GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 156

UNIT 7 approximate value of Ka 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
CONTROL SYSTEMS

2013 ONE MARK (A) 1 (B) 5

The Bode plot of a transfer function G ^s h is shown in the figure

(C) 10 (D) 100
7.1

below. 7.3 The signal flow graph for a system is given below. The transfer
Y ^s h
U ^s h
function for this system is

GATE Electronics & Communication (A) s1 (B) s1

5s2  6s  2 s 2  6s  2
by RK Kanodia
(C) 2 s  1 (D) 2 1
Now in 3 Volume s  4s  2 5s  6s  2
Purchase Online at maximum discount from online store
and get POSTAL and Online Test Series Free Statement for Linked Answer Questions 4 and 5:
visit www.nodia.co.in The state diagram of a system is shown below. A system is
o  AX  Bu ;
described by the state-variable equations X
y  CX  Du

7.4 The state-variable equations of the system shown in the figure above
are
o  > 1 0 H X  > 1H u
X Xo  > 1 0 H X  > 1H u
(A) 1 1 1 (B) 1 1 1
y  61  1@X  u y  6 1  1@X  u
o  > 1 0 H X  > 1H u
X Xo  > 1  1H X  > 1H u
(C) 1 1 1 (D) 0 1 1
y  6 1  1@X  u y  61  1@ X  u
7.5 The state transition matrix eAt of the system shown in the figure
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above is
et 0 et
(A) > t tH (B) > H
0
The gain _20 log G ^s h i is 32 dB and  8 dB at 1 rad/s and te e  tet et
10 rad/s respectively. The phase is negative for all X. Then G ^s h is
(A) 39.8 (B) 392.8 et 0 et  tet
s s (C) > t tH (D) > H
e e 0 et
(C) 32 (D) 322
s s
2012 ONE MARK

2013 TWO MARKS (s  9) (s  2)

2
7.6 A system with transfer function G (s) 
(s  1) (s  3) (s  4)
7.2 The open-loop transfer function of a dc motor is given as is excited by sin (Xt). The steady-state output of the system is zero
X ^s h
 10 . When connected in feedback as shown below, the
Va ^s h 1  10s
at
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 157

(A) X  1 rad/s (B) X  2 rad/s

(C) X  3 rad/s (D) X  4 rad/s

2012 TWO MARKS

7.7 The feedback system shown below oscillates at 2 rad/s when

2011 TWO MARKS

7.11 The block diagram of a system with one input u and two outputs y1
(A) K  2 and a  0.75 (B) K  3 and a  0.75 and y2 is given below.
(C) K  4 and a  0.5 (D) K  2 and a  0.5

7.8 The state variable description of an LTI system is given by

Jxo1N J 0 a1 0NJx1N J0N
K O K OK O K O
Kxo2O  K 0 0 a2OKx2O  K0O u
Kxo O Ka 0 0OKx 3O K 1O
3 3
L P L PL P L P
Jx1N
K O
y  _1 0 0iKx2O SPECIAL EDITION ( STUDY MATERIAL FORM )
Kx 3O At market Book is available in 3 volume i.e. in 3 book binding
L P
where y is the output and u is the input. The system is control-
form. But at NODIA Online Store book is available in 10 book
lable for binding form. Each unit of Book is in separate binding.
Available Only at NODIA Online Store
(A) a1 ! 0, a2  0, a 3 ! 0 (B) a1  0, a2 ! 0, a 3 ! 0
(C) a1  0, a 3 ! 0, a 3  0 (D) a1 ! 0, a2 ! 0, a 3  0 Click to Buy
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2011 ONE MARK

7.9 The root locus plot for a system is given below. The open loop
transfer function corresponding to this plot is given by

s (s  1)
(A) G ^s h H ^s h  k A state space model of the above system in terms of the state vec-
(s  2) (s  3)
tor x and the output vector y  [y1 y2]T is
(s  1)
(B) G ^s h H ^s h  k (A) xo  [2] x  [1] u ; y  [1 2] x
s (s  2) (s  3) 2
(B) xo  [ 2] x  [1] u; y  > H x
1
(C) G ^s h H ^s h  k 1
2
s (s  1) (s  2) (s  3)
2 0
(C) xo  >
0  2H
x  > H u ; y  81 2B x
(s  1) 1
(D) G ^s h H ^s h  k 1
s (s  2) (s  3)
(D) xo  > H x  > H u ; y  > H x
7.10 For the transfer function G (jX)  5  jX , the corresponding Nyquist 2 0 1 1
plot for positive frequency has the form 0 2 1 2

Common Data For Q. 7.4 & 7.5

The input-output transfer function of a plant H (s)  100 .
s (s  10) 2
The plant is placed in a unity negative feedback configuration as
shown in the figure below.
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 158

7.12 The gain margin of the system under closed loop unity negative
feedback is
(A) 10s  1 (B) 100s  1
(A) 0 dB (B) 20 dB 0. 1 s  1 0. 1 s  1
(C) 26 dB (D) 46 dB (C) 100s (D) .1s  1
0
10s  1 10s  1
7.13 The signal flow graph that DOES NOT model the plant transfer
function H (s) is
2010 TWO MARKS

7.17 A unity negative feedback closed loop system has a plant with the
transfer function G (s)  2 1 and a controller Gc (s) in the
s  2s  2
feed forward path. For a unit set input, the transfer function of the
controller that gives minimum steady state error is
(A) Gc (s)  s  1 (B) Gc (s)  s  2
s2 s1
(s  1) (s  4)
GATE Electronics & Communication (C) Gc (s)  (D) Gc (s)  1  2  3s
(s  2) (s  3) s
by RK Kanodia
Now in 3 Volume Common Data For Q. 7.10 & 7.11 :
Purchase Online at maximum discount from online store The signal flow graph of a system is shown below:
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7.18 The state variable representation of the system can be

1 1
xo  >
 1 0H
x > Hu
0
(A) x  > 1 0H x  >2H u
1 1 0
o
(B) 2
yo  [0 0.5] x yo  80 0.5B x
1
xo  >
 1 0H
x > Hu xo  >
0H
x > Hu
1 1 0 1 0
(C) 2 (D) 1 2
2010 ONE MARK yo  80.5 0.5B x yo  80.5 0.5B x
7.19 The transfer function of the system is
The transfer function Y (s) /R (s) of the system shown is
(A) s2 1 s1
7.14
(B)
s 1 s21
(C) 2 s  1 (D) s1
s s1 s2s1

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1
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(A) 0 (B)
s1 2009 ONE MARK

(C) 2 (D) 2 7.20 The magnitude plot of a rational transfer function G (s) with real
s1 s3
coefficients is shown below. Which of the following compensators
Y (s)
7.15 A system with transfer function  s has an output has such a magnitude plot ?
Q X (s) s  p
y (t)  cos a2t  k
3
for the input signal x (t)  p cos a2t  Q k. Then, the system param-
2
eter p is
(A) 3 (B) 2/ 3
(C) 1 (D) 3 /2

7.16 For the asymptotic Bode magnitude plot shown below, the system (A) Lead compensator (B) Lag compensator
transfer function can be
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 159

(C) PID compensator (D) Lead-lag compensator (C) 6 dB and 90c

(D) 3 dB and 90c
7.21 Consider the system
dx  Ax  Bu with A  =1 0G and B  = p G
dt 0 1 q 2008 ONE MARKS
where p and q are arbitrary real numbers. Which of the following 7.26 Step responses of a set of three second-order underdamped systems
statements about the controllability of the system is true ? all have the same percentage overshoot. Which of the following
(A) The system is completely state controllable for any nonzero diagrams represents the poles of the three systems ?
values of p and q
(B) Only p  0 and q  0 result in controllability
(C) The system is uncontrollable for all values of p and q
(D) We cannot conclude about controllability from the given data

2009 TWO MARKS

7.22 The feedback configuration and the pole-zero locations of

G (s)  s2  2s  2
2

s  2s  2
are shown below. The root locus for negative values of k , i.e. for
SPECIAL EDITION ( STUDY MATERIAL FORM )
 3  k  0 , has breakaway/break-in points and angle of depar-
ture at pole P (with respect to the positive real axis) equal to At market Book is available in 3 volume i.e. in 3 book binding
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(A) ! 2 and 0c (B) ! 2 and 45c

(C) ! 3 and 0c (D) ! 3 and 45c

7.23 The unit step response of an under-damped second order system

has steady state value of -2. Which one of the following transfer
functions has theses properties ?
(A) 2  2.24 (B) 2  3.82
s  2.59s  1.12 s  1.91s  1.91
(C) 2  2.24 (D) 2  382
s  2.59s  1.12 s  1.91s  1.91

Common Data For Q. 7.16 and 7.17 :

The Nyquist plot of a stable transfer function G (s) is shown in the
figure are interested in the stability of the closed loop system in
the feedback configuration shown.

7.27 The pole-zero given below correspond to a

7.24 Which of the following statements is true ?

(A) G (s) is an all-pass filter
(B) G (s) has a zero in the right-half plane
(C) G (s) is the impedance of a passive network
(D) G (s) is marginally stable
(A) Law pass filter (B) High pass filter
7.25 The gain and phase margins of G (s) for closed loop stability are
(A) 6 dB and 180c (C) Band filter (D) Notch filter
(B) 3 dB and 180c
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 160

2008 TWO MARKS

7.28 Group I lists a set of four transfer functions. Group II gives a list
of possible step response y (t). Match the step responses with the
corresponding transfer functions.

(A) P  3, Q  1, R  4, S  2 (B) P  3, Q  2, R  4, S  1
GATE Electronics & Communication (C) P  2, Q  1, R  4, S  2 (D) P  3, Q  4, R  1, S  2
by RK Kanodia
7.29 A signal flow graph of a system is given below
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The set of equalities that corresponds to this signal flow graph is

C  H 0 x1
  


x1 0 0
  


   

(A) d x2  H B 0 x2  0 1 e o

u1 
 

 



dt u2
 
 

   

 B C 0 x3
 
 

x3


1 0 



0 B H x1
  


x1 1 0

  

 
 

(B) d x2  0  B  H x2  0 1 e o

u1 
 

 
 

dt u2
 
 

   

0 C  C x3

  

x3



0 0 



 B C 0 x1
 
 

x1 1 0
 
 

   

(C) d x2   C  H 0 x2  0 1 e o

u1 
 

 
 

dt u2
 
 

   

B H 0 x3
 
 

x3

0 0




 B 0 C x1
  


x1 1 0
 
 

   

(D) d x2  H 0 B x2  0 1 e o

u1 



 
 

dt u2

  

   

 C 0  B x3
 
 

x3

0 0 



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7.30 A certain system has transfer function
G (s)  2 s  8
s  Bs  4
where B is a parameter. Consider the standard negative unity
feedback configuration as shown below

Which of the following statements is true?

(A) The closed loop systems is never stable for any value of B
(B) For some positive value of B, the closed loop system is stable,
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 161

but not for all positive values. T (s)  5

(C) For all positive values of B, the closed loop system is stable. (s  5)( s2  s  1)
The second-order approximation of T (s) using dominant pole con-
(D) The closed loop system stable for all values of B, both positive cept is
and negative. 1 5
(A) (B)
(s  5)( s  1) (s  5)( s  1)
7.31 The number of open right half plane of
10 (C) 2 5 (D) 2 1
G (s)  5 is s s1 s s1
s  2s 4  3s3  6s2  5s  3 7.37 The open-loop transfer function of a plant is given as G (s)  s 1- 1 .
(A) 0 (B) 1 2

If the plant is operated in a unity feedback configuration, then the

(C) 2 (D) 3 lead compensator that an stabilize this control system is
10 (s  1) 10 (s  4)
7.32 The magnitude of frequency responses of an underdamped second (A) (B)
s2 s2
order system is 5 at 0 rad/sec and peaks to 10 at 5 2 rad/sec.
3 10 (s  2) 2 (s  2)
The transfer function of the system is (C) (D)
s  10 s  10
(A) 2 500 (B) 2 375
s  10s  100 s  5s  75 7.38 A unity feedback control system has an open-loop transfer function
(C) 2 720 (D) 2 1125 K
G (s) 
s  12s  144 s  25s  225 s (s2  7s  12)
7.33 Group I gives two possible choices for the impedance Z in the diagram.
The gain K for which s  1  j1 will lie on the root locus of this
The circuit elements in Z satisfy the conditions R2 C2  R1 C1. The
SPECIAL EDITION ( STUDY MATERIAL FORM )
transfer functions V0 represents a kind of controller.
Vi At market Book is available in 3 volume i.e. in 3 book binding
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Match the impedances in Group I with the type of controllers in
Group II system is
(A) 4 (B) 5.5
(C) 6.5 (D) 10

7.39 The asymptotic Bode plot of a transfer function is as shown in the

figure. The transfer function G (s) corresponding to this Bode plot is

(A) Q  1, R  2 (B) Q  1, R  3
(C) Q  2, R  3 (D) Q  3, R  2

2007 ONE MARK

7.34 If the closed-loop transfer function of a control system is given as

s5 (A) 1 (B) 1
T (s) , then It is (s  1)( s  20) s (s  1)( s  20)
(s  2)( s  3)
(A) an unstable system (B) an uncontrollable system (C) 100 (D) 100
s (s  1)( s  20) s (s  1)( 1  0.05s)
(C) a minimum phase system (D) a non-minimum phase sys-
tem
7.40 The state space representation of a separately excited DC servo
motor dynamics is given as
dX
> di H  = 1  10G=ia G  =10Gu
dt 1 1 X 0
2007 TWO MARKS o
dt
7.35 A control system with PD controller is shown in the figure. If the where X is the speed of the motor, ia is the armature current and
velocity error constant KV  1000 and the damping ratio [  0.5 , X (s)
then the value of KP and KD are u is the armature voltage. The transfer function of the motor
U (s)
is

(A) 10 (B) 1
s2  11s  11 s2  11s  11
(C) 2 10s  10 (D) 2 1
(A) KP  100, KD  0.09 (B) KP  100, KD  0.9 s  11s  11 s  s  11
(C) KP  10, KD  0.09 (D) KP  10, KD  0.9

7.36 The transfer function of a plant is Statement for linked Answer Question 8.33 & 8.34 :
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 162

Consider a linear system whose state space representation is 7.47 The transfer function of a phase lead compensator is given by
x (t)  Ax (t). If the initial state vector of the system is x (0)  = G,
1 Gc (s)  1  3Ts where T  0 The maximum phase shift provide by
2 1  Ts
2x such a compensator is
then the system response is x (t)  > H. If the itial state vector
e
 2e2t (A) Q (B) Q
2 3
of the system changes to x (0)  = G, then the system response
1
2 (C) Q (D) Q
et
becomes x (t)  > tH
4 6
e
7.48 A linear system is described by the following state equation

Xo (t)  AX (t)  BU (t), A  =

 1 0G
7.41 The eigenvalue and eigenvector pairs (Mi vi) for the system are 0 1

(A) e 1 = Go and e 2 = Go (B) e 1, = Go and e2, = Go

1 1 1 1
1 2 1 2 The state transition matrix of the system is
 cos t sin t
(A) =
 sin t cos t G
(B) =
 sin t  cos t G
cos t sin t
(C) e 1, = Go and e 2, = Go (D) e 2 = Go and e1, = Go
1 1 1 1
1 2 1 2
 cos t  sin t cos t  sin t
(C) =
 sin t cos t G
(D) =
cos t sin t G
7.42 The system matrix A is
(A) =
 1 1G
(B) =
 1  2G
0 1 1 1

(C) =
 1  1G
(D) =
 2  3G
2 1 0 1 Statement for Linked Answer Questions 7.41 & 7.42 :
Consider a unity - gain feedback control system whose open - loop
transfer function is : G (s)  as  1
GATE Electronics & Communication s2
by RK Kanodia 7.49 The value of a so that the system has a phase - margin equal to Q
4
is approximately equal to
Now in 3 Volume (A) 2.40 (B) 1.40
Purchase Online at maximum discount from online store (C) 0.84 (D) 0.74
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visit www.nodia.co.in 7.50 With the value of a set for a phase - margin of Q , the value of unit
4
- impulse response of the open - loop system at t  1 second is equal
2006 ONE MARK to
(A) 3.40 (B) 2.40
7.43 The open-loop function of a unity-gain feedback control system is
given by (C) 1.84 (D) 1.74

G (s)  K
(s  1)( s  2)
2005 ONE MARK
The gain margin of the system in dB is given by
(A) 0 (B) 1 7.51 A linear system is equivalently represented by two sets of state
equations :
(C) 20 (D) 3
Xo  AX  BU and Wo  CW  DU
The eigenvalues of the representations are also computed as [M]
2006 TWO MARKS
and [N]. Which one of the following statements is true ?
7.44 Consider two transfer functions G1 (s)  2 1 and (A) [M]  [N] and X  W (B) [M]  [N] and X ! W
s  as  b (C) [M] ! [N] and X  W (D) [M]  [N] and X ! W
G2 (s)  2 s .
s  as  b
The 3-dB bandwidths of their frequency responses are, respectively 7.52 Which one of the following polar diagrams corresponds to a lag
(A) a2  4b , a2  4b (B) a2  4b , a2  4b
(C) a2  4b , a2  4b (D) a2  4b , a2  4b
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7.45 The Nyquist plot of G (jX) H (jX)for a closed loop control system,
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passes through ( 1, j0) point in the GH plane. The gain margin of
the system in dB is equal to network ?
(A) infinite (B) greater than zero
(C) less than zero (D) zero

7.46 The positive values of K and a so that the system shown in the
figures below oscillates at a frequency of 2 rad/sec respectively are

(A) 1, 0.75 (B) 2, 0.75

(C) 1, 1 (D) 2, 2
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 163

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

2005 TWO MARKS

7.54 The polar diagram of a conditionally stable system for open loop
gain K  1 is shown in the figure. The open loop transfer function
of the system is known to be stable. The closed loop system is stable
for
Statement for Linked Answer Question 40 and 41 :
The open loop transfer function of a unity feedback system is given
by
2s
G (s)  3e
s (s  2)
7.59 The gain and phase crossover frequencies in rad/sec are, respectively

SPECIAL EDITION ( STUDY MATERIAL FORM )

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binding form. Each unit of Book is in separate binding.
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2 8 8 2 Click to Buy
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8 8
7.55 In the derivation of expression for peak percent overshoot (A) 0.632 and 1.26 (B) 0.632 and 0.485
 QY (C) 0.485 and 0.632 (D) 1.26 and 0.632
Mp  exp e o # 100%
1  Y2
7.60 Based on the above results, the gain and phase margins of the
Which one of the following conditions is NOT required ?
system will be
(A) System is linear and time invariant
(A) 7.09 dB and 87.5c (B) 7.09 dB and 87.5c
(B) The system transfer function has a pair of complex conjugate
(C) 7.09 dB and  87.5c (D)  7.09 and  87.5c
poles and no zeroes.
(C) There is no transportation delay in the system.
(D) The system has zero initial conditions. 2004 ONE MARK

7.61 The gain margin for the system with open-loop transfer function
7.56 A ramp input applied to an unity feedback system results in 5%
2 (1  s)
steady state error. The type number and zero frequency gain of the G (s) H (s)  , is
system are respectively s2
(A) 3 (B) 0
(A) 1 and 20 (B) 0 and 20
1 (C) 1 (D)  3
(C) 0 and (D) 1 and 1
20 20
7.62 Given G (s) H (s)  K .The point of intersection of the
7.57 A double integrator plant G (s)  K/s , H (s)  1 is to be compensated
2
s (s  1)( s  3)
asymptotes of the root loci with the real axis is
to achieve the damping ratio [  0.5 and an undamped natural
(A)  4 (B) 1.33
frequency, Xn  5 rad/sec which one of the following compensator
Ge (s) will be suitable ? (C)  1.33 (D) 4
(A) s  3 (B) s  99
s  99 s3
2004 TWO MARKS
(C) s  6 (D)  6
s
s  8.33 s 7.63 Consider the Bode magnitude plot shown in the fig. The transfer
K (1  s) function H (s) is
7.58 An unity feedback system is given as G (s)  .
s (s  3)
Indicate the correct root locus diagram.

(s  10) 10 (s  1)
(A) (B)
(s  1)( s  100) (s  10)( s  100)
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 164

(D) controllable and observable

102 (s  1) 103 (s  100)
(C) (D)
(s  10)( s  100) (s  1)( s  10)
Given A  =
0 1G
1 0
7.71 , the state transition matrix eAt is given by
7.64 A causal system having the transfer function H (s)  1/ (s  2) is
excited with 10u (t). The time at which the output reaches 99% of 0 et
(A) > t H (B) = G
et 0
its steady state value is
e 0 0 et
(A) 2.7 sec (B) 2.5 sec
et 0
(C) > H (D) = t G
0 et
(C) 2.3 sec (D) 2.1 sec
0 et e 0
7.65 A system has poles at 0.1 Hz, 1 Hz and 80 Hz; zeros at 5 Hz, 100
Hz and 200 Hz. The approximate phase of the system response at 2003 ONE MARK
20 Hz is
7.72 Fig. shows the Nyquist plot of the open-loop transfer function
(A)  90c (B) 0c
G (s) H (s) of a system. If G (s) H (s) has one right-hand pole, the
(C) 90c (D)  180c closed-loop system is
7.66 Consider the signal flow graph shown in Fig. The gain x5 is
x1

(A) always stable

GATE Electronics & Communication (B) unstable with one closed-loop right hand pole
by RK Kanodia (C) unstable with two closed-loop right hand poles
Now in 3 Volume (D) unstable with three closed-loop right hand poles
Purchase Online at maximum discount from online store 7.73 A PD controller is used to compensate a system. Compared to the
and get POSTAL and Online Test Series Free uncompensated system, the compensated system has
visit www.nodia.co.in (A) a higher type number (B) reduced damping
(C) higher noise amplification (D) larger transient overshoot
1  (be  cf  dg) bedg
(A) (B)
abcd 1  (be  cf  dg)
abcd 1  (be  cf  dg)  bedg 2003 TWO MARKS
(C) (D)
1  (be  cf  dg)  bedg abcd 7.74 The signal flow graph of a system is shown in Fig. below. The
2 2
If A  =
1  3G
7.67 , then sin At is transfer function C (s)/ R (s) of the system is

sin ( 4t)  2 sin ( t)  2 sin ( 4t)  2 sin ( t)

(A) 1 = G
3  sin ( 4t)  sin ( t) 2 sin ( 4t)  sin ( t)
sin ( 2t) sin (2t)
(B) =
sin (t) sin ( 3t)G
(A) 6 (B) 6s
sin (4t)  2 sin (t) 2 sin ( 4t)  2 sin ( t)
(C) 1 =
2 sin (4t)  sin (t) G
s2  29s  6 s2  29s  6
3  sin ( 4t)  sin (t)
cos ( t)  2 cos (t) 2 cos ( 4t)  2 cos ( t) s (s  2) s (s  27)
(D) 1 = G
(C) (D)
3  cos ( 4t)  cos ( t)  2 cos ( 4t)  cos (t) s2  29s  6 s2  29s  6
The root locus of system G (s) H (s)  K has the break-
7.68 The open-loop transfer function of a unity feedback system is 7.75
s (s  2)( s  3)
G (s)  K For more GATE Resources, Mock Test and
s (s2  s  2)( s  3) Study material join the community
The range of K for which the system is stable is
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(A) 21  K  0 (B) 13  K  0
4
away point located at
(C) 21  K  3 (D)  6  K  3
4 (A) ( 0.5, 0) (B) ( 2.548, 0)
7.69 For the polynomial P (s)  s  s  2s3  2s2  3s  15 the number
2 4
(C) ( 4, 0) (D) ( 0.784, 0)
of roots which lie in the right half of the s plane is
(A) 4 (B) 2 7.76 The approximate Bode magnitude plot of a minimum phase system
is shown in Fig. below. The transfer function of the system is
(C) 3 (D) 1

7.70 The state variable equations of a system are : xo1  3x1  x2  u, xo2  2x1
and y  x1  u . The system is
(A) controllable but not observable
(B) observable but not controllable
(C) neither controllable nor observable
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 165

7.82 The phase margin of a system with the open - loop transfer function
(1  s)
G (s) H (s) 
(1  s)( 2  s)
(A) 0c (B) 63.4c
(C) 90c (D) 3

7.83 The transfer function Y (s)/ U (s) of system described by the state
equation xo (t)  2x (t)  2u (t) and y (t)  0.5x (t) is
(s  0.1) 3 (s  0.1) 3
(A) 108 (B) 107 (A) 0.5 (B) 1
(s  10) (s  100)
2 (s  10)( s  100) (s  2) (s  2)

(C)
(s  0.1) 2
(D)
(s  0.1) 3 (C) 0.5 (D) 1
(s  2) (s  2)
(s  10) (s  100)
2
(s  10)( s  100) 2
7.77 A second-order system has the transfer function
2002 TWO MARKS
C (s)
 2 4
R (s) s  4s  4 7.84 The system shown in the figure remains stable when
With r (t) as the unit-step function, the response c (t) of the system (A) k   1 (B)  1  k  3
is represented by (C) 1  k  3 (D) k  3

7.85 The transfer function of a system is G (s)  100 . For a

(s  1)( s  100)
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unit - step input to the system the approximate settling time for 2%
criterion is

(A)100 sec (B) 4 sec

7.78 The gain margin and the phase margin of feedback system with
8 (C) 1 sec (D) 0.01 sec
G (s) H (s)  are
(s  100) 3 7.86 The characteristic polynomial of a system is
(A) dB, 0c (B) 3, 3
q (s)  2s5  s 4  4s3  2s2  2s  1
(C) 3, 0c (D) 88.5 dB, 3
The system is
7.79 The zero-input response of a system given by the state-space equation (A) stable (B) marginally stable
=xo G  =1 1G=x G and =x (0)G  = 0 G is
xo1 1 0 x1 x1 (0) 1 (C) unstable (D) oscillatory
2 2 2
7.87 The system with the open loop transfer function
(A) = G (B) = G
tet et
G (s) H (s)  1 has a gain margin of
t t
(A)  6 db s (s  s  1)
2
(B) 0 db
(C) = t G (D) = t G
et t
te te (C) 35 db (D) 6 db

2002 ONE MARK 2001 ONE MARK

7.80 Consider a system with transfer function G (s)  2s  6 . Its 7.88 The Nyquist plot for the open-loop transfer function G (s) of a unity
damping ratio will be 0.5 when the value of k is ks  s  6 negative feedback system is shown in the figure, if G (s) has no pole
(A) 2 (B) 3 in the right-half of s plane, the number of roots of the system
6 characteristic equation in the right-half of s plane is
(C) 1 (D) 6 (A) 0 (B) 1
6
(C) 2 (D) 3
7.81 Which of the following points is NOT on the root locus of a system
with the open-loop transfer function G (s) H (s)  k 7.89 The equivalent of the block diagram in the figure is given is
s (s  1)( s  3)
(A) s  j 3 (B) s  1.5
(C) s  3 (D) s  3
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 166

7.90 If the characteristic equation of a closed - loop system is s2  2s  2  0

, then the system is Z3 (s)  Z3 (s)
(A) ,
(A) overdamped (B) critically damped Z1 (s)  Z3 (s)  Z4 (s) Z1 (s)  Z3 (s)
(C) underdamped (D) undamped  Z3 (s)  Z3 (s)
(B) ,
Z2 (s)  Z3 (s)  Z4 (s) Z1 (s)  Z3 (s)
7.91 The root-locus diagram for a closed-loop feedback system is shown Z3 (s) Z3 (s)
(C) ,
in the figure. The system is overdamped. Z2 (s)  Z3 (s)  Z4 (s) Z1 (s)  Z3 (s)
 Z3 (s) Z3 (s)
(D) ,
Z2 (s)  Z3 (s)  Z4 (s) Z1 (s)  Z3 (s)
7.93 The open-loop DC gain of a unity negative feedback system with
closed-loop transfer function 2 s  4 is
s  7s  13
GATE Electronics & Communication (A) 4 (B) 4
by RK Kanodia 13 9
(C) 4 (D) 13
Now in 3 Volume
Purchase Online at maximum discount from online store 7.94 The feedback control system in the figure is stable
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(A) for all K $ 0 (B) only if K $ 0

(C) only if 0 # K  1 (D) only if 0 # K # 1

2000 ONE MARK

7.95 An amplifier with resistive negative feedback has tow left half plane
poles in its open-loop transfer function. The amplifier
(A) will always be unstable at high frequency
(B) will be stable for all frequency
(C) may be unstable, depending on the feedback factor
(D) will oscillate at low frequency.

2000 TWO MARKS

(A) only if 0 # k # 1 (B) only if 1  k  5 For more GATE Resources, Mock Test and
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2001 TWO MARK 7.96 A system described by the transfer function H (s)  3 1
is stable. The constraints on B and k are. s  Bs2  ks  3
7.92 An electrical system and its signal-flow graph representations are
shown the figure (A) and (B) respectively. The values of G2 and H (A) B  0, Bk  3 (B) B  0, Bk  3
, respectively are (C) B  0, Bk  3 (D) B  0, Bk  3

1999 ONE MARK

7.97 For a second order system with the closed-loop transfer function
T (s)  2 9
s  4s  9
the settling time for 2-percent band, in seconds, is
(A) 1.5 (B) 2.0
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 167

(C) 3.0 (D) 4.0 the eigen values of the closed-loop system will be
(A) 0,  1,  2
7.98 The gain margin (in dB) of a system a having the loop transfer
(B) 0,  1,  3
function
(C)  1,  1,  2
G (s) H (s)  2 is
s (s  1) (D) 0,  1,  1
(A) 0 (B) 3
(C) 6 (D) 3 1998 ONE MARK

7.99 The system modeled described by the state equations is 7.105 The number of roots of s  5s  7s  3  0 in the left half of the s
3 2

-plane is
X >
2  3H
x  > Hu
0 1 0
1 (A) zero (B) one
Y  81 1B x (C) two (D) three

(A) controllable and observable 7.106 The transfer function of a tachometer is of the form
(B) controllable, but not observable (A) Ks (B) K
s
(C) observable, but not controllable
(C) K (D) K
(D) neither controllable nor observable (s  1) s (s  1)
7.100 The phase margin (in degrees) of a system having the loop transfer
SPECIAL EDITION ( STUDY MATERIAL FORM )
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s (s  1)
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1999 TWO MARKS
7.107 Consider a unity feedback control system with open-loop transfer
7.101 An amplifier is assumed to have a single-pole high-frequency transfer function G (s)  K .
function. The rise time of its output response to a step function input s (s  1)
is 35 n sec . The upper 3 dB frequency (in MHz) for the amplifier to The steady state error of the system due to unit step input is
as sinusoidal input is approximately at (A) zero
(A) 4.55 (B) K
(B) 10 (C) 1/K
(C) 20 (D) infinite
(D) 28.6 7.108 The transfer function of a zero-order-hold system is
7.102 If the closed - loop transfer function T (s) of a unity negative feedback (A) (1/s) (1  esT )
system is given by (B) (1/s) (1  esT )
T (s)  n an  1 s  an (C) 1  (1/s) esT
s  a1 sn  1  ....  an  1 s  an (D) 1  (1/s) esT
then the steady state error for a unit ramp input is
(A) an (B) an 7.109 In the Bode-plot of a unity feedback control system, the value of
an  1 an  2
phase of G (jX) at the gain cross over frequency is  125c. The phase
a
(C) n  2 (D) zero margin of the system is
an  2
(A)  125c
7.103 Consider the points s1  3  j4 and s2  3  j2 in the s-plane. (B)  55c
Then, for a system with the open-loop transfer function
(C) 55c
G (s) H (s)  K 4 (D) 125c
(s  1)
(A) s1 is on the root locus, but not s2
7.110 Consider a feedback control system with loop transfer function
(B) s2 is on the root locus, but not s1
K (1  0.5s)
(C) both s1 and s2 are on the root locus G (s) H (s) 
s (1  s) (1  2s)
(D) neither s1 nor s2 is on the root locus The type of the closed loop system is
(A) zero
7.104 For the system described by the state equation
R 0 1 0V R0V (B) one
S W S W (C) two
xo  S 0 0 1W x  S0W u
SS0.5 1 2WW SS1WW (D) three
T X T X
If the control signal u is given by u  [ 0.5  3  5] x  v , then
7.111 The transfer function of a phase lead controller is 1  3Ts . The
1  Ts
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 168

maximum value of phase provided by this controller is (C) 0

(A) 90c (D) None of the above
(B) 60c
(C) 45c ***********

(D) 30c

7.112 The Nyquist plot of a phase transfer function g (jX) H (jX) of a system
encloses the (1, 0) point. The gain margin of the system is
(A) less than zero
(B) zero
(C) greater than zero
(D) infinity

7.113 The transfer function of a system is 2s 2  6s  5

(s  1) 2 (s  2)
The characteristic equation of the system is
(A) 2s2  6s  5  0
(B) (s  1) 2 (s  2)  0
(C) 2s2  6s  5  (s  1) 2 (s  2)  0

GATE Electronics & Communication

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(D) 2s2  6s  5  (s  1) 2 (s  2)  0

7.114 In a synchro error detector, the output voltage is proportional to

[X (t)] n, where X (t) is the rotor velocity and n equals
(A) 2
(B) 1
(C) 1
(D) 2

1997 ONE MARK

7.115 In the signal flow graph of the figure is y/x equals

(A) 3 For more GATE Resources, Mock Test and

(B) 5
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(C) 2
(D) None of the above

7.116 A certain linear time invariant system has the state and the output
equations given below
Xo1 1  1 X1
> o H  >0 1 H>X H  >1H u
0
X2 2

y  81 1B: X1 D
X2
dy
If X1 (0)  1, X2 (0)  1, u (0)  0, then is
dt t0
(A) 1
(B) 1
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 169

Pk2  ^1 h^s1h^1 h^1 h  s1

SOLUTIONS since, all the loops are touching to the paths Pk1 and Pk2 so,
% k 1  %k 2  1
Now, we have
%  1  (sum of individual loops)
7.1 Option (B) is correct.  (sum of product of nontouching
From the given plot, we obtain the slope as loops)
20 log G2  20 log G1
Slope  Here, the loops are
log w2  log w1
L1  ^ 4h^1 h  4
L2  ^ 4h^s1h  4s1
From the figure
20 log G2  8 dB
20 log G1  32 dB L 3  ^ 2h^s1h^s1h  2s2
and X1  1 rad/s L 4  ^ 2h^s1h^1 h  2s1
X2  10 rad/s As all the loop L1, L2, L 3 and L 4 are touching to each other so,
So, the slope is %  1  ^L1  L2  L 3  L 4h
Slope   8  32  1  ^ 4  4s1  2s2  2s1h
log 10  log 1
 40 dB/decade  5  6s1  2s2
Therefore, the transfer function can be given as SPECIAL EDITION ( STUDY MATERIAL FORM )
G ^s h  k2 At market Book is available in 3 volume i.e. in 3 book binding
S
at X  1
form. But at NODIA Online Store book is available in 10 book
G ^ jXh  k 2  k
w binding form. Each unit of Book is in separate binding.
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32

k  10  39.8
20
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Hence, the Transfer function is
G ^s h  k2  392.8 From Masons gain formulae
Y ^s h
s s
Option (C) is correct.  4Pk %k
U ^s h
7.2

Given, open loop transfer function %

G ^s h  10Ka  Ka 1  s2  s1
1  10s s  10 5  6s1  2s2
 2s1
By taking inverse Laplace transform, we have
g ^ t h  e t 5s  6s  2
1
10

Comparing with standard form of transfer function, Aet/U , we get 7.4 Option (A) is correct.
the open loop time constant, For the shown state diagram we can denote the states x1 , x2 as below
Uol  10
Now, we obtain the closed loop transfer function for the given
system as
G ^s h
H ^s h   10Ka
1  G ^s h 1  10s  10Ka
 Ka
s  ^Ka  101 h So, from the state diagram, we obtain
By taking inverse Laplace transform, we get xo1  x1  u
h ^ t h  ka .e^k  ht xo2  x2  ^1 h^ 1h^1 h^ 1h u  ^ 1h^1 h^ 1h x1
1
a 10

So, the time constant of closed loop system is obtained as

xo2  x2  x1  u
Ucl  1 1
ka  10 and y
or, Ucl  1  ^ 1h^1 h x2  ^ 1h^1 h^ 1h x1  ^1 h^ 1h^1 h^ 1h^1 h u
ka
(approximately)  x1  x 2  u
Now, given that ka reduces open loop time constant by a factor of Hence, in matrix form we can write the state variable equations
xo1  1 0 x1 1
> o H  > 1  1H >x H  > 1 H u
100. i.e.,
Ucl  Uol x2 2
100
y  81  1B > H  u
x1
or, 1  10 and
ka 100 x2
Hence, ka = 10 which can be written in more general form as
1 0 1
Xo  >
1  1H
Option (A) is correct. X > H
7.3

For the given SFG, we have two forward paths 1

Pk1  ^1 h^s1h^s1h^1 h  s2 y  81  1B X  u
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 170

7.5 Option (A) is correct. Auxiliary equation A (s)  as2  (k  1)  0

s2  k  1   k  1 (k  2)  (k  2)
From the obtained state-variable equations
We have a (k  1)
1 0
A >
1  1H
s  j k2
jX  j k  2
S1 0
SI  A  >
 1 S  1H
X  k2  2 (Oscillation frequency)
So,
k 2

^SI  Ah1  1 >S  1 0 H and a  2  1  3  0.75

22 4
^S  1h2 1 S  1
and
R 1 V 7.8 Option (D) is correct.
S 0 W
S1 General form of state equations are given as
S 1 1 W
W
S xo  Ax  Bu
S^S  1h S  1W
2

T X yo  Cx  Du
Hence, the state transition matrix is obtained as For the given problem
e  L ^SI  Ah
At 1 1
R 0 a 0V R0V
ZR 1 V_ S 1 W S W
]]S
1 S S  1
0 Wbb A  S 0 0 a2W, B  S0W
W SSa
L [
S 1 1 W` 0 0WW SS1WW
]S^S  1h2 S  1Wb
3
RT 0 a 0VXR0V R 0VT X
\T Xa S 1 WS W S W
AB  S 0 0 a2WS0W  Sa2W
GATE Electronics & Communication SSa 0 0WWSS1WW SS 0WW
3
RT 0 0XT aX1 a2VWTRS0XVW RSa1 a2VW
by RK Kanodia S
Now in 3 Volume A2 B  Sa2 a 3 0 0WS0W  S 0W
SS 0 a a 0 WWSS1WW SS 0WW
3 1
Purchase Online at maximum discount from online store T XT X T X
For controllability it is necessary that following matrix has a tank
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visit www.nodia.co.in R0 0 a a V
S 1 2W
U  6B : AB : A B@  S0 a2
2
0W
SS1 0 0WW
e1 0
 > t tH So, a2 ! 0 T X
te e
a1 a 2 ! 0 & a1 ! 0 a 3 may be zero or not.
7.6 Option (C) is correct.
(s2  9) (s  2) 7.9 Option (B) is correct.
G (s)  For given plot root locus exists from  3 to 3, So there must be odd
(s  1) (s  3) (s  4)
( X2  9) (jX  2) number of poles and zeros. There is a double pole at s  3
 poles  0,  2,  3,  3
(jX  1) (jX  3) (jX  4) Now
The steady state output will be zero if zeros  1
G (jX)  0 k (s  1)
Thus transfer function G (s) H (s) 
X 2  9  0 & X  3 rad/s s (s  2) (s  3) 2
7.10 Option (A) is correct.
7.7 Option (A) is correct.
We have G (jX)  5  jX
K (s  1)
Y (s)  [R (s)  Y (s)] Here T  5 . Thus G (jX) is a straight line parallel to jX axis.
s3  as2  2s  1
K (s  1) K (s  1) Option (B) is correct.
Y (s) ;1  3
s  as2  2s  1E s3  as2  2s  1
7.11
 R (s)
For more GATE Resources, Mock Test and
Y (s) [s3  as2  s (2  k)  (1  k)]  K (s  1) R (s)
Y (s)
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R (s)
K (s  1)
 3 dy
s  as2  s (2  k)  (1  k) Here x  y1 and xo  1
dx
Routh Table :
y  > H  > H  > Hx
y1 x 1
y2 2x 2
Now y1  1 u
s2
y1 (s  2)  u
yo1  2y1  u
xo  2x  u
a (2  K)  (1  K)
For oscillation, 0 xo  2x  u
a
xo  [ 2] x  [1] u
a  K1
K2 Drawing SFG as shown below
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 171

Y (s)
 1
R (s) s  1
7.15 Option (B) is correct.
Transfer function is given as
Y (s)
xo1  [ 2] x1  [1] u H (s)   s
Thus X (s) s  p
y1  x1 ; y2  2x1 jX
H (jX) 
jX  p
y  > H  > H x1
y1 1
Amplitude Response
y2 2
H (jX)  X
Here x1  x X2 p 2
7.12 Option (C) is correct. Phase Response Rh (X)  90c  tan1 a X k
100 p
We have G (s) H (s)  Q
s (s  10) 2 Input x (t)  p cos a2t  k
100 2
Now G (jX) H (jX) 
jX (jX  10) 2 Output y (t)  H (jX) x (t  Rh)  cos a2t  Q k
3
If Xp is phase cross over frequency +G (jX) H (jX)  180c
H (jX)  p  X
Xp
 180c  100 tan1 0  tan1 3  2 tan1 a
10 k X2 p 2
Thus
SPECIAL EDITION ( STUDY MATERIAL FORM )
or  180c  90  2 tan1 (0.1Xp)
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G (jX) H (jX)  100
Now
X (X2  100)
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G (jX) H (jX)  100  1 1  2
10 (100  100) 20 , (X  2 rad/ sec)
p 4p2
Gain Margin  20 log 10 G (jX) H (jX) or 4p 2  4  p 2 & 3p 2  4
 20 log 10 b 1 l or p  2/ 3
20
 26 dB Alternative :
7.13 Option (D) is correct. Rh  9 Q  a Q kC  Q
3 2 6
From option (D) TF  H (s)
Q  Q  tan1 X
100 100 So,
6 2 apk
 !
s (s2  100) s (s  10) 2
tan1 a X k  Q  Q  Q
7.14 Option (B) is correct. p 2 6 3
X  tan Q  3
a3k
From the given block diagram
p
2  3, (X  2 rad/ sec)
p
or p  2/ 3
7.16 Option (A) is correct.
Initial slope is zero, so K  1
At corner frequency X 1  0.5 rad/ sec , slope increases by  20 dB/
H (s)  Y (s)  E (s) $ 1 decade, so there is a zero in the transfer function at X 1
s1 At corner frequency X 2  10 rad/ sec , slope decreases by  20 dB/
E (s)  R (s)  H (s) decade and becomes zero, so there is a pole in transfer function at
 R (s)  Y (s) 
E (s) X2
(s  1)
K a1  s k
X1
E (s) :1 
s  1D
1  R (s)  Y (s) Transfer function H (s) 
a X2k
1  s
sE (s)
 R (s)  Y (s) ...(1)
(s  1) 1 a1  s k (1  10s)
0.1
E (s)  
(1  0.1s)
a 0.1 k
Y (s)  ...(2) 1  s
s1
7.17 Option (D) is correct.
From (1) and (2) sY (s)  R (s)  Y (s)
Steady state error is given as
(s  1) Y (s)  R (s) sR (s)
Transfer function eSS  lim
s"0 1  G (s) GC (s)
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 172

R (s)  1 (unit step unit) %  1  [L1  L2]  1  : 1  12 D  1  1  12

s s s s s
eSS  lim 1 %1  1, %2  2
1  G (s) GC (s)
s"0
 P1 % 1  P2 % 2
Y (s)
1 So, H (s) 
 lim U (s) %
s"0 G (s)
1 2 C 1 :11:1
s  2s  2 2 s (1  s)
s  2
eSS will be minimum if lim GC (s) is maximum
s"0 1  1  12 (s  s  1)
In option (D) s s

lim GC (s)  lim 1  2  3s  3 Option (C) is correct.

7.20

s"0 s"0 s This compensator is roughly equivalent to combining lead and lad
So, 1
eSS  lim  0 (minimum) compensators in the same design and it is referred also as PID
s"0 3
compensator.
7.18 Option (D) is correct. 7.21 Option (C) is correct.
Assign output of each integrator by a state variable
A =
1G
and B  = G
1 0 p
Here
0 q

AB  =
1G=q G =q G
1 0 p p

0

S  8B AB B  =
q pG
p q
GATE Electronics & Communication
by RK Kanodia S  pq  pq  0
Since S is singular, system is completely uncontrollable for all val-
Now in 3 Volume
ues of p and q .
Purchase Online at maximum discount from online store
7.22 Option (B) is correct.
and get POSTAL and Online Test Series Free The characteristic equation is
visit www.nodia.co.in 1  G (s) H (s)  0
K (s2  2s  2)
or 1 0
s 2  2s  2
or s  2s  2  K (s  2s  2)  0
2 2

K  s2  2s  2
2
or
s  2s  2
For break away & break in point differentiating above w.r.t. s we
have
xo1  x1  x2
dK  (s  2s  2)( 2s  2)  (s  2s  2)( 2s  2)  0
2 2
xo2  x1  2u
ds (s  2s  2) 2
2
y  0.5x1  0.5x2
Thus (s2  2s  2)( 2s  2)  (s2  2s  2)( 2s  2)  0
State variable representation
or s ! 2
1 1
xo  >
 1 0H
x  > Hu
0
2 Let Rd be the angle of departure at pole P , then
yo  [0.5 0.5] x
7.19 Option (C) is correct.
By massons gain formula
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Study material join the community
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Transfer function
H (s) 
Y (s)

/ PK %K
U (s) %
Forward path given
P1 (abcdef )  2 # 1 # 1 # 0.5  12
s s s
1
P2 (abcdef )  2 # # 1 # 0.5
3
Loop gain L1 (cdc)  1  Rd  Rp1  Rz1  Rz2  180c
s
 Rd  180c  ( Rp1  Rz1  R2)
L2 (bcdb)  # #  1  21
1 1
s s s
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 173

 180c  (90c  180  45c)  45c 1  G (s) H (s) 0

7.23 Option (B) is correct. 1 s8 0
For under-damped second order response s 2  Bs  4
or s 2  Bs  4  s  8 0
T (s)  kXn2 where Y  1
s  2YXn s  Xn2 2 or s2  (B  1) s  4 0
Thus (A) or (B) may be correct This will be stable if (B  1)  0 
B   1. Thus system is stable
For option (A) Xn  1.12 and 2YXn  2.59 
Y  1.12 for all positive value of B.
For option (B) Xn  1.91 and 2YXn  1.51 
Y  0.69 7.31 Option (C) is correct.
7.24 Option (B) is correct. The characteristic equation is
The plot has one encirclement of origin in clockwise direction. Thus 1  G (s)  0
G (s) has a zero is in RHP. or s5  2s 4  3s3  6s2  5s  3  0
7.25 Option (C) is correct. Substituting s  z1 we have
The Nyzuist plot intersect the real axis ate - 0.5. Thus 3z 5  5z 4  6 z 3  3 z 2  2 z  1  0
G. M.  20 log x  20 log 0.5  6.020 dB The routh table is shown below. As there are tow sign change in
And its phase margin is 90c. first column, there are two RHS poles.
7.26 Option (C) is correct.
Transfer function for the given pole zero plot is:
(s  Z1)( s  Z2) SPECIAL EDITION ( STUDY MATERIAL FORM )
(s  P1)( s  P2) At market Book is available in 3 volume i.e. in 3 book binding
From the plot Re (P1 and P2 )>(Z1 and Z2 ) form. But at NODIA Online Store book is available in 10 book
So, these are two lead compensator. binding form. Each unit of Book is in separate binding.
Hence both high pass filters and the system is high pass filter. Available Only at NODIA Online Store
7.27 Option (C) is correct. Click to Buy
Percent overshoot depends only on damping ratio, Y .
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Mp  e YQ 1  Y
2

If Mp is same then Y is also same and we get

Y  cos R z5 3 6 2
4 5 3 1
Thus R  constant z
3 21 7
The option (C) only have same angle. z 5 5

7.28 Option (D) is correct. z2 4

3 3
P  2 25 2YXn  0, Y  0 Undamped Graph 3 z 1
 7
4
s  25


z0 1
Q 62 2YXn  20, Y  1 Overdamped Graph 4
s2  20s  62


7.32 Option (C) is correct.

R 2 62 2YXn  12, Y  1 Critically Graph 1
For underdamped second order system the transfer function is
s  12s  62 KXn2


T (s) 
7 2 s  2YXn s  Xn2
2
S 2YXn  7, Y  1 underdamped Graph 2
s2  7s  72 It peaks at resonant frequency. Therefore


7.29 Option (C) is correct. Resonant frequency Xr  Xn 1  2Y2

We labeled the given SFG as below : and peak at this frequency
Nr  5
2Y 1  Y2

We have Xr  5 2 , and Nr  10 . Only options (A) satisfy these

3
values.
Xn  10, Y  1
2
From this SFG we have
xo1  Hx1  Cx3  N1 where Xr  10 1  2` 1 j  5 2
4
xo2  Hx1  Bx3 and Nr  1 5 1  10 Hence satisfied
xo3  Cx1  Bx3  u2 22 1 4 3
 H 0 C x1 Option (B) is correct.
  


x1 0 1
     
 

7.33

x2  H 0 B x2  0 0 e o
 


u1 
   

Thus The given circuit is a inverting amplifier and transfer function is

u2
     
 

 C 0  B x3 Vo   Z   Z (sC1 R1  1)
    
  

x3 

1 0


Vi R 1
R1
7.30 Option (C) is correct. sC R + 1
1 1

The characteristic equation of closed lop transfer function is

 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 174

(sC2 R2  1)
For Q , Z 
sC2 7.38 Option (D) is correct.
Vo (sC2 R2  1) (sC1 R1  1) For ufb system the characteristics equation is
 # PID Controller
Vi sC2 R1 1  G (s)  0
For R, Z  R2 K
(sC2 R2  1) or 1 0
s (s2  7s  12)
Vo R2 (sC1 R1  1) s (s  7s  12)  K  0
2
 # or
Vi (sC2 R2  1) R1
Point s  1  j lie on root locus if it satisfy above equation i.e
Since R2 C2  R1 C1, it is lag compensator.
( 1  j)[(  1  j) 2  7 ( 1  j)  12)  K]  0
7.34 Option (D) is correct.
or K  10
In a minimum phase system, all the poles as well as zeros are on the
left half of the s plane. In given system as there is right half zero 7.39 Option (D) is correct.
(s  5), the system is a non-minimum phase system. At every corner frequency there is change of -20 db/decade in slope
which indicate pole at every corner frequency. Thus
7.35 Option (B) is correct.
G (s)  K
We have Kv  lim sG (s) H (s) s (1  s)`1  s j
20

s 0

(Kp  KD s) 100 Bode plot is in (1  sT) form

or 1000  lim s  Kp
s (s  100)

20 log K
s 0
 60 dB = 1000
Now characteristics equations is X X = 0. 1
Thus K 5
GATE Electronics & Communication 100
Hence G (s) 
by RK Kanodia s (s  1)( 1  .05s)
Now in 3 Volume 7.40 Option (A) is correct.
dX
> di H = = 1  10G=in G + =10Gu
Purchase Online at maximum discount from online store dt 1 1 X 0
We have a

and get POSTAL and Online Test Series Free dt

dX  X  i
visit www.nodia.co.in or
dt n ...(1)

1  G (s) H (s)  0 and dia  X  10i  10u ...(2)

a
(Kp  KD s) 100 dt
1000  lims

0s  Kp
s (s  100) Taking Laplace transform (i) we get
Now characteristics equation is sX (s)  X (s)  Ia (s)
1  G (s) H (s)  0 or (s  1) X (s)  Ia (s) ...(3)
(100  KD s) 100 Taking Laplace transform (ii) we get
or 1 0 Kp  100
s (s  10) sIa (s)  X (s)  10Ia (s)  10U (s)
or s2  (10  100KD) s  10 4  0 or X (s)  ( 10  s) Ia (s)  10U (s)
Comparing with s2  2YXn  Xn2  0 we get  ( 10  s)( s  1) X (s)  10U (s) From (3)
2YXn  10  100KD or X (s)  [s2  11s  10] X (s)  10U (s)
or KD  0.9 or (s  11s  11) X (s)  10U (s)
2

Option (D) is correct. X (s)

 2 10
7.36
or
We have T (s)  5 U (s) (s  11s  11)
(s  5)( s2  s  1) 7.41 Option (A) is correct.
 5  2 1
5`1  s j (s2  s  1) s s1 For more GATE Resources, Mock Test and
5
In given transfer function denominator is (s  5)[( s  0.5) 2  43 ] Study material join the community
. We can see easily that pole at s  0.5 ! j 23 is dominant then http://www.facebook.com/gateec2014
pole at s  5 . Thus we have approximated it.
We have xo (t)  Ax (t)
Option (A) is correct.
A =
r sG
7.37
p q
1  1 Let
G (s) 
s  1 (s  1)( s  1)
2

For initial state vector x (0)  = G the system response is

1
The lead compensator C (s) should first stabilize the plant i.e. 2
e2t
x (t)  > H
remove 1 term. From only options (A), C (s) can remove this
(s  1)  2e2t
term
e2t
> d ( 2e2t)H =
r s G= 2G
d
dt p q 1
1 10 (s  1) Thus
Thus G (s) C (s)  # dt
(s  1)( s  1) (s  2) t0

10 2 (0)

> 4e2 (0) H  =r s G= 2G

 Only option (A)  2e p q 1
(s  1)( s  2) or
satisfies.
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 175

2 p  2q
= 4 G  = r  2s G
Given system is 2nd order and for 2nd order system G.M. is infinite.
7.44 Option (D) is correct.
We get p  2q  2 and r  2s  4 ...(i) 7.45 Option (D) is correct.
For initial state vector x (0)  = G the system response is
1
If the Nyquist polt of G (jX) H (jX) for a closed loop system pass
1
et through ( 1, j0) point, the gain margin is 1 and in dB
x (t)  > tH GM  20 log 1
e
t  0 dB
> d ( et)H =
r s G= 1G
d
dt e p q 1
Thus
dt
7.46 Option (B) is correct.
t0
The characteristics equation is
 e (0)
> e (0) H  =r s G= 1G
p q 1 1  G (s) H (s)  0

1 pq K (s  1)
= 1G  = r  s G
1 0
s3  as2  2s  1
We get p  q  1 and r  s  1 ...(2) s3  as2  (2  K) s  K  1  0
Solving (1) and (2) set of equations we get The Routh Table is shown below. For system to be oscillatory

=r s G  = 2  3G
p q 0 1 stable

The characteristic equation SPECIAL EDITION ( STUDY MATERIAL FORM )

MI  A  0 At market Book is available in 3 volume i.e. in 3 book binding
M 1
form. But at NODIA Online Store book is available in 10 book
0 binding form. Each unit of Book is in separate binding.
2 M3
Available Only at NODIA Online Store
or M (M  3)  2  0
Click to Buy
or M  1,  2
Thus Eigen values are  1 and  2
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Eigen vectors for M1  1
(M1 I  A) X1  0 a (2  K)  (K  1)
0
a
M1  1 x11
or = 2 M  3G=x G  0 or a  K1 ...(1)
1 21 K2
 1  1 x11
= 2 2 G=x G  0 Then we have
21
as2  K  1  0
or  x11  x21  0 At 2 rad/sec we have
or x11  x21  0 s  jX s2  X2  4 ,
We have only one independent equation x11  x21.


Thus  4a  K  1  0 ...(2)
Let x11  K , then x21  K , the Eigen vector will be Solving (i) and (ii) we get K  2 and a  0.75 .
=x G  = K G  K = 1G
x11 K 1
21 s3 1 2K
Now Eigen vector for M2  2 s 2
a 1K
(M2 I  A) X2  0 s1 (1  K) a  (1  K)
a
M2  1 x12
or = 2 M  3G=x G  0 s0 1K
2 22
 2  1 x11
or = 2 1 G=x G  0 7.47 Option (D) is correct.
21 The transfer function of given compensator is
 x11  x21  0
Gc (s)  1  3Ts
or
T0
or x11  x21  0 1  Ts
Comparing with
We have only one independent equation x11  x21.
Gc (s)  1  aTs we get a  3
Let x11  K, then x21  K , the Eigen vector will be 1  Ts

=x G  = 2K G  K = 2G
x12 K 1 The maximum phase sift is
Gmax  tan1 a  1
22

7.42 Option (D) is correct. 2 a

1 3  1
As shown in previous solution the system matrix is  tan  tan1 1
2 3 3
A =
 2  3G
0 1
or Gmax  Q
6
7.43 Option (D) is correct. 7.48 Option (A) is correct.
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 176

s 1 T (s)  1  sT
(sI  A)  =
0 s G = 1 0G =1 s G
s 0 0 1 C  1; T  0
  1  sCT
1  X2 T2
1 =s  1G  >s  1 H
s 1
s2  1 T (jX) 
(sI  A) 1 
2

1 s 1 s 1  X2 C2 T2
s 1
2
s 1 2
s 1
2

and +T (jX)  tan1 (XT)  tan1 (XCT)

G (t)  eAt  L1 [(sI  A)] 1  =
 sin t cos t G
cos t sin t
At X  0 , T (jX)  1
7.49 Option (C) is correct. At X  0 , +T (jX)  tan1 0  0
We have G (s)  as  1 At X  3 , T (jX)  1
C
s2
+G (jX)  tan1 (Xa)  Q At X  3 , +T (jX)  0
Since PM is Q i.e. 45c, thus 7.53 Option (A) is correct.
4 Despite the presence of negative feedback, control systems still have
Q  Q  + G (j X ) X Gain cross over Frequen-
g g
4 

problems of instability because components used have nonlinearity.

cy There are always some variation as compared to ideal characteristics.

or Q  Q  tan1 (X a)  Q 7.54 Option (B) is correct.

g
4
7.55 Option (C) is correct.
or Q  tan1 (X a)
4 g The peak percent overshoot is determined for LTI second order
closed loop system with zero initial condition. Its transfer function
GATE Electronics & Communication is
T (s)  Xn2
by RK Kanodia
s2  2YXn s  Xn2
Now in 3 Volume Transfer function has a pair of complex conjugate poles and zeroes.
Purchase Online at maximum discount from online store 7.56 Option (A) is correct.
and get POSTAL and Online Test Series Free For ramp input we have R (s)  12
visit www.nodia.co.in s
Now ess  lim sE (s)


s 0
or aXg  1 R (s) 1
 lim s  lim
At gain crossover frequency G (jXg)  1 1  G (s) s 0 s  sG (s)
 

s 0

1 + a2 Xg2 or ess  lim 1  5%  1 Finite

Thus 1 s 0 sG (s) 20


Xg2
But kv  1  lim sG (s)  20
ess


or 1  1  Xg2 (as aXg  1) s 0

or Xg  (2)
1
4 kv is finite for type 1 system having ramp input.

7.50 Option (C) is correct.

7.57 Option (A) is correct.
For a  0.84 we have 7.58 Option (C) is correct.
Any point on real axis of s  is part of root locus if number of OL
G (s)  0.84s2  1
s poles and zeros to right of that point is even. Thus (B) and (C) are
Due to ufb system H (s)  1 and due to unit impulse response possible option.
R (s)  1, thus The characteristics equation is
C (s)  G (s) R (s)  G (s) 1  G (s) H (s)  0

 0.84s2  1  12  0.84 For more GATE Resources, Mock Test and

s s s
Taking inverse Laplace transform Study material join the community
c (t)  (t  0.84) u (t) http://www.facebook.com/gateec2014
At t  1, c (1 sec)  1  0.84  1.84 K (1  s)
or 1 0
7.51 Option (C) is correct. s (s  3)
K  s  3s
2
We have Xo  AX  BU where M is set of Eigen values or
o 1s
and W  CW  DU where N is set of Eigen values
For break away & break in point
If a liner system is equivalently represented by two sets of state
dK  (1  s)( 2s  3)  s2  3s  0
equations, then for both sets, states will be same but their sets of ds
Eigne values will not be same i.e. or  s2  2s  3  0
X  W but M ! N which gives s  3 ,  1
7.52 Option (D) is correct. Here  1 must be the break away point and 3 must be the break in
The transfer function of a lag network is point.
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 177

G (jX) H (jX)  2 1  X2  3
7.59 Option (D) is correct.
2s
X2
G (s)  3e
s (s  2) Thus gain margin is  1  0 and in dB this is  3 .
2jX 3
or G (jX)  3e Option (C) is correct.
jX (jX  2)
7.62

3 Centroid is the point where all asymptotes intersects.

G (jX)  4Real of Open Loop Pole  4Real Part of Open Loop Pole
X X2  4 T =
4No.of Open Loop Pole  4No.of Open Loop zero
Let at frequency Xg the gain is 1. Thus
3   1  3  1.33
1 3
Xg (Xg2  4)
7.63 Option (C) is correct.
or Xg4  4Xg2  9  0 The given bode plot is shown below
or Xg2  1.606
or Xg  1.26 rad/sec
Now +G (jX)  2X  Q  tan1 X
2 2
Let at frequency XG we have +GH  180c
X
 Q  2XG  Q  tan1 G
2 2
1 XG Q SPECIAL EDITION ( STUDY MATERIAL FORM )
or 2XG  tan 
2 2 At market Book is available in 3 volume i.e. in 3 book binding
XG 1 XG 3 form. But at NODIA Online Store book is available in 10 book
or 2XG  c  ` jm  Q
2 3 2 2 binding form. Each unit of Book is in separate binding.
5XG XG3 Available Only at NODIA Online Store
or  Q
2 24 2 Click to Buy
5XG
. Q www.nodia.co.in
2 2
or XG  0.63 rad At X  1 change in slope is +20 dB  1 zero at X  1
At X  10 change in slope is  20 dB 1 poles at X  10
Option (D) is correct.


7.60
At X  100 change in slope is  20 dB 1 poles at X  100
The gain at phase crossover frequency XG is


K (s  1)
G (jXg)  3  3 Thus T (s)  s
1
( 10  1)( 100
s
 1)
XG (XG2  4) 0.63 (0.632  4) 2

or G (jXg)  2.27 Now 20 log10 K  20 K  0.1

0.1 (s  1) 100 (s  1)
G.M.  20 log G (jXg) Thus T (s)  s 
( 10  1)( 100
s
 1) (s  10)( s  100)
 20 log 2.26  7.08 dB
7.64 Option (C) is correct.
Since G.M. is negative system is unstable.
We have r (t)  10u (t)
The phase at gain cross over frequency is
X or R (s)  10
+G (jXg)  2Xg  Q  tan1 g s
2 2
Q Now H (s)  1
 2 # 1.26   tan1 1.26 s2
2 2
C (s)  H (s) $ R (s)  1 $ 10 10
or  4.65 rad or  266.5c s  2 s s (s  2)
PM  180c  +G (jXg)  180c  266.5c  86.5c or C (s)  5  5
s s2
7.61 Option (D) is correct.
c (t)  5 [1  e2t]
The open loop transfer function is
2 (1  s) The steady state value of c (t) is 5. It will reach 99% of steady
G (s) H (s)  state value reaches at t , where
s2
Substituting s  jX we have 5 [1  e2t]  0.99 # 5
2 (1  jX) or 1  e2t  0.99
G (jX) H (jX)  ...(1)
 X2 e2t  0.1
+G (jX) H (jX)  180c  tan1 X
or  2t  ln 0.1
The frequency at which phase becomes  180c, is called phase
or t  2.3 sec
crossover frequency.
7.65 Option (A) is correct.
Thus  180  180c  tan1 XG Approximate (comparable to 90c) phase shift are
or tan1 XG  0 Due to pole at 0.01 Hz  90c 

or XG  0 Due to pole at 80 Hz  90c

The gain at XG  0 is Due to pole at 80 Hz 0 

Due to zero at 5 Hz 90c


 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 178

Due to zero at 100 Hz 0

1  2 x12


Due to zero at 200 Hz 0 

or = 1 2 G=x G  0
Thus approximate total  90c phase shift is provided. 22

We have only one independent equation x12  2x22

7.66 Option (C) is correct.
Mason Gain Formula Let x22  K , then x12  2K . Thus Eigen vector will be

=x G  = K G  K = 1 G
4pk 3 k x12 2K 2
T (s) 
3 22

In given SFG there is only one forward path and 3 possible loop. Digonalizing matrix
1 2
M =
x21 x22 G = 1 1G
p1  abcd x11 x12

31  1
1 2
3 1  (sum of indivudual loops) - (Sum of two non touching Now M1  `  1 j= G
3 1 1
loops)
Now Diagonal matrix of sin At is D where
 1  (L1  L2  L3)  (L1 L3)
sin (M1 t) sin ( 4t)
D =
sin (M2 t)G = sin (M2 t)G
0 0
Non touching loop are L1 and L3 where 
0 0
L1 L2  bedg
Now matrix B  sin At  MDM1
C (s) p1 3 1  1 2 sin ( 4t) 1 2

` 1 j=
1 1G= sin ( t)G= 1  1G
Thus 0
R (s) 1  (be  cf  dg)  bedg
3 0

GATE Electronics & Communication

 sin ( 4t)  2 sin ( t) 2 sin ( 4t)  2 sin ( t)
` 1 j=
sin ( 4t)  2 sin (t)  2 sin ( 4t)  sin ( t)G
by RK Kanodia
3
Now in 3 Volume
Purchase Online at maximum discount from online store  sin ( 4t)  2 sin ( t) 2 sin ( 4t)  2 sin ( t)
` 1 j=
and get POSTAL and Online Test Series Free 3 sin ( 4t)  sin ( t)  2 sin ( 4t)  2 sin ( t)G
visit www.nodia.co.in
sin ( 4t)  2 sin ( t)  2 sin ( 4t)  2 sin ( t)
 ` 1 j= Gs
abcd 3  sin ( 4t  sin ( t) 2 sin ( 4t)  sin ( t)

1  (be  cf  dg)  bedg 7.68 Option (A) is correct.
7.67 Option (A) is correct. For ufb system the characteristic equation is
2 2
A =
1  3G
1  G (s)  0
We have
1 K1 + G (s) 0
Characteristic equation is s (s2  2s  2)( s  3)
[MI  A]  0 s  4s  5s  6s  K  0
4 3 2

The routh table is shown below. For system to be stable,

M  2 2
or 0 (21  4K)
1 M  3 0  K and 0 
2/7
or (M + 2)( M + 3)  2  0 21
This gives 0K
or M2  5M  4  0 4
Thus M1  4 and M2  1
s4 1 5 K
Eigen values are  4 and  1.
Eigen vectors for M1  4
For more GATE Resources, Mock Test and
(M1 I  A) X1  0
M1  2  2 x11
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or = 1 M  3G=x G  0 http://www.facebook.com/gateec2014
1 21
 2  2 x11
= 1  1G=x G  0 s3 4 6 0
21
2 7
or  2x11  2x21  0 s 2 K
21  4K
or x11  x21  0 s 1
7/2
0
We have only one independent equation x11  x21. s 0
K
Let x21  K , then x11  K , the Eigen vector will be
K 1 Option (B) is correct.
=x G  = K G  K = 1 G
7.69
x11
21 We have P (s)  s5  s 4  2s3  3s  15
Now Eigen vector for M2  1 The routh table is shown below.
If F 0+ then 2F +F 12 is positive and 15F2F 2412F  144 is negative. Thus
2

(M2 I  A) X2  0 

M2  2  2 x12
=  1 M  3G=x G  0
there are two sign change in first column. Hence system has 2 root
or
2 22
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 179

on RHS of plane. where L1 and L3 are non-touching

C (s)
s5 1 2 3 This
R (s)
p1 3 1
s 4 1 2 15 
1  (loop gain) + pair of non  touching loops
F  12 ^ s s27 h
3 0
s ^
s  27
h
 
1 ^  24s  s2 h  s2 . s3
s

2 2F + 12 3 1  29s  s62
s F 15 0 s
s (s  27)
15F2  24F  144 
s1 2F  12 s2  29s  6
s0 0 7.75 Option (D) is correct.
We have
7.70 Option (D) is correct.
1  G (s) H (s)  0
 3  1 x1
=x G = = 2 0 G=x G + = 0 Gu
x1 1
We have or 1 K 0
2 2 s (s  2)( s  3)
Y  [1 0]= G  = G u
x1 1 or K  s (s2  5s2  6s)
and
x2 2 dK  (3s2  10s  6)  0
3 1
A =
2 0G
, B  = G and C  [1 0]
1 ds
Here
0
The controllability matrix is which gives s   10 ! 100  72  0.784,  2.548
6
1 3
QC  [B AB ]  =
0 2G SPECIAL EDITION ( STUDY MATERIAL FORM )
det QC ! 0 Thus controllable At market Book is available in 3 volume i.e. in 3 book binding
The observability matrix is form. But at NODIA Online Store book is available in 10 book
Q0  [CT AT CT ]
binding form. Each unit of Book is in separate binding.
Available Only at NODIA Online Store
1 3
=
0  1G Click to Buy
!0

det Q0 ! 0 Thus observable www.nodia.co.in

7.71 Option (B) is correct.
The location of poles on s  plane is
s1 0
(sI  A)  =
0 s G =0 1G = 0 s  1G
s 0 1 0
 

(s  1)
2= G > H
1
1 0 s1 0
(sI  A) 1 
(s  1) 0 (s  1) 0 1
s1

eAt  L1 [(sI  A)] 1  = G

et 0 Since breakpoint must lie on root locus so s  0.748 is possible.
0 et
7.76 Option (A) is correct.
7.72 Option (A) is correct. The given bode plot is shown below
Z  PN
N Net encirclement of ( 1  j0) by Nyquist plot,


P Number of open loop poles in right hand side of s  plane



Z Number of closed loop poles in right hand side of s  plane



Here N  1 and P  1
Thus Z 0
Hence there are no roots on RH of s plane and system is always
stable. At X  0.1 change in slope is  60 dB 3 zeroes at X  0.1


7.73 Option (C) is correct. At X  10 change in slope is  40 dB 2 poles at X  10



PD Controller may accentuate noise at higher frequency. It does not At X  100 change in slope is  20 dB 1 poles at X  100 

effect the type of system and it increases the damping. It also reduce K ( 0s.1  1) 3
Thus T (s)  s
the maximum overshoot. ( 10  1) 2 ( 100
s
 1)
7.74 Option (D) is correct. Now 20 log10 K  20
Mason Gain Formula or K  10
4pk 3 k 10 ( 0s.1  1) 3 108 (s  0.1) 3
T (s)  Thus T (s)  s 
3 ( 10  1) ( 100  1)
2 s
(s  10) 2 (s  100)
In given SFG there is only forward path and 3 possible loop. 7.77 Option (B) is correct.
p1  1 The characteristics equation is
31  1  3  24  s  27 s2  4s  4  0
s s s
Comparing with
L1   2 , L2   24 and L3   3
s s s s2  2YXn  Xn2  0
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 180

we get 2YXn  4 and Xn2  4 Y (s)  0.5X (s)

Thus Y 1 Critically damped 0.5 # 2U (s)
or Y (s) 
s2
ts  4  4  2
YXn 1#2 Y (s) 1
or 
U (s) (s  2)
7.78 Option (B) is correct.
7.84 Option (D) is correct.
7.79 Option (C) is correct.
From Mason gain formula we can write transfer function as
We have
Y (s) K
K
xo1  s

=xo G  =1 1G=x2 G
and =
x2 (0)G = 0 G
1 0 x1 x1 (0) 1
 R (s) 1  ( s3  sK ) s  3 (3  K)
2
For system to be stable (3  K)  0 i.e. K  3
A =
1G
1 0
1 7.85 Option (B) is correct.
s1 0 The characteristics equation is
(sI  A)  =
s G =1 1G =  1 s  1G
s 0 1 0
  (s  1)( s  100)  0
0
s2  101s  100  0
1 >(s  1) H> H
1
0 s1 0
(sI  A) 1  1
Comparing with s  2YXn  Xn2  0 we get
2
(s  1) 2  1 (s  1)
1
(s  1)2 s1
2YXn  101 and Xn2  100
L1 [(sI  A) 1]  eAt  = t t G
et 0
te e Thus Y  101 Overdamped
20
x (t)  eAt # [x (t0)]  = t t G= G  = t G
et 0 1 et
te e 0 te For overdamped system settling time can be determined by the
GATE Electronics & Communication dominant pole of the closed loop system. In given system dominant
by RK Kanodia pole consideration is at s  1. Thus
Now in 3 Volume 1 1 and Ts  4  4 sec
Purchase Online at maximum discount from online store T T
and get POSTAL and Online Test Series Free 7.86 Option (B) is correct.
visit www.nodia.co.in Routh table is shown below. Here all element in 3rd row are zero, so
system is marginal stable.

7.80 Option (C) is correct. s5 2 4 2

The characteristics equation is s 4 1 2 1
ks2  s  6  0 3
s 0 0 0
or s2  1 s  6  0 2
K K s
Comparing with s2  2YXn s  Xn2  0 we have s1
we get 2YXn  1 and Xn2  6 s0
K K
or 2 # 0.5 # 6 KX = 1 Given Y  0.5 7.87 Option (B) is correct.
K
The open loop transfer function is
or 6  2 & K 1
1
K 6 G (s) H (s)  1
K
s (s2  s  1)
7.81 Option (B) is correct. Substituting s  jX we have
Any point on real axis lies on the root locus if total number of poles 1
and zeros to the right of that point is odd. Here s  1.5 does not G (jX) H (jX) 
jX ( X2  jX  1)
lie on real axis because there are total two poles and zeros (0 and
For more GATE Resources, Mock Test and
 1) to the right of s  1.5 .
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7.82 Option (D) is correct.
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From the expression of OLTF it may be easily see that the maximum
+G (jX) H (jX)  Q  tan1 X 2
magnitude is 0.5 and does not become 1 at any frequency. Thus gain
cross over frequency does not exist. When gain cross over frequency 2 (1  X )
does not exist, the phase margin is infinite. The frequency at which phase becomes  180c, is called phase
7.83 Option (D) is correct. crossover frequency.
We have xo (t)  2x (t)  2u (t) ...(i) XG
Thus  180  90  tan1
Taking Laplace transform we get 1  X2G
XG
sX (s)  2X (s)  2U (s) or  90  tan1
1  X2G
or (s  2) X (s)  2U (s)
or 1  X2G  0
2U (s)
or X (s)  XG  1 rad/sec
(s  2)
The gain margin at this frequency XG  1 is
Now y (t)  0.5x (t)
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 181

GM  20 log10 G (jXG) H (jXG)

Comparing (1) and (6) we have
 20 log10 (XG (1  XG2 ) 2  XG2
Z3 (s)
 20 log 1  0 H 
Z1 (s)  Z3 (s)
7.88 Option (A) is correct. 7.93 Option (B) is correct.
Z  PN For unity negative feedback system the closed loop transfer function
N  Net encirclement of ( 1  j0) by Nyquist plot, is
P Number of open loop poles in right had side of s  plane G (s) s4
CLTF   , G (s) OL Gain


1  G (s) s2  7s  13


Z Number of closed loop poles in right hand side of s  plane



1  G (s)
 s  7s  13
2
Here N  0 (1 encirclement in CW direction and other in CCW) or
G (s) s4
and P  0
or 1  s2  7s  13  1  s2  6s  9
Thus Z  0 G (s) s4 s4
Hence there are no roots on RH of s  plane.
or G (s)  2 s  4
s  6s  9
7.89 Option (D) is correct. For DC gain s  0 , thus
Take off point is moved after G2 as shown below
Thus G (0)  4
9
SPECIAL EDITION ( STUDY MATERIAL FORM )
At market Book is available in 3 volume i.e. in 3 book binding
form. But at NODIA Online Store book is available in 10 book
7.90 Option (C) is correct.
binding form. Each unit of Book is in separate binding.
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The characteristics equation is
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s2  2s  2  0
Comparing with s2  2YXn  Xn2  0 we get
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2YXn  2 and Xn2  2
7.94 Option (C) is correct.
From the Block diagram transfer function is
2 Xn 
G (s)
Y  1 T (s) 
and 1  G (s) H (s)
2
K (s  2)
Since Y  1 thus system is under damped Where G (s) 
(s  2)
7.91 Option (D) is correct. and H (s)  (s  2)
If roots of characteristics equation lie on negative axis at different
The Characteristic equation is
positions (i.e. unequal), then system response is over damped.
1  G (s) H (s)  0
From the root locus diagram we see that for 0  K  1, the roots
are on imaginary axis and for 1  K  5 roots are on complex plain. K (s  2)
1 (s  2)  0
For K  5 roots are again on imaginary axis. (s  2) 2
Thus system is over damped for 0 # K  1 and K  5 . or (s  2)  K (s  2) 2  0
2

or (1  K) s  4 (1  K) s  4K  4  0
2
7.92 Option (C) is correct.
From SFG we have Routh Table is shown below. For System to be stable 1  k  0 ,
I1 (s)  G1 Vi (s)  HI2 (s) ...(1) and 4  4k  0 and 4  4k  0 . This gives  1  K  1
I2 (s)  G2 I1 (s) ...(2) As per question for 0 # K  1
V0 (s)  G3 I2 (s) ...(3) 1k 4  4k
s2
Now applying KVL in given block diagram we have
Vi (s) = I1 (s) Z1 (s) + [I1 (s)  I2 (s)] Z3 (s) ...(4)
1
s 4  4k 0
0  [I2 (s)  I1 (s)] Z3 (s)  I2 (s) Z2 (s)  I2 (s) Z4 (s) ...(5) s0 4  4k
From (4) we have
or Vi (s)  I1 (s)[ Z1 (s)  Z3 (S)]  I2 (s) Z3 (S) 7.95 Option (B) is correct.
1 Z3 (s) It is stable at all frequencies because for resistive network feedback
or I1 (s)  Vi  I2 ...(6)
Z1 (s)  Z3 (s) Z1 (s)  Z3 (s) factor is always less than unity. Hence overall gain decreases.
From (5) we have 7.96 Option (B) is correct.
I1 (s) Z3 (S)  I2 (s)[ Z2 (s)  Z3 (s)  Z4 (s)] ...(7) The characteristics equation is s2  Bs2  ks  3  0
I1 (s) Z3 (s) The Routh Table is shown below
or Is (s) 
Z3 (s)  Z2 (s)  Z4 (s)
For system to be stable B  0 and BK  3  0
Comparing (2) and (7) we have B
Z3 (s) Thus B  0 and BK  3
G2 
Z3 (s)  Z2 (s)  Z4 (s)
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 182

s3 1 K The observability matrix is

Q0  [CT AT CT ]  =
1  2G
1 2
s 2
B 3 !0
BK  3
s1 B 0 det Q0 ! 0 Thus observable
0 3
s 7.100 Option (D) is correct.
we have G (s) H (s)  2 3
s (s  1)
7.97 Option (B) is correct. 2 3
or G (jX) H (jX) 
Closed loop transfer function is given as jX (jX  1)
Gain cross over frequency
T (s)  2 9
s  4s  9 G (jX) H (jX) at X  X  1
by comparing with standard form we get natural freq. g

X 9
2
A or 2 3 1
X X2  1
Xn  3
12  X2 (X2  1)
2YXn  4
X  X  12  0
4 2

damping factor 4  2/3

Y 
2#3 (X  4) (X2  3)  0
2

for second order system the setting time for 2-percent band is given X2  3 and X2  4
by which gives X1, X2  ! 3
GATE Electronics & Communication Xg  3
by RK Kanodia G (X) at X  X  90  tan1 (Xg)
g

Now in 3 Volume  90  tan1 3  90  60  150

Purchase Online at maximum discount from online store Phase margin  180  G (X) at X  X
g

and get POSTAL and Online Test Series Free  180  150  30c
visit www.nodia.co.in 7.101 Option (B) is correct.
7.102 Option (C) is correct.
ts  4  4 4 2 Closed-loop transfer function is given by
YXn 3 # 2/3 2
T (s)  n an  1 s  an
7.98 Option (D) is correct. s  a1 sn  1  ...  an  1 s  an
Given loop transfer function is an  1 s  an
n1
 s  a1 s  ...an  2 s
n 2

G (s) H (s)  2
s (s  1) 1 n a n1 s  a n
s  a1 sn  1  ...an  2 s2
G (jX) H (jX)  2
jX (jX  1) Thus G (s) H (s)  an  1 s  an
Phase cross over frequency can be calculated as sn  a1 sn  1  ....an  2 s2
For unity feed back H (s)  1
G (X) at X  X  180c
p
Thus G (s)  an  1 s  an
So here G (X)  90c  tan1 (X) sn  a1 sn  1  ....an  2 s2
1
 90c  tan (Xp)  180c Steady state error is given by
tan1 (Xp)  90c E (s)  lim R (s) 1
s"0 1  G (s) H (s)
Xp  3
for unity feed back H (s)  1
Gain margin
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20 log 10 =
G (jX) H (jX) G
1 at X  Xp
Study material join the community
G.M.  20 log 10 e
G (jX) H (jXp) o
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G (jXp) H (jXp)  2 0 Here input R (s)  12 (unit Ramp)

Xp
X2p  1 s
1 1
so G.M.  20 log 10 b 1 l  3 so E (s)  lim 2
s " 0 s 1  G (s)
0 n1
 lim 12 s  a1 s  ....  an  2 s
n 2

7.99 Option (A) is correct. s"0 s sn  a1 sn  1  ....  an

A =
2  3G
, B  = G and C  [1 1]
0 1 0
Here  an  2
1 an
The controllability matrix is 7.103 Option (B) is correct.

QC  [B AB ]  =
1  3G
0 1 7.104

7.105 Option (A) is correct.

det QC ! 0 Thus controllable
7.106 Option (A) is correct.
 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 183

By applying Rouths criteria Phase is

s3  5s2  7s  3  0 G (X)  tan1 (3TX)  tan1 (TX)
G (X)  tan1 ; 3TX  T X
s3 1 7 1  3T 2 X2 E
s2 5 3 G (X)  tan1 ; 2TX2 2 E
7#53
1  3T X
s1
 32
0
5 5 For maximum value of phase
s0 3 dG (X)
0
dX
There is no sign change in the first column. Thus there is no root or 1  3T 2 X2
lying in the left-half plane.
TX  1
7.107 Option (A) is correct. 3
Techometer acts like a differentiator so its transfer function is of the So maximum phase is
form ks . Gmax  tan1 ; 2TX2 2 E at TX  1
1  3T X 3
7.108 Option (A) is correct. R V
2 1
Open loop transfer function is S W
; 3E
 tan1 S 3 W  tan1 1  30c
K 1
G (s) 
s (s  1) SS1  3 # 3 WW
T X
Steady state error
SPECIAL EDITION ( STUDY MATERIAL FORM )
sR (s)
E (s)  lim At market Book is available in 3 volume i.e. in 3 book binding
s " 0 1  G (s) H (s)

Where R (s)  input H (s)  1 (unity feedback)

form. But at NODIA Online Store book is available in 10 book
binding form. Each unit of Book is in separate binding.
R (s)  1 Available Only at NODIA Online Store
s
s1
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s s (s  1)
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s"0 s  s  K
s"0
1 K
s (s  1)
7.109 Option (B) is correct. 7.113 Option (A) is correct.
Fig given below shows a unit impulse input given to a zero-order hold G (jX) H (jX) enclose the ( 1, 0) point so here G (jXp) H (jXp)  1
circuit which holds the input signal for a duration T & therefore, Xp  Phase cross over frequency
the output is a unit step function till duration T . Gain Margin  20 log 10 1
G (jXp) H (jXp)
so gain margin will be less than zero.
7.114 Option (B) is correct.
The denominator of Transfer function is called the characteristic
equation of the system. so here characteristic equation is
(s  1) 2 (s  2)  0

h (t)  u (t)  u (t  T)
7.115 Option (C) is correct.
In synchro error detector, output voltage is proportional to [X (t)],
Taking Laplace transform we have
where X (t) is the rotor velocity so here n  1
H (s)  1  1 esT  1 61  esT @ Option (C) is correct.
s s s 7.116

By massons gain formulae

7.110 Option (C) is correct.
y

/ %k Pk
Phase margin  180c  Rg where Rg  value of phase at gain x %
crossover frequency.
Forward path gain P1  5 # 2 # 1  10
Here Rg  125c
%  1  (2 #  2)  1  4  5
so P.M  180c  125c  55c
%1  1
7.111 Option (B) is correct. y
so gain  10 # 1  2
Open loop transfer function is given by x 5
K (1  0.5s)
G (s) H (s)  7.117 Option (C) is correct.
s (1  s) (1  2s)
By given matrix equations we can have
Close looped system is of type 1.
It must be noted that type of the system is defined as no. of poles Xo1  dx1  x1  x2  0
dt
lies at origin in OLTF.
lying o dx
X2  2  0  x2  N
dt
7.112 Option (D) is correct.
y  [1 1] > H  x1  x2
Transfer function of the phase lead controller is x1
1  (3TX) j x2
T.F  1  3Ts 
1s 1  (TX) j dy
 dx1  dx2
dt dt dt
GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 184

dy
 x1  N
dt
dy
 x1 (0)  N (0)
dt t0

 10  0

GATE Electronics & Communication

by RK Kanodia
Now in 3 Volume
Purchase Online at maximum discount from online store
and get POSTAL and Online Test Series Free
visit www.nodia.co.in

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Study material join the community
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 GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 184

UNIT 8 The optimum threshold to achieve minimum bit error rate (BER) is
8.6

(A) 1 (B) 4
2 5

COMMUNICATION SYSTEMS (C) 1 (D) 3

2

2012 ONE MARK

8.7 The power spectral density of a real process X (t) for positive
2013 ONE MARK
frequencies is shown below. The values of E [X 2 (t)] and E [X (t)] ,
8.1 The bit rate of a digital communication system is R kbits/s . The respectively, are
modulation used is 32-QAM. The minimum bandwidth required for
ISI free transmission is
(A) R/10 Hz (B) R/10 kHz
(C) R/5 Hz (D) R/5 kHz

2013 TWO MARKS

8.2 Let U and V be two independent zero mean Gaussain random

variables of variances 1 and 1 respectively. The probability
4 9 (A) 6000/Q, 0 (B) 6400/Q, 0
GATE Electronics & Communication (C) 6400/Q, 20/ (Q 2 ) (D) 6000/Q, 20/ (Q 2 )
by RK Kanodia
Now in 3 Volume 8.8 In a baseband communications link, frequencies upto 3500 Hz are
used for signaling. Using a raised cosine pulse with 75% excess
Purchase Online at maximum discount from online store bandwidth and for no inter-symbol interference, the maxi mum
and get POSTAL and Online Test Series Free possible signaling rate in symbols per second is
visit www.nodia.co.in (A) 1750 (B) 2625
P ^3V F 2U h is (C) 4000 (D) 5250

(A) 4/9 (B) 1/2 8.9 A source alphabet consists of N symbols with the probability of the
(C) 2/3 (D) 5/9 first two symbols being the same. A source encoder increases the
probability of the first symbol by a small amount F and decreases
8.3 Consider two identically distributed zero-mean random variables U that of the second by F. After encoding, the entropy of the source
and V . Let the cumulative distribution functions of U and 2V be (A) increases (B) remains the same
F ^x h and G ^x h respectively. Then, for all values of x (C) increases only if N  2 (D) decreases
(A) F ^x h  G ^x h # 0 (B) F ^x h  G ^x h $ 0
(C) ^F (x)  G (x)h .x # 0 (D) ^F (x)  G (x)h .x $ 0 Two independent random variables X and Y are uniformly
8.10

distributed in the interval 6 1, 1@. The probability that max 6X, Y @

8.4 Let U and V be two independent and identically distributed random is less than 1/2 is
variables such that P ^U  1h  P ^U  1h  1 . The entropy (A) 3/4 (B) 9/16
H ^U  V h in bits is
2
(C) 1/4 (D) 2/3
(A) 3/4 (B) 1
(C) 3/2 (D) log 2 3 2012 TWO MARKS

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Bits 1 and 0 are transmitted with equal probability. At the re-
ceiver, the pdf of the respective received signals for both bits are http://www.facebook.com/gateec2014
as shown below.
8.11 A BPSK scheme operating over an AWGN channel with noise
power spectral density of N 0 /2, uses equiprobable signals
s1 (t)  2E sin (Xc t) and s2 (t)  2E sin (Xc t) over the symbol
T T
interval (0, T). If the local oscillator in a coherent receiver is ahead
in phase by 45c with respect to the received signal, the probability
of error in the resulting system is
(A) Q c 2E m (B) Q c E m
N0 N0
(C) Q c
2N 0 m
(D) Q c
4N 0 m
8.5 If the detection threshold is 1, the BER will be E E
(A) 1 (B) 1
2 4 8.12 A binary symmetric channel (BSC) has a transition probability of
(C) 1 (D) 1 1/8. If the binary symbol X is such that P (X  0)  9/10, then the
8 16