terms as zero.
1m+>
Example 5.6 : A regulator system has the plant
(VJ'U: J•n.!Feb.-2006)
i) Compute K so that the control law 11 = - Kx + ru), r(I) = reference input, places the closed loop poles at -2 ± jM,-5. (81 ii) Design an observer such that tne eigen values of the observer are located al -2±j./f2,-s. (61 iii) Draw a block diagram implementation of the control conjiguration. (JI iv) Obtain the stale model of the observer based stale feed back control system. (31
Modem Control Theory
Solution:
x
y A
Pole Placement Technique
5 ·41
0 0
[~ ~] [r ~]; (0 0 1) x
-.
0 0
B
C
= [~}
=
[O 0 1]
i) Let us first check the controllability matrix of the system the controllability matrix for n 3 is given by,
=
o, B
[B:AB:A2B]
0
m;
[~ ~][~] m [r ~][~ ~]= [~
AB=
0
0 0
A·A
0 0
-6 -11
-6
:i 25
-6 -11
[~ EJ[~l = m [~ ~] = = -6
0 1
IOcl
t(t)
l;t
o
0
Rank of Q, = 3 The system is completely state controllable as the rank of controllability matrix QC: n: 3. Now the desired characteristics equation is, [s-(-5)][s-(-2+jJ12)][
s-( -2-;./U))
(s+5)((s+2)-j"12) (s+5)[(s+2)2+12] (s+5)[s2
+4s+16]
[ (s+2)+
iJU)
5-46
Modem Control Theory
Pole Placement Technique
[~ ~s][~] [~sl [~ ;:] 0
A2B =
-6
=
30 19
1
19
J
=
1co-1i
0
QC Rank of
Oc
.I -5
=
IQd
;
= -1 ... o
3
The system is completely state controllable as rank of matrix Oc = n. Hence it is possible to place the poles arbitrarily. Let us now find the stale feedback gain matrix by use of transformation matrix T.
The characteristics equation for the system is,
s ls(s+5)+ 6 )-(-1 )(O] s(s2 +5s+6]
= s3 +5s2 +6s
Comparing the above equation with s3+a1s2 a1
=5
;
a2
=6
;
a3
+a 2s+a
3
=0
=0
The desired characteristic equation is,
[s-(-1 +ii)] [ s-(-1-jl)] [s-(-5)]
[(s+ 1 )-jl )[(s + 1) + jl J[s+5]
[Cs+1)2 +1J(s+SJ (s2 +2s+2)(s+5)
o3 we have o1 = 7; o2 = 12; o3 = 10 K = (1>3 -a 3 1>2 -a 2 61 -a iJ T-1 state equation is given in controllable form T = I T"1 = I
Comparing above equation with s3+ ots2 + o2s +
As the given
= s3 +7s2 +12s+!O
Modem Control Theory
K
5 -47
=
Pole Placement Technique
[!0-0 12-6 7-5]1 [10 6
21[~
01 OJ 0 = [10 6 2)
0 0 I
The state feedback gain matrix, K = ,_.
Example 5.8 :
Y(.-)
u'(,:)
Co11,;ider a
[to
6 2]
linear system described by
tlie
transfer function
lO Design a feedback controller roitlt a state fudback so I/tat closed loop = >(>+l)(s+ 2)
poles are placed at - Z, - 1 ± j.
(VTU : July/Aug.-2006)
Solution : The transfer furlction is given as ,
Y(s) U(s)
10 s(s+l)(s+2)
10
s(sz
+3s+2)
10 s2+3s2+2s
The denominator is decomposed as below, s3 +3s2 +2s = ((s+3)s+2]s Let us simulate this denominator
o(t)
Fig. 5.9 The complete slate diagram is as shown in the following Fig. 5.10 after simulating the numerator by shifting take off point.
y(t)
Fig. 5.10
Modem Control Theory
Pole Placement Technique
5-49
Now the desired characteristics equation is, ls-(-2)][ s-(-1 + jl)] [ s-(-1
-jl)]
(s+2} (s+l +jl) (s+l-jl) [s+2J[(s+1}2
+1]
s3 +2s2 +2s+2s2
= (s+2)(s2
+2s
•2)
+4s+4
s3 +4s2 +6s+4 Comparing above equation with s3+ o1s2 + o2s + &3 we have, &1 = 4;
o2
o3
= 6;
=4
Let us now find state feedback gain matrix by direct substitution desired state feedback gain matrix K be, K
Consider,
Isl
-A+ BK!
(K1
method. Let the
K3)
Ki
I
I[~ ~]-[~ -2 I3J+m[Ki K2 K3) I[~ ~]-[~ I3]+[11 JJI 0
0
1·
0
l[il
0
1
0
0
0 -2
Kz
2:~
0
s+f:KJ
2
s[ s( s+ 3+ K3 )-(-1)(2+K2)+1
[Ki]]
Equating coefficients of above equation with those from desired equation, 3 + K3
=4;
K3
=1
2 + Kz
., State feedback gain matrix,
=6 K
.'. Kz = 4
= [4
4 1]
K1
=4
Modem Control Theory
,,_.
Example
A
-1
"lr 1
2
J
Polo Placement
Consider tire 5ystem described l>y Ilic state model x = Ax. Y
5.9:
Jl
5-50
A
(~l
~l
Let us check the controllability
o,
[c'
C=[l
OJ
matrix of dual of given system for n
A'C°l
[~I ~][~]=[~I]
·A"C'
Rank of Qc = 2 = Number of state variables The dual system is completely state controllable :.Original system is observable.
=
[~cl]
be observer gain matrix
<2
The desired characteristics equation is, [s-(-5)Jfs-(-5)[
Cr where
(VTU:July/Aug.·2006)
Solution:
K0
=
0 I Design a full order stat( oltserver. The tlesi"·d eigL'>I talue» for the
C = [l
obseroer matrix arc 111 = -5, u2 = -5.
Let
Technique
= (s+5)(s+5)
= s2 +10s+25
Consider,
![s+ 1 + K,.1
-1
-l+Kc2
s+2
J
]I
=2
5. 51
Modom Control Theory
Polo Placoment Technique
( s+ 1+ Kc1 )(s+2)-(-I ){-1 + K,.2) ( s+I + Kcl )(s+2)+1
(-t
+ K02}
s2 +s+K01s+2s+2+2K01-l
+K02
s2 +(l+K,1 +2)s+(1+2Kc1+Kc2) s2 +(J+Kc1)s+(l
+2K01 +Kci)
Comparing the coefficients in above equation with those of desired characteristic equation, 3+Kd l + 2K,.1 + Kc2
10;
K01 =7
25;
K,,i=25-t-2K"'=24-2(7)"1(l
The observer gain matrix K0 is given as,
l
Example 5.10 : 0 A= 0 [ -]
l 0 -5
0 I -6
Consider I/re systtm drfined by
.=
.r
Ax + 811, toher«
Ol
J By 11si11g statt feedback control
/l = ~
[
=
lltt closed loop poft<s at s =-2±j4 ands "K" lJy any one method.
11
= -Kt, it ts desired to hat'I'
-10. Determin« lilt stale fe£dbackgoi11 matrix (VTU : J~n.fFeb.· 2008)
Solution : We have
Let us check the controllability matrix of the system. Herc n = 3. The controllability matrix • Q< =
B =
[0101:
All
(n:
All : A 21l]
=[~
~5
-1
lj
I A2 = A ·A
=
(~1
0
-s
:][~]=[:] 1
:J[~t q 0
-5 -6
n =[~I 6
-5
I ] -6
29
.}t
Modern
Control
5.52
Theory
0 l 1 -6 -6 31
l
Pole Placement Technique
:.IQcl
= 1[31-(36)]=
-5 ~
o
Rank of matrix Qc is 3. The system is completely state controllable as rank of
matrix
Oc = n.
Let us now find state feedback gain matrix by direct substitution method. The desired characteristic equation is,
f s-(-2 + j·l)]f s-(-2 -j4) J ls-(-10)]
= [(s + 2)-i4 J[(s + 2)+
[(s+2)2 +16] [s+IO}
i4]Ls + 10]
= (s2 +4s+20J[s+10)
s3 +14s2 +60s+200
s,
= 14,
02 = 60,
03= 200 K2 K3j
K = [ K1
let the desired feedback gain matrix K be Consider.
Jsl-A+BKj
I[~ ~]-[~ 0
0 s
I[~ ~]-[
:J+m lK1 0
1
0
0 s
I
0 -1 -5
~ ]+[K,~
0 -1 -5 -6
~
0
K2
-1
l[l+~KI 5+JS
s+6~KJI
s[s(
s+ 6+ K3) +(S+ JS)]+
s[s2
+(6+K3)s+(5+K2)]+(1+K1)
1 (1
+ K1)
s3 +(6+K3)s2 +(S+JS )s+( l+
K1)
Kz
K3)1
JJI
5 .57
Modem Control Theory
Ke
=
Pole Placement Technique
«•>[c~'r m [74 ~] [~ ~im [74 ~J m [~1l -18 -42
25 41 -95
0 0
25
K., = -18
-42
41
-95
This is the required observer feedback gain matrix by Ackermann's formula.
Review Questions J. Whal do you 17U'an be pole p/aremrnt probkm? 2. Prow the necessarya11d s11fflcinil rondilionfor arbitrary polt plactmtnl? 3 Write 5horl note of <1JOluation of stalef«dbiick gain matrix. 4. What art' the diffemtt tnrlhods of <1JOluating stolt fttdbackgain matrix? Explain any ont mtlhcd in dttnil. 5. lks
7. What do you mun by state obserwrs? 8. Write a note on full onler slate obstrwr? 9. What i• dual probltm? IO. What is nt«SS11ry and suffidtnl conditionforstate observation.
11. Wh41 art different mtlhods of <1JOluation of sl#lt
obstn,'l'f
goin rnalrix K, ?
12. Write• note on stltction of suiloble wlut of ~r g•in mmrix K., 13. Wrltt n 5hort nott 011 obstrotr bllstd controller. 14. Derite 11~ transfer [unction of observer hosed controlltr. 15. Con.sid" 11.. systtm defined hy x A
=
Ax+Bu 0
1
0
0
~l
[ -1 -5 -6
r
By using the statt fudbnck control u = -Kx, it is tksirtd to havt I/it dosed loop poles at s = -2 i j4 ands = -10. Dettrmine the state feedback gai11 matrix K. I Ans.: K = [ 199 SS 8 )J
5. 59
Modem Control Theory
Pole PlacementTechnique
22. Consider II.- systom r<pr'5ented by
x
y
r~ :l\Tm" -~l
= [1
0
o] .r
Dt.
000
Controllers 6.1 Background The concept of a control system is to sense deviation of the output from the desired value and correct it. till the desired output is achieved. The deviation of the actual output form its desired value is called an error. The measurement of error is possible because of feedback. The feedback allows us to compare the actual output with its desired value to generate the error. The error is denoted as e(t). The desired value of the output is also called reference input or a set point. The error obtained is required to be analysed to take the proper corrective action. The controller is an clement which accepts the error in some form and decides the proper corrective action. The output of the controller is then applied to the process or final control clement. This brings the output back to its desired set point value. The controller is the heart of a control system. The accuracy of the entire system depends on how sensitive is the controller to the error detected and how it is manipulating such an error. The controller has it' own logic lo handle the error. Now a days for better accuracy, the digital controllers such as microprocessors, microcontrollers, computers are used. Such controllers execute certain algorithm to calculate the manipulating signal. This chapter explains the basic discontinuous controllers such as on-off controller, continuous controllers such as proportional, integral etc. and composite controllers such "5 proportional plus integral, proportional plus derivative etc. Let us study first the general properties of the controller.
6.2 Properties of Controller Consider a control system shown in the Fig. 6.1 which includes a controller. The actual output is sensed by a sensor and converted to a proper feedback signal b(t) using a feedback clement. The set point value is the reference input r(t). For example the actual output variable may be temperature but using the thermocouple as the feedback element, the feedback signal b(t) is an electrical voltage. This is then compared with reference input which i~ also an electrical voltage. The thermocouple senses the output temperature and produces the corresponding electrical e.m.I as the feedback signal. Hence actual output variable sensed and the feedback signal may be having different forms. (6 • 1)
Modem Control Theory
8·2 Controller output (p)
•
l
ett)
r(I)
Controllers
Controller
Process lo be conlrotled
Controlledoutpul ~
Errot detector
c(t)
Sensing
'--/-~------! Feedback element 1-------' Feedback signal
Fig. 6.1 Basic control systam
6.2.1 Error The error detector compares the feedback signal b(t) with the reference input r(t) to generate an error. e{t)
=
r(t) - b(t)
This gives an absolute indication of an error.
For example if the set point for a range of 5 mV to 20 mV is 12 mV and the feedback signal is 11.8 mV then error is 0.2 mV. But actual vairable to be controlled may be different such as temperature, pressure etc. Hence to obtain correct information from the error, it is expressed in percentage forrn related to the controller operation. It is expressed as the percentage of the measured variable range. The range of the measured variable b(t) is also called span. span = b""" -bnun
Thus
Hence error can be expressed as percent of span as,
r-b er
cP = error as % of span
Where ••
b,ou. -bmin x 100
Example 6.1 : The rattgt of measured variable for a certoi11 co11trol system is 2 m V to 12 mV and a sttpoinl of 7 mV. Firrd the error as percent of span wlu"rl tire measured tariabl« is 6.5 m V.
Solution :
1.>11,,.,
= 12 mV, el' =
bmin
b
= 2 mV,
r-~ m.u: -
5%
xlOO min
b = 6.5 mV, 7-6.5 100 12-2 x
r = 1 mV
Modem Control Theory
6-3
Controllers
6.2.2 Variable Range In practical systems, the controlled variable has a range of values within which the control is required to be maintained. This range is specified as the maximum and minimum values allowed for the controlled variable. It can be specified as some nominal values and plus-minus tolerance allowed about this value. Sur.h range is important for the design of contollers.
6.2.3 Controller Output Range Similar to the controlled variable, a range is associated with a controller output variable. It is also specified intcrms of the maximum and minimum values. But oltcn the controller output is expressed as a percentage where minimum controller output is 0% and maximum controller output is 100 %. But 0 % controller output docs not mean, zero output. For example it is necessary requirement or the system that a steam flow corresponding to ,V.1h opening or the valves should be mininum. Thus 0% controller output in such case corresponds to the
}'.;th
opening of the valve.
The controller output as a percent of full scale when the output changes within the specified range is expressed as,
p = Where
u-umin
xlOO
um4x -umin
p
controller output as a percent of full scale
u
value of the output
u •"-'x
maximum value of controlling variable
u min
minimum value of controlling variable
6.2.4 Control Lag The control system can have a lag associated with it. The control lag is the time required by the process and controller loop to make the necessary changes to obtain the output at Its setpoinL The control lag must be compared with the process lag while designing the controllers. For example in a process a valve is required to be open or closed for controlling the output variable. Physically the action of opening or closing of the valve is very slow and is the part of the process lag. Jn such a case there is no point in designing a fast controller than the process lag.
6.2.5 Dead Zone Many a times a dead zone is associated with a process control loop. The time corresponding to dead zone is called dead time. The time elapsed between the instant when error occurs and the instant when first corrective action occurs is called dead time.
Modem Control Theory
Controllers
Nothing happens in the system, during this time though the error occurs. This part is also called dead band. The effect of such dead time must be considered while the design of the controllers.
6.3 Classification of Controllers The classification of the controllers is based on the response of the controller and mode of operation of the controller. For example in a simple temperature control of a room, the heater is to be controlled. It should be switched on or off by the controller when temperature crosses its setpoint. Such an operation of the controller is called discontinuous operation nnd the mode of operation is called disconti.nuous mode of controller. But in some process control systems, simple on/off decision is not sufficient. For example, controlling the steam flow by opening or closing the valve. In such case a smooth opening or closing of valve is necessary. The controller in such a case is said to be operating in a continuous mode. Thus the controllers are basically classified as discontinuous controllers and continuous controllers. The discontinuous mode controllers arc further classified as ON-OFF controllers and multiposition controllers. The continuous mode controllers arc further classified as proportional controllers, integral controllers and derivative controllers. Some continuous mode controllers can be combined to obtain composite controller mode. The examples of such composite controllers are Pl, PD and PIO controllers. The most of the controllers arc placed in the forward path of control system. But in some cases, input to the controller is controlled through a feedback path. The example of such a controller is rate feedback controller. In this chapter, only continuous and composite controllers are discussed.
6.4 Continuous Controller Modes In the discontinuous controller mode, the output of th" controller is discontinuous and not smoothly varying. But in the continuous controller mode, the controller output varies smoothly proportional to the error or proportional to some form of the error. Depending upon which form of the error is used as the input to the controller to produce the continuous controller output, these controllers arc classified as, L Proportional control mode 2. Integral control mode 3. Derivative control mode Let. us discuss these control modes in detail.
6-5
Modem Control Theory
Controllers
6.5 Proportional Control Mode In this control mode, the output of the controller is simple proportional to the error e(t). The relation between the error e(t) and the controller output p is determined by constant called proportional gain constant denoted as KP. The output of the controller is a linear function of the error e(t). Thus each value of the error has a unique value of the controller output. The range of the error which covers 0 % to 100 % controller output is called proportional band. Now though there exists linear relation between controller output and the error, for a zero error the controller output should not be zero, otherwise the process will come to halt. Hence there exists some controller output Po for the zero error. Hence mathematically the proportional control mode is expressed as, ... (!)
Proportional gain constant
Where Po
Controller output with zero error
The direct and reverse action is possible in the proportional control mode. The error may be positive or negative because error is r-b and b can be less or greater than reference setpoint r. If the controlled variable i.e. input to the controller increases, causing increase in the controller output, the action is called direct action. For example the output valve is to be controlled to maintain the liquid level in a tank. So if the level increases, the valve should be opened more to maintain the level. If the controlled variable decreases, causing increase in the controller output or increase in the controlled variable, causing decrease in the controller output, the action is called reverse action. For example simple heater control for maintaining temperature. If the temperature increases, the drive to the heater must be decreased.
So if e(t) is negative then Kpe(t) gets subtracted from p0 and c(t) is positive, then JS,e(t) gets added to Po- this Is reverse action. So equation (!) represents the reverse action. But using negative sign to the correction term Kpe(I), the direct action proportionnl controller can be achieved. The proportional controller depends on the proper design of the gain K.,. For fixed Peif gain KP is high then large output results for small error but narrow error band can be handled. Beyond these limit' of the error, output will be saturated. If the gain is small then the response is smaller but large error band can be handled. This is shown in the Fig. 6.2.
Conlrollcr
output
High gmn
100%t--------~~~~~~--· 50(14
0 ...
I
f
---
-· .
'
----~---..J----------
__ ..
-
Controllers
6-6
Modem Control Theory
I
'
1 I I I I
' 'I 0 '
Po= 50 %
--------r---1---- ·----,__
,
Q
'
Narrow
~
I,..
. !
t,.
Error%
error band
Wider error band
Fig. 6.2 The proportional band is mathematically defined by,
6.5.1 Characteristic of Proportional Mode The various characteristics of the proportional mode are, 1. When the error is zero, the controller output is constant equal to Po2. If the error occurs, then for every 1 % of error the correction of Kr % is achieved. If error is positive, KP% correction gets added to p0 and if error is negative, Kr% correction gets subtracted from Po· 3. The band of error exists for which the output of the controller is between 0 % to 100 % without saturation. 4. The gain KP and the error band PB are inversely proportional to each other.
6.5.2 Offset The major disadvantage of the proportional control mode is that it produces an offset error in the output. When the load changes, the output deviates from the setpoinl Such a deviation is called offset error or steady state error. Such an offset error is shown in the Fig. 6.3. The offset error depends upon the reaction rate of the controller. Slow reaction rate produces small offset error while fast reaction rate produces large offset error.
Modem Control Theory
6·7
Controllers
Oull)Ul
Fig. 6.3 Offset error in porportlonal mode The dead time or transfer lag present in the system further worsens the result. It produces not only the large offset at the output but the time required to achieve steady state is also large. The offset error can be minimized by the large proportional gain KP which reduces the proportional band. If KP is made very large, the proportional band becomes so small that it acts as an ON/OFF controller producing oscillations about the setpoint instead of an offset error.
6.5.3 Applications The proportional controller can be suitable where, 1. Manual reset of the operating point is possible. 2. Load changes arc small. 3. The dead time exists in the system is small.
6.6 Integral Control Mode In the proportional produces an offset error. another control mode is of the errors. This mode
control mode, error reduces but can not go to zero. It finally It can not adapt with the changing load conditions. To avoid this, oftenly used in the control systems which is based on the history is called integral mode or reset action controller.
In such a controller, the value of the controller output p(t) is changed at a rate which is proportional to the actuating error signal e(t). Mathematically it is expressed as, d p(t) dr Where
K;
=
K e(I) I
: Constant relating error and rate
The constant K, iq also called Integral constant. Integrating the above equation, actual controller output at any time t can be obtained as,
Modem
Control Theory
6•8
K, j c{t)dt.;. p(O)
p(t) Where
Controllers
... (2)
p(O) = Controller output when integral action starts i.e. at t = 0.
The output signal from the controller, at any instant is the area under the actuating error signal curve up to that instant. If the value of the error is doubled, the value of p(t) varies twice as fast i.c. rate of the controller output change also doubles. If the error is zero. the controller output is not changed. The control signal p{t) can have nonzero value when the error signal <'{t) is zero. This is because the output depends on the history of the error and not on the instantaneous value of the error. This is shown in the Fig. 6.4. e(l)
p(t)
(a)Error
(bl Controller output
Fig. 6.4 Integral mode The scale factor or constant K, expresses the scaling between error and the controller output. Thus a large value of K1 means that a small error produces a large rate of change of p(t) and vicevcrsa. This is shown in the Fig. 6.5. If there is positive error, the controller output begins to ramp up.
(+)
dp(t) di
0 (-) -'-------''-------Error 0
Fig. 6.5
Modem Control Thoory
6·9
Controllers
6.6.1 Step Response of Integral Mode e(t)
The step response of the Integral control mode is shown in the Fig. 6.6.
A
The integration lime constant is the time taken for the output to change by an amount equal to the input error step. This is shown in the Fig. 6.6. p(t)
A
I.___,' r,
ll can be seen that when error is positive, the outpul p(t) ramps up. For zero error, there is no change in the output. And when error is negative, the output p(t) ramps down.
Fig. 6.6 Step response
6.6.2 Characteristics of Integral Mode The integrating controller is relatively slow controller. It changes its output at a rate which is dependent on the integrating time constant, until the error signal is cancelled. Compared lo the proporlional control, the integral control requires lime lo build up an appreciable output. However it continues lo act till the error signal disappears. This corrects the problem of the offset error in the proportional controller. For example, Jct us assume that the integral controller is used to control the armature current of a d.c. motor and to keep its value constant ot 500 A. As long as the armature current is less U10n 500 A, the armature voltage, controlled by the controller, will increase. TI1us the output of the controller will increase and will continue to do so till the error becomes zero i.e. armature current becomes 500 A. Then the controller output will remain at that value reached. This is possible because the output of the controller can remain al any value within its r.mr,c. if the input is zero. 'The controller must not be overdriven as it will not then be effective. Thus for an iJllC!,'Tlll mode, I. If error is zero, the output remains at a fixed value equal to what is was, when the error became ·~ro. 2. If the error is not zero, then the output begins to increase or decrease, at a rate K; % per second for every ± l % of error. In some cases, the inverse of K, called integral time is specified, denoted as 1;.
Modem Control Theory
Controllers
6 -10
Ti : ..!....
Ki
:
Integral time
It is expressed in minutes instead of seconds.
6.6.3 Applications The comparison of proportional and integral mode behaviour at the time of occurrence of an error signal is tabulated below, Controller
Initial behaviour
p
Acts immediately. Action according to Kr.
Ofbet error always present. Larger the KP smaller lhe error.
I
Acts slowly. It Is the time integral of the error signal.
Error signal always
Steady state behaviour
becomes zero.
Table 6.1 It can be S(.'Cn that proportional mode is more favourable at the start while the integral is better for steady state response. In pure integral mode, error can oscillate about zero and can be cyclic. Hence in practice integral mode is never used alone but combined with the proportional mode, to enjoy the advantages of both the modes.
6.7 Derivative Control Mode In practice the error is Junction of time and at a particular instant it can be zero. But it may not remain zero forever after that instant. Hence some action is required corresponding to the rate at which the error is changing. Such a controller is called derivative controller. In this mode, the output of the controller depends on the time rate of change of the actual errors. Hence it is also called rate action mode or anticipatory action mode. The mathematical equation for the mode is. p(t) Where
K d c(t) J di Derivative gain constant.
The derivative gain constant indicates by how much % the controller output must change for every % per sec rate of change of the error. Generally Kd is expressed in minutes. The important feature of this type of control mode i.~ that for a given rate of change of error signal, there is a unique value of the controller output.
Modem Control Theory
Controllers
6·11
The advantage of the derivative control action is that ii responds to the rate of change of error and can produce the significant correction before the magnitude of the actuating error becomes too large. Derivative control thus anticipates the actuating error, initiates an early corrective action and tends to increase stability of the system improving the transient response.
6.7.1 Characteristics of Derivative Control Mode The Fig. 6.7 shows how derivative mode changes the controller output for the various rates of change of the error. e(t)
Error
0
Timet
p(%)
100% Conltoller output 50%
0%
Timet
Fig. 6.7 The controller output is SO% for the zero error. When error starts increasing, the controller output suddenly jumps to the higher value. It further jumps to a higher value for higher rate of increase of error. Then error becomes constant, the output returns to 50 %. When error is decreasing i.e. having negative slope, controller output decreases suddenly to a lower value.
Modem Control Theory
6-12
Controllers
The various characteristics of the derivative mode arc, I. For a given rate of change of error signal, there is a unique value of the controller output. 2. When the error is zero, the controller output is zero. 3. When the error is constant i.e. rate of change of error is zero, the controller output is zero. 4. When error is changing, the controller output changes by Kd % for even I % per second rate of change of error.
6.7.2 Applications When the error is zero or a constant, the derivative controller output is zero. Hence it is never used alone. Its gain should be small because faster rate of change of error can cause very large sudden change of controller output. This may lead to the instability of the system.
6.8 Composite Control Mo.des As mentioned earlier, due to offset error proportional mode is not used alone. Similarly integral and derivative modes are also not used individually in practice. Thus to tnke the advantages of various modes together, the composite control modes are used. TI1e various composite control modes are, I. Proportional + Integral mode (Pf) 2. Proportional + Derivative mode (PD) 3. Proportional + Integral + Derivative mode (PIO) let us sec U1e characteristics of these three modes.
6.9 Proportional + Integral Mode (Pl Control Mode) This is a composite control mode obtained by combining the proportional mode and the integral mode. The mathematical expression for such a composite control is, p(t)
=
I
Kr e(t)+KpK;J
e(t)dt+p(O) 0
Where
p(O)
Initial value of the output at t = 0
Modem Control Theory
Controllers
6 -13
The important advantage of this control is that one to one correspondence of proportional mode is available while the offset gets eliminated due to integral mode, the integral part of such a composite control provides a reset of the zero error output after a load change occurs. Consider the load change occurring at t.., t 1 and due to which error varies as shown in the Fig. 6.8. The controller output changes suddenly by amount VP due to the proportional action. After that the controller output changes linearly with respect to time at a rate KP I T1• The reset rate is defined as the reciprocal of T;.
e(I)
Error
p(I) Conlrollor output ------~--------' :
--- }------J: V1
} VP _ _.'--------
0
ti
1,
-----
--· .
Riso in output duo to intcg1et action between limes 11 and ti Immediate rise due to prol)0'1ionul action Tome
'---v--' Integral acton time
Fig. 6.8 Behaviour of Pl controller The response shown in the Fig. 6.8 is for the direct action of the controller. The response of composite Pl control mode for the reverse action is shown in the Fig. 6.9. In the reverse action, the proportional part is the image of the error. The sum of proportional plus Integral action finally leaves the error to zero.
Modem Control Theory
Controllers
6 -15
6.9.2 Applications The composite Pl mode completely removes the offset problems of proportional mode. Such a mode can be used in the systems with the frequent or large load changes. But the process must have relatively slow changes in the load, to prevent the oscillations.
6.10 Proportional + Derivative Mode (PD Control Mode) The series combination of proportional and derivative control modes gives proportional plus derivative control mode. The mathematical expression for the PD composite control is,
The behaviour of such a PD control to a romp type of the input is shown in the Fig. 6.10. Error e(I)
Ramp
lime
0 P(I) Controller output
:}v;-- --- -- ----. ------ fv;·--~--------------·
.
I
..
Rlso in signal due to proportional action between Um°' t1 :ind
tz
Immediate rtse due to derivative action
I
... 0t----..t.t1---, 2--------
Time
Fig. 6.10 Bohavlour of PD controllor The ramp function of error occurs at t = t 1. The derivative mode causes a step Vd at t 1 and proportional mode causes a rise of Vr equal lo Vd at t 2• This is for direct action PD control.
Modem Control Theory
Controllers
6 -19 I
0.02(3t+7j+0.02x0.04
J
(3t+7)dt+0.5
()
(0.06t]+0.14+8xl0-4[
3:2
+7{
+0.5
(0.06xl.5]+ (0.14}+ Bx 10-4[~+(1.5)2+7x1.5 0.09 + 0.14 + 0.0111 + 0.5 Le.
u•
=
]+ 0.5
0.7411
74.11 % after t = 1.5 minutes.
p(t)
Example 6.3: Suppose the error shown i11 the Fig. 6.14 is applitd lo a PD controller with KP = 6 and K,1 = 0.4 sec with p(O) = 25 %. Dr11w the graph of the controller output. e(t){%)
Fig. 6.14
Solutlon : The output of the PD controller is p(t)
=
d c(t) KP e(t)+K.,K,1 (i!+25%
It is necessary to obtain output response, dividing c(t) into Uu~'C sections, Section I)
0 e(t)
Cit p1(l)
s
t
s2
2t
... slope= 2
2
6x2t+6x0.4x2+25
= 12t + 29.8
Thus there is an instantaneous change of 29.8 - 25 ,. 4.8 % produced by this error in lhe output at t = 0.
Modem Control Theory
Controllers
6 -23
i
J\K
+
p
= 0 for the step
input
Lim s G(s)H(s) = ~;
• -o
2<;
e,.., = So there is a finite error for the ramp input of magnitude A, which depends on ~and For good time response, the system demands. 1. less settling time 2. Less overshoot 3. Less rise time 4. Smallest steady stale error. Increasing Kv, the steady state error for the ramp input can be reduced but it increases overshoot and settling time. This m:1y lead to instability of the system. So to keep steady state error and overshoot well within the acceptable limits, the various composite controllers arc used. Let us see the effect of PD, Pl, PID and rate feedback controller on the time rl'Sponsc of the second order system under consideration.
6.14 PD Type of Controller A controller in the forward path, which changes the controller output corresponding to proportional plus derivative ol error sit,'1lal is called PD controller.
i.e.
Output of controller Taking Laplace
=
= K e(t) + Td
K E(s) + ~TJ E(s)
=
de(l)
di
E(s) [ K + s1'0)
The T.F. of such controller is [K + sT,iJ. This can be realized as shown in the Fig. 6.17.
--"'-"-s(s + ~••nl
Ptanl
Fig. 6.17
Assuming K
Controllers
6·24
Modem Control Theory
= 1, we can write, (I .. sT,1)ro~ s(s+ 21',oo,,)
G(s)
C(s)
and
R(s)
Comparing denominator with standard form, oo., L~ same as in the previous P type controller. 2 ( Oln
and
=
21; oo,,
.. ~
Td
( = <, .. oon2Td Because of this controller, damping ratio increases by factor Lim
KP
= s _.... O
K,
= 5 -.a" s G(s)H(s) = 2~
G(s)H(s)
ro T -¥·
= ..
Lim
oo
As there is no change in coefficients, error also will remain same. Key Point: Hence PD co11troller has folluwing effects on system. i)
ii)
It increases damping ratio.
·oo,,'
for system remains unchanged.
ill) 'TYPE' of the system remains unchanged. iv) It reduces peak overshoot. v)
It reduces settling time.
vi) Steady state error remains unchanged. Key Point : In generRI P.O. controller improves transient part unihout affecting steady
state. 6.15 Pl Type of Controller A controller in the forward path, which changes the controller output corresponding to the proportional plus integral of the error signal is called Pl eontroller.
Modem Control Theory
I
Le,
6-25 K e(t) + K;
Output of controller
Taking Laplace
= K E(s) + ~
J
Controllers c(t) dt
~i]
E(s) = E(s) [ K+
:. The T.F. of such controller is [ K + ~i
)
and can be realised as shown in the Fig. 6.18.
__.,,,__ 2
s (s + 2~w.) Plant
Fig. 6.18 Assuming K = 1, we can write,
~~
G(s)
s(s + 2~0),,) (K, +sl%
s2 (s+ 2~ro0) i.e, system becomes TYPE 2 in nature. and
C(s) R(s)
(K, + s) ro~
=
s 3 + 2~ Cl),, s 2 + s
% + K; w~
i.e. it becomes third order. Now as order increases by one, system relatively becomes less stable as K; must be designed in such a way that system will remain in stable condition. Second order system is always stable. Key Point:
Hence lransitnl rtsponse gets a/Jtcled badly
if
controller is not dtsig11£d
properly. Lim
While
..... 5
K, =
0
G(s)H(s) = ...
c.,. = 0
Lim ..... s G(s)H(s) = - , c.,. = O 0
5
Key Point : Hence as type is increased by ont, error becomes zero for ram.~ t.VPe of inputs i.e. steady state of system gets improvtdand system bteomes more accurate in nature.
Modern
Control
6 -26
Theory
Controllers
Hence l'I controller has following effects : i) It increases order of the system. ii) It mcrcascs
TYPE of the system.
iii) Design of K, must be proper to maintain stability of system. So it makes system relatively less stable. iv) Steady state error reduces tremendously for same type of inputs. Koy Point: /11 gcr:ernl this controller improves steady state part ajfedi11g tlte transisnt part.
6.16 PIO Type of Controller As PD improves transien! and Pl improves steady state, combination of two may be used to improve overall time response of the system. This can be realized as shown in the Fig. 6.19(a).
Plant and
Contror.e<
Fig. 6.19(a) TI1e design of such controller is complicated in practice.
6.17 Rate Feedback Controller (Output Derivative Controller) This is achieved by fct.'Cling back the dertvativc of output signal internally using a tachogcnerator and comparing with slgnat proportional to error as shown. This is called minor loop feedback compcns.ition. Output of controller Output of the controller This can be realized as shown in the Fig. 6.19(b).
=
. dc(t) K l:(t) - K1 -dt
= K F.(s) -
sK1 C(s) E(sl
KE(s)
Assuming K = 1, lei us study its effect on same system which is considered earlier >
with G(s) = - --~~-- -·. >(> + 2.,
Fig. 6.19 (b)
Modern Control Theory
Controllers
6 ·30
Key Point: 1 Time co11sta11t 'T' is tire time required by tire system output to reach 63.2 of its final value d11ri11g the first attempt.
%
The equation for the actual response c(t) is,
... For unit step
Where
=
(l)d
(l)n J1 -1:,2 Damped frequency of oscillations tan- 1
a
and
{Ji
~1:,2 } radians.
Note : As ~ is Ute function of I:, alone, if ~ is known, the equation of ~ solved to obtain the corresponding value of damping ratio I:,.
can be
6.19 Steady State Error Consider a simple closed loop system using negative feedback as shown in the Fig. 6.21. C(s)
Fig. 6.21 Where
E(s) = Error signal,
and
Now,
E(s)
But
B(s)
C(s) · H(s)
R(s) - B(s)
E(s)
R(s) - C(s)H(s)
and
C(s)
E(s) · G(s)
E(s)
R(s) - E(s)G(s)H(s)
E(s) + E(s)G(s)H(s)
=
R(s)
B(s)
Feedback signal
6. 31
Modem Control Thoory R(s)
E(s)
E(s)
Controllers
1 + C(s)H(s) for nonunity feedback
=
R(s)
for unity feedback
I+ C(s)
This E(s) is the error in Laplace domain and is expression in "s'. We want to calculate the error value. In time domain, corresponding error will be e(t). Now steady state of the system is that state which remains as t ~ ... Steady state error, e..
. . ..
= Um L'(t)
Now we can relate this in Laplace domain by using final value theorem which states that,
,Lim ...
F(t)
Where F(s) = L I f(t) I
Lim s F(s)
s-+O
Lim e(t) = Lim s E(s)
Therefore,
I
-t•
Where E{s) is L I c(t) I
s-10
Substituting E(s) from the expression derived, we can write = Lim e,_
sR(s)
• _, o l + G{s)H(s)
For negative feedback systems use positive sign in denominator while use negative sign in denominator if system uses positive feedback. From the above expression it can be concluded that steady state error depends on, i) R{s) i.e. reference input, its type and magnitude. ii) C(s)H(s) i.e. open loop transfer furu:tion. iii) Dominant nonlinearities present if any.
Examples with Solutions ••
Example 6.4: Thefigure slurnrs PD controller used for the system. Determine the value of T4 so tltat system will be critically damped. Calculate its settling timt.
... , Solution:
G(s) = (I.- sTd)4 s(s+ 1.6) Cts) R(s)
Controllers
6. 32
Modern Control Theory
11( ) · s
I
=
(l.,. sTd)4 s (s + 1.6)
(1+s'fd)4
l+(l+sTd)4 s(s + 1.6)
s2+1.65+4'Tds+4
Comparing denominator with standard form, w~
4,
ron
=2
and
2!'; (!)11
=
1.6 + 4 Td
C. = 1.6+ 4Td • 4 Now system required is critically damped, i.e. E, = 1 1.6+ 4 Td --4-4
1.6+ 4T0 . . an d sett I mg time
0.6 T, ,._.
4
2X1=2
=
4
E,ro.,
sec.
Example 6.5: A 1111ity feedbacksystem is show11 in tirefollowingfigure. i) In tile abse11a of derivative feedback controller (K0 if E, is lo be modified lo 0.5 by use of controller.
= 0). Find E,
and ron' ii) Find K0,
C{s)
Solution:
G(s)
ior Ko = 0
50x-s(s+1- 3)
H(s)
C(s) R(s)
G(s) 1 + G(s)H(s)
50
:. w,.
_
50
- 82 + 3s + 50
= ./50 = 7.071 rad/sec.
Modern
,. ·Controne;s
6. 34
Control Theory
14
JM. = 3.7416
w,.
=
rad/sec.
14 K1.,. J.6 14K
+ l.o
~=0.5given 2x"H 0.1529
w,.
H
~
= 0.9695 sec.
= 3.7416 Ji -(0.5)2 =
3.2403 rad/sec
(l)d
e-•JP T,
11.+
4
i; w,,
=
x 100=16.3 'Yo
2.1381 sec.
Example 6.7: For tlu system shown determine % M1, a11d T,, when it is excited by 1111it ~tep i11p11t. If fbr the same system, I'D controller liaving constant T0 ~ 1/30 is used in fbr«Jard path, determine new values of damping ratio, MP andT,. Draw rtsptclive waveforms.
Solution : Without controller,
JOO
G(s) •
s(s+ 12)' H(s) =I
C(s)
100 s2 + 12s + 100
R(s) 0)2 n
100
:. Oln = 10 rad/sec.
~ = 0.6
12 OJ0
e
J1 - ~2
-·VJ·-~2 4
)
= 10x 0.8 = 8 rad/sec. = 9.47 'l'o
i;
1
I
•
6. 35
Modem Control Theory
Controllers
With controller 100
S('S'+i2i
G(s)
(1 +i )too
(s+ 30) x 3.33
s(s+ 12)
s(s+ 12)
H(s)
=1
(s+ 30) 3.33 C(s)
s(s + 12)
3.33(s ... 30)
3.33(s + 30)
R(s)
+ (s + 30) 3.33 1 s(s+ 12)
s2 + 12s + 3.33s + 100
s 2 + 15.33s + 100
100
:. ro.,= 10 rad/sec,
2sro.,
15.33
;
;:·3
:o
:. Sis improvcd,Old % MP
•ro,, =
= 0.7665
n
e-•VR
= 10J1 -(0.7665)2 xlOO
=
= 6.4224 rad/sec.
2.353 %
Overshoot decreased to 2.3 % from 9.47 %. T5 = ~
~
= 0.5218 sec.
Comparison : Following figure shows comparison between system with controller and system without controller. 0.47 %
t) 2.353%
Withoutcontroller
0.5218 sec With conttofler
tlmo
6-37
and
Controllers
2 ~
0.3162
This is the required damping ratio. For error calculation : C(s)H(s)
;
5
s (l + O.Ss)
For ramp input magnitude is unity i.e, A=I lim s G(s)H(s)3
•-o Cs~
lim -11 s.SO-:o-)= 5
.-us
+ ~•
=
Case bl Tbe derivative feedback is introduced in the system. The system becomes. Cts) =
K,, s [0.Ss + I + K1
and H(>)=l
I
1
~ 1+
C(sJ
$Ki
----
s ( 1•0.5s)
0.5s2 • s ( 1 + K1)
C(s) R(s)
KA s [0.5s + 1 + K1 I+
J
K,, slO.Ss+ I+ K1
K,..
J
0.5s2 +(1 + K1) HK,,
Modern
6 -40
Control Theory
=W
Ji -1;
2
Old
211
6.8676 rad I sec
Bui
w,, K,
and
0.4037
~ =
'"*
w:;
0
= 47.165
'.\+KP 2xro., 2.5449
Example 6.10 : An integral controller is used for tem,,....aturecontrol witlrin a range 40 - 60 •c. 17u: st/ point is 48 •c. Th« controller output is initially 12 % when error 15 uro. Tire intrgral constant K1 = - 0.2 % controlleroutput per ~nd per percent11geerror. If tire temperature increases to 54 •c, calcublte II~ controller output •fltr 2 s« for" constant error.
Solution : For integral controller,
... EP =error
p(I) p(O)
12 %,
K1
= - 0.2 % I
sec I "•error
Ep = Constnnt = b.,,.,.r-b 100 -bmm x Set point = 48 °C,
Now
60 EP JEl'dt
Al I= 2,
•c,
bmin = 40
48-54
--xlOO 60-40 ~ t
b
oe
= Actual
temperature
= 54 "C
= - 30%
as error is constant
p(t)
(- 0.2) (- 30) (t) + 12
p(t)
(0.2 x 30 x 2) + 12
= 24 %
This is the controller output after 2 sec.
111•
Example 6.11 ; For a proportional controller,tire amtrolltd variable is 11 tnnpemturc willr ra11gt of 50 to 130 'C with a st/ point of 73.5 •c. TM controlltr output is 50 % ftw uro error. The offset error is corresponding to a lead clumgt which causes 55 % c:mitrolkr 011tp11t. If tire proportional gain is 2, find tlit % control/tr output if thr ltmptral11rt is 11
61
-c.
Solutlun : For a proportional controller, p(t) Where
Kp 6r(t) + Po Kp
= 2,
Po
= 50 'l'o
Modem Control Theory
6-43
Controllers
After end of 2 sec, integral term gets accumulated to, P1(2) For 2 S t S 4, Ep2(t)
2xl.,f~t2]
1
+40=48.8% 1~2
-2.51+7
t
t
j Ep2(t)
dt
!(-25t+7)dt=[-2
2
+7{
- 1.25 (t2 - 4) + 7(t - 2) 2(-2.5t +7)+4.-l[-1.25(12
-4) +7(1-2)1+48.8
- St+ 14 - 5.5~ + 22 + 30.81 - 61.6 + 48.8 - 5.512 + 25.Bt + 23.2
for 2 S I S 4
... (2)
This is plotted for 2 - 4 sec. After end of 4 sec, the integral term gels accumulated to, P1(4) For 4 S t S 6, Ep3(1) !Ep:i(t)dt
-t.25(t2 -4)+7(t-2t%4 +48.8= 47.8% 1.51 - 9 = !(1.5t-9)dt
·;t
= [1
2
-911
0.75 (12 - 16) - 9 (t - 4) p3(I)
2[l .5t-9)+4.4[0.75(t2
-16)-9(1-4))+47.8
31 - 18 + 3.3t2 - 52.8 - 39.6t + 158.4 + 47.8 3.3t2 - 36.6 I + 135.4 This is plollcd for 4 - 6 sec. After 6 integral response of p3(t~1%6 34.6 %
=
S<.'C
for 4 S t S 6
... (3)
as error is zero hence output is accumulated
The graph of p(t) against time is shown in the Fig. 6.23. (Sec Fig. 6.23 on next Page) ,,_.
Example 6.14 : A proportional controlkr is employedfor the control of temperature i11 the range 50 •c - 130 •c •oith a set point of 73.5 •c. The zero error controller output is 50 %. Wlrat will be the offset mar re>uiting from a change in t/Je controller output to 55 % ? The proportionalgai11 is 2 % I %. Find tire offset in •c.
Solution : For proportional control mode
Modern Control Theory
6. 44
Controllers
Fig. 6.23
p Where
KpEP +Po
Po = Controller output with no error = 50 % 2%per%
p-po = 55-50 = l ..S % Kr 2 The offset error is 2.5 %.
= 130 "C and bmm = 50 "C.
The rani;c of 50 "C to 130 "C so b'"'" r-b
c = bnl,\\
2.5 b
xlOO
-blllHl
73.5-b 00 130_50x1 71•.s
·c
This is temperature corresponding to offset error.
Where r = set point
Modom Control
••
Example
Pi(OJ
6. 45
Thoory
Controllors
=
6.15 : A PID co11trnller lras K,. = 2.0. K1 2.2 scc"1• Ko= 2 sec and ~O %. Druw th« plot of controller 01111'111 for error of Fig. 6.24. ErmrE0k 4
2
0-'---+--_, _ _,.,,....
.,_
t Sec
-2 -4
Fig. 6.24 Solution : Kr= 2. K1 = 2.2
S«-1,
K0 = 2 sec, P1(0) = 40 %
from the i,.;vcn error plot, for O - 2 sec, Er= m1l
i;, For 2 - 4
St'C,
2-0
Where m1 =Slope=
2_0
=1
for 0 - 2 sec.
Ep
Two points on the line are (2, 2) and (4, -3). Yz-Y1
At (2, 2),
For 4 - 6 sec.
=-3-2,,_25 ·1-2
mi
x2 -XI
Ep
- 2.St + C2
2
,
- 2.5 x 2 + C2,
Ep
- 2.St + 7
EP
m3t + C3
for 2 - 4 sec.
Two points on the line arc (4. - 3) and (6, 0). 0-(-3) ~=+1.5 Ep At (4, -3),
-3
+l.5t+<; 1.5 x 4 + C3, ... 1.5 t -9
for 4 - 6 S« .
.: <; = 7
Modem Control Theory
Controllers
6 -46
Three mode equation for PID controller is,
dE
I
p = KrEp + KpKi} Epdl + KpKo -af + P1(0) 0
dE 2EP+4.4JEPdt+4-af+40 I
p
0
For 0 - 2 sec,
d
I
Pt
2t
+4.4J l di +4di(t)+40 0
... (!)
2t+2.2t2 +44
This is plotted for 0 - 2 sec. At the end of 2 sec, the integral term has accumulated to,
P1(2) = 4..J!tdl+40=4.4x[t;J:
For 2 - 4 sec,
+41l=48.8%
I d 2(-2.5t +7) +4.4J (-2.St +7) dt +4 di(-2.St+7)+
P2
48.8
2 &
-5t+l4+4.4[-1.25t2
+7t]~ -4x2.5+48.8
-5t + 14 +4.4{-t2S(t2
-4)+7(1-2)}-10
+48.8
-5.5t2+25.St+13.2
... (2)
This is plotted for 2 - 4 sec. At the end of 2 S<.'C, the integral term has accumulated to, P1(4) = 4.4{-1.25(12 -4)+7(t-2)}11".1+48.8
= 44.4 %
For 4 - 6 sec, I
p3
2(1.St-9)+.JAJ
I (1.St-9) dt +4Ft(l.5t-9)+44
..J
4
3t-18+4.4[075t2 3t-18+4.4{01s(12 3.3t2 -36.6t +138 This is plotted for 4 - 6 sec.
-9t)'.1 +.Jxl.5+44.4 -t6)-9(t-4)}+6+44.4 •.. (3)
Modern
Control Theory
6-47
Controllers
After 6 Sl'C, error is zero hence the output will simply be the accumulated integral response providing a constant output.
P1(6)
·l.4{075(t2
-16)-9(t-4)}1,.6
+~4.4
31.2 %
,,...
Example 6.16 : A temperature control system has tire block diagram given in Fig. 6.25. The input signal is a voltage and represents tire desired temperature 0, is a unit step
function and
i) D(s)
=1
ii) D{s)
= 1 + Pf
iii) D(s)
= 1 + 0.3 s. What is tire effect of
the integral term in the Pl controller a•rd the derivative term in PD controller on tire steady stale error ? (VTU: July/Aug.-2006, 8 Marks)
O(r)
~
200 Plant
Fig. 6.25
0
Controllers
6 ·48
Modem Control Theory
Solution : For the given system. G(sl =
200D(s) (s+ll(s ... 2)'
H() 002 s " · '
Lim sE(>)
Lim sl~s) ,.~o I +G(s)H(s)
1->.-,tl
i)
= 0.02 s
D(s} l.im ~-H 1 +
ii)
R(s)
D(s)
0.02 sx-s l()O x0.02 (s+l)(s+2)
0.02
;:20
&.66x10·3
2
I+ 0.1
s sx 0.02 s
Lim <-O
oo(i+¥Lom]
=0
2 1+
(s+l)(s+2) iii)
D(s)
1 + 0.3 s 0.02 sx--
l".,..
------"''----- = 6.66x10·3 :.-0 Lim
1+(200(1 +0.1s) x0.02) (>+l)(s~2)
Due to Pl controller, the steady state error reduces drastically while PD controller has no effect on the steady state error.
Review Questions l.
Whal is co11troller ? Explain its function i11 a systrrn.
2. Gill< /ht classification of controllers. 3. Stalt and txplain tlu: various J!TOJXrlks of controller. 4. State tlie various contin11ous controller mod.ts. 5. Explain tht prof}"'lional co11tro/ modr. State its d111racttristics. 6. Explain the integro!control mode, State its cbamctrristics. 7. Explain ti~ derivatiw control mode. State its tharocteristics. 8. Explain how constant K, •JJ
Modem Control Theory 11.
Controllers
6 -49
Explain the Pl control mod<, stating its characteristics.
(July·2005)
Explain the PD control mode, slatmg its chanwmslics. 13. Writ< a not« on tlzru moat control/er. 14. Discuss tht effrr:t of following controllers 011 th< second order con!rol systtm, al Pl controller bl PD <011trofler 12.
15. Explain 11" effrr:t of Pl controller on typical second order systtm. 16. E;rplain the ef!tct of PD controller on typical second order •yst<m.
(July·2005)
I 7. A fetdbnck systtm whid1 11$<$ a rate fetdlxic:k controller is slwrv11 in th< figure.
0.5s1 + s( I+
Kil
al For KA a in in absence of derioative fttdbtuk (Kh a 0) dtlmnint 11" damping ratio and naluralfrtqutncy of oscillations. Also find IN s s trror for unit ramp input. bl Determine tire constant K1, if th< damping foe/or nquired is 0.6, with KA = 10. With this oolue of Kh, dtter111inc ss error for writ ramp input. (Ans.: 0.316, 3.16 rad/sec, 0.2, 1.8, 0.38)
ODD
(6. 50)
Nonlinear Systems
7.1 Introductionto Nonlinear Systems It has been mentioned earlier that a control system is said lo be linear if it obeys law of superposition. Most of the control systems are nonlinear in nature and are treated to be linear, under certain approximations, from ease of analysis point of view. Let us discuss now the properties of nonlinear systems. ln practice nonlinearities may exist in the systems inherently or may be purposely introduced in the systems, to improve the performance. Hence knowledge of properties of nonlinear systems and various nonlinearities is important.
7.2 Properties of Nonlinear Systems The various characteristics of nonlinear systems arc, 1. The most important characteristics of a nonlinear system is that it docs not obey the law of superposition. Hence its behaviour with respect to standard test inputs can not be used as base to analyse its behaviour with respect to other inputs. Its response is different for different amplitudes of input signals. Hence while doing the analysis of nonlinear system, alongwith the mathematical model of the system, it is necessary to have information about amplitudes of the probable inputs, initial conditions etc. This makes the analysis of the nonlinear system difficult. 2. Unear system gives sinusoidal output for a sinusoidal input, may be introducing a phase shift. But nonlinear system produces higher harmonics and sometimes the subharmonics. Hence for sinusoidal input, the output of a nonlinear system is generally nonsinusoidal. The output consists of frequencies which are multiples of the input frequency i.e. harmonics. The subharmonics means the presence of frequencies which are lower than the input frequency. The input and output relations are not linear. 3. In linear systems, the sinusoidal oscillations depend on the input amplitude and the initial conditions. But in a nonlinear system, the periodic oscillations may exist
(7 • 1)
Nonlinear Systems
7-2
Modem Control Theory
which are not dependent on the applied input and other system parameter variations. ln nonlinear system, such periodic oscillations arc nonsinusoidal having fixed amplitude and frequency. Such oscillations are called limit cycles in case of nonlinear system. 4. Another important phenomenon which exists only in case of nonlinear system is jump resonance. nus can be explained by considering a frequency response. The Fig. 7.1 (a} shows the frequency response of a linear system which shows that output varies continuously as the frequency changes, Similarly though frequency is increased or decreased, the output travels along the same curve again and again. But in case of a nonlinear system, if frequency is increased, the output shows discontinuity l.e. it jumps at a certain frequency. And if frequency is decreased, it jumps back but at different frequency. nus is shown in the Fig. 7.1 (b).
0Utpu1
Peak resonant
Output
~ (a) Linear system
(b) Nonlinear system
Fig. 7.1 Jump reaonanc:e 5. There is no definite criterion for judging the stability of the nonlinear system. inc analysis and design techniques of linear systems cannot be applied to the nonlinear system.
7.3 Classification of Nonllnearitles Basically the nonlinearities are classified based on the fact that whether they are inherent or purposely introduced in the system. Depending upon this the two classes of nonlinearities are, 1. Inherent nonlinearities.
2. Intentional nonlinearities.
7.4 Inherent Nonlinearltles The nonlinearities which are unavoldeble in the control systems are called inherent nonlinearities. Let us discuss various types of inherent nonlinearities existing in practice.
7.5
Modem Control Theory
Nonlinear Systems
Ouiput
Output
(a) o,,_off with doadzone
(b) On·off with doadzone and hysteresis
Fig. 7.7
7.4.5 Nonlinear Friction In any practical system, where there is a relative motion between the two moving surfaces, there exists a friction. There are various Output types of such friction. All of them arc nonlinear Viscous except viscous friction. The nonlinear friction Is 0.()15 called Coulomb friction. The coulomb friction is a force acting in opposite direction of motion but it is constant in magnitude irrespective of velocity.
Coulomb friction
The viscous and coulomb friction are shown in the Fig. 7.8. It can be seen that viscous friction is linear while coulomb friction is constant irrespective of the input.
Fig. 7.8 Friction
The example of coulomb friction is the friction existing between the brushes resting against the commutator in an electrical motor.
7.4.6 Backlash This type of nonlinearity is generally found in the mechanical linkage where coupling is not perfect. A common example Is the existence of free play between teeth of drive gear and driven gear of a gear train. This is shown in the Fig. 7.9 (a). The driven ge..r docs not work until there is contact between drive and driven gear. So input member has to travel the distance B to have the output, as indicated by input-output characteristics shown in the Fig. 7.9 (b). Once the contact is made output follows the input. If Ule motion is to be reversed, then input member has lo travel distance of 2B in reverse direction, as shown in the Fig. 7.9 (b),
Nonlinear Systems
7-6
Modem Control Theory
(a}
(b}
Fig. 7.9 Backlash This is the most complex nonlinearity from modelling point of view. This is also memory dependent nonlinearity. It is a double valued nonlinearity like on-off with hysteresis. A process control valve with hysteresis is another example of backlash type of nonlinearity.
7.4.7 Nonlinear Spring A linear spring has a linear equation for its force and displacement. Linear spring force
... (1)
kx
where
k
positive spring constant
and
x
spring displacement
The nonlinear spring is not governed displacement are related by the equation. Nonlinear spring force
by the equation
(1)
but
kx+k'x3
where
k
Positive spring constant
and
k'
Nonlinear spring constant
its force and ... (2)
The k' can be positive or negative. If k' is positive, the spring is called hard spring while if k' is negative, the spring is called soft spring. If k' = 0, the spring is no longer a nonlinear but behaves in a linear fashion. The degree of nonlinearity in the spring is totally controlled by the magnitude of k'. If such a spring is used in the system consisting of mass and friction then it is experimentally observed that the frequency of the free oscillations increases or decreases as the amplitude decreases depending upon the value of k',
7.7
Modem Control Theory
Nonlinear Systems
Frequency k'>O Hard spring
k'
Amplitude Fig. 7.10 Frequency-amplitude curvesfor n7l1near spring The frequency-amplitude curves for a system with nonlinear spring arc shown in the Fig. 7.10. For k' > 0, as the amplitude decreases, frequency also decreases. This is hard spring. For k' < 0, as amplitude decreases, the frequency increases. This is soft spring. For a linear spring such characteristics is a straight line. These curves are often used to find out whether them exists a nonlinearity in the spring and also to find the degree of nonlinearity.
7.5 Absolute Value Nonlinearity In some systems, though the direction of the input is changed, the output direction remains same. The value of output is absolute irrespectiveof the direction of the input applied. The best example of such a nonlinearity is a relay operated by the solenoid. Irrespective of the direction of the current through the solenoid, relay gets operated. This is shown in the Fig. 7.ll (a) and (b). Output
· :]~~__..
Relay (a)
(b)
Fig. 7 .11 Absolute value nonlinearity Thus the output is not affected by the direction of the input applied.
Phase Plane Method
8.1 Introduction The phase plane method is an important method used for the analysis of nonlinear systems. The phase plane method is applied normally to first order and second order of linear and nonlinear systems. The system behaviour is qualitatively analysed ;ilong with design of system parameters so as to get the desired response from the system. The periodic oscillations in nonlinear systems called limit cycle can be identified with this method which helps in investigating the stability of the system.
8.2 Limit Cycle In case of Ilncar lime invariant systems a periodic oscillation is sinusoidal. The amplitude of such oscillations is a function of excitation applied and initial conditions. IJ the system parameters arc changed slightly i.e, if the system poles arc shifted from the imllginary axis of complex s plane, then the oscillations will no longer exist. In case of nonlinear systems, periodic oscillations are present but arc not dependent on amplitude of applied excitation. Also the oscillations arc not much sensitive to parameter variations. A periodic oscillation is case of a nonlinear system is called limit cycle. Normally the limit cycles are non-sinusoidal. The limit cycles which arc having fixed amplitude and time period can be observed over a finite range of system parameters. The limit cycle is also called self excited oscillation which is observed in certain nonlinear systems, Consider a system described by following equation, M;;-µ(l-x2);+kx=O Herc M, µ and k are positive quantities, This is a nonlinear equation. For small values of x, the damping will be negative which indicates the energy is actually put in the system. For large values ol -. thP dam;,1ng is positive which removes the energy from the system. Thus it can be expected that such system may exhibit a sustained oscillation. Now as it is not a forced system, these oscillations arc called self excited oscillations or limit cycle. (8. 1)
Modem Control Theory
8-2
Phase Plane Method
8.3 Jump Resonance The phenomenon of jump resonance is seen only in case of non linear systems. This can be explained by considering the frequency response plots. The Fig. 8.1 shows frequency n.'Sponse plot for a linear system while the Fig. 8.2 shows frequency response plot for nonlinear system.
t
Response
I\
.' [f\ : t:
~
.,_
Fig. 8.1 Llnoar system
Fig. 8.2 Nonlinear system
Consider a nonlinear system excited by sinusoidal input of constant amplitude. If the input frequency is increased then jump (or disrontinuity) occurs in the response amplitude. With decrease in frequency also, jump can be observed but at a different frequency. A nonlinear system is one which behaves qulte differently for various input functions. Consequently the logical design of a nonlinear system requires a complete description of the input signals. This dependence of the system behaviour on the actual input functions is demonstrated by the jump resonance phenomenon which is observed in certain closed loop systems with saturation.
Let us consider a nonlinear equation describing a certain system. M ~+ f=+ k1x+k2x3
= f cos rot
If the magnitude of the response for above system L~ plotted against frequency, then corresponding frequency response plot for k2 > 0 is shown in the Fig. 8.3. 8
F0tk,>O
Fig. 8.3 Jump resonance
8-3
Modem Control Theory
Phase Plane Method
Due to the presence of a nonlinear term k2 x 3, the resonant peak is bent towards the higher frequencies. As the input frequency is gradually increased from zero with input amplitude fixed, the measured response follows the curve through the points A, B and C. But at C, an increment in frequency results in a discontinuous jump down to the point D, after which with further increase in frequency, the rcsponsc curve follows through DE. If the frequency is decreased, the response follows the curve EDF with jump to B occurring at F and then move to A. For certain range of frequency the response function is double valued. For k, < 0, the resonant peak bends towards lower frequencies which is shown in the Fig. 8.4. similar Jump resonance takes place with the difference that there is upward jump when frequency is increased and downward jump when frequency is decreased. Response D F
Fork2< 0
w-
Fig. 8.4 Jump resonance Tile overall curve shows the familiar hysteresis or jump resonance phenomenon. It is well known in nonlinear mechanics and is frequently observed in high gain servo systems. In such cases, the response is multivalued and depends not only on the present value of the input but also on the past history.
8.4 The Phase Plane Method The equation of a second order nonlinear system having free motion has the following form,
.. . . . y-h(y,y)y+g(y,y)y
=0
At any particular moment,. the state of the system can be represented with the help of a point having coordinates (y, y) in rectangular coordinate system. This coordinate plane is called a 'phase plane'.
8-5
Modem Control Theory
Phase Plane Method
t,
Fig. 8.5 The starting point A with initial condilion but no initial velocity. the system comes to equilibrium i.e. to the origin with damped oscillatory behaviour.
8.5 Basic Concept of Phase-Plane Method Consider the diffcrentinl equation for a linear second order control system d2y2(t) +« dyd(t) +Jly(t) = u(t) dt t Herc u is input while y is outpuL The phase plane method is basicall y a graphical method of solving S<."COnd order linear as weU as non linear system equations. The co-ordinate plane whose axes
.
correspond to dependent variable y(t) and its first derivative y(t) is called phase plane. In phase plane plot y(t) and
y(t) arc plotted in a two axes plane. The trace of the phase
plane plot as time t increases is called trnjcctory. The starting point of the trajectory depends on the initial conditions imposed on the system. For different sets of initial conditions, various trajectories can be plotted. The complete dynamic behaviour of the system for various initial conditions can be obtained from the trajectories plotted in phase plane. The dynamic performance and analysis of given second order system differential equation can be made based on destination point of trajectories. The method has limitation of its application restricted to second order as for higher order system equations, three or higher dimensional plane is needed which is not much convenient to construct and visualise. But transient response of second order system excited by step input can be studied by this method. This method is thus applicable to linear time functions and can not be applied to systems excited by sinusoidal signals. The phase plane method can be considered as a special case of state space. With the use of slate variables known as phase variables, the state of second order system can be defined.
8-7
Modem Control Theory
Phase Plano Method
In phase plane analysis, the trajectories can be constructed analytically, graphically or experimentally. The analytical method is straight but not useful from practical point of view. The graphical methods are used when it is not possible to solve the given differential equation analytically. The graphical methods arc applicable to both linear and non linear equations.
8.6 Isocline Method for Construction of Phase Trajectory The isocline method i~ a graphical approach for determination of phase portrait. Consider the slate equation for a second order system.
111e slope m of trajectory at any point in the phase plane is given by, dx2
m = f2(xpx2)
~
f1("1·"2)
The point (x1,x2.' of the phase plane has with it associated slope of the trajectory except at singular point where the slope is indeterrniante. lsodines are the lines in the phase plane corresponding lo slopes of the phase portrait. In order to construct the phase trajectory, it is necessary to integrate the slope equation. This is very difficult except for few simple cases. We have, (~ ( x~.x2
)=
m1f1(x1.x2)
The above equation gives locus of all points in the phase plane where the slope of trajectory is m1. Such a locus is called isocline. TI1c direction of phase traiectroy at any point (x1,x2) can be obtained from the sign or ax1 and ax2 for small increment in at.
Consider once .1gain second order system equation. d2>•(1) --
dt2
dy(t) dt
-e- n-:-+(ly(t)
= u (t}
Subject to initial conditions y(t0)
= Yo
Modem Control Theory
Phase Plano Method
8-9
Based on various slopes of trajectories, isoclines arc drawn as shown in the Fig. 8.7
Fig. 8.7 conditions, the starting point is 0 which is the point on slope m1. The phase trajectory which is drown from point on m1. The phase trajectory which is drawn from point 0 on the to that having slope m2 is a straight line whose slope is the m +rn average of m1 and m2 i.e. -1-2--"This 2• line intersects isocline 2 at point P which has a From the given initial isocline 1 corresponding to the isocline whose slope is isodine whose slope is m1
slope of m2• Thus we get line segment OP. This procedure is continued at point P to get subsequent line segments PQ QR and so on as shown in the Fig. 8.7 Now consider undamped second order system such that ~ = 0 d2y(t) +2~(00 dy(t) +~ dt2 dt For simplicity let us select Uln d2v(t)
--·-+y(t) dt2 Let
\IS
=
y(t) =
=I
0
now select the state variables XI y(t) X2
The state equations arc,
.
y(t)
0
I
The above equation is
Modern Control Theory
Phase Plane Method
8 • 10
Integrating and rearranging above equation.
xi-x~
=
C
Here C is arbitrary constant obtained frorn initial conditions. The phase portrait in phase plane is as shown in the Fig. 8.8. The above equation represents a circle. The phase portrait consists of concentric circles depending on difforent initial conditions
(x1to x2lo x3to etc)
Initial oolldltion ~~~~--11---t-~--1--.~1----i~-t
Fig. 8.8
8. 7 Application of Phase-Plane Method to Linear Control System Before: studying the application of phase-plane method lo non-linear systems, let us see its application in linear systems. Let us consider second order linear control system as shown in the Fig.8.9 1------
Fig. 8.9
C(s)
8·11
Modem Control Theory
Phase Plane Method
The transfer function is given as,
C(s)
[~]
R(s) = [
1+---
(~)
K ] = s(T1s+l)+K
K
s(T,s+l)
Jn the forward path, the transfer function is C(s)
K
E(s)
!i(T1s+1) KE(s) KE(s)
Taking inverse Laplace transform, T1 c+c
Ke
e
r-c
I!
= r-<:
e
"
But
r-
..
c = r-c
c .. C=
r- f.'
Let the input be step input r (t)
I{
r(I)
r(I)
.
..
=0
The above equation becomes 0 Now At
e (OJ
fort>
0
r(O) - c (0)
I "' 0, c = 0 i.e. c (0) = 0
r (0) = R
Modem Control Theory
Phase Plane Mothod
8-13
When the value of K is adjusted in such a way as to get the system overdampcd then the phase plane trajectory is as shown in the Fig. 8.11
(a) Step response (Overdampod case)
(b) Corresponding phase portrait
Fig. 8.11 For the linear systems, where analytical solution can be computed for c and c then it is not required to draw phase portrait. It can be effectively applied in case of non-linear systems.
8.8 Second Order Nonlinear System on Phase Plane Let us consider the Van dcr Pol's differential equation,
..
.
.,
.
y-µ(1-y•)y+y
=0
.
This equation describes the behaviour in many nonlinear systems. Let us select the state variables as x 1 "' y and x2
= y.
TI\e stale equations are given as,
x2
)' .. µ(l-xi)x2-'1
The equilibrium point of this system is origin of the phase plane. Now there are two cases viz. µ > 0 and µ < 0. The phase portraits for both these cases is shown in the Fig. 8.12. When µ > 0 and initial value of x1 [i.e. x1 (0)1 is large then the system response is damped. The amplitude of x1(t)[ .. y1(1)] decreases till the system enters the limit cycle. This is shown by the outer trajectory. For small initial values of xi' the damping is negative.
Modem Control Theory
-
(a)~~>
Phase Plane Method
8·14
Stable llmilcycie
O
(b)~
Fig. 8.12 The amplitude of x1(t)[=y1(t)] increases till the system stale enters the limit cycle which is shown by inner trajectory. The limit cycle is stable as the paths in its neighbourhood arc converging. The limit cycle for µ < 0 is unstable. With increase in order of the system, the complexity is involved in applying phase plane method. The method of phase trajectories is restricted normally lo second order systems and ii is special case of phase space defined for nth order system. ln case of lime invariant systems, the total phase-plane contains trajectories with only one curve passing through every point of the plane except for some critical points. Through these critical points either none of the trajectories or infinite number of trajectories pass. For lime varying systems, two or more trajectories may pass through a single point. The phase portrait then becomes complex and not easy to interpret. Hence the method of phase-plane is mostly applied lo second order systems having constant parameters with no or constant input. Some simple lime varying systems can be analysed by this method.
8.9 Different Types of Phase Portraits In this section. the phase portraits of some commonly known linear systems is presented. Thl'SC portraits can be used for analysis of piecewise linear systems.
8.9.1 Phase Portraits for Type 0 System Consider a transfer function for a type 0 system. C(s)
ro;
R(s) = s2 +2sCJ\,s
+
Let us consider the input r(t) to be zero i.e. unforced system ls2 +2soons+ro~
JC(s) .. O
Modem Control Theory
c+2;col\ C+o>~C Let x1
a
C and x2
Phase Plane Method
8·15
a
;l = X2i ;2::
=
0
C. Therefore the state model is given by, -2~COnX2-(0~X]
Eliminating the time variable from the two equations by dividing the equations, dx2
-2~Ctl0Xz -C1)~x,
~
X2
dx
If we consider the point at which x1 = x2 = 0 then the slope al this point dx~ =
O
0
which is Indeterminate. This point is called singular point. We will study the behaviour of trajectories which are dosed to this singular point. To find out the singular points which arc nothing but the roots of characteristic equation s2+2;rons+oT,; (s-y1)(s-y2) = 0. Depending on the values of Yt and y2, six
=
different types of singular points arc obtained. These arc described as below. 8.9.1.1 Stable System with Complex Roots Let y1
= -a+jf1
y2 = -o.-jf1
a>O and ll>O
The output response in this case is c(t) = C1e-"1 sin
x2
=
C. The typical phase lrafcctory obtained is a logarithmic spiral into the
singular point which is shown in the Fig. 8.13. This type of singular point is called a stable focus.
jo>axis
~-----
ill
-a•
oaxis-
I I I I I
*----(a) Stable focus
-jjl
(b) Nature of roots Ins plane
Fig. 8.13
Modern Control Theory
Phase Plane Method
8 -17
t
jw jp
o-
-JP (a) Center
(b) Nature of roots Fig. 8.15
8.9.1.4 Stable System with Real Roots Let y1 and y2 be two real and distinct roots located in left half of s-plane. The response in this case is given by, C(t) = Ctcrt• +C2crz• The response obtained from the system is ovcrdamped. The singular point obtained after plotting the phase trajectory on (x 1 = y, x2 = y) plane for this case is called a stable node. The phase portrait contains two straight line trajectories having equations x2(t)=y1x1(t); x2(t) y2x1(t) which satisfies the differential equation of the given system.
=
Assuming that the term eT2' decays faster in the transient response than the term cT 11. As time t increases, x1 ~c,er11 ~a and x2 ~11C1cY1' ~a. All the trajectories are tangential at the origin to the straight line trajectory given for initial conditions x2(0) = /2x1(0).
2(t)=)'1x1(t).
The other trajectory is
All the trajectories arc asymptotic to this straight line trajectory. If the roots are repeated for stable systems then there is a single trajectory whose slope depends on value of root.
Modem Control
Theory
Phase Plane Method
8·18
t
jN
Fig. 8.17 Nature of roots
Fig. 8.16 Stable node
The phase trajectory and nature of roots is shown in the Fig. S.16 and Fig. S.17.
8.9.1.S Unstable System with Positive Real Roots Let y1 and r2 be two real and distinct roots localed in right half of s plane r1 is a smaller root. The singular point in this case is called unstable node. The phase portrait on
.
(x1 = y, x2 = y) plane is shown in the Fig. 8.18. Jh)
Fig. 8.18 Unstable node
Fig. 8.19 Nature of roots
The trajectories emerge from the singular point and go to the infinity. The trajectories arc tangential to the straight line trajectory x2(t) 11x1(t) al origin and asymptotic al infinity to other straight line trajectory. For the repeated roots, there is a single trajectory.
=
Modem Control Theory
Phase Plane Method
8 -19
8.9.1.6 Unstable System with One Positive and One Negative Real Root ln this case we have two straight line trajectories which have slopes dependent on values of root. The singular point obtained in this case is called a saddle. The stralghtli.ne obtained due to negative root gives trajectory entering into the singular point while due to positive part gives trajectory which leaves the singular point.
The other trajectories approach towards the singular point close to incoming straight line then takes a curve away and leaves the singular point in its vicinity. It approaches the second straight line asymptotically. The phase portrait on (x 1 plane is shown in the Fig. 8.20.
Fig. 8.20 Saddlo point
=
y, x2
=
.
y)
Fig. 8.21 Naturo of roots
8.9.2 Forced Second Order Type O System Let us consider a second order system excited by constant input with magnitude A. C(s)ls2 ·•2~o>ns+ro~
I = A~
The above differential equation can also be written in following form. d2 d • .z di2(C-A)+2~w0 dt(C-A)+w,,(C-A) = 0 The equation is similar lo that obtained for unforced system wilh the difference that the singular point is shifted from origin to point C A on the Y axis.
=
8.9.3 Phase Portraits for Type 1 System Consider the following transfer function which represents Type 1 system. This system is excited by constant input of magnitude A.
8-21
Modem Control Theory
Phase Plane Method
The total equation becomes,
l1 x 1(1) ~(x1
-
x1(0)J
=A - X2(t)-A/1~A - x2(t)]+
= -fx2
1 lx
1(0)-A
+x2(0)+A/1~A - x2(0)1
:;to)]
- x2(0)1-Al1{ AA_
The corresponding phase portrait for A > 0 is shown in the l'ig. 8.22.
A
Fig. 8.22 Phase portrait for A > 0 If initial conditions are zero i.e. x1(0) = x2(0) = 0 then ~XI
= -X7-A11{A~X2]
The corresponding phase trajectory is shown in the Fig. 8.22. The trajectory is asymptotic to the line x2 = A. If the initial point is different (xi (0), Xz(O)] then the trajectory will have same shape but shifted horizontally by k units so as to pass through the point (x1 (0), x2(0H The equation of trajectory in this case is given by, ~(x1-K)
=
-x2-Aln(A~xz)
Modem Control Theory
Phase Plane Method
8 ·22
The phase portrait for A < 0 is shown in the Fig. 8.23. The value of A = 0 is a special case which is unforced system and the phase portrait consists of number of straight lines ' 1 having the slope of - 'T. Fig. 8.23 Phase portrait for A < 0.
A
Fig. 8.23 Phase portraits for A < 0
8.9.4 Phase Portraits for Type 2 System Consider the transfer function of linear system representing type 2 system C(s) R(s)
= .l.. Js2
Let the input applied be of constant differential equation is given by, JC=
A
Let the state variable be x 1 = C, x2 = C The corresponding state model is given as, Xl Xz
c : Xz c =1A
magnitude, r(t)
A. The corresponding
8 ·23
Modem Control Theory
Phase Plane Method
By eliminating the time variable we get,
dx2
ii;z; tx2dx2
A
Jx2 A dx1
Integrating the above equation,
Where C is constant of Integration obtained from initial conditions.
J x~(t)
21\+C
J x~(O)
+C
2A C = X1(0)-
J x~(O) 2A
The nature of trajectory ls parabolic for an initial state point (x1(0), x2(0)1 passing through a point x 1 = C on the x-axis where C is given by above equation. This Is shown in the Fig. 8.24 for A < 0 and A > 0.
"1 =v
C
(a) For A>
o
(b) For A
Fig. 8.24
Phase Plane Method
8 -24
Modem Control Theory
With increase in time t, each trajectory is shown in clockwise direction. The direction of phase trajectory is obtained from the relationship;
1 ..
x2• As x1 increases with time in
upper half of phase plane, the state point hence moves from left to right in the lower half of phase plane. When x1 decreases with time, the state point moves from right to left
=
The time interval between the two points of a trajectory Is given by 6t 6x 1 I x2.tv. With the aid of this equation the time scale can be given to the trajectories. But the phase portrait is generally used to get an idea about transients in the system. A
The phase portrait or A < 0 is shown in the Fig. 8.24 (b). The unforced system with case. -,
= 0 is a special
dx2 dx';-
0
..·
dx 2- - 0
Integrating above equation
J dx
o
2
x2(t)
C
Where C is constant of integration obtained from initial conditions. C
=
x2(0)
:. We have equation of trajectory as x2(t) = x2(0). This equation represents the trajectories of straight lines parallel to x 1 axis. ,...
Example 8.1 : Determine the kind of singularity for eaclt of tltt following differt11tial
. ...
equations. i) y+3y+2y=O
= 34
ii) y-By+17y
(VTU: Jan/Feb.·2005)
Solution : i) y+ Jy+ly = 0 Let the state variables be x1 = y and x2 =
. .
.
The corresponding state model is
x1=y=x2;
x2=y=-3x2-2x1
Eliminating the time variable, dx2
-3x2 - 2x1
dXi'
X7
y
Modem Control Theory
8. 25
Phase Plane Method
The characteristic equation is given by, s2 +3s+2
0
(s+2)(s+t)
0
s = -2; s =-I We have, y1 = -1 and y2 = -2 The nature of phase trajectory is as shown in the Fig. 8.25 and the type of singular point obtained is stable node.
o--
(a)
(b)
Fig. 8.25 The response in this case is given by y(t) =
c1eYJ1
+C2cY21.
.
The transient term eY2•
decays at faster rate than eYt1• As time t-+~, x1 -.C1eY:1 -+0; x2
.
singular point node obtained in this case (0, OJ is stable in (y, y) plane . ii)
y-8y+17y
.. .
y-8y+t7y-34
34 0
The above equation can be rewritten as, d2y dy --8-+17y-34 dt2 dt d2(}'-2) d dt2-8dt(y-2)+17(y-2)
0 0
=
'Yt<'Y11 -+0. The
8 ·26
Modem Control Theory
Phase Plane Method
The characteristic equation is given by, 0
(s-2)2 -8(s-2)+17(s-2)
0
(s2 -4s+4)-8s+t6+17s-34
= 0
s2+5s-14 (s+7)(s-2) Y1
0 -7; y2
=2
There arc two roots one positive and one negative root. The singular point obtained In this case is called saddle point which is shifted from origin by 2 units. The slopes C1f the two straight line trajectories with slopes dependent on root values. The nature of phase trajectory is shown in the Fig. 8.26.
t
joo
Fig. 8.26
8.10 Singular Points of a Nonlinear System Singular point is a point on the phase plane where the slope of the trajectory i.' indeterminate. In the state model of a given system, when time variable is eliminated then we get,
By equating the numerator and denominator equations to zero i.e, solutions of the equations f(x1, x2) 0 and x2 0 gives the locations of the singular points along with Its number.
=
=
It is also Important lo know the behaviour of the trajectories in the vicinity of a singular point of a nonlinear system. The nonlinear equations are Ilnearized at the singular
8 -27
Modern Control Theory
Phase Plane Method
point using say Taylor series expansion and then the nature of phase trajectory around the singular point is determined by linear system analysis which is described earlier.
8.11 Delta Method In isocline method, entire phase plane must be filled with line segments of different possible slopes. If only a single solution curve is needed, only a few of these lines are put to use, thus wasting a lot of effort. A technique of construction known as the delta method leads more directly to the solution. This method applies to the solution of equations of the type,
.
x+f(x, x, t) ;; 0 Where f must be continuous dependent.
and single valued
.•• (1)
l>ut
may be nonlinear and time
The equation (1) is written as ... (2) The constant
ooij
may be determined from the equation (1) itself or m~y have to chosen
= °'o t
from other information. Once again it is convenient to introduce the definitions 1 dx . an d v = dt so as to give,
ooij vdx+ dv ooij x+(f(v,x,,t)-Cl>(jx) , dv dx
or where
li(v, x, t)
0
... (3)
_Ix+ ll(v, x, 1)1 v
1 x, t)-x 2f(v,
.•. (4)
... (5)
(l1j
·' Equation (4) is similar to that one used for the case of isocline method. Because of the . Jay, in which it is set up, the S method is most immediately applicable to the equations with oscillatory solutions, although it is not limited to this class of equation. Function o of equation (5) depends upon v, x and 1 but it is assumed constant for small changes in these variables. Under this assumption, the variables ol equation (4) can be separated and integrated to give, v2 +(x+5)2
= constant= R2 (say)
... (6)
Equation (6) represents a circle of radius R and centre at (-ll, 0). Thus for a suitable increment, the solution curve is the arc of a circle having these properties. The construction
Modem Control Theory
: I ·28
Phase Plana Method
is shown in Fig. 8.21 where [x(O), v(O)) represents a point on the solutiof\ curve at the time 1(0). These values used in equation (S) allow the ~kulatlon. of 6. This values of 6 determines the centre of the drcular arc located on the x-axis. The radius R Is 1utomatically fixed. A short drcular arc represents • portion of 10lulion curve. Actually it is more accurate to use the average values of v, x, 't for that Increment. Again, the allowable length of the arc is a rompromiae.
Fig. 8.27
Examples with Solutions ,,_.
Example 8.2 : A linear secottd order snw is described by tht! equation C+2l;w0 C+w~C = 0 where i; = 0.15, ~ = 1 rod/sec, CCO) = 1.5 and C ~ 0. Determine the singular point. Construct the phase trajectory, using the method of isoclines CVTU: July/Aug.·2006)
Solution : Consider the equation
C+21;w0 C+w~C Given !;= 0.15,
W0
=
0
= 1, C (0) = 1.5, C = 0 c+2(o.15H1>c+12c
o
C+0.3 C+C
0
Taking Laplace transform, s2Qs1+0.35 Qs)+C(s) (s2 +0.35+1] C(s) Consider
s2+0.35+1
=
0 0 0 -0.3±Jco.3)2 -4{t)(J) -0.3±i(t 2 = 2 -0.15±j0·9886
·9mJ
Modem
Control Theory
Phase Plane Method
8 ·29
The roots arc complex conjugate and located in left half of s plane. In time domain the response will be underdampcd consisting of oscillations and the singular point is thus sta bel focus. Consider again the differential equation d2qt) +0.3 dQt) +Qt) dt2 dt
. .
Let
:. XI = Qt) = X2,
X] = C(t),
0
.
X2 "' Qt)
X2 = Qt) = -
.
C(t) - 0.3 C(t) " -X1~.3X2 m
dx21 dx ,
=
dx2fdt dx1/dt
=
-x1 -0.3x2 x2
-X1~-3X2
-x,
-[m~o.3]x1
The above equation is an equation of isocline. At initial point (1.5,0) 0
--1-(1-5) m+0.3
m
-0.3--
= --
Due to this trajectory moves downwards Using the method of isocline, the phase portrait can be drawn as shown in the
Fig.8.28 Considering various values of m m
Isocline equation
1
•2 = -
5 10
-
0.7692 .,
xz. - 0.1886 •2
=-
9=tan-1m 45'
.,
78.69'
0.0970 •1
84.28°
•2.
0
-1
Xz =
1.4285 .,
-45'
-5
•2 • 0.2127 "'
-78.69'
-10
Xz
= 0.1030
- 84.28'
-
•2 c 0
90-
X1
-90'
.
PhasePlane Method
8 ·32
Modem Control Theory
The type of singular point obtained is stable focus . ii) Consider,
y + 3 y+ 2y
=0 y and
Let the state variables be x 1 =
. .
.
x2 = y
.
The corresponding state model is x1 =y=x2;
x2 = y=-3x2-2x1
Eliminating the time variable,
The characteristic equation is given by, s2+3s+2
0
(s+2)(s+l)
0 -2; s
= -1
We have, y1 =-I and y2 = - 2
t j(J)
x,
(a)
=y
(b)
Fig. 8.30 The nature of phase trajectory is as shown in the Fig. 8.30 and the type of singular point obtained is stable node. The response in this case is given by y(t)
= C1e111
+C2el21•
decays at faster rate than cl1'. As time t ~ -, x1 ~ C1cYJ1
~
The transient term eY2• O; x2 = y1c111
singular point node obtained in this case (0, 0) is stable in (y, y) plane.
~
0. The
8-33
Modem Control Theory Iii) Consider,
.
Phase Plane Method
=0 be x1 = y ; x2 = y
y + 3 y - 10
Let the state variables
The corresponding state model Is,
X2
.
Y =-3y+10=-3X2.,_JQ
Eliminating the time variable dx2
-3x2+10
di<";" :
X2
The .characteristic equation is given by, s2+3s-10
0
s2 +Ss-2s-10
0
s(s+5)-2(s+5)
0
(s + 5) (s - 2)
0
Y1 = -5; Y2 = 2 There are two roots one positive and one negative. The singular point obtained in this case is called saddle point. The slopes of the two straight line trajectories with slopes dependent on root values. The nature of phase trajectory Is shown In the Fig. 8.31.
v, =-5
x,
Fig. 8.31
=y
Modem Control Theory
8. 35
Phase Plane Method
Fig 8.32 C
I
M
Modem
m
lsoclino equation
1
Y2 • - 0.825y1
45•
2
Y2 • - 0.3846y1
63.43°
5
Y2 c
0.1785 Y1
78.69°
Y2 • - 0.0943y1
84.28°
10
-
O=tan-1 m
90"
_,
Y2 • 0
-2
Y2 • 0.7142Y1
-63.43"
-5
Y2 • 0.2272Y1
-78.69"
-10
Y2 • 0.1063y1
-84.28°
M
-45'
Y2 • 2.5y1
-
i•
Phase Plane Method
8 ·36
Control Theory
-90'
Y2 • 0
The phase-plane plot is shown in the Fig. 8.32.(See Fig. 8.32. on previous page)
Draw IM phase-plane trajtctoryfor lht following equation using isocline method : '.;+2!;(1);+w2x=O Given, 1;=·0.5, w2 l, Initial point (0, 6) Example 8.5 :
(VTU: JanJ Feb.-2007)
Solution: Consider the equation
'x"+21:,(1)11 ;+~x Given that I:,
= 0.5, Ci>n = 1 , x = 6, ~:
x+ x+x Let
= 0
Yt
=0
0
x
Y2 = x Yt
x = Y2 ; y 2 = x
= -x
m
~ = dyifdt = _y_2_
dy 2
-y l
dy ifdt
- x = -y2 -y I - )' 2
8 .37
Modem Control Theory
Phase Plane Method
(m+l)y2 Y2 The above equation is an equation of isocline. m+t
=
_l'..J. Y2
m
=
_l'..J._1 Y2
ll t initial point (6, 0).
:. The trajectory moves downwards Using the method of isocline, the phase portrait can be drawn considering various values of m. m
Isocline equation
1
Y2
0.5 Y1
45°
0.33 Y1
63.430
0.166 Y1
78.69"
Y2 • - 0.090 Y1
84.28'
2
Y2
5
Y2
10
~
Y2
-2
===-
O· tan-1m
=0
90'
Y2 = 1Y1
-45°
-5
Y2 • 0.25 Y1
-10
Y2 = 0.1111
-20
Y2
-
Y2 = 0
= 0.0526
-63,43' Y1
-78.69"
Y1
~.28'
The phase-plane plot is as shown in the Fig. 8.33.
-90"
Modem Control
Theory
8. 38
Fig. 8.33
Phase Plane Method
(8. 40)
Liapunov's Stability Analysis 9.1 Introduction
.
11'e state equation for a general time invariant system has the form x
=
f [x, u). If the
input u is constant then the equation will have form x = F(x). For this system, the points, at which derivetives of all state variables arc zero, arc the singular points. These singular point' are nothing but equilibrium points where the system stays if it is undisturbed when the system is placed at these points. The stability of such a system is defined in two different ways. If the input to the system is zero with arbitrary initial conditions, the resulting trajectory in phase-plane, discussed in earlier chapter, tends towards the equilibrium state. If U1e input to the system is provided then the stability is defined as for bounded input, the system output is also bounded. For linear systems with non-zero eigen values, there is only one equllibrium state and the behaviour of such systems about this equilibrium state totally determines the qualitative behaviour in the entire state space. In cruse of nonlinear systems, the behaviour for small deviations about the equilibrium point is different from that for large deviations. Hence local stability for such systems does not indicate the overall stability in the state space. Also the non-linear systems having multiple equilibrium states, the trajectories move from one equilibrium point and tend to other with time. Thus stability in case of non-linear system is always referred to equilibrium slate instead of global term stability which is the total stability of the system. Jn case of linear control systems, many of the stability criteria such as Routh's stability test. Nyquist stability criterion etc. arc available. But these cannot be applied for non-linear systems. The second method of Liapunov which is also called direct method of Liapunov is the most common method for obtaining the stability of non-linear systems. nus method is equally applicable to time varying systems. stability analysis of linear, time in variant systems and for solving quadratic optimal control problem.
(9 • 1)
Llapunov's Stability Analysis
9·2
Modem Control Theory
9.2 Stability in the Sense of Llapunov
.
Consider a system defined by the state equation x = f(x, t). Let us assume that this system has a unique solution starting at the given initial condition. Let us consider this solution as F(t : Xe- 10) where x = x0 at t = lo and t is the observed time. F(t0
:
xo- to)
=
Xo
=
If we consider a state x., for system with equation ; f (x, t) in such a way that ((•,~ ti e. 0 for all t then this x0 is called equilibrium state. For linear, time invariant systems having A non-singular, there is only one equilibrium state while there are one or more equilibrium states if A is singular. In case of non-linear systems as we have seen previously there are more than one equilibrium slates. The isolated equilibrium states that is isolated from each other can be shifted lo origin i.e, f(O, l) = 0 by properly shifting the coordinates. TilCSC equihbrium stales can be obtained from the solution oi equation f(x,,,. t) = 0. Now we will consider the stab!Uty analysis of equilibrium states at the origin. We "ill consider a spherical region of radius R about an equilibrium state x., such that
!Ix -
Xe
:J
S
R
U x - xc II is called Euclidean norm and is defined as
II
x-
•.II =
J2 + (x2 - x'k)1+ ...
[x1 - x1
(X,, -x,,il2
Let S(o) consists of all points such that 11 "o - x. II S Ii and let S(E) conststs of all points such that II F (t : >
=
Any equilibrium state x0 of the system ; f(x, t) is said lo be stable in the sense of Liapunov if corresponding to each S(t) there is S(~ such that trajectories starting in S(li) do not leave S(E) as time t Increases indefinitely. The real number I) depends on E and in general also depends on lo· If Ii does not depend on to- the equilibrium state is said to be uniformly stable. The region S(e) must be selected first and for each S(c), there must be a region S(ll) in such a way that the trajectories starting within S(o5) do not leave S(e) as time t progresses. There arc many types of stability definitions such as asymptotic stability, asymptotic stability in large. We will also see the definition of instability along with definitions of these types of stability.
Modem Control Theory
9.3
Liapunov'sStabilityAnalysis
9.3 Asymptotic Stability An equilibrium state xe is said to be asymptotically stable if it is stable in the sense of Liapunov and every solution starting within S(o) converges without leaving S(t) to x, as t increases indefinitely. The asymptotic stability is more important than mere stability. The asymptotic stability is a local stability. Hence establishing asymptotic stability does not indicate the proper operation of the system. The size of the largest region of the asymptotic stability is required. This region is called domain of attraction which is part of state space where asymptotically stable trajectories originate.
9.4 Asymptotic Stability in the Large I( the asymptotic stability holds for all states that is all points in state space, from which trajectories originate, the equilibrium state is said to be asymptotically stable in the large. Alternatively we can say that the system is asymptotically stable in the large if it is asymptotically stable for every initial state regardless of how near or how far it is from the origin.
The equilibrium state Xe of the system is said to be asymptotically stable in the large if it is stable and if every solution converges to x0 as t increases indefinitely. The necessary condition for asymptotic stability is to be determined which is normally different. Practically it stability in large is that there must be one equilibrium state in the whole state space. In control system problems, asymptotic stability in large is required in the absence of which, the largest region of asymptotic stability is to be determined which is normally difficult. Practically, it is sufficient to get a region of asymptotic stability large enough so that no disturbance will exceed it.
9.5 Instability An equilibrium state "• is said lo be unstable if for some real number E > 0 and any real number Ii > 0, irrespective of how small it is, there is always a state "o in S(o) such that the trajectory starting at this slates leaves S(e).
9.6 Graphlcal Representation The graphical representation of stability, asymptotic stability and instability is shown in the Fig. 9.1 (a), (b) and (c) respectively. The region S(o) is binding the initial state "o and the region S(E) corresponds to the boundary for the trajectory starting at "o· The correct region of allowable initial conditions are not specified for the definitions of stability that we have seen previously. Thus these
Modem Control Theory
9-4
Llapunov's Stability Analysls
definitions apply to the neighbourhood of the equilibrium slate only if S(I'.) is not corresponding to the entire state plane. S{t)
S(t)
S(6)
S(6)
(a) Stablllty
(b) Asymptotic stablllty
(c) tnstablllty
Flg. 9.1 In Fig. 9.1 (c), the trajectory is leaving S(E) and shows that the equilibrium state is unstable. It can not be stated that the trajectory will go to infinity as it may approach a limit cycle outside the region S(e). In case of linear, time invariant system if it is unstable trajectories starting near the unstable equilibrium state will go to infinity which is not necessarily true in case of non linear systems. The definitions of stability that we have seen so for are not the only ones defining the stability or equilibrium state. There arc other ways of defining st4bility also. In classical control theory, the systems which arc asymptotically stable are called stable systems and those which are stable in the sense or Liapunov but are not asymptotically stable are called unstable.
9.7 Some Important Definitions · In this section WC will consider some important definitions understanding Liapunov's stability criterion.
which are useful in
9.7.1 Positive Definiteness A scalar function F(x) is said to be positive definite in a particular region which includes the origin of state space if F(x) > 0 for all non-zero states x in that region and F(O)" 0. e.g. F(x) " xi +2x~
9.7.2 Negative Definiteness A scalar function F(x) is said to be negative definite if - F(x) is positive definite. e.g.
F(x)
= - xf-(3x
1
+2x2)2
Modem Control Theory
9.5
Liapunov's Stabllity Analysis
9.7.3 Positive Semidefiniteness A scalar function F(x) is said to be positive semidefinite if it is positive at all states in the particular region except al the origin and at certain other states where it is zero.
e.g. 9.7.4 Negative Semidefinite A scalar function F(x) is said to be negative semidefinite semidefinite.
if - F(x) is positive
9.7.5 Indefiniteness A scalar function F(x) is said to be indefinite in the particular region if it assumes both positive and negative values irrespective how small the region is e.g.
9.8 Quadratic Form A class of scalar functions which plays important role in the stability analysis based on Liapunov's second method is the quadratic form.
e.g.
P is real symmetric matrix and x is a real vector.
9.9 Hennltian Fonn If x is a complex n vector and P is a Hermitian matrix then the complex quadratic form is called Hermitian form.
e.g.
F(x)
The stability analysis in state space the Hermitian form is commonly used than the quadratic from as it is more general. The positive definiteness of the quadratic form or Hermitian form F(x) can be determined by Sylvester's criterion which states the necessary and sufficient conditions for quadratic or Hermitian form to be positive definite is
Modem Control Theory
9-6
Liapunov's Stability Analysis
=
F(x) x" Px is positive semidefinite ii P is singular and all the principle minors arc non-negative. F(x) is negative definite if - F(x) is positive definite. Similarly F(x) is negative semidefinite if - F(x) is positive semidefinite.
9.10 Liapunov's Second Method A system which is vibrating is stable lf Its total energy is continuously decreasing. This indicates that the time derivative o( the total cneri,>y must be negative. The energy is decreased till an equilibrium state is reached. The total energy ts a positive definite function.
This foct obtained from classical mechanics theory is generalised in Liapunov's second method. If the system has an asymptotically stable equilibrium state then the stored t'nergy decays with increase in time till it attains minimum value at the equilibrium stale. But there is no simple way for defining an energy function. For purely mathematical system. This difficulty was overcome as Uapunov introduced Liapunov function method which is fictitious energy function. Llapunov functions depend on x1, x1, ... xn and I. It is given as F(x1, x2, .•• Xn• t) or as F(x, t). In Liapunov's second method, the sign behaviour of F(x, t) and its time derivative F(x, t) " dF(x, t)/dt gives us information about stability, asymptotic stabil.ity or instability of an equilibrium state without requiring to solve the equations directly to get the solution.
9.11 Llapunov's Stability Theorem Consider a scalar (unction V(x), where x is n vector and is positive definite, then the states x that satisfy V(x) C, where C is a positive constant, lie on a closed hyper surface in n dimensional stale space at least in the neighbourhood of origin. This is shown in the Fig. 9.2.
=
If V(x) is a positive definite function obtained for a given system such that its time derivative taken along the trajectory is always negative then V(x) becomes smaller and smaller In terms o( C and finally reduced to zero as x reduces to zero. This indicates asymptotic stability of the origin. Liapunov's main stability theorem is based on this and gives a sufficient condition for asymptotic stability.
Modem Control
Liapunov's Stability Analysis
9-7
Theory
Llapunov's stability theorem is as given below.
.
Consider a system described by equation
=
x = f(x, t) where f(o, t) 0 for all t, If there exists a scalar function V(x, t) having continuous first partial derivatives and satisfying the conditions such as V(x, t) is positive definite and V(x, t) is negative definite then the equilibrium state at the origin is uniformly asymptotically stable. lf V(x, t) -+ oo as II x II -+ oo where II x II is norm of x, then the equillbrium state at the origin is uniformly asymptotically stable in the large.
Vincrease
Fig. 9.2
The above theorem is a basic theorem of
.=
second method. The theorem 2 is stated below . Consider the system described by x f(x, t) where f (0, t) = 0 for all t
.
.
positive definite, V (x, t) is negative semidefinite V (o (t : "& to), I) does not vanish Identically in t
.
definite In some region about the origin and U (x, t) is positive definite in the same region.
9.12 Stability of Linear and Nonlinear Systems 1f the equilibrium state in case of linear, time invariant system is asymptotically stable locally then it is asymptotically stable in the large. But in case of a nonlinear system. the equilibrium state has to be asymptotically stable in the large for the state to· be locally asymptotically stable. Hence the asymptotic stability of the equilibrium state of linear, time invariant systems and those of nonlinear systems Is different, lf it is required to tc~l thr. asymptotic stability of any equilibrium state for a nonlinear system then the stability anruysis of linearised models of non-linear systems is totally insufficient. The nonlinear systems arc to be tested without making them linearized. There are various method based on Llapunov's second method such as Krasovskii's method
9-8
Modem Control Theory
liapunov'sStabilityAnalysis
which can be used to test sufficient conditions for asymptotic stability of nonlinear systems. We shall study in detail Krasovskii's method for asymptotic stability test.
9.13 Construction of Liapunov's Functions for Nonlinear Systems by Krasovskli's Method Liapunov's direct method is a most powerful tool available for doing the stability analysis of nonlinear systems. The stability analysis by Liapunov's method includes determination of a positive definite function V(x) called Liapunov's function. But unfortunately there is no universal technique or method for selecting Liapunov's function which will be unique for a special problem. Some of the Liapunov's function will give better results than other functions, There are various techniques available for construction of Liapunov's functions. Similarly for a given function V(x), there is no general method which allows us to confirm that the function is positive definite. But if V(x) is in the quadratic form in x1's then Sylvester's theorem can be applied to test the positive definiteness of the function. If it is not possible to obtain the Liapunov's function of required type then it docs not indicate that the system is unstable. The theorems that we have seen so far just provides sufficient conditions for stability. The indication from the fact that Liapunov's function of required type is not obtained is that the attempt to establish the stability of the system is failed. Krasovskii's method gives a sufficient condition for the equilibrium state to be asymptotically stable. Consider the system described by following equation. x = f(x); f(O) = 0. The origin is a singular point.
Lei us consider Liapunov's function as, V = fT Pf Herc P is a symmetric positive definite matrix. Now,
v (If
ax_ JI
~·Ti-
9-9
Modem Control Theory
I
Llapunov's Stability Analysls
is called Jacobian Matrix given by
ar,
i)f,
ilf1
";ix1
~
ilx,.
i1f2
i)f2
i)f2
~
CJx;-
~
.
.
Substituting ( in expression for V we have, fT JT pf + fT PJf
V
= Let
(T (IT P +PI)
f
Q
As V is positive definite, for the system to be asymptotically stable, Q should be negative definite. If in addition V(x) is tending to infinity as norm of x tends to infinity (II x II~-) the system is asymptotically stable in the large. Alternatively if Liapunov function is taken as V(x)
a
fT(x) . f(x)
J(x) is the [acobian Matrix given earlier ilf1
rx;l(x)
ilf2
ilf2 ()~2
-n
f (x)
Now we have,
V (x)
ilf1
()7;
()~I i)(
We have
i)(l
~
ilf2
~:
i)fn
";ix2
() f(xl Clx
"'"Tx di
iJf,,
I
;rx;; -
• = l(x) x
ee
l(x) . f(x)
i
= f~ (x) · f(x) + fT(x) · (x) fT(x) ·
= [ J(x)
· f(x)IT f(x) + fT (x) · )(x) · f(x)
f (x) · f(x) + (T (x) · I(x) · f(x)
{f(x) ( JT(x) + J(x)] f(x)
Let j(x) : JT
Llapunov'sStabilityAnalysis
9·10
Modem Control Theory
(x) + J(x). lf j(x) is negative definite then
Vex)
will also be negative
definite. Therefore V(x) is Liapunov function and the origin is asymptotically stable. If V(x) is tending to infinity as stable in large.
I!
x
II
-1 -
then the equilibrium state is asymptotically
The Krasovskii's theorem is different from the normal lincari:zation technique. It is not limited to small departures from the equilibrium stale. The Liapunov function V(x) and V(x) arc expressed interms of f(x) or ~ rather than in terms of x, Krasovskii method gives sufficient condition for asymptotic stability of non-linear systems and necessary and sufficient condition in case of linear systems.
An equilibrium state of a non-linear system may be stable even if the conditions stated above are not satisfied. So comments on stability of non-linear systems cannot be vigoursly made by applying this method. Let us have an illustration of Krasovskii's method consider a second order non-linear system described by following equations
x1 + bx2
X:z
Let us consider that g1(0) We will also assume that
= g2(0) = 0 and g1
lg1 (x1) + g2 {x2)J2 + (x1 + bxiJ2 -> ec as
II x II
(x1) and g2("2) arc real and differentiable. -1 -
.
Now we will obtain the sufficient conditions for asymptotic stability of the equilibrium slate x = 0 f1 = x, = f (x,) • g, (x1) + 112 {x2) f2
=
"2 = f(xi) = x1 + bx2
Let J(x) be the Jacobian matrix given as
J(x)
* ~1=1
0~1
~
iJf2 UX2
ox,
[g'1~X1)
[:,(x,)+g7(X7)] a;;-!x1 +bx2) I
gl~2)]
Here
iJg,(x,) d '( .\ iJg2(X2) ax;an S2 xv = ~
Now we have, J(x)
jf {x)
+ J(x)
Modern Control Theory
9 -11
[:~::~~ ~Hg'1;x1)
Llapunov's
Stability
Analysis
s2~2>J
g] lx1l+s'1 (x1) 1+g'.zC•2 )] [ g'ztx2)+ I b+ b
By Krasovskii's theorem if i(x) is negative definite then the equilibrium state x "' O of the system considered is asymptotically stable in the large. Hence if g'1(~1) < 0 for all x1"/.0 and 4b g'1(x1)- [1 + s2(x2))2 > 0 for all Xt "/. 0, x2 7' 0 then the equilibrium state at x : 0 is asymptotically stable in the large. This two conditions arc sufficient for asymptotic stability.
9.14 The Direct Method of Liapunov and the Linear System For linear systems, Liapunov's direct method proves to be a simple method for stability analysis, Use of Liapunov's method for linear systems is helpful in extending the thinking towards nonlinear systems. Consider linear system described by state equation x = Ax The linear system described by above equation is asymptotica.lly stable in the large at the origin if and only if for any symmetric, positive definite matrix Q there exists a symmetric positive definite matrix r which is the unique solution A Tp + PA = - Q. The proof of above theorem can be given. For this we will assume the symmetric positive definite matrix P exists which is the unique solution of the equation V(x) xT Px.
=
Consider the scalar function, V(x) = x1 Px Herc V(x) > 0 for x We have,
7'
0 and V(O)
=0
V(x)
.
V(x)
A{x)1 Px + x1 PA x
x1 (ATP+ PA) x -XT
.
Qx
As Q is positive definite, V(x) is negative definite
Llapunov'sStabilityAnalysis
9 ·12
Modern Control Theory
Let norm of x define as
II x II
(xT Px)l/2
V(x)
II x
112
V(x) --> -
as II x II --> ee
The system is therefore asymptotically stable in the large at the origin. The result is also necessary. To prove this, assume that the system is asymptotically stable and P is negative definite. V(x)
xT Px
V(x)
-[~1Px+xrp;) -( - XT Qx)
> 0
XT Qx
=
This is the contradiction as V(x) xTPx satisfies instability theorem. Hence the conditions for the positive definiteness of P arc necessary and sufficient for asymptotic stability of the system. The Liapunov's direct method applied to linear time invariant systems is same as the Hurwitz stability criterion.
Examples with Solutions
n•
Example 9.1 : Show that tlze following q1111draticform is positive definite
V(x) = Sx~ +xi +4x~
+2X1X2 -4.l'1:t3 -2x2:t3
Solution : The above given V(s) can be written as
l
rs V(x) =
XT
p x
=
[x1 X2 X3J
1 -2] 1 -J [xXz
J 1 -2 -1
x3
4
l
Applying Sylvester's criterion we have, 8
> O;I~ ~I= 7>0;1~-2
1I -1
-21 -I = 20 > 0 4
As o.11 the successive principal minors of the matrix P arc positive, V(x) is positive definite.
9 -13
Modem Control Thoory
'"*
Example 9.2 :
X2
=-
Liapunov's Stability Analysis
Determine tire stability of a non-linear system governed by tire eouations
X2
Solution : Let us select the Llapunov's function to be V =xi +x~
We have,
xf +x~ i)V dx ,
Fx";.. dt i)V
Now,
i)V dx2
+ dXz".
dt
i)V
dXz" = 2x2
~
2x1;
dxt
• 2 Xt = - XI + 2Xl X2
dt
dx2
dt
X2=-X2
Substituting,
dV
dt
- 2xi (1 - 2 x1 Thus
~y
xv - 2x~
is negative definite if 1 - 2x1 x2 > 0. Thus Liapunov's criterion is satisfied
and hence the origin of the system is asymptotically stable.
11*
Determine the stab;Jity of tire system described by tire followingequation
Example 9.3 :
.
x=Ax;A=
[-1 1] -2 -4
Solution : To check the stability of the system described by equation ~
=
Ax we will solve the equation ATP+ PA= - Q where for any symmetric, positive definite matrix Q. there exists a symmetric positive definite matrix P. We can find out matrix P for any arbitrary choice of positive, definite real symmetric matrix Q. Let us select Q = I, where l is identity matrix
-Q
ATP+PA
Let
p
=
[Pn
P12
P21
P22
l
J
9 ·14
Modem Control Theory
Liapunov's StabilityAnalysis
o]
-1 [ 0 -1
But P is symmetric matrix ., P12 = P21 :.[
-2p11-4P12 P11 -5P12 -2P22
r11-Sr12-2Pn]=[-1 2P12 -Sp~2 0
-2 P11 - 4 P12 P11 - 5 P12 - 2 P22 2p12 - 5 P22
OJ -1
- 1 0
- I
Solving the three equations using Cramer's rule,
1-2
['. = 11
I
-I OI =-2(25+4)+~(-5) -5 -2 =-2(29)- 20 = -78 0 2 -5
-~ ~~ -1
1-2
I
I
01
=-1(25+4)+4(0-2) -2 =-29-8 2 -5 =-37
-1
01=-2(0-2)+1(-5)
1 0 -2 =4-5=-1 0 -1 -5
-~ ~ -~1 0
2
-1
rt-2(5)+4-1(2) =-10-4-2 =-16
Pu 6Pt2
P12
Pit
-1
1
= -ll- = -78 = 78
Modem Control Theory
9·15
Liapunov's Stability Analysis
Using Sylvester's criterion it can be seen U1J1l P is positive definite. Therefore origin of the sytcm under consideration is asymptotically stable in the large.
i•
Example 9. 4 : Use Krasouskii's theorem to show that the equilibrium state x = 0 of tllr system describedIn;
.
x1 = - 3
x1 = x1
-
X1
+ X2
x2 - xl
is asymptotically stable in the large.
(VTU: Jan./Feb.-2005)
Solution : We have, ;I : 1, (x) = - 3x1 + X2
~ =
f2(x)
= x1
-
x2
-
xi
The Jacobian matrix is given by
a12 ~
ar,
; ~=-1-3xi
1
,
2
(-3 1
1
-1-3x~
Let
Q
)TP ... PJ
Let
p
[Pn
P21
]
P12]~[1 ~]be identity matrix P22 0
Q
1 ][I OJ [I 0 ][-3 1 ] [~3 -1-3x~ 0 1 +0 l 1 -1-3xi
Q
I ] [-3 l ] [~ [~3 -l-3xi + 1 -l-3x~ = 2
2 ] -2-6x~
Modem
Llapunov's Stability Analysis
9·16
Control Theory
For system to be asymptotically stable Q must be negative definite i.e. - Q must be positive definite
-Q
= - [-{>2
2 ] [ 6 -2 ] -2-6"i = -2 2+6x~
Using Sylvester's criterion, 6 > 0 and 12 + 36 x~ - 4 > 0 i.e, 12 (1 + 3x~) > 4 then the equilibrium state x = 0 is asymptotically stable in the large.
of the follmving
Example 9.5 : lnwstigate th« stability method of Liaupnov.
,,...
• Xz
= - X1 - Xl2 X2
Solution : Given that ~1
no11-/i11erar
(VTU: July
sysl
using direct
/Aug.· 2005,J•nJFeb.-2005)
= "2• ~2 = - x1 - x~ x2
Let the Uapunov function be, V = x~ + xi
.
.
.
2 XI XI + 2 X2 "z
V
2 Xl
"2
+ 2 X2 (- Xl - X~ X2)
2x1 X2 - 2x1 X2 - 2 xf x~ - 2 xi xi It ran be seen that V < 0 for all non-zero values of x1 and x2. Hence the function is negative definite. Therefore the origin of the system is asymptotically stable in large.
"*
Example 9.6 : A smmd order systmi is reprtsnrttd by
·
x=AxwhereA=
[o 1] -1
-I
Assuming matrix Q to bt identity matrix, solo« for matrix P in the equation ATP+ PA= - Q. Use Liapunov theorem a11d detemrint the stability of tire origin of the system. Write th« Liapunou function V(x). (VTU: July/Aug.·2005, JanJFeb.-2007) Solution : ;
Let
/
=
Ax
p
P12} [P11 P21 P22
Q-['
- 0
CJ1
- 2 P12
P11 - P12 - P22 2 (P12 - Pn)
Uapunov's Stability Analysis
9 ·17
Modem Control Theory
= + 1/2
-1
:. P12
0
:. Pn - P22 = P12 = 2
-1
:. P12 - P22
I
I
2
1
P12 +
P11
P12 + Pzz = 2 + 1 = 2
=
2
l
P22
+
2
=-
I
2
= 1
1
3
' [! '.] Using Sylvester's criterion, P can be tested for positive definiteness
• P is positive definite. Hence the equilibrium state at origin is asymptotically stable in the large. The Liapunov's function is
V(•) • .'p
V(x)
>"
('> '>I
[f '.] [:;]
Modem Control Theory
''*
Llapunov's Stability Analysis
9·18
Example 9. 7 : A system is describtd by tM following equation : (VT\J: Jan/Feb.· 2008)
;=Ax Where A=[ ~I ~} Ass1m1i11g matrix Q to be the ide11tify matrix, sol!.lf! for matrix P and commmt 011 the stability of the system using the equation ATP + PA = -Q.
Solution : To check the stability of the system described by equation x =Ax. We will solve the equation, /\ T P+ PA =-Q where for any symmetric, positive definite matrix Q, there exists a symmetric, positive definite matrix Q, there exists a symmetric positive definite matrix P. We can find out matrix P for any arbitrary choice of positive, definite, real symmetric matrix Q. Let us select Q = I. where I is Identity matrix. A TP+PA =-Q. p
Let
=
[Pn
P12] P22
P21
A
=
[-1 -2] I
-I
,
AT=[=~
~]
But Pis symmetric matrix :.p12 =p21 -2pu +P12 +P21 [-2p11 -Sp12 + P22 -2p11+P12+P21
-1
-2p11 -SP12 +r22
0
-2P12 -2r21 -8P22
-2p11 -5p12 +P22 ]=[-I -2p12 -2P21 ··Sp22 0
-1
Solving these three equations using Cramer's rule
0] -1
Modem Control Theory
1
6 = [-2 -2 -5
is]
-2 -2
[-I 0
-5
-1
-2
[-2
-1
6p II =
"'Pu = -2 0 -2 -1
is]
~sl
~I]
[-2 6p22 = -2 -5 -2 -2 -1
9 -19
=-lliQ+fl-1(16+ =-2(42)-18-6
~+ 1(4-10)
=-84-18-6=-108 =-1(~0+ ~::!{.!)+ 1(-5) =-80-1-5 =-86
=-2(0+ 1) +!.{!§_+~+ =-2+11i+2 ill
!fn
=-?:.{5!-1(2)-1(4-10) =-10-2+6=-6
P11
-86
86
-6- = -108
108
6P12
P12
1.8
P21
tipll
Llapunov's Stability Analysis
-ti- = -108
6P22
-6
-ti- = -108 p -
86
6
108
-18]
108 106 [ -18 6 108 108
Using Sylvester's criterion, it can be seen that P is positive definite. Therefore origin of the system under consideration is asymptotically stable in the large. ..
>m+
•
,3
Example !Lii ; Consider lire system willr differential equalio11 c +kc+ k 1 c + ( = 0. Examine Lite stability by Uapu110t•'s met/rad, gi1Je11 tlrat k > 0 and k1 > 0. CVTU: July/Aug.-2007)
Solution: Consider the given equation, -
•
.J
c+ kc +k1 e +c=O,k > 0, k1 > 0
Let
e
">
M
.. Modom Control Thoory
9. 20
Liapunov's StabilityAnalysis
As there is no specific procedure for selecting Liapunov's function but it has to be selected from experience. Let us select f = x~ + x}ns Liapunov's function df
dt
The above equation is negative semidefinite and hence the system is stable.
I»*
Exami11e the stability of ti~ system d~cribed lly tlte foll1r.ui11g equation by Krasovaskii's theorem. Example 9.9 : XI= X2
-X1
= X1 -X2 -X~
(VTU: July/Aug.·2007!
Solution: We have,
The Jacobian matrix is given by,
r~l L
Let
Q
Let
p
Q
()
-1-3x~
]
)T P+P)
[P11 P12 ]=[1 P21 P22 0
[~l r-1 LO
~] be identity matrix
r
1 ][' OJ 0][-1 -1-3x~ 0 1 + 0 1 1
I
]
[-1
-1-3x~ + 1
0
-1-3x~
]
0 } -1-:Jx~
= [-21
l
-2-6x~
]
Modem Control Theory
Q
9 ·21
=
Llapunov's Stability Analysis
-2 1 ] [ 1 -2(1+3xn
For system to be asymptotically stable Q must be negative definite i.e. - Q must be positive definite. -2
- Q
= -[ 1
1 ] [2 -1 -2(1+3xl) = -1 2(1+3xn
]
Using Sylvester's criterion, 2>0 and 4+ 12x~ -1 >0 i.e. 4+12xi > 1 i.e. 4(1+3xD > l then equilibrium state x = 0 is asymptotically stable in the large. Example 9.10 : Deiereminewlitthu or not fall
,,_.
Solution : Consider the given equation Q(x,,x2)
= lOxi +4xi +x~ +2X1X2 -2X2X3 -4X1X3
Consider,
=
P11Xi +p22xi +p33x3 +2P12X1X2
+2p23x2X3
+2p13X1x3
Equating coefficients of above equation with those from given equation P11 =10 ;
P12 =4;
2P12 =2;
P12=1
2p23 =-2;
P23 =-1
2p13 =-4;
P13 =-2
P33=1
The given equation can be rewritten as,
4 -1
=~] [:~1 1
x3
9 ·22
Modern Control Thoory
Llapunov's StabllltyAnalysis
Applying Sylvester's criterion we have, 10 > 0;
1
10 1
~I
= 39 >
o;
10 1 -2] 1 4 -1 [-2 -1 I
10(-1-1)-1(1-2)-2(-1+8) 30+1-14 =17 >0 As all the successive principal minors of the matrix P are positive, Q(x1,x:z) is positive definite.
I"*
: Using Lyap1111ov's direct met/rod, find tire range of K lo guarani~ stability of the system slrown in the Fig. 9.3. (VfU: JanJFeb.· 2006) Example 9.11
Solution:
Fig. 9.3 The above diagram can be redrawn as,
~----------
9. 23
Modem Control Theory
Liapunov's Stability Analysis
In matrix from above equations arc written as,
.
XI X2 X3
Here
A
[!
0 -1
[~
0 -1
-:][:;l -1
X3
-:i
:.AT=[~
-K
-1
-1 0
IJ
To find the range of values of K for stability, we will solve the equation A r P+PA =-Q.
Let us select
Q
This selection of Q is valid as this docs not make xT Qx identically equal to zero except at the origin.
-l
0
l [::; ::: :~ l
~
-1
PtJ
P23
P33
-:1-[~ ~ ~1
P12
+ [::;
Pn
P1;
p23
:~] [~ ~1 p33 0 0 -1
Solving the equation we get 1 J«2-K) p
0 -1 2(2-K)
0 l 2(2-K) 1 2(2-K)
-1 2(2-K) 1 2(2-K) 1 (2-K)
0 0 -I
9 -24
Modem Control Theory
For
r
Liapunov'sStabilityAnalysis
to be positive definite it is necessary and sufficient that K(2-l<) >0 K>0;2-K>0,2>K
Thus for 0 < K < 2, the system is asymptotically stable.
Review Questions 1. Wrilt a note on stability in the sense of Liapunot1. 2. Define ii stability iii Asymptotic sta/lility iill Asymptotic stability in the large. J. Shotv the graphical rrpres
semirkfi11ittnt55vi inrkfinitmtss 5. Disc11ss Quadraticform mid Hermitian form. 6. Erplai11 Unpunov's second method and /Jap1111ov's stability theortm. 7. Write a nott 011 stability of lintJJr and nonlinear systems. 8. Write a short note on Kra5ovskli's mtlhod of constnicting Uapunov's functions for nonlinMr systems. 9. Examine th« stability of ti~ origin of the fo/lOtving systtm
- 6 XI - 5 X2
X2
JO. Write a Liapuno»function for the system
rl ;2;, Lr 1] [ x, J' J l 2 -3 -i
•i
Dtlermi"' the stabi/ity of ti.- origin of ll1t system. 11. Determine /ht stability of the equilibriumstale of the following syslmf. X1
- X1 - 2Xl:
.\'2
X1-4xz-]
+2
12. Statt and explainLiapunoo's thtoremson iJ Asymptotic stability
ii)
Global a.•ymptotic stability and iii) Instability. = 0 of system dtscribed by
13. Ust Krnsousk'ii's theorem to shaw that tht equilibrium slate x
x 1:::: - 3x1.,. x1 x2 ; >:z - x2 - x~
9 ·25
Modem Control Theory
Llapunov's StabilityAnalysis
14. Investigate tl"' stability of folluu1ing non-lineer system using direct mtthod Uap1111cro. Xl
= X2
X2=-x1-xl
X2
15. A 5'QJnd ordrr system is rrprtstnted by ~=Axrvl1LrtA=f0
L
-I
']·
-I
Assuming matrix Q lo be U!entity matrix, sol<>e for matrix p in the equation ATP + PA ~ - Q. Usr Liapuno» theorem and dtlem1i11e I/it stability of tht origin of systrm. Write tilt Uapunov func!iDn V(x).
DOD
(9.
26)
Modern Control Theory ·-=
~
•.
.........
Chapterwise University Questions with Answer Jon./Feb. • 2005
July/Aug. • 2005 Jon./Feb. • 2006
July/Aug.• 2006 Jon./Feb. • 2007
July/Aug.• 2007 Jon./Feb. • 2008
(P • 1)
CD
State Variable Analysis and Design
Q.1
Mention tlie diS11dlHl11tages of conr>enlionAJ amtTol lhtory and nplain how ~ art overco= in modern control lhtory with particulilr refermce lo (i) Non·lilWlr systems (ii) Time varying system (iii) Analysis (iv) DtSign and (11) Computerapplications.
Ans. :
Reier section 1.1.
Q.2
Explain the concepts of state NriAbk, state and state model of 11 lirmtr system.
Ans. :
Refer section 1.2.
Q.3
Define the concept of i) Stott ii) S14te variJlbles iii) Stolt space tu) State vector.
Ans. :
Refer section 1.2.
Q.4
Compare clsissical control theory against modern control theory.
(Ja.nJFeb.-2005,10 Marks)
(JanJFeb.·2006, 6 Marb)
(July/Aug.-2006,Jan./Feb.-2008,6 Marks; Jan.lFeb.-2007, 10 Marks)
(July/Aug.·2006,4 Marks) Ans. : · Refer section 1.1. Q.5
Ust advantages of modern control theory over conventional controt theory.
Ans. :
Refer section l.1.2.
Q.6
Drove Ille equationof the vector model differential state equation.
Ans. :
Refer section 1.3.
Q. 7
Mention tlie differences betwt't11 stalt space tedmiq11tSll1ld classicalapproach.
Ana. :
Refer section i.1.
(July/Aug.- 2007, 4 M•rla)
(July/Aug.- 2007, 1 Maries)
(Jan./Feb.-2008,6 Marks)
ODO
(P • 2)
p.4
Modem Control Theory
State Space Representation
Unmriu the fol/{1Wingequation in the neighbourhood of the origin.
Q.6
"29 =lnn-129-3sin9+211eul2 dt2
+113
Obtain tht approximate response O(t) for u equilibrium.
0.02, with the system initially at (JanJFcb.·2006, 6 Marlcsl
Ans.:
sin 0
.... in the neighbourhood of origin (20)3 (20)5 20+-3-+-5-+ 20-30+2ueu12 11
cu/2
(11/ 2)2
•29
+u3 (u/ 2)3
I+2+-2-!-+-3-!-+
d20
-0+2u
dtT
= - 0 + 2u = -
.
... neglecting higher order terms
d2a +9 = 2u dt2
X2(t)
in the neighbourhood of origin
... lineanzed equation
X1(t) + 2u
Thus the state model is,
[~::::] ~ [~1 ~][~~~:~]+[~]
u
p.5
Modem Control Theory
[sl-A)
Adj [sl-A[
[sI
s[l
=
OJ-[o-1
0 1
State Space Representation
']=(s -1]
0
1
s
[+1s -l]T s [s 1]s =
-I
:L[s/+1
-Ar'
Adj[sl-AJJ~l Isl-A I 52 +I
2_
s2 +I ZSR
L-1{q{s)BU(s)I where U(s)
L-1
![s~ l s :+ 5 1
s +1 s 2
2+1
l
l
52 s+I
l
= O.OZ s
[~)[0.~2] =L-1
[s(~s~~)l+1 l 2
L-1 {~+ Os+C} s +1
L-1 { s(~~~l)} As2 + A + Bs2 + sC
2
s2 +I I
52
0.04
i.e, A + 8 = 0, C = 0, A = 0.04, 8 = - 0.04.
0.04 - 0.04 cos t 0.04 - 0.04 cos I Q, 7
... u
= 0.02
Clioosi11g appropriate pliysica/ i.-oriables as slate variables, oblai11 tlu: slate model for I/re electric circuit sliown ill Fig. 2. (JanJFeb.·2006, 8 Marks) 1 F
1 H
10
y{I)
Fig. 2 Ans. :
Refer example 2.29.
M
P-6
Modern Control Theory Q.8
.
)'(.<)
State Space Representation
>(>+2)(.<+3)
For ll1e tmnsfcr [unction -- - = -··-·,--·-·-R(.<) (> + 1)· {; H) O/Jtai11 the stat« mode!
111
ii Phase oorwble canonical form iii lordun cm1011ical form
UanJFel>.-2006, 8 Marks)
Ans. :
Refer example 2.30.
Q.9
l'ig. 3 s/1ows tire />lock diagrcm of a speed co11lro/ system willi slate variable feedback. Th« driue motor ;5 mt annature controlled de motor with armnturt resistanr« R0, '1rmature inductance L,., motor torque constant Kr, inertia referred to motor shaft /, viscous friction coefficient referred to Ilic motor slrafl B, back emf constant Kb and tachometer K1. '111< applied armature voltage is controlled by a llrrce phase [ull-conoerter. c, is conlrol 110/tage, is armature voltage. e, is lite reference t'Oltage correspo11di11g lo the desired speeil. Taking X 1 = rn (speed) and Xz = ia (an11alure current) tis t/11• Mak variables. u = e, as Ifie input, and y = (O as tl1e output, derive 11 state vanable mode for the fe<>dbnck system. Uuly/Aug.·2006, 6 Marks)
•a
Fig. 3 Ans. :
Refer example 2.31.
Q.10
For a RLC network sho11111 i11 Fig. 4, write tire stale model in matrix 11ot11tio11 cl1oosing X1(t) = i(I) and X2(tl vcW where X1(1) and X2W are stale variabfrs, Vc(t) is 011tp11t, v(t) is input. Uuly/Aug.-2006, 8 Marks)
=
~T
~,, :u Tl Fig. 4 Ans.
Refer example 2.1..
M
p.7
Modem Control Theory Q.11
For a transfer ftmction gioen l1y G(s)
State Space Representation
= -,--2--write the stalt model i11 tltt ii Phase s·+JF+2
variable form ii) Diagonal form.
(Ju.ly/Aug.-2006, 8 Mari...)
Ans. :
Refer example 2.32.
Q.12
Stott
Ans. :
Refer section 2.3.1.
Q.13
Obtain the state space representation model for Giw11 R = I MO and C = l µI'.
tire
properties of [ordan matrix,
(July/Aug.·2006, 6 Marks)
tiff
follcnving electrical circuit i11 Fig. 5. (JanJFeb.·2007,10 Mades)
R
R
u(t)
y(t)
Fig. 5 Ans. : Q.14
Refer example 2.22.
... .. .
Obtain tit;, stale space representation of the folluwing system and draw its phase variable dillgram: Y +6 Y +II Y +6Y=6tL (JanJFeb.· 2007, 10 Marks; Ju.ly/Aug.·2007, 6 Marks)
Ans. : Taking Laplace transform of both sides and neglecting initial conditions we get, Y(s) 6 U(s)- s3+6s2 +lls+6· Then refer example 2.15. Q.15
Obtai11 the state modd of the electrical network shown in Fig. 6 selecting 'V', 'i1 ·and as state variables and voltage across R2 and cuntnt 12 through R2 are the output variables Y1 and ¥2. (July/Aug.·2007, 8 Marks)
·;2•
Fig. 6 Ans. : Refer example 2.19 for the procedure. The direction of i1 is opposite in this example. Thus the matrices A and B are,
.I
P-8
Modem Control Theory
RI -r; A
+-
0
-I-; i
1
Li
L1
R2
0
State Space Representation
+...!... L2
-c -c:
0
B
=
0 0
There are two outputs,
Eo(tl
Y2
i2 = X2
c Q.16
=
yI
[~
i2R2
=
R2X2
~2]
For tire network show11 in Fig. 7, dwosing i1(t) = X1(tl and iz(t) = X1(t} uariables. obtain tlie stale eq11atio11 and output equatio11 i11 vector matrix form.
as
state
I
(JanJFeb.·2008, 8 Marks)
1 H
Hl
IH l~t) u{t)
~------i i_t_)l~---~ 10
1n
y(t)
Fig. 7 Ans. :
Refer example 2.33.
Q.17
Ob1JZi11 the stat~ model in plrase variableform and write I~ block diagramfor t/1e system represented by,
(JanJFcb.·2008, 6 Marks)
,,Jy .i2y dy +6-+11-+10y=311(I). di 3 1112 dt Ans. :
Refer example 2.16.
Q.18
For tire following transferfunction obtain t~ state modtl in catwnicolform: (JanJFeb.·2008, 8 Marla)
G(sJ = Y(s) U(s) Ans. :
=
6 s3 + 6s2 + lls + 6
Refer example 2. I 5.
ODO
Matrix Algebra and Derivation of Transfer function
® Q.1
Determine lite transfer matrixfor lite system
QanJFeb.·2005, 10 Mules)
[x,x ] [-3 l][x' n 4 6]o ["'] :[YI ]=[J -1] [X1] 1
=
-2 0 X2
-5
u1
yz
B
1
Xi
Ans.:
Refer example 3.20.
Q.2
Find t11e transformation matrix P I/tat transforms lht matrix A into diago11al or Jordan formwhmA=
rl14 -1I -2] I
0
2. 3
QanJfeb.·2005, 10 Marks)
Ans . : Refer example 3.15. Q.3
What are ge11raliud eigen vectors? How are they determined? Quly/Aug.·2005, 5 Marks)
Ans. :
Refer section 3.10.
Q.4
Convert the followi11gstate model into canonicalform
QulyJAug.-2005,8 Marks)
x =[~ ~]x+[~ ]11 1
Y
= [1
OIX
Ans. :
Refer example 3.21.
Q.5
Convert the following square matrix A into Jordan canonical form using a suitable non-singular transformatio11 matrix P. QulyJAug.·2005, 7 Marks)
A=f~L-4 -9~ Ans. :
~i
-6
~for example 3.22.
(P • 9)
Modern Control Q.6
Matrix Algebra & Derivation of Transfer Function
p • 10
Theory
Consider the matrix
~i
A=[~1 -:3
-1 i) Find the eigen val11es and eigen vectors of A ii) Write tlte modal matrix iii) Show that lite modal matrix indeed diagonalizes A. Ans.:
Refer example 3.23.
Q.7
Given the stale model X " AX - Bu, y " CX toher« A
=[ ~
~
-1
-2
~i'
-3
B
= [~]
and C
=
(Jan.JFeb.-2006, 12 Marks)
[1 0 OJ
l
i) Sim11/nle and find tire transfer function ~~~ using Mason's gai11 formula. ii) Determine the transfer f1111ctio11 from tlie state model formulation. (JanJFeb.·2.006, 7 Marks) Ans. : Q.8
Refer example 3.24. ti« vector [_~] is an eigm wetor of A = [-~ ~ -I -2
~1
Find tire cigen value of A
0
Ans. :
corresponding to tlu vector given. (July/ Aug.·2006, 5 Marks) Refer example 3.25. Show that tlU' cliaracteristic equation and eigm Mluts of a sytem matrix art 111wariant under linear transformation. (July/Aug.·2006, 8 Marks) Refer section 3.9.
Q.10
Delennine tire transferJunction of the giwn state vector differentialequation below :
Ans. : Q.9
[~~]=[~ ~ X3
0 -4
~1[;:]+[~1]11
-5
X3
5 (July/Aug.·2007, 8 Marks)
Modem Control Theory
Matrix Algebra & Derivation of Transfer Function
p • 12 1 (;
s+5 s(s+-l)(s+l)
1 s+5 5 s - s(s+4)(s+1) + s(s+4)(s+1) 1 -s-5+5 s+ s (s+-1) (s+l)
=
s s (s+4) (s+l)
(s+4) (s+l) - s s (s+4) (s+l)
(s+.t)(s+l) T.F.
I
= ;;-
s2 +5s+4 -s s(s+4)(s+l)
s2 +4s+4 s (s+4) (s+l)
Obtai11 eigen values, eigm vectors a11d stale model in canonical form for a system described by the following stale model : (July/Aug.·2007, 10 Marks)
Q.11
~ ~ ~] [~;:]3 = [-12 -7 -6
[=:] .T3
+ [~] w11d y 2
= [1 o o J
[=:] X3
Ans. : Refer example 3.14 for obtaining M. Once M is obtained find M""1 and the new matrices B = M"1B and C =CM
[
Q.12
l
1 2 -3.1
M= -I -4 -1 1
[4.5 2.5-2 -1I } B= [6.-55}
M""1= -3
3
2.5
1
C=(I
-12 -7
Refer example 3.14.
~1
-6
2 1].
4.5
Obtain tht model matrix for tire matrix givt11 btlow :
A-[~~ Ans. :
1.5
(JanJFcb.·2008, 4 Marks)
Solution of State Equations
Q.1
For a o1Jslem represerrted by X = AX tire respc~ lo one set of i11itial co11diti1ms L~ X(I) 2''-41]
= [ c-1•
.r initia ... I cond.ition ' a11d anot I1er set oJ ts. X( I)
the state transition matrix OW.
=
c-it]
[4c_21
. matix . . D etermine
A and
(JanJFeb.-2005, 10 Marks)
Ans.:
Refer example 4.19.
Q.2
Mr11tio11 ilu: ccmditions for complete controllability and complete obsavability of conti11ous time systems. Using these, explain Ille principle of duality between controllability and observability.
Ans.:
Refer sections 4.12 and 4.13.
Q.3
Use controllability and observability matrices to determine wlretlier tlie system represented by tlie flow grapli show11 i11 Fig. 1 is completely co11trollable and completely obstroablt.
<J•nJfeb.-2005,JanJFeb,-2008 10 Morks,;July/Aug.-2007 6 Marks)
-1
y
u
Fig. 1 (JanJfeb.-2005, 10 Marks) Ans. :
Refer example 4.33.
Q.4
Given thr time i11varia11t system :
(P - 14)
M
Modern
p -15
Control Theory
=
Solution of State Equations
e 1 and
and 111111 11(tJ y(I) = 2-nk I, find X1(1) and X2(t). Find also XifOJ a11d X2(0). Wlmt li.1ppe11s if n = O? (JanJFcb.·2005,10 Marks! Ans. :
Refer example 4.34.
Q.S
What is a stat« transition matrix ? I.isl tire properties of state transitio11 matrix. (July/Aug.·2005, July/Aug.·2007,Jan.fcb.·2008,6 Marks; July/Aug.·2006, 5 Marks, Jan./Feb.·2007, 10 Mari<s)
Ans. :
Refer sections 4.3 ,111d 4.4.
Q.6
Gi11!'11
the stale model tif a sy~tem : X
[~
Y
:5] X
+[~]II
(1 OJ X
- [!.}
with initial condillo11s X(O) -
Determine : i)
Tile state transition matrix.
ii)
The stale trnnsitio11 equation X(t) and 011tp11t Y(t) for an 1111il step illput.
iii) lliverse state tra11sitio11 matrix.
Quly/Aug.·2005, 14 Mark51
Ans. :
Refer example 4.35.
Q.7
F.rp/ai11 tilt concep! of co11trollabi/ity and ob!:croabi/ity. Quly/Aug.·2005, )uly/Aug.·2006, 6 Marksl
Ana. :
Refer section 4.11.
Q.8
Determine tire controllability and observability of tire following stale model.
y
ftO 5 l)X
Quly/Aug.·2005, 8 Marksl
Modem
P-18
Control Theory
Solutionof State Equations
Ans. :
Refer example 4.36.
Q.9
A system represented by following statt model is controllable but not obsnvablt. Show t/ull tire no11-<Jbstruability is due to 11 pok-:uro ettnetllation in C/sl-Af18.
[! :Jx +mu
x
-~1
=
Y
[1 I O]X
(July/Aug.-2005, 6 M..001
Ans.:
Refer example 4.37.
Q.10
List 11tleast thrte imporl11nl propenie» of th« state transition matrix.
Ans.:
Refer section 4.4.
Q.11
Consider tlie lwmogenevus equation X = AX, wltere A is a 3 " 3 matrix. The following three solulio11 for three different initial comfitions art 111N1i111ble,
(JanJFeb.-2006, 3 Marks)
i:: lJ' [ .-1
[
i)
,-21 _;~-21
.
l[ l 2r31 ' -6~-31
Identify the initial conditio11s
ii) Find the stale transition matrix iii) Hence or otherwise find the system matrix A.
CJanJFeb.·2006, 10 Mulu)
Ans. :
Refer example 4.38.
Q.12
Obt11i11 the time response y(I) of lite system give.. btlow by first transforming the state model into a 'Canonical model'.
X=[ ~
~
-6
-11
~1
-6
X + [~] u, Y 2
u is a unit step function and Ans. :
Refer example 4.39.
= [1
xT (0) = [O
0 OJ X 0 2]
(JanJFeb.-2006, 12 Marb)
p ·18
Modem Control Theory
Solution of State Equations
Obtain the state transition matrix using :
Q.17
i)
l.llplace transformation method and Cayley-Ha111i/to11mtlhod. For tire system describe by,
ii)
X(I)= [~
~]X(O)
(JanJFeb.-2007, 10 Marks)
Ans. :
Refer example 4.43.
Q.18
State the conditions for C-Ompfetely controllability and complei« o/Jser.x1bility. Determin«
•;: :"[t~ [
c~nrro_;•]b[i;~l+[ai1]d[ulobseniability -11
4)
,J
6 X3
of
the
SIJSlem
described
by,
1
(JanJFeb.·2007, 10 M3rks) Ans. : Refer section 4.11 and example, 4.36 for the procedure. The system is completely controllable and completely observable. Q.19
Obtain the olJS<'rmble phas« variable slate model of T.F.
=
T(s)
=
)'(>) U(s)
= bo>3
+bis2 +bi> +b3. Draw tire signal .flow grap/1 of Tis). >3 +•1>2 ... azs +n3 (July/Aug.·2007, 8 Marks)
Ans. :
Refer example 4.46.
Q.20
Obtain the controllable phase t1ariable form of tlie transferfunction 'r.t:
a
Tts)
=
Y(sj
= bo> 3 +b1>2
0(>) Ans. :
Refer example 4.47.
Q.21
Compute ell• for the given matrix : A-
1 -
+b3.
... bzs s3 +n1s2 +a2s+a3
ro]
6 01 ·1h- [ o ·A[0 6} " - -C•) 0 '
Quly/Aug.·2007,6 Marks)
-l-(r 6l)
ooJ 6
(July/Aug.-2007,6 Marks)
Modem
Ans. :
Refer examples 4.27 and 4.28.
Q.26
Dd<•ri11111t
.X (I}=
t/w compfrtr tim1•
f[-2O
']s(t),
Solution of State Equations
p ·20
Control Theory
rt'S)JOllSC
of t/1r $ystrm i:1w11 /1y :
lnnd
uihrre X
(I
Ans. :
Refer example 4.7.
Q.27
Deter111i11r tile co11trol/ability dtSCl'ibed by.
IJ
Y(t)
=fl
-l I X(IJ. (JanJFel>.·2008,
12 Marks)
a11d obsernability 11si11K K11l111a11 's ltst for tire systn11
A (J;m./Fcb.-2008,
10 Marks)
Ans. : Refer example 4.36 for the procedure and the given system is completely controllable and observable.
DOD
Modern Control Theory
Q.11
P-23
Omsider the systcn defined by ;
Pole Placement Technique
= Ax + Bu, w/1ere A=[~ ~ ~] -1
B"'
-5 -6
[~.J1
IJy
l
usitrg slate feedback rontrol u a -Kx, ii is desired to /rave the c/OS
Refer example S.10
Q.12
Consider the systtm x =Ax+ Uu a11d Y = C><. w/rerr
A=[~ ~ -6
1111d C = [t 0 0 ]. Determine lite obsm.'fr gai11 matrix by the
US<'
-11
o
l J
I -6
rol B
=
loJ' L1
of:
The direct substitution me/hod. ii) Ackermann's formula. i)
Assume that µI =-2+j3.4641 Ans. :
the desired Eige11 values of tlrt and µ2 =-2-j3.4641, µ 3 =-S.
obserwr gai11 matrix arc <J•nJFeb.-2008, 8 M•rksl
Refer example 5.11.
ODD
'®
Controllers
I
Q.1
l:xplain effects of a Pl controller on the static and dynamic responseof a system.
Ans. :
Refer section 6.9.
Q.2
Consider a typical second order, type one system with u11ityfeedback, being controlled l1y a PD controller and stun» that i) Damping increase with PD control.
Ans. :
Refer section 6.14.
Q.3
A temperature control system has th« block diagram given i11 Fig.1. The i11p11t signal is a voltage a11d represents the desired temperature er Find the steady-state error of the
(July/Aug.·2005, 5 Marks)
(JulylAug.·2005, 4 Marks)
system whm
!+ is a unit step function a1ul i) D(s)
= 1 i1) D(s)= I+
0/
=
iii) Dis) 1 + 0.3 s. What is tl:e effect of the i11tegral lmn i11 lite Pl controllerand tl:t derivative term i11 PD controller on the steady state error ? (July/Aug.·2007, 8 Marks) O(r)
~
200
9
Plant
Fig. 1 Ans. :
Refer example 6.16.
Q.4
Wl1J1t is a controller ? £xplai11 P, l, Pl and PIO controllers.
Ans. :
Refer sections 6.1, 6.5, 6.6, 6.9 and 6.11
Q.5
Define controller. Explain properties of P, Pl and PIO controllers with llie help of block diagram. (JulylAug.-2007, 8 Marks)
Ans. :
Refer sections 6.1, 6.5, 6.9 and 6.11
Q.6
Deft11e co11tro/ler. Explain PD, Pl and PID coturollers. What are the advantages of PID controller ? (JanJfeb.2008, 6 Marks)
Ans. :
Refer sections 6.1, 6.10, 6.9 and 6.11 (P • 24)
(JanJfeb.-2007, 10 Marks)
ono M
jG)
Nonlinear Systems I
Q.1
£xplai11 the folluwingbehaviour of 11011-lincar systems : Iii Frequency amplitudedependence . (ii) Multivalued responses and jump resonances.
Ans. :
Refer section 7.2.
Q.2
Discuss tl1e basic featuresof the followingnon-linearities i)
Non-linear friction
ii)
On--0ff controllers
iii) Back lash
(Jan/Feb.-2005,10 Marks)
(Jan.lFcb.-2006, 9 Marks)
Ans. :
Refer sections 7.4.5 and 7.4.6.
Q.3
What are inherent nonlinearities ? Explai11 1my three of thmdJuly/Aug.-2006, 6 Marks)
Ans. :
Refer section 7.4.
Q.4
Sketch the following non-linearities : Ideal relay ii) Reilly with dead zone iii) Reilly with dead zone and lrystcrisis io) Relay with hysterisis v) Dead zone. (July/Aug.-2006,4 Marks) i)
Ans.: Q.5
Discuss tlte basic features of t/1t Back/as/1 non-linearities wit/1 suital>/efigures.
Ans. :
Refer section 7.4.6.
(July/Aug.-2007,4 Marks)
Q.6
Explain any three non-linearities in control systems.
Ans. :
Refer section 7.4.
QanJFeb.-2008,6 Marks)
000
(P - 25)
'® Q.1
Phase Plane Method
I
Determine the kind of si11gularity for tJJch of the fo/1111ui11gdiffermtioleq11atio11s. (i) y-3y+2y=O
(ii) y-8y+17y~34 Ans. : Q.2
(JanJfeJ>.·2005,10 Marks)
Refer example 8.1.
What is a phase 11/a11e 11/ot? Describe delta method or any other method of drawi11g phase plane trajectories.
Ans. : Q.3
(JanJFeb.·2005,10 Maries)
Refer sections 8.1 and 8.5.
Witlr rejerenc«to 11011-lineur system explain : (i) /1111111 r.•sona11ce (ii)
Limit cycles
(July/Aug.·2005,6 Maries)
Ans. :
Refer sections 8.3 and 8.2.
Q.4
Whal arc si11g11lar points? Explain tire classification of singular points based 011 the location of dgm values of tire system. (July/Aug.·2005; JanJfcb.·2008,8 Mark$)
Ans. :
Refer section 8.9.
Q.5
/'ig. 1 slunv« pl111se portraits for typt' • 0 systons. Classify tlrem i11to the categories. Stable focus. stable node. saddle point and w-on. (See Fig. 1 011 next page.)
Ans. :
Stable Iocus
(JanJfcb.·2006;July/Aug.-2007, 6 Marks)
Fig. 2 Stable focus (P • 26)
p ·29
Modem Control Theory
Phase Plane Method
Unstable node
x, ~y
Fig. 6 Unstable node Saddle point
Fig. 7 Saddle point Q.6
Explain l/1e concept of jump resonance with a suitable example. (JanJFcb.·2006, 5 Marks)
Ana. :
Refer section 8.3.
Q.7
Explain th« delta 1r1ttlrod of constructing phas» trajtctorie.s.
Ana. :
Refer section 8.11.
Q.8
Using isocline method, draw lite phas<e trajectory for tltt system. d2x+o.6'.!!.+x dt2 di with x = 1 a11d dx dt
Ans. :
Refer section 8.4.
OanJFcb.·2006, 7 Morks)
.. o
= 0 as initial amditia11.
(JanJFcb.-2006,8 Marks)
Modorn Control
Q.9
p ·30
Thoory
Phaso Plano Method
.
A linfar second order sen» is described by the equation C+2E,ll\,C+or,c
= 0 wliere
E, = 0.15. 10• = l rad/sec, CIOJ = l.5 and C = 0. Detmnirtt tile singular point. Constriid lite phase trajectory, using the method of isodine». (July/Aug.·2006,10 Marks) Ans. :
Refer example 8.2.
Q.10
Define singular point on a pnase plm1e. &plain diffoent types of sit1gular points · (July/Aug.-2006, 10 Marks)
Ans. :
Reier section 8.9.
Q.11 Ans. :
What are singular points? Explain differc1t singular points adopted in 1wn-li11ear control ;-ystems. (JanJFcb.-2007,8 Marks) Refer section 8.9.
Q.12
Find out singular points for tM followingsystems : i) x +0.5x+2x=O
. .
ii) .v+3y +2y=O iii) y +3y -10=0
(JanJFcb.·2007,12 Marks)
Ans. :
Refer example 8.3.
Q.13
Draw tlte phase-plane trajectoryfor thefollowittg equation using Isodine method: x+2E,ro;+w2x=U
Given, f,=0.5.
lll"'l,
Initial point
(0, 6).
QanJFcb.-2007,U Marks)
Ans. :
Refer example 8.5.
Q.14
Discuss lllC basic features of the jt••np rtSOna11ce 11011-linearities wit/1 suitable figures : /ump n~o111111cr. (July/Aug.·2007, 8 Marks)
Ans. :
Refer section 8.3.
Q, 15
&plain th« construction of a pllflSt trajectory by Delta method.
Ans. :
Refer section S.S.
QanJFeb.-2008, 6 Markli)
ODD
Liapunov's Stability Analysis
Q.1
State and explain Uapu11ov's theorems 011 Ii) Asymptotic stability (ii) Global asymptotic stability und (iii) l11stabilily. Qan./Feb.-2005;July/Aug.-2006;JanJFeb.·2008, 10 Marks)
Ans. :
Refer sections 9.3, 9.4 and 9.5.
Q.2
Use Krasovskii's theorem lo show that lite equilibrium slate x = 0 of lite system described by
.r
1
=-3x, +x2
.t2 =x1
-x2
-xi
is asymptotically stale i11 I/re large.
(Jan./Feb.·2005, 10 Marks)
Ans. :
Refer example 9.4.
Q.3
Define: Ii) Stability, Iii) Asymlolic stability (iii) Asymptotic stability in the large. Quly/Aug.·2005, 5 Marks, July/Aug.-2007, 6 Marks)
Ans. :
Refer sections 9.29.3, 9.4 and 9.5.
Q.4
Investigate I/re stability
of tlie following non-linear system
using direct method of
.
Liapunou. x1 =x2 -~2 =-.rl
-xfx2
(July/Aug.-2005,S Marks)
Ans. :
Refer example 9.5.
Q.5
A seco11d order system is represe11led by, • (l x=Ax where A= [ _1
1 ] -:
Assuming matrix Q to be identity matrix, solve for matrix P in tire eqw1lion ATP + PA Q. Use Liapunou theorem and determine I/re stability of tire origi11 of the system. Write tire Liapunot» [unction V(x). (July/Aug.-2005, 10 Marks)
=-
Ans. :
Refer example 9.6.
(P • 31)
- .. ,. . p -32
Modern Control Theory Q.6
Liapunov's Stability Analysis
Using Liaprmovs direct 111etl1od, find tlrr range of K lo guarantee sta/1ility of tire system show11 i11 Fig. I.
. _:~•~Ghl-f~~I
•O
s
Fig. 1 (JonJFeb.-2006,14 Marks) Ans. :
Refer example 9.11.
Q. 7
Choos« an appropriate Liapunov fimction and check tire stability of tire equilibrium state of tire system described by
X2 = - X7 - X~ X2
Ans. : Q,8
O•nJFeb.-2006, 6 Muks)
Refer example 9.5.
Determine whether or not folll1toing quadraticform is positive definite : Q (.q,x2) = 10xr +.ix~ +x~ + 2x1x2 -2xzx3 -4x1x3
(Jan.lFeb.-2007, 10 Marks)
Ans. :
Refer example 9.10.
Q.9
Explain with a11 example • i) Liapunov main stability thrown ii) Liapunop second method and iii) Krasooskii's tlrcorem. (JanJFcb.·2007,10 Marks)
Ans. :
Refer sections 9.10, 9.11 and 9.13.
Q.10
Fi11d the Liapwrovfunctionfor the system: X(I)
= [~ ~1]x.
Ans. :
Refer example 9.6.
Q.11
Explain Llap1111011's stability criterion.
Ans. :
Refer section 9.1 I.
(JanJFeb.·2007,8 Marks)
(July/Aug.-2007,6 Marks)
~
Modern Control Theory
Q.12
P-33
Liapunov's
Stability Analysis
Consider tlie system 1uitl1 diffmmt111/equation.
= U Examine tire stnbility
::+k;.+k1e.J +c
by Uap1111ov's method, gioen thot K > 0
and K 1 > 0.
(July/Aug.-2007,6 Marks]
Ans. :
Refer example 9.8.
Q.13
Exnmme ti~ stability of the system described by th« following equation by Krasovdskii's.
Ans. :
Refer example 9.9.
Q.14
A system is described by tire followingequation :
Ouly/Aug.-2007,8 Marks)
;=Ax Where A
=[-t 1
2].
-4
Q to be tire identify matrix, solve for matrix P and comment on the stability of tile system 11si11g the equation ATP+ PA -Q.
Assuming matrix
=
Qan/Fcb.- 2008, 10 Marks) Ans. :
Refer example 9.7.
ODO
(P • 34)
Related Documents
Ag Ua
May 2020 13
Ua-iowafbnotes09
June 2020 13
Ua 37.3
June 2020 7
Ua-es
June 2020 11
Rpp-sm1biologi Ua
December 2019 22
Delta Rectifier Settings-ua
November 2019 14More Documents from ""
Mybtechproject-180429062425.pdf
June 2020 0
[book] Modern_control_theory [u.a Bakshi].pdf
July 2020 2
Circular_array_antenna_2.docx
June 2020 3