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Control Systems
10ES43
10ES43 CONTROL SYSTEMS (Common to EC/TC/EE/IT/BM/ML) Subject Code:10ES43 No. of Lecture Hrs./Week : 04 Total No. of Lecture Hrs.:52
IA Marks: 25 Exam Hours : 03 Exam Marks : 100
PART – A UNIT 1: Modeling of Systems: Introduction to Control Systems, Types of control systems, Effect of feedback systems, Differential equations of physical systems – Mechanical systemsFriction, Translational systems (Mechanical accelerometer, Levered systems excluded), Rotational systems, Gear trains. Electrical systems, Analogous systems. 6 Hours UNIT 2: Block diagrams and signal flow graphs: Transfer functions, Block diagrams, Signal Flow graphs (Statevariable formulation excluded). 7 Hours UNIT 3: Time Response of feed back control systems: Standard test signals, Unit step response of First and second order systems, Time response specifications, Time response specifications of second order systems, steady – state errors and error constants. 7Hours
UNIT 4: Stability analysis: Concepts of stability, Necessary conditions for Stability, Routh-Hurwitz stability criterion, Relative stability analysis; Special cases of RH criterion. 6 Hours PART – B UNIT 5: Root–Locus Techniques: Introduction, basic properties of root loci, Construction of root loci. 6 Hours UNIT 6: Stability analysis in frequency domain: Introduction, Mathematical preliminaries, Nyquist Stability criterion, (Inverse polar plots excluded), Assessment of relative stability using Nyquist criterion, (Systems with transportation lag excluded). 7Hours Department of EEE, SJBIT
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UNIT 7: Frequency domain analysis: Correlation between time and frequency response, Bode plots, All pass and minimum phase systems, Experimental determination of transfer functions, Assessment of relative stability using Bode Plots. 7 Hours UNIT 8: Introduction to State variable analysis: Concepts of state, state variable and state models for electrical systems, Solution of state equations. 6 Hours TEXT BOOK : 1. Control Systems Engineering, I. J. Nagarath and M.Gopal, New Age International (P) Limited, 4 Edition – 2005 2 Modern Control Engineering, K. Ogata, PHI, 5th Edition, 2010.
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10ES43 Table of contents
Sl .no
Contents UNIT 1:Modeling of Systems: Introduction to Control Systems
1
Page number
Types of control systems, Effect of feedback systems,
Differential equations of physical systems – Mechanical systems- Friction
5 to 12
Translational systems (Mechanical accelerometer, Levered systems excluded)
Rotational systems, Gear trains. Electrical systems, Analogous systems.
UNIT 2: Block diagrams and signal flow graphs Transfer functions 2
Block diagram
13 to 23
Signal Flow graphs (Statevariable formulation excluded). UNIT 3: Time Response of feed back control systems: Standard test signals 3
Unit step response of First and second order systems,
Time response specifications
24 to 46
Time response specifications of second order systems, steady state errors and error constants. UNIT 4: Stability analysis: Concepts of stability
Necessary conditions for Stability
4
47 to 65
Routh-Hurwitz stability criterion
Relative stability analysis
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Special cases of RH criterion.
UNIT 5: Root–Locus Techniques: Introduction 5
basic properties of root loci
66 to 84
Construction of root loci. UNIT 6: Stability analysis in frequency domain: Introduction 6
Mathematical preliminaries
Nyquist Stability criterion, (Inverse polar plots excluded)
85 to 102
Assessment of relative stability using Nyquist criterion,
UNIT 7: Frequency domain analysis: Correlation between time and frequency response
7
Bode plots
All pass and minimum phase systems
Experimental determination of transfer functions
Assessment of relative stability using Bode Plots.
103 to 118
UNIT 8: Introduction to State variable analysis: Concepts of state 8
state variable and state models for electrical systems
119 to 127
Solution of state equations
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UNIT-1 Modeling of Systems Introduction to control systems A system is an arrangement of or a combination of different physical components connected or related in such a manner so as to form an entire unit to attain a certain objective Control system is an arrangement of different physical elements connected in such a manner so as to regulate, director command itself to achieve a certain objective Requirements of good control system are accuracy, sensitivity, noise, stability, bandwidth, speed, oscillations
Types of control systems A system in which the control action is totally independent of the output of the system is called as open loop system
Example: Automatic hand driver, automatic washing machine, bread toaster, electric lift, traffic signals, coffee server, theatre lamp etc.
A system in which the control action is somehow dependent on the output is called as closed loop system The elements of closed loop system are command, reference input, error detector, control element controlled system and feedback element
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Elements of closed loop system are: 1. Command : The command is the externally produced input and independent of the feedback control system. 2. Reference Input Element: It is used to produce the standard signals proportional to the command. 3. Error Detector : The error detector receives the measured signal and compare it with reference input. The difference of two signals produces error signal. 4. Control Element : This regulates the output according to the signal obtained from error detector. 5. Controlled System : This represents what we are controlling by feedback loop. 6. Feedback Element : This element feedback the output to the error detector for comparison with the reference input. Example: Automatic electric iron, servo voltage stabilizer, sun-seeker solar system, water level controller, human perspiration system.
Feedback system is that in which part of output is feeded back to input In feedback system corrective action starts only after the output has been affected Advantages of closed loop system: 1. Accuracy is very high as any error arising is corrected. 2. It senses changes -in output due to environmental or parametric change, internal disturbance etc. and corrects the same. 3. Reduce effect of non-linearities. 4. High bandwidth. 5. Facilitates automation. Disadvantages 1. Complicated in design and maintenance costlier. 2. System may become unstable. Department of EEE, SJBIT
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Advantages of open loop system: 1. They are simple in construction and design. 2. They are economic. 3. Easy for maintenance. 4. Not much problem of stability. 5. Convenient to use when output is difficult to measure. Disadvantages of open loop system 1. Inaccurate and unreliable because accuracy is dependent on accuracy of calibration. 2. Inaccurate results are obtained with parameter variations, internal disturbances. 3. To maintain quality and accuracy, recalibration of controller is necessary from time to time.
Feed forward system is that in which the corrective action is initiated without waiting for the effect of disturbance to show up in the output System having multiple inputs and multiple outputs is known as multiple output (MIMO) control system A servomechanism is a power amplifying feedback control system in which the controlled variable is mechanical position or its time derivative such as velocity, acceleration A regulator or regulating control system is a feedback control system in which the reference input remains constant for long periods/entire intervals of operation An adaptive control system is one that continuously and automatically measures the dynamic characteristics of the plant. The system which follows the principle of superposition and proportionality is called a linear system.
The motion take place along a straight line is known as translational motion. Rotational motion of a body is the motion about a fixed axis. The elements of rotational system are inertia (J), damping coefficient (B) and torsional stiffness (K). Mechanical system Department of EEE, SJBIT
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For every mechanical system, there is analogous electrical system. DAlembert‘s principle states that, the algebric sum of externally applied forces and the forces resisting motion in any given direction is zero. For mechanical network, analogous electrical network can be obtained by using f-v and f-i analogy. Force-voltage analogy: In this method force is analogous to voltage. Similarly,
displacement n charge q.
Force-current analogy: In this method force is analogous to current.
Mechanical rotational system: (a) Force-voltage analogy:
(b) Force-current analogy :
Mechanical coupling: Laplace transform of signal x (t) is denoted by X (s)
Example problems: Q 1. Draw the f-1 analogous mechanical system for the electrical circuit of fig. below:
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Ans. f (t) is analogous to e (t) f (t) is analogous to R. f (t) frictional force f is analogous to r. Spring constant K is analogous to reciprocal Mass M is analogous to inductance L
Q. 2. Draw the mathematical model of the following system and obtain the transfer function.
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Ans. Using torque-voltage analogy, we have
The equation is
Q. 3. Write force-current analogous quantities. Ans. Force analogous to current
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Q. 4. Draw the mathematical model of the following system and obtain the transfer function.
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Ans. Writting nodal equation At node x
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UNIT-2 Block diagrams and signal flow graphs Transfer function The ratio of laplace transform of the output to the laplace transform of input under the assumption of zero initial conditions is defined as the transfer function Of system. It is denoted by G(s).
Importance : Transfer function is highly important because of following reasons 1. It is used to give the gain of given blocksystem. 2. The system poles/zeros can be found from transfer function. 3. Stability can be determined from characteristic equation. 4. The system differential equation can be obtained from transfer function by replacing. s-variable with linear differential operator Significance of Transfer Function
Where
C(s) is laplace transform of output R(s) is laplace transform of input. Transfer function gives the gain of the given block system. Properties of Transfer Function 1. The transfer function is independent of the inputs to the system. 2. The transfer function of a system is the laplace transform of its impulse response for zero initial conditions. 3. The system poles/zeros can be found out from transfer function. 4. The transfer function is defined only for linear invariant systems. It is not defined for non-linear systems. Limitations of transfer function are listed below 1 Transfer function is valid only for linear time invariant system. 2 It does not take into account the initial conditions initial conditions loose its significance. 3 It does not give any idea about how the present output is progressing.
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Poles are the value of ‗s‘ which when substituted in the denominator of a transfer function, make the transfer function value as infinity Zeros are the value of ‗s‘ which when substituted in the numerator of transfer function, make the transfer function value as zero The characteristic equation can be obtained by equating the denominator polynomial of the transfer function to zero . The highest power of ‗s‘ in the characteristic equation is called the order o system The number of poles at the origin defines the type of system Block diagram algebra Block diagram gives a pictorial representation of a control system by way of short handing the transfer function Signal flow graph further shortens the representation of a control system by eliminating summing symbol take-off point and block This elimination is achieved by way of representing the variables by points called ―nodes‖ A pictorial representation of the relationship between input and output of a system is termed as block diagram. The direction of flow of signal from one block to other is indicated by an arrow. The point in a block diagram at which signal can be added or subtracted is termed as summing point. Gain is the ratio of laplace transform of output to laplace transform of input . Blocks in series are algebraically combined by multiplication. The lines drawn between the blocks to indicate the connections between the blocks are termed as branches. The point from which a signal is taken for the feedback purpose is called as take-off point. The order of summing point can be changed if two or more summing points are in series. Signal Flow Graphs Department of EEE, SJBIT
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A signal flow grow is a pictorial representation of a system and it displays graphically, the transmission of signal in system Node: It is a point from where branches originate or terminate or pass through. Branch : It is connecting link between two nodes. Path : The time traced by connecting two or more node is called path. Loop : It is a path that originates and terminates on same node and along which node other node is traversed more than once.
Mason s gain formula is used to find the gain of signal flow graph According to Mason‘s gain formula
where
i = Number of forward path = Gain of ith forward path = System determinant
= 1 — (sum of all individual loops) + (sum of all gain products of two nontouching loops) - (sum of all gain product of three non-touching loops) +…. The gain associated with each branch is called branch transmittance The independent and dependent variable of a control system are represented by small circles as nodes. The relationship between nodes is represented by drawing a line between two nodes Such l‘ns are called branches. Example problems:
Q. 1. Find out the 2010
for the system shown in the following block diagram.
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Ans. First draw the signal flow graph.
Step I Obtain total number of forward paths There is only one forward path
Step II. Total number of single loop There are two loops. Thus
Step III. Value of
As there is one forward path which touch all the loops Step IV. Obtain transfer function
Q. 2. The transfer function of a system is Calculate the phase shift at
.
Ans.
Hence here is phase shift of zero corresponding to Department of EEE, SJBIT
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Q. 3. Laplace transform of a function f (t) is given by
Find out the initial and final values of f (t). Ans.
Applying final value theorem
Q. 4. Find out the inverse Laplace transform of the function
.
Ans.
The term
can be factorized as, (s + 2) (s + 3)
Using partial fraction expansion
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Q. 5. Represent the following set of equations by a signal 2011 flow graph and determine the overall gain relating
10ES43
June/July .
Ans. Given equations are :
required signal flow graph is
Step I. Obtain the total number of forward paths
Step II. Obtain the number of single loops
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Step lll. Obtain the number of two non-touching loop
Step IV. Number of three non-touching loops --no— Step V. Find the value of
Applying Mason‘s gain formula Overall transform function
Q. 6. Simplify the block diagram in fig and obtain the transfer 2010 function relating C(s) and R(s).
June/July
Ans.
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There are no loops.
Q. 21. From the block diagram shown in the figure below draw the corresponding signal flow graph and evaluate close loop transfer function relating Dec/Jan 2006 the output and input.
Ans. Required signal flow graph is:
By Mason‘s gain formula:
Individual loop, Non-touching loops = 0
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Q. 7. The transfer function of a system is given by
Determine the state model in canonical form using parallel decomposition method. Ans.
Using partial fractions
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State diagram representation in diagonal canonical form by parallel decomposition
Q. 8. Determine the transfer function of the system given in fig.
Ans.
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UNIT 3 Time Response of feed back control systems
Time-domain Analysis of Control Systems In time-domain analysis the response of a dynamic system to an input is expressed as a function of time. It is possible to compute the time response of a system if the nature of input and the mathematical model of the system are known. Usually, the input signals to control systems are not known fully ahead of time. In a radar tracking system, the position and the speed of the target to be tracked may vary in a random fashion. It is therefore difficult to express the actual input signals mathematically by simple equations. The characteristics of actual input signals are a sudden shock, a sudden change, a constant velocity, and constant acceleration. The dynamic behavior of a system is therefore judged and compared under application of standard test signals – an impulse, a step, a constant velocity, and constant acceleration. Another standard signal of great importance is a sinusoidal signal. The time response of any system has two components: transient response and the steadystate response. Transient response is dependent upon the system poles only and not on the type of input. It is therefore sufficient to analyze the transient response using a step input. The steady-state response depends on system dynamics and the input quantity. It is then examined using different test signals by final value theorem. Standard test signals a) Step signal:
r (t ) Au(t ).
b) Ramp signal:
r (t ) At; t 0.
c) Parabolic signal: r (t ) At 2 / 2; t 0. d) Impulse signal: r (t ) (t ).
Time-response of first-order systems Let us consider the armature-controlled dc motor driving a load, such as a video tape. The objective is to drive the tape at constant speed. Note that it is an open-loop system.
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G( s)
10ES43
kk kk ak1km W ( s) a ak k 1 m ; If r (t ) au(t ) , W ( s) 1 m 1 m R( s ) m s 1 ms 1 s s s 1/ m
w(t ) ak1km ak1kmet / m ; wss (t ) lim w(t ) ak1km t
wss (t ) is the steady-state final speed. If the desired speed is wr , choosing a
wr the motor k1km
will eventually reach the desired speed. We are interested not only in final speed, but also in the speed of response. Here, m is the time constant of motor which is responsible for the speed of response. The time response is plotted in the Figure in next page. A plot of et / m is shown, from where it is seen that, for t 5 m the value of et / m is less than 1% of its original value. Therefore, the speed of the motor will reach and stay within 1% of its final speed at 5 time constants.
Figure: Time responses
Let us now consider the closed-loop system shown below.
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Here, T ( s)
10ES43
k1km / ( m s 1) k1km kk W ( s) 1 o R( s) 1 k1k2 km / ( m s 1) m s (1 k1k2 km ) o s 1 where, ko
km m and o . 1 k1k2 km 1 k1k2 km
If r (t ) a , the response would be, w(t ) ak1ko ak1ko e
t / o
.
If a is properly chosen, the tape can reach a desired speed. It will reach the desired speed in 5 o seconds. Here, o m . Thus, we can control the speed of response in feedback system. Although the time-constant is reduced by a factor (1 k1k2 km ) , in the feedback system, the motor gain constant is also reduced by the same factor. In order to compensate for this loss of gain, the applied reference voltage must be increased by the same factor. Ramp response of first-order system Let, k1k0 1 for simplicity. Then, T ( s)
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1 W ( s) . Also, let, r (t ) tu(t ) . o s 1 R( s )
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o2 1 1 o Then, W ( s) 2 ; s ( o s 1) s 2 s o s 1
w(t ) tu(t ) o (1 et / o )u(t ) The error signal is, e(t ) r (t ) w(t ) Or, e(t ) o (1 et / o )u(t )
ess (t ) o Thus, the first-order system will track the unit ramp input with a steady-state error o , which is equal to the time-constant of the system.
Time-response of second-order systems
Consider the antenna position control system. Its transfer function from r to y is,
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k1k2 km k1k2 km / m n2 Y ( s) R( s) m s 2 s k1k2 km s 2 1 s k k k / s 2 2n s n2 1 2 m m
m
where, we define,
n2 k1k2 km / m and
2n
1
m
. The constant is called the damping
ratio and n is called the natural frequency. The system above is in fact a standard second order system. The transfer function T ( s) has two poles and no zero. Its poles are, s1 , s2 n jn 1 2 jd .
Here, is called the damping factor, d is called damped or actual frequency. The location of poles for different are plotted in Figure below. For 0 , the two poles
jn are purely imaginary. If 0 1 , the two poles are complex conjugate. All possible cases are described in a table shown below.
Unit step response of second-order systems
n2 s 2n 1 1 1 2 Suppose, r (t ) u(t ), R( s) ; Y ( s) 2 2 s s 2n s n s s 2n s n2 s
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Or, Natural frequency, n The natural frequency of a second order system is the frequency of oscillation of the system without damping.
s 2n s n n 1 1 Y ( s) 2 2 2 s ( s n ) n (1 ) s ( s n )2 (n 1
Performing inverse Laplace transform,
Damping ratio, The damping ratio is defined as the ratio of the damping factor, to the natural frequency n .
y (t ) 1 ent cos(n 1 2 )t ent sin(n 1 2 )t
b . T (s) 2 s as b
Suppose,
nt
e 2 ) 1 standard equation, 1 2 cos(an 21 n and )t sin(n 1 2 )t or, y(twith Comparing 1 2
b. 2 n
y (t ) 1
or,
d n 1 2 and tan 1
ent 1
2
sin(d t ) ,where,
1 2 / cos 1
n t e sin(d t ) d ………………………………………………… …….(01) y (t ) 1
or,
The plot of e t sin(d t ) is shown in Figure. The steady-state response is, yss (t ) lim y(t ) 1 t
Thus, the system has zero steady-state error. The pole of T ( s) dictates the response, e t sin(d t ) . Department of EEE, SJBIT
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The response y (t ) for different is shown in Figure below.
Time response specifications Control systems are generally designed with damping less than one, i.e., oscillatory step response. Higher order control systems usually have a pair of complex conjugate poles with damping less than unity that dominate over the other poles. Therefore the time response of second- and higher-order control systems to a step input is generally of damped oscillatory nature as shown in Figure next (next page). In specifying the transient-response characteristics of a control system to a unit step input, we usually specify the following: 1. Delay time, t d 2. Rise time, tr 3. Peak time, t p 4. Peak overshoot, M p 5. Settling time, t s 6. Steady-state error, ess Department of EEE, SJBIT
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1. Delay time, t d : It is the time required for the response to reach 50% of the final value in first attempt. 2. Rise time, tr : It is the time required for the response to rise from 0 to 100% of the final value for the underdamped system. 3. Peak time, t p : It is the time required for the response to reach the peak of time response or the peak overshoot. 4. Settling time, t s : It is the time required for the response to reach and stay within a specified tolerance band ( 2% or 5%) of its final value. 5. Peak overshoot, M p : It is the normalized difference between the time response peak and the steady output and is defined as,
%M p
c(t p ) c() c ( )
100%
6. Steady-state error, ess : It indicates the error between the actual output and desired output as ‗t‘ tends to infinity. ess lim[r (t ) c(t )] . t
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Let us now obtain the expressions for the rise time, peak time, peak overshoot, and settling time for the second order system. 1.
Rise
time,
tr :
Put
t tr , sin(d tr ) 0 sin , tr
; cos1 . d
2.
Put
0
Peak
time,
tp :
dy 0 and dt
y(t ) 1
at
for
t tp ;
solve
n t e sin(d t ) n e t cos(d t ) . d tan(d t p )
1 2 d n 1 2 tan , d t p k n
k 0,1, 2,
Peak overshoot occurs at k = 1. t p / d / n 1 2 . 3. Settling time, t s : For 2% tolerance band,
n t 4 e 0.02 , ts 4T . d s
4. Steady-state error, ess : It is found previously that steady-state error for step input is zero. Let us now consider ramp input, r (t ) tu(t ) . Then, ess lim s{R( s) Y ( s)} lim s{ s 0
s 0
n2 1 1 } s 2 s 2 s 2 2n s n2
2 2 2 n2 1 1 s 2n s n n ess lim {1 2 } lim s 0 s s 0 s s 2n s n2 s 2 2n s n2
Therefore, the steady-state error due to ramp input is
2
n
2n 2 . 2 n n
.
Steady-state error and error constants The steady-state performance of a stable control system is generally judged by its steadystate error to step, ramp and parabolic inputs. For a unity feedback system,
E ( s)
R( s ) sR( s) , ess lim sE ( s) lim . s 0 s 0 1 G ( s ) 1 G( s)
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It is seen that steady-state error depends upon the input R( s) and the forward transfer function G( s) . The steady-state errors for different inputs are derived as follows: 1. For unit-step input: r (t ) u (t ), R( s)
ess lim sE ( s) lim s 0
s 0
1 s
1 1 1 ; k p is called position error constant. 1 G( s) 1 G(0) 1 k p
2. For unit-ramp input: r (t ) tu (t ), R( s)
ess lim sE ( s) lim s 0
s 0
1 s2
1 1 1 lim ; s 1 G( s) s 0 sG( s) kv
kv is
called
velocity
error
constant. 3. For unit-parabolic input: r (t ) t 2 / 2, R( s)
ess lim sE ( s) lim s 0
s 0
1 s3
1 1 1 lim 2 ; k a is called acceleration error s 0 s 1 G( s) s G ( s ) ka 2
const.
Types of Feedback Control System The open-loop transfer function of a system can be written as,
G( s)
K ( s z1 )( s z2 )( s z3 ) s n ( s p1 )( s p2 )( s p3 )
K (Tz1s 1)(Tz 2 s 1)(Tz 3 s 1) s n (Tp1s 1)(Tp 2 s 1)(Tp 3s 1)
If n = 0, the system is called type-0 system, if n = 1, the system is called type-1 system, if n = 2, the system is called type-2 system, etc. Steady-state errors for various inputs and system types are tabulated below.
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The error constants for non-unity feedback systems may be obtained by replacing G(s) by G(s)H(s). Systems of type higher than 2 are not employed due to two reasons: 1. The system is difficult to stabilize. 2. The dynamic errors for such systems tend to be larger than those types-0, -1 and -2.
Effect of Adding a Zero to a System Let a zero at s = -z be added to a second order system. Then we have,
( s z )n2 / z n2 n2 C (s) s 2 . 2 2 2 2 2 R( s) s 2n s n s 2n s n z s 2n s n The multiplication term is adjusted to make the steady-state gain of the system unity. This gives css = 1 when the input is unit step. Let cz(t) be the response of the system given by the above equation and c(t) is the response without adding the pole. Manipulation of the above equation gives, cz (t ) c(t )
1d c(t ). z dt
The effect of added derivative term is to produce a pronounced early peak to the system response which will be clear from the figure in the next page. Closer the zero to origin, the more pronounce the peaking phenomenon. Due to this fact, the zeros on the real axis near the origin are generally avoided in design. However, in a sluggish system the artful introduction of a zero at the proper position can improve the transient response. We can see from equation (03) that as z increases, i.e., the zero moves further into the left half of the splane, its effect becomes less pronounced.
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Design Specifications of Second-order Systems A control system is generally required to meet three time response specifications: steady-state accuracy, damping factor (or peak overshoot, Mp) and settling time ts. Steady-state accuracy requirement is met by suitable choice of Kp, Kv, or Ka depending on the type of the system. For most control systems in the range of 0.7 – 0.28 (or peak overshoot of 5 – 40%) is considered acceptable. For this range of , the closed-loop pole locations are restricted to the shaded region of the s-plane as shown in Figure. For
the
system, n k1k2 km / m ;
antenna 1 2n m
; ess
ramp
position 2
n
; ts
4
n
control
. Here, k 2 is only the adjustable
parameter. If we increase k 2 , n will increase and thus settling time will decrease. At the same time, will decrease, this indicates the increase in peak overshoot. Thus by merely increasing gain, we cannot improve both transient and steady-state error specifications. We need to add additional components to the system. These are called compensators. It will allow improvement of both transient and steady-state specifications.
Introduction to PID Controler
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The classical three-term PID controller proportional feedback control can reduce error responses but that it still allows a non-zero steady-state error for a proportional system. In addition, proportional feedback increases the speed of response but has a much larger transient overshoot. When the controller includes a term proportional to the integral of the error, then the steady-state error can be eliminated. But this comes at the expense of further deterioration in the dynamic response. Addition of a term proportional to the derivative of the error can damp the dynamic response. Combined, these three kinds of actions form the classical PID controller, which is widely used in industry. This principle mode of action of the PID controller can be explained by the parallel connection of the P, I and D elements shown in Figure 3.1 From this diagram the transfer function of the PID controller is
(3.1)
Figure 3.1: Block diagram of the PID controller The controller variables are Gain integral action time derivative action time
Eq. (3.1) can be rearranged to give
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These three variables , and are usually tuned within given ranges. Therefore, they are often called the tuning parameters of the controller. By proper choice of these tuning parameters a controller can be adapted for a specific plant to obtain a good behaviour of the controlled system. If follows from Eq. (3.2) that the time response of the controller output is
(3.3)
Using this relationship for a step input of , i.e. , the step response of the PID controller can be easily determined. The result is shown in Figure 3.2a. One has to observe that the length of the arrow the impulse.
of the D action is only a measure of the weight of
Figure 3.2: Step responses (a) of the ideal and (b) of the real PID controller In the previous considerations it has been assumed that a D behaviour can be realised by the PID controller. This is an ideal assumption and in reality the ideal D element cannot be realised . In real PID controllers a lag is included in the D behaviour. Instead of a D element in the block diagram of Figure 3.1 a
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element with the transfer function
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is introduced. From this the transfer function of the real PID controller or more precisely of the
controller follows as (3.5)
Introducing the controller tuning parameters and
it follows (3.6)
The step response of the controller is shown in Figure 3.2b. This response from gives a large rise, which declines fast to a value close to the P action, and then migrates into the slower I action. The P, I and D behaviour can be tuned independently. In commercial controllers the 'D step' at can often be tuned 5 to 25 times larger than the 'P step'. A strongly weighted D action may cause the actuator to reach its maximum value, i.e. it reaches its 'limits'. As special cases of PID controllers one obtains for: a) the PI controller with transfer function (3.7)
b) the ideal PD controller with the transfer function Department of EEE, SJBIT
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and the
controller with the transfer function (3.9)
c) and
the P controller with the transfer function (3.10)
The step responses of these types of controllers are compiled in Figure3.3. A pure I controller may also be applied and this has the transfer function (8.11)
Figure 3.3: Step responses of the PID controller family
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Advantages and disadvantages of the different types of controllers
In addition, the different cases should be compared with respect to the normalised maximum overshoot
.
The different cases are discussed below: a) The P controller shows a relatively high maximum overshoot settling time
as well as a steady-state error
, a long
.
b) The I controller has a higher maximum overshoot than the P controller due to the slowly starting I behaviour, but no steady-state error. c) The PI controller fuses the properties of the P and I controllers. It shows a maximum overshoot and settling time similar to the P controller but no steady-state error. d) The real PD controller according to Eq. (3.9) with has a smaller maximum overshoot due to the 'faster' D action compared with the controller types mentioned under a) to c). Also in this case a steady-state error is visible, which is smaller than in the case of the P controller. This is because the PD controller generally is tuned to have a larger gain due to the positive phase shift of the D action. For the results shown in Figure 3.5 the gain for the P controller is and for the PD controller
. The plant has a gain of
.
e) The PID controller according to Eq. (3.6) with fuses the properties of a PI and PD controller. It shows a smaller maximum overshoot than the PD controller and has no steady state error due to the I action. The qualitative concepts of this example are also relevant to other type of plants with delayed proportional behaviour. This discussion has given some first insights into the static and dynamic behaviour of control loops.
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Figure 3.5: Behaviour of the normalised controlled variable at the input to the plant
;
for step disturbance for different types of
controllers Example problems: Q. 1. Consider the system having transfer function
Calculate the settling time for 2% tolerance band, for the unit step response. Ans.
Comparing with
we get
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setting time for 2% tolerance band, we get setting time
Q. 2. Calculate the natural frequency of a second order system described by the differential equation
Ans. The given equation is
…(1) Now standard equation of a second order system is given as …(2) so comparing eq (1) and (2), we get
Q.3. For a second order system the roots of its characteristic equation are underdamped natural frequency of the system will be. (a) 4 rad/sec (b) 3 rad/sec (c) 5 rad/sec (d) 7 rad/sec
the
Ans.
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Q. 4. What will be the response of a first order system with unit step input?
Ans. The transfer function of a first order system is given as
With unit step input
The response is shown graphically in fig.
Q.5. What is the unit step response of the transfer function?
Ans. We have
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Q.6. Measurements conducts on a servomechanism show that response to be when subjected to a unit step input. Determine the undamped natural frequency and damping ratio of the system.
Ans.
Unit step input means
Comparing, we get
Natural frequency
Q. 7. The open-loop transfer function of a unity feedback control system is given by
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Determine the value of K and B such that the closed loop until step response has w = 3 rad/sec and Ans. The characteristics equation of the system is 1 + G (s) = 0
since, b = 1.2, and
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UNIT 4 Stability Analysis Concepts of stability Because of its feedback structure a control system can become unstable, e.g. oscillations with increasing amplitudes in the signals can occur . A linear time-invariant system is called ( asymptotically) stable, if its weighting function decays to zero, i.e. if (4.1)
is valid. If the modulus of the weighting function increases with increasing to infinity, the system is called unstable. A special case is a system where the modulus of the weighting function does not exceed a finite value as or for which it approaches a finite value. Such systems are called critically stable. Examples are undamped
S and I elements
This definition shows that stability is a system property for linear systems. If Eq. (4.1) is valid, then there exists no initial condition and no bounded input signal which drives the output to infinity. This definition can be directly applied to the stability analysis of linear systems by determining the value of the weighting function for . If this value exists, and if it is zero, the system is stable. However, in most cases the weighting function is not given in an explicit analytic form and therefore it is costly to determine the final value. The transfer function
of a system is often known and as it is the Laplace transform of the
weighting function , there is an equivalent stability condition for according to Eq. (4.1). The analysis of this condition shows that for the stability analysis it is sufficient to check the poles of the transfer function characteristic equation
of the system, that is the roots
of its
(4.2)
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Necessary conditions for Stability The following necessary and sufficient stability conditions can be formulated: a) Asymptotic stability A linear system is only asymptotically stable, if for the roots equation
of its characteristic
for all is valid, or in other words, if all poles of its transfer function lie in the left-half plane. b) Instability A linear system is only unstable, if at least one pole of its transfer function lies in the right-half plane, or, if at least one multiple pole (multiplicity imaginary axis of the plane.
) is on the
c) Critical stability A linear system is critically stable, if at least one single pole exists on the imaginary axis, no pole of the transfer function lies in the right-half plane, and in addition no multiple poles lie on the imaginary axis. It has been shown above that the stability of linear systems can be assessed by the distribution of the roots of the characteristic equation in the plane (Figure 5.2). For control problems there is often no need know these root with high precision. For a stability analysis it is interesting to know whether all roots of the characteristic equation lie in the left-half plane or not. Therefore simple criteria are available for easily checking stability, called stability criteria. These are partly in algebraic, partly in graphical form.
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Figure 4.2: Stability of a linear system discussed by the distribution of the roots of the characteristic equation in the plane
Routh criterion For given coefficients of the characteristic equation the method of Routh, which is an alternative to the method of Hurwitz, can be applied,. Here the coefficients will be arranged in the first two rows of the Routh schema, which contains rows:
The coefficients first two rows according to
in the third row are the results from cross multiplication the
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Building the cross products one starts with the elements of the first row. The calculation of these values will be continued until all remaining elements become zero. The calculation of the values are performed accordingly from the two rows above as follows:
From these new rows further rows will be built in the same way, where for the last two rows finally
and
follows. Now the Routh criterion is: A polynomial
is Hurwitzian, if and only if the following three conditions are valid:
a)
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10ES43 are positive,
b) all coefficients
in the first column of the Routh schema are positive.
Example 1
The Routh schema is:
For proving instability it is sufficient to build the Routh schema only until negative or zero value occurs in the first column. In the example given above the schema could have been stopped at the 5th row. Another interesting property of the Routh scheme says, that the number of roots with positive real parts is equal to the number of changes of sign of the values in the first column. Example 2 Determine the stability of the system whose characteristics equation given by a(s) s 6 4s 5 3s 4 2s 3 s 2 4s 4. The above polynomial satisfies the necessary condition for stability since all the coefficients are positive and nonzero. Writing the Routh array, we have
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s6 s5
1 4
s4
5
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3 2
4.3 1.2
2
1 4
0
4 .2 4.0 2 2 5
2
5
s2
3
s1
12 3( ) 4.2 76 5 3
2
s0
0
2 5
5 12 2.0 ( )
4.4 1.0 4
.4 4.4 12 2 5 5
2
15
4
4 5
5
s3
4.1 4.1
4 0
2.4 .0 2 4 2
0
76
.4 0 4 15 76 15
We conclude that the system has roots in the right half plane, since the elements of the first column are not all positive. In fact there are two roots in the right half plane, since there are two sign changes. In other words two closed loop poles of the system lie in the right half plane and hence the system is unstable.
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Example 3. Determine the stability of the following polynomial. a(s) s 5 5s 4 11s 3 23s 2 28s 12. Writing the Routh array, we have s5 1 11 s4
28
5
23
12
6.4
25.6
0
3
12
0
0
s3 s
2
s1
Since the entire row is zero, we construct an auxiliary equation by taking the coefficients of the previous row, i.e., a1 (s) 3s 2 12. Differentiating the above equation with respect to ‗s‘, we get da1 (s)
6s.
ds
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(1)
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So the Routh array is continued by taking the coefficients of equation (1).
s1 6 0 0 s 12 Since there are no sign changes in the first column of the Routh array, there are no roots in the right half plane. However, since one entire row in the Routh array was zero, there are roots in the imaginary axis. The roots in the imaginary axis can be obtained by solving the auxiliary equation. Therefore, 3s 2 12 0, s 2 4 0, s j2 Example 4. Consider the system shown below. The stability properties of the system are a function of the proportional feedback gain ‗k‘. Determine the range of ‗k‘ over which the system is asymptotically stable.
r
+
k
s 1
y
s(s 1)(s 6)
-
The characteristics equation for the system is given by
1 k
s 1 s(s 1)(s 6)
0,
s 3 5s 2 (k 6)s k 0. Department of EEE, SJBIT
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Therefore the corresponding Routh array is s3 s2 s1
1 5
k-6 k
(4k 30) 5
s0
k
For the system to be stable, it is necessary that all the elements in the first column of the Routh array must be positive. Therefore, 4k 30
0
5 k 7.5
and and
k 0, k 0,
k 7.5
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The state-space representation of a system is given as .
7 12 1 x x u, 0 0 1 y 1 2x.
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For the system to be stable, the poles of the system should lie in the left half plane. In other words all the real poles should be negative or the real parts of complex poles must be negative. The poles of the system are nothing but the eigenvalues of the ‗A‘ matrix of the system. The MATLAB code is shown below a = [-7 -12; 1 0]; [v,d] = eig(a)
The result is v = -0.9701 0.2425
0.9487 -0.3162
d = -4 0
0 -3
The diagonal elements of the matrix ‗d‘ are eigenvalues of the system and columns of the matrix ‗v‘ represent the corresponding eigenvectors. Example problem: Q. 1. Calculate the value of k for which the unity feedback system
crosses the imaginary axis.
Ans.
For the point of intersection on imaginary axis, construct Routh array
To cross imaginary axis Department of EEE, SJBIT
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so at K = 48, it will cross imaginary axis. Q. 2. A closed loop system is shown in the following fig. Find out the largest possible value of
for which this system would be stable.
Ans. As we know
Now
or characteristic equation becomes Now the Routh array can be written as
For stability all elements of column should be positive So,
The largest possible value for the system to be, stable. Q. 3. By means of Routh criterion, determine the stability of the system represented by the following equation Ans. This equations states
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There is one row that becomes zero. So it is being replaced by
equation can be written as:
Imaginary roots make system unstable and response is continuous oscillatory. Therefore system is unstable. Q. 4. Consider the closed loop feedback system shown in the figure below. Determine the range of K for which the system is stable.
Ans.
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This is the range of K for the system to be stable.
Q. 5. Discuss the Routh — Hurwitz criteria for determining the stability of a control system and calculate the range of K for stable operation of following characteristic equation
Ans. Routh-Hurwitz criterion helps in determining relative stability of a control system. From the characteristic equation of control system Routh‘s array is constructed. In case there is no change of sign in first column and the system is stable. Given characteristic equation is
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Q. 6. Discuss the Routh-Hurwitz criteria for determining the stability of a control system and calculate the range of K for stable operation of following characteristic equation. Ans. Routh-Hurwitz criterion helps in determining relative stability of a control system. From the characteristic equation of control system Routh‘s array is constructed. In case there is no change of sign in first column and the system is stable. Given characteristic equation is
Q. 7. Determine the stability of system having characteristic equation: using Routh Hurwitz criterion.
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Ans. Given characteristic equation is
Since row becomes zero, we formula auxiliary equation Differentiate auxiliary equation
Since the last element is negative, there is a sign change in first column, the system is unstable.
Q. 8. Determine the stability of a system with characteristic equation
Ans. Given characteristic equation is
Since terms corresponding to S3 are zero, we will form -auxiliary equation corresponding to
differentiating equation we get
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Consider auxiliary equation
Differentiating 2s = 0
As there is no sign change system may be stable. Solving auxiliary equation
Thus system is unstable. Q. 9. Find the range of K for stability of Ans. Given characteristic equation is Routh‘s array for given equation is
- For system to be stable all elements in first column should be greater than zero.
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so, value of K for stability will be 0 < K < 0.44.
Q. 10. Determine the stability of the system whose characteristic equation is given by
Ans. Routh‘s array is
As their is sign change in first column thus system is unstable.
Q. 11. A unity feedback control system is characterized by open loop transfer function Using Routh’s criterion, calculate the range of values of K for the system to be stable. Ans. The characteristic equation is 1 + G(s). H(s), H(s)=1 1 + G (s)
The Routh‘s array is
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For stable system. there should be no sign change in first column. For s° K + 1 > 0 K > —1
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UNIT 5 Root-locus techniques Introduction A designer can determine whether his design for a control system meets the specifications if he knows the desired time response of the controlled variable. By deriving the differential equations for the control system and solving them, an accurate solution of the system's performance can be obtained, but this approach is not feasible for other than simple systems. It is not easy to determine from this solution just what parameters in the system should be changed to improve the response. A designer wishes to be able to predict the performance by an analysis that does not require the actual solution of the differential equations. The first thing that a designer wants to know about a given system is whether or not it is stable. This can be determined by examining the roots obtained from the characteristic equation (5.1)
of the closed loop. The work involved in determining the roots of this equation can be avoided by applying the Hurwitz or Routh criterion. Determining in this way whether the system is stable or unstable does not satisfy the designer, because it does not indicate the degree of stability of the system, i.e., the amount of overshoot and the settling time of the controlled variable for a step input. Not only must the system be stable, but the overshoot must be maintained within prescribed bounds and transients must die out in a sufficiently short time. The root-locus method described in this section not only indicates whether a system is stable or unstable but, for a stable system, also shows the degree of stability. The root locus is a plot of the roots of the characteristic equation of the closed loop as a function of the gain. This graphical approach yields a clear indication of the effect of gain adjustment with relatively small effort. With this method one determines the closed-loop poles in the plane - these are the roots of Eq.(5.1) - by using the known distribution of the poles and zeros of the open-loop transfer function . If for instance a parameter is varied, the roots of the characteristic equation will move on certain curves in the plane as shown by the example in Figure5.1. On these curves lie all
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Figure 5.1: Plot of all roots of the characteristic equation
10ES43
for
.
Values of are red and underlined. possible roots of the characteristic equation for all values of the varied parameter from zero to infinity. These curves are defined as the root-locus plot of the closed loop. Once this plot is obtained, the roots that best fit the system performance specifications can be selected. Corresponding to the selected roots there is a required value of the parameter which can be determined from the plot. When the roots have been selected, the time response can be obtained. Since the process of finding the root locus by calculating the roots for various values of a parameter becomes tedious, a simpler method of obtaining the root locus is desired. The graphical method for determining the root-locus plot is shown in the following. An open-loop transfer function with poles at the origin of the plane is often described by (5.2)
where is the gain of the open loop. In order to represent this transfer function in terms of the open-loop poles and zeros it is rewritten as
(5.3)
or
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(5.4)
with
and
. The relationship between the factor
and the open-loop gain
is
(5.5)
The characteristic equation of the closed loop using Eq. (5.3) is (5.6)
or (5.7)
All complex numbers root locus.
, which fulfil this condition for
, represent the
From the above it can be concluded that the magnitude of must always be unity and its phase angle must be an odd multiple of . Consequently, the following two conditions are formalised for the root locus for all positive values of
from zero to infinity:
a) Magnitude condition: (5.8)
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b) Angle condition (5.9) for
In a similar manner, the conditions for negative values of ( determined. The magnitude conditions is the same, but the angle must satisfy the c) Angle condition
) can be
(5.10) for
Apparently the angle condition is independent of
. All points of the plane that fulfil the
angle condition are the loci of the poles of the closed loop by varying
. The calibration of
the curves by the values of is obtained by the magnitude condition according to Eq. (5.8). Based upon this interpretation of the conditions the root locus can constructed in a graphical/numerical way. Once the open-loop transfer function has been determined and put into the proper form, the poles and zeros of this function are plotted in the plane.
The plot of the locus of the closed loop poles as a function of the open loop gain K, when K is varied from 0 to +00. When system gain K is varied from 0 to +oo, the locus is called direct root locus. When system gain K is varied from -oo to 0, the locus is called as inverse root locus. The root locus is always symmetrical about the real axis i.e. x-axis. The number of separate branches of the root locus equals either the number of open loop poles are number of open-loop zeros whichever is greater. A section of root locus lies on the real axis if the total number of open-loop poles and zeros to the right of the section is odd. If the root locus intersects the imaginary axis then the point of intersection are conjugate. From the open loop complex pole the root locus departs making an angle with the horizontal line. The root locus starts from open-loop poles. The root locus terminates either on open loop zero or infinity. The number of branches of roots locus are:
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and M if P No. of poles ‗P‘ M —> No. of zeros ‗Z‘
Centroid is the centre of asymptotes. It is given by (an)
Angle of asymptotes is denoted by ‗p
Angle of departure is. tangent to root locus at complex pole
Based on the pole and zero distributions of an open-loop system the stability of the closedloop system can be discussed as a function of one scalar parameter. The root-locus method shown in this module is a technique that can be used as a tool to design control systems. The basic ideas and its relevancy to control system design are introduced and illustrated. Ten general rules for constructing root loci for positive and negative gain are shortly presented such that they can be easily applied. This is demonstrated by some discussed examples, by a table with sixteen examples and by a comprehensive design of a closed-loop system of higher order.
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Example Problems: Q.1. Consider the example
with
,
are the roots
and
and
. The poles of the closed-loop transfer function
of the characteristic equation
and are given by
As
and
it can be seen that for
the poles of the closed loop
transfer function are identical with those of the open-loop transfer function values a)
. For other
the following two cases are considered: :
Both roots
and
and
are real and lie on the real axis in the range of
;
b) :
The roots
and
are conjugate complex with the real part
which does not depend on , and the imaginary part Im The curve has two branches as shown in Figure 6.2.
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, .
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Figure 5.2: Root locus of a simple second-order system At is the breakaway point of the two branches. Checking the angle condition the condition
must be valid. The complex numbers
and
have the angles
and
and the
magnitudes and . The triangle ( ) in Figure 6.2 yields the angle condition. Evaluating the magnitude condition according to Eq. (6.8)
one obtains the value
The value of
on the root locus. E.g. for
at the breakaway point
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the gain of the open loop is
is
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Table 5.1 shows further examples of some 1st- and 2nd-order systems. Table 5.1: Root loci of 1st- and 2nd-order systems root locus
root locus
General rules for constructing root loci To facilitate the application of the root-locus method for systems of higher order than 2nd, rules can be established. These rules are based upon the interpretation of the angle condition and the analysis of the characteristic equation. The rules presented aid in obtaining the root locus by expediting the manual plotting of the locus. But for automatic plotting using a computer these rules provide checkpoints to ensure that the solution is correct. Though the angle and magnitude conditions can also be applied to systems having dead time, in the following we restrict to the case of the open-loop rational transfer functions according to Eq. (5.3)
(5.11)
or (5.12)
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As this transfer function can be written in terms of poles and zeros )
and
(
;
can be represented by their magnitudes and angles
or
(5.13)
From Eq. (5.8) the magnitude condition
(5.14)
and from Eq. (5.9) the angle condition (5.15) for
follows. Here and denote the angles of the complex values and , respectively. All angles are considered positive, measured in the counterclockwise sense. If for each point the sum of these angles in the plane is calculated, just those particular points that fulfil the condition in Eq. (5.15) are points on the root locus. This principle of constructing a root-locus curve - as shown in Figure 5.3 - is mostly used for automatic rootlocus plotting.
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Figure 5.3: Pole-zero diagram for construction of the root locus In the following the most important rules for the construction of root loci for
are listed:
Rule 1 Symmetry As all roots are either real or complex conjugate pairs so that the root locus is symmetrical to the real axis. Rule 2 Number of branches The number of branches of the root locus is equal to the number of poles of the open-loop transfer function. Rule 3 Locus start and end points The locus starting points (
) are at the open-loop poles and the locus ending
points ( ) are at the open-loop zeros. branches end at infinity. The number of starting branches from a pole and ending branches at a zero is equal to the multiplicity of the poles and zeros, respectively. A point at infinity is considered as an equivalent zero of multiplicity equal to . Rule 4 Real axis locus If the total number of poles and zeros to the right of a point on the real axis is odd, this point lies on the locus. Rule 5 Asymptotes There are
asymptotes of the root locus with a slope of (5.16)
For Figure 5.4.
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and 4 one obtains the asymptote configurations as shown in
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Figure 5.4: Asymptote configurations of the root locus
Rule 6 Real axis intercept of the asymptotes The real axis crossing
of the asymptotes is at (5.17)
Rule 7 Breakaway and break-in points on the real axis At least one breakaway or break-in point exists if a branch of the root locus is on the real axis between two poles or zeros, respectively. Conditions to find such real points are based on the fact that they represent multiple real roots. In addition to the characteristic equation (6.1) for multiple roots the condition (5.18)
must be fulfilled, which is equivalent to (5.19)
for . If there are no poles or zeros, the corresponding sum is zero. Rule 8 Complex pole/zero angle of departure/entry The angle of departure of pairs of poles with multiplicity
is (6.20)
and the angle of entry of the pairs of zeros with multiplicity Department of EEE, SJBIT
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(5.21)
Rule 9 Root-locus calibration The labels of the values of
can be determined by using (5.22)
For the denominator is equal to one. Rule 10 Asymptotic stability The closed loop system is asymptotically stable for all values of for which the locus lies in the left-half plane. From the imaginary-axis crossing points the critical values
can be determined.
The rules shown above are for positive values of
. According to the angle condition of
Eq. (5.10) for negative values of some rules have to be modified. In the following these rules are numbered as above but labelled by a *. Rule 3* Locus start and end points The locus starting points (
) are at the open-loop poles and the locus ending
points ( ) are at the open-loop zeros. branches end at infinity. The number of starting branches from a pole and ending branches at a zero is equal to the multiplicity of the poles and zeros, respectively. A point at infinity is considered as an equivalent zero of multiplicity equal to . Rule 4* Real axis locus If the total number of poles and zeros to the right of a point on the real axis is even including zero, this point lies on the locus. Rule 5* Asymptotes There are
asymptotes of the root locus with a slope of (5.23)
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Rule 8* Complex pole/zero angle of departure/entry The angle of departure of pairs of poles with multiplicity
is (5.24)
and the angle of entry of the pairs of zeros with multiplicity (5.25)
The root-locus method can also be applied for other cases than varying
. This is possible as
long as can be rewritten such that the angle condition according to Eq. (5.15) and the rules given above can be applied. This will be demonstrated in the following two examples. Q.2. Given the closed-loop characteristic equation
the root locus for varying the parameter rewritten as
is required. The characteristic equation is therefore
This form then correspondents to the standard form
to which the rules can be applied. Q.3.Given the closed-loop characteristic equation
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Using the rules 1 to 10 one can easily predict the geometrical form of the root locus based on the distribution of the open-loop poles and zeros. Table 6.2 shows some typical distributions of open-loop poles and zeros and their root loci. Table 6.2: Typical distributions of open-loop poles and zeros and the root loci
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For the qualitative assessment of the root locus one can use a physical analogy. If all openloop poles are substituted by a negative electrical charge and all zeros by a commensurate positive one and if a massless negative charged particle is put onto a point of the root locus, a movement is observed. The path that the particle takes because of the interplay between the repulsion of the poles and the attraction of the zeros lies just on the root locus. Comparing the root locus examples 3 and 9 of Table 5.2 the 'repulsive' effect of the additional pole can be clearly seen. The systematic application of the rules from section 5.2 for the construction of a root locus is shown in the following non-trivial example for the open-loop transfer function
(5.26)
The degree of the numerator polynomial is one zero (
. This means that the transfer function has
). The degree of the denominator polynomial is
and we have the
four poles ( , , , 2). First the poles (x) and the zeros (o) of the open loop are drawn on the plane as shown in Figure 5.5. According to rule 3 these poles are just
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Figure 5.5: Root locus of underlined. those points of the root locus where
. Values of
and the zeros where
are in red and
. We have
branches that go to infinity and the asymptotes of these three branches are lines which intercept the real axis according to rule 6. From Eq. (5.17) the crossing is at (5.27)
and the slopes of the asymptotes are according to Eq. (5.16) (5.28)
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i.e.
The asymptotes are shown in Figure 5.5 as blue lines. Using Rule 4 it can be checked which points on the real axis are points on the root locus. The points
with
and
belong to the root locus, because to the right of them the number of poles and zeros is odd. According to rule 7 breakaway and break-in points can only occur pairwise on the real axis to the left of -2. These points are real solutions of the Eq. (5.19). Here we have (5.29)
or
This equation has the solutions roots
and
,
and
. The real
are the positions of the breakaway and the break-in point.
The angle of departure of the root locus from the complex pole at determined from Figure 5.6 according to Eq. (5.20):
can be (5.30)
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With this specifications the root locus can be sketched. Using rule 9 the value of can be determined for some selected points. The value at the intersection with the imaginary axis is
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UNIT 6 Frequency Domain Analysis Bode plot If
the
absolute
value
and
the
phase
are separately plotted over the frequency
of
the
frequency
response
, one obtains the
Figure 6.1: Plot of a frequency response: (a) linear, (b) logarithmic presentation ( logarithmic scale) (Bode plot)
on a
amplitude response and the phase response. Both together are the frequency response characteristics.
and
are normally drawn with a logarithm and
This representation is called a Bode diagram or Bode plot. Usually decibels [dB] By definition this is
The logarithmic representation of the amplitude response scale in this diagram and is called the magnitude.
with a linear scale. will be specified in
has consequently a linear
The logarithmic representation has some advantages for series connections of transfer functions. For complicated frequency responses, e.g. with
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for , it can be represented as series connections of the frequency responses of simple elements of the form
and
for
From this it follows that
with for
From the representation
and
respectively, one obtains the logarithmic characteristic of the magnitude
and the phase characteristic Department of EEE, SJBIT
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with for and for . Thus, the frequency response of a series connection is obtained by addition of the individual frequency response characteristics. A further advantage of this logarithmic representation is for the determination of the inverse of a frequency response, that is for
. Here
and
are valid, the curves of line) and
and
need only to be mirrored at the axes
(0-
.
Because of the double logarithmic and of the single logarithmic scale of
and
,
respectively, the curve of and that of can be approximated by line segments. This approximation by lines allows the analysis and synthesis of control systems using simple geometric constructions. They are important concepts for the control engineer. Improtant points:
The magnitude and phase relationship between sinusoidal input and steady state output of a system is known as frequency response. The polar plot of a sinusoidal transfer function G (jw) is plot of the magnitude of G (jw) versus the phase angle of G (jw) on polar coordinates as ‗co‘ varied from zero to infinity. The phase margin is that amount, of additional phase lag at the gain crossover frequency required to bring the system to the verge of instability. The gain margin is the reciprocal of the magnitude l G(jw) l at the frequency at which the phase angle as _1800. The inverse polar plot at G (jw) is a graph of 1/G (jw) as a function of w. Bode plot is a graphical representation of the transfer function for determining the stability of control system. Bode plot is a combination of two plot - magnitude plot and phase plot.
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The transfer function having no poles and zeros in the right -half s-plane are called minimum phase transfer function. System with minimum phase transfer function are called minimum phase systems. The transfer function having poles and zeros in the right half s-plane are called nonminimum phase transfer functions systems with non-minimum phase transfer function. are called non-minimum phase system. In bode plot the relative stability of the system is determined from the gain margin and phase margin. . If gain cross frequency is less than phase cross over frequency then gain margin and phase margin both are positive and system is stable. If gain cross over frequency is greater than the phase crossover frequency than both gain margin and‘phase margin are negative. It gain cross over frequency is equal to me phase cross over trequency me gain marg and phase margin are zero and system is marginally stable. The maximum value of magnitude is known as resonant peak. The magnitude of resonant peak gives the information about the relative stability of the system. The frequency at which magnitude has maximum value is known as resonant frequency. Bandwidth is defined a the range of frequencies in which the magnitude of closed loop does not drop —3 db.
Example Problems:
with
is rearranged into the form
This system can now be decomposed into an integrator, two PD and two is
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elements, that
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From this simple analysis the Bode plot can be determined by adding the Bode plots of the elements
to
. In this figure the variable quantities
Figure: Representation of a dynamic system by two frequency response diagrams: (a) Bode plot, (b) Nyquist plot
in the terms for and are the breakpoint frequencies in the Bode plot. Also shown in Figure 4.20 is the Nyquist plot of the frequency response. Both representations of Figure 4.20 basically contain the same information about the system.
Based on the example given above the procedure for constructing a Bode plot of a given system can be recapitulated:
a) The given transfer function must be put into the form
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with where possible poles of according to their multiplicity . b) Then for
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at
will be specially considered
the asymptotes of the elements will be used to approximate
and
. c) If necessary corrections of the approximations can be performed.
Introduction to lead, lag and lead-lag compensation networks
Compensation technique is used to make the unstable system stable by introducing the poles and or zeros at suitable place. In control system design, if the designed specifications do not satisfy the requirements of the system or leads to expensive and conflicting demands, then it is required to insert an additiohal component within the structure of feedback system. This adjustment is called compensation. Compensators are of three types: (a) Lag phase compensator(b) Lead phase compensator (c) Lead-lag compensator. Depending on the location the compensation is divided in following types: (a) Series or cascade compensation (b) Parallel or feedback compensation (c) Load compensation. Load compensations is provided to damp out oscillations in the system having mechanical output. Primary function of the lead compensator is to reshape the frequency response curve to provide sufficient phase lead angle to offset the excessive phase lag associated with the components of fixed systems.
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The primary function of lag compensator is to provide attenuation in the high frequency range to give a system sufficient phase margin.
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Phase lag network is shown in figure below. The primary function of lag compensator is to provide attenuation in high frequency range to give a system sufficient phase margin. Various design steps of phase lag network are
Example Problems: Q1. The asymptotic magnitude Bode plot of a system is given in the figure below Find the transfer function of the system analytically It is known that the system is minimum phase system.
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Q 2. Sketöh the Bode Plot for the transfer function given by, and from Plot find (a) Phase and Gain cross rer frequencies (b) Gain Margin and Phase Margin. Is this System Stable?
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The gain crosses 0db axis at co = 1.24 rad/sec, the gain crossover frequency is co = 1.24 rad/sec. The phase crosses —180° line at co = 0.4 rad/sec, therefore phase crossover frequency is co = 0.4 rad/sec. At phase cross over the gain is 20 dB, therefore gain margin is —20 dB. At gain crossover the phase angle is 2150, the phase margin is 180° + (—215°) = —35°. As both gain and phase margins are negative, the system is unstable. Q3. Sketch the bode plot for the transfer function given by
and from plot find gain margin and phase margin. Ans. On 0)-axis mark the point at 23.7 rad/sec. since in denominator (jw) term is having power one, from 23.7 draw a line of slope —20 db/decade to meet y-axis. This will be the starting point. Step 1. From the starting point to I corner frequency (0.33) the slope of the line is —20 db/decade. From I corner frequency (0.33) to second corner frequency (1) the slope of the line will be — 20 ÷ (—20) = —40 db/decade. From II corner frequency to IV corner frequency (2) the slope of the line be —40 + (÷20) = —20 db/decade. From III corner frequency to IV corner frequency, the slope of line will be —20 + (—20) = —40 db/decade. From IV corner frequency (5) to V corner frequency the slope will be —40 ÷ (+20) = —20 db/decade. After V corner frequency, the slope will be (—20) ÷ (—20) = —40 db/decade. Step 2. Draw the phase plot. Step 3. From graph Phase margin = +34° Gain margin =infinity
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Q 4. State the advantages of bode plots. Determine the value of K in the transfer function given below such that (a) Time gain margin is 20 dB (b) The phase margin is 30°
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On the phase plot, we made this poirtt and extend the line tO magnitude plot where it intersects. For required phase margin shift magnitude plot upwards Total shift = 19 dB (from graph) 20 log K = 19 K =8.91.1 Q.5.Design a suitable compensator such that the system will have K= 10 and phase margin = 50°.
The open loop transfer function is thus,
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The bode prot is plotted in fig. The frequency range selected in rad/sec.
rad/sec to
From bode plot it is noted that uncompensated system is stable having a phase margin of 30° and the gain crossover occurs at o = 6.75 rad/sec. To increase the phase margin to 50° the gain crossover point is to be shifted to a higher value of frequency and this is possible by introducing a phase lead compensation network. The required phase lead is:
The value of a parameter of the phase read network is given by:
The maximum phase lead (network)
is contributed at mid frequency
compensation network. Allowing an amplification of compensation network transfer function at
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ofphase lead
the magnitude of phase lead
is determined below:
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UNIT 7 Stability in frequency domain Nyquist criterion This graphical method, which was originally developed for the stability analysis of feedback amplifiers, is especially suitable for different control applications. With this method the closed-loop stability analysis is based on the locus of the open-loop frequency response . Since only knowledge of the frequency response practical approach for the following cases: a) For many cases are known.
is necessary, it is a versatile
can be determined by series connection of elements whose parameters
b) Frequency responses of the loop elements determined by experiments or considered directly.
can be
c) Systems with dead time can be investigated. d) Using the frequency response characteristic of not only the stability analysis, but also the design of stable control systems can be easily performed.
Nyquist criterion using Nyquist plots To derive this criterion one starts with the rational transfer function of the open loop
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Figure: Poles of the open and closed loop in the their multiplicity)
plane (multiple poles are counted according to
To determine , the locus can be drawn on the Nyquist diagram and the phase angle checked. Expediently one moves this curve by 1 to the left in the plane. Thus for stability analysis of the closed loop the locus according to Figure 5.5 has to be drawn.
Figure : Nyquist diagrams of Here
of the open loop
and
is the continuous change in the angle of the vector from the so called critical point (-1,j0)
to the moving point on the locus of
for
. Points where the locus passes through
the point (-1,j0) or where it has points at infinity correspond to the zeros and poles of on the imaginary axis, respectively. These discontinuities are not taken into account for the derivation of . Figure shows an example of a
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Figure: Determination of continuous changes in the angle where two discontinuous changes of the angle occur. Thereby the continuous change of the angle consists of three parts
The rotation is counter clockwise positive.
As the closed loop is only asymptotically stable for the Nyquist criterion follows:
, then from the general case of
The closed loop is asymptotically stable, if and only if the continuous change in the angle of the vector from the critical point (-1,j0) to the moving point of the locus
For the case with a negative gain the case with a positive time in the open loop.
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of the open loop is
of the open loop the locus is rotated by 180 relative to
. The Nyquist criterion remains valid also in the case of a dead
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Simplified forms of the Nyquist criterion It follows from that for an open-loop stable system, that is Therefore the Nyquist criterion can be reformulated as follows:
and
, then
.
If the open loop is asymptotically stable, then the closed loop is only asymptotically stable, if the frequency response locus of the open loop does neither revolve around or pass through the critical point (-1,j0). Another form of the simplified Nyquist criterion for 'left-hand rule':
with poles at
is the so called
The open loop has only poles in the left-half plane with the exception of a single or double pole at
(P, I or
behaviour). In this case the closed loop is only stable, if the critical
point (-1,j0) is on the left hand-side of the locus .
in the direction of increasing values of
This form of the Nyquist criterion is sufficient for most cases. The part of the locus that is significant is that closest to the critical point. For very complicated curves one should go back to the general case. The left-hand rule can be graphically derived from the generalised locus The orthogonal ( a curve with hand side of
)-net is observed and asymptotic stability of the closed loop is given, if passes through the critical point (-1,j0). Such a curve is always on the left.
The Nyquist criterion using Bode plots Because of the simplicity of the graphical construction of the frequency response characteristics of a given transfer function the application of the Nyquist criterion is often more simple using Bode plots. The continuous change of the angle critical point (-1,j0) to the locus of response of
of the vector from the
must be expressed by the amplitude and phase
. From figure
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Figure : Positive (+) and negative (-) intersections of the locus the left-hand side of the critical point
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with the real axis on
it can be seen that this change of the angle is directly related to the count of intersections of the locus with the real axis on the left-hand side of the critical point between . The Nyquist criterion can therefore also represented by the count of these intersections if the gain of the open loop is positive. Regarding the intersections of the locus of with the real axis in the range , the transfer from the upper to the lower half plane in the direction of increasing values are treated as positive intersections while the reverse transfer are negative intersections (Figure 5.7). The change of the angle is zero if the count of positive intersections is equal to the count of negative intersections . The change of the angle depends also on the number of positive and negative intersections and if the open loop does not have poles on the imaginary axis, the change of the angle is
In the case of an open loop containing an integrator, i.e. a single pole in the origin of the complex plane ( ), the locus starts for at , where an additional change of the angle. For proportional and integral behaviour of the open loop
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is added to the
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is valid. In principle this relation is also valid for
, but the locus starts for
at
(Figure 5.8), and this intersection would be counted
Figure : Count of the intersections on the left-hand side of the critical point for open loop as a negative one if
, i.e. if the locus for small
behaviour of the
is in the upper half plane of the real axis. But
de facto there is for (and accordingly ) no intersection. This follows from the detailed investigation of the discontinuous change of the angle, which occurs at . As only a continuous change of the angle is taken into account and because of reason of symmetry the start of the locus at is counted as a half intersection, positive for and negative for analogous to the definition given above For continuous changes of the angle
The open loop with the transfer function
has
poles in the left-half
, which is
plane and possibly a
single ( ) or double pole ( ) at . If the locus of has positive and negative intersections with the real axis to the left of the critical point, then the closed loop is only asymptotically stable, if
is valid. For the special case, that the open loop is stable ( negative intersections must be equal.
,
), the number of positive and
From this it follows that the difference of the number of positive and negative intersections in the case of that for
is an integer and for the number
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is even, for
not an integer. From this follows immediately, the number
is uneven and therefore in Page 106
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is an even number, such that the closed loop is asymptotically stable. This is only .
The Nyquist criterion can now be transferred directly into the representation using frequency response characteristics. The magnitude response
, which corresponds to the locus
, is always positive at the intersections of the locus with the real axis in the range of . These points of intersection correspond to the crossings of the phase response with lines
,
etc., i.e. a uneven multiple of 180 . In the case of a positive
intersection of the locus, the phase response at the lines crosses from below to top and reverse from top to below on a negative intersection as shown in Figure 5.9. In the following these crossings
Figure : Frequency response characteristics of and negative (-) crossings of the phase response
and definition of positive (+) with the -180 line
will be defined as positive (+) and negative (-) crossings of the phase response
over the
particular lines, where may be valid. If the phase response starts at 180 this point is counted as a half crossing with the corresponding sign. Based on the discussions above the Nyquist criterion can be formulated in a form suitable for frequency response characteristics: The open loop with the transfer function has poles in the right-half plane, and possibly a single or double pole at . are the number of positive and of negative crossings of the phase response over the lines in the frequency range where valid. The closed loop is only asymptotically stable, if Department of EEE, SJBIT
is
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is valid. For the special case of an open-loop stable system (
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,
)
must be valid. Table 7.1: Examples of stability analysis using the Nyquist criterion with frequency response characteristics
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Finally the 'left-hand rule' will be given using Bode diagrams, because this version is for the most cases sufficient and simple to apply. The open loop has only poles in the left-half plane with the exception of possibly one single or one multiple pole at if
(P, I or
has a phase of
behaviour). In this case the closed loop is only asymptotically stable, for the crossover frequency
at
.
This stability criterion offers the possibility of a practical assessment of the 'quality of stability' of a control loop. The larger the distance of the locus from the critical point the Department of EEE, SJBIT
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farther is the closed loop from the stability margin. As a measure of this distance the terms gain margin and phase margin are introduced according to Figure below
Figure : Phase and gain margin Bode diagram
and
or
, respectively, in the (a) Nyquist diagram and (b)
Example Problems: Q1 The polar plot of the open-loop transter of feedback control system intersects the real axis at—2 Calculate gain margin (in dB) of the system.
Q2. What is the gain margin of a system in decibels if its Nyquist plot cuts the negative real axis at — 0.7? Ans. a = —0.7
Q4. Consider a feed lock system with the open-loop transfer function. Given by
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Examine the stability of the closed-loop system. Using Nyquist stability theory .
Q 5. Draw the Nyquest plot for the open loop transfer function given below:
and obtain the gain margin and phase margin.
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Q6. Consider a feed lock system with the open-loop transfer function. Given by
Examine the stability of the closed-loop system. Using Nyquist stability theory.
Q7. Sketch the Nyquist plot for the system with the open loop transfer function
and determine the range of K for which the system is
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To get point of intersection on real axis, equate imaglnary part to zero.
Q.8. Sketch the Nyquist plot for system with
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Q 9. How is it possible to make assessment of relative stability using Nyquist criterion? Construct Nyquist plot for the system whose open loop transfer function is
Find the range of K for stability. Ans.
Nyquist critierion can be used to make assessment of relative stability. Using the characteristic equation the Nyquist plot is drawn. A feedback system is sable if and only if, the i.e. contour in the G (s) plane does not encircle the (—1, 0) point when the number of poles of G(s) in the right hand s plane is zero. If G (a) has P poles in the right hand plane, then the number of anticlockwise encirciements of the (—1, 0) point must be equal to P for a stable system, N=—P0 where N = No of clockwise encirclements about (—1, 0) point in C (s) plane P0 = No of poles G (s) in RHP 0
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UNIT 8 Introduction to State variable analysis State-variale representation of single-input-single-output systems In the following a short introduction into the representation of systems using state-variable techniques is given. For this purpose, the example of the network from Figure is used. The dynamical behaviour of this network is completely defined for
initial conditions
, if the
,
and the
input variable
for
are known. For these specifications the variables
and
can be determined for all
. The variables and characterise the 'state' of the network and are therefore called state variables of the network. The differential equations describe the dynamical behaviour of this network. Inserting into one obtains the 2nd-order differential equation according to which completely describes the system with respect to the input-output behaviour. But one can also use the two original differential equations and can write them in vector notation so that the 1st-order vector differential equation
(8.1) with the initial condition
is obtained. This linear 1st-order vector differential equation describes the connection between the input variable and the state variables. To complete a state-space system, one needs an additional equation that describes the dependence of the output variable on the state variables. In this example, it is the direct relationship Introducing the state vector
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into above eq, with the vectors
and with the matrix
and with the scalar variables and one obtains the general state-space representation of a linear time-invariant single-inputsingle-output system: (8.2)
initial condition
(8.3) The Eq. (8.2) is the state equation, and in the general case it is a linear system of 1st-order differential equations of
state variables
, which are combined in the state
vector . Eq. (8.3) is the output equation, which maps the states and inputs linearly to the output. This is an algebraic equation, whereas the state equation is a differential equation.
State-space representation of multi-input-multi-output systems The Eqs. (8.2) and (8.3) describe an th-order linear time-invariant single-input-singleoutput system. For linear multi-input-multi-output systems of order with inputs and outputs these equations become
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with the initial condition
(8.5) where the following notation is used:
state vector
(
) vector
input vector
(
) vector
output vector
(
) vector
system matrix
(
) matrix
input matrix
(
) matrix
output matrix
(
) matrix
feedthrough matrix
(
) matrix
It goes without saying that the general representation of Eqs. (8.4) and (8.5) also includes the single-input-single-output case. The matrices and have constants elements. If these elements are time-varying, the matrices of the corresponding time-varying system are substituted by matrix functions of time, e.g.
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.
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The relationship between transfer functions and the state-space representation In the following, the Eqs. (8.4) and (8.5) will be transformed into the domain using the Laplace transform, which will be done analogously to the scalar case . For this, the operator notation
from section 2.1 is adopted and when applying it to Eq. (8.4), one obtains
or rearranged
The solution of the state equation in the domain is then given by (8.6) with (8.7) Similarly, for Eq. (12.5) yields
Substituting
from Eq. (12.6), the system output in the domain is
To obtain the relationship with transfer functions, the initial condition has to be set to zero. For a single-input-single-output system according to Eqs. (8.2) and (8.3) the system output is
Comparing this equation with Eq. (8.3) the transfer function is given by (8.8)
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from Eq. (8.7) is a matrix of rational functions of , which can always be
(8.9)
where
is a matrix with polynomial elements in . From Eq. (8.2) it is obvious that (8.10)
and (8.11) which is the characteristic polynomial of the system. The zeros of this polynomial are the poles of the transfer function and at the same time eigenvalues of the system matrix . If the system in the state-space representation is fully controllable and observable (see section 8.6), then the number of poles are equal to the number of eigenvalues.
State-space vs transfer function approach The key advantage of transfer functions is in their compactness, which makes them suitable for frequency-domain analysis and stability studies. However, the transfer function approach suffers from neglecting the initial conditions. Not only does state-space representation serve as an alternative to transfer functions, but also it is not limited to linear and time-invariant systems and it has the following advantages: 1. Single-input-single-output and multi-input-multi-output systems can be formally treated equal. 2. The state-space representation is best suited both for the theoretical treatment of control systems (analytical solutions, optimisation) and for numerical calculations. 3. The determination of the system response in the homogeneous case with the initial condition is very simple. 4. This representation gives a better insight into the inner system behaviour. General system properties, for example, the system controllability or observability can be defined and determined.
Uniqueness of the state variables
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Initially it sounds paradoxical that the choice of the state variables is not unique. This means that for one and the same system with the input , the output and state variables, there exist an infinite number of state-space representations. For each value of time one gets the state in the -dimensional state space. The values are the cartesian coordinates of the state
where the unit vectors are -dimensional linear independent vectors. Their elements are besides the th element, which has the value of 1 - all zero. For describing the state also other basis vectors can be used. Candidates are all linear independent and -dimensional vectors . Therefore, it is always possible, to write the state as (8.12)
After introducing the new state
and the quadratic matrix
one can rewrite Eq. (8.12) as (8.13)
The constant and regular matrix is a so-called transformation matrix. Now, instead of the state the new state can be used and its behaviour analysed. Describing the system in the new coordinates the state from Eq. (8.13) is inserted into Eqs. (8.4) and (8.5) and one obtains (8.14)
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(8.15)
with the initial condition
The benefits of the transformation of systems into different state-space representations are:
Most system properties do not depend on the choice of the state variables. They remain unchanged after a regular transformation and may be analysed in an appropriate representation form. The computational determination and analysis of system properties can be tremendously simplified if the representation form is specifically selected. In particular certain canonical forms are of interest.
Example 12.5.1 In order to demonstrate a transformation, the example from Eq. (8.1) is used with the system parameters , and . The initial condition is assumed to be zero and therefore omitted. With these values one obtains the state equation as
and the output equation as
For the regular transformation matrix
one obtains the matrices for Eqs. (8.14) and (8.15) as
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The new representation
consists of two decoupled differential equations with respect to the state variables
and
.
The analysis and treatment of a system in such a structured representation form, as shown in the example above, is doubtless more simple. As the representation form must be specifically selected depending on the type of analysis or synthesis problem, the different representation forms, for example, the canonical forms, are not discussed separately and are introduced in the following sections when they are needed.
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10ES43 CONTROL SYSTEMS (Common to EC/TC/EE/IT/BM/ML) Subject Code:10ES43 No. of Lecture Hrs./Week : 04 Total No. of Lecture Hrs.:52
IA Marks: 25 Exam Hours : 03 Exam Marks : 100
PART – A UNIT 1: Modeling of Systems: Introduction to Control Systems, Types of control systems, Effect of feedback systems, Differential equations of physical systems – Mechanical systemsFriction, Translational systems (Mechanical accelerometer, Levered systems excluded), Rotational systems, Gear trains. Electrical systems, Analogous systems. 6 Hours UNIT 2: Block diagrams and signal flow graphs: Transfer functions, Block diagrams, Signal Flow graphs (Statevariable formulation excluded). 7 Hours UNIT 3: Time Response of feed back control systems: Standard test signals, Unit step response of First and second order systems, Time response specifications, Time response specifications of second order systems, steady – state errors and error constants. 7Hours
UNIT 4: Stability analysis: Concepts of stability, Necessary conditions for Stability, Routh-Hurwitz stability criterion, Relative stability analysis; Special cases of RH criterion. 6 Hours PART – B UNIT 5: Root–Locus Techniques: Introduction, basic properties of root loci, Construction of root loci. 6 Hours UNIT 6: Stability analysis in frequency domain: Introduction, Mathematical preliminaries, Nyquist Stability criterion, (Inverse polar plots excluded), Assessment of relative stability using Nyquist criterion, (Systems with transportation lag excluded). 7Hours Department of EEE, SJBIT
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UNIT 7: Frequency domain analysis: Correlation between time and frequency response, Bode plots, All pass and minimum phase systems, Experimental determination of transfer functions, Assessment of relative stability using Bode Plots. 7 Hours UNIT 8: Introduction to State variable analysis: Concepts of state, state variable and state models for electrical systems, Solution of state equations. 6 Hours TEXT BOOK : 1. Control Systems Engineering, I. J. Nagarath and M.Gopal, New Age International (P) Limited, 4 Edition – 2005 2 Modern Control Engineering, K. Ogata, PHI, 5th Edition, 2010.
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10ES43 Table of contents
Sl .no
Contents UNIT 1:Modeling of Systems: Introduction to Control Systems
1
Page number
Types of control systems, Effect of feedback systems,
Differential equations of physical systems – Mechanical systems- Friction
5 to 12
Translational systems (Mechanical accelerometer, Levered systems excluded)
Rotational systems, Gear trains. Electrical systems, Analogous systems.
UNIT 2: Block diagrams and signal flow graphs Transfer functions 2
Block diagram
13 to 23
Signal Flow graphs (Statevariable formulation excluded). UNIT 3: Time Response of feed back control systems: Standard test signals 3
Unit step response of First and second order systems,
Time response specifications
24 to 46
Time response specifications of second order systems, steady state errors and error constants. UNIT 4: Stability analysis: Concepts of stability
Necessary conditions for Stability
4
47 to 65
Routh-Hurwitz stability criterion
Relative stability analysis
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Special cases of RH criterion.
UNIT 5: Root–Locus Techniques: Introduction 5
basic properties of root loci
66 to 84
Construction of root loci. UNIT 6: Stability analysis in frequency domain: Introduction 6
Mathematical preliminaries
Nyquist Stability criterion, (Inverse polar plots excluded)
85 to 102
Assessment of relative stability using Nyquist criterion,
UNIT 7: Frequency domain analysis: Correlation between time and frequency response
7
Bode plots
All pass and minimum phase systems
Experimental determination of transfer functions
Assessment of relative stability using Bode Plots.
103 to 118
UNIT 8: Introduction to State variable analysis: Concepts of state 8
state variable and state models for electrical systems
119 to 127
Solution of state equations
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UNIT-1 Modeling of Systems Introduction to control systems A system is an arrangement of or a combination of different physical components connected or related in such a manner so as to form an entire unit to attain a certain objective Control system is an arrangement of different physical elements connected in such a manner so as to regulate, director command itself to achieve a certain objective Requirements of good control system are accuracy, sensitivity, noise, stability, bandwidth, speed, oscillations
Types of control systems A system in which the control action is totally independent of the output of the system is called as open loop system
Example: Automatic hand driver, automatic washing machine, bread toaster, electric lift, traffic signals, coffee server, theatre lamp etc.
A system in which the control action is somehow dependent on the output is called as closed loop system The elements of closed loop system are command, reference input, error detector, control element controlled system and feedback element
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Elements of closed loop system are: 1. Command : The command is the externally produced input and independent of the feedback control system. 2. Reference Input Element: It is used to produce the standard signals proportional to the command. 3. Error Detector : The error detector receives the measured signal and compare it with reference input. The difference of two signals produces error signal. 4. Control Element : This regulates the output according to the signal obtained from error detector. 5. Controlled System : This represents what we are controlling by feedback loop. 6. Feedback Element : This element feedback the output to the error detector for comparison with the reference input. Example: Automatic electric iron, servo voltage stabilizer, sun-seeker solar system, water level controller, human perspiration system.
Feedback system is that in which part of output is feeded back to input In feedback system corrective action starts only after the output has been affected Advantages of closed loop system: 1. Accuracy is very high as any error arising is corrected. 2. It senses changes -in output due to environmental or parametric change, internal disturbance etc. and corrects the same. 3. Reduce effect of non-linearities. 4. High bandwidth. 5. Facilitates automation. Disadvantages 1. Complicated in design and maintenance costlier. 2. System may become unstable. Department of EEE, SJBIT
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Advantages of open loop system: 1. They are simple in construction and design. 2. They are economic. 3. Easy for maintenance. 4. Not much problem of stability. 5. Convenient to use when output is difficult to measure. Disadvantages of open loop system 1. Inaccurate and unreliable because accuracy is dependent on accuracy of calibration. 2. Inaccurate results are obtained with parameter variations, internal disturbances. 3. To maintain quality and accuracy, recalibration of controller is necessary from time to time.
Feed forward system is that in which the corrective action is initiated without waiting for the effect of disturbance to show up in the output System having multiple inputs and multiple outputs is known as multiple output (MIMO) control system A servomechanism is a power amplifying feedback control system in which the controlled variable is mechanical position or its time derivative such as velocity, acceleration A regulator or regulating control system is a feedback control system in which the reference input remains constant for long periods/entire intervals of operation An adaptive control system is one that continuously and automatically measures the dynamic characteristics of the plant. The system which follows the principle of superposition and proportionality is called a linear system.
The motion take place along a straight line is known as translational motion. Rotational motion of a body is the motion about a fixed axis. The elements of rotational system are inertia (J), damping coefficient (B) and torsional stiffness (K). Mechanical system Department of EEE, SJBIT
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For every mechanical system, there is analogous electrical system. DAlembert‘s principle states that, the algebric sum of externally applied forces and the forces resisting motion in any given direction is zero. For mechanical network, analogous electrical network can be obtained by using f-v and f-i analogy. Force-voltage analogy: In this method force is analogous to voltage. Similarly,
displacement n charge q.
Force-current analogy: In this method force is analogous to current.
Mechanical rotational system: (a) Force-voltage analogy:
(b) Force-current analogy :
Mechanical coupling: Laplace transform of signal x (t) is denoted by X (s)
Example problems: Q 1. Draw the f-1 analogous mechanical system for the electrical circuit of fig. below:
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Ans. f (t) is analogous to e (t) f (t) is analogous to R. f (t) frictional force f is analogous to r. Spring constant K is analogous to reciprocal Mass M is analogous to inductance L
Q. 2. Draw the mathematical model of the following system and obtain the transfer function.
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Ans. Using torque-voltage analogy, we have
The equation is
Q. 3. Write force-current analogous quantities. Ans. Force analogous to current
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Q. 4. Draw the mathematical model of the following system and obtain the transfer function.
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Ans. Writting nodal equation At node x
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UNIT-2 Block diagrams and signal flow graphs Transfer function The ratio of laplace transform of the output to the laplace transform of input under the assumption of zero initial conditions is defined as the transfer function Of system. It is denoted by G(s).
Importance : Transfer function is highly important because of following reasons 1. It is used to give the gain of given blocksystem. 2. The system poles/zeros can be found from transfer function. 3. Stability can be determined from characteristic equation. 4. The system differential equation can be obtained from transfer function by replacing. s-variable with linear differential operator Significance of Transfer Function
Where
C(s) is laplace transform of output R(s) is laplace transform of input. Transfer function gives the gain of the given block system. Properties of Transfer Function 1. The transfer function is independent of the inputs to the system. 2. The transfer function of a system is the laplace transform of its impulse response for zero initial conditions. 3. The system poles/zeros can be found out from transfer function. 4. The transfer function is defined only for linear invariant systems. It is not defined for non-linear systems. Limitations of transfer function are listed below 1 Transfer function is valid only for linear time invariant system. 2 It does not take into account the initial conditions initial conditions loose its significance. 3 It does not give any idea about how the present output is progressing.
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Poles are the value of ‗s‘ which when substituted in the denominator of a transfer function, make the transfer function value as infinity Zeros are the value of ‗s‘ which when substituted in the numerator of transfer function, make the transfer function value as zero The characteristic equation can be obtained by equating the denominator polynomial of the transfer function to zero . The highest power of ‗s‘ in the characteristic equation is called the order o system The number of poles at the origin defines the type of system Block diagram algebra Block diagram gives a pictorial representation of a control system by way of short handing the transfer function Signal flow graph further shortens the representation of a control system by eliminating summing symbol take-off point and block This elimination is achieved by way of representing the variables by points called ―nodes‖ A pictorial representation of the relationship between input and output of a system is termed as block diagram. The direction of flow of signal from one block to other is indicated by an arrow. The point in a block diagram at which signal can be added or subtracted is termed as summing point. Gain is the ratio of laplace transform of output to laplace transform of input . Blocks in series are algebraically combined by multiplication. The lines drawn between the blocks to indicate the connections between the blocks are termed as branches. The point from which a signal is taken for the feedback purpose is called as take-off point. The order of summing point can be changed if two or more summing points are in series. Signal Flow Graphs Department of EEE, SJBIT
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A signal flow grow is a pictorial representation of a system and it displays graphically, the transmission of signal in system Node: It is a point from where branches originate or terminate or pass through. Branch : It is connecting link between two nodes. Path : The time traced by connecting two or more node is called path. Loop : It is a path that originates and terminates on same node and along which node other node is traversed more than once.
Mason s gain formula is used to find the gain of signal flow graph According to Mason‘s gain formula
where
i = Number of forward path = Gain of ith forward path = System determinant
= 1 — (sum of all individual loops) + (sum of all gain products of two nontouching loops) - (sum of all gain product of three non-touching loops) +…. The gain associated with each branch is called branch transmittance The independent and dependent variable of a control system are represented by small circles as nodes. The relationship between nodes is represented by drawing a line between two nodes Such l‘ns are called branches. Example problems:
Q. 1. Find out the 2010
for the system shown in the following block diagram.
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Ans. First draw the signal flow graph.
Step I Obtain total number of forward paths There is only one forward path
Step II. Total number of single loop There are two loops. Thus
Step III. Value of
As there is one forward path which touch all the loops Step IV. Obtain transfer function
Q. 2. The transfer function of a system is Calculate the phase shift at
.
Ans.
Hence here is phase shift of zero corresponding to Department of EEE, SJBIT
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Q. 3. Laplace transform of a function f (t) is given by
Find out the initial and final values of f (t). Ans.
Applying final value theorem
Q. 4. Find out the inverse Laplace transform of the function
.
Ans.
The term
can be factorized as, (s + 2) (s + 3)
Using partial fraction expansion
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Q. 5. Represent the following set of equations by a signal 2011 flow graph and determine the overall gain relating
10ES43
June/July .
Ans. Given equations are :
required signal flow graph is
Step I. Obtain the total number of forward paths
Step II. Obtain the number of single loops
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Step lll. Obtain the number of two non-touching loop
Step IV. Number of three non-touching loops --no— Step V. Find the value of
Applying Mason‘s gain formula Overall transform function
Q. 6. Simplify the block diagram in fig and obtain the transfer 2010 function relating C(s) and R(s).
June/July
Ans.
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There are no loops.
Q. 21. From the block diagram shown in the figure below draw the corresponding signal flow graph and evaluate close loop transfer function relating Dec/Jan 2006 the output and input.
Ans. Required signal flow graph is:
By Mason‘s gain formula:
Individual loop, Non-touching loops = 0
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Q. 7. The transfer function of a system is given by
Determine the state model in canonical form using parallel decomposition method. Ans.
Using partial fractions
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State diagram representation in diagonal canonical form by parallel decomposition
Q. 8. Determine the transfer function of the system given in fig.
Ans.
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UNIT 3 Time Response of feed back control systems
Time-domain Analysis of Control Systems In time-domain analysis the response of a dynamic system to an input is expressed as a function of time. It is possible to compute the time response of a system if the nature of input and the mathematical model of the system are known. Usually, the input signals to control systems are not known fully ahead of time. In a radar tracking system, the position and the speed of the target to be tracked may vary in a random fashion. It is therefore difficult to express the actual input signals mathematically by simple equations. The characteristics of actual input signals are a sudden shock, a sudden change, a constant velocity, and constant acceleration. The dynamic behavior of a system is therefore judged and compared under application of standard test signals – an impulse, a step, a constant velocity, and constant acceleration. Another standard signal of great importance is a sinusoidal signal. The time response of any system has two components: transient response and the steadystate response. Transient response is dependent upon the system poles only and not on the type of input. It is therefore sufficient to analyze the transient response using a step input. The steady-state response depends on system dynamics and the input quantity. It is then examined using different test signals by final value theorem. Standard test signals a) Step signal:
r (t ) Au(t ).
b) Ramp signal:
r (t ) At; t 0.
c) Parabolic signal: r (t ) At 2 / 2; t 0. d) Impulse signal: r (t ) (t ).
Time-response of first-order systems Let us consider the armature-controlled dc motor driving a load, such as a video tape. The objective is to drive the tape at constant speed. Note that it is an open-loop system.
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G( s)
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kk kk ak1km W ( s) a ak k 1 m ; If r (t ) au(t ) , W ( s) 1 m 1 m R( s ) m s 1 ms 1 s s s 1/ m
w(t ) ak1km ak1kmet / m ; wss (t ) lim w(t ) ak1km t
wss (t ) is the steady-state final speed. If the desired speed is wr , choosing a
wr the motor k1km
will eventually reach the desired speed. We are interested not only in final speed, but also in the speed of response. Here, m is the time constant of motor which is responsible for the speed of response. The time response is plotted in the Figure in next page. A plot of et / m is shown, from where it is seen that, for t 5 m the value of et / m is less than 1% of its original value. Therefore, the speed of the motor will reach and stay within 1% of its final speed at 5 time constants.
Figure: Time responses
Let us now consider the closed-loop system shown below.
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Here, T ( s)
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k1km / ( m s 1) k1km kk W ( s) 1 o R( s) 1 k1k2 km / ( m s 1) m s (1 k1k2 km ) o s 1 where, ko
km m and o . 1 k1k2 km 1 k1k2 km
If r (t ) a , the response would be, w(t ) ak1ko ak1ko e
t / o
.
If a is properly chosen, the tape can reach a desired speed. It will reach the desired speed in 5 o seconds. Here, o m . Thus, we can control the speed of response in feedback system. Although the time-constant is reduced by a factor (1 k1k2 km ) , in the feedback system, the motor gain constant is also reduced by the same factor. In order to compensate for this loss of gain, the applied reference voltage must be increased by the same factor. Ramp response of first-order system Let, k1k0 1 for simplicity. Then, T ( s)
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1 W ( s) . Also, let, r (t ) tu(t ) . o s 1 R( s )
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o2 1 1 o Then, W ( s) 2 ; s ( o s 1) s 2 s o s 1
w(t ) tu(t ) o (1 et / o )u(t ) The error signal is, e(t ) r (t ) w(t ) Or, e(t ) o (1 et / o )u(t )
ess (t ) o Thus, the first-order system will track the unit ramp input with a steady-state error o , which is equal to the time-constant of the system.
Time-response of second-order systems
Consider the antenna position control system. Its transfer function from r to y is,
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k1k2 km k1k2 km / m n2 Y ( s) R( s) m s 2 s k1k2 km s 2 1 s k k k / s 2 2n s n2 1 2 m m
m
where, we define,
n2 k1k2 km / m and
2n
1
m
. The constant is called the damping
ratio and n is called the natural frequency. The system above is in fact a standard second order system. The transfer function T ( s) has two poles and no zero. Its poles are, s1 , s2 n jn 1 2 jd .
Here, is called the damping factor, d is called damped or actual frequency. The location of poles for different are plotted in Figure below. For 0 , the two poles
jn are purely imaginary. If 0 1 , the two poles are complex conjugate. All possible cases are described in a table shown below.
Unit step response of second-order systems
n2 s 2n 1 1 1 2 Suppose, r (t ) u(t ), R( s) ; Y ( s) 2 2 s s 2n s n s s 2n s n2 s
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Or, Natural frequency, n The natural frequency of a second order system is the frequency of oscillation of the system without damping.
s 2n s n n 1 1 Y ( s) 2 2 2 s ( s n ) n (1 ) s ( s n )2 (n 1
Performing inverse Laplace transform,
Damping ratio, The damping ratio is defined as the ratio of the damping factor, to the natural frequency n .
y (t ) 1 ent cos(n 1 2 )t ent sin(n 1 2 )t
b . T (s) 2 s as b
Suppose,
nt
e 2 ) 1 standard equation, 1 2 cos(an 21 n and )t sin(n 1 2 )t or, y(twith Comparing 1 2
b. 2 n
y (t ) 1
or,
d n 1 2 and tan 1
ent 1
2
sin(d t ) ,where,
1 2 / cos 1
n t e sin(d t ) d ………………………………………………… …….(01) y (t ) 1
or,
The plot of e t sin(d t ) is shown in Figure. The steady-state response is, yss (t ) lim y(t ) 1 t
Thus, the system has zero steady-state error. The pole of T ( s) dictates the response, e t sin(d t ) . Department of EEE, SJBIT
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The response y (t ) for different is shown in Figure below.
Time response specifications Control systems are generally designed with damping less than one, i.e., oscillatory step response. Higher order control systems usually have a pair of complex conjugate poles with damping less than unity that dominate over the other poles. Therefore the time response of second- and higher-order control systems to a step input is generally of damped oscillatory nature as shown in Figure next (next page). In specifying the transient-response characteristics of a control system to a unit step input, we usually specify the following: 1. Delay time, t d 2. Rise time, tr 3. Peak time, t p 4. Peak overshoot, M p 5. Settling time, t s 6. Steady-state error, ess Department of EEE, SJBIT
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1. Delay time, t d : It is the time required for the response to reach 50% of the final value in first attempt. 2. Rise time, tr : It is the time required for the response to rise from 0 to 100% of the final value for the underdamped system. 3. Peak time, t p : It is the time required for the response to reach the peak of time response or the peak overshoot. 4. Settling time, t s : It is the time required for the response to reach and stay within a specified tolerance band ( 2% or 5%) of its final value. 5. Peak overshoot, M p : It is the normalized difference between the time response peak and the steady output and is defined as,
%M p
c(t p ) c() c ( )
100%
6. Steady-state error, ess : It indicates the error between the actual output and desired output as ‗t‘ tends to infinity. ess lim[r (t ) c(t )] . t
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Let us now obtain the expressions for the rise time, peak time, peak overshoot, and settling time for the second order system. 1.
Rise
time,
tr :
Put
t tr , sin(d tr ) 0 sin , tr
; cos1 . d
2.
Put
0
Peak
time,
tp :
dy 0 and dt
y(t ) 1
at
for
t tp ;
solve
n t e sin(d t ) n e t cos(d t ) . d tan(d t p )
1 2 d n 1 2 tan , d t p k n
k 0,1, 2,
Peak overshoot occurs at k = 1. t p / d / n 1 2 . 3. Settling time, t s : For 2% tolerance band,
n t 4 e 0.02 , ts 4T . d s
4. Steady-state error, ess : It is found previously that steady-state error for step input is zero. Let us now consider ramp input, r (t ) tu(t ) . Then, ess lim s{R( s) Y ( s)} lim s{ s 0
s 0
n2 1 1 } s 2 s 2 s 2 2n s n2
2 2 2 n2 1 1 s 2n s n n ess lim {1 2 } lim s 0 s s 0 s s 2n s n2 s 2 2n s n2
Therefore, the steady-state error due to ramp input is
2
n
2n 2 . 2 n n
.
Steady-state error and error constants The steady-state performance of a stable control system is generally judged by its steadystate error to step, ramp and parabolic inputs. For a unity feedback system,
E ( s)
R( s ) sR( s) , ess lim sE ( s) lim . s 0 s 0 1 G ( s ) 1 G( s)
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It is seen that steady-state error depends upon the input R( s) and the forward transfer function G( s) . The steady-state errors for different inputs are derived as follows: 1. For unit-step input: r (t ) u (t ), R( s)
ess lim sE ( s) lim s 0
s 0
1 s
1 1 1 ; k p is called position error constant. 1 G( s) 1 G(0) 1 k p
2. For unit-ramp input: r (t ) tu (t ), R( s)
ess lim sE ( s) lim s 0
s 0
1 s2
1 1 1 lim ; s 1 G( s) s 0 sG( s) kv
kv is
called
velocity
error
constant. 3. For unit-parabolic input: r (t ) t 2 / 2, R( s)
ess lim sE ( s) lim s 0
s 0
1 s3
1 1 1 lim 2 ; k a is called acceleration error s 0 s 1 G( s) s G ( s ) ka 2
const.
Types of Feedback Control System The open-loop transfer function of a system can be written as,
G( s)
K ( s z1 )( s z2 )( s z3 ) s n ( s p1 )( s p2 )( s p3 )
K (Tz1s 1)(Tz 2 s 1)(Tz 3 s 1) s n (Tp1s 1)(Tp 2 s 1)(Tp 3s 1)
If n = 0, the system is called type-0 system, if n = 1, the system is called type-1 system, if n = 2, the system is called type-2 system, etc. Steady-state errors for various inputs and system types are tabulated below.
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The error constants for non-unity feedback systems may be obtained by replacing G(s) by G(s)H(s). Systems of type higher than 2 are not employed due to two reasons: 1. The system is difficult to stabilize. 2. The dynamic errors for such systems tend to be larger than those types-0, -1 and -2.
Effect of Adding a Zero to a System Let a zero at s = -z be added to a second order system. Then we have,
( s z )n2 / z n2 n2 C (s) s 2 . 2 2 2 2 2 R( s) s 2n s n s 2n s n z s 2n s n The multiplication term is adjusted to make the steady-state gain of the system unity. This gives css = 1 when the input is unit step. Let cz(t) be the response of the system given by the above equation and c(t) is the response without adding the pole. Manipulation of the above equation gives, cz (t ) c(t )
1d c(t ). z dt
The effect of added derivative term is to produce a pronounced early peak to the system response which will be clear from the figure in the next page. Closer the zero to origin, the more pronounce the peaking phenomenon. Due to this fact, the zeros on the real axis near the origin are generally avoided in design. However, in a sluggish system the artful introduction of a zero at the proper position can improve the transient response. We can see from equation (03) that as z increases, i.e., the zero moves further into the left half of the splane, its effect becomes less pronounced.
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Design Specifications of Second-order Systems A control system is generally required to meet three time response specifications: steady-state accuracy, damping factor (or peak overshoot, Mp) and settling time ts. Steady-state accuracy requirement is met by suitable choice of Kp, Kv, or Ka depending on the type of the system. For most control systems in the range of 0.7 – 0.28 (or peak overshoot of 5 – 40%) is considered acceptable. For this range of , the closed-loop pole locations are restricted to the shaded region of the s-plane as shown in Figure. For
the
system, n k1k2 km / m ;
antenna 1 2n m
; ess
ramp
position 2
n
; ts
4
n
control
. Here, k 2 is only the adjustable
parameter. If we increase k 2 , n will increase and thus settling time will decrease. At the same time, will decrease, this indicates the increase in peak overshoot. Thus by merely increasing gain, we cannot improve both transient and steady-state error specifications. We need to add additional components to the system. These are called compensators. It will allow improvement of both transient and steady-state specifications.
Introduction to PID Controler
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The classical three-term PID controller proportional feedback control can reduce error responses but that it still allows a non-zero steady-state error for a proportional system. In addition, proportional feedback increases the speed of response but has a much larger transient overshoot. When the controller includes a term proportional to the integral of the error, then the steady-state error can be eliminated. But this comes at the expense of further deterioration in the dynamic response. Addition of a term proportional to the derivative of the error can damp the dynamic response. Combined, these three kinds of actions form the classical PID controller, which is widely used in industry. This principle mode of action of the PID controller can be explained by the parallel connection of the P, I and D elements shown in Figure 3.1 From this diagram the transfer function of the PID controller is
(3.1)
Figure 3.1: Block diagram of the PID controller The controller variables are Gain integral action time derivative action time
Eq. (3.1) can be rearranged to give
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These three variables , and are usually tuned within given ranges. Therefore, they are often called the tuning parameters of the controller. By proper choice of these tuning parameters a controller can be adapted for a specific plant to obtain a good behaviour of the controlled system. If follows from Eq. (3.2) that the time response of the controller output is
(3.3)
Using this relationship for a step input of , i.e. , the step response of the PID controller can be easily determined. The result is shown in Figure 3.2a. One has to observe that the length of the arrow the impulse.
of the D action is only a measure of the weight of
Figure 3.2: Step responses (a) of the ideal and (b) of the real PID controller In the previous considerations it has been assumed that a D behaviour can be realised by the PID controller. This is an ideal assumption and in reality the ideal D element cannot be realised . In real PID controllers a lag is included in the D behaviour. Instead of a D element in the block diagram of Figure 3.1 a
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element with the transfer function
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is introduced. From this the transfer function of the real PID controller or more precisely of the
controller follows as (3.5)
Introducing the controller tuning parameters and
it follows (3.6)
The step response of the controller is shown in Figure 3.2b. This response from gives a large rise, which declines fast to a value close to the P action, and then migrates into the slower I action. The P, I and D behaviour can be tuned independently. In commercial controllers the 'D step' at can often be tuned 5 to 25 times larger than the 'P step'. A strongly weighted D action may cause the actuator to reach its maximum value, i.e. it reaches its 'limits'. As special cases of PID controllers one obtains for: a) the PI controller with transfer function (3.7)
b) the ideal PD controller with the transfer function Department of EEE, SJBIT
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and the
controller with the transfer function (3.9)
c) and
the P controller with the transfer function (3.10)
The step responses of these types of controllers are compiled in Figure3.3. A pure I controller may also be applied and this has the transfer function (8.11)
Figure 3.3: Step responses of the PID controller family
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Advantages and disadvantages of the different types of controllers
In addition, the different cases should be compared with respect to the normalised maximum overshoot
.
The different cases are discussed below: a) The P controller shows a relatively high maximum overshoot settling time
as well as a steady-state error
, a long
.
b) The I controller has a higher maximum overshoot than the P controller due to the slowly starting I behaviour, but no steady-state error. c) The PI controller fuses the properties of the P and I controllers. It shows a maximum overshoot and settling time similar to the P controller but no steady-state error. d) The real PD controller according to Eq. (3.9) with has a smaller maximum overshoot due to the 'faster' D action compared with the controller types mentioned under a) to c). Also in this case a steady-state error is visible, which is smaller than in the case of the P controller. This is because the PD controller generally is tuned to have a larger gain due to the positive phase shift of the D action. For the results shown in Figure 3.5 the gain for the P controller is and for the PD controller
. The plant has a gain of
.
e) The PID controller according to Eq. (3.6) with fuses the properties of a PI and PD controller. It shows a smaller maximum overshoot than the PD controller and has no steady state error due to the I action. The qualitative concepts of this example are also relevant to other type of plants with delayed proportional behaviour. This discussion has given some first insights into the static and dynamic behaviour of control loops.
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Figure 3.5: Behaviour of the normalised controlled variable at the input to the plant
;
for step disturbance for different types of
controllers Example problems: Q. 1. Consider the system having transfer function
Calculate the settling time for 2% tolerance band, for the unit step response. Ans.
Comparing with
we get
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setting time for 2% tolerance band, we get setting time
Q. 2. Calculate the natural frequency of a second order system described by the differential equation
Ans. The given equation is
…(1) Now standard equation of a second order system is given as …(2) so comparing eq (1) and (2), we get
Q.3. For a second order system the roots of its characteristic equation are underdamped natural frequency of the system will be. (a) 4 rad/sec (b) 3 rad/sec (c) 5 rad/sec (d) 7 rad/sec
the
Ans.
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Q. 4. What will be the response of a first order system with unit step input?
Ans. The transfer function of a first order system is given as
With unit step input
The response is shown graphically in fig.
Q.5. What is the unit step response of the transfer function?
Ans. We have
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Q.6. Measurements conducts on a servomechanism show that response to be when subjected to a unit step input. Determine the undamped natural frequency and damping ratio of the system.
Ans.
Unit step input means
Comparing, we get
Natural frequency
Q. 7. The open-loop transfer function of a unity feedback control system is given by
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Determine the value of K and B such that the closed loop until step response has w = 3 rad/sec and Ans. The characteristics equation of the system is 1 + G (s) = 0
since, b = 1.2, and
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UNIT 4 Stability Analysis Concepts of stability Because of its feedback structure a control system can become unstable, e.g. oscillations with increasing amplitudes in the signals can occur . A linear time-invariant system is called ( asymptotically) stable, if its weighting function decays to zero, i.e. if (4.1)
is valid. If the modulus of the weighting function increases with increasing to infinity, the system is called unstable. A special case is a system where the modulus of the weighting function does not exceed a finite value as or for which it approaches a finite value. Such systems are called critically stable. Examples are undamped
S and I elements
This definition shows that stability is a system property for linear systems. If Eq. (4.1) is valid, then there exists no initial condition and no bounded input signal which drives the output to infinity. This definition can be directly applied to the stability analysis of linear systems by determining the value of the weighting function for . If this value exists, and if it is zero, the system is stable. However, in most cases the weighting function is not given in an explicit analytic form and therefore it is costly to determine the final value. The transfer function
of a system is often known and as it is the Laplace transform of the
weighting function , there is an equivalent stability condition for according to Eq. (4.1). The analysis of this condition shows that for the stability analysis it is sufficient to check the poles of the transfer function characteristic equation
of the system, that is the roots
of its
(4.2)
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Necessary conditions for Stability The following necessary and sufficient stability conditions can be formulated: a) Asymptotic stability A linear system is only asymptotically stable, if for the roots equation
of its characteristic
for all is valid, or in other words, if all poles of its transfer function lie in the left-half plane. b) Instability A linear system is only unstable, if at least one pole of its transfer function lies in the right-half plane, or, if at least one multiple pole (multiplicity imaginary axis of the plane.
) is on the
c) Critical stability A linear system is critically stable, if at least one single pole exists on the imaginary axis, no pole of the transfer function lies in the right-half plane, and in addition no multiple poles lie on the imaginary axis. It has been shown above that the stability of linear systems can be assessed by the distribution of the roots of the characteristic equation in the plane (Figure 5.2). For control problems there is often no need know these root with high precision. For a stability analysis it is interesting to know whether all roots of the characteristic equation lie in the left-half plane or not. Therefore simple criteria are available for easily checking stability, called stability criteria. These are partly in algebraic, partly in graphical form.
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Figure 4.2: Stability of a linear system discussed by the distribution of the roots of the characteristic equation in the plane
Routh criterion For given coefficients of the characteristic equation the method of Routh, which is an alternative to the method of Hurwitz, can be applied,. Here the coefficients will be arranged in the first two rows of the Routh schema, which contains rows:
The coefficients first two rows according to
in the third row are the results from cross multiplication the
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Building the cross products one starts with the elements of the first row. The calculation of these values will be continued until all remaining elements become zero. The calculation of the values are performed accordingly from the two rows above as follows:
From these new rows further rows will be built in the same way, where for the last two rows finally
and
follows. Now the Routh criterion is: A polynomial
is Hurwitzian, if and only if the following three conditions are valid:
a)
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b) all coefficients
in the first column of the Routh schema are positive.
Example 1
The Routh schema is:
For proving instability it is sufficient to build the Routh schema only until negative or zero value occurs in the first column. In the example given above the schema could have been stopped at the 5th row. Another interesting property of the Routh scheme says, that the number of roots with positive real parts is equal to the number of changes of sign of the values in the first column. Example 2 Determine the stability of the system whose characteristics equation given by a(s) s 6 4s 5 3s 4 2s 3 s 2 4s 4. The above polynomial satisfies the necessary condition for stability since all the coefficients are positive and nonzero. Writing the Routh array, we have
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s6 s5
1 4
s4
5
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3 2
4.3 1.2
2
1 4
0
4 .2 4.0 2 2 5
2
5
s2
3
s1
12 3( ) 4.2 76 5 3
2
s0
0
2 5
5 12 2.0 ( )
4.4 1.0 4
.4 4.4 12 2 5 5
2
15
4
4 5
5
s3
4.1 4.1
4 0
2.4 .0 2 4 2
0
76
.4 0 4 15 76 15
We conclude that the system has roots in the right half plane, since the elements of the first column are not all positive. In fact there are two roots in the right half plane, since there are two sign changes. In other words two closed loop poles of the system lie in the right half plane and hence the system is unstable.
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Example 3. Determine the stability of the following polynomial. a(s) s 5 5s 4 11s 3 23s 2 28s 12. Writing the Routh array, we have s5 1 11 s4
28
5
23
12
6.4
25.6
0
3
12
0
0
s3 s
2
s1
Since the entire row is zero, we construct an auxiliary equation by taking the coefficients of the previous row, i.e., a1 (s) 3s 2 12. Differentiating the above equation with respect to ‗s‘, we get da1 (s)
6s.
ds
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(1)
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So the Routh array is continued by taking the coefficients of equation (1).
s1 6 0 0 s 12 Since there are no sign changes in the first column of the Routh array, there are no roots in the right half plane. However, since one entire row in the Routh array was zero, there are roots in the imaginary axis. The roots in the imaginary axis can be obtained by solving the auxiliary equation. Therefore, 3s 2 12 0, s 2 4 0, s j2 Example 4. Consider the system shown below. The stability properties of the system are a function of the proportional feedback gain ‗k‘. Determine the range of ‗k‘ over which the system is asymptotically stable.
r
+
k
s 1
y
s(s 1)(s 6)
-
The characteristics equation for the system is given by
1 k
s 1 s(s 1)(s 6)
0,
s 3 5s 2 (k 6)s k 0. Department of EEE, SJBIT
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Therefore the corresponding Routh array is s3 s2 s1
1 5
k-6 k
(4k 30) 5
s0
k
For the system to be stable, it is necessary that all the elements in the first column of the Routh array must be positive. Therefore, 4k 30
0
5 k 7.5
and and
k 0, k 0,
k 7.5
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The state-space representation of a system is given as .
7 12 1 x x u, 0 0 1 y 1 2x.
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For the system to be stable, the poles of the system should lie in the left half plane. In other words all the real poles should be negative or the real parts of complex poles must be negative. The poles of the system are nothing but the eigenvalues of the ‗A‘ matrix of the system. The MATLAB code is shown below a = [-7 -12; 1 0]; [v,d] = eig(a)
The result is v = -0.9701 0.2425
0.9487 -0.3162
d = -4 0
0 -3
The diagonal elements of the matrix ‗d‘ are eigenvalues of the system and columns of the matrix ‗v‘ represent the corresponding eigenvectors. Example problem: Q. 1. Calculate the value of k for which the unity feedback system
crosses the imaginary axis.
Ans.
For the point of intersection on imaginary axis, construct Routh array
To cross imaginary axis Department of EEE, SJBIT
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so at K = 48, it will cross imaginary axis. Q. 2. A closed loop system is shown in the following fig. Find out the largest possible value of
for which this system would be stable.
Ans. As we know
Now
or characteristic equation becomes Now the Routh array can be written as
For stability all elements of column should be positive So,
The largest possible value for the system to be, stable. Q. 3. By means of Routh criterion, determine the stability of the system represented by the following equation Ans. This equations states
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There is one row that becomes zero. So it is being replaced by
equation can be written as:
Imaginary roots make system unstable and response is continuous oscillatory. Therefore system is unstable. Q. 4. Consider the closed loop feedback system shown in the figure below. Determine the range of K for which the system is stable.
Ans.
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This is the range of K for the system to be stable.
Q. 5. Discuss the Routh — Hurwitz criteria for determining the stability of a control system and calculate the range of K for stable operation of following characteristic equation
Ans. Routh-Hurwitz criterion helps in determining relative stability of a control system. From the characteristic equation of control system Routh‘s array is constructed. In case there is no change of sign in first column and the system is stable. Given characteristic equation is
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Q. 6. Discuss the Routh-Hurwitz criteria for determining the stability of a control system and calculate the range of K for stable operation of following characteristic equation. Ans. Routh-Hurwitz criterion helps in determining relative stability of a control system. From the characteristic equation of control system Routh‘s array is constructed. In case there is no change of sign in first column and the system is stable. Given characteristic equation is
Q. 7. Determine the stability of system having characteristic equation: using Routh Hurwitz criterion.
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Ans. Given characteristic equation is
Since row becomes zero, we formula auxiliary equation Differentiate auxiliary equation
Since the last element is negative, there is a sign change in first column, the system is unstable.
Q. 8. Determine the stability of a system with characteristic equation
Ans. Given characteristic equation is
Since terms corresponding to S3 are zero, we will form -auxiliary equation corresponding to
differentiating equation we get
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Consider auxiliary equation
Differentiating 2s = 0
As there is no sign change system may be stable. Solving auxiliary equation
Thus system is unstable. Q. 9. Find the range of K for stability of Ans. Given characteristic equation is Routh‘s array for given equation is
- For system to be stable all elements in first column should be greater than zero.
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so, value of K for stability will be 0 < K < 0.44.
Q. 10. Determine the stability of the system whose characteristic equation is given by
Ans. Routh‘s array is
As their is sign change in first column thus system is unstable.
Q. 11. A unity feedback control system is characterized by open loop transfer function Using Routh’s criterion, calculate the range of values of K for the system to be stable. Ans. The characteristic equation is 1 + G(s). H(s), H(s)=1 1 + G (s)
The Routh‘s array is
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For stable system. there should be no sign change in first column. For s° K + 1 > 0 K > —1
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UNIT 5 Root-locus techniques Introduction A designer can determine whether his design for a control system meets the specifications if he knows the desired time response of the controlled variable. By deriving the differential equations for the control system and solving them, an accurate solution of the system's performance can be obtained, but this approach is not feasible for other than simple systems. It is not easy to determine from this solution just what parameters in the system should be changed to improve the response. A designer wishes to be able to predict the performance by an analysis that does not require the actual solution of the differential equations. The first thing that a designer wants to know about a given system is whether or not it is stable. This can be determined by examining the roots obtained from the characteristic equation (5.1)
of the closed loop. The work involved in determining the roots of this equation can be avoided by applying the Hurwitz or Routh criterion. Determining in this way whether the system is stable or unstable does not satisfy the designer, because it does not indicate the degree of stability of the system, i.e., the amount of overshoot and the settling time of the controlled variable for a step input. Not only must the system be stable, but the overshoot must be maintained within prescribed bounds and transients must die out in a sufficiently short time. The root-locus method described in this section not only indicates whether a system is stable or unstable but, for a stable system, also shows the degree of stability. The root locus is a plot of the roots of the characteristic equation of the closed loop as a function of the gain. This graphical approach yields a clear indication of the effect of gain adjustment with relatively small effort. With this method one determines the closed-loop poles in the plane - these are the roots of Eq.(5.1) - by using the known distribution of the poles and zeros of the open-loop transfer function . If for instance a parameter is varied, the roots of the characteristic equation will move on certain curves in the plane as shown by the example in Figure5.1. On these curves lie all
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Figure 5.1: Plot of all roots of the characteristic equation
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for
.
Values of are red and underlined. possible roots of the characteristic equation for all values of the varied parameter from zero to infinity. These curves are defined as the root-locus plot of the closed loop. Once this plot is obtained, the roots that best fit the system performance specifications can be selected. Corresponding to the selected roots there is a required value of the parameter which can be determined from the plot. When the roots have been selected, the time response can be obtained. Since the process of finding the root locus by calculating the roots for various values of a parameter becomes tedious, a simpler method of obtaining the root locus is desired. The graphical method for determining the root-locus plot is shown in the following. An open-loop transfer function with poles at the origin of the plane is often described by (5.2)
where is the gain of the open loop. In order to represent this transfer function in terms of the open-loop poles and zeros it is rewritten as
(5.3)
or
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(5.4)
with
and
. The relationship between the factor
and the open-loop gain
is
(5.5)
The characteristic equation of the closed loop using Eq. (5.3) is (5.6)
or (5.7)
All complex numbers root locus.
, which fulfil this condition for
, represent the
From the above it can be concluded that the magnitude of must always be unity and its phase angle must be an odd multiple of . Consequently, the following two conditions are formalised for the root locus for all positive values of
from zero to infinity:
a) Magnitude condition: (5.8)
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b) Angle condition (5.9) for
In a similar manner, the conditions for negative values of ( determined. The magnitude conditions is the same, but the angle must satisfy the c) Angle condition
) can be
(5.10) for
Apparently the angle condition is independent of
. All points of the plane that fulfil the
angle condition are the loci of the poles of the closed loop by varying
. The calibration of
the curves by the values of is obtained by the magnitude condition according to Eq. (5.8). Based upon this interpretation of the conditions the root locus can constructed in a graphical/numerical way. Once the open-loop transfer function has been determined and put into the proper form, the poles and zeros of this function are plotted in the plane.
The plot of the locus of the closed loop poles as a function of the open loop gain K, when K is varied from 0 to +00. When system gain K is varied from 0 to +oo, the locus is called direct root locus. When system gain K is varied from -oo to 0, the locus is called as inverse root locus. The root locus is always symmetrical about the real axis i.e. x-axis. The number of separate branches of the root locus equals either the number of open loop poles are number of open-loop zeros whichever is greater. A section of root locus lies on the real axis if the total number of open-loop poles and zeros to the right of the section is odd. If the root locus intersects the imaginary axis then the point of intersection are conjugate. From the open loop complex pole the root locus departs making an angle with the horizontal line. The root locus starts from open-loop poles. The root locus terminates either on open loop zero or infinity. The number of branches of roots locus are:
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and M if P
Centroid is the centre of asymptotes. It is given by (an)
Angle of asymptotes is denoted by ‗p
Angle of departure is. tangent to root locus at complex pole
Based on the pole and zero distributions of an open-loop system the stability of the closedloop system can be discussed as a function of one scalar parameter. The root-locus method shown in this module is a technique that can be used as a tool to design control systems. The basic ideas and its relevancy to control system design are introduced and illustrated. Ten general rules for constructing root loci for positive and negative gain are shortly presented such that they can be easily applied. This is demonstrated by some discussed examples, by a table with sixteen examples and by a comprehensive design of a closed-loop system of higher order.
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Example Problems: Q.1. Consider the example
with
,
are the roots
and
and
. The poles of the closed-loop transfer function
of the characteristic equation
and are given by
As
and
it can be seen that for
the poles of the closed loop
transfer function are identical with those of the open-loop transfer function values a)
. For other
the following two cases are considered: :
Both roots
and
and
are real and lie on the real axis in the range of
;
b) :
The roots
and
are conjugate complex with the real part
which does not depend on , and the imaginary part Im The curve has two branches as shown in Figure 6.2.
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, .
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Figure 5.2: Root locus of a simple second-order system At is the breakaway point of the two branches. Checking the angle condition the condition
must be valid. The complex numbers
and
have the angles
and
and the
magnitudes and . The triangle ( ) in Figure 6.2 yields the angle condition. Evaluating the magnitude condition according to Eq. (6.8)
one obtains the value
The value of
on the root locus. E.g. for
at the breakaway point
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the gain of the open loop is
is
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Table 5.1 shows further examples of some 1st- and 2nd-order systems. Table 5.1: Root loci of 1st- and 2nd-order systems root locus
root locus
General rules for constructing root loci To facilitate the application of the root-locus method for systems of higher order than 2nd, rules can be established. These rules are based upon the interpretation of the angle condition and the analysis of the characteristic equation. The rules presented aid in obtaining the root locus by expediting the manual plotting of the locus. But for automatic plotting using a computer these rules provide checkpoints to ensure that the solution is correct. Though the angle and magnitude conditions can also be applied to systems having dead time, in the following we restrict to the case of the open-loop rational transfer functions according to Eq. (5.3)
(5.11)
or (5.12)
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As this transfer function can be written in terms of poles and zeros )
and
(
;
can be represented by their magnitudes and angles
or
(5.13)
From Eq. (5.8) the magnitude condition
(5.14)
and from Eq. (5.9) the angle condition (5.15) for
follows. Here and denote the angles of the complex values and , respectively. All angles are considered positive, measured in the counterclockwise sense. If for each point the sum of these angles in the plane is calculated, just those particular points that fulfil the condition in Eq. (5.15) are points on the root locus. This principle of constructing a root-locus curve - as shown in Figure 5.3 - is mostly used for automatic rootlocus plotting.
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Figure 5.3: Pole-zero diagram for construction of the root locus In the following the most important rules for the construction of root loci for
are listed:
Rule 1 Symmetry As all roots are either real or complex conjugate pairs so that the root locus is symmetrical to the real axis. Rule 2 Number of branches The number of branches of the root locus is equal to the number of poles of the open-loop transfer function. Rule 3 Locus start and end points The locus starting points (
) are at the open-loop poles and the locus ending
points ( ) are at the open-loop zeros. branches end at infinity. The number of starting branches from a pole and ending branches at a zero is equal to the multiplicity of the poles and zeros, respectively. A point at infinity is considered as an equivalent zero of multiplicity equal to . Rule 4 Real axis locus If the total number of poles and zeros to the right of a point on the real axis is odd, this point lies on the locus. Rule 5 Asymptotes There are
asymptotes of the root locus with a slope of (5.16)
For Figure 5.4.
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and 4 one obtains the asymptote configurations as shown in
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Figure 5.4: Asymptote configurations of the root locus
Rule 6 Real axis intercept of the asymptotes The real axis crossing
of the asymptotes is at (5.17)
Rule 7 Breakaway and break-in points on the real axis At least one breakaway or break-in point exists if a branch of the root locus is on the real axis between two poles or zeros, respectively. Conditions to find such real points are based on the fact that they represent multiple real roots. In addition to the characteristic equation (6.1) for multiple roots the condition (5.18)
must be fulfilled, which is equivalent to (5.19)
for . If there are no poles or zeros, the corresponding sum is zero. Rule 8 Complex pole/zero angle of departure/entry The angle of departure of pairs of poles with multiplicity
is (6.20)
and the angle of entry of the pairs of zeros with multiplicity Department of EEE, SJBIT
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(5.21)
Rule 9 Root-locus calibration The labels of the values of
can be determined by using (5.22)
For the denominator is equal to one. Rule 10 Asymptotic stability The closed loop system is asymptotically stable for all values of for which the locus lies in the left-half plane. From the imaginary-axis crossing points the critical values
can be determined.
The rules shown above are for positive values of
. According to the angle condition of
Eq. (5.10) for negative values of some rules have to be modified. In the following these rules are numbered as above but labelled by a *. Rule 3* Locus start and end points The locus starting points (
) are at the open-loop poles and the locus ending
points ( ) are at the open-loop zeros. branches end at infinity. The number of starting branches from a pole and ending branches at a zero is equal to the multiplicity of the poles and zeros, respectively. A point at infinity is considered as an equivalent zero of multiplicity equal to . Rule 4* Real axis locus If the total number of poles and zeros to the right of a point on the real axis is even including zero, this point lies on the locus. Rule 5* Asymptotes There are
asymptotes of the root locus with a slope of (5.23)
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Rule 8* Complex pole/zero angle of departure/entry The angle of departure of pairs of poles with multiplicity
is (5.24)
and the angle of entry of the pairs of zeros with multiplicity (5.25)
The root-locus method can also be applied for other cases than varying
. This is possible as
long as can be rewritten such that the angle condition according to Eq. (5.15) and the rules given above can be applied. This will be demonstrated in the following two examples. Q.2. Given the closed-loop characteristic equation
the root locus for varying the parameter rewritten as
is required. The characteristic equation is therefore
This form then correspondents to the standard form
to which the rules can be applied. Q.3.Given the closed-loop characteristic equation
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Using the rules 1 to 10 one can easily predict the geometrical form of the root locus based on the distribution of the open-loop poles and zeros. Table 6.2 shows some typical distributions of open-loop poles and zeros and their root loci. Table 6.2: Typical distributions of open-loop poles and zeros and the root loci
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For the qualitative assessment of the root locus one can use a physical analogy. If all openloop poles are substituted by a negative electrical charge and all zeros by a commensurate positive one and if a massless negative charged particle is put onto a point of the root locus, a movement is observed. The path that the particle takes because of the interplay between the repulsion of the poles and the attraction of the zeros lies just on the root locus. Comparing the root locus examples 3 and 9 of Table 5.2 the 'repulsive' effect of the additional pole can be clearly seen. The systematic application of the rules from section 5.2 for the construction of a root locus is shown in the following non-trivial example for the open-loop transfer function
(5.26)
The degree of the numerator polynomial is one zero (
. This means that the transfer function has
). The degree of the denominator polynomial is
and we have the
four poles ( , , , 2). First the poles (x) and the zeros (o) of the open loop are drawn on the plane as shown in Figure 5.5. According to rule 3 these poles are just
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Figure 5.5: Root locus of underlined. those points of the root locus where
. Values of
and the zeros where
are in red and
. We have
branches that go to infinity and the asymptotes of these three branches are lines which intercept the real axis according to rule 6. From Eq. (5.17) the crossing is at (5.27)
and the slopes of the asymptotes are according to Eq. (5.16) (5.28)
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i.e.
The asymptotes are shown in Figure 5.5 as blue lines. Using Rule 4 it can be checked which points on the real axis are points on the root locus. The points
with
and
belong to the root locus, because to the right of them the number of poles and zeros is odd. According to rule 7 breakaway and break-in points can only occur pairwise on the real axis to the left of -2. These points are real solutions of the Eq. (5.19). Here we have (5.29)
or
This equation has the solutions roots
and
,
and
. The real
are the positions of the breakaway and the break-in point.
The angle of departure of the root locus from the complex pole at determined from Figure 5.6 according to Eq. (5.20):
can be (5.30)
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With this specifications the root locus can be sketched. Using rule 9 the value of can be determined for some selected points. The value at the intersection with the imaginary axis is
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UNIT 6 Frequency Domain Analysis Bode plot If
the
absolute
value
and
the
phase
are separately plotted over the frequency
of
the
frequency
response
, one obtains the
Figure 6.1: Plot of a frequency response: (a) linear, (b) logarithmic presentation ( logarithmic scale) (Bode plot)
on a
amplitude response and the phase response. Both together are the frequency response characteristics.
and
are normally drawn with a logarithm and
This representation is called a Bode diagram or Bode plot. Usually decibels [dB] By definition this is
The logarithmic representation of the amplitude response scale in this diagram and is called the magnitude.
with a linear scale. will be specified in
has consequently a linear
The logarithmic representation has some advantages for series connections of transfer functions. For complicated frequency responses, e.g. with
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for , it can be represented as series connections of the frequency responses of simple elements of the form
and
for
From this it follows that
with for
From the representation
and
respectively, one obtains the logarithmic characteristic of the magnitude
and the phase characteristic Department of EEE, SJBIT
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with for and for . Thus, the frequency response of a series connection is obtained by addition of the individual frequency response characteristics. A further advantage of this logarithmic representation is for the determination of the inverse of a frequency response, that is for
. Here
and
are valid, the curves of line) and
and
need only to be mirrored at the axes
(0-
.
Because of the double logarithmic and of the single logarithmic scale of
and
,
respectively, the curve of and that of can be approximated by line segments. This approximation by lines allows the analysis and synthesis of control systems using simple geometric constructions. They are important concepts for the control engineer. Improtant points:
The magnitude and phase relationship between sinusoidal input and steady state output of a system is known as frequency response. The polar plot of a sinusoidal transfer function G (jw) is plot of the magnitude of G (jw) versus the phase angle of G (jw) on polar coordinates as ‗co‘ varied from zero to infinity. The phase margin is that amount, of additional phase lag at the gain crossover frequency required to bring the system to the verge of instability. The gain margin is the reciprocal of the magnitude l G(jw) l at the frequency at which the phase angle as _1800. The inverse polar plot at G (jw) is a graph of 1/G (jw) as a function of w. Bode plot is a graphical representation of the transfer function for determining the stability of control system. Bode plot is a combination of two plot - magnitude plot and phase plot.
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The transfer function having no poles and zeros in the right -half s-plane are called minimum phase transfer function. System with minimum phase transfer function are called minimum phase systems. The transfer function having poles and zeros in the right half s-plane are called nonminimum phase transfer functions systems with non-minimum phase transfer function. are called non-minimum phase system. In bode plot the relative stability of the system is determined from the gain margin and phase margin. . If gain cross frequency is less than phase cross over frequency then gain margin and phase margin both are positive and system is stable. If gain cross over frequency is greater than the phase crossover frequency than both gain margin and‘phase margin are negative. It gain cross over frequency is equal to me phase cross over trequency me gain marg and phase margin are zero and system is marginally stable. The maximum value of magnitude is known as resonant peak. The magnitude of resonant peak gives the information about the relative stability of the system. The frequency at which magnitude has maximum value is known as resonant frequency. Bandwidth is defined a the range of frequencies in which the magnitude of closed loop does not drop —3 db.
Example Problems:
with
is rearranged into the form
This system can now be decomposed into an integrator, two PD and two is
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elements, that
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From this simple analysis the Bode plot can be determined by adding the Bode plots of the elements
to
. In this figure the variable quantities
Figure: Representation of a dynamic system by two frequency response diagrams: (a) Bode plot, (b) Nyquist plot
in the terms for and are the breakpoint frequencies in the Bode plot. Also shown in Figure 4.20 is the Nyquist plot of the frequency response. Both representations of Figure 4.20 basically contain the same information about the system.
Based on the example given above the procedure for constructing a Bode plot of a given system can be recapitulated:
a) The given transfer function must be put into the form
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with where possible poles of according to their multiplicity . b) Then for
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at
will be specially considered
the asymptotes of the elements will be used to approximate
and
. c) If necessary corrections of the approximations can be performed.
Introduction to lead, lag and lead-lag compensation networks
Compensation technique is used to make the unstable system stable by introducing the poles and or zeros at suitable place. In control system design, if the designed specifications do not satisfy the requirements of the system or leads to expensive and conflicting demands, then it is required to insert an additiohal component within the structure of feedback system. This adjustment is called compensation. Compensators are of three types: (a) Lag phase compensator(b) Lead phase compensator (c) Lead-lag compensator. Depending on the location the compensation is divided in following types: (a) Series or cascade compensation (b) Parallel or feedback compensation (c) Load compensation. Load compensations is provided to damp out oscillations in the system having mechanical output. Primary function of the lead compensator is to reshape the frequency response curve to provide sufficient phase lead angle to offset the excessive phase lag associated with the components of fixed systems.
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The primary function of lag compensator is to provide attenuation in the high frequency range to give a system sufficient phase margin.
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Phase lag network is shown in figure below. The primary function of lag compensator is to provide attenuation in high frequency range to give a system sufficient phase margin. Various design steps of phase lag network are
Example Problems: Q1. The asymptotic magnitude Bode plot of a system is given in the figure below Find the transfer function of the system analytically It is known that the system is minimum phase system.
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Q 2. Sketöh the Bode Plot for the transfer function given by, and from Plot find (a) Phase and Gain cross rer frequencies (b) Gain Margin and Phase Margin. Is this System Stable?
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The gain crosses 0db axis at co = 1.24 rad/sec, the gain crossover frequency is co = 1.24 rad/sec. The phase crosses —180° line at co = 0.4 rad/sec, therefore phase crossover frequency is co = 0.4 rad/sec. At phase cross over the gain is 20 dB, therefore gain margin is —20 dB. At gain crossover the phase angle is 2150, the phase margin is 180° + (—215°) = —35°. As both gain and phase margins are negative, the system is unstable. Q3. Sketch the bode plot for the transfer function given by
and from plot find gain margin and phase margin. Ans. On 0)-axis mark the point at 23.7 rad/sec. since in denominator (jw) term is having power one, from 23.7 draw a line of slope —20 db/decade to meet y-axis. This will be the starting point. Step 1. From the starting point to I corner frequency (0.33) the slope of the line is —20 db/decade. From I corner frequency (0.33) to second corner frequency (1) the slope of the line will be — 20 ÷ (—20) = —40 db/decade. From II corner frequency to IV corner frequency (2) the slope of the line be —40 + (÷20) = —20 db/decade. From III corner frequency to IV corner frequency, the slope of line will be —20 + (—20) = —40 db/decade. From IV corner frequency (5) to V corner frequency the slope will be —40 ÷ (+20) = —20 db/decade. After V corner frequency, the slope will be (—20) ÷ (—20) = —40 db/decade. Step 2. Draw the phase plot. Step 3. From graph Phase margin = +34° Gain margin =infinity
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Q 4. State the advantages of bode plots. Determine the value of K in the transfer function given below such that (a) Time gain margin is 20 dB (b) The phase margin is 30°
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On the phase plot, we made this poirtt and extend the line tO magnitude plot where it intersects. For required phase margin shift magnitude plot upwards Total shift = 19 dB (from graph) 20 log K = 19 K =8.91.1 Q.5.Design a suitable compensator such that the system will have K= 10 and phase margin = 50°.
The open loop transfer function is thus,
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The bode prot is plotted in fig. The frequency range selected in rad/sec.
rad/sec to
From bode plot it is noted that uncompensated system is stable having a phase margin of 30° and the gain crossover occurs at o = 6.75 rad/sec. To increase the phase margin to 50° the gain crossover point is to be shifted to a higher value of frequency and this is possible by introducing a phase lead compensation network. The required phase lead is:
The value of a parameter of the phase read network is given by:
The maximum phase lead (network)
is contributed at mid frequency
compensation network. Allowing an amplification of compensation network transfer function at
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ofphase lead
the magnitude of phase lead
is determined below:
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UNIT 7 Stability in frequency domain Nyquist criterion This graphical method, which was originally developed for the stability analysis of feedback amplifiers, is especially suitable for different control applications. With this method the closed-loop stability analysis is based on the locus of the open-loop frequency response . Since only knowledge of the frequency response practical approach for the following cases: a) For many cases are known.
is necessary, it is a versatile
can be determined by series connection of elements whose parameters
b) Frequency responses of the loop elements determined by experiments or considered directly.
can be
c) Systems with dead time can be investigated. d) Using the frequency response characteristic of not only the stability analysis, but also the design of stable control systems can be easily performed.
Nyquist criterion using Nyquist plots To derive this criterion one starts with the rational transfer function of the open loop
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Figure: Poles of the open and closed loop in the their multiplicity)
plane (multiple poles are counted according to
To determine , the locus can be drawn on the Nyquist diagram and the phase angle checked. Expediently one moves this curve by 1 to the left in the plane. Thus for stability analysis of the closed loop the locus according to Figure 5.5 has to be drawn.
Figure : Nyquist diagrams of Here
of the open loop
and
is the continuous change in the angle of the vector from the so called critical point (-1,j0)
to the moving point on the locus of
for
. Points where the locus passes through
the point (-1,j0) or where it has points at infinity correspond to the zeros and poles of on the imaginary axis, respectively. These discontinuities are not taken into account for the derivation of . Figure shows an example of a
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Figure: Determination of continuous changes in the angle where two discontinuous changes of the angle occur. Thereby the continuous change of the angle consists of three parts
The rotation is counter clockwise positive.
As the closed loop is only asymptotically stable for the Nyquist criterion follows:
, then from the general case of
The closed loop is asymptotically stable, if and only if the continuous change in the angle of the vector from the critical point (-1,j0) to the moving point of the locus
For the case with a negative gain the case with a positive time in the open loop.
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of the open loop is
of the open loop the locus is rotated by 180 relative to
. The Nyquist criterion remains valid also in the case of a dead
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Simplified forms of the Nyquist criterion It follows from that for an open-loop stable system, that is Therefore the Nyquist criterion can be reformulated as follows:
and
, then
.
If the open loop is asymptotically stable, then the closed loop is only asymptotically stable, if the frequency response locus of the open loop does neither revolve around or pass through the critical point (-1,j0). Another form of the simplified Nyquist criterion for 'left-hand rule':
with poles at
is the so called
The open loop has only poles in the left-half plane with the exception of a single or double pole at
(P, I or
behaviour). In this case the closed loop is only stable, if the critical
point (-1,j0) is on the left hand-side of the locus .
in the direction of increasing values of
This form of the Nyquist criterion is sufficient for most cases. The part of the locus that is significant is that closest to the critical point. For very complicated curves one should go back to the general case. The left-hand rule can be graphically derived from the generalised locus The orthogonal ( a curve with hand side of
)-net is observed and asymptotic stability of the closed loop is given, if passes through the critical point (-1,j0). Such a curve is always on the left.
The Nyquist criterion using Bode plots Because of the simplicity of the graphical construction of the frequency response characteristics of a given transfer function the application of the Nyquist criterion is often more simple using Bode plots. The continuous change of the angle critical point (-1,j0) to the locus of response of
of the vector from the
must be expressed by the amplitude and phase
. From figure
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Figure : Positive (+) and negative (-) intersections of the locus the left-hand side of the critical point
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with the real axis on
it can be seen that this change of the angle is directly related to the count of intersections of the locus with the real axis on the left-hand side of the critical point between . The Nyquist criterion can therefore also represented by the count of these intersections if the gain of the open loop is positive. Regarding the intersections of the locus of with the real axis in the range , the transfer from the upper to the lower half plane in the direction of increasing values are treated as positive intersections while the reverse transfer are negative intersections (Figure 5.7). The change of the angle is zero if the count of positive intersections is equal to the count of negative intersections . The change of the angle depends also on the number of positive and negative intersections and if the open loop does not have poles on the imaginary axis, the change of the angle is
In the case of an open loop containing an integrator, i.e. a single pole in the origin of the complex plane ( ), the locus starts for at , where an additional change of the angle. For proportional and integral behaviour of the open loop
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is added to the
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is valid. In principle this relation is also valid for
, but the locus starts for
at
(Figure 5.8), and this intersection would be counted
Figure : Count of the intersections on the left-hand side of the critical point for open loop as a negative one if
, i.e. if the locus for small
behaviour of the
is in the upper half plane of the real axis. But
de facto there is for (and accordingly ) no intersection. This follows from the detailed investigation of the discontinuous change of the angle, which occurs at . As only a continuous change of the angle is taken into account and because of reason of symmetry the start of the locus at is counted as a half intersection, positive for and negative for analogous to the definition given above For continuous changes of the angle
The open loop with the transfer function
has
poles in the left-half
, which is
plane and possibly a
single ( ) or double pole ( ) at . If the locus of has positive and negative intersections with the real axis to the left of the critical point, then the closed loop is only asymptotically stable, if
is valid. For the special case, that the open loop is stable ( negative intersections must be equal.
,
), the number of positive and
From this it follows that the difference of the number of positive and negative intersections in the case of that for
is an integer and for the number
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is even, for
not an integer. From this follows immediately, the number
is uneven and therefore in Page 106
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is an even number, such that the closed loop is asymptotically stable. This is only .
The Nyquist criterion can now be transferred directly into the representation using frequency response characteristics. The magnitude response
, which corresponds to the locus
, is always positive at the intersections of the locus with the real axis in the range of . These points of intersection correspond to the crossings of the phase response with lines
,
etc., i.e. a uneven multiple of 180 . In the case of a positive
intersection of the locus, the phase response at the lines crosses from below to top and reverse from top to below on a negative intersection as shown in Figure 5.9. In the following these crossings
Figure : Frequency response characteristics of and negative (-) crossings of the phase response
and definition of positive (+) with the -180 line
will be defined as positive (+) and negative (-) crossings of the phase response
over the
particular lines, where may be valid. If the phase response starts at 180 this point is counted as a half crossing with the corresponding sign. Based on the discussions above the Nyquist criterion can be formulated in a form suitable for frequency response characteristics: The open loop with the transfer function has poles in the right-half plane, and possibly a single or double pole at . are the number of positive and of negative crossings of the phase response over the lines in the frequency range where valid. The closed loop is only asymptotically stable, if Department of EEE, SJBIT
is
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is valid. For the special case of an open-loop stable system (
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,
)
must be valid. Table 7.1: Examples of stability analysis using the Nyquist criterion with frequency response characteristics
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Finally the 'left-hand rule' will be given using Bode diagrams, because this version is for the most cases sufficient and simple to apply. The open loop has only poles in the left-half plane with the exception of possibly one single or one multiple pole at if
(P, I or
has a phase of
behaviour). In this case the closed loop is only asymptotically stable, for the crossover frequency
at
.
This stability criterion offers the possibility of a practical assessment of the 'quality of stability' of a control loop. The larger the distance of the locus from the critical point the Department of EEE, SJBIT
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farther is the closed loop from the stability margin. As a measure of this distance the terms gain margin and phase margin are introduced according to Figure below
Figure : Phase and gain margin Bode diagram
and
or
, respectively, in the (a) Nyquist diagram and (b)
Example Problems: Q1 The polar plot of the open-loop transter of feedback control system intersects the real axis at—2 Calculate gain margin (in dB) of the system.
Q2. What is the gain margin of a system in decibels if its Nyquist plot cuts the negative real axis at — 0.7? Ans. a = —0.7
Q4. Consider a feed lock system with the open-loop transfer function. Given by
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Examine the stability of the closed-loop system. Using Nyquist stability theory .
Q 5. Draw the Nyquest plot for the open loop transfer function given below:
and obtain the gain margin and phase margin.
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Q6. Consider a feed lock system with the open-loop transfer function. Given by
Examine the stability of the closed-loop system. Using Nyquist stability theory.
Q7. Sketch the Nyquist plot for the system with the open loop transfer function
and determine the range of K for which the system is
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To get point of intersection on real axis, equate imaglnary part to zero.
Q.8. Sketch the Nyquist plot for system with
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Q 9. How is it possible to make assessment of relative stability using Nyquist criterion? Construct Nyquist plot for the system whose open loop transfer function is
Find the range of K for stability. Ans.
Nyquist critierion can be used to make assessment of relative stability. Using the characteristic equation the Nyquist plot is drawn. A feedback system is sable if and only if, the i.e. contour in the G (s) plane does not encircle the (—1, 0) point when the number of poles of G(s) in the right hand s plane is zero. If G (a) has P poles in the right hand plane, then the number of anticlockwise encirciements of the (—1, 0) point must be equal to P for a stable system, N=—P0 where N = No of clockwise encirclements about (—1, 0) point in C (s) plane P0 = No of poles G (s) in RHP 0
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UNIT 8 Introduction to State variable analysis State-variale representation of single-input-single-output systems In the following a short introduction into the representation of systems using state-variable techniques is given. For this purpose, the example of the network from Figure is used. The dynamical behaviour of this network is completely defined for
initial conditions
, if the
,
and the
input variable
for
are known. For these specifications the variables
and
can be determined for all
. The variables and characterise the 'state' of the network and are therefore called state variables of the network. The differential equations describe the dynamical behaviour of this network. Inserting into one obtains the 2nd-order differential equation according to which completely describes the system with respect to the input-output behaviour. But one can also use the two original differential equations and can write them in vector notation so that the 1st-order vector differential equation
(8.1) with the initial condition
is obtained. This linear 1st-order vector differential equation describes the connection between the input variable and the state variables. To complete a state-space system, one needs an additional equation that describes the dependence of the output variable on the state variables. In this example, it is the direct relationship Introducing the state vector
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into above eq, with the vectors
and with the matrix
and with the scalar variables and one obtains the general state-space representation of a linear time-invariant single-inputsingle-output system: (8.2)
initial condition
(8.3) The Eq. (8.2) is the state equation, and in the general case it is a linear system of 1st-order differential equations of
state variables
, which are combined in the state
vector . Eq. (8.3) is the output equation, which maps the states and inputs linearly to the output. This is an algebraic equation, whereas the state equation is a differential equation.
State-space representation of multi-input-multi-output systems The Eqs. (8.2) and (8.3) describe an th-order linear time-invariant single-input-singleoutput system. For linear multi-input-multi-output systems of order with inputs and outputs these equations become
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with the initial condition
(8.5) where the following notation is used:
state vector
(
) vector
input vector
(
) vector
output vector
(
) vector
system matrix
(
) matrix
input matrix
(
) matrix
output matrix
(
) matrix
feedthrough matrix
(
) matrix
It goes without saying that the general representation of Eqs. (8.4) and (8.5) also includes the single-input-single-output case. The matrices and have constants elements. If these elements are time-varying, the matrices of the corresponding time-varying system are substituted by matrix functions of time, e.g.
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.
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The relationship between transfer functions and the state-space representation In the following, the Eqs. (8.4) and (8.5) will be transformed into the domain using the Laplace transform, which will be done analogously to the scalar case . For this, the operator notation
from section 2.1 is adopted and when applying it to Eq. (8.4), one obtains
or rearranged
The solution of the state equation in the domain is then given by (8.6) with (8.7) Similarly, for Eq. (12.5) yields
Substituting
from Eq. (12.6), the system output in the domain is
To obtain the relationship with transfer functions, the initial condition has to be set to zero. For a single-input-single-output system according to Eqs. (8.2) and (8.3) the system output is
Comparing this equation with Eq. (8.3) the transfer function is given by (8.8)
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from Eq. (8.7) is a matrix of rational functions of , which can always be
(8.9)
where
is a matrix with polynomial elements in . From Eq. (8.2) it is obvious that (8.10)
and (8.11) which is the characteristic polynomial of the system. The zeros of this polynomial are the poles of the transfer function and at the same time eigenvalues of the system matrix . If the system in the state-space representation is fully controllable and observable (see section 8.6), then the number of poles are equal to the number of eigenvalues.
State-space vs transfer function approach The key advantage of transfer functions is in their compactness, which makes them suitable for frequency-domain analysis and stability studies. However, the transfer function approach suffers from neglecting the initial conditions. Not only does state-space representation serve as an alternative to transfer functions, but also it is not limited to linear and time-invariant systems and it has the following advantages: 1. Single-input-single-output and multi-input-multi-output systems can be formally treated equal. 2. The state-space representation is best suited both for the theoretical treatment of control systems (analytical solutions, optimisation) and for numerical calculations. 3. The determination of the system response in the homogeneous case with the initial condition is very simple. 4. This representation gives a better insight into the inner system behaviour. General system properties, for example, the system controllability or observability can be defined and determined.
Uniqueness of the state variables
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Initially it sounds paradoxical that the choice of the state variables is not unique. This means that for one and the same system with the input , the output and state variables, there exist an infinite number of state-space representations. For each value of time one gets the state in the -dimensional state space. The values are the cartesian coordinates of the state
where the unit vectors are -dimensional linear independent vectors. Their elements are besides the th element, which has the value of 1 - all zero. For describing the state also other basis vectors can be used. Candidates are all linear independent and -dimensional vectors . Therefore, it is always possible, to write the state as (8.12)
After introducing the new state
and the quadratic matrix
one can rewrite Eq. (8.12) as (8.13)
The constant and regular matrix is a so-called transformation matrix. Now, instead of the state the new state can be used and its behaviour analysed. Describing the system in the new coordinates the state from Eq. (8.13) is inserted into Eqs. (8.4) and (8.5) and one obtains (8.14)
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(8.15)
with the initial condition
The benefits of the transformation of systems into different state-space representations are:
Most system properties do not depend on the choice of the state variables. They remain unchanged after a regular transformation and may be analysed in an appropriate representation form. The computational determination and analysis of system properties can be tremendously simplified if the representation form is specifically selected. In particular certain canonical forms are of interest.
Example 12.5.1 In order to demonstrate a transformation, the example from Eq. (8.1) is used with the system parameters , and . The initial condition is assumed to be zero and therefore omitted. With these values one obtains the state equation as
and the output equation as
For the regular transformation matrix
one obtains the matrices for Eqs. (8.14) and (8.15) as
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The new representation
consists of two decoupled differential equations with respect to the state variables
and
.
The analysis and treatment of a system in such a structured representation form, as shown in the example above, is doubtless more simple. As the representation form must be specifically selected depending on the type of analysis or synthesis problem, the different representation forms, for example, the canonical forms, are not discussed separately and are introduced in the following sections when they are needed.
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