GATE Objective & Numerical Type Questions · Using final value theorem steady state output can be written as, 0 lim lim ( ) ts yt sYs 00 23 3 lim lim 0 ss ss32 3 ss ys ss s Hence,

CONTROLSYSTEMSChapter 1 : Basics of Control Systems

GATE Objective & Numerical Type Questions Question 1 [Practice Book] [GATE IN 1992, IIT-Delhi : 2 Marks]

A system has a complex pole pair of ( 1 2) j and a real zero of (– 3). The steady state output to a unit step input is 2. The transfer function of the system is _______________.

Sol. Given : Poles are at 1 2 s j and zero is at 3 s

Transfer function can be written as,

( ) ( 3)

( ) ( 1 2 ) ( 1 2 )

C s K s

R s s j s j

2 2

( ) ( 3)( )

( ) ( 1) 2

C s K s

G sR s s

For unit step input,

1

( ) ( ), ( ) r t u t R ss

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

0

3lim ( ) lim ( ) 2

5 ss

t s

Kc c t sC s

10

3K

2

10 3( )

3 2 5

s

G ss s

Question 12 [Practice Book] [GATE IN 1999 IIT-Bombay : 1 Mark]

A transfer function has two zeros at infinity. Then the relation between the numerator degree (N) and the denominator degree (M) of the transfer function is

(A) N = M + 2 (B) N = M – 2 (C) N = M + 1 (D) N = M – 1 Ans. (B) Sol. For two zeros to exist at s . It indicate there are two extra poles compare to zero. Therefore, the

numerator degree, N should be less than the denominator degree, M by 2 i.e. N = M – 2 Hence, the correct option is (B).

Question 16 [Practice Book] [GATE EE 2002 IISc-Bangalore : 2 Marks]

A single input single output system with y as output and u as input, is describe by

2

22 10 5 3

d y dy duy u

dt dt dt

For an input ( )u t with zero initial conditions the above system produces the same output as with no

input and with initial conditions (0 ) 4, (0 ) 1 d

y ydt

Input ( )u t is

(A) (3/5)1 7( ) ( )

5 25 tt e u t (B) 31 7

( ) ( )5 25

tt e u t

(C) (3/5)7( )

25 te u t (D) None of these

Ans. (A)

GATE ACADEMY ® 1 - 2 Basics of Control Systems

Sol. Given : 2

22 10 5 3

d y dy duy u

dt dt dt ….. (i)

Where u is the input and y is the output. The response with no input ( ) 0u t

But with initial conditions (0) 1, '(0) 4 y y

Using Laplace of equation (i), 2 ( ) (0) '(0) 2 ( ) (0) 10 ( )s Y s sy y sY s y Y s 5 ( ) (0) 3 ( ) sU s u U s

….(ii)

2( 2 10) ( ) 4 2 0 s s Y s s

2

2( )

2 10

s

Y ss s

…… (iii)

With input ( ) ?u t and zero initial conditions

From equation (ii), we get

2( 2 10) ( ) (5 3) ( ) s s Y s s U s

2 2 10

( ) ( )5 3

s sU s Y s

s

For the same response as equation (iii), we get

2 1 (5 3) 7

( )5 3 5 5 3

s s

U ss s

1 7 1

( )5 5 5 3

U s

s

1 7 1

( )35 255

U ss

Using inverse Laplace transform, we get

1 1 1 7 1( ) ( )

35 255

u t L U s Ls

(3/5)1 7( ) ( )

5 25 tu t t e

Hence, the correct option is (A).

Question 17 [Practice Book] [GATE EE 2003 IIT-Madras : 1 Mark]

A control system is defined by the following mathematical relationship

2

22

6 5 12(1 ) td x dxx e

dt dt

The response of the system as t is (A) 6x (B) 2x

(C) 2.4x (D) 2 x

Ans. (C) Sol. Given : The mathematical model of control system

2

22

6 5 12(1 ) td x dxx e

dt dt

Where x(t) is the response of the system. Laplace transform with initial condition assumed to be zero. (0) 0 and '(0) 0 x x


2 1 1( ) (0) '(0) 6 ( ) (0) 5 ( ) 12

2

s X s s x x sX s x X ss s

2 2( 6 5) ( ) 12

( 2)

s s X s

s s

2

24( )

( 2) ( 6 5)

X s

s s s s

Applying final value theorem,

20 0

24lim ( ) lim ( ) lim

( 2) ( 6 5)t s sx t sX s

s s s

( ) 2.4 x Hence, the correct option is (C). Question 5 [Work Book] [GATE IN 2003 IIT-Madras : 1 Mark]

In the feedback scheme shown in figure, the time-constant of the closed system will be

(A) A (B) (1 )A

(C) (D) (1 )

A

Ans. (D) Sol. Given block diagram is shown below.

Transfer function of system is given by,

( ) 1( ) 11

1

AC s As

AR s A ss

( ) '(1 )( ) 1 '1

1

AC s AA

sR s sA

where '1

AA

A

closed loop dc gain of system

'1 A

closed loop time constant of system

Hence, the correct option is (D). Question 19 [Practice Book] [GATE EC 2004 IIT-Delhi : 2 Marks]

A system described by the following differential equation 2

23 2 ( )

d y dyy x t

dt dt is initially at rest.

For input ( ) 2 ( )x t u t , the output ( )y t is

(A) 2(1 2 ) ( )t te e u t (B) 2(1 2 2 ) ( )t te e u t

(C) 2(0.5 1.5 ) ( )t te e u t (D) 2(0.5 2 2 ) ( )t te e u t

1� �A

s

�

r c

1� �A

s

�

r c


Ans. (A)

Sol. Given : 2

23 2 ( )

d y dyy x t

dt dt and ( ) 2 ( )x t u t

Where ( )x t is the input and ( )y t is the output. Taking Laplace transform,

2

( )X ss

Using Laplace transform with zero initial conditions.

2 ( ) 3 ( ) 2 ( ) ( )s Y s sY s Y s X s

2 2( 3 2) ( )s s Y s

s

2

( )( 2) ( 1)

Y ss s s

By applying partial fraction, we get

2

( 2) ( 1) 2 1

A B C

s s s s s s

1, 1 and 2A B C

Hence, 1 1 2

( )2 1

Y ss s s

Taking inverse Laplace transform, we get 2( ) 1 2 ( )t ty t e e u t Hence, the correct option is (A).

Question 6 [Work Book] [GATE IN 2004 IIT-Delhi : 1 Mark]

Consider the following systems

System 1 :1

( ) ,(2 1)

G ss

System 2 :1

( )(5 1)

G ss

The true statement regarding the system is (A) Bandwidth of system 1 is greater than the bandwidth of system 2. (B) Bandwidth of system 1 is lower than the bandwidth of system 2. (C) Bandwidth of both the systems are the same. (D) Bandwidth of both system are infinite. Ans. (A)

Sol. Given : System 1 :1

( ) ,(2 1)

G ss

System 2 :1

( )(5 1)

G ss

For first order system, bandwidth is reciprocal of time constant. Standard representation of first order system is given by,

1

( )1

G ss

where time constant

1

( ) , 2sec(2 1)

G ss

1 1

BW (system 1)2

1

( ) , 5sec(5 1)

G ss

1 1

BW (system 2)5

Bandwidth (system 1) is greater than bandwidth (system 2). Hence, the correct option is (A).


Question 21 [Practice Book] [GATE EC 2005 IIT-Bombay : 1 Mark]

Despite the presence of negative feedback, control systems still have problems of instability because the

(A) Components used have non-linearity. (B) Dynamic equations of the subsystems are not known exactly. (C) Mathematical analysis involves approximations. (D) System has large negative phase angle at high frequencies. Ans. (A) Sol. Despite the presence of negative feedback, control systems still have problems of instability because

the components used have non-linearity which introduces instability in the system. Hence, the correct option is (A).

Question 23 [Practice Book] [GATE IN 2007 IIT-Kanpur : 1 Mark]

A feedback control system with high gain K, is shown in the figure below.

Then the closed loop transfer function is (A) Sensitive to perturbations in ( )G s and ( )H s .

(B) Sensitive to perturbations in ( )G s but not to perturbations in ( )H s .

(C) Sensitive to perturbation in ( )H s but not to perturbations in ( )G s .

(D) Insensitive to perturbations in ( )G s and ( )H s .

Ans. (C) Sol. The given figure is shown below.

Transfer function is given by,

1

C KG

R KGH

1

KGC R

KGH

If K is very high then 1 KGH KGH

1

1

KG KGC R R R

KGH KGH H

It is clear from the above equation that when gain K is very high output of system is not affected by G but it is inversely proportional to feedback path gain H.

Hence, the correct option is (C).

Question 26 [Practice Book] [GATE EE 2009 IIT-Roorkee : 1 Mark]

The measurement system shown in the figure uses three sub-systems in cascade whose gains are

specified as 1 2,G G and 3

1

G. The relative small errors associated with each respective subsystem

1 2,G G and 3G are 1 2, and 3 . The error associated with the output is

(A) 1 23

1

(B) 1 2

3

(C) 1 2 3 (D) 1 2 3

K ( )G s ( )C s( )R s

( )H s

K G CR

H

1G 2G3

1

GInput Output


Ans. (C) Sol. Given :

Overall transfer function,

1 23

( ) 1( )

( )

C sG s G G

R s G

Output 1 2

3

G G

C RG

1 2 3ln ln ln ln ln C G G G R

Differentiating,

1 2 31 2 3

1 1 1 1 dC dG dG dG

C G G G

As no error in R so dR = 0

31 2

1 3 3

dGdG dGdC

C G G G

1 2 3

So relative error in output 1 2 3 Hence, the correct option is (C).

Question 27 [Practice Book] [GATE EE 2011 IIT-Madras : 2 Marks]

The response ( )h t of a linear time invariant system to an impulse ( )t , under initially relaxed condition

is 2( ) t th t e e . The response of this system for a unit step input ( )u t is

(A) 2( ) t tu t e e (B) 2( ) ( )t te e u t

(C) 2(1.5 0.5 ) ( )t te e u t (D) 2( ) ( )t te t e u t

Ans. (C)

Sol. Given : Impulse response 2( ) t th t e e

Transfer function of a system is Laplace transform of its impulse response. Transfer function = L(Impulse response)

2( ) ( )t tH s L e e

1 1

( )1 2

H ss s

( ) 1 1

( )( ) 1 2

C sH s

R s s s

1

( )R ss

(step input)

1 1 1

( ) ( ) ( )1 2

C s R s H ss s s

1 1

( )( 1) ( 2)

C ss s s s

1 1 1 1 1

( )1 2 2

C ss s s s

1.5 1 0.5

( )1 2

C ss s s

1G 2G3

1

GInput Output

R C


Step Response :

1 1.5 1 0.5( )

1 2c t L

s s s

2( ) (1.5 0.5 ) ( )t tc t e u t Hence, the correct option is (C).

Question 1 [Work Book] [GATE EC 2014 IIT-Kharagpur (Set 3) : 1 Mark]

The input 23 ( ),te u t where ( )u t is the unit step function, is applied to a system with transfer function

2.

3

s

s

If the initial value of the output is 2, then the value of the output at steady state is _________.

Ans. 0

Sol. Given : 2( ) 3 ( ),tr t e u t ( ) 2

( )( ) 3

Y s sH s

R s s

and initial value of the output is 2.

Taking Laplace transform of r(t), we get 3

( )2

R ss

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

0

lim ( ) lim ( )t s

y t sY s

0 0

2 3 3lim lim 0

3 2 3ss s s

s sy s

s s s

Hence, the answer is 0.

Question 8 [Work Book] [GATE IN 2014 IIT-Kharagpur : 1 Mark]

A plant has an open-loop transfer function,

20( )

( 0.1)( 2)( 100)pG ss s s

The approximate model obtained by retaining only one of the above poles, which is closest to the frequency response of the original transfer function at low frequency is

(A) 0.1

0.1s (B)

2

2s

(C) 100

100s (D)

20

0.1s Ans. (A)

Sol. Given : Open-loop transfer function

20( )

( 0.1)( 2)( 100)pG ss s s

At low frequency ω 0 and considering dominant pole concept the above transfer function can be written as,

0

20( ) lim

( 0.1)( 2)( 100)ps

G ss s s

( )pG s0

20 0.1lim

( 0.1)(2)(100) ( 0.1)s s s

2( ) 3 ( )tr t e u t� � 2

3

s

s

��

Steady stateoutput = ?


Dominant pole : The pole which dominates the step response of a system. These are the poles which are closest to the j axis. Dominant pole may be in the right half of s-plane.


Question 9 [Work Book] [GATE EE 2015 IIT-Kanpur (Set-2) : 1 Mark]

The unit step response of a system with the transfer function 1 2

( )1

sG s

s

is given by which one of

the following waveforms? (A) (B)

(C) (D)

Ans. (A)

Sol. Given : 1 2

( )1

sG s

s

( ) ( ) ( )Y s G s U s

(1 2 ) 1

( )(1 )

sY s

s s

( )1

A BY s

s s

1, 3A B

1 3

( )1

Y ss s

0.1�2�

j

�100�

Dominantpole

y

1

0

2�

5t

y

1

0

2

5t

y

t

1

0

2

0.75�

5

y

1

0

2�

5t

( )Y s( )G s

1

1( )U s

s�

( )u t

t


Taking inverse Laplace transform, we get

( ) ( ) 3 ( )ty t u t e u t

( ) (1 3 ) ( )ty t e u t


Question 10 [Work Book] [GATE EC 2016 IISc-Bangalore (Set-1) : 2 Marks]

A first-order low-pass filter of time constant T is excited with different input signals (with zero initial conditions up to t = 0). Match the excitation signals X, Y, Z with the corresponding time responses for t ≥ 0

X : Impulse P : /1 t Te

Y : Unit step Q : /(1 )t Tt T e

Z : Ramp R : /t Te (A) X → R, Y → Q, Z → P (B) X → Q, Y → P, Z → R (C) X → R, Y → P, Z → Q (D) X → P, Y → R, Z → Q Ans. (C) Sol. For first order system

1

( ) ; ( ) 1G s H ssT

Fig. First order system

Closed loop transfer function is given by,

( ) ( )

( ) 1 ( ) ( )

Y s G s

R s G s H s

( )

( ) ( )1 ( ) ( )

G sY s R s

G s H s

For impulse response ( ) 1R s

( ) 1 /

( )11 ( ) ( ) 1

G s sTY s

G s H ssT

1

( )1

Y ssT

/1( ) t Ty t e

T for 0t

For step response 1

( )R ss

1 (1 ) ( )

( )(1 ) (1 )

sT sTY s

s sT s sT

1 1

( )1(1 )

T TY s

s sT s T sT

y

1

0

2�

5t

1

sT( )R s ( )Y s


/( ) (1 )t Ty t e for 0t

For ramp response 2

1( )R s

s

2 2

1 1( )

1(1 ) 1

T TY s

s sT s sT

/( ) (1 )t Ty t t T e for 0t Hence, the correct option is (C).

Trick : You can directly check by

d

dt (ramp response) = step response

d

dt (step response) = impulse response

Question 3 [Work Book] [GATE EC 2016 IISc-Bangalore (Set-2) : 1 Mark]

The response of the system 2

( )( 1)( 3)

sG s

s s

to the unit step input ( )u t is ( )y t .

The value of dy

dt at 0t is ________.

Ans. 1

Sol. Given : ( 2)

( )( 1) ( 3)

sG s

s s

and ( ) ( )r t u t

Hence, 1

( )R ss

Laplace transform of the output is given by, ( ) ( ) ( )Y s G s R s

1 2

( ) ( )( 1) ( 3)

sY s G s

s s s s

(0)y initial value Hence apply initial value theorem (0) lim ( )

sy sY s

21

2(0) lim

1 3( 1) ( 3) 1 1s

s sy

s s ss s

(0) 0y

( ) (0)dy

L sY s ydt

( 2) ( 2)

( )( 1) ( 3) ( 1) ( 3)

dy s s sL sY s

dt s s s s s

0

lims

t

dy dysL

dt dt

0

21

( 2)lim 1

1 3( 1) ( 3)1 1

st

dy s s sdt s s s

s s

Hence, the correct answer is 1.


Question 30 [Practice Book] [GATE IN 2016 IISc-Bangalore : 2 Marks]

The relationship between the force ( )f t and the displacement ( )x t of a spring-mass system (with mass

M, viscous damping D and spring constant K) is

2

2

( ) ( )( ) ( )

d x t dx tM D Kx t f t

dt dt

( )X s and ( )F s are the Laplace transforms of ( )x t and ( )f t respectively. With 0.1,M 2,D 10K

in appropriate units, the transfer function( )

( )( )

X sG s

F s is

(A) 2

10

20 100s s (B) 2 20 100s s (C)

2

2

10

20 100

s

s s (D)

2 20 100

s

s s

Ans. (A)

Sol. 2 ( ) ( )

( ) ( )d x t dx t

M D Kx t f tdt dt

2

( ) 1

( )

X S

F S MS DS K

2

( ) 10

( ) 20 100

X S

F S S S

ESE Objective Type Questions Question 11 [Work Book] [ESE EC 1993]

The pole zero plot of open transfer function system shown in figure and the steady state gain is 2, the transfer function ( )G s will be given by,

(A) 2

2( 1)

4 5

s

s s

(B) 2

5( 1)

4 5

s

s s

(C) 2

10( 1)

4 5

s

s s

(D) 2

10( 1)

( 2)

s

s

Ans. (C) Sol. Given : Steady state gain = 2 Given pole zero plot is shown in figure below.

– 1

– j

Re– 2

Im

j

– 1

– j

Re– 2

Im

j


Poles are 2s j and zero at s = – 1.

Transfer function can be written as,

2

( 1) ( 1)( )

( 2 )( 2 ) ( 4 5)

K s K sT s

s j s j s s

2

10( 1)( )

( 4 5)

sT s

s s


Question 15 [Practice Book] [ESE EE 1997]

If the unit step response of a network is (1 ),te then its unit impulse response will be

(A) te (B) 1

e (C)

1 te

(D) (1 ) te

Ans. (A) Sol. Given : The unit step response of network is

( ) 1 tc t e

d

dt (Unit step response) Impulse response

(1 )t tde e

dt


Question 24 [Practice Book] [ESE EC 2002]

Consider the following single-loop feedback structure illustrating the return difference

The return difference for A is

(A) 1 A (B) 1 A (C) 1

A

A

(D) 1

A

A

Ans. (B)

Sol. Overall transfer function ( )

( )1 ( ) ( )

G sM s

G s H s

Sensitivity by with respect to ( )G s

MA

MM AMS

A A MM

1 1

MA

AA AS

A A A

2

1[1 ]

[1 ]

A AA

A

1

1 A

The return difference for A is called desensitivity.

Desensitivity 1

1Sensitivity

A

Hence, the correct option is (B).


Question 5 [Work Book] [ESE EC 2005]

A linear network has the system function

( )

( ) ( )

H s c

s a s b

The outputs of the network with zero initial conditions for two different inputs are tabled as

Input x(t) Output y(t)

u(t) 32 t tDe Ee 2 ( )te u t 3t tFe Ge

Then the values of c and H are, respectively (A) 2 and 3 (B) 3 and 2 (C) 2 and 2 (D) 1 and 3 Ans. (A)

Sol. Given : ( )

( )( ) ( )

s cT s H

s a s b

…… (i)

When input is u(t) output is,

32 t tDe Ee

When input is 2 ( )te u t output is,

3t tFe Ge Using equation (i), when input is u(t) output is,

1( )

( ) ( )

KH s c D E

s s a s b s s a s b

Taking inverse Laplace transform, we get

32 t tDe Ee So, a = 1 and b = 3 Using final value theorem,

3

0

( )lim lim 2

( ) ( )t t

s t

s H s cDe Ee

s s a s b

2 and 6Hc

Hcab

Using equation (i) when input is 2 ( )te u t output is,

( )

( 2) ( ) ( )

H s c

s s a s b

Only two terms are present in the response. Hence 2s c s 2c 3H ( 6)Hc


Question 34 [Practice Book] [ESE EC 2007]

For the system given below, the feedback does not reduce the closed-loop sensitivity due to variation of which one of the following?

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

KA

s � �

( )Y s( )R s �

�

�


Ans. (C)

Sol. Given system is shown below

( )KA

G ss

… Transfer function without feedback

( ) ( )

( )1 ( ) ( ) 1

( )

KAG s KAs

F sKAG s H s s KA

s

… T.F. with Feedback

Sensitivity of ( )G s with respect to ‘A’

( )2

( ) ( ) 0.

( ) ( )( )

G sA

A G s A s K KAS

KAG s A ss

2

( ) ( )1

( )

s s K

K s

Sensitivity of ( )F s with respect to ‘A’

( )2

( ) ( ) ( ).

( ) ( )( )

F sA

A F s A s KA K KA KS

KAF s A s KAs KA

2

( ) ( )

( )

s KA K s KA KA

K s KA

( )F sA

sS

s KA

( ) ( )G s F sA AS S i.e. sensitivity is reduced due to feedback

Similarly ( ) ( )G s F sK KS S

( ) 1G sKS ( )F s

K

sS

s K A

Sensitivity of ( )G s with respect to

( )2

( ) ( ) 0 0.

( ) / ( ) ( )G s G s s KA

SG s KA s s

( ) 0G sS

Sensitivity of ( )F s with respect to

( )2

( ) ( ) 0 ( ).

( ) ( )( )

F s F s s K A KA KAS

KAF s s K As K A

2 2

2

( )( )

( )

s K A K A

KA s K A

( )

KA

s K A

Hence, the sensitivity is reduced due to feedback.

( )R s KA

s � �( )s�

�

�

�


Sensitivity of ( )G s with respect to K

( ) ( )

( ) ( ) / ( )G sK

K G s K KAS

G s K KA s K s

( )2

( ) ( )0 0

( )G sK

s s KAS

A s

( ) 0G sKS

Sensitivity of ( )F s with respect to K

( ) ( )

( ) ( ) / ( ) ( )F sK

K F s K KAS

F s K KA s KA K s KA

( )2

( ) ( ) 0 0

( )F sK

s KA s KA KAS

A s KA

( ) 0F sKS

Hence, ( ) ( )G s F sK KS S

So, the sensitivity with respect to K does not reduce due to feedback.


Question 12 [Work Book] [ESE EC 2012]

The sensitivity TKS of transfer function

1 2

3 4

KT

K

with respect to the parameter K is given by

(A) 23

K

K (B)

2

3

2 4

K

K K (C)

2

2

3 10 8

K

K K (D)

2

4

2 5 7

K

K K Ans. (C)

Sol. Given : 1 2

3 4

KT

K

Sensitivity of ' 'T with respect to changes in ' 'K is T

KS and can be written as,

%changein

%changeinTK

TT T KTS

KK K TK

1 2

(3 4 )3 4 (1 2 )

TK

K KS K

K K K

2

(3 4 )2 (1 2 )4

(3 4 )TK

K KS

K

(3 4 )

(1 2 )

KK

K

2

(6 8 4 8 ) 2

(3 4 ) (1 2 ) 8 10 3TK

K K K KS

K K K K



Consider the following statements regarding advantages of closed loop negative feedback control systems over open loop systems.

1. The overall reliability of the closed loop system is more than that of open loop system. 2. The transient response in a closed loop system decays more quickly than in open loop system. 3. In an open loop system, closing of the loop increases the overall gain of the system. 4. In the closed loop system, the effect of variation of component parameters on its performance is

reduced.


Which of these statements are correct? (A) 1 and 2 (B) 1 and 3 (C) 2 and 4 (D) 3 and 4 Ans. (A) Sol. (i) For negative feedback control system, overall gain reduces by factor (1 ).GH So option (B) and

(D) cannot be correct. (ii) The transient response in a closed loop system decay more quickly because time constant for

closed-loop system is less than open loop system. (iii) Overall reliability of closed-loop system is more than open loop system. Hence, the correct option is (A). Question 54 [IES EE 2013]

A forcing function 2( 2 ) ( 1)t t u t is applied to a linear system. The Laplace transform of the forcing

function is

(A) 23

2 sse

s

(B) 21 ss

es

(C) 22

1 1s se es s

(D) 2

3

2 sse

s

Ans. (D)

Sol. Given : 2( ) ( 2 ) ( 1)f t t t u t 2( 1) 1 ( 1)t u t

2( ) ( 1) ( 1) ( 1)f t t u t u t Taking Laplace transform, we get

3

2( )

s se eF s

s s

2

3

2 sse

s

Hence, the correct option is (D).


The unit impulse response of a system given as 2( ) 4 6t tc t e e . The step response of the same

system for 0t equal to

(A) 23 4 1t te e (B) 23 4 1t te e (C) 23 4 1t te e (D) 23 4 1t te e

Ans. (C) Sol. Given : 2( ) 4 6t tc t e e Step response can be calculated as, Step response impulse response

Step response 2( 4 6 )t te e dt

Step response 24 3t te e C At 0,t step response is zero. 0 4 3 C 1C

Step response 24 3 1t te e Hence, the correct option is (C).

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GATE Objective & Numerical Type Questions · Using final value theorem steady state output can be written as, 0 lim lim ( ) ts yt sYs 00 23 3 lim lim 0 ss ss32 3 ss ys ss s Hence,