1 

CONTROLSYSTEMSChapter 3 : Time Response Analysis

 

GATE Objective & Numerical Type Solutions

Question 4 [Practice Book] [GATE EC 1991 IIT-Madras : 2 Mark]

A unity feedback control system has the open loop transfer function.

If the input to the system is a unit ramp, the steady state error will be (A) 0 (B) 0.5 (C) 2 (D) Ans. (A)

Sol. Given : , and

  Taking Laplace transform of r(t), we get

Steady state error is given by,

 

       sse

 

Hence, the correct option is (A).

Alternatively, Velocity error coefficient is given by,

Steady state error for ramp input is given by,

For type - 2 system steady state error due to ramp input will be zero.

Question 6 [Practice Book] [GATE EE 1991 IIT-Madras : 2 Mark]

A first order system and its response to a unit step input are shown in figure below, the system parameters a and K are respectively

(A) 5, 10 (B) 10, 5 (C) 2, 10 (D) 10, 2

2

4(1 2 )( )

( 2)

sG s

s s

2

4 1 2

2

sG s

s s

( ) ( )r t t u t

2

1( )R s

s

0 0

( )lim ( ) lim

1 ( ) ( )sss s

sR se sE s

G s H s

2

0 0

2

11

lim lim 04(1 2 ) 4(1 2 )

1( 2) ( 2)

s s

ss

s ss

s s s s

0limvs

K

20

4(1 2 )( ) ( ) lim

( 2)s

s ssG s H s

s s

1 10ss

v

eK

t (sec)0.2

( )c t

2.0

0

K

s a�( )r t ( )c t

  2 

Ans. (A)

Sol. For first order system loop transfer function is and its response to a unit step input are shown

in figure below.

For the given transfer function and its response.

From the figure, time constant is 0.2 sec.

On comparing with above standard equation, we get, .

Using final value theorem steady state output can be written as,

ssc

From figure,  

                        

Hence, the correct option is (A). Question 7 [Practice Book] [GATE EC 1992 IIT-Delhi : 8 Marks]

Block diagram model of a position control system is shown in figure.

(a) In absence of derivative feedback determine damping ratio of the system for amplifier gain

Also find the steady state error to unit ramp input.

(b) Find suitable values of the parameters and so that damping ratio of the system is increased to

0.7 without affecting the steady-state error as obtained in part (a).

( )

( ) 1

C s K

R s s

t (sec)�

( )c t

K

0

/( ) (1 )tc t K e� �� �

( )

( ) 1

KC s a

sR sa

t (sec)0.2

( )c t

2.0

0

K

s a�( )r t ( )c t

15a

a

0lim ( ) lim ( )sst s

c c t s C s

0 0

1lim ( ) lims s

K K Ks R s s

s a s a s a

2ssc

2 10K

Ka

1

(0.5 1)s s �( )Y s( )R s AK

tsK

AmplifierMotor

( 0),tK 5.AK

AK tK

  3 

Sol. Given : The given block diagram of a positional control system is shown in figure below.

(a) Given :

and

Closed-loop transfer function for negative feedback is given by,

….. (i)

Transfer function for second-order system with unit step input is given by,

               …..(ii)

where, damping ratio, natural angular frequency

On comparing equation(i) and (ii), we get

and

Steady state error is given by,

 

      sse           Ans. 

(b) The open loop transfer function can be written as,

The closed loop transfer function can be written as,

…..(i)

Transfer function for second-order system with unit step input is given by,

               ….. (ii)

where, damping ratio, natural angular frequency

1

(0.5 1)s s �( )Y s( )R s AK

tsK

AmplifierMotor

0tK

5 10( )

(0.5 1) ( 2)G s

s s s s

( ) 1H s

( )( )

1 ( ) ( )

G sT s

G s H s

2

1010( 2)

10 2 101( 2)

s s

s ss s

2

2 2

( )

( ) 2n

n n

C s

R s s s

n

10 rad/sec.n 2 2n

1

10

0 0

( )lim ( ) lim

1 ( ) ( )ss s s

sR se sE s

G s H s

2

0 0

11

lim lim 0.210 10

1( 2) ( 2)

s s

ss

ss s s

2

2OLTF

2 2A

t

K

s s sK

2

[ (2 2 )]A

t

K

s s K

2

2( )CLTF

( ) (2 2 ) 2A

t A

KY s

R s s s K K

2

2 2

( )

( ) 2n

n n

C s

R s s s

n

  4 

Velocity error coefficient is given by,

vK

Steady state error is given by,

 

….. (iii)

On comparing equation (i) and (ii), we get

1 0.9899t A AK K K ….. (iv)

From equation (iii) and equation (iv), we get

 and . Ans.

Question 8 [Practice Book] [GATE IN 1992 IIT-Delhi : 1 Marks]

For what values of ‘a’ does the system shown in figure have a zero steady state error [i.e., ] for a

step input?

(A) a = 0 (B) a = 2 (C) (D) No value of ‘a’

Ans. (A) Sol. The given system is shown in figure.

DC gain of feedback ( )H s is HK

0 0

1 1lim ( ) lim

4 4H s sK H s

s

0lim ( ) ( )vs

K s G s H s

0

2 2lim

[ (2 2 )] 2 2A A

st t

K Ks

s s K K

1ss

v

eK

10.2

22 2

A

t

KK

0.2 1A tK K

2n AK

2 2 2t nK

2 2 1

2 2t t

n A

K K

K

10.7

2t

A

K

K

25AK 4tK

lim ( )t

E t

2

1

5

s

s s a

�

� �

1

4s �

E t( )

4a

2

1

5

s

s s a

�

� �

1

4s �

E t( )

  5 

Closed-loop transfer function is given by,

2

2

1( ) 5( )

( 1)1 ( ) ( ) 1( 5 ) ( 4)

sG s s s aM s

sG s H ss s a s

2 3 2 2

( 1) ( 4) ( 1) ( 4)( )

( 5 ) ( 4) ( 1) 4 5 20 4 1

s s s sM s

s s a s s s s s s as a s

3 2

( 1) ( 4)( )

9 (21 ) (4 1)

s sM s

s s a s a

Steady state error is given by,

0 0

1lim ( ) lim ( ) ( )sss s

H

e sE s R s M sK

3 20

( 1) ( 4)lim 4 ( )

9 (21 ) (4 1)sss

s se s R s

s s a s a

For unit step input 1

( )R ss

3 20

( 1) ( 4)lim 4

9 (21 ) (4 1)sss

s se

s s a s a

4

0 44 1a

4

44 1a

16 4 4a 0a Hence, the correct option is (A).

Question 9 [Practice Book] [GATE IN 1993 IIT-Bombay : 2 Marks]

A unit step is applied at to a first order system without time delay. The response has a value of 1.264 units at min and 2 units at steady state. The transfer function of the system is

(A) (B) (C) (D)

Ans. (A) Sol. The first order system step response is given by,

( ) 1t

c t k e

At  

 

                

 

At  

                    

                  

 

First order transfer function is given by,

           

Ans.

0t 10t

2

1 600s2

600 s1

1 600s1

600 s

,t 2 1k e 2k

10 min,t 10

1.264 1k e

10min 600sec.

2( )

1 1 600

kG s

s s

  6 

Question 12 [Practice Book] [GATE IN 1994 IIT-Kharagpur : 1 Mark]

A unity feedback closed loop second order system has a transfer function2

81

0.6 9s s and is excited by a

step input of 1 unit. The steady state error of the output is

(A) 10 (B) 0.0 (C) 1.0 (D) 0.1

Ans. (B)

Sol. Given : 2

2

2

9981 ( )0.6( )

90.6 9 1 ( )10.6

KG ss sT ss s G s

s s

…..(i)

Closed-loop transfer function for negative feedback is given by,

( )

( )1 ( )

G sT s

G s

…..(ii)

On comparing equation (i) and (ii), we get

2

9( )

0.6G s

s s

Steady state error is given by,

0 0

( )lim ( ) lim

1 ( )sss s

sR se sE s

G s

2

1

09

10.6

ss

s s

  Hence, the correct option is (B).

Assignment 11 [Work Book] [GATE EC 1997 IIT-Madras : 5 Marks]

The figure shows the block diagram representation of a control system. The system in block A has an impulse response ( ) ( ).t

Ah t e u t The system in block B has an impulse response 2( ) ( ).tBh t e u t The

block ‘k’ is an amplifier by a factor k. For the overall system the input is ( )x t and output ( )y t .

(a) Find the transfer function ( )

( )

Y s

X s when 1.k

(b) Find the impulse response when 0.k (c) Find the values of k for which the system becomes unstable.

Sol. Given : Impulse response ( ) ( )tAh t e u t

Taking Laplace transform, we get

1

( )1AH s

s

Impulse response 2( ) ( )tBh t e u t

Taking Laplace transform, we get

1

( )2BH s

s

A B

k

( )x t ( )y t

  7 

(a) 1k , ( ) ( ) ( )A BG s H s H s

Closed-loop transfer function for negative feedback is given by,

( ) ( )

( )( ) 1 ( ) ( )

Y s G sT s

X s G s H s

1

( 1) ( 2)( )

11

( 1) ( 2)

s sT s

s s

2

1

3 3s s

Ans.

(b) 0k

( ) 1

( )( ) ( 1) ( 2)

Y sH s

X s s s

1 2

A B

s s

1A and 1B

1 1

( )1 2

H ss s

Taking inverse Laplace transform, we get Impulse response 2( ) ( ) ( )t th t e e u t Ans.

(c) Value of k for unstable system

2

( )3 2

kT s

s s k

Characteristic equation 2 3 2 0s s k Routh Tabulation :

2s   1  2 + k 1s   3  00s   2 + k   

For instability 2 0k

2k system is unstable. Ans.

Question 8 [Work Book] [GATE EC 1999 IIT-Bombay : 2 Marks]

If the closed-loop transfer function T(s) of a unity negative feedback system is given by

11

1 1.....n n

n nn n

a s aT s

s a s a s a

then the steady state error for a unit ramp input is

(A) 1

n

n

a

a

(B) 2

n

n

a

a

(C) 1

2

n

n

a

a

(D) zero

Ans. (D)

Sol. Given : 11

1 1.....n n

n nn n

a s aT s

s a s a s a

and r(t) = ( )t u t

11 2

1 2

11 2

1 2

.....( )

1.....

n nn n

n

n nn n

n

a s a

s a s a sT s

a s a

s a s a s

…... (i)

  8 

Closed-loop transfer function for negative feedback is given by,

( )

( )1 ( ) ( )

G sT s

G s H s

…... (ii)

On comparing equation (i) and (ii), we get

11 2

1 2

( ).....

n nn n

n

a s aG s

s a s a s

Taking Laplace transform of r(t), we get

2

1( )R s

s

The steady state error due to a unit ramp input is given by,

1

ssv

eK

Velocity error coefficient is given by,

11 20

1 2

( )lim ( ) ( )

.....n n

v n nsn

s a s aK sG s H s

s a s a s

So, that 1 1

0ssv

eK

 

Hence, the correct option is (D).

Question 31 [Practice Book] [GATE EC 2000 IIT-Kharagpur : 5 Marks]

The block diagram of a feedback system is shown in the figure.

(a) Find the closed loop transfer function. (b) Find the minimum value of G for which the step response of the system would exhibit an overshoot,

as shown in figure. (c) For G equal to twice the minimum value, find the time period T indicated in the figure. Sol. (a) The given block diagram of a feedback system is shown below.

( 3)

G

s s �

0G �

Input Output

0 ( )V t

0.1

T

t

Stepresponse

( 3)

G

s s �

0G �

Input Output

  9 

Closed-loop transfer function for negative feedback is given by,

( )

( )1 ( )

G sT s

G s

( 3)

1( 3)

Gs s

Gs s

( )T s 2( 3) 3

G G

s s G s s G

Ans.

(b) Given : Maximum peak overshoot is given by,

21MPO 0.1e

Damping ratio is 0.6

For the given response transfer function is

2

( )3

GT s

s s G

…..(i)

Transfer function for second-order system is given by,

2

2 2

( )

( ) 2n

n n

C s

R s s s

             …..(ii)

where, damping ratio, n natural angular frequency

On comparing equation (i) and equation (ii), we get

2 3n and 2n G

3

2n G

3

2 0.6G

6.25G Ans. (c) ' 2 2 6.25 12.5G G

2 'n G 12.5 3.53n

2 3n 0.424

Damped frequency of oscillation is given by,

21d n

223.53 1 (0.424) 3.197

T

1.96 sec.T Ans.

Question 37 [Practice Book] [GATE IN 2002 IISc-Bangalore : 2 Marks]

The forward path transfer function of an unity feedback system is given by,

1 5 1 10 1 2

1 1 8 1 20

s s sG s

s s s

If e t is the error to a unit impulse input the value of the performance index 0

J e t dt

is equal to

(A) zero (B) infinity (C) – 12 (D) 0.5

Ans. (D)

  10 

Sol.

( ) ( ) ( )E s R s B s where LT( ) ( )e t E s

( ) ( ) ( ) ( )E s R s C s H s

( ) ( ) ( ) ( ) ( )E s R s E s G s H s

( ) 1 ( ) ( ) ( )E s G s H s R s

( )( )

1 ( ) ( )

R sE s

G s H s

Given : ( ) ( ), ( ) 1, ( ) 1r t t R s H s

Therefore, 1

( )(1 5 ) (1 10 ) (1 2 )

1(1 ) (1 8 ) (1 20 )

E ss s s

s s s

(1 ) (1 8 )(1 20 )

( )(1 ) (1 8 )(1 20 ) (1 5 )(1 10 )(1 2 )

s s sE s

s s s s s s

For causal ( )e t :

0

( ) ( ) stE s e t e dt

Given 0

( )J e t dt

0

( )s

J E s

0

(1 ) (1 8 )(1 20 )lim

(1 )(1 8 )(1 20 ) (1 5 )(1 10 )(1 2 )s

s s sJ

s s s s s s

1 1 1

0.5(1 1 1) (1 1 1)

J

Ans.

Question 11 [Work Book] [GATE EC 2003 IIT-Madras : 2 Marks]

A second-order system has the transfer function 2

( ) 4

( ) 4 4

C s

R s s s

with ( )r t

as the unit-step function,

the response ( )c t of the system is represented by

(A) (B)

( )R s ( )G s

( )H s

( )C s( )E s

( )B s

Step Response

Am

pli

tude

Time (sec)

1.5

1

0.5

00 2 4 6

Step Response

Am

pli

tude

Time (sec)

1.0

0.5

00 2 4 6

  11 

(C) (D)

Ans. (B)

Sol. Given : 2

( ) 4

( ) 4 4

C s

R s s s

         …..(i)

Transfer function for second-order system is given by,

2

2 2

( )

( ) 2n

n n

C s

R s s s

             …..(ii)

where, damping ratio, n natural angular frequency

2 4n 2 rad/secn

2 4n

1 Since 1 , system is critically damped. The final value can be calculated using final value theorem,

20

4 ( )lim

4 4sss

R sc s

s s

20

14

lim 14 4ss

s

ssc

s s

For 2% tolerance band settling time is given by,

4 4

2sec2s

n

T

This mean that the response ( )c t will be settle to its final value after 2 sec.

Hence, the correct option is (B). Question 13 [Work Book] [GATE IN 2004 IIT-Delhi : 2 Marks]

A certain system exhibited an overshoot of 16% when subjected to an input of 2 ( )u t , where ( )u t is a step

input. The damping ratio and decay ratio respectively are (A) (0.8, 0.0810) (B) (0.5, 0.0256) (C) (1.0, 0.1626) (D) (1.1, 0.0089) Ans. (B) Sol. Given : MPO = 16 % [By default 1st peak overshoot] Percentage MPO is given by,

% MPO21 100e

2116 100e

2

ln(0.16)1

Step Response

Am

pli

tud

e

Time (sec)

2.0

1.0

0.5

00

1.5

5 10 15 20 25

Step Response

Am

pli

tud

e

Time (sec)

1.0

0.5

00 5 10

  12 

2 1

3.939

0.5

Damping ratio ( ) 0.5

Concept of decay ratio :

Decay ratio nd

st

2 peak overshoot

1 peak overshoot

The 2nd peak overshoot is given by,

21nd%MPO(2 ) 100

3

e

The decay ratio is given by,

Decay ratio 22

2 2

0.5

1 (0.5)1

0.5

1 1 (0.5)

0.004330.027 0.025

0.16

33

e e

e e

Hence, the correct option is (B).

Question 51 [Practice Book] [GATE EE 2007 IIT-Kanpur : 2 Marks]

RLC circuit shown in figure. For a step input ie , the overshoot in the output 0

e will be

(A) 0 %, since the system is not under damped (B) 5 %

(C) 16 % (D) 48 %

Ans. (C)

1

t

c t( )

0

cmax

pt

1 peak overshootst

2 peak overshootnd

10R � � 1mHL �

ie

0e10 FC � �

  13 

Sol. The given RLC circuit is shown below.

Transform domain :

 

      0

1

( ) ( )1i

CsE s E sR Ls

Cs

    [By VDR]

  0 2

( )( )

1iE s

E sRCs LCs

 

      0

2

( ) 11( )i

E sRE s LC s sL LC

 

  Characteristic equation is given by,

  2 10

Rs s

L LC       …(i)

Standard characteristic equation for second order system is given by,

  2 22 0n ns s       …(ii)

On comparing equation (i) and equation (ii),

 1

nLC

  

and 2 n

R

L  

     1

2 2 2n

R R LC R C

L L L

 

     6

3

10 10 100.5

2 1 10

 

  Maximum peak overshoot is given by,

21MPO e

2

0.5

1 (0.5)e

 

      MPO 0.163 or 16.3% 16%  

  Hence, the correct option is (C).

10R � � 1mHL �

ie

0e10 FC � �

R Ls

( )i

E s0( )E s

1

Cs

  14 

Question 52 [Practice Book] [GATE EE 2007 IIT-Kanpur : 2 Marks]

If the above step response is to be observed on a non-storage CRO, then it would be best have the ie as a

(A) step function (B) square wave of 50 Hz

(C) square wave of 300 Hz

(D)

square wave of 2 kHz

Ans. (C)

Sol. The resonance frequency is given by,

1

nLC

3 6

1

1 10 10 10

 

    410 rad/secn  

  Settling time for 2% tolerance band is given by,

4

sn

t 4

40.8 msec.

0.5 10

 

  To observe the transient response, input must be applied atleast upto 0.8msecst .

Peak time is given by,

2 4 21 10 1 (0.5)

p

n

t

0.36 msec  

  To observe peak overshoot, the input must be applied atleast upto 0.36 msec.pt

Case 1 : If step input is applied.

The given CRO is non-storage which means it can’t record the response. So it is not possible to read the

transient response practically as it will appear only for one st i.e., 0.8 msec. After 0.8 msec CRO only

displays constant value 1.

Therefore, the constant input 1 1e should be applied again and again (i.e., square wave) so that transient

response appears always on CRO screen.

Case 2 : If square wave input is applied.

ei

1

0t

e0

0t

e0ss =1

tp =0.36 ms ts=0.8 ms

ei

t0

  15 

For obtaining transient response,

( and )2 p s

Tt t  

  Option (B) :

For 1 50 Hzf  

    1 110 ms

2 2 50 s

Tt

 

10 ms 0.8 ms  

  The output of the CRO is constant in between 0.8 ms to 10 ms as shown below,

Option (C) :

For 2 300 Hzf  

    2 11.67 ms

2 2 300 s

Tt

 

  The output of the CRO is constant in between 0.8 ms to 1.67 ms which is very less as shown below,

Option (D) :

For 3 2 kHzf  

    33

10.25 ms

2 2 2 10 s

Tt

 

  Since, 0.25 ms is very much less than 0.8 ms hence, the output waveform will not appear on the CRO screen.

Hence 300 Hz frequency square wave is the most suitable as 1.672

T msec because only for this

frequency transient response will always appear on CRO screen.

Therefore, it would be best to have ie as a square wave of 300 Hz.

Hence, the correct option is (C).

Question 53 [Practice Book] [GATE EE 2007 IIT-Kanpur : 2 Marks]

Consider the feedback system shown below which is subjected to a unit step input. The system is stable and as following parameters 4, 10

p iK K , 500 rad/sec and 0.7 . The steady state value of Z is

e0

t (ms)0 0.8 10

0 1sse �

e0

t (ms)0 0.8

0 1sse �

1.67

  16 

(A) 1 (B) 0.25 (C) 0.1 (D) 0

Ans. (A)

Sol. Given : 1

( )R ss

The given feedback system is shown below.

2

2 2( )

2n

n n

G ss s

The equivalent representation can be drawn as shown in figure below.

Closed-loop transfer function for negative unity feedback is given by,

( ) '( )

( )( ) 1 '( )

C s G sT s

R s G s

( )

( )

( )1 ( )

ip

ip

KK G s

C s sKR s

K G ss

From the figure, the error signal can be written as,

( ) ( ) ( )E s R s C s ( )

( ) 11 ( )

ip

ip

KK G s

sR s

KK G s

s

1 1

( )1 ( )i

p

E sKs

K G ss

From the figure,

( ) ( )iKZ s E s

s 2 ( ) ( )

i

p i

K s

s s sK K G s

Kp 2 22s s

� � �

Z

0

1 iK

s

Kp 2 22s s

� � �

Z

0

1 iK

s

( )R s

( )E s( )C s

( )i

p

KK G s

s

� �� �

� �( )C s( )R s

( )E s

  17 

( )Z s2

2 2

1

( )2

i

np i

n n

K

ss sK K

s s

( )Z s 2 2

2 2 2

2

2 ( )i n n

n n n p i

K s s

s s s s sK K

Steady state value can be calculated as,

0

lim ( ) lim ( )ss t sz z t s Z s

2

21n

in i

KK

Hence, the correct option is (A).

Question 63 [Practice Book] [GATE EE 2011 IIT-Madras : 2 Marks]

A two-loop position control system is shown below.

The gain k of the Tacho-generator influences mainly by (A) peak overshoot. (B) natural frequency of oscillation. (C) phase shift of the closed loop transfer function at very low frequency 0 .

(D) phase shift of the closed loop transfer function at very high frequency .

Ans. (A) Sol. Given : A two-loop position control system is shown below.

On solving the inner loop, we get

2

11( 1)(1 )1

( 1)

s sks s s k

s s

Now, the overall transfer function can be written as,

2

( ) 1

( ) ( 1) 1

Y s

R s s k s

        …..(i)

( )R s1

( 1)s s �( )Y s

ks

Tacho-generator

Motor

( )R s1

( 1)s s �( )Y s

ks

Tacho-generator

Motor

  18 

Transfer function for second-order system with unit step input is given by,

2

2 2

( )

( ) 2n

n n

C s

R s s s

             …..(ii)

where, damping ratio, n natural angular frequency

On comparing equation (i) and equation (ii), we get

ω 1n and 2ξ ω 1n k

So 1

ξ2

k

…..(iii)

Maximum peak overshoot is given by,

21MPO 0 MPO 1, 0 1e

Peak overshoots depends on damping factor ξ and ξ is proportional to gain from equation (iii). So, gain

k of the Tacho-generator influences mainly by peak overshoot.

Hence, the correct option is (A).

Question 20 [Work Book] [GATE EC/EE/IN 2013 IIT-Bombay : 2 Marks]

The open-loop transfer function of a dc motor is given as ( ) 10

( ) 1 10a

s

V s s

. When connected in feedback

as shown below, the approximate value of aK that will reduce the time constant of the closed loop system

by one hundred times as compared to that of the open-loop system is

(A) 1 (B) 5 (C) 10 (D) 100

Ans. (C)

Sol. Given : closed loop open loop

1

100 where represents time constant and

( ) 10

( ) 1 10a

s

V s s

For first order system loop transfer function is ( )

( ) 1

C s K

R s s

comparing with

( ) 10

( ) 1 10a

s

V s s

open loop 10

Closed-loop transfer function for negative unity feedback is given by,

( )

( )1 ( )

G sT s

G s

Here

10( )

1 10aG s Ks

10( ) 1 10

10( )1

1 10

a

a

Ks s

R sK

s

10 10

1 10 10 10 (10 1)a a

a a

K K

s K s K

10

1 10s�+–

( )R s ( )sa

K( )

aV s

10

1 10s�+–

( )R s ( )sa

K( )

aV s

  19 

Dividing numerator and denominator by 10 1aK

10

10 1( )

( ) 101

10 1

a

a

a

K

Ks

R ss

K

For first order system loop transfer function is ( )

( ) 1

C s K

R s s

. On comparing with

( )

( )

s

R s

we get

closed loop

10

10 1aK

We have closed loop open loop

1

100

10 1

1010 1 100aK

10 1 100aK

10 99aK

9.9 10aK

Hence, the correct option is (C). Question 67 [Practice Book] [GATE EC 2015 (Set-03) IIT-Kanpur : 2 Marks]

The position control of a DC servo-motor is given in the figure. The values of the parameters are

1 N-m/A,TK 1 ,aR 0.1H,aL 25 kg-m ,J 1 N-m/(rad/sec)B and 1 V/(rad/sec).bK

The steady-state position response (in radians) due to unit impulse disturbance torque dT is ______.

Ans. – 0.5

Sol. Given : 1 N-m/A,TK 1 ,aR ,

For unit impulse

( )aV s T

a a

K

R L s�

( )dT s

1

Js B�1

s( )s�

bK

0.1H,aL 25 kg-m ,J 1 N-m/(rad/sec)B 1 V/(rad/sec)bK

( )aV s T

a a

K

R L s�

( )dT s

1

Js B�1

s( )s�

bK

( ) 1dT s

1( )

1( ) 1( ) ( )

b TD

a a

X s Js BK KT s

Js B R L s

  20 

( )( )

( ) ( )D

a a b T

T ss

s Js B R L s K K

( ) 1DT s

Steady state response can be calculated using final value theorem. Applying final value theorem,

Hence, the correct answer is – 0.5. Question 68 [Practice Book] [GATE EC 2016 (Set-02) IISc-Bangalore : 2 Marks]

In the feedback system shown below 2

1( )

( 2 )G s

s s

.

The step response of the closed -loop system should have minimum settling time and have no overshoot.

The required value of gain K to achieve this is ________.

Ans. 1

Sol. Given : 2

1( )

2G s

s s

, '( ) ( )G s KG s

The second order closed loop transfer function with negative unity feedback is given by,

2

2

2

( ) '( ) 2( ) 1 '( ) 21

2

KY s G s Ks s

KR s G s s s Ks s

……. (i)

Minimum settling time and no overshoot

1

From equation (i),

n K

And 2 2n

2 2K

1

1K

1K

1K

( ) 1

( ) ( ) ( )D a a b T

X s

T s Js B R L s K K

1

( )( ) ( )a a b T

ss Js B R L s K K

0 0

1(0) lim ( ) lim

( ) ( )s sa a b T

s sJs B R L s K K

1 1(0) 0.5

1 1a b TBR K K

( )G sKr y

  21 

IES Objective Solutions

Question 4 [Practice Book] [IES EE 1992]

Damping factor and un-damped natural frequency for the position control system is given by

(A) 2 ,KJ KJ respectively (B) ,2

K K

fJ J respectively

(C) ,2

f K

JKJ respectively (D) ,

2

JKJ

Kf respectively

Ans. (C)

Sol. Characteristic equation of position control system is given by,

2 0f K

s sJ J

….(i)

Standard form of second order characteristic equation is given by,

2 22 0n ns s ….(ii)

On comparing equation (i) and (ii), we get

2 n

f

J 2

n

K

J

2

f J

J K rad/secn

K

J

2

f

KJ

Hence, the correct option is (C).

Question 14 [Practice Book] [IES EE 1995]

Consider a system shown in the given figure.

If the system is distributed so that (0) 1,c then ( )c t for a unit step input will be

(A) 1 t (B) 1 t (C) 1 2t (D) 1 2t

Ans. (C)

Sol. Given :

( ) 2

( )

C s

U s s

2

( ) ( )C s U ss

2

2( )C s

s

Taking inverse Laplace transform, we get

12

2( ) 2c t L t

s

( ) 2 (0) 2 1c t t c t

Hence, the correct option is (C).

2

s

  22 

Question 22 [Work Book] [IES EC 2003]

The unit impulse response of a system having transfer function K

s is shown above. The value of is:

(A) 1t (B) 1

1

t (C) 2t (D)

2

1

t

Ans. (D)

Sol. Given : ( )

( )

C s K

R s s

and ( ) 1R s

( )K

C ss

Taking inverse Laplace transform, we get

( ) tc t Ke

Time constant 1

t

Time constant is the time at which

1( ) 0.37c t Ke K

So, 2

1t

2

1

t

Hence, the correct option is (D).

Question 54 [Practice Book] [IES EE 2005]

Which one of the following is the steady state error of a control system with step error, ramp error and

parabolic error constants ,p vK K and aK respectively for the input 2(1 )3 ( )t u t ?

(A) 3 3

1 2p aK K

(B)

3 3

1 p aK K

(C)

3 3

1 p aK K

(D)

3 6

1 p aK K

Ans. (D)

Sol. Given : 2( ) 3(1 ) ( )r t t u t

Taking Laplace transform, we get

3

3 6( )R s

s s

Unit Impulse Response

K

0.63K

0.37K

1t 2tt

  23 

Steady state error is given by,

3

0 0

3 6( )

lim lim1 ( ) 1 ( )ss s s

ssR s s s

eG s G s

2 20 0

3 6lim lim

1 ( ) ( )sss s

eG s s s G s

2

0 0

3 6

1 lim ( ) lim ( )ss

s s

eG s s G s

3 6

1ssp a

eK K

Hence, the correct option is (D). Question 72 [Practice Book] [IES EE 2007]

For a unity feedback control system with forward path transfer function ( ) ,5

KG s

s

what is error

transfer function ( )eW s used for determination of error coefficients?

(A) 5

K

s (B)

5

K

s K (C)

5

5

s

s K

(D) ( 5)

5

K s

s K

Ans. (C)

Sol. Given : ( ) , ( ) 15

KG s H s

s

( ) 1

( ) 1 ( )eW s

R s G s

( ) 1

( ) 15

eW sKR s

s

( ) 5

( ) 5eW s s

R s s K

Hence, the correct option is (C). Question 77 [Practice Book] [IES EE 2008]

In the time domain analysis of feedback control systems which one pair of the following is not correctly matched?

(A) Under damped : Minimize the effect of nonlinearities (B) Dominant poles : Transients die out more rapidly (C) Far away poles to the left half of s–plane : Transients die out more rapidly (D) A pole near to the left of dominant complex poles and near a zero : Magnitude of transient is small Ans. (B) Sol. Dominant Pole : The poles that are close to the imaginary axis in the left-half s-plane give rise to transient

responses that will decay relatively slowly, whereas the poles that are far away from the axis (relative to the dominant poles) correspond to fast-decaying time responses.

Hence, the correct option is (B).

5

K

s �R s( ) C s( )

( )eW s

–

  24 

Question 89 [Practice Book] [IES EE 2009]

In a fluid flow system two fluids are mixed in appropriate proportion. The concentration at the mixing point is y(t) and it is reproduced without change, dT seconds later at the monitoring point as b(t). What is

the transfer function between b(t) and y(t)? (Where S is distance between monitoring point and mixing point)

(A) dTe (B) dT se (C) dT se (D) dTe Ans. (C)

Sol. Given : ( ) ( )dy t b t T

Taking Laplace transform, we get

( ) ( )dT sY s e B s

( )

( )dT sY s

eB s

Hence, the correct option is (C). Question 97 [Practice Book] [IES EC 2011, 2001]

When two identical first order systems have been cascaded non-interactively the unit step response on the system will be

(A) Over-damped (B) Under-damped (C) Un-damped (D) Critically-damped

Ans. (D)

Sol. Cascading of two first order system with non-interactively

02

( ) 1( )

( ) (1 )i

V sT s

V s sRC

The pole-zero diagram of above transfer function

Hence, the correct option is (D). Question 108 [Practice Book] [IES EC 2012]

Assertion (A) : A second order system subjected to a unit impulse oscillates at its natural frequency. Reason (R) : Impulse input contains frequencies from to .

Codes : (A) Both A and R are individually true and R is the correct explanation of A.

Amplifierwith

gain = 1iV 0V

RR

CC

1

1 RCs�1

1 RCs�( )iV s

0 ( )V s

j

�2 poles

Two repeated poles.

So, system is critical damped ( = 0).�

–1/RC

  25 

(B) Both A and R are individually true but R is not the correct explanation of A.

(C) A is true but R is false.

(D) A is false but R is true.

Ans. (D)

Sol. For 0 , a second order system subjected to a unit impulse oscillates at its natural frequency but it is not

always true.

0 Un-damped

Marginal stable

Imaginary

 

The impulse response contains all the frequency components having frequency response as

Hence, the correct option is (D).

Question 131 [Practice Book] [IES EE 2014]

The dominant poles of a servo-system are located at ( 2 2)s j . The damping ratio of the system is

(A) 1 (B) 0.8 (C) 0.707 (D) 0.6

Ans. (C)

Sol. Given : Pole are located at 2 2s j …. (i)

Poles of second-order transfer function is given by,

21n ns j …. (ii)

On comparing equation (i) and (ii), we get

21 2n

2 2 2 2 2(1 ) 4 4n n n …. (iii)

and 2n

2 2 4n …. (iv)

From equation (iii) and (iv), we get

2 8n

2 4 1

8 2

0.707

Hence, the correct option is (C) .

t

c t( )

0

j

�

1

� �( )L t�

f

  26 

Question 135 [Practice Book] [IES EE 2014]

A unity feedback second order control system is characterized by the open loop transfer function

( ) , ( ) 1( )

KG s H s

s Js B

J = moment of inertia, B = damping constant and K = system gain The transient response specification which is not affected by system gain variation is (A) Peak overshoot (B) Rise time (C) Settling time (D) Time to peak overshoot Ans. (C)

Sol. Given : ( )( )

KG s

s Js B

and ( ) 1H s

Characteristics equation is given by, 1 ( ) ( ) 0G s H s

2 0Js Bs K    2 0Bs K

sJ J

…. (i)

Characteristics equation for standard second-order system is given by,

2 22 0n ns s …. (ii)

On comparing equation (i) and (ii), we get

n

K

J and 2 n

B

J

1

2

B

K J

Peak overshoot is given by,

21

pM e

Since depends on gain K so peak overshoot affected by K.

Rise time is given by,

rd

t

Sine d depends on gain K so rise time affected by K, settling time is given by,

4

sn

t

4 8

2

JB BJ

Settling time is independent of gain K so st will not be affected by K.

Hence, the correct option is (C).