CONTROL SYSTEMS Chapter 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) Cs Ks R s s j s j 2 2 () ( 3) () () ( 1) 2 Cs Ks Gs Rs s For unit step input, 1 () ( ), () rt ut Rs s Steady state value can be calculated using final value theorem. Applying final value theorem, 0 3 lim ( ) lim () 2 5 ss t s K c ct sC s 10 3 K 2 10 3 () 3 2 5 s Gs s 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 2 2 10 5 3 dy dy du y u dt dt dt For an input () ut 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 y dt Input () ut is (A) (3/5) 1 7 () () 5 25 t t e ut (B) 3 1 7 () () 5 25 t t e ut (C) (3/5) 7 () 25 t e ut (D) None of these Ans. (A)
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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,
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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,
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).
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).
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).
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.
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 = ?
GATE ACADEMY ® 1 - 8 Basics of Control Systems
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.
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
GATE ACADEMY ® 1 - 10 Basics of Control Systems
/( ) (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).
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
GATE ACADEMY ® 1 - 12 Basics of Control Systems
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
Hence, the correct option is (C).
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
Hence, the correct option is (A).
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).
GATE ACADEMY ® 1 - 13 Basics of Control Systems
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
Hence, the correct option is (A).
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 �
�
�
GATE ACADEMY ® 1 - 14 Basics of Control Systems
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�
�
�
�
GATE ACADEMY ® 1 - 15 Basics of Control Systems
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.
Hence, the correct option is (C).
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
Hence, the correct option is (C).
Question 51 [Practice Book] [ESE EE 2013]
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.
GATE ACADEMY ® 1 - 16 Basics of Control Systems
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).
Question 56 [Practice Book] [ESE EE 2014]
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).