CWT–04
Booklet No.:
Booklet Series:
03092014
Control Systems
Electronics Engineering
A
Student Name:
Roll Number:
Duration: 90 Minutes
PAPER
MAXIMUM MARKS: 60
INSTRUCTIONS
1.
IMMEDIATELY AFTER THE COMMENCEMENT OF THE EXAMINATION, YOU SHOULD CHECK
THAT THIS TEST BOOKLET DOES NOT HAVE ANY UNPRINTED OR TORN OR MISSING PAGES
OR ITEMS ETC. IF SO, GET IT REPLACED BY A COMPLETE TEST BOOKLET.
2.
This Test Booklet contains 30 questions. Each question comprises four responses (answers).
You will select the response which you want to mark on the Answer Sheet. In case you feel that
there is more than one correct response, mark the response which you consider the best. In any
case, choose ONLY ONE response for each item.
3.
You have to mark all your response ONLY on the separate Answer Sheet provided.
4.
All items carry equal marks.
5.
Before you proceed to mark in the Answer Sheet the response to various items in the Test
Booklet, you have to fill in some particulars in the Answer Sheet as per instructions.
6.
Each questions 2 marks and 2/3 negative mark is assigned for the wrong answer.
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P h ( 0 1 1 ) - 2 6 1 9 4 8 6 9 , C e l l : 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7 : E - ma i l : q h e n g i n e e r z o n e @ g ma i l . c o m
Control Systems
(1.)
A system is shown in the given figure the value of K which gives steady error of 25% to a unit step
input is given by
(a.) 20
(b.) 5
(c.) 3
(d.) 4
Ans: (c)
(2.)
For the block diagram shown below, the limiting value of K for stability of inner loop is found to be
x  k  y the other all system will be stable if and only if
(a.) 4x  k  4y
(b.)
x
y
k 
2
2
(c.) 2x  k  2y
(d.) x  k  y
Ans: (b)
(3.)
The open loop transfer function of a ufb control system is given by G s  
K
for the
s sT1  1 sT2  1
system to be stable the range of K is
1 1
(a.) 0  k    
 T1 T2 
1 1
(b.) k    
 T1 T2 
(c.) 0  k  T1 T2
(d.) k  T1 T2
Ans: (a)
Common Data for 4 to 6
Consider the network shown below
2
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
(4.)
The state variable may be
(a.) i2 ,i4 ,v0
(b.) i2 ,i4
(c.) i1 , i3
(d.) i1 , i3 , i5
Ans: (a)
(5.)
In state space representation matrix A is
 2
 3

1
(a.)  
 3

 1
 3
 2
 3

1
(b.) 
 3

 1
 3
 1
 3

2
(c.) 
 3

 1
 3
 1
 3

2
(d.)  
 3

 1
 3
1
3
2

3
2

3
1 
3 

2 
3 

1
 
3 
1
3
2
3
2

3
1
3

2
3

1
3 


1
2
 
3
3

2
1


3
3

2
1

 
3
3 

1
3
2

3
2

3

2 
3 

2 
3 

1
 
3 
Ans: (a)
(6.)
The matrix B is
 2 
 3 


1
(a.)   
 3


 1 
 3 
3
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
2
3
 
1
(b.)  
3
 
1 
 3 
 1
 3 


1
(c.)   
 3


 1 
 3 
2
3
 
1
(d.)  
3
 
2
 3 
Ans: (b)
(7.)
Consider the speed control system of figure when in the inner loop correspond to motor back e.m.f
the controller is an integrator with gain K observe that the load is inertia only
1

Determine the value of K for which steady-state error to unit ramp Vr  3  2  input in less than
s 

01 rad/sec
(a.) 100
(b.) 10
(c.) 1
(d.) 0.01
Ans: (b)
(8.)
A units feedback system having or open loop given G  s  H  s   K
1  s 
1 s
be comes stable when
(a.) K  1
(b.) K  1
(c.)
K 1
(d.) K  1
Ans: (c)
4
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
(9.)
The root locus for a control system with G  s   K
s  1 ,
s 1
(a.) Is the circle with centre at  1,1 and radius
2
(b.) Is the circle with centre at  1,0  and radius
2
H s   1
(c.) Is the circle with centre at  1,1 and radius 2
(d.) Is the circle with centre at  1,0  and radius 2
Ans: (b)
(10.)
When subjected to a unit ramp input the closed loop control system shown in the figure will have a
steady state error
(a.) 0.5
(b.) –0.5
(c.) 0
(d.) 2/3
Ans: (d)
(11.)
Signal flow graph of above transfer function
(a.)
(b.)
(c.)
(d.) None of these
5
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
Ans: (a)
(12.)
Transfer function
(a.)
(b.)
(c.)
X s 
is
U s 
s 2 3  1 3  2  s  1
s 2  1 s  2  2 3
s 2 3  1 3  2  s  2 3
s 2  1 s  2  1
s 2 3  1 3  2  s   1  2 3

s 2  1 s  2
(d.) None of these
Ans: (c)
(13.)
Consider the mechanical system shown in the given figure if the system is set into motion by unit
impulse force, the equation of the resulting oscillation will be
(a.) x t   2 sin t
(b.) x t   2sin t
(c.) x t  
(d.) x t  
1
sin t
2
1
sin t
2
Ans: (c)
(14.)
The signal flow graph of a system is shown in the figure the transfer function
C s 
of the system
R s 
is
6
(a.)
s s  29
s 2  29s  6
(b.)
s  s  27 
s 2  29s  6
(c.)
6s
s 2  29s  6
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
(d.) None of these
Ans: (b)
(15.)
The root locus flat of a control system is given as
The value of damping factor and natural frequency for the system at gain K  10 is
(a.) 0.5 2 &1
(b.) 0.5&1
(c.) 0.25&1
(d.) 0.25& 2
Ans: (b)
(16.)
Maximum possible sensitivity of CLTF w.r.t OLTF at   1 is
(a.) At K  5, SGT
max
(b.) At K  4, SGT
max
(c.) K  3, SGT
 2
max
 2
 2
(d.) None of these
Ans: (a)
Common Data for 17 & 18
0 1 
1 
T
x t     u t  with the initial X  0   1, 3 condition and
A state variable system x t   

0

3
0


 
the unit step input u t  has
(17.)
The state transition matrix
1 t

0
e  e 3t 
(a.) 
3


e t
0

7
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
1 t

1
e  e 3t 
(b.) 
3


e 3 t
0

1
(c.) 
0
1  e 
t
e t

1


1
1  e 3t  

(d.) 
3


e 3 t
0

Ans: (d)
(18.)
The state transition equation
t  e 3t 
(a.) X t   
3t 
 3e 
t  e 3t 
(b.) X t    t 
 e

1  e 3t 
(c.) X t   

t
 e

t  e t 
(d.) X t   
t 
 3e 
Ans: (a)
(19.)
Consider the system
1 0 
p
dx
and B    where p and q are arbitrary real
 Ax  Bu with A  

dt
0 1 
q 
numbers. Which controllability of the system is true
(a.) Only P  0 and q  0 result in controllability
(b.) The system is uncontrollable for all values of p and q
(c.) The system is completely state controllable for any non zero values of p and q
(d.) None of these
Ans: (b)

(20.)
s
e 2
A unity feedback control system has a forward loop T.F as
the phase cross cover frequency is
1 s
given by
(a.) 2  tan 
(b.)   cot

(c.)   tan
(d.)   tan
2

2

2
Ans: (c)
Common Data for 21 & 2
8
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
The transfer function of a compensator is given as Gc s  
(21.)
sx
s y
Gc s  is a lead compensator if
(a.) x  3, y  1
(b.) x  3, y  2
(c.) x  1, y  2
(d.) x  3, y  1
Ans: (c)
(22.)
The phase of the above lead compensator is maximum at
(a.)
6 rad/sec
(b.)
2 rad/sec
(c.)
1
rad/sec
3
(d.)
3 rad/sec
Ans: (b)
(23.)
Consider the Bode magnitude plot shown in the figure the transfer function H s  is
(a.)
102 s  1
s  10s  100
(b.)
10 s  1
s  10s  100
(c.)
103  s  100 
s  1s  10 
(d.)
s  10 
s  1s  100
Ans: (a)
(24.)
The system with OLTF G s  H s 

1
has a gain margin of
s  s  s  2
2
(a.) 2 dB
(b.) –6 dB
(c.) 6 dB
(d.) –2 dB
9
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
Ans: (c)
(25.)
The root locus of the system G s  H s  
(a.)
 2.55, 0
(b.)
 0.784, 0
(c.)
 2.04, 0
(d.)
 0.784, 0
K
has the break away point located at
s  s  2  s  3 
Ans: (b)
(26.)
For the system shown
The number of poles in R.H.S on j -axis
(a.) 2 & 0
(b.) 3 & 0
(c.) 0 & 2
(d.) 0 & 3
Ans: (a)
Common Data for 27 to 28
(27.)
The controllability matrix for this system is
0 1 1
(a.) 1 6 1
1 4 4
0 1 2
(b.) 1 1 1 
1 2 4 
 10 10 10 
(c.)  10
0
20
 10 10 40 
10
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
 10 10 10 
(d.)  10
0
20 

 10 10 40 
Ans: (b)
(28.)
The observability matrix is
0 1 1
(a.) 1 6 1


1 4 4
0 1 2
(b.) 1 1 1 
1 2 4 
 10 10 10 
(c.)  10
0
20

 10 10 40 
 10 10 10 
(d.)  10
0
20 

 10 10 40 
Ans: (c)
(29.)
A lead compensating network
A. Improves response time
B. Stabilizes the system with low phase margin
C. Enables moderate increase in gain without affecting stability
D. Increases resonant frequency
in the above correct are
(a.) (A), (C), and (D)
(b.) (A) and (C)
(c.) (A) and (D)
(d.) All
Ans: (d)
(30.)
The frequency response of a linear time invariant system is given by H  f  
5
. The step
1  j 10 f
response of the system is
(a.) 5 1  e 5t  u t 
(b.) 5 1  e t /5  u t 
(c.)
1
1  e 5t  u t 
5
(d.)
1
1
s  5  s  1
Ans: (b)
11
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
12
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
Answer Key
Civil Engineering
(1.)
(d)
(9.)
(a)
(17.)
(b)
(25.)
(b)
(2.)
(d)
(10.)
(b)
(18.)
(c)
(26.)
(b)
(3.)
(c)
(11.)
(d)
(19.)
(c)
(27.)
(c)
(4.)
(a)
(12.)
(a)
(20.)
(b)
(28.)
(c)
(5.)
(a)
(13.)
(d)
(21.)
(a)
(29.)
(b)
(6.)
(b)
(14.)
(a)
(22.)
(b)
(30.)
(b)
(7.)
(b)
(15.)
(a)
(23.)
(d)
(8.)
(a)
(16.)
(b)
(24.)
(d)
13
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
(1.)
A system is shown in the given figure the value of K which gives
steady error of 25% to a unit step input is given by
(a.) 20
(b.)5
(c.) 3
(d.)4
Ans: (c)
(2.)
For the block diagram shown below, the limiting value of K for
stability of inner loop is found to be x < k < y the other all system
will be stable if and only if
(a.) 4x  k  4y
(b.)
x
y
k 
2
2
(c.) 2x  k  2y
(d.) x  k  y
Ans: (b)
(3.)
The open loop transfer function of a ufb control system is given by
G s  
14
K
for the system to be stable the range of K is
s  sT1  1  sT2  1
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
1 1
(a.) 0  k    
 T1 T2 
1 1
(b.) k    
 T1 T2 
(c.) 0  k  T1 T2
(d.) k  T1 T2
Ans: (a)
(4.)
Consider the speed control system of figure when in the inner loop
correspond to motor back e.m.f the controller is an integrator with
gain K observe that the load is inertia only
Determine the value of K for which steady-state error to unit ramp
1

Vr  3  2  input in less than 01 rad/sec
s 

(a.) 100
(b.)10
(c.) 1
(d.)0.01
Ans: (b)
(5.)
A
units
feedback
G s  H s   K
1  s 
1 s
system
having
or
open
loop
given
be comes stable when
(a.) K  1
(b.) K  1
15
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
(c.) K  1
(d.) K  1
Ans: (c)
(6.)
The root locus for a control system with G s   K
s  1 ,
(a.) Is the circle with centre at (–1, 1) and radius
2
(b.)Is the circle with centre at (–1, 0) and radius
2
s 1
H s   1
(c.) Is the circle with centre at (–1, 1) and radius 2
(d.)Is the circle with centre at (–1, 0) and radius 2
Ans: (b)
(7.)
When subjected to a unit ramp input the closed loop control system
shown in the figure will have a steady state error
(a.) 0.5
(b.)–0.5
(c.) 0
(d.)2/3
Ans: (d)
(8.)
Signal flow graph of above transfer function
(a.)
16
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
(b.)
(c.)
(d.)None of these
Ans: (a)
(9.)
Transfer function
(a.)
(b.)
(c.)
X s 
U s 
is
s 2 3  1 3  2  s  1
s 2  1 s   2   2 3
s 2 3  1 3  2  s   2 3
s 2  1 s   2  1
s 2 3  1 3  2  s   1   2 3

s  1 s   2
2
(d.)None of these
Ans: (c)
17
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
(10.)
Consider the mechanical system shown in the given figure if the
system is set into motion by unit impulse force, the equation of the
resulting oscillation will be
(a.) x t   2 sin t
(b.) x t   2sin t
(c.) x t  
(d.) x t  
1
sin t
2
1
sin t
2
Ans: (c)
(11.)
The signal flow graph of a system is shown in the figure the transfer
function
C s 
R s 
(a.)
s s  29
s 2  29s  6
(b.)
s s  27 
s 2  29s  6
(c.)
6s
s 2  29s  6
of the system is
(d.)None of these
18
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
Ans: (b)
(12.)
The root locus flat of a control system is given as
The value of damping factor and natural frequency for the system at
gain K = 10 is
(a.) 0.5 2 &1
(b.) 0.5 &1
(c.) 0.25 &1
(d.) 0.25 & 2
Ans: (b)
(13.)
Maximum possible sensitivity of CLTF w.r.t OLTF at   1 is
(a.) At K  5, SGT
max
(b.)At K  4, SGT
max
(c.) K  3, SGT
 2
max
 2
 2
(d.)None of these
19
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
Ans: (a)
(14.)
Consider the system
1 0 
p
dx
B

 Ax  Bu with A  
and

q 
dt
0 1 
 
where p and q are arbitrary real numbers. Which controllability of
the system is true
(a.) Only p = 0 and q = 0 result in controllability
(b.)The system is uncontrollable for all values of p and q
(c.) The system is completely state controllable for any non zero
values of p and q
(d.)None of these
Ans: (b)

(15.)
A unity feedback control system has a forward loop T.F as
s
2
e
the
1 s
phase cross cover frequency is given by
(a.) 2  tan 
(b.)   cot

(c.)   tan
(d.)   tan
2

2

2
Ans: (c)
(16.)
Consider the Bode magnitude plot shown in the figure the transfer
function H s  is
20
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
102  s  1
(a.)
s  10 s  100 
(b.)
10  s  1
s  10 s  100 
103  s  100 
(c.)
s  1s  10 
(d.)
s  10 
s  1s  100 
Ans: (a)
(17.)
The system with OLTF
G s  H s  
1
has a gain margin
s  s 2  s  2
of
(a.) 2 dB
(b.)–6 dB
(c.) 6 dB
(d.)–2 dB
Ans: (c)
(18.)
The root locus of the system G  s  H  s  
K
s  s  2  s  3 
has the
break away point located at
(a.)  2.55, 0 
(b.)  0.784, 0 
21
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
(c.)  2.04, 0 
(d.)  0.784, 0 
Ans: (b)
(19.)
For the system shown
The number of poles in R.H.S on j  -axis
(a.) 2 & 0
(b.)3 & 0
(c.) 0 & 2
(d.)0 & 3
Ans: (a)
(20.)
A lead compensating network
1. Improves response time
2. Stabilizes the system with low phase margin
3. Enables moderate increase in gain without affecting stability
4. Increases resonant frequency
In the above correct are
(a.) 1, 3 and 4
(b.) 1 and 3
(c.) 1 and 4
(d.) All of above
Ans: (d)
22
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
(21.)
The frequency response of a linear time invariant system is given by
H f  
5
. The step response of the system is
1  j 10 f
(a.) 5 1  e 5t  u t 
(b.) 5 1  e t /5  u t 
(c.)
1
1  e 5t  u t 

5
(d.)
1
1
s  5  s  1
Ans: (b)
Common Data for Questions 22 to 24:
Consider the network shown below
(22.)
The state variable may be
(a.) i2 , i4 ,v 0
(b.) i2 , i 4
(c.) i1 , i3
(d.) i1 , i3 , i5
Ans: (a)
(23.)
23
In state space representation matrix A is
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
1 1 
 2
 3  3 3 


1
2 2 


(a.) 
 3
3 3 
 1
2
1 



 3
3
3 
 2
 3

1
(b.) 
 3
 1

 3
1
3
2
3
2

3
 1
 3

2
(c.) 
 3
 1

 3
1
2
 
3
3

2
1


3
3
2
1 


3
3 

1
3

2
3
1 
3 

1 2 
 1


 3
3 3 


2
2 2 


(d.) 
 3
3 3 
 1

2
1


 
3
3 
 3
Ans: (a)
(24.)
The matrix B is
 2 
 3 


1
(a.)   
 3
 1
 
 3 
24
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
2
3
 
1
(b.)  
3
1 
 
 3 
 1
 3 


1
(c.)   
 3
 1 


 3 
2
3
 
1
(d.)  
3
2
 
 3 
Ans: (b)
Common Data for Questions 25 & 26:
The transfer function of a compensator is given as Gc  s  
(25.)
sx
sy
Gc s  is a lead compensator if
(a.) x  3, y  1
(b.) x  3, y  2
(c.) x  1, y  2
(d.) x  3, y  1
Ans: (c)
(26.)
25
The phase of the above lead compensator is maximum at
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
(a.) 6 rad/sec
(b.) 2 rad/sec
(c.)
1
rad/sec
3
(d.) 3 rad/sec
Ans: (b)
Common Data for Questions 27 to 28 :
(27.)
The controllability matrix for this system is
0 1 1
(a.) 1 6 1


1 4 4
0 1 2
(b.) 1 1 1 


1 2 4 
 10 10 10 
(c.)  10 0
20 


 10 10 40 
26
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
 10 10 10 
(d.)  10 0
20 


 10 10 40 
Ans: (b)
(28.)
The observability matrix is
0 1 1
(a.) 1 6 1


1 4 4
0 1 2
(b.) 1 1 1 


1 2 4 
 10 10 10 
(c.)  10 0
20 


 10 10 40 
 10 10 10 
(d.)  10 0
20 


 10 10 40 
Ans: (c)
Common Data for Questions 29 & 30 :
0 1 
1 
x t     u t  with the initial
A state variable system x t   

0 3
0 
X  0   1, 3 condition and the unit step input u t  has
T
(29.)
The state transition matrix
1 t

3t 
0
e

e


(a.) 
3


e t
0

27
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Control Systems
1 t


1
e  e 3t  


(b.)
3


e 3t
0

1
(c.) 
0
1  e 
t
e t

1


1
1  e 3t  


(d.)
3


e 3t
0

Ans: (d)
(30.)
The state transition equation
t  e 3t 
(a.) X t   
3t 
 3e

t  e 3t 
(b.) X t   

t
 e

1  e 3t 
(c.) X t   

t
 e

t  e t 
(d.) X t   
t 
 3e 
Ans: (a)
28
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m