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MAHALAKSHMI

ENGINEERING COLLEGE

TIRUCHIRAPALLI-621213.

QUESTION BANK

DEPARTMENT: ECE SEMESTER - III

SUBJECT CODE: EC2255 SUBJECT NAME: CONTROL SYSTEMS

UNIT-1

SYSTEMS AND THEIR REPRESENTATION

1. Write Mason’s gain formula. (May’ 12)

Masons Gain Formula

The relationship between an input variable and an output variable of a signal flow

graph is given by the net gain between input and output nodes and is known as overall gain

of the system. Masons gain formula is used to obtain the overall gain (transfer function) of

signal flow graphs.

Gain P is given by

k

kkPP1

Where, Pk is gain of kth forward path,

∆ is determinant of graph

∆=1-(sum of all individual loop gains)+(sum of gain products of all possible

combinations of two nontouching loops – sum of gain products of all possible combination of

three nontouching loops) + ∙∙∙

∆k is cofactor of kth forward path determinant of graph with loops touching kth forward

path. It is obtained from ∆ by removing the loops touching the path Pk.

2. What is mathematical model of a system? (May’12,06)

The weight of mechanical translational system is represented by the element Mass(M)

The elastic deformation of the body can be represented by a Spring ( K)

The frictional existing in rotating mechanical system can be represented by Dash pot (B)

3. What is control system? (May’11)

A control system is an interconnection of components or devices so that the output of

the overall system will follow as closely as possible a desired signal.

4. What are the two major types of control systems? (Nov’10)

Control systems are basically classified as –

Open-loop control system

Closed-loop control system

5. Why negative feedback is preferred in control systems? (Nov’10)

Negative Feedback systems have the following features:

- reduced effect of nonlinearities and distortion

- Increased accuracy

- Increased bandwidth

- Less sensitivity to variation of system parameters

- Tendency towards oscillations

- Reduced effects of external disturbance

6. Distinguish between open loop and closed loop systems (Nov’11)

In open-loop system the control action is independent of output. In closed-loop

system control action is somehow dependent on output. Each system has at least two things

in common, a controller and an actuator (final control element). The input to the controller is

called reference input. This signal represents the desired system output.

Closed-loop systems are also called feedback control systems. Feedback is the

property of the closed-loop systems which permits the output to be compared with the input

of the system so that appropriate control action may be formed as a function of inputs and

outputs.

7. What is linear system? (Nov’11)

8. Define transfer function. (May’09)

The transfer function of an LTI system is the ratio of Laplace transform of the output

variable to the Laplace transform of the input variable assuming zero initial conditions.

9. Write force balance equation of ideal spring, ideal mass. (Dec’09)

10. What is an analogous system? (Dec’09,05)

Analogous system may have entirely different physical appearances. For example, a

given electrical circuit consisting of resistance, inductance and capacitance may be

analogous to a mechanical system consisting of a suitable combination of mass, dashpot

and spring.

1. Write the differential equations governing the Mechanical system shown in fig and

determine the transfer function. (16) (May’12, Nov’11, May’06)

Solution:

Transfer function = )(

)(

sF

sX

Step 1:

Free Body diagram for M1

The opposing force acting on Mass M1 are fm1, fb, fk1, fb and fk

w.k.t. 2

1

2

11dt

Xdmfm

dt

dXBfb

111

111 Xkf k

dt

XXdBfb

1

XXkf k 1

By Newton’s 2nd Law:

Applied force = Opposing Force

0111 kkbbm fffff - - - - - (1)

011

111

12

1

2

1 XXKdt

XXdBXK

dt

dXB

dt

XdM - - - - - (2)

Taking Laplace transform for equation (2)

0))()(()())()(()()( 1111111

2

1 sXsXKsXKsXsXBSsSXBsXM

0))()()())()()()( 1111111

2

1 sKXsKXsXKsBsXsBsXssXBsXsM

])[()]()()[( 11

2

11 KBssXKKsBBsMsX

)()(

)()()(

11

2

1

1KKsBBsM

KBssXsX - - - - - (3)

Step 2:

Free Body Diagram for M2

The opposing force acting on Mass M2 are fm2, fb, fk, fb2

w.k.t. 2

2

22dt

XdMfm

)( 1XXdt

dBfb

)( 1XXKf k

dt

dXBfb 22

By Newton’s 2nd Law:

)(22 tfffff bkbm

)(2112

2

2 tfdt

dxBXXKXX

dt

dB

dt

XdM - - - - - (4)

Taking Laplace transform:

)()(())()(())()(()( 211

2

2 sFsXsBsXsXKsXsXBssXsM

Subs X1(s) value in the above equation:

)()()(

)()(])()[(

11

2

1

2

2

2

2 sFKKsBBsM

KBssXKsBBsMsX

)()()(

)()()(][)([)(

11

2

1

11

2

12

2

2 sFKKsBBsM

KBsKKsBBsMKsBBsMsX

2

2

2

211

2

1

11

2

1

)(])()][()([

)()(

)(

)(

kBsKsBBsMKKsBBsM

KKsBBsM

sF

sX

Result:

The differential equations governing the system are

1. 011

111

12

1

2

1 XXKdt

XXdBXK

dt

dXB

dt

XdM

2. )(2112

2

2 tfdt

dxBXXKXX

dt

dB

dt

XdM

Transfer function of the system is

1. 2

2

2

211

2

1

11

2

1

)(])()][()([

)()(

)(

)(

kBsKsBBsMKKsBBsM

KKsBBsM

sF

sX

2.Explain Synchros and its types. (10) (May’12)

Figure: Synchro error detector

A synchro is an electromagnetic transducer that is used to convert angular shaft

position into an electric signal. The basic element of a synchro is a synchro transmitter

whose construction is very similar to that of the 3- alternator.

An ac voltage is applied to the rotor winding through slip-rings. The schematic diagram of

synchro synchro transmitter- control transformer pair is shown above.

( ) sinr r cv t V t ;

1

2

3

( ) sin cos( 120 )

( ) sin cos

( ) sin cos( 240 )

s r c

s r c

s r c

v t KV t

v t KV t

v t KV t

;

1 2

2 3

3 1

( ) 3 sin sin( 240 )

( ) 3 sin sin( 120 )

( ) 3 sin sin

s s r c

s s r c

s s r c

v t KV t

v t KV t

v t KV t

When 0 , 3 1( ) 0s sv t and maximum voltage is induced on S2 coil. This position of the rotor

is defined as the “electrical zero” of the transmitter and used as reference position of the

rotor.

The output of the synchro transmitter is applied to the stator winding of a “synchro control

transformer”. Circulating current of the same phase but of different magnitude flows through

the two sets of stator coil. The pair acts as an error detector.

The voltage induced in the control transformer rotor is proportional to the cosine of the angle

between the two rotors and is given by,

( ) sin cos ; where, 90r ce t K V t .

( ) sin cos(90 ) sin sin( ) ( )sinr c r c r ce t K V t K V t K V t …(01)

The equation above holds for small angular displacement.

Thus the synchro transmitter-control transformer pair acts as an error detector which gives a

voltage signal at the rotor terminal of the control transformer proportional to the angular

difference between the shaft positions.

Equation (01) is represented graphically in figure below for an arbitrary time variation of (

).

It is seen that the output of the synchro error detector is a modulated signal, where the ac

signal applied to the rotor of synchro transmitter acts as carrier and the modulating signal is,

( ) ( ); m s s re t K K K V

3.Write the rules for block diagram reduction techniques. (May’12, Nov’06)

Block diagram reduction technique

Block diagram:

A block diagram is a short hand, pictorial representation of cause and effect

relationship between the input and the output of a physical system. It characterizes the

functional relationship amongst the components of a control system.

Lower case letters usually represent functions of time. Upper case letters

denote Laplace transformed quantities as a function of complex variable s or Fourier

transformed quantity i.e. frequency function as function of imaginary variable jωs .

Summing point: It represents an operation of addition and / or subtraction.

Negative feedback: Summing point is a subtractor.

Positive feedback: Summing point is an adder.

Stimulus: It is an externally introduced input signal affecting the controlled output.

Take off point: In order to employ the same signal or variable as an input to more than

block or summing point, take off point is used. This permits the signal to proceed unaltered

along several different paths to several destinations.

4. (i) Derive the transfer function for Armature controlled DC motor. (May’11, Nov’11)

A. Transfer Function of Armature Controlled DC Motor

; ab b a a a b

dide K L R i e e

dt dt

In Laplace domain,

2

0

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

b b

a a a b

T a

E s K s s

L s R I s E s E s

Js f s s K I s

;

0

( ) ( )

( ) ( )( )

T

a a T b

KsG s

E s s R sL Js f K K

Neglecting aL ,

2

0

/ /( )

( / ) ( ) ( 1)

T a T a m

T b a m

K R K R KG s

Js s f K K R s Js f s s;

where, 0 /T b af f K K R and / ; / .m T a mK K R f J f

mK and m are called the motor gain and time constant respectively. These two parameters

are usually supplied by the manufacturer.The block diagram model is,

1;f f M f f a T aK i T K K i i K i

2

02 M T a

d dJ f T K i

dt dt

B. Transfer Function of a Field-controlled DC Motor

Figure 1

'

1 1

2

2

TM a f f a f

f

f f f

M T f

T K i K K i i K i

diL R i e

dt

d dJ f T K i

dt dt

;

2

( ) ( ) ( )

( ) ( ) ( ) ( )

f f f

M T f

L s R I s E s

Js fs s T s K I s

We obtain,

( )( )

( ) ( )( )

T

f f

KsG s

E s s L s R Js f

5. Explain the working of AC servomotor in control systems.

A.C. Servomotor

For low power application a.c. motors are preferred, because of their light weight,

ruggedness and no brush contact. Two phase induction motors are mostly used in control

system.

The motor has two stator windings displaced 90 electrical degree apart. The voltages

applied to the windings are not balanced. Under normal operating conditions a fixed voltage

from a constant voltage source is applied to one phase. The other phase, called control

phase, is energized by a voltage of variable magnitude which is 90 out of phase w.r.t. the

voltage of fixed phase.

The torque speed characteristic of the motor is different from conventional motor. X / R ratio

is low and the curve has negative slope for stabilization.

The torque-speed curve is not linear. But we assume it as linear for the derivation of transfer

function. The troque is afunction of both speed and the r.m.s. control voltage, ie.,

( , )MT f E .

Using Tailor series expansion about the normal operating point0 0 0( , , )MT E we get,

0 0

0 0

0 0 0( ) ( )M MM M

E E E E

T TT T E E

E

For position control system, 0 0E ,0 0 , 0MT

Thus, the above equation may be simplified as,

0 ;MT kE m J f where,

0

0

M

E E

Tk

Eand

0

0

M

E E

Tm .

Performing Laplace transform, 2

0( ) ( ) ( ) ( )kE s ms s Js s f s

Or, 2

0

( )( )

( ) ( ) ( 1)

m

m

Ks kG s

E s Js f m s s s; where,

0

m

kK

f mand

0

m

J

f m.

Since ‘m’ is negative the transient part is decaying as m is positive. If ‘m’ would positive and

0m f the transient part will increase with time and the system would be unstable. k and m

are the slope of the torque-voltage and torque-speed curve.

6. Determine the overall transfer function C(S)/R(S) for the system shown in fig.

(Nov’10,05)

To simplify the inner feedback loop to obtain the following block diagram.

To combine the two series blocks into one.

To obtain the transfer function for the standard feedback system.

7.Find the transfer function of the closed-loop system below.

To simplify the inner feedback blocks.

To get the following block diagram.

To combine the two sets of series blocks.

Calculate the overall transfer function of the system

8. Find the overall gain C(s) / R(s) for the signal flow graph shown below

(May’09)

Figure 2 Signal flow graph of example

G1 C(s) R(s)

G7 G6

-H1

G2 G3 G4 G5

-H2

X1 X2 X3 X4 X5

1

There are three forward paths.

The gain of the forward path are: P1=G1G2G3G4G5

P2=G1G6G4G5

P3= G1G2G7

There are four loops with loop gains:

L1=-G4H1,

L2=-G2G7H2,

L3= -G6G4G5H2 ,

L4=-G2G3G4G5H2

There is one combination of Loops L1 and L2 which are non-touching with loop gain product

L1L2=G2G7H2G4H1

∆ = 1+G4H1+G2G7H2+G6G4G5H2+G2G3G4G5H2+ G2G7H2G4H1

Forward path 1 and 2 touch all the four loops.

Therefore ∆1= 1, ∆2= 1.

Forward path 3 is not in touch with loop1.

Hence, ∆3= 1+G4H1.

The transfer function T =

2174225432254627214

14721654154321332211

1

1

HHGGGHGGGGHGGGHGGHG

HGGGGGGGGGGGGGPPP

sR

sC

9. Draw the analogous electric circuit of the system below using f-i analogy.

10. Find the overall gain C(s) / R(s) for the signal flow graph shown below

(May’09)

Find the gains 1

3

2

5

1

6 ,,X

X

X

X

X

X for the signal flow graph shown in Fig.4.

Figure 3 Signal flow graph of MIMO system

b

a d c f e

-h

-g

-i

X1 X6 X5

X4 X3 X2

Case 1: 1

6

X

X

There are two forward paths.

The gain of the forward path are: P1=acdef

P2=abef

There are four loops with loop gains:

L1=-cg,

L2=-eh,

L3= -cdei,

L4=-bei

There is one combination of Loops L1 and L2 which are nontouching with loop gain product

L1L2=cgeh

∆ = 1+cg+eh+cdei+bei+cgeh

Forward path 1 and 2 touch all the four loops.

Therefore ∆1= 1, ∆2= 1.

The transfer function

T = cgehbeicdeiehcg

abefcdefPP

X

X

1

2211

1

6

Case 2: 2

5

X

X

The modified signal flow graph for case 2 is shown in Fig.5.

Figure 4 Signal flow graph of example 4 case 2

The transfer function can directly manipulated from case 1 as branches a and f are removed

which do not form the loops. Hence,

The transfer function

T=cgehbeicdeiehcg

becdePP

X

X

1

2211

2

5

Case 3: 1

3

X

X

The signal flow graph is redrawn to obtain the clarity of the functional relation as shown in

Fig.6.

b

1 d c 1 e

-h

-g

-i

X2 X5 X5

X4 X3 X2

Figure 5 Signal flow graph of example 4 case 3

There are two forward paths.

The gain of the forward path are: P1=abcd

P2=ac

There are five loops with loop gains:

L1=-eh, L2=-cg, L3= -bei, L4=edf, L5=-befg

There is one combination of Loops L1 and L2 which are nontouching with loop gain product

L1L2=ehcg

∆ = 1+eh+cg+bei+efd+befg+ehcg

Forward path 1 touches all the five loops. Therefore ∆1= 1.

Forward path 2 does not touch loop L1. Hence, ∆2= 1+ eh

The transfer function T = ehcgbefgefdbeicgeh

ehacabefPP

X

X

1

12211

1

3

a e b f

-h

-g

-i

X1 X5

X4 X3

X2

c

d

X3 1

MAHALAKSHMI

ENGINEERING COLLEGE

TIRUCHIRAPALLI-621213.

QUESTION BANK

DEPARTMENT: EEE SEMESTER - III

SUBJECT CODE: EE2253 SUBJECT NAME: CONTROL SYSTEMS

UNIT-2

TIME DOMAIN ANALYSIS

PART – A

1. What is transient and steady state response? (May’ 12) 1. Transient Response:

It is the response of the system when the input changes from one state to

another

2. Steady State Response:

It is the response as time t approaches infinity.

2. Name the test signals used in time response analysis. (May’ 12)

Name of the Signal Time domain r(t) Laplace Transform

Step A A / s

Unit Step 1 1 / s

Ramp At A / s2

Unit Ramp t 1 / s2

Parabolic At2 / 2 A / s3

Unit Parabolic t2 / 2 1 / s3

Impulse δ (t) 1

3. How is system classified depending on the value of damping? (May’11) Definition:

It is defined as the ratio of the actual damping to the critical damping

The response c(t) of second order system depends on the value of damping

ratio.

The second order system can be classified into four types depending upon

the value of damping ratio

1. Undamped system ζ = 0

2. Under damped system 0 < ζ < 1

3. Critically damped system ζ = 1

4. Over damped system ζ > 1

4. Sketch the response of a second order under damped system. (May’11)

r (t) c(t)

1 1

Input t Output Response t

5. List the time domain specifications. (Nov’11)

The performance characteristics of a controlled system are specified in terms of the

transient response to a unit step i/p since it is easy to generate & is sufficiently drastic.

The transient response of a practical C.S often exhibits damped oscillations before

reaching steady state. In specifying the transient response characteristic of a C.S to unit

step i/p, it is common to specify the following terms.

1) Delay time (td)

2) Rise time (tr)

3) Peak time (tp)

4) Max over shoot (Mp)

5) Settling time (ts)

6. What are generalized error & static error constants? (Nov’11) Static Position Error Constant

The steady state error of the system for a unit step input is

0G1

1

s

1

sG1

slime

0sss

The static position error constant Kp is defined by

0GsGlimK0sp

Thus, the steady state error in terms of the static position error constant Kp is given

by

p

ss K1

1e

For a type 0 system,

K

1sp

T1s2

T1s1

T

1sm

T1sb

T1sa

TK

limK0s

p

and

K1

1e

ss

For a type 1 or higher system

1N

1sp

T1s2

T1s1

Ts

1sm

T1sb

T1sa

TK

limKN

0sp

and ss

e

Response of a feedback control system to a step input involves a steady state error if

there is no integration in the feed forward path. If a zero steady state error for a step input is

desired, the type of the system must be one or higher.

7. Define position, velocity error constants. (Nov’10)

Type of

system

Error constants Steady state error ess

Kp Kv Ka Unit step

input

Unit ramp

input

Unit parabolic

input

0 K 0 0 1/(1+K)

1 K 0 0 1/K

2 K 0 0 1/K

3 0 0 0

8. Define damping ratio (May’10) It is defined as the ratio of the actual damping to the critical damping

9. Define Peak overshoot & Peak time. (May’10)

Peak time :- (tp)

It is the time required for the response to reach the 1st of peak of the overshoot.

Maximum over shoot :- (MP)

It is the maximum peak value of the response curve measured from unity. The

amount of max over shoot directly indicates the relative stability of the system.

10. Consider a first order system given by G(S)=C(S)/R(S) = K/S+a .Plot its unit step response. (May’09) C(t)

r(t) 0.95

0.865

0.632

t=0 t 0 T 2T 3T t

1. Draw the response of second order system for under damped case and when

input is unit step. (May’ 12)

Response of Under Damped Second Order System:

For Unit Step Input:

The standard form of closed loop transfer function of second order system is

22

2

2)(

)(

nn

n

sssR

sC

For under damped system, 0 < ζ < 1 and root of the denominator are complex

conjugate.

The roots of the denominator are

12

nns

Since ζ < 1, ζ2 is also less than 1

Therefore, the damped frequency of oscillation, 21nd

dn js

The response in s – domain, 22

2

2)()(

nn

n

sssRsC

For unit step input, r(t) = 1 and R(s) = 1/s

)2(

)(22

2

nn

n

ssssC

)2()2(

)(2222

2

nnnn

n

ss

CBs

s

A

ssssC

scBsssA nnn )()2( 222

1. Put s = 0

22

nn A

A = 1

2. Equating the co-efficient of s2.

CsBsssA nnn

2222 )2(

0 = A + B

0 = 1 + B

B = -1

3. Equating co-efficient of s

0 = 2Aωnζ + C

= 2ωnζ + C

C = - 2ωnζ

22 2

21)(

nn

n

ss

s

ssC

Let us add and subtract 22

n to denominator of the second term

222222 2

21)(

nnnn

n

ss

s

ssC

222222 2

21

nnnn

n

ss

s

s

2222)(

21

nnn

n

s

s

s

)1()(

21222

nn

n

s

s

s

22)(

21

dn

n

s

s

s

22 )()(

1

dn

n

dn

n

ss

s

s

Multiply and divide by ωd

2222 )()(

1)(

dn

d

d

n

dn

n

ss

s

ssC

The response in time domain is given by

2222

11

)()(

1)}({)(

dn

d

d

n

dn

n

ss

s

sLsCLtc

tete d

t

d

nd

t nn sincos1

tte d

d

nd

tn sincos1

tte d

n

nd

tn sin1

cos12

tte dd

tn sin1

cos12

tte

tc dd

tn

sincos11

1)( 2

2

Note:

On constructing on right angle triangle with ζ and21 , we get

1 21

1

ζ

sin θ = 21

cos θ = ζ

2-1tan

θ

tt

etc dd

tn

sincoscossin1

1)(2

)sin(

11)(

2t

etc d

tn

21 1

tan

For closed loop under damped second order system:

Unit step response = )sin(

1 2t

ed

tn

Step Response = )sin(

1 2t

eA d

tn

r (t) c(t)

1 1

Input t Output Response t

The response of the under damped second order system for unit step input sketched and

observed that the response oscillates before settling to a final value. The oscillation depends

on the value of damping ratio.

2. Derive the expressions for Rise time, Peak time, Peak overshoot, delay time

(May’ 12)

The performance characteristics of a controlled system are specified in terms of the

transient response to a unit step i/p since it is easy to generate & is sufficiently drastic.

The transient response of a practical C.S often exhibits damped oscillations before

reaching steady state. In specifying the transient response characteristic of a C.S to unit

step i/p, it is common to specify the following terms.

1) Delay time (td)

2) Rise time (tr)

3) Peak time (tp)

4) Max over shoot (Mp)

5) Settling time (ts)

Response curve

1) Delay time :- (td)

It is the time required for the response to reach 50% of its final value for

the 1st time.

2) Rise time :- (tr)

It is the time required for the response to rise from 10% and 90% or 0%

to 100% of its final value. For under damped system, second order system the 0 to

100% rise time is commonly used. For over damped system, the 10 to 90% rise time is

commonly used.

3) Peak time :- (tp)

It is the time required for the response to reach the 1st of peak of the overshoot.

4) Maximum over shoot :- (MP)

It is the maximum peak value of the response curve measured from unity. The

amount of max over shoot directly indicates the relative stability of the system.

5) Settling time :- (ts)

It is the time required for the response curve to reach & stay with in a range about

the final value of size specified by absolute percentage of the final value (usually 5% to

2%). The settling time is related to the largest time const., of C.S.

3. A unity feedback control system has an open loop transfer function G(S)=

10/S(S+2).Find the rise time, percentage over shoot, peak time and settling

time.

4.For a unity feedback control system the open loop transfer function

G(S) = 10(S+2)/ S2(S+1).Find (a) position, velocity and acceleration error constants.

(Nov’11)

5. Explain P, PI, PID, PD controllers (Nov’09,May’09)

An automatic controller compares the actual value of the system output with the reference

input (desired value), determines the deviation, and produces a control signal that will

reduce the deviation to zero or a small value. The manner in which the automatic controller

produces the control signal is called the control action. The controllers may be classified

according to their control actions as

1) Two position or on-off controllers

2) Proportional controllers

3) Integral controllers

4) Proportional-plus- integral controllers

5) Proportional-plus-derivative controllers

6) Proportional-plus-integral-plus-derivative controllers

A two position controller has two fixed positions usually on or off.

A proportional control system is a feedback control system in which the output forcing

function is directly proportional to error.

A integral control system is a feedback control system in which the output forcing

function is directly proportional to the first time integral of error.

A proportional-plus-integral control system is a feedback control system in which the

output forcing function is a linear combination of the error and its first time integral.

A proportional-plus-derivative control system is a feedback control system in which

the output forcing function is a linear combination of the error and its first time

derivative.

A proportional-plus-derivative-plus-integral control system is a feedback control

system in which the output forcing function is a linear combination of the error, its first

time derivative and its first time integral.

Many industrial controllers are electric, hydraulic, pneumatic, electronic or their

combinations. The choice of the controller is based on the nature of plant and operating

conditions. Controllers may also be classified according to the power employed in the

operation as

1) Electric controllers

2) Hydraulic controllers

3) Pneumatic controllers

4) Electronic controllers.

The block diagram of a typical controller is shown in Fig.1. It consists of an automatic

controller, an actuator, a plant and a sensor. The controller detects the actuating error signal

and amplifies it. The output of a controller is fed to the actuator that produces the input to the

plant according to the control signal. The sensor is a device that converts the output variable

into another suitable variable to compare the output to reference input signal. Sensor is a

feedback element of the closed loop control system.

Figure 6 Block diagram of control system

Two Position Control Actions

In a two position control action system, the actuating element has only two positions which

are generally on and off. Generally these are electric devices. These are widely used are

they are simple and inexpensive. The output of the controller is given by Eqn.1.

0

0

2

1

teU

teUtu ….. (1)

Where, U1 and U2 are constants (U2= -U1 or zero)

The block diagram of on-off controller is shown in Fig. 2

−

Actuating

Error signal e(t) Reference

Input r(t)

Error

Detector

Feed back

signal b(t)

Controller

Output u(t)

−

Output c(t)

automatic controller

actuating

error signal e(t) reference

input r(t)

error

detector Amplifier Actuator Plant

Sensor feed back

signal b(t)

Controller

output u(t)

Proportional Control Action

The proportional controller is essentially an amplifier with an adjustable gain. For a

controller with proportional control action the relationship between output of the controller

u(t) and the actuating error signal e(t) is

teKtup

..... (2)

Where, Kp is the proportional gain.

Or in Laplace transformed quantities

pK

sE

sU ….. (3)

Whatever the actual mechanism may be the proportional controller is essentially an

amplifier with an adjustable gain. The block diagram of proportional controller is shown in

Fig.3.

Figure 7 Block diagram of a proportional controller

Integral Control Action

The value of the controller output u(t) is changed at a rate proportional to the actuating error

signal e(t) given by Eqn.4

t

0ii

dtteKtuor teKdt

tdu …..(4)

Where, Ki is an adjustable constant.

The transfer function of integral controller is

s

K

sE

sU i ….(5)

−

Actuating


Input r(t)

Error

Detector

Feed back

signal b(t)

Controller

Output u(t)

Kp

If the value of e(t) is doubled, then u(t) varies twice as fast. For zero actuating error,

the value of u(t) remains stationary. The integral control action is also called reset control.

Fig.4 shows the block diagram of the integral controller.

Figure 8 Block diagram of an integral controller

Proportional-plus-Integral Control Action

The control action of a proportional-plus-integral controller is defined by

t

0i

p

pdtte

T

KteKtu …..(6)

The transfer function of the controller is

sT

11K

sE

sU

i

p .....(7)

Where, Kp is the proportional gain, Ti is the integral time which are adjustable.

The integral time adjusts the integral control action, while change in proportional gain

affects both the proportional and integral action. The inverse of the integral time is called

reset rate. The reset rate is the number of times per minute that a proportional part of the

control action is duplicated. Fig.5 shows the block diagram of the proportional-plus-integral

controller. For an actuating error of unit step input, the controller output is shown in Fig.6.

−

Actuating


Input r(t)

Error

Detector

Feed back

signal b(t)

Controller

Output u(t)

s

Ki

Figure 9 Block diagram of a proportional-plus-integral control system

Figure 10 Response of PI controller to unit actuating error signal

Proportional-plus-Derivative Control Action

The control action of proportional-plus-derivative controller is defined by

dt

tdeTKteKtu

dpp …..(8)

The transfer function is

sT1KsE

sUdp

…..(9)

Where, Kp is the proportional gain and Td is a derivative time constant.

Both, Kp and Td are adjustable. The derivative control action is also called rate

control. In rate controller the output is proportional to the rate of change of actuating error

signal. The derivative time Td is the time interval by which the rate action advances the effect

of the proportional control action. The derivative controller is anticipatory in nature and

t 0

Kp

u(t)

2Kp

proportional only

P-I control action

Ti

−

Actuating


Input r(t)

Error

Detector

Feed back

signal b(t)

Controller

output u(t)

sT

sT1K

i

ip

amplifies the noise effect. Fig.7 shows the block diagram of the proportional-plus-derivative.

For an actuating error of unit ramp input, the controller output is shown in Fig.8

Figure 11 Block diagram of a proportional-plus-derivative controller

Figure 12 Response of PD controller to unit actuating error signal

Proportional-plus-Integral-plus-Derivative Control Action

It is a combination of proportional control action, integral control action and derivative control

action. The equation of the controller is

dt

tdeTKdtte

T

KteKtu

dp

t

oi

p

p …..(10)

or the transfer function is

sTsT

11K

sE

sUd

i

p …..(11)

t 0

u(t)

proportional only

P-D control action

Td

−

actuating

error signal e(t) reference

input r(t)

error

detector

feed back

signal b(t)

Controller

output u(t)

ST1KDP

Where, Kp is the proportional gain, Ti is the integral time, and Td is the derivative time. The

block diagram of PID controller is shown in Fig.9. For an actuating error of unit ramp input,

the controller output is shown in Fig.10.

Figure 13 Block diagram of a proportional-plus-integral-plus-derivative controller

Figure 14 Response of PID controller to unit actuating error signal

11. Consider a open loop transfer function G(S) of second order system with unity

feedback system. Express damping ratio of the closed loop system in terms of

a, wn, and δ, where a is a constant, wn is the natural frequency of oscillations

and δ is the damping ratio of open loop system.(nov 11

t 0

u(t)

proportional only

PD control action PID control action

−

actuating

error signal e(t)

reference

input r(t)

error

detector

feed back

signal b(t)

Controller

output u(t)

sT

sTTsTK

d

diip

21

UNIT – III

FREQUENCY DOMAIN ANALAYSIS

PART – A

1. What is frequency response analysis? (May’ 12)

The Frequency response is the steady state response (output) of a system when the

input to the system is a sinusoidal signal. The frequency response of a system is normally

obtained by varying the frequency of the input signal by keeping the magnitude of the input

signal at a constant value.

2. Define gain cross over frequency? (Nov’11) The phase cross over frequency is the frequency at which the magnitude of the open loop transfer function is unity.

3. Define gain cross over frequency? (Nov’11) The phase cross over frequency is the frequency at which the phase of the open loop

transfer function is 180.

4. Define Phase Margin? (Nov’10) The phase margin, γ is the amount of additional phase lag at the gain cross

over frequency, Wgc required to bring the system to the verge of instability. It is given

by 180 gc

5. Define Gain Margin? (Nov’10)

The gain margin, Kg is defined as the reciprocal of the magnitude of open loop transfer function at phase cross over frequency, Wpc

Gain Margin Kg=1

(pcat

G j

6. How do you calculate the gain margin from the polar plot? (May’10)

Gain Margin (GM): o The gain margin is the reciprocal of magnitude at the frequency at which the

phase angle is -1800

.

o In terms of dB

10 10 10

120log 20log | ( ) | 20log ( )

| ( ) |GM in dB G jwc x

G jwc

7. How do you calculate the gain margin from the polar plot? (May’10) Phase margin is that amount of additional phase lag at the gain crossover

frequency required to bring the system to the verge of instability (marginally stabile)

Φm=180

0

+Φ

Where

Φ=∠G(jωg)

if Φm>0 => +PM (Stable System)

if Φm<0 => -PM (Unstable System)

8. How do you find the stability of the system by using polar plot? (May’10) Stable: If critical point (-1+j0) is within the plot as shown, Both GM & PM are +ve

Unstable: If critical point (-1+j0) is outside the plot as shown, Both GM & PM are -ve

Marginally Stable System: If critical point (-1+j0) is on the plot as shown, Both GM & PM are ZERO

9. What is cut off rate? (Nov’09) The slop of the log magnitude curve near the cut off frequency is called cut –

off rate.

10. Consider the first order system G(S) = 1/1+ST .Draw its polar plot. (Nov’09)

PART B

1. Define resonant peak and resonant frequency. (May’09)

Peak response Mp :

The peak response Mp is defined as the maximum value of M( ) that is given in

Eq.(1.4). In general, the magnitude of Mp gives an indication of the relative stability of

a feed back control system. Normally, a large Mp corresponds to a large peak

overshoot in the step response. For most design problems it is generally accepted

that an optimum value Mp of should be somewhere between 1.1 & 1.5.

Resonant frequency p :

The resonant frequency p is defined as the frequency at which the peak resonance

Mp

2. Sketch bode plot for the following transfer function and determine the

system gain K for the gain cross over frequency to be 5rad/sec

G(S) = Ks2/(1+0.2s)(1+0.02s)

Solution:

The sinusoidal transfer function G(j ω) is obtained by replacing S by j ω in the given S-

domain transfer function

G(j ω)=K(j ω)2/(1+0.2 j ω)(1+0.002 j ω)

Let K=1, G(j ω)= (j ω)2/(1+0.2 j ω)(1+0.002 j ω)

MAGNITUDE PLOT

The corner frequencies are ωc1=1/0.2=5rad/sec and ωc2=1/0.02=50rad/sec

The various values of G(j ω) are below tabulation , in the increasing order of their corner

frequencies. also slope contributed by each term and the change in slope at the corner

frequency

TREM CORNER

FREQUENCY

SLOPE IN db/dec CHANGE IN

SLOPE db/dec

(jω)2 ----- +40

1/(1+j0.2) ωc1=1/0.2=5 -20 40-20=20

ωc2=1/0.02=50 -20 20-20

Choose a low frequency ωl such that ωl< ωc1& Choose a high frequency ωh such that

ωh> ωc2.Let ωl=0.5 rad/sec and ωh=100 rad/sec

Let A=|G(jω)| in db

Let us calculate A at ωl, , ωc1, ωc2, ωh

At ω= ωl, A=20 log|(jω)2| =20 log(ω2)=20 log(0.5)2= -12db

At ω= ωc1, A=20 log|(jω)2|=20log(5)2=28db

At ω= ωc2, A=[slope from ωc1 to ωc2×log ωc2/ωc1]+A(at ω= ωc1)

=20log50/5+28=48db

At ω= ωh,A=[slope from ωc2 to ωh×log ωh/ωc2]+A(at ω= ωc2)

=0×log (100/50)+48=48db

Let points a,b,c,d be the points corresponding to frequencies ωl, , ωc1, ωc2, ωh respectively on

the magnitude plot.

PHASE PLOT

The phase angle G(jω) as a function of ω is given by

Φ= G(jω)= 180˚-tan-10.2ω-tan-10.02 ω

The phase angle of G(jω) are calculated for various values of ω

ω rad/sec Tan-1(0.2 ω) deg Tan-1(0.02 ω) deg Φ= G(jω)

0.5 5.7 0.6 173.7≈174

1 11.3 1.1 167.6≈168

5 45 5.7 129.3≈130

10 63.4 11.3 105.3≈106

50 84.3 45 50.7≈50

100 87.1 63.4 29.5≈30

CALCULATION OF K

Given that the gain crossover frequency is 5rad/sec

At ω=5,gain= 28db

At ω=5,frequency the db gain should be zero

20log K=-28db

Log K=-28/20

K=10-28/20=0.0398

NOTE: The frequency ω rad/sec is a corner frequency .hence in the exact plot the db gain

at ω=5rad/sec will be 3db less than the approximate plot.therefore for exact plot the

20logK=-25

LogK=-25/20=>K=10-25//20=0.056

3. Construct the polar given transfer function

Example

Problem : Construct the polar plot for the (critically damped system)

defined by :

Solution :

Limiting conditions :

(i)

(ii)

(iii)

0j1)j(T:0

o

2

180j

1

e0)j(T.e.i

)j(T:

2

j)j(T:1

Re-0.2 0 0.2 0.4 0.6 0.8 1

-0.6

-0.4

-0.2

Im

2)1s(

1)s(T

2)j1(

1)j(T

w 0 0.25 0.5 0.75 1.0 1.5 2.0 3.0 5.0

T(jw) 1

+j0

0.83

-j0.44

0.48

-j0.64

0.18

-j0.61

0

-j0.5

-0.12

-j0.28

-0.12

-j0.16

-0.08

-j0.06

-0.04

-j0.01

4. Draw the bode plot given transfer function.

5. For the following T.F draw the Bode plot and obtain Gain cross over

frequency (wgc) ,Phase cross over frequency , Gain Margin and Phase

Margin.

G(s) = 20 / [s (1+3s) (1+4s)]

The sinusoidal T.F of G(s) is obtained by replacing s by jw in the given T.F

G(jw) = 20 / [jw (1+j3w) (1+j4w)]

Corner frequencies: wc1= 1/4 = 0.25 rad /sec ;

wc2 = 1/3 = 0.33 rad /sec

Choose a lower corner frequency and a higher Corner frequency

wl= 0.025 rad/sec ; wh = 3.3 rad / sec

Calculation of Gain (A) (MAGNITUDE PLOT)

wl ; A= 20 log [ 20 / 0.025 ] = 58 .06 dB

wc1 ; A = [Slope from wl to wc1 x log (wc1 / wl ] + Gain (A)@wl

= - 20 log [ 0.25 / 0.025 ] + 58.06

= 38.06 dB

wc2 ; A = [Slope from wc1 to wc2 x log (wc2 / wc1 ] + Gain (A)@ wc1

= - 40 log [ 0.33 / 0.25 ] + 38

= 33 dB

Awh ; A = [Slope from wc2 to wh x log (wh / wc2 ] + Gain (A) @ wc2

= - 60 log [ 3.3 / 0.33 ] + 33

= - 27 dB

Calculations of Gain cross over frequency

The frequency at which the dB magnitude is Zero wgc = 1.1 rad / sec

Calculations of Phase cross over frequency

The frequency at which the Phase of the system is - 180o

wpc = 0.3 rad / sec

Gain Margin

The gain margin in dB is given by the negative of dB magnitude of G(jw) at phase cross over

frequency

UNIT – IV

STABILITY ANALAYSIS

PART – A

1. What is root locus? (May’ 12, 09)

The roots of the closed-loop characteristic equation define the system characteristic

responses Their location in the complex s-plane lead to prediction of the

characteristics of the time domain responses in terms of:damping ratio,

natural frequency, wn

damping constant, first-order modes

Consider how these roots change as the loop gain is varied from 0 to

2. State Nyquist stability Criterion. (May’ 12)

If the open loop system is unstable, then for the closed loop system to be

stable, the number of poles of G H(s) lying in PHP and number of encirclement

should be in the counter clock wise direction in the G(s) Plane.

3. What is the necessary condition for stability? (May’11)

A system is bounded input bounded output stable if all the roots of

characteristics equation lie in the left half of the complex S plane.

4. What is characteristic equation? (May’11)

Consider an nth-order system whose the characteristic equation (which is also the

denominator of the transfer function) is

5. Define stability. (Nov’11)

A linear time invariant system is said to be stable if following conditions are satisfied.

1. When system is excited by a bounded input, output is also bounded & controllable.

2. In the absence of input, output must tend to zero irrespective of the initial conditions.

6. How the roots of characteristic are related to stability?

7. What do you mean by dominant pole? (Nov’10)

8. What are break away points? (Nov’10, May’09)

To obtain the breakaway point of the locus, differentiate the characteristics

polynomial with respect to S and put 0dk

ds. Solve foe S which give the breakaway points.

There are more than one break away point and they are real and complex conjugate also.

9. How will you find the root locus on real axis?

On a given section of the real axis, root axis is found in the section only if the total

number of poles and zeros of GH(s) to the right of the section is odd if the feedback is –Ve

and K is +Ve, root locus is found on the real axis where even number of poles and zeros

occurred to the right of the concerned point.

10. What is absolute stability and conditional stability of a system with respect to a parameter? (Nov’09)

1. Using Routh criterion determine the stability of the system whose characteristics equation is s4 + 10 s3 + 36s2 + 70s + 75 (May’ 12)

Determine the location of roots with respect to s = -2 given that

F(s) = s4 + 10 s3 + 36s2 + 70s + 75

Sol: shift the origin with respect to s = -2

s = s1 – 2

(s – 2) 4 + 10 (s – 2)3 + 36(s – 2)2 + 70 (s –2) + 75 = 0

S 4 + 2s 3 + 0s 2 + 14s + 15 = 0

S 4 1 0 15

S 3 2 14 0

S 2 -7 15 0

S 1 18.28 0 0

S 0 15

Two sign change, there are two roots to the right of s = -2 & remaining ‘2’ are to the left

of the line s = -2. Hence the system is unstable.

2. s6 + 4s5 +3s4 – 16s2- 64s – 48 = 0 Find the number of roots of this equation with

positive real part, zero real part & negative real part

Sol: S6 1 3 -16 -48

S5 4 0 -64 0

S4 3 0 -48 0

S3 0 0 0

dA

A(s) = 3S4 – 48 = 0 = 12s3

ds

S6 1 3 -16 -48

S5 4 0 -64 0

S4 3 0 -48 0

S3 12 0 0 0

S2 ( )0 -48 0 0

S1 576 0 0 0

S0 -48

Lim 576

0 +

Therefore One sign change & system is unstable. Thus there is one root in R.H.S of the s –

plane i.e. with positive real part. Now

solve A(s) = 0 for the dominant roots

A(s) = 3s4 – 48 =0

Put S2 = Y

3Y2 = 48 Y2 =16, Y = 16 = 4

S2 = + 4 S2 = -4

S = 2 S = 2j

So S = 2j are the two parts on imaginary axis i.e. with zero real part. Root in R.H.S.

indicated by a sign change is S = 2 as obtained by solving A(s) = 0. Total there are 6 roots

as n = 6.

Roots with Positive real part = 1

Roots with zero real part = 2

Roots with negative real part = 6 –2 – 1 = 3

3. Explain the step by step procedure rules for constructing root locus

1) The root locus is symmetrical about real axis. The roots of the characteristic equation

are either real or complex conjugate or combination of both. Therefore their locus

must be symmetrical about the real axis.

2) As K increases from zero to infinity, each branch of the root locus originates from an

open loop pole (n nos.) with K= 0 and terminates either on an open loop zero (m

nos.) with K = along the asymptotes or on infinity (zero at ). The number of

branches terminating on infinity is equal to (n – m).

3) Determine the root locus on the real axis. Root loci on the real axis are determined

by open loop poles and zeros lying on it. In constructing the root loci on the real axis

choose a test point on it. If the total number of real poles and real zeros to the right of

this point is odd, then the point lies on root locus. The complex conjugate poles and

zeros of the open loop transfer function have no effect on the location of the root loci

on the real axis.

4) Determine the asymptotes of root loci. The root loci for very large values of s must be

asymptotic to straight lines whose angles are given by

)12(1-mn0,1,2,;mn

1)(2q180asymptotesofAngle

A

q

5) All the asymptotes intersect on the real axis. It is denoted by a , given by

)13(mn

)zz(z)pp(p

mn

zerosofsumpolesofsumσ

m21n21

a

6) Find breakaway and break-in points. The breakaway and break-in points either lie on

the real axis or occur in complex conjugate pairs. On real axis, breakaway points

exist between two adjacent poles and break-in in points exist between two adjacent

zeros. To calculate this polynomial 0ds

dK must be solved. The resulting roots are

the breakaway / break-in points. The characteristic equation given by Eqn.(7), can be

rearranged as

)z(s)z)(szK(sA(s)

and )p(s)p)(sp(s B(s) where,

(14)0KA(s)B(s)

m21

n21

The breakaway and break-in points are given by

)15(0Bds

dABA

ds

d

ds

dK

Note that the breakaway points and break-in points must be the roots of Eqn.(15),

but not all roots of Eqn.(15) are breakaway or break-in points. If the root is not on

the root locus portion of the real axis, then this root neither corresponds to

breakaway or break-in point. If the roots of Eqn.(15) are complex conjugate pair, to

ascertain that they lie on root loci, check the corresponding K value. If K is positive,

then root is a breakaway or break-in point.

7) Determine the angle of departure of the root locus from a complex pole

)16()zerosotherfromquestioninpolecomplexatovectorsofanglesof(sum

poles)otherfromquestioninpolecomplexatovectorsofanglesof(sum

180pcomplexafromdepartureofAngle

8) Determine the angle of arrival of the root locus at a complex zero

(17)poles)otherfromquestioninzerocomplexatovectorsofanglesof(sum

zeros)otherfromquestioninzerocomplexatovectorsofanglesof(sum

180zerocomplexatarrivalofAngle

9) Find the points where the root loci may cross the imaginary axis. The points where

the root loci intersect the j axis can be found by

a) use of Routh’s stability criterion or

b) letting s = j in the characteristic equation , equating both the real part and

imaginary part to zero, and solving for and K. The values of thus found

give the frequencies at which root loci cross the imaginary axis. The

corresponding K value is the gain at each crossing frequency.

10) The value of K corresponding to any point s on a root locus can be obtained using

the magnitude condition, or

)18(zerostopointsbetweenlengthofproduct

polestopointsbetweenlengthsofproductK

PHASE MARGIN AND GAIN MARGIN OF ROOT LOCUS

Gain Margin

It is a factor by which the design value of the gain can be multiplied before the closed

loop system becomes unstable.

(19)KofvalueDesign

overcrossimaginaryatKofValueMarginGain

The Phase Margin

Find the point j 1 on the imaginary axis for which 1jHjG for the design

value of K i.e. design

Kj/AjB .

The phase margin is

(20))H(jωjωargG180φ11

4. Sketch the root locus of a unity negative feedback system whose forward path

transfer function is s

KG(s) .

Solution:

1) Root locus is symmetrical about real axis.

2) There are no open loop zeros(m = 0). Open loop pole is at s = 0 (n = 1). One branch

of root locus starts from the open loop pole when K = 0 and goes to asymptotically

when K .

3) Root locus lies on the entire negative real axis as there is one pole towards right of

any point on the negative real axis.

4) The asymptote angle is A = .01,)12(180

mnqmn

q

Angle of asymptote is A = 180 .

5) Centroid of the asymptote is mn

zeros)of(sumpoles)of(sumσA

0.01

0

6) The root locus does not branch. Hence, there is no need to calculate the break

points.

7) The root locus departs at an angle of -180 from the open loop pole at s = 0.

8) The root locus does not cross the imaginary axis. Hence there is no imaginary axis

cross over.

The root locus plot is shown in Fig.1

Figure 15 Root locus plot of K/s

Comments on stability:

The system is stable for all the values of K > 0. Th system is over damped.

5. The open loop transfer function is 21)(s

2)K(sG(s) . Sketch the root locus plot

Solution:


2) There is one open loop zero at s=-2.0(m=1). There are two open loop poles at

s=-1, -1(n=2). Two branches of root loci start from the open loop pole when

K= 0. One branch goes to open loop zero at s =-2.0 when K and other goes to

(open loop zero ) asymptotically when K .

3) Root locus lies on negative real axis for s ≤ -2.0 as the number of open loop poles

plus number of open loop zeros to the right of s=-0.2 are odd in number.

4) The asymptote angle is A = .01,)12(180

mnqmn

q

Angle of asymptote is A = 180 .



0.01

)2()11(

6) The root locus has break points.

The root loci brakes out at the open loop poles at s=-1, when K =0 and breaks in onto

the real axis at s=-3, when K=4. One branch goes to open loop zero at s=-2 and

other goes to along the asymptotically.

7) The branches of the root locus at s=-1, -1 break at K=0 and are tangential to a line

s=-1+j0 hence depart at 90 .

8) The locus arrives at open loop zero at 180 .

9) The root locus does not cross the imaginary axis, hence there is no need to find the

imaginary axis cross over.

The root locus plot is shown in Fig.2.

Figure 16 Root locus plot of K(s+2)/(s+1)2

4K3,s0;K1,s

02)(s

1)(s2)1)(s2(s

0ds

dKbygivenispointBreak

2)(s

1)(sK

21

2

2

2


System is stable for all values of K > 0. The system is over damped for K > 4. It is

critically damped at K = 0, 4.

6. The open loop transfer function is 2)s(s

4)K(sG(s) . Sketch the root locus.

Solution


2) There are is one open loop zero at s=-4(m=1). There are two open loop poles at

s=0, -2(n=2). Two branches of root loci start from the open loop poles when K= 0.

One branch goes to open loop zero when K and other goes to infinity

asymptotically when K .

3) Entire negative real axis except the segment between s=-4 to s=-2 lies on the root

locus.

4) The asymptote angle is A = .01,1,0,)12(180

mnqmn

q

Angle of asymptote are A = 180 .



0.21

)4()2(

6) The brake points are given by dK/ds =0.

7) Angle of departure from open loop pole at s =0 is 180 . Angle of departure from

pole at s=-2.0 is 0 .

8) The angle of arrival at open loop zero at s=-4 is 180

9) The root locus does not cross the imaginary axis. Hence there is no imaginary cross

over.

The root locus plot is shown in fig.3.

11.7K6.828,s

0.343;K1.172,s

04)(s

2s)(s4)2)(s(2s

ds

dK

4)(s

2)s(sK

2

1

2

2

Figure 3 Root locus plot of K(s+4)/s(s+2)


System is stable for all values of K.

0 > K > 0.343 : > 1 over damped

K = 0.343 : = 1 critically damped

0.343 > K > 11.7 : < 1 under damped

K = 11.7 : = 1 critically damped

K > 11.7 : >1 over damped.

7. The open loop transfer function is 3.6)(ss

0.2)K(sG(s)

2. Sketch the root locus.

Solution:


2) There is one open loop zero at s = -0.2(m=1). There are three open loop poles at

s = 0, 0, -3.6(n=3). Three branches of root loci start from the three open loop poles

when K= 0 and one branch goes to open loop zero at s = -0.2 when K and other

two go to asymptotically when K .

3) Root locus lies on negative real axis between -3.6 to -0.2 as the number of open loop

poles plus open zeros to the right of any point on the real axis in this range is odd.

4) The asymptote angle is A = 1,01,)12(180

mnqmn

q

Angle of asymptote are A = 90 , 270 .



7.12

)2.0()6.3(

6) The root locus does branch out, which are given by dK/ds =0.

The root loci brakeout at the open loop poles at s = 0, when K =0 and breakin onto

the real axis at s=-0.432, when K=2.55 One branch goes to open loop zero at s=-0.2

and other goes breaksout with the another locus starting from open loop ploe at s= -

3.6. The break point is at s=-1.67 with K=3.66. The loci go to infinity in the complex

plane with constant real part s= -1.67.

7) The branches of the root locus at s=0,0 break at K=0 and are tangential to imaginary

axis or depart at 90 . The locus departs from open loop pole at s=-3.6 at 0 .

8) The locus arrives at open loop zero at s=-0.2 at 180 .

9) The root locus does not cross the imaginary axis, hence there is no imaginary axis

cross over.


ly.respective 3.662.55,0,Kand1.670.432,0,s

01.44s4.8s2s

0.2)(s

)3.6s(s0.2)7.2s)(s(3s

ds

dK

0.2s

)3.6s(s-K

23

2

232

23


System is stable for all values of K. System is critically damped at K= 2.55, 3.66. It is

under damped for 2.55 > K > 0 and K >3.66. It is over damped for 3.66 > K >2.55.

8. The open loop transfer function is 25)6ss(s

KG(s) . Sketch the root locus.

Solution:


2) There are no open loop zeros (m=0). There are three open loop poles at s=-0,

-3 j4(n=3). Three branches of root loci start from the open loop poles when K= 0

and all the three branches go asymptotically when K .

3) Entire negative real axis lies on the root locus as there is a single pole at s=0 on the

real axis.

4) The asymptote angle is A = .2,1,01,1,0,)12(180

mnqmn

q

Angle of asymptote are A = 60 , 180 , 300 .



0.23

)33(

6) The brake points are given by dK/ds =0.

j18.0434K

j2.0817and2s

02512s3sds

dK

25s)6s(s25)6ss(sK

1,2

1,2

2

232

For a point to be break point, the corresponding value of K is a real number greater

than or equal to zero. Hence, S1,2 are not break points.

7) Angle of departure from the open loop pole at s=0 is 180 . Angle of departure from

complex pole s= -3+j4 is

zeros) from inquestion polecomplex a to vectorsof angles theof (sum

poles)other fromquestion in polecomplex a to vectorsof angles theof (sum

180p

87.36)903

4tan180(180 1

p

Similarly, Angle of departure from complex pole s= -3-j4 is

36.87or323.13)270(233.13180φp

8) The root locus does cross the imaginary axis. The cross over point and the gain at

the cross over can be obtained by

Rouths criterion

The characteristic equation is 0K25s6ss 23. The Routh’s array is

For the system to be stable K < 150. At K=150 the auxillary equation is 6s2+150=0.

s = ±j5.

or

substitute s= j in the characteristic equation. Equate real and imaginary parts to zero.

Solve for and K.

The plot of root locus is shown in Fig.5.

6s6

K150s

K6s

251s

0

1

2

3

1500,Kj50,ω

025ωjωK)6ω(

0Kjω25jω6jω

0K25s6ss

22

23

23


System is stable for all values of 150 > K > 0. At K=150, it has sustained oscillation of

5rad/sec. The system is unstable for K >150.

9. Sketch the root locus of a unity negative feedback system whose forward path

transfer function is j)3j)(s31)(s(s

2)K(sG(s)H(s) . Comment on the

stability of the system.


2) There is one open loop zero at s = -2 (m = 1). There are three open loop poles at

s = -1, -3 ± j (n=3). All the three branches of root locus start from the open loop

poles when K = 0. One locus starting from s = -1 goes to zero at s = -2 when

K , and other two branches go to asymptotically (zeros at ) when K .

3) Root locus lies on the negative real axis in the range s=-1 to s= -2 as there is one

pole to the right of any point s on the real axis in this range.

4) The asymptote angle is A = .0,11mnq,mn

1)(2q180

Angle of asymptote is A = 90 , 270 .



2.51

2)(3)31(

6) The root locus does not branch. Hence, there is no need to calculate break points.

7) The angle of departure at real pole at s=-1 is 180 . The angle of departure at the

complex pole at s=-3+j is 71.57 .



180p

The angle of departure at the complex pole at s=-3-j is -71.57 .

8) The root locus does not cross the imaginary axis. Hence there is no imaginary axis

cross over.

The root locus plot is shown in Fig.1

Figure 1 Root locus plot of K(s+2)/(s+1)(s+3+j)(s+3-j)


The system is stable for all the values of K > 0.

57.71135)90(153.43180

900

2tanθ,135or -45

1-

1tan

153.43)atan2(-2,1θ

153.43or 57.262-

1tanθ

p

1

3

1

1

1

1

57.71522)270(206.57180p

10. The open loop transfer function is10)0.6s0.5)(ss(s

KG(s)H(s)

2 Sketch

the root locus plot. Comment on the stability of the system.

Solution:


2) There are no open loop zeros (m=0). There are four open loop poles (n=4) at s=0,

-0.5, -0.3 ± j3.1480. Four branches of root loci start from the four open loop poles

when K= 0 and go to (open loop zero at infinity) asymptotically when K .

3) Root locus lies on negative real axis between s = 0 to s = -0.5 as there is one pole to

the right of any point s on the real axis in this range.

4) The asymptote angle is A = .3,2,1,01,)12(180

mnqmn

q

Angle of asymptote is A = 45 , 135 , 225 , ±315 .



275.04

)3.03.05.0(

The value of K at s=-0.275 is 0.6137.

6) The root locus has break points.

K = -s(s+0.5)(s2+0.6s+10) = -(s4+1.1s3+10.3s2+5s)

Break points are given by dK/ds = 0

0520.6s3.3s4sds

dK 23

s= -0.2497, -0.2877 j 2.2189

There is only one break point at -0.2497. Value of K at s = -0.2497 is 0.6195.

7) The angle of departure at real pole at s=0 is 180 and at s=-0.5 is 0 . The angle of

departure at the complex pole at s = -0.3 + j3.148 is -91.8



180p

The angle of departure at the complex pole at s = -0.3 - j3.148 is 91.8

8) The root locus does cross the imaginary axis, The cross over frequency and gain is

obtained from Routh’s criterion.

The characteristic equation is

s(s+0.5)(s2+0.6s+10)+K =0 or s4+1.1s3+10.3s2+5s+K=0

The Routh’s array is

Ks5.75

1.1K-28.75s

K5.75s

51.1s

K10.31s

0

1

2

3

4

8.91)9086.4(95.4180

900

6.296tanθ, 4.68

0.2

3.148tan

4.95or6.840.3-

3.148tanθ

p

1

3

1

2

1

1

91.8 )270273.6(264.6180p

The system is stable if 0 < K < 26.13

The auxiliary equation at K 26.13 is 5.75s2+26.13 = 0 which gives s = ± j2.13 at

imaginary axis crossover.


Figure 17 Root locus plot of K/s(s+0.5)(s2+0.6s+10)


System is stable for all values of 26.13 >K > 0. The system has sustained oscillation at

= 2.13 rad/sec at K=26.13. The system is unstable for K > 26.13.

11. The open loop transfer function is 20)4s)(s4s(s

KG(s)

2. Sketch the root

locus.

Solution:


2) There are no open loop zeros (m=0). There are three open loop poles (n=3) at s = -0,

-4, -2 j4. Three branches of root loci start from the three open loop poles when K=

0 and to infinity asymptotically when K .

3) Root locus lies on negative real axis between s = 0 to s = -4.0 as there is one pole to

the right of any point s on the real axis in this range.

4) The asymptote angle is A = 3,2,1,01,)12(180

mnqmn

q

Angle of asymptote are A = 45 , 135 , 225 , 315 .



0.24

)0.40.20.2(

6) The root locus does branch out, which are given by dK/ds =0.

The root loci brakeout at the open loop poles at s = -2.0, when K = 64 and breakin and

breakout at s=-2+j2.45, when K=100

7) The angle of departure at real pole at s=0 is 180 and at s=-4 is 0 . The angle of

departure at the complex pole at s = -2 + j4 is -90 .



180p

The angle of departure at the complex pole at s = -2 – j4 is 90

100Kj2.45,2.0s

64;K2.0,s

40)16s2)(4s(s

08040s32s16s8s4s

08072s24s4s

0ds

dKbygivenispointBreak

80s)36s8s(s

20)4s4)(ss(sK

2

1

2

223

23

234

2

90)90.463(116.6-180

900

8tanθ,4.36

2

4tanθ

116.6)atan2(4,-2θ

116.6or63.42-

4tanθ

p

1

3

1

2

1

1

1

90270

)270296.6(243.4-180p

8) The root locus does cross the imaginary axis, The cross over point and gain at cross

over is obtained by either Routh’s array or substitute s= j in the characteristic

equation and solve for and gain K by equating the real and imaginary parts to zero.

Routh’s array

The characteristic equation is 0K80s36s8ss 234

For the system to be stable K > 0 and 2080-8K > 0. The imaginary crossover is given

by 2080-8K=0 or K = 260.

At K = 260, the auxiliary equation is 26s2+260 = 0. The imaginary cross over occurs

at s= j 10.

or


Ks26

8K2080s

K26s

808s

K361s

isarrayRouthsThe

0

1

2

3

4

260K0K36ωω

10js;10j0,ω080ω8ω

zerotopartsimaginaryandrealEquate

080ω8ωjK36ωω

0Kjω80jω36jω8jω

jωsput

0K80s36s8ss

24

3

324

234

234

Figure 18 Root locus plot of K/s(s+4)(s2+4s+20)


For 260 > K > 0 system is stable

K = 260 system has stained oscillations of 10 rad/sec.

K > 260 system is unstable.

12. Given that G(S) H(S) = K/S(S+2) (S+10) = K/(Sx2(S/2+1)x10(S/10+1))

= 0.05 K/S(1+0.5S)(1+0.1S)

The open loop transfer function has a pole at origin. Hence choose the Nyquist contour

on S-plane enclosing the entire right half plane except the origin

The Nyquist contour has four sections C1, C2, C3, C4 the mapping of each section is

performed separately and the overall Nyquist plot is obtained by combining the individual

sections

Mapping of section C1

In section c1,ω varies from 0 to ∞. The mapping of section C1 is given by the locus

G(jω)H(jω ) as ω is varied from 0 to ∞. This polar plot of G(jω)H(jω)

G(S)H(S)= 0.05K/S(1+0.5S)(1+0.1S)

Let S=jω

G(jω)H(jω S)= 0.05K/ jωS(1+0.5 jω )(1+0.1 jω )

= 0.05K/-0.6ω2+jω(1-0.05ω2)

when the locus of G(jω) H(jω ) crosses real axis the imaginary term will be zero and the

corresponding frequency is the phase crossover frequency ‘

ωPC.

At ω=ωpc ωpc (1-0.05 ωpc2)=0 1-0.05 ωpc

2=0

At ω=ωpc = 4.427 rad/sec

G(jω)H(jω) = 0.05K/-0.6ω2 = -0.00417 K

The open loop system is type -1 and third order system. Also it is a minimum phase system

with all poles. Hence the polar plot of G(jω)H(jω) starts at -90˚axis at infinity crosses real

axis at -0.00417 K and ends at origin in second quadrant. The section C1 and its mapping

show below diagram A and B

ω=∞

ω=0

σ

R∞

C1

C3

C4

σ

ω=+∞

ω=0

ω=-∞

ω=∞

-0.00417K

jv

Section c1 in s-plane

Mapping of section c1 in G(S)H(S)


The mapping of section C2 from S-plane to G(S)H(S) –Plane is obtained by letting

Lt

S=R∞ Rejθ in G(S)H(S) and varying θ from +∏/2 to -∏/2. Since S Rejθ and R, the

G(S)H(S) can be approximately as shown below [ ie (1+ST)≈ ST ]

G(S)H(S) = 0.05K/S(1+0.5S)(1+0.1S) ≈ 0.05K/S x 0.5S x 0.1S = K/S3

Lt

Let S=R∞ Rejθ

G(S)H(S)| S=R∞ Rejθ K/S3| = = 0e-j3θ

when θ=+∏/2 , G(S)H(S) = 0e-j3∏/2-----------------1a

when θ=-∏/2 , G(S)H(S) = 0e+j3∏/2--------------------------1b

From the equation 1a and 1b we can study that section c2 in S-plane fig1a

is mapped as circular are of zero radius around origin in G(S)H(S) plane with argument

(phase varying from -3∏/2 and +3∏/2 as shown in fig 1b


R∞

R∞

C2 S=Plane jω

Section c2 in S-Plane

jv

u

G(S)H(S)

plane

K

Lt

R∞ (Rejθ)3

In Section C3, ω varies from -∞ to 0. The mapping of section C3 is given by the locus of

G(jω)H(jω) as a ω is varied from -∞ to 0. This locus is the inverse polar plot of G(jω)H(jω)

The inverse polar plot is given by the mirror image of polar plot with respect to real axis.

The section C3 in S-Plane and its corresponding contour in G(S)H(S) plane shown in fig 2a

and fig 2b


The mapping of section c4 from s-plane to G(S)H(S) plane is obtained by letting

Lt

S=R∞ Rejθ in G(S)H(S) and varying θ from +∏/2 to -∏/2. Since S Rejθ and R, the

G(S)H(S) can be approximately as shown below [ ie (1+ST)≈ 1 ]

0.05K

G(S)H(S) = ----------------------- = 0.05K/S

S(1+0.5S)(1+0.1S)

Lt

Let S= R0 Rejθ

G(S)H(S) = αe-jθ

when θ=+∏/2 , G(S)H(S) = αe-j3∏/2-----------------2a

when θ=∏/2 , G(S)H(S) = αe+j3∏/2--------------------------2b

From equation above we can say that sec c4 in S-Plane fig 3a is mapped as a circular are

of infinity radius with argument(phase) varying from +∏/2 to-∏/2 shown in figure

σ

jω

ω=0

ω=-∞

S-Plane

jv

ω=0

ω=-∞

-0.00417k

Complete Nyquist plot

The entire Nyquist plot in G(S)H(S) plane can be obtained by combining the mappings

of individual sections shown

R0

-0.00417K

-1+j0 for

K<240 -1+j0K>240

G(S) H(S) Contour

K

G(S)H(S) = ----------------

S(S+2)(S+10)

R->∞

jv

R->0

C4

S-Plane

G(S)H(S)

plane

Mapping of section c4 Section C4 in s-plane

STABILITY ANALYSIS

When -0.00417K =-1, the contour passes through (-1+j0) point and corresponding value

of K is the limiting value for K for stability

Limiting value of K = 1/0.00417 = 240

When K<240

When K is less than 240, the contour crosses real axis at appoint between 0 & -1+j0 . On

traveling through Nyquist plot along the indicated direction if it found that the point -1+j0 is

not encircled. Also the open loop transfer function has no poles on the right half of S-Plane.

Therefore the closed loop system is stable.

When K>240

When K is graeter than 240, the contour crosses real axis at a point B/W -1+j0 and -∞.

On traveling through Nyquist plot along the indicated direction it is found that the point -1+j0

is encircled in clockwise direction two time. [ since there are two clockwise encirclement and

no right half open loop poles, the closed loop system has 2 poles on right half of S-Plane.

Therefore the closed loop system is unstable

RESULT

The value of K for stability is 0<K<240

UNIT – V

COMPENSATORS AND DESIGN

1. What are the two methods of designing a control system? (May’ 12)

The two methods of designing a control system are design using root locus and

design using bode plot.

2. What is compensator? (May’ 12)

The compensation is the design procedure in which the system behavior is altered to meet the desired specification, by introducing additional device is called compensator. 3. What are the different types of compensator available? (May’11)

i. Lag Compensator. ii. Lead Compensator. iii. Lag-Lead Compensator.

4. What is series compensation? (Nov’11)

In series compensation a compensator is introduced in series with plant to alter the system behavior and to meet the desired specification.

5. What is feedback compensation? (Nov’11)

In feedback or parallel compensation, the compensator is placed in the inner

feedback path around one or more components to meet the desired specification.

6. When lag/lead/lag-lead compensation is employed? (Nov’10)

Lag compensation is employed for a suitable system for improvement in

steady state performance. Lead compensation is employed for stable as well as for

unstable system for improvement in transient performance. Lag lead compensation is

employed for stable and unstable system for improvement in both steady state and

transient response.

7. What are the uses of lead compensator? (May’10)

The lead compensating network contributes phase lead frequency response

characteristics which appreciably improve the transient response and to a small

extent the steady state performance.

8. Draw the electrical lead compensator network? (May’10)

9. Draw the bode plot of lead compensator? (Nov’09)

10. What is lag-lead compensator? (May’09)

A compensator having the characteristics of lead lag network is called lag-

lead compensator.

11. Draw electrical lag-lead compensator network? (May’09)

PART B

1. Design a Lead Compensator.

2. Design A lead Compensator.

3. Design lags lead compensator.

4. Explain the procedure for lead compensation and lag compensation? (May’ 12)

Lag Compensator Design

In the previous lecture we discussed lead compensator design. In this lecture we would see how to design a phase lag compensator

Phase lag compensator

The essential feature of a lag compensator is to provide an increased low frequency gain, thus decreasing the steady state error, without changing the transient response significantly. For frequency response design it is convenient to use the following transfer function of a lag compensator.

where,

The above expression is only the lag part of the compensator. The overall compensator is

Typical objective of lag compensator design is to provide an additional gain of α in the low frequency region and to leave the system with sufficient phase margin. The frequency response of a lag compensator, with α=4 and τ=3, is shown in Figure 1 where

the magnitude varies from dB to 0 dB.

Figure 1: Frequency response of a lag compensator

Since the lag compensator provides the maximum lag near the two corner frequencies, to maintain the PM of the system, zero of the compensator should be chosen such that ω = 1/ τ is much lower than the gain crossover frequency of the uncompensated system.

Consider the following system

Design a lag compensator so that the phase margin (PM) is at least 50° and steady state

error to a unit step input is . The overall compensator is

where,

When Steady state error for unit step input is

Thus, Now let us modify the system transfer function by introducing K with the original system. Thus the modified system becomes

5. Problem Design a Compensator.

PM of the closed loop system should be 50°. Let the gain crossover frequency of the uncompensated system with K be ωg .

Required PM is 50°. Since the PM is achieved only by selecting K, it might be deviated from this value when the other parameters are also designed. Thus we put a safety margin of 5° to the PM which makes the required PM to be 55°.

To make ωg = 2.8 rad/sec, the gain crossover frequency of the modified system, magnitude at ωg should be 1. Thus

Putting the value of ωg in the last equation, we get K = 5.1. Thus,

The only parameter left to be designed is τ. Since the desired PM is already achieved with gain K, We should place ω = 1/ τ such that it does not much effect the PM of the modified system with K. If we place 1/ τ one decade below the gain crossover frequency, then

or,

The overall compensator is

With this compensator actual phase margin of the system becomes 52.7°, as shown in Figure 2, which meets the design criteria.

Lag -lead Compensator

When a single lead or lag compensator cannot guarantee the specified design criteria, a lag-lead compensator is used. In lag-lead compensator the lag part precedes the lead part. A continuous time lag-lead compensator is given by

where,

The corner frequencies are , , , . The frequency response is shown in Figure 1.

http://nptel.iitm.ac.in/courses/108103008/module5/lec7/3.html#m5l7f2

Figure 1: Frequency response of a lag-lead compensator

In a nutshell,

• If it is not specified which type of compensator has to be designed, one should first check the PM and BW of the uncompensated system with adjustable gain K.

• If the BW is smaller than the acceptable BW one may go for lead compensator. If the BW is large, lead compensator may not be useful since it provides high frequency amplification.

• One may go for a lag compensator when BW is large provided the open loop system is stable.

• If the lag compensator results in a too low BW (slow speed of response), a lag-lead compensator may be used.

6. Lag-lead compensator design Consider the following system with transfer function

Design a lag-lead compensator C(s) such that the phase margin of the compensated system is at least 45° at gain crossover frequency around 10 rad/sec and the velocity error constant Kv is 30. The lag-lead compensator is given by

where,

When

Thus K = 30 . Bode plot of the modified system KG(s) is shown in Figure 2. The gain crossover frequency and phase margin of KG(s) are found out to be 9.77 rad/sec and -17.2° respectively.

Figure 2: Bode plot of the uncompensated system for Example 1

Since the PM of the uncompensated system with K is negative, we need a lead compensator to compensate for the negative PM and achieve the desired phase margin. However, we know that introduction of a lead compensator will eventually increase the gain crossover frequency to maintain the low frequency gain. Thus the gain crossover frequency of the system cascaded with a lead compensator is likely to be much above the specified one, since the gain crossover frequency of the uncompensated system with K is already 9.77 rad/sec. Thus a lag-lead compensator is required to compensate for both. We design the lead part first. From Figure 2, it is seen that at 10 rad/sec the phase angle of the system is -198°.

Since the new ωg should be 10 rad/sec, the required additional phase at ωg, to maintain the specified PM, is 45 - (180 - 198) = 63° . With safety margin 2°,

And

which gives . However, introducing this compensator will actually increase the gain crossover frequency where the phase characteristic will be different than the designed one. This can be seen from Figure 3.

Figure 3: Frequency response of the system in Example 1 with only a lead compensator

The gain crossover frequency is increased to 23.2 rad/sec. At 10 rad/sec, the phase angle is -134° and gain is 12.6 dB. To make this as the actual gain crossover frequency, lag part should provide an attenuation of -12.6 dB at high frequencies.

At high frequencies the magnitude of the lag compensator part is . Thus ,

which gives . Now, should be placed much below the new gain crossover

frequency to retain the desired PM. Let be 0.25. Thus

The overall compensator is

The frequency response of the system after introducing the above compensator is shown in Figure 4, which shows that the desired performance criteria are met.

7. Problem Find The stability of given system

8. Find the stability of given System Using Bode plot.

9. Design compensator for given System.

10. Design a Lag Compensator of Given System.

mahalakshmimahalakshmiengineeringcollege.com/pdf/ece/ivsem... · graph is given by the net gain...

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