Control Systems

 

CONTROL SYSTEM NOTES (For Bachelor of Engineering)    Notes by: 

PROF. SHESHADRI G. S   Soft Copy material designed by: 

KARTHIK KUMAR H P Your feedbacks can be mailed to: 

[email protected]   

     

INDEX  (1) Introduction to Control system  (2) Mathematical model of linear systems  (3) Transfer functions  (4) Block diagram  (5) Signal Flow Graphs  (6) System Stability  (7) Root Locus Plots  (8) Bode Plots 

 

Control Systems

Introduction to Control Systems By: Sheshadri.G.S.

CIT, Gubbi. Control System means any quantity of interest in a machine or mechanism is maintained or altered in accordance with desired manner. OR A system which controls the output quantity is called a control system.

Definitions: 1.

Controlled Variable: It is the quantity or condition that is measured & controlled.

2. Controller: Controller means measuring the value of the controlled variable of the system & applying the manipulated variable to the system to correct or to limit the deviation of the measured value to the desired value. 3. Plant: A plant is a piece of equipment, which is a set of machine parts functioning together. The purpose of which is to perform a particular operation. Example: Furnace, Space craft etc., 4. System: A system is a combination of components that works together & performs certain objective. 5. Disturbance: A disturbance is a signal that tends to affect the value of the output of a system. If a disturbance is created inside the system, it is called internal. While an external disturbance is generated outside the system. 6. Feedback Control: It is an operation that, in the presence of disturbance tends to reduce the difference between the output of a system & some reference input. 7. Servo Mechanism: A servo mechanism is a feedback controlled system in which the output is some mechanical position, velocity or acceleration. 8. Open loop System: In an Open loop System, the control action is independent of the desired output. OR When the output quantity of the control system is not fed back to the input quantity, the control system is called an Open loop System. 9. Closed loop System: In the Closed loop Control System the control action is dependent on the desired output, where the output quantity is considerably controlled by sending a command signal to input quantity. By: Sheshadri.G.S. HPK Kumar

1

2

Introduction to Control System 10. Feed Back: Normally, the feed back signal has opposite polarity to the input signal. This is called negative feed back. The advantage is the resultant signal obtained from the comparator being difference of the two signals is of smaller magnitude. It can be handled easily by the control system. The resulting signal is called Actuating Signal This signal has zero value when the desired output is obtained. In that condition, control system will not operate.

Effects of Feed Back: Let the system has open loop gain Input signal . Then the feed back signal is,

E(S)

R(S) B(S)

-

feed back loop gain

Output signal

&

G(S)

C(S)

&

–

Hence,

H(S) = With this eqn.

(1) , we can write the effects of feed back as follows.

(a)

Overall Gain: Eqn. shows that the gain of the open loop system is reduced by a factor in a feed back system. Here the feed back signal is negative. If the feed back gain has positive value, the overall gain will be reduced. If the feed back gain has negative value, the overall gain may increase.

(b)

Stability: If a system is able to follow the input command signal, the system is said to be Stable. A system is said to be Unstable, if its output is out of control. In eqn. , if the output of the system is infinite for any finite input. This shows that a stable system may become unstable for certain value of a feed back gain. Therefore if the feed back is not properly used, the system can be harmful.

(c)

Sensitivity: This depends on the system parameters. For a good control system, it is desirable that the system should be insensitive to its parameter changes. Sensitivity, SG =

This function of the system can be reduced by increasing the value of . This can be done by selecting proper feed back. By: Sheshadri.G.S.

HPK Kumar

Control Systems (d)

Noise: Examples are brush & commutation noise in electrical machines, Vibrations in moving system etc.,. The effect of feed back on these noise signals will be greatly influenced by the point at which these signals are introduced in the system. It is possible to reduce the effect of noise by proper design of feed back system.

Classification of Control Systems The Control System can be classified mainly depending upon, (a) (b) (c) (d)

Method of analysis & design, as Linear & Non- Linear Systems. The type of the signal, as Time Varying, Time Invariant, Continuous data, Discrete data systems etc., The type of system components, as Electro Mechanical, Hydraulic, Thermal, Pneumatic Control systems etc., The main purpose, as Position control & Velocity control Systems.

1.

Linear & Non - Linear Systems: In a linear system, the principle of superposition can be applied. In non - linear system, th is p rin cip le ca n ’t b e a p p lied . T h erefo re a lin ea r system is th a t w h ich o b eys su p erp o sitio n principle & homogeneity.

2. Time Varying & Time Invariant Systems: While operating a control system, if the parameters are unaffected by the time, then the system is called Time Invariant Control System. Most physical systems have parameters changing with time. If this variation is measurable during the system operation then the system is called Time Varying System. If there is no non-linearity in the time varying system, then the system may be called as Linear Time varying System. 3. Discrete Data Systems: If the signal is not continuously varying with time but it is in the form of pulses. Then the control system is called Discrete Data Control System. If the signal is in the form of pulse data, then the system is called Sampled Data Control System. Here the information supplied intermittently at specific instants of time. This has the advantage of Time sharing system. On the other hand, if the signal is in the form of digital code, the system is called Digital Coded System. Here use of Digital computers, µp, µc is made use of such systems are analyzed by the Z- transform theory. 4. Continuous Data Systems: If the signal obtained at various parts of the system are varying continuously with time, then the system is called Continuous Data Control Systems. 5. Adaptive Control systems: In some control systems, certain parameters are either not constant or vary in an unknown manner. If the parameter variations are large or rapid, it may be desirable to design for the capability of continuously measuring them & changing th e compensation, so that the system performance criteria can always satisfied. This is called Adaptive Control Systems. By: Sheshadri.G.S. HPK Kumar

3

4

Introduction to Control System

R(s)

E(s)

Compensator

Identification & Parameter adjustment

System

+

C(s)

B(s) H(S) 6. Optimal Control System: Optimal Control System is obtained by minimizing and/or maximizing the performance index. This index depends upon the physical system & skill. 7. Single Variable Control System: In simple control system there will be One input & One output such systems are called Single variable System (SISO – Single Input & Single Output). 8. Multi Variable Control System: In Multivariable control system there will be more than one input & correspondingly more output’s (M IM O - Multiple Inputs & Multiple Outputs).

Comparison between Open loop & Closed loop Gain Open Loop System 1. An open loop system has the ability to perform accurately, if its calibration is good. If the calibration is not perfect its performance will go down. 2.

It is easier to build.

Closed Loop System 1. A closed loop system has got the ability to perform accurately because of the feed back.

2. It is difficult to build.

3. In general it is more stable as the feed back is absent.

3. Less Stable Comparatively.

4. If non- lin ea rity’s a re present; the system operation is not good.

4. Even under the presence of nonlin ea rity’s th e system o p era tes b etter th a n o p en loop system.

5. Feed back is absent. Example: (i) Traffic Control System. (ii) Control of furnace for coal heating. (iii) An Electric Washing Machine.

5. Feed back is present. Example: (i) Pressure Control System. (ii) Speed Control System. (iii) Robot Control System. (iv) Temperature Control System.

Note: Any control system which operates on time basis is an Open Loop System.

By: Sheshadri.G.S. HPK Kumar

Control Systems

5

Block Diagram of Closed Loop System: Controller

E(S) Ref. i/p

Control Elements

Plant

Controlled o/p

Feed Back elements

Actuator

Thermometer Block Diagram of Temperature Control System: Thermo meter

A/D Converter

Interface

Electric Furnace Heater

Relay

Amplifier

Interface

Programmed i/p

Temperature Control of Passenger Compartment Car: Sun

Sensor Radiation Heat Sensor

Desired Temperature i/p

Controller

Air Conditioner

Ambient Temperature

Passenger Car

O/p

Sensor ************************************************************************

****************

By: Sheshadri.G.S. HPK Kumar

 

Control Systems  1

Mathematical Models of Linear Systems By: Sheshadri.G.S.

CIT, Gubbi. A physical system is a collection of physical objects connected together to serve an objective. An idealized physical system is called a Physical model. Once a physical model is obtained, the next step is to obtain Mathematical model. When a mathematical model is solved for various i/p conditions, the result represents the dynamic behavior of the system. Analogous System:  The concept of analogous system is very useful in practice. Since one type of system may be easier to handle experimentally than another. A given electrical system consisting of resistance, inductance & capacitances may be analogous to the mechanical system consisting of suitable combination of Dash pot, Mass & Spring. The advantages of electrical systems are, 1. 2.

Many circuit theorems, impedance concepts can be applicable. An Electrical engineer familiar with electrical systems can easily analyze the system under study & can predict the behavior of the system. 3. The electrical analog system is easy to handle experimentally. Translational System:     It has 3 types of forces due to elements. 1. Inertial Force:  Due to inertial mass, 

 

     

M

F(t)

 

 

Fm t

M. a

2 2

.  

Where,

. .

 

2. Damping Force [Viscous Damping]:  Due to viscous damping, it is proportional to velocity & is given by, D     .     Damping force is denoted by either D or B or F       3. Spring Force:   Spring force is proportional to displacement.      .  .            

Fk  

Designed By:

HPK Kumar ([email protected])

 

 

2

 

Mathematical Models of Linear Systems  Where,

Rotational system: 

 

 

 

1. Inertial Torque:

 

 

   

2. Damping Torque: 3. Spring Torque

.    

:

   

 

 

 

 

  

 

Analogous quantities in translational & Rotational system:   

The electrical analog of the mechanical system can be obtained by, (i) (ii)

Sl. No.

Force Voltage analogy: (F.V) Force Current analogy: (F.I)

Mechanical Translational System

Mechanical Rotational System

F.V Analogy

F.I Analogy

1.

Force (F)

Torque (T)

Voltage (V)

Current (I)

2.

Mass (M)

Moment of Inertia (M)

Inductance (L)

Capacitance (C)

3.

Viscous friction (D or B or F)

Viscous friction (D or B or F)

Resistance (R)

Conductance (G)

4.

Spring stiffness (k)

5.

Linear displacement ( )

Torsional spring stiffness Reciprocal of Capacitance (1/C) ( ) Charge (q) Angular displacement ( )

Reciprocal of Inductance (1/L) Flux ( )

6.

Linear velocity ( )

Angular Velocity (w)

Voltage (v)

Current (i)

D’Alemberts Principle:    

The static equilibrium of a dynamic system subjected to an external driving force obeys the following principle,

“For any body, the algebraic sum of externally applied forces resisting motion in any given direction is zero”. 

Example Problems:  (1) Obtain the electrical analog (FV & FI analog circuits) for the Machine system shown & also write the equations.

Free Body diagram D2

2 2

 

 

D1 1 1

 

M1 F t

M1

F t  

 

ßÞà Þßà  ßÞà Þßà ßÞà Þßà  ßÞà Þßà ßÞà Þßà  ßÞà Þßà ßÞà Þßà  ßÞà   Designed By:    

HPK Kumar ([email protected]) 

 

Control Systems  1

Transfer Functions By: Sheshadri.G.S.

CIT, Gubbi. The input- output relationship in a linear time invariant system is defined by the transfer function. The features of the transfer functions are, (1) (2) (3) (4) (5)

It is applicable to Linear Time Invariant system. It is the ratio between the Laplace Transform of the o/p variable to the Laplace Transform of the i/p variable. It is assumed that initial conditions are zero. It is independent of i/p excitation. It is used to obtain systems o/p response. An equation describing the physical system has integrals & differentials, the step involved in obtaining the transfer function are; (1) Write the differential equation of the system.   

by ‘S’ &

by 1/S. 

(2)

Replace the terms

(3)

Eliminate all the variables except the desired variables.  Impulse Response of the Linear System: 

                    

R(S)    

G(S)

C(S)

In a control system, when there is a single i/p of unit impulse function, then there will be some response of the Linear System. The Laplace Transform of the i/p will be R(S) = 1

G(S) = 

 

.  

   

 1 

 

.1  

i.e., the Laplace Transform of the system o/p will be simply the ‘Transfer function’ of the system.

                        

 

 

 

 

 

 

 

 

Taking L-1  

Here G(t) will be impulse response of the Linear System. This is called Weighing Function. Hence LT of the impulse response is the Transfer function of the system itself.

R

PROBLEMS:  (1) Obtain the Transfer‐Function(TF) of the circuit shown in circuit 1.0  Solution:

R

Laplace Transformed network

i(S)

1

Circuit 1.0  

Designed By:

HPK Kumar ([email protected])

 

C

i

Contd……

2

 

Transfer Functions   

  . .,       

 .

 

 

 .

1

 

             &             

 

 

   

  

 

1

    

   

 

 

1

 

 

 

1 1  

 

Where,

= RC

(2) Obtain the TF of the mechanical system shown in circuit 2. 

Designed By:    

HPK Kumar ([email protected]) 

 

Control Systems  3 (3) Transfer Function of an Armature Controlled DC Motor in circuit 3.0: 

La

Ra

 

If = Constant

ia

 

V  i

‘Ra’ Æ Resistance of armature in Ω’s.

ia

Eb

F

Vf

‘La’ Æ Inductance of armature in H’s. ‘ia’ Æ Armature current. & ‘if’Æ Field current.

 

 

 

‘Vi’ Æ Applied armature voltage.

circuit 3.0  

   

Let,

 

 

Tm

J,

 

     

‘Eb’ Æ Back e.m.f in volts. ‘Tm’ Æ Torque developed by the motor in N-m.

F

 Æ Angular displacement of motor shaft in radians

 

J Æ Equivalent moment of inertia of motor & load referred to the motor shaft.

 

F Æ Equivalent Viscous friction co-efficient of motor & load referred to the motor shaft. The air gap flux ‘ ’ is proportional to the field current. i.e.,     

   .

 

Where, ‘Kf’ is a constant.

The torque developed by the motor ‘Tm’ is proportional to the product of the arm current & the air gap flux.  

   

Where, Ka & Kf are the constants.       

Since the field current is constant,

Where, KT is Motor – torque constant.

The motor back e.m.f is proportional to the speed & is given by,

  

 

Where, Kb is back e.m.f constant.

The differential equation of the armature circuit is,

    

 

The torque equation is,

   2 2

 

 

 

 

Taking LT for above equation, we get  

 

     

      

    

  

    

 

------------------------------- (1) ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐  (A)  

      

Taking LT for the torque equation & equating, we get  

    

  

 

      

Designed By:

HPK Kumar ([email protected])

 

4

 

Transfer Functions   

  

2

      

------------------------------ (2)  

  

 

-------------------------- (B)

 

Taking LT for back e.m.f equation, we get   

    

------------------------------- (C)

Substituting the values of Ia (S) & Eb (S) from equation (C) & (2) in equation (1), we get   

 

 

      

  

       

 

     

 

 

    

   .

 

 

    

 

 

The block diagram representation of armature controlled DC Motor can be obtained as follows,  From equation (A),

Ia(S)

1  

Vi(S)

-

 

Eb(S) From equation (B),

Ia(S)  

   

 

(S)

From equation (C),

(S)

   

Eb(S)

The complete block diagram is as shown below, 1

Vi(S)

Ia(S)  

2

 

(S)

Eb(S)

Designed By:    

HPK Kumar ([email protected]) 

 

Control Systems  5 (4) Transfer function of Field Controlled DC Motor in circuit 4.0:  Let,

Lf

Rf

Ia = Constant

Rf Æ Field winding resistance. Lf Æ Field winding inductance.

Vf

if

Vf Æ Field control voltage.

Tm

If Æ Field current.  

Tm Æ Torque developed by motor. J Æ Equivalent moment of inertia of motor & load referred to the motor shaft.

circuit 4.0

F Æ Equivalent Viscous friction co-efficient of motor & load referred to the motor shaft.

J, F

Æ Angular displacement of motor shaft. In the field controlled DC motor, the armature current is fed from a constant current source.  

   

Where, Ka & Kf are the constants.

The KVL equation for the field circuit is,     

 

 

On Laplace Transform,   

    

    

    

     .

    

    

   

   

---------------------------------- (1) --------------------------------- (A)        

The torque equation is ,   

 

Where, KT is Motor – torque constant.

  

On Laplace Transform,     

  

 

       

 

      

       .

 

 

2

       .

    

     

Substituting the value of  

--------------------------- (2)

------------------------- (B) from equation (2) in equation (1), we get

Designed By:

HPK Kumar ([email protected])

 

6

 

Transfer Functions 

  

2

 

  

       .

2

 

 

    

       

     

    

The block diagram representation of field controlled DC Motor can be obtained as follows,  From equation (A), 1  

Vf (S)

If (S)

 

 

   

 

From equation (B),

If (S)  

(S)

The complete block diagram is as shown below, 1  

Vf (S)

(5) Obtain the TF  

If (S)

 

 

 

R

 for the network shown in circuit 5.0: 

Laplace Transformed network

Solution:

(S)

 

Vi

R

C

1

R

circuit 5.0 I(S)

Vi (S)

C

I1(S) R

I2(S) 1

Vo (S)

Applying KVL to this circuit,  

    

   

 

----------------

------------------ (1)

       1  

           

 

.

1

 

1

  

    

 

 .

    1

    

2

 

 

1

  Let,   

τ 

τ     2 1 τ   τ 

Designed By:    

HPK Kumar ([email protected]) 

Vo

 

Control Systems  7

1M Ω

100 k  Ω (6) Find the TF  

 for the network shown in circuit 6.0: 

Solution:

1

10  F

Vi

Laplace Transformed network 10 10  

F

Circuit 6.0 10

 

Vi (S)

Loop 1

10

 

 V0 (S)

Loop 2

Writing KVL for loop (1), we get

 

 

    1

5

  10        

 105

1

 .   105  

   

1

2

1

5

  10 

2

  

------------------------------------------- (1)

Writing KVL for loop (2), we get

 

 

    

 

    10 1

    2

 

       

  

 

 

   

0

-------------------------------------------- (2)

11

.

2

  10

 

106

    

    

 

2

  

 .     10

6

--------------------------------- (3)

Substituting for I1 (S) from equation (2) in (1), we get     2

.   10 

  

    

5

10  

1

11  

2

 

10 

 .   5

10  

11  

   

1 1

2

1

  

 1

1

From equation (3) the above equation becomes,    5

10  

    

 .     10

  

 

6

  

   10 

2

 21 

10

10 10 

 21 

Vi(S) 10  

1

 

10

 

+

I2(S)

1

I1(S) 10

11

 

1

Designed By:

 

 

1   HPK Kumar

([email protected])  

106

V0(S)

8

 

Transfer Functions 

ßÞà Þßà  ßÞà Þßà ßÞà Þßà  ßÞà Þßà ßÞà Þßà  ßÞà Þßà ßÞà Þßà  ßÞà  

Designed By:    

HPK Kumar ([email protected]) 

Control Systems

Block Diagrams By:

Sheshadri.G.S. CIT, Gubbi.

It is a representation of the control system giving the inter-relation between the transfer function of various components. The block diagram is obtained after obtaining the differential equation & Transfer function of all components of a control system. The arrow head pointing towards the block indicates the i/p & pointing away from the block indicates the o/p. If

is the TF,

After obtaining the block diagram for each & every component, all blocks are combined to obtain a complete representation. It is then reduced to a simple form with the help of block diagram algebra. The following block diagram reduction algebra is used, (1) Blocks in Cascade [Series] :

G2(S) (2) Combining blocks in Parallel:

(3) Eliminating a feed back loop:

(2)

(4) Moving a take-off point beyond a block: (3) (6) (2) C(S) G(S)

(3)

(6)

(2)

(2) (5) Moving a Take-off point ahead of a block: (3)

(2) G(S)

(3)

(6) (2)

(6)

(6)

(2)

(6)

1

2

Block Diagrams P ROBLEMS : Reduce the Block Diagrams shown below: (1)

+ -

-

-

Solution: By eliminating the feed-back paths, we get

+ -

Combining the blocks in series, we get

+

C(S)

-

Eliminating the feed back path, we get

Control Systems (2)

R(S)

Solution:

-

C(S)

-

Shifting the take-off ‘ ’ beyond the block ‘ ’, we get

R(S)

-

-

Combining R(S)

and eliminating

-

(feed back loop), we get

C(S)

-

Eliminating the feed back path

R(S)

C(S)

, we get

C(S)

-

Combining all the three blocks, we get R(S)

C(S)

3

4

Block Diagrams

R(S)

Solution:

C(S)

-

-

-

(3)

Re-arranging the block diagram, we get R(S)

-

Eliminating

C(S)

-

-

loop & combining, we get C(S)

R(S)

-

-

Eliminating feed back loop R(S)

C(S)

-

Eliminating feed back loop R(S)

, we get C(S)

Control Systems

Signal Flow Graphs By:

Sheshadri.G.S. CIT, Gubbi.

For complicated systems, Block diagram reduction method becomes tedious & time consuming. An alternate method is that signal flow graphs developed by S.J. Mason. In these graphs, each node represents a system variable & each branch connected between two nodes acts as Signal Multiplier. The direction of signal flow is indicated by an arrow.

Definitions: Node: A node is a point representing a variable. Transmittance: A transmittance is a gain between two nodes. Branch: A branch is a line joining two nodes. The signal travels along a branch. Input node [Source]: It is a node which has only out going signals. Output node [Sink]: It is a node which is having only incoming signals. Mixed node: It is a node which has both incoming & outgoing branches (signals). Path: It is the traversal of connected branches in the direction of branch arrows. Such that no node is traversed more than once. 8. Loop: It is a closed path. 9. Loop Gain: It is the product of the branch transmittances of a loop. 10. Non-Touching Loops: Loops are Non-Touching, if they do not possess any common node. 11. Forward Path: It is a path from i/p node to the o/p node w hich doesn’t cross any node m ore than once. 12. Forward Path Gain: It is the product of branch transmittances of a forward path. 1. 2. 3. 4. 5. 6. 7.

M ASO N ’S GAIN FO RM U LA: The relation between the i/p variable & the o/p variable of a signal flow graphs is given by the net gain between the i/p & the o/p nodes and is known as Overall gain of the system. Mason’s gain form ula for the determ ination of overall system gain is given by, Where,

–

Path gain of

forward path.

Determinant of the graph.

The value of the T

for that part of the graph not touching the

Overall gain of the system.

forward path.

1

2

Signal Flow Graphs Problems: (1) Obtain the closed loop TF,

by using M ason’s gain form ula.

C(S)

R(S)

Solution: M ason’s gain form ula is, No. of forward paths:

Gain Products of all possible combinations of two non-touching loops:

No. of individual loops:

No. of three non-touching loops = 0.

(2) Obtain the closed loop TF,

by using M ason’s gain form ula.

C(S)

R(S)

Solution: M ason’s gain form ula is, Forward Paths:

Contd......

Control Systems No. of individual loops:

Two Non-touching loops:

(3) Construct a signal flow graph from the following equations. Obtain overallTF using M ason’s gain form ula.

Where

is i/p variable &

is o/p variable.

Solution:

No. of forward paths:

Individual loops:

Two non-touching loops:

Three non-touching loops = 0 M ason’s gain form ula is,

(4) Obtain

by Block Diagram Reduction method & verify the result by signal flow graph.

+ R(S)

C(S)

+ +

Contd......

3

4

Signal Flow Graphs Solution:

Re-arranging the summing points,

R(S)

-

C(S)

-

C(S)

R(S)

Signal flow graphs: R(S)

C(S)

No. of forward paths:

No. of individual loops: M ason’s gain form ula is,

(5) Obtain the TF & Verify by signal flow graph.

+ R(S)

-

C(S)

-

Solution: Shifting the take-off point

ahead of the block

. The BD reduces to,

+ R(S)

-

-

Contd......

C(S)

Control Systems

R(S)

5

C(S)

-

R(S)

C(S)

C(S)

R(S)

Signal flow graph:

R(S)

C(S)

No. of forward paths:

No. of individual loops:

(6) Reduce the Block Diagram shown.

-

R(S)

-

Solution: Shifting

C(S)

+

-

beyond

, we get

-

R(S)

-

-

C(S)

+

6

Signal Flow Graphs Eliminating feed back loop

, we get

R(S)

C(S)

-

Eliminating feed back loop

+

-

, we get

R(S)

C(S)

-

+

R(S)

C(S)

-

Eliminating the another feed back loop

+

, we get

R(S)

C(S)

+

R(S)

C(S)

Signal flow graph:

R(S)

C(S)

Contd......

Control Systems No. of forward paths:

No. of individual loops:

(7) Obtain the closed loop TF

by using M ason’s gain form ula.

R(S)

C(S)

Solution: No. of forward paths:

No. of individual loops:

Two non-touching loops:

(8) Obtain the TF of the closed loop control system represented by the Block Diagram shown below using block diagram reduction method.

-

-

Solution: Shifting the take off point of

-

-

beyond block

& Simplifying for the blocks

, we get

7

8

Signal Flow Graphs Eliminating

loop, we get

-

(9) U sing M ason’s gain rule, obtain the overall TF of a control system represented by the signal flow graph shown below.

Solution: No. of forward paths:

Individual loops:

Two non-touching loops = 0

(10)

Construct signal flow graph from the following equations & obtain the overall TF.

Contd......

Control Systems Solution:

No. of forward paths:

No. of individual loops:

Two non-touching loops:

Three non-touching loops:

Four non-touching loops = 0

(11)

O btain the TF using M ason’s gain form ula.

Contd......

9

10

Signal Flow Graphs Solution: No. of forward paths:

No. of individual loops:

Two non-touching loops:

Three non-touching loops = 0

(12)

Obtain the TF using M ason’s gain form ula.

Solution: No. of forward paths:

No. of individual loops:

Two non-touching loops:

(13)

O btain the TF using M ason’s gain form ula.

Contd......

Control Systems Solution: No. of forward paths:

No. of individual loops:

Two non-touching loops:

Three non-touching loops = 0

(14)

Draw the signal flow graph for the Block Diagram shown in fig. Hence obtain M ason’s gain form ula.

, Using

Solution:

No. of forward paths:

No. of individual loops:

–

Two non-touching loops:

–

Contd......

11

12

Signal Flow Graphs Three non-touching loops = 0

(15)

Obtain TF,

using block diagram algebra & also by using Masons Gain Formula. Hence Verify

the TF in both the methods.

-

Solution: Same block diagram can be re-arranged as shown below.

Shifting the take-off points beyond

we get

Control Systems

Substituting ‘x’ value in the block diagram . T he block diagram becom es,

Signal flow graph:

No. of forward paths: No. of individual loops:

(16)

Obtain TF,

Two non-touching loops = 0

using block diagram algebra & also by using Masons Gain Formula. Hence Verify

the TF in both the methods. Contd......

13

14

Signal Flow Graphs

Solution: Same Block Diagram can be written as,

Substituting the value of ‘x’

Signal flow Graph:

Control Systems No. of forward paths:

No. of individual loops:

(17)

Two non-touching loops = 0

Find the output

Solution: (i)

Let

then we can find

No. of forward paths:

No. of individual loops:

Two non-touching loops:

Three non-touching loops = 0

15

16

Signal Flow Graphs (ii)

Let

Determine

No. of forward paths:

No. of individual loops: ‘ ’ rem ains sam e.

(iii)

Let

Determine

No. of forward paths: ‘ ’rem ains sam e.

(iv) Let Determine (i.e., R esponse at ‘2’ w hen source ‘2’ is acting). (figure is in next page) No. of forward paths: ‘ ’rem ains sam e.

Control Systems

Hence,

17

Control Systems

System Stability By:

Sheshadri.G.S. CIT, Gubbi.

While considering the performance specification in the control system design, the essential & desirable requirement will be the system stability. This means that the system must be stable at all times during operation. Stability may be used to define the usefulness of the system. Stability studies include absolute & relative stability. Absolute stability is the quality of stable or unstable performance. Relative Stability is the quantitative study of stability. The stability study is based on the properties of the TF. In the analysis, the characteristic equation is of importance to describe the transient response of the system. From the roots of the characteristic equation, some of the conclusions drawn will be as follows, (1) When all the roots of the characteristic equation lie in the left half of the S-plane, the system response due to initial condition will decrease to zero at time Thus the system will be termed as stable. (2) When one or more roots lie on the imaginary axis & there are no roots on the RHS of Splane, the response will be oscillatory without damping. Such a system will be termed as critically stable. (3) When one or more roots lie on the RHS of S-plane, the response will exponentially increase in magnitude; there by the system will be Unstable. Some of the Definitions of stability are, (1) A system is stable, if its o/p is bounded for any bounded i/p. (2) A system is stable, if it‟s response to a bounded disturbing signal vanishes ultimately as time „t‟ approaches infinity. (3) A system is un stable, if it‟s respon se to a bounded disturbing signal results in an o/p of infinite amplitude or an Oscillatory signal. (4) If the o/p response to a bounded i/p signal results in constant amplitude or constant amplitude oscillations, then the system may be stable or unstable under some limited constraints. Such a system is called Limitedly Stable system. (5) If a system response is stable for a limited range of variation of its parameters, it is called Conditionally Stable System. (6)

If a system response is stable for all variation of its parameters, it is called Absolutely Stable

system.

Routh-Hurwitz Criteria: A designer has so often to design the system that satisfies certain specifications. In general, a system before being put in to use has to be tested for its stability. Routh-Hurwitz stability criteria may be used. This criterion is used to know about the absolute stability. i.e., no extra information can be obtained regarding improvement. As per Routh-Hurwitz criteria, the necessary conditions for a system to be stable are, (1) None of the co-efficient‟ of the Characteristic equation should be missing or zero. (2) All the co-efficient‟ should be real & should have the sam e sign.

1

2

System Stability A sufficient condition for a system to be stable is that each & every term of the column of the Routh array must be positive or should have the same sign. Routh array can be obtained as follows. The Characteristic equation is of the form, … … … … … … … … … Where,

: 0

0

0 0 0

0

0

:

Similarly we can evaluate rest of the elements,

The following are the limitations of Routh-Hurwitz stability criteria, (1) It is valid only if the Characteristic equation is algebraic. (2) If any co-efficient of the Characteristic equation is complex or contains power of „ ‟, this criterion cannot be applied. (3) It gives information about how many roots are lying in the RHS of S-plane; values of the roots are not available. Also it cannot distinguish between real & complex roots. Special cases in Routh-Hurwitz criteria: (1) When the term in a row is zero, but all other terms are non-zeroes then substitute a small positive number for zero & proceed to evaluate the rest of the elements. When the column term is zero, it means that there is an imaginary root. (2) All zero row: In the case, write auxiliary equation from preceding row, differentiate this equation & substitute all zero row by the co-efficient‟ obtained by differentiating the auxiliary eq uation. This case occurs when the roots are in pairs. The system is limitedly stable. Problems: COMMENT ON THE STABILITY OF THE SYSTEM WHOSE CHARACTERISTIC EQUATION IS GIVEN BELOW: (1) 1

21

20

6

36

0

15

20

0

28

0

0

20

0

0

The no. of sign changes in the column = zero. No roots are lying in the RHS of S-plane. The given System is Absolutely Stable.

Control Systems

3

(2) 4

2

3

5

-4.66

0

5

0

The no. of sign changes in the column = 2 Two roots are lying in the RHS of S-plane. The given System is unstable.

(3) 1

1

5

2

4

0

-1

5

0

14

0

0

5

0

0

The no. of sign changes in the column = 2 Two roots are lying in the RHS of S-plane. The given System is unstable.

(4) 1

2

3

+ve

2

4

1

+ve

2.5

0

+ve

1

0

-ve

Two roots are lying in the RHS of S-plane.

0

0

+ve

The given System is unstable.

0

0

+ve

0

1

The no. of sign changes in the column = 2

(5) 1

2

3

2

4

6 All Zero row.

2

6

0

-16

0

0

6

0

0

A.E. is

The no. of sign changes in the column = 2 Two roots are lying in the RHS of S-plane. The given System is unstable & limitedly stable.

(6) 1

8

20

16

2

12

16

0

2

12

16

0

A.E. is

The no. of sign changes in the 6

16

0

0

2.66

0

0

0

16

0

0

0

column = 0

The System is limitedly stable. (Because, in the zero).

row all the elements are

4

System Stability (7)

. Find the restriction on ‘K’, So that

The open loop TF of a unity feed back system is

the closed loop is stable. Solution:

For Stable system, (i) 10

1

6.5

K

(ii)

0 K

0 For,

the closed loop system is stable.

(8) Solution: 1

4

K

For Stable system,

3

1

0

(i)

K

0

(i)

0

0

0

0

K

For, (9) D eterm ine the value of‘K & b’,So that the system open loop T.F. frequency of 2 radians.

( Assuming unity feed back i.e., H(S)=1 )

Solution: The characteristic equation is

1

(3+K)

b

(1+K) 0

(1+K)

0

From equations (9.1) & (9.2) Either from (9.1) or (9.2)

the closed loop system is stable oscillates at a

Control Systems (10)

5

The open-loop TF of a unity feed back system is given by the above expression.Find the value of‘K’for w hich the system is just stable.

Solution: The characteristic equation is

1

23

2K

9

(15+K)

0

2K

0

0

0

0

0

2K

(i) (ii)

K>0 192 – K > 0 K < 192

(iii)

(192 – K)(15+K) – 162K > 0 (for the max. value of K)

From this evaluate for K, Using, Considering the positive value of „K‟, So, 0 < K < 61.68

When the value of „K ‟ is 61.68 the system is just stable. (11)

Using Routh-Hurw itz criteria,find out the range of‘K’for w hich the system is stable.The characteristic equation is Solution: (i) K>0 1 (2K+3) (ii) 5K

10 0 Considering the positive value of „K‟,

10 T he range of K is „ (12)

0 ‟

A proposed control system has a system & a controller as shown. Access the stability of the system by a suitable m ethod.W hat are the ranges of‘K’for the system to be stable?

Solution:

The characteristic equation is

16

(1+K)

8

K 0

K (13)

0

(i) (ii)

K>0

The range of K is,

6

System Stability

No. of sign changes = 2, Two roots lie on RHS of S-plane. The system is Unstable.

(14)

Auxiliary Equation:

No sign changes in the column. T he system is “ Limitedly Stable” .

(15)

(i) (ii)

For the system to be stable the range of the K is,

(16)

The open-loop TF of a control system is given by

Find the range of

the gain constant ‘K’for stability using Routh-Hurwitz criteria. Solution: The characteristic equation is

(i) (ii) (iii) (iv)

The Range of K is the system to be stable.

for

Control Systems (17) 1

4

6

2

5

2

1.5

5

0

(i) (ii)

-1.666

2

0

(iii)

6.8

0

0

2

0

0

No. of sign changes = 2. Two roots lie on RHS of S-plane. The system is Unstable.

(18) Solution: 1

11

6

6

10

0

6

0

No sign changes. The system is Absolutely Stable.

(19) Solution: No. of sign changes = 1 The system is Unstable.

1

-5

2

-6

-2

0

-6

0

(20) +ve

1

2

4

+ve +ve

1

2

1

3

0

-ve

1

0

+ve

0

0

0

0

+ve

1

No. of sign changes = 2. The system is Unstable.

(21) +ve

2

6

1

+ve

1

3

1

-1

0

1

0

0

0

0

0

+ve +ve -ve +ve (22)

1

No. of sign changes = 2. The system is Unstable.

7

8

System Stability 1

-2

-7

-4

1

-3

-4

0

1

-3

-4

0

Auxiliary equation:

-1.5

-4

0

0

-16.66

0

0

0

-4

0

0

0

No. of sign changes = 1. The system is Unstable & Limitedly Stable.

1

5

8

4

2

8

8

0

1

4

4

0

(23) Auxiliary equation:

Auxiliary equation: 2

4

0

0

4

0

0

0

No. of sign changes = 0 The system is Limitedly Stable.

(24) 1

2

0.5

0(4)

0(4)

0(0)

1

0.5

0

2

0

0

0.5

0

0

Auxiliary Equation: No. of sign changes = 0 The system is Limitedly Stable.

(25) 1

2

2

Auxiliary Equation:

0.8

1.6

0

No. of sign changes = 2

-4

2

0

2

0

0

2

0

0

The system is Unstable.

(26) 1 34.5

7500 7500K

(i) (ii)

0 7500K

0

For a stable closed loop system there should not be any sign change among the elements of the Hurwitz table. This requires & T he range of „K ‟ is for the system to be stable.

column in the Routh-

Control Systems (27)

Given

,

9

.

Solution: The characteristic equation is 1

(21+K)

10

13K

(i) (ii)

0 13K (28)

The range of K is to be stable.

0

, for the system

,

Solution: The Characteristic equation is 1

15

2K

7

25+K

0

2K

0

2K

0

0

0

0

(i) (ii) (iii) To solve for the K value, By simplifying, Hence, the range of K is

(29) Solution: 1

5

15

20K

10

0

15

0

0

0

0

0

15 (30)

(i) (ii) (iii)

T he gain „K ‟ alw ays should be a real quantity. Hence this system is always Unstable.

D eterm ine the values of‘K’& ‘a’.Such that the system oscillates at a frequency of 2 rad/sec.

Solution:

The characteristic equation is , 1

(2+K)

a

(1+K)

becomes

0 (1+K)

(31)

0

From (1) & (2),

&

10

System Stability 1

2

3

-2

0

5

0

0

0

0

0

No. of sign changes = 2

1

5

The system is Unstable.

(32) 9

10

-9

No. of sign changes = 3

0

The system is Unstable.

0 0

0

0

0

1 9 21.33 15

24 24 21.33 15

24 24 21.33 15

23 15 0

7.5

15

0

0

0

0

0

15

0

0

0

1 5 7.2

9 9 0

4 20 28.8

36 36 0

9

0

36

36

0

0

0

0

0

0

0

0

(33) Auxiliary equation:

No. of sign changes = 2 The system is Unstable.

(34)

36

Solution:

No. of sign changes = 2

,

(35)

The Characteristic equation is 1 3

2 K

(i) (ii)

0 K

0

Auxiliary equation:

The range of K is

The system is Unstable.

Control Systems (36)

The open-loop transfer function of a unity feed back control system is given by, , using Routh-Hurwitz criteria. Discuss the stability of the closed loop-

controlsystem .D eterm ine the value of‘K’w hich w illcause sustained oscillations in the closed loop system. What are the corresponding oscillating frequencies? Solution: The characteristic equation is 1

69

12

198

(i) (ii)

0

52.5

(iii)

0 0

0

0

0

The Auxiliary equation for the

row is

When

Hence, (37)

A feed back system has open-loop transfer function

Determine the

maximum value of‘K’for stability ofthe closed-loop system. Solution: Generally control systems have very low Band width which implies that it has very low frequency range of operations. Hence for low frequency ranges, the term can be replaced by . i.e.,

The characteristic equation is

,

1 5

K 0

K

0

(i) (ii) The range of K is stable.

for the system to be

11

 

Control Systems  1

Root Locus Plots By: Sheshadri.G.S.

CIT, Gubbi. It gives complete dynamic response of the system. It provides a measure of sensitivity of roots to the variation in the parameter being considered. It is applied for single as well as multiple loop system. It can be defined as follows, It is the plot of the loci of the root of the complementary equation when one or more parameters of the open-loop Transfer function are varied, mostly the only one variable available is the gain ‘K’ The negative gain has no physical significance hence varying ‘K’ from ‘0’ to ‘∞’ , the plot is obtained called the “Root Locus Point”. Rules for the Construction of Root Locus  (1) The root locus is symmetrical about the real axis. (2) The no. of branches terminating on ‘∞’ equals the no. of open-loop pole-zeroes. (3) Each branch of the root locus originates from an open-loop pole at ‘K = 0’ & terminates at open-loop zero corresponding to ‘K = ∞’. (4) A point on the real axis lies on the locus, if the no. of open-loop poles & zeroes on the real axis to the right of this point is odd. (5) The root locus branches that tend to ‘∞’, do so along the straight line. Asymptotes making angle with the real axis is given by P = No. of poles

&

  180

0

,

Where, n=1,3,5,…………………

Z =No. of zeroes.

(6) The asymptotes cross the real axis at a point known as Centroid.

i.e.,

∑

∑

(7) The break away or the break in points [Saddle points] of the root locus or determined from the roots of the equation

0.

(8) The intersection of the root locus branches with the imaginary axis can be determined by the use of RouthHurwitz criteria or by putting ‘     ’ in the characteristic equation & equating the real part and imaginary to zero. To solve for ‘ ’ & ‘K’ i.e., the value of ‘ ’ is intersection point on the imaginary axis & ‘K’ is the value of gain at the intersection point. 180 (9) The angle of departure from a complex open-loop pole( ) is given by,

ßÞà Þßà  ßÞà Þßà ßÞà Þßà  ßÞà Þßà ßÞà Þßà  ßÞà Þßà ßÞà Þßà  ßÞà  

Designed By:

HPK Kumar ([email protected])

 

Control Systems

Bode Plots By: Sheshadri.G.S.

CIT, Gubbi. Sinusoidal transfer function is commonly represented by Bode Plot. It is a plot of magnitude against frequency. i.e., angle of transfer function against frequency. The following are the advantages of Bode Plot, (1) Plotting of Bode Plot is relatively easier as compared to other methods. (2) Low & High frequency characteristics can be represented on a single diagram. (3) Study of relative stability is easier as parameters of analysis of relative stability are gain & phase margin which are visibly seen on sketch. (4) If modification of an existing system is to be studied, it can be easily done on a Bode Plot. Initial Magnitude:

,

If

, , , , , , Phase Plot:

Magnitude Plot:

GCF

PCF

+ve PM

-ve GM line GCF

0 dB line PCF

GCF

PCF line

-ve PM GCF

0 dB line +ve GM PCF

1

2

Bode Plots Problems: (1) To find the angle for the quadratic term Solution:

Put,

(2) Determine the transfer function. Whose approximate plot is as shown. dB B C 40 dB A

D

Solution: The corner frequencies are

&

at

Therefore there must be a factor

, the slope changes from

. Since the initial slope is

to

, There must be a pole at the

origin i.e., at The slope at

changes from

to

due to a factor

To find the value of‘K’: B 20

A

C

(3) Find the open-loop TF of a system, whose approximate plot is as shown. dB A B C

.

Open-loop TF is,

.

Control Systems Solution: Corner frequencies are To find the value of‘K’: but

Since the Plot between

has a slope of &

factor is

, the corresponding factor is

, It is having a slope of

. Therefore at

, there must be a zero & the

in the numerator.

At

,the slop is changing from

to

& the corresponding factor is

in the Numerator. At

, the slope has changed from

to

due to a factor

in the denominator. Therefore the open-loop transfer function is,

(4) Find the open-loop TF of a system, whose approximate plot is as shown. dB

Solution:

(1)

Slope of the first line =

(2)

At

slope changes from

indicating the term to

indicating a term

in the

numerator. (3)

At

slope changes to

(4)

At

slope changes to

(5)

At

slope changes to

indicating a term indicating a term indicating a term

in the Numerator. in the Denominator. in the

Denominator. (6)

At

slope changes to

Denominator.

indicating a term

in the

3

Bode Plots

4

The Open-loop transfer function is,

(5) Find the open-loop TF of a system, whose approximate plot is as shown. dB A

B

Solution: To find the value of‘K’: Let

, be the origin.

(6) Derive the Transfer function of the system from the data given on the Bode diagram given below. dB

[[

Solution: dB,

Between

there is a

decrease of there is a decrease of

To find the value of‘K’:

Calculation of

:

Therefore, the Open-loop transfer function is,

Calculation of

:

Control Systems Exam ination Problem (M ar/Apr’99): (7) The sketch given shows the Bode Magnitude plot for a system. Obtain the Transfer function. dB

D

40 A B

(Z) E

(P)

(P) (DZ)

C Solution:

Since the initial slope is

there must be zero at the origin. &

Examination Problem (Sep/O ct’99): (8) Estimate the Transfer function for the Bode Magnitude plot shown in figure. dB

(Z) Solution:

5

6

Bode Plots Examination Problem (97): (9) The bode plot (magnitude) of a unity feed back control system is as shown in the fig. Obtain the phase plot. dB

Solution: