GATE SOLVED PAPER - EC CONTROL SYSTEM 2013 ONE MARK Q. 1 The Bode plot of a transfer function Gs ^h is shown in the figure below. The gain log Gs 20 ^ _ h i is 32 dB and 8 dB - at 1 / rad s and 10 / rad s respectively. The phase is negative for all w . Then Gs ^h is (A) . s 39 8 (B) . s 39 8 2 (C) s 32 (D) s 32 2 2013 TWO MARKS Q. 2 The signal flow graph for a system is given below. The transfer function Us Ys ^ ^ h h for this system is (A) s s s 5 6 2 1 2 + + + (B) s s s 6 2 1 2 + + + (C) s s s 4 2 1 2 + + + (D) s s 5 6 2 1 2 + + Statement for Linked Answer Questions 3 and 4: The state diagram of a system is shown below. A system is described by the state- variable equations u X AX B = + o ; y u CX D = + mywbut.com
Transcript
GATE SOLVED PAPER - ECCONTROL SYSTEM
2013 ONE MARK
Q. 1 The Bode plot of a transfer function G s^ h is shown in the figure below.
The gain log G s20 ^_ h i is 32 dB and 8 dB- at 1 /rad s and 10 /rad s respectively. The phase is negative for all w . Then G s^ h is
(A) .s
39 8 (B) .s
39 82
(C) s32 (D)
s32
2
2013 TWO MARKS
Q. 2 The signal flow graph for a system is given below. The transfer function U sY s^
^
h
h for
this system is
(A) s s
s5 6 2
12 + +
+ (B) s s
s6 2
12 + +
+
(C) s s
s4 2
12 + +
+ (D) s s5 6 2
12 + +
Statement for Linked Answer Questions 3 and 4:The state diagram of a system is shown below. A system is described by the state-variable equations uX AX B= +o ; y uCX D= +
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Q. 3 The state-variable equations of the system shown in the figure above are
(A) u
y u
X X
X
11
01
11
1 1
=-
- +-
= - +
o > >
6
H H
@
(B) u
y u
X X
X
11
01
11
1 1
=-- - +
-
= - - +
o > >
6
H H
@
(C) u
y u
X X
X
11
01
11
1 1
=-- - +
-
= - - -
o > >
6
H H
@
(D) u
y u
X X
X
10
11
11
1 1
=- -
- +-
= - -
o > >
6
H H
@
Q. 4 The state transition matrix eAt of the system shown in the figure above is
(A) ete e
0t
t t
-
- -> H (B) ete e
0t
t t-
-
- -> H
(C) ee e
0t
t t
-
- -> H (D) e te
e0
t t
t
-- -
-> H
Q. 5 The open-loop transfer function of a dc motor is given as 1 1010
V ss
sa=w
+^
^
h
h . When connected in feedback as shown below, the 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
(A) 1 (B) 5
(C) 10 (D) 100
2012 ONE MARK
Q. 6 A system with transfer function ( )( )( )( )
( )( )G s
s s ss s1 3 4
9 22
= + + ++ +
is excited by ( )sin tw . The steady-state output of the system is zero at(A) 1 /rad sw = (B) /rad s2w =(C) /rad s3w = (D) /rad s4w =
2012 TWO MARKS
Q. 7 The feedback system shown below oscillates at 2 /rad s when
Q. 8 The state variable description of an LTI system is given by
xxx
1
2
3
o
o
o
J
L
KKK
N
P
OOO
0
a
aa
xxx
u00 0
0 0
0013
1
2
1
2
3
= +
J
L
KKK
J
L
KKK
J
L
KKK
N
P
OOO
N
P
OOO
N
P
OOO
y xxx
1 0 01
2
3
=
J
L
KKK
_
N
P
OOO
i
where y is the output and u is the input. The system is controllable for(A) 0, 0, 0a a a1 2 3! != (B) 0, 0, 0a a a1 2 3! !=(C) 0, 0, 0a a a1 3 3!= = (D) 0, 0, 0a a a1 2 3! ! =
2011 ONE MARK
Q. 9 The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by
(A) ( )( )
( )G s H s k
s ss s
2 31= + +
+^ ^h h (B)
( )( )( )
G s H s ks s s
s2 3
12=
+ ++
^ ^h h
(C) ( )( )( )
G s H s ks s s s1 2 3
1= - + +^ ^h h (D) ( )( )
( )G s H s k
s s ss2 3
1= + ++
^ ^h h
Q. 10 For the transfer function ( )G j j5w w= + , the corresponding Nyquist plot for positive frequency has the form
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2011 TWO MARKS
Q. 11 The block diagram of a system with one input u and two outputs y1 and y2 is given below.
A state space model of the above system in terms of the state vector x and the output vector [ ]y y y T
1 2= is(A) [2] [1] ; [ ]x x u y x1 2= + =o
(B) [ ] [ ] ;x x u y x2 112= - + =o > H
(C) ;x x u y x2
002
11 1 2=
-- + =o > > 8H H B
(D) ;x x u y x20
02
11
12= + =o > > >H H H
Common Data For Q. 12 and 13
The input-output transfer function of a plant ( )( )
H ss s 10
1002=
+.
The plant is placed in a unity negative feedback configuration as shown in the figure below.
Q. 12 The gain margin of the system under closed loop unity negative feedback is(A) 0 dB (B) 20 dB
(C) 26 dB (D) 46 dB
Q. 13 The signal flow graph that DOES NOT model the plant transfer function ( )H s is
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2010 ONE MARK
Q. 14 The transfer function ( )/ ( )Y s R s of the system shown is
(A) 0 (B) s 11+
(C) s 12+ (D) s 3
2+
Q. 15 A system with transfer function ( )( )
X sY s
s ps= + has an output ( ) cosy t t2 3
p= -a k
for the input signal ( ) cosx t p t2 2p= -a k. Then, the system parameter p is
(A) 3 (B) 2/ 3
(C) 1 (D) /3 2
Q. 16 For the asymptotic Bode magnitude plot shown below, the system transfer function can be
(A) . ss
0 1 110 1
++ (B) . s
s0 1 1100 1
++
(C) ss
10 1100
+ (D) .ss
10 10 1 1
++
2010 TWO MARKS
Common Data For Q. 17 and 18 The signal flow graph of a system is shown below:
Q. 17 The state variable representation of the system can be
(A)
[ . ]
x x u
y x
11
10
02
0 0 5
= - +
=
o
o
> >H H (B)
.
x x u
y x
11
10
02
0 0 5
=-- +
=
o
o
> >
8
H H
B
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GATE SOLVED PAPER - EC CONTROL SYSTEM
(C)
. .
x x u
y x
11
10
02
0 5 0 5
= - +
=
o
o
> >
8
H H
B
(D)
. .
x x u
y x
11
10
02
0 5 0 5
=-- +
=
o
o
> >
8
H H
B
Q. 18 The transfer function of the system is
(A) ss
11
2 ++ (B)
ss
11
2 +-
(C) s s
s1
12 + +
+ (D) s s
s1
12 + +
-
Q. 19 A unity negative feedback closed loop system has a plant with the transfer function ( )G s 2 2s s
12= + + and a controller ( )G sc in the feed forward path. For a unit
set input, the transfer function of the controller that gives minimum steady state error is
(A) ( )G s ss
21
c = ++ (B) ( )G s s
s12
c = ++
(C) ( )( )( )( )( )
G ss ss s
2 31 4
c = + ++ +
(D) ( )G s s s1 2 3c = + +
2009 ONE MARK
Q. 20 The magnitude plot of a rational transfer function ( )G s with real coefficients is shown below. Which of the following compensators has such a magnitude plot ?
(A) Lead compensator (B) Lag compensator
(C) PID compensator (D) Lead-lag compensator
Q. 21 Consider the system
dtdx Ax Bu= + with A
10
01= = G and B
pq= = G
where p and q are arbitrary real numbers. Which of the following statements
about the controllability of the system is true ?(A) The system is completely state controllable for any nonzero 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
Q. 22 The feedback configuration and the pole-zero locations of
( )G s s ss s
2 22 2
2
2=
+ +- +
are shown below. The root locus for negative values of k , i.e. for k 0< <3- , has
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breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) equal to
(A) 2! and 0c (B) 2! and 45c
(C) 3! and 0c (D) 3! and 45c
Q. 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) . .
.s s2 59 1 12
2 242 + +
- (B) . .
.s s1 91 1 91
3 822 + +
-
(C) . .
.s s2 59 1 12
2 242 - +
- (D) . .s s1 91 1 91
3822 - +
-
Common Data For Q. 24 and 25 :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.
Q. 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
Q. 25 The gain and phase margins of ( )G s for closed loop stability are(A) 6 dB and 180c (B) 3 dB and 180c
(C) 6 dB and 90c (D) 3 dB and 90c
2008 ONE MARKS
Q. 26 Step responses of a set of three second-order underdamped systems all have the same percentage overshoot. Which of the following diagrams represents the poles of the three systems ?
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Q. 27 The pole-zero given below correspond to a
(A) Law pass filter (B) High pass filter
(C) Band filter (D) Notch filter
2008 TWO MARKS
Q. 28 A signal flow graph of a system is given below
The set of equalities that corresponds to this signal flow graph is
(A) dtd
xxx
xxx
uu
000
001
010
1
2
3
1
2
3
1
2
bga
gab
=-
-+
J
L
KKK
J
L
KKK
e
N
P
OOO
N
P
OOO
o
R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW
(B) dtd
xxx
xxx
uu
000
100
010
1
2
3
1
2
3
1
2
aab
ggb
= - --
+
J
L
KKK
J
L
KKK
e
N
P
OOO
N
P
OOO
o
R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW
(C) dtd
xxx
xxx
uu
000
100
010
1
2
3
1
2
3
1
2
aba
bgg
=-- - +
J
L
KKK
J
L
KKK
e
N
P
OOO
N
P
OOO
o
R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW
(D) dtd
xxx
xxx
uu
000
100
010
1
2
3
1
2
3
1
2
agb
baa
=-
- -+
J
L
KKK
J
L
KKK
e
N
P
OOO
N
P
OOO
o
R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW
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GATE SOLVED PAPER - EC CONTROL SYSTEM
Q. 29 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.
where a 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 a(B) For some positive value of a, the closed loop system is stable, but not for
all positive values.
(C) For all positive values of a, the closed loop system is stable.
(D) The closed loop system stable for all values of a, both positive and negative.
Q. 31 The number of open right half plane of
( )G s s s s s s2 3 6 5 3
105 4 3 2=+ + + + +
is
(A) 0 (B) 1
(C) 2 (D) 3
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Q. 32 The magnitude of frequency responses of an underdamped second order system
is 5 at 0 rad/sec and peaks to 3
10 at 5 2 rad/sec. The transfer function of the system is
(A) s s10 100
5002 + +
(B) s s5 75
3752 + +
(C) s s12 144
7202 + +
(D) s s25 225
11252 + +
Q. 33 Group I gives two possible choices for the impedance Z in the diagram. The circuit elements in Z satisfy the conditions R C R C>2 2 1 1. The transfer functions
VV
i
0 represents a kind of controller.
Match the impedances in Group I with the type of controllers in Group II
Q. 34 If the closed-loop transfer function of a control system is given as ( )( )( )
T ss s
s2 3
5+ +
-
, then It is
(A) an unstable system
(B) an uncontrollable system
(C) a minimum phase system
(D) a non-minimum phase system
2007 TWO MARKS
Q. 35 A control system with PD controller is shown in the figure. If the velocity error constant K 1000V = and the damping ratio .0 5z = , then the value of KP and KD are
Q. 36 The transfer function of a plant is ( )T s ( )( )s s s5 1
52=
+ + +The second-order approximation of ( )T s using dominant pole concept is
(A) ( )( )s s5 1
1+ +
(B) ( )( )s s5 1
5+ +
(C) s s 1
52 + +
(D) s s 1
12 + +
Q. 37 The open-loop transfer function of a plant is given as ( )G s 1s1
2= - . If the plant is operated in a unity feedback configuration, then the lead compensator that an stabilize this control system is
(A) ( )s
s2
10 1+-
(B) ( )s
s2
10 4++
(C) ( )
ss
1010 2
++
(D) ( )s
s10
2 2++
Q. 38 A unity feedback control system has an open-loop transfer function
( )( )
G ss s s
K7 122=
+ +
The gain K for which s j1 1= + will lie on the root locus of this system is(A) 4 (B) 5.5
(C) 6.5 (D) 10
Q. 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) ( )( )s s1 20
1+ +
(B) ( )( )s s s1 20
1+ +
(C) ( )( )s s s1 20
100+ +
(D) ( )( . )s s s1 1 0 05
100+ +
Q. 40 The state space representation of a separately excited DC servo motor dynamics is given as
dtd
dtdio
w
> H i u11
110
010a
w=
-- - += = =G G G
where w is the speed of the motor, ia is the armature current and u is the
armature voltage. The transfer function ( )( )
U ssw
of the motor is
(A) s s11 11
102 + +
(B)s s11 11
12 + +
(C) s s
s11 11
10 102 + +
+ (D) s s 11
12 + +
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GATE SOLVED PAPER - EC CONTROL SYSTEM
Statement for linked Answer Question 41 and 42 :Consider a linear system whose state space representation is ( ) ( )x t Ax t= . If
the initial state vector of the system is ( )x 012= -= G, then the system response is
( )x te
e2
x
t
2
2= -
-
-> H. If the itial state vector of the system changes to ( )x 012= -= G, then
the system response becomes ( )x tee
t
t= -
-
-> H
Q. 41 The eigenvalue and eigenvector pairs ( )vi il for the system are
(A) 111- -e o= G and 2
12- -e o= G (B) ,1
11- -e o= G and ,2
12-e o= G
(C) ,111- -e o= G and ,2
12- -e o= G (D) 2
11- -e o= G and ,1
12-e o= G
Q. 42 The system matrix A is
(A) 01
11-= G (B)
11
12- -= G
(C) 21
11- -= G (D)
02
13- -= G
2006 ONE MARK
Q. 43 The open-loop function of a unity-gain feedback control system is given by
( )G s ( )( )s s
K1 2
=+ +
The gain margin of the system in dB is given by(A) 0 (B) 1
(C) 20 (D) 3
2006 TWO MARKS
Q. 44 Consider two transfer functions ( )G ss as b
11 2=
+ + and ( )G s
s as bs
2 2=+ +
.
The 3-dB bandwidths of their frequency responses are, respectively(A) ,a b a b4 42 2- + (B) ,a b a b4 42 2+ -
(C) ,a b a b4 42 2- - (D) ,a b a b4 42 2+ +
Q. 45 The Nyquist plot of ( ) ( )G j H jw w for a closed loop control system, passes through ( , )j1 0- point in the GH plane. The gain margin of the system in dB is equal to(A) infinite (B) greater than zero
(C) less than zero (D) zero
Q. 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
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Q. 47 The transfer function of a phase lead compensator is given by ( )G sTsTs
11 3
c =++
where T 0> The maximum phase shift provide by such a compensator is
(A) 2p (B)
3p
(C) 4p (D)
6p
Q. 48 A linear system is described by the following state equation
( )X to ( ) ( ),AX t BU t A01
10= + = -= G
The state transition matrix of the system is
(A) cossin
sincos
tt
tt-= G (B)
cossin
sincos
tt
tt
-- -= G
(C) cossin
sincos
tt
tt
--
-= G (D)
coscos
sinsin
tt
tt
-= G
Statement for Linked Answer Questions 49 and 50:Consider a unity - gain feedback control system whose open - loop transfer
function is : ( )G ss
as 12= +
Q. 49 The value of a so that the system has a phase - margin equal to 4p is approximately
equal to(A) 2.40 (B) 1.40
(C) 0.84 (D) 0.74
Q. 50 With the value of a set for a phase - margin of 4p , the value of unit - impulse
response of the open - loop system at t 1= second is equal to(A) 3.40 (B) 2.40
(C) 1.84 (D) 1.74
2005 ONE MARK
Q. 51 Which one of the following polar diagrams corresponds to a lag network ?
Q. 52 A linear system is equivalently represented by two sets of state equations :
Xo AX BU= + and W CW DU= +o
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The eigenvalues of the representations are also computed as [ ]l and [ ]m . Which
one of the following statements is true ?(A) [ ] [ ]l m= and X W= (B) [ ] [ ]l m= and X W!
(C) [ ] [ ]!l m and X W= (D) [ ] [ ]l m= and X W!
Q. 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
Q. 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
(A) K 5< and K21
81< < (B) K
81< and K
21 5< <
(C) K81< and K5 < (D) K
81> and K5 >
Q. 55 In the derivation of expression for peak percent overshoot
Mp %exp1
1002
#x
px=-
-e o
Which one of the following conditions is NOT required ?(A) System is linear and time invariant
(B) The system transfer function has a pair of complex conjugate poles and no zeroes.
(C) There is no transportation delay in the system.
(D) The system has zero initial conditions.
Q. 56 A ramp input applied to an unity feedback system results in 5% steady state error. The type number and zero frequency gain of the system are respectively(A) 1 and 20 (B) 0 and 20
(C) 0 and 201 (D) 1 and
201
Q. 57 A double integrator plant ( ) / , ( )G s K s H s 12= = is to be compensated to achieve the damping ratio .0 5z = and an undamped natural frequency, 5nw = rad/sec
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which one of the following compensator ( )G se will be suitable ?
(A) ss
993
++ (B)
ss
399
++
(C) .s
s8 336
+- (D)
ss 6-
Q. 58 An unity feedback system is given as ( )( )( )
G ss sK s
31=+-
. Indicate the correct root locus diagram.
Statement for Linked Answer Question 59 and 60The open loop transfer function of a unity feedback system is given by
( )G s ( )s s
e2
3 s2
= +-
Q. 59 The gain and phase crossover frequencies in rad/sec are, respectively(A) 0.632 and 1.26
(B) 0.632 and 0.485
(C) 0.485 and 0.632
(D) 1.26 and 0.632
Q. 60 Based on the above results, the gain and phase margins of the system will be(A) -7.09 dB and .87 5c
(B) .7 09 dB and .87 5c
(C) .7 09 dB and .87 5c-
(D) .7 09- and .87 5c-
2004 ONE MARK
Q. 61 The gain margin for the system with open-loop transfer function
( ) ( )G s H s ( )
ss2 1
2= +, is
(A) 3 (B) 0
(C) 1 (D) 3-
Q. 62 Given ( ) ( )G s H s ( 1)( 3)s s sK= + + .The point of intersection of the asymptotes of the
root loci with the real axis is(A) 4- (B) .1 33
(C) .1 33- (D) 4
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2004 TWO MARKS
Q. 63 Consider the Bode magnitude plot shown in the fig. The transfer function ( )H s is
(A) ( )( )
( )s s
s1 100
10+ +
+ (B)
( )( )( )
s ss
10 10010 1
+ ++
(C) ( )( )
( )s s
s10 100
10 12
+ ++
(D) ( )( )
( )s s
s1 10
10 1003
+ ++
Q. 64 A causal system having the transfer function ( ) 1/( 2)H s s= + is excited with ( )u t10 . The time at which the output reaches 99% of its steady state value is
(A) 2.7 sec (B) 2.5 sec
(C) 2.3 sec (D) 2.1 sec
Q. 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 20 Hz is(A) 90c- (B) 0c
(C) 90c (D) 180c-
Q. 66 Consider the signal flow graph shown in Fig. The gain xx
1
5 is
(A) ( )
abcdbe cf dg1 - + +
(B) ( )be cf dg
bedg1 - + +
(C) ( )be cf dg bedg
abcd1 - + + +
(D) ( )
abcdbe cf dg bedg1 - + + +
Q. 67 If A21
23=
--= G, then sinAt is
(A) ( ) ( )
( ) ( )( ) ( )( ) ( )
sin sinsin sin
sin sinsin sin
t tt t
t tt t3
1 4 24
2 4 22 4
- + -- - + -
- - + -- + -= G
(B) ( )
( )( )
( )sin
sinsin
sint
ttt
2 23
--= G
(C) ( ) ( )( ) ( )
( ) ( )( ) ( )
sin sinsin sin
sin sinsin sin
t tt t
t tt t3
1 4 24
2 4 22 4
+- - +
- - -+= G
(D) ( ) ( )( ) ( )
( ) ( )( ) ( )
cos coscos cos
cos coscos cos
t tt t
t tt t3
1 24
2 4 22 4
- +- - + -
- + -- - += G
Q. 68 The open-loop transfer function of a unity feedback system is
( )G s ( )( )s s s s
K2 32=
+ + +The range of K for which the system is stable is
(A) K421 0> > (B) K13 0> >
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(C) K421 < < 3 (D) K6 < < 3-
Q. 69 For the polynomial ( )P s s s s s s2 2 3 152 4 3 2= + + + + + the number of roots which lie in the right half of the s -plane is(A) 4 (B) 2
(C) 3 (D) 1
Q. 70 The state variable equations of a system are : ,x x x u x x3 21 1 2 2 1=- - = =o o and y x u1= + . The system is(A) controllable but not observable
(B) observable but not controllable
(C) neither controllable nor observable
(D) controllable and observable
Q. 71 Given A10
01= = G, the state transition matrix eAt is given by
(A) e
e00t
t
-
-
> H (B) e
e00t
t= G
(C) e
e00t
t
-
-> H (D) e
e00t
t
= G
2003 ONE MARK
Q. 72 Fig. shows the Nyquist plot of the open-loop transfer function ( ) ( )G s H s of a system. If ( ) ( )G s H s has one right-hand pole, the closed-loop system is
(A) always stable
(B) unstable with one closed-loop right hand pole
(C) unstable with two closed-loop right hand poles
(D) unstable with three closed-loop right hand poles
Q. 73 A PD controller is used to compensate a system. Compared to the uncompensated system, the compensated system has(A) a higher type number (B) reduced damping
Q. 74 The signal flow graph of a system is shown in Fig. below. The transfer function ( )/ ( )C s R s of the system is
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(A) s s29 6
62 + +
(B) s s
s29 66
2 + +
(C) ( )
s ss s
29 62
2 + ++
(D) ( )
s ss s
29 627
2 + ++
Q. 75 The root locus of system ( ) ( )G s H s ( )( )s s s
K2 3
=+ +
has the break-away point located at(A) ( . , )0 5 0- (B) ( . , )2 548 0-(C) ( , )4 0- (D) ( . , )0 784 0-
Q. 76 The approximate Bode magnitude plot of a minimum phase system is shown in Fig. below. The transfer function of the system is
(A) ( ) ( )
( . )s s
s10
10 1000 18
2
3
+ ++
(B) ( )( )
( . )s s
s10
10 1000 17
3
+ ++
(C) ( ) ( )
( . )s s
s10 100
0 12
2
+ ++
(D) ( )( )
( . )s s
s10 100
0 12
3
+ ++
Q. 77 A second-order system has the transfer function
( )( )
R sC s
s s4 4
42=+ +
With ( )r t as the unit-step function, the response ( )c t of the system is
represented by
Q. 78 The gain margin and the phase margin of feedback system with
( ) ( )G s H s ( )s 100
83=
+ are
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(A) ,0dB c (B) ,3 3
(C) ,03 c (D) .88 5 dB, 3
Q. 79 The zero-input response of a system given by the state-space equationxx
xx
11
01
1
2
1
2=
o
o= = =G G G and ( )( )
xx
00
10
1
2== =G G is
(A) tet
t
= G (B) et
t
= G
(C) ete
t
t= G (D) t
tet= G
2002 ONE MARK
Q. 80 Consider a system with transfer function ( )G sks s
s6
62=+ ++ . Its damping ratio
will be 0.5 when the value of k is
(A) 62 (B) 3
(C) 61 (D) 6
Q. 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 ( )( )s s s
k1 3
=+ +
(A) s j 3=- (B) 1.5s =-(C) s 3=- (D) s 3=-
Q. 82 The phase margin of a system with the open - loop transfer function
( ) ( )G s H s ( )( )
( )s s
s1 2
1=+ +
-
(A) 0c (B) .63 4c
(C) 90c (D) 3
Q. 83 The transfer function ( )/ ( )Y s U s of system described by the state equation ( ) ( ) ( )x t x t u t2 2=- +o and ( ) . ( )y t x t0 5= is
(A) ( )
.s 20 5-
(B) ( )s 2
1-
(C) ( )
.s 20 5+
(D) ( )s 2
1+
2002 TWO MARKS
Q. 84 The system shown in the figure remains stable when(A) k 1<- (B) k1 3< <-(C) k1 3< < (D) k 3>
Q. 85 The transfer function of a system is ( )G s ( 1)( 100)s s100= + + . For a unit - step input to
the system the approximate settling time for 2% criterion is
(A)100 sec (B) 4 sec
(C) 1 sec (D) 0.01 sec
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Q. 86 The characteristic polynomial of a system is
( )q s s s s s s2 4 2 2 15 4 3 2= + + + + +The system is(A) stable
(B) marginally stable
(C) unstable
(D) oscillatory
Q. 87 The system with the open loop transfer function ( ) ( )( )
G s H ss s s 1
12=+ +
has a gain margin of(A) 6- db (B) 0 db
(C) 35 db (D) 6 db
2001 ONE MARK
Q. 88 The Nyquist plot for the open-loop transfer function ( )G s of a unity negative feedback system is shown in the figure, if ( )G s has no pole in the right-half of s -plane, the number of roots of the system characteristic equation in the right-half of s -plane is(A) 0 (B) 1
(C) 2 (D) 3
Q. 89 The equivalent of the block diagram in the figure is given is
Q. 90 The root-locus diagram for a closed-loop feedback system is shown in the figure. The system is overdamped.
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(A) only if k0 1# # (B) only if k1 5< <(C) only if k 5> (D) if k0 1<# or k 5>
Q. 91 If the characteristic equation of a closed - loop system is s s2 2 02 + + = , then the system is(A) overdamped (B) critically damped
(C) underdamped (D) undamped
2001 TWO MARK
Q. 92 An electrical system and its signal-flow graph representations are shown the figure (A) and (B) respectively. The values of G2 and H , respectively are
(A) ( ) ( ) ( )
( ),
( ) ( )( )
Z s Z s Z sZ s
Z s Z sZ s
1 3 4
3
1 3
3
+ + +-
(B) ( ) ( ) ( )
( ),
( ) ( )( )
Z s Z s Z sZ s
Z s Z sZ s
2 3 4
3
1 3
3
- +-
+-
(C) ( ) ( ) ( )
( ),
( ) ( )( )
Z s Z s Z sZ s
Z s Z sZ s
2 3 4
3
1 3
3
+ + + (D)
( ) ( ) ( )( )
,( ) ( )
( )Z s Z s Z s
Z sZ s Z s
Z s2 3 4
3
1 3
3
- +-
+Q. 93 The open-loop DC gain of a unity negative feedback system with closed-loop
transfer function s s
s7 13
42 + +
+ is
(A) 134 (B)
94
(C) 4 (D) 13
Q. 94 The feedback control system in the figure is stable
(A) for all 0K $ (B) only if 0K $
(C) only if 0 1K <# (D) only if 0 1K# #
2000 ONE MARK
Q. 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.
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2000 TWO MARKS
Q. 96 A system described by the transfer function ( )H s s s ks 3
13 2a
=+ + +
is stable. The constraints on a and k are.(A) , k0 3> <a a (B) , k0 3> >a a(C) , k0 3< >a a (D) , k0 3> <a a
1999 ONE MARK
Q. 97 For a second order system with the closed-loop transfer function
( )T s s s4 9
92=+ +
the settling time for 2-percent band, in seconds, is(A) 1.5 (B) 2.0
(C) 3.0 (D) 4.0
Q. 98 The gain margin (in dB) of a system a having the loop transfer function
( ) ( )G s H s ( )s s 1
2= + is
(A) 0 (B) 3
(C) 6 (D) 3
Q. 99 The system modeled described by the state equations is
X x u02
13
01= - +> >H H
Y x1 1= 8 B
(A) controllable and observable (B) controllable, but not observable
(C) observable, but not controllable (D) neither controllable nor observable
Q. 100 The phase margin (in degrees) of a system having the loop transfer function
( ) ( )G s H s ( )s s 12 3= + is
(A) 45c (B) 30c-(C) 60c (D) 30c
1999 TWO MARKS
Q. 101 An amplifier is assumed to have a single-pole high-frequency transfer function. The rise time of its output response to a step function input is 35 secn . The upper 3 dB frequency (in MHz) for the amplifier to as sinusoidal input is approximately at(A) 4.55 (B) 10
(C) 20 (D) 28.6
Q. 102 If the closed - loop transfer function ( )T s of a unity negative feedback system is given by
( )T s ....s a s a s a
a s an n
n n
n n
11
1
1=+ + + +
+-
-
-
then the steady state error for a unit ramp input is
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(A) aan
n
1- (B) a
an
n
2-
(C) aa
n
n 2- (D) zero
Q. 103 Consider the points s j3 41 =- + and s j3 22 =- - in the s-plane. Then, for a system with the open-loop transfer function
( ) ( )G s H s ( )s
K1 4=
+(A) s1 is on the root locus, but not s2
(B) s2 is on the root locus, but not s1
(C) both s1 and s2 are on the root locus
(D) neither s1 nor s2 is on the root locus
Q. 104 For the system described by the state equation
xo .
x u00
0 5
101
012
001
= +
R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW
If the control signal u is given by [ . ]u x v0 5 3 5= - - - + , then the eigen values of the closed-loop system will be(A) , ,0 1 2- - (B) , ,0 1 3- -(C) , ,1 1 2- - - (D) , ,0 1 1- -
1998 ONE MARK
Q. 105 The number of roots of s s s5 7 3 03 2+ + + = in the left half of the s -plane is(A) zero (B) one
(C) two (D) three
Q. 106 The transfer function of a tachometer is of the form
(A) Ks (B) sK
(C) ( )s
K1+ (D)
( )s sK
1+
Q. 107 Consider a unity feedback control system with open-loop transfer function
( )( )
G ss s
K1
= + .
The steady state error of the system due to unit step input is(A) zero (B) K
(C) /K1 (D) infinite
Q. 108 The transfer function of a zero-order-hold system is(A) ( / )( )s e1 1 sT+ - (B) ( / )( )s e1 1 sT- -
(C) ( / )s e1 1 sT- - (D) ( / )s e1 1 sT+ -
Q. 109 In the Bode-plot of a unity feedback control system, the value of phase of ( )G jw at the gain cross over frequency is 125c- . The phase margin of the system is(A) 125c- (B) 55c-(C) 55c (D) 125c
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Q. 110 Consider a feedback control system with loop transfer function
( ) ( )( )( )
( . )G s H s
s s sK s1 1 2
1 0 5= + ++
The type of the closed loop system is(A) zero (B) one
(C) two (D) three
Q. 111 The transfer function of a phase lead controller is TsTs
11 3
++ . The maximum value
of phase provided by this controller is(A) 90c (B) 60c
(C) 45c (D) 30c
Q. 112 The Nyquist plot of a phase transfer function ( ) ( )g j H jw w 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
Q. 113 The transfer function of a system is ( ) ( )s s
s s1 2
2 6 52
2
+ ++ + . The characteristic equation
of the system is(A) s s2 6 5 02 + + = (B) ( ) ( )s s1 2 02+ + =(C) ( ) ( )s s s s2 6 5 1 2 02 2+ + + + + = (D) ( ) ( )s s s s2 6 5 1 2 02 2+ + - + + =
Q. 114 In a synchro error detector, the output voltage is proportional to [ ( )] , ( )wheret tnw w is the rotor velocity and n equals(A) –2 (B) –1
(C) 1 (D) 2
1997 ONE MARK
Q. 115 In the signal flow graph of the figure is /y x equals
(A) 3 (B) 25
(C) 2 (D) None of the above
Q. 116 A certain linear time invariant system has the state and the output equations given below
XX
1
2
o
o> H XX u
10
11
01
1
2=
-+> > >H H H
y XX1 1
2
1= 8 :B D, (0) 1, (0) 1, (0) 0,If then isX X u dtdy
t1 2
0= =- =
=
(A) 1 (B) –1
(C) 0 (D) None of the above
***********
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SOLUTIONS
Sol. 1 Option (B) is correct.From the given plot, we obtain the slope as
Slope log loglog log
w wG G20 20
2 1
2 1= --
From the figure logG20 2 8 dB=- logG20 1 32 dB=and 1w 1 /rad s= 2w 10 /rad s=So, the slope is
Slope log log8 3210 1
= -- -
40 /dB decade=-Therefore, the transfer function can be given as
G s^ h Sk
2=at 1w =
G jw^ h wk k2= =
In decibel, log G j20 w^ h logk20 32= =
or, k .10 39 832
20= =Hence, the Transfer function is
G s^ h .sk
s39 8
2 2= =
Sol. 2 Option (A) is correct.For the given SFG, we have two forward paths
Pk1 s s s1 11 1 2= =- - -^ ^ ^ ^h h h h
Pk2 s s1 1 11 1= =- -^ ^ ^ ^h h h h
since, all the loops are touching to the paths Pk1and Pk2 so,
k1D 1k2D= =Now, we have
D 1= - (sum of individual loops)
+ (sum of product of nontouching loops)
Here, the loops are
L1 4 1 4= - =-^ ^h h
L2 s s4 41 1= - =- -^ ^h h
L3 s s s2 21 1 2= - =-- - -^ ^ ^h h h
L4 s s2 1 21 1= - =-- -^ ^ ^h h h
As all the loop , ,L L L1 2 3 and L4 are touching to each other so,
D L L L L1 1 2 3 4= - + + +^ h
s s s1 4 4 2 21 2 1= - - - - -- - -^ h
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s s5 6 21 2= + +From Mason’s gain formulae
U sY s^
^
h
h Pk k
DS D=
s ss s
5 6 21 2
2 1
=+ +
+- -
- -
s ss
5 6 21
2=+ ++
Sol. 3 Option (A) is correct.For the shown state diagram we can denote the states x1, x2 as below
So, from the state diagram, we obtain
x1o x u1=- - x2o x u x1 1 1 1 1 1 12 1=- + - - + - -^ ^ ^ ^ ^ ^ ^h h h h h h h
x2o x x u2 1=- + +and y x x u1 1 1 1 1 1 1 1 1 12 1= - + - - + - -^ ^ ^ ^ ^ ^ ^ ^ ^ ^h h h h h h h h h h
x x u1 2= - +Hence, in matrix form we can write the state variable equations
xx
1
2
o
o> H xx u
11
01
11
1
2=
-- +
-> > >H H H
and y xx u1 1
1
2= - +8 >B H
which can be written in more general form as
Xo X1
101
11=
-- +
-> >H H
y X u1 1= - +8 B
Sol. 4 Option (A) is correct.From the obtained state-variable equationsWe have
A 1
101=
--> H
So, SI A- S
S1
10
1=+- +> H
and SI A 1- -^ h
S
SS1
1 11
012=
++
+^ h> H
S
S S
11
11
0
11
2
=+
+ +^ h
R
T
SSSS
V
X
WWWW
Hence, the state transition matrix is obtained as eAt L SI A1 1= -- -
^ h
LS
S S
11
11
0
11
1
2
=+
+ +
-
^ h
R
T
SSSS
V
X
WWWW
Z
[
\
]]
]]
_
`
a
bb
bb
ete e
0t t
1
=-
- -> H
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Sol. 5 Option (C) is correct.Given, open loop transfer function
G s^ h sK
sK
1 1010 a a
101= + =
+By taking inverse Laplace transform, we have
g t^ h e t101
= -
Comparing with standard form of transfer function, Ae /t t- , we get the open loop time constant, olt 10=Now, we obtain the closed loop transfer function for the given system as
H s^ h 1 10 10G sG s
s KK
110
a
a=+
= + +^
^
h
h
s KK
a
a
101=
+ +^ hTaking inverse Laplace transform, we get h t^ h .k ea
k ta 101
= - +^ h
So, the time constant of closed loop system is obtained as
clt k
1a 10
1=+
or, clt k1a
= (approximately)
Now, given that ka reduces open loop time constant by a factor of 100. i.e.,
clt 100olt=
or, k1a 100
10=
Hence, ka 10=
Sol. 6 Option (C) is correct.
( )G s ( )( )( )
( )( )s s s
s s1 3 4
9 22
= + + ++ +
( )( )( )
( )( )j j j
j1 3 4
9 22
w w ww w= + + +
- + +
The steady state output will be zero if
( )G jw 0= 92w- + 0= & w 3 /rad s=
Sol. 7 Option (A) is correct.
( )Y s ( )
[ ( ) ( )]s as s
K sR s Y s
2 11
3 2=+ + +
+ -
( )( )
Y ss as s
K s1
2 11
3 2++ + +
+; E
( )( )
s as sK s
R s2 11
3 2=+ + +
+
( ) [ ( ) ( )]Y s s as s k k2 13 2+ + + + + ( 1) ( )K s R s= +
Transfer Function, ( )( )( )
H sR sY s=
( ) ( )( )
s as s k kK s
2 11
3 2=+ + + + +
+
Routh Table :
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For oscillation, ( ) ( )
aa K K2 1+ - +
0=
a KK
21= +
+
Auxiliary equation ( )A s ( )as k 1 02= + + =
s2 ak 1=- +
( )( )
kk k
11 2= +
- + + ( )k 2=- +
s j k 2= +
jw j k 2= +
w k 2 2= + = (Oscillation frequency)
k 2=
and a .2 22 1
43 0 75= +
+ = =
Sol. 8 Option (D) is correct.General form of state equations are given as
xo x uA B= + yo x uC D= +For the given problem
A 0
,a
aa
00 0
0 03
1
2=
R
T
SSSS
V
X
WWWW B
001
=
R
T
SSSS
V
X
WWWW
AB 0
a
aa a
00 0
0 0
001
0
03
1
2 2= =
R
T
SSSS
R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW
V
X
WWWW
A B2 a aa a
a a a a0
0
00 0
0
001
00
2 3
3 1
1 2 1 2
= =
R
T
SSSS
R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW
V
X
WWWW
For controllability it is necessary that following matrix has a tank of n 3= .
U : :B AB A B2= 6 @ 0
aa a0
01 0
00
2
1 2
=
R
T
SSSS
V
X
WWWW
So, a2 0!
a a1 2 0! a 01& ! a3 may be zero or not.
Sol. 9 Option (B) is correct.For given plot root locus exists from 3- to 3, So there must be odd number of poles and zeros. There is a double pole at s 3=-Now poles , , ,0 2 3 3= - - - zeros 1=-
Thus transfer function ( ) ( )G s H s ( )( )
( )s s s
k s2 3
12=
+ ++
Sol. 10 Option (A) is correct.
We have ( )G jw j5 w= +Here s 5= . Thus ( )G jw is a straight line parallel to jw axis.
Sol. 11 Option (B) is correct.
Here x y1= and xo dxdy1=
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y yy
xx2
1
2= => >H H x
12= > H
Now y1 s u21= +
( )y s 21 + u= y y21 1+o u= x x2+o u= xo x u2=- + xo [ 2] [1]x u= - +Drawing SFG as shown below
Thus x1o [ ] [ ]x u2 11= - + y1 x1= ; y x22 1=
y yy x
12
1
21= => >H H
Here x1 x=
Sol. 12 Option (C) is correct.
We have ( ) ( )G s H s ( )s s 10
1002=
+
Now ( ) ( )G j H jw w ( )j j 10
1002w w
=+
If pw is phase cross over frequency ( ) ( )G j H j 180c+ w w =
Thus 180c- 100 0 2tan tan tan 10p1 1 13
w= - -- - -a k
or 180c- 90 2 (0.1 )tan p1 w=- - -
or 45c (0.1 )tan p1 w= -
or tan45c 0.1 1pw =or pw 10 /rad se=
Now ( ) ( )G j H jw w ( )100
1002w w
=+
At pw w=
( ) ( )G j H jw w ( )10 100 100
100201= + =
Gain Margin ( ) ( )log G j H j20 10 w w=-
log20 201
10=- b l
26 dB=
Sol. 13 Option (D) is correct.
From option (D) TF ( )H s=
( ) ( )s s s s100
10010
1002 2!=+ +
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Sol. 14 Option (B) is correct.From the given block diagram
( )H s ( ) ( )Y s E s s 11$= - +
( )E s ( ) ( )R s H s= -
( ) ( )( )
( )R s Y s
sE s
1= - + +
( )E s s1 11- +: D ( ) ( )R s Y s= -
( )
( )ssE s
1+ ( ) ( )R s Y s= - ...(1)
( )Y s ( )
sE s
1= + ...(2)
From (1) and (2) ( )sY s ( ) ( )R s Y s= - ( ) ( )s Y s1+ ( )R s=Transfer function
( )( )
R sY s
s 11= +
Sol. 15 Option (B) is correct.Transfer function is given as
( )H s ( )( )
X sY s
s ps= = +
( )H jw j pj
ww= +
Amplitude Response
( )H jw p2 2w
w=+
Phase Response ( )hq w 90 tan p1c w= - -a k
Input ( )x t cosp t2 2p= -a k
Output ( )y t ( ) ( ) cosH j x t t2 3hw q p= - = -a k
( )H jw pp2 2w
w= =+
p1
4, ( 2 / )secrad
p2
2w=
+=
or 4p 2 4 3 4p p2 2&= + =
or p /2 3=Alternative :
hq 3 2 6p p p= - - - =a k9 C
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So, 6p tan p2
1p w= - -a k
tan p1 w-a k 2 6 3
p p p= - =
pw tan 3 3p= =a k
p2 , ( 2 / )secrad3 w= =
or p /2 3=
Sol. 16 Option (A) is correct.Initial slope is zero, so K 1=At corner frequency 0.5 / secrad1w = , slope increases by 20+ dB/decade, so there is a zero in the transfer function at 1wAt corner frequency 10 / secrad2w = , slope decreases by 20- dB/decade and becomes zero, so there is a pole in transfer function at 2w
Transfer function ( )H s s
K s
1
1
2
1
w
w=+
+
a
a
k
k
.
.( . )( )
s
s
ss
1 0 1
1 1 0 11 0 11 10=
+
+= +
+
a
a
k
k
Sol. 17 Option (D) is correct.Assign output of each integrator by a state variable
x1o x x1 2=- + x2o x u21=- + y . .x x0 5 0 51 2= +State variable representation
xo x u11
10
02=
-- +> >H H
yo [ . . ]x0 5 0 5=
Sol. 18 Option (C) is correct.By masson’s gain formula
Transfer function
( )H s ( )( )
U sY s PK K
DD= = /
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Forward path given
( )P abcdef1 .s s s2 1 1 0 5 1
2# # #= =
( )P abcdef2 .2 31 1 0 5# # #=
Loop gain ( )L cdc1 s1=-
( )L bcdb2 s s s1 1 1 1
2# #= - = -
D [ ]L L1 1 2= - + s s s s1 1 1 1 1 1
2 2= - - - = + +: D
11D = , 22D =
So, ( )H s ( )( )
U sY s= P P1 1 2 2
DD D= +
( )
( )
s s
s ss s
s
1 1 1
1 1 1 1
11
2
2
2
: :=
+ +
+=
+ ++
Sol. 19 Option (D) is correct.Steady state error is given as
eSS ( ) ( )( )
limG s G ssR s
1s C0= +"
( )R s s1= (unit step unit)
eSS ( ) ( )
limG s G s1
1s C0
= +"
( )
lim
s sG s
12 2
1s C0
2
=+
+ +"
eSS will be minimum if ( )limG ss
C0"
is maximumIn option (D)
( )limG ss
C0"
lim s s1 2 3s 0
3= + + ="
So, eSS lim 1 0s 0 3
= ="
(minimum)
Sol. 20 Option (C) is correct.This compensator is roughly equivalent to combining lead and lad compensators in the same design and it is referred also as PID compensator.
Sol. 21 Option (C) is correct.
Here A 10
01= = G and B
pq= = G
AB pq
pq
10
01= == = =G G G
S B ABpq
qp= =8 =B G
S pq pq 0= - =Since S is singular, system is completely uncontrollable for all values of p and q .
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Sol. 22 Option (B) is correct.The characteristic equation is
( ) ( )G s H s1 + 0=
or ( )s s
K s s1
2 22 2
2
2
++ +- +
0=
or ( )s s K s s2 2 2 22 2+ + + - + 0=
or K s ss s
2 22 2
2
2=-
- ++ +
For break away & break in point differentiating above w.r.t. s we have
dsdK
( )( )( ) ( )( )
s ss s s s s s
2 22 2 2 2 2 2 2 2
2 2
2 2
=-- +
- + + - + + - 0=
Thus ( )( ) ( )( )s s s s s s2 2 2 2 2 2 2 22 2- + + - + + - 0=
or s 2!=Let dq be the angle of departure at pole P , then
d p z z1 1 2q q q q- - + + 180c= dq- ( )180 p z1 1 2c q q q= - - + + ( )180 90 180 45c c c= - + - 45c=-
Sol. 23 Option (B) is correct.For under-damped second order response
( )sI sa ( ) ( ) ( )s I s U s10 10aw=- - +or ( )sw ( ) ( ) ( )s I s U s10 10a= - - + ( )( ) ( ) ( )s s s U s10 1 10w= - - + + From (3)
or ( )sw [ ] ( ) ( )s s s U s11 10 102 w=- + + +or ( ) ( )s s s11 112 w+ + ( )U s10=
or ( )( )
U ssw
( )s s11 11
102=+ +
Sol. 41 Option (A) is correct.
We have ( )x to ( )Ax t=
Let A pr
qs= = G
For initial state vector ( )x 012= -= G the system response is ( )x t
ee2
t
t
2
2= -
-
-> H
Thus ( )e
e2dtd t
dtd t
t
2
2
0-
-
-
=
> H pr
qs
12= -= =G G
or e
e2
4
( )
( )
2 0
2 0
- -
-> H pr
qs
12= -= =G G
24
-= G
p qr s
22=
--= G
We get p q2- 2=- and r s2 4- = ...(i)
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For initial state vector ( )x 011= -= G the system response is ( )x t
ee
t
t= -
-
-> H
Thus ( )ee
dtd t
dtd t
t 0-
-
-
=
> H pr
qs
11= -= =G G
e
e
( )
( )
0
0
- -
-> H pr
qs
11= -= =G G
11
-= G
p qr s=--= G
We get p q- 1=- and r s 1- = ...(2)
Solving (1) and (2) set of equations we get
pr
qs= G
02
13= - -= G
The characteristic equation
I Al - 0=
213
ll-+ 0=
or ( )3 2l l+ + 0=or l ,1 2=- -Thus Eigen values are 1- and 2-Eigen vectors for 11l =- ( )I A X1 1l - 0=
or xx2
13
1
1
11
21
ll-+= =G G 0=
xx
12
12
11
21
- -= =G G 0=
or x x11 21- - 0=or x x11 21+ 0=We have only one independent equation x x11 21=- .
Let x K11 = , then x K21 =- , the Eigen vector will be
xx11
21= G
KK K
11= - = -= =G G
Now Eigen vector for 22l =- ( )I A X2 2l - 0=
or xx2
13
2
2
12
22
ll-+= =G G 0=
or 22
11
- -= G
xx11
21= G 0=
or x x11 21- - 0=or x x11 21+ 0=We have only one independent equation x x11 21=- .
Let ,x K11 = then x K21 =- , the Eigen vector will be
xx12
22= G
KK K2
12= - = -= =G G
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Sol. 42 Option (D) is correct.As shown in previous solution the system matrix is
A 02
13= - -= G
Sol. 43 Option (D) is correct.Given system is 2nd order and for 2nd order system G.M. is infinite.
Sol. 44 Option (D) is correct.
Sol. 45 Option (D) is correct.If the Nyquist polt of ( ) ( )G j H jw w for a closed loop system pass through ( , )j1 0- point, the gain margin is 1 and in dB
GM log20 1=- 0= dB
Sol. 46 Option (B) is correct.The characteristics equation is
( ) ( )G s H s1 + 0=
( )
s as sK s
12 11
3 2++ + +
+ 0=
( )s as K s K2 13 2+ + + + + 0=The Routh Table is shown below. For system to be oscillatory stable
( ) ( )
aa K K2 1+ - +
0=
or a KK
21=
++ ...(1)
Then we have
as K 12 + + 0=At 2 rad/sec we have
s j s 42 2"w w= =- =- ,
Thus a K4 1- + + 0= ...(2)
Solving (i) and (ii) we get K 2= and .a 0 75= .
s3 1 K2 +
s2 a K1 +
s1 ( ) ( )a
K a K1 1+ - +
s0 K1 +
Sol. 47 Option (D) is correct.The transfer function of given compensator is
( )G sc TsTs
11 3=
++ T 0>
Comparing with
( )G sc TsaTs
11=
++ we get a 3=
The maximum phase sift is
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maxf tana
a2
11= -- tan tan2 33 1
311 1= - =- -
or maxf 6p=
Sol. 48 Option (A) is correct.
( )sI A- s
s00 0
110= - -= =G G
ss11
=-
= G
( )sI A 1- - s
ss1
11
12=+
-= G s
s
s
s
ss
1
11
11
1
2
2
2
2
= +
+-
+
+> H
( )tf [( )]e L sI AAt 1 1= = -- - cossin
sincos
tt
tt= -= G
Sol. 49 Option (C) is correct.
We have ( )G s s
as 12= +
( )G j+ w ( )tan a1 w p= --
Since PM is 4p i.e. 45c, thus
4p ( )G j g g "+p w w= + Gain cross over Frequency
or 4p ( )tan ag
1p w p= + --
or 4p ( )tan ag
1 w= -
or a gw 1=At gain crossover frequency ( )G j 1gw =
Thus 1 a
g
g2
2 2+
ww
1=
or 1 1+ g2w= (as )a 1gw =
or gw ( )2 4
1
=
Sol. 50 Option (C) is correct.
For .a 0 84= we have
( )G s .ss0 84 12= +
Due to ufb system ( )H s 1= and due to unit impulse response ( )R s 1= , thus
( )C s ( ) ( ) ( )G s R s G s= =
. .ss
s s0 84 1 1 0 84
2 2= + = +
Taking inverse Laplace transform
( )c t ( . ) ( )t u t0 84= +At t 1= , ( )secc 1 . .1 0 84 1 84= + =
Sol. 51 Option (D) is correct.The transfer function of a lag network is
( )T s s TsT
11
b=
++ ;T1 0> >b
( )T jw T
T11
2 2 2
2 2
w bw=
++
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and ( )T j+ w ( ) ( )tan tanT T1 1w wb= -- -
At 0w = , ( )T jw 1=At 0w = , ( )T j+ w 0 0tan 1=- =-
At 3w = , ( )T jw 1b
=
At 3w = , ( )T j+ w 0=
Sol. 52 Option (C) is correct.
We have Xo AX BU= + where l is set of Eigen values
and Wo CW DU= + where m is set of Eigen values
If a liner system is equivalently represented by two sets of state equations, then
for both sets, states will be same but their sets of Eigne values will not be same
i.e.
X W= but !l m
Sol. 53 Option (A) is correct.Despite the presence of negative feedback, control systems still have problems of instability because components used have nonlinearity. There are always some variation as compared to ideal characteristics.
Sol. 54 Option (B) is correct.
Sol. 55 Option (C) is correct.The peak percent overshoot is determined for LTI second order closed loop system with zero initial condition. It’s transfer function is
( )T s s s2 n n
n2 2
2
xw ww=
+ +Transfer function has a pair of complex conjugate poles and zeroes.
Sol. 56 Option (A) is correct.
For ramp input we have ( )R ss12=
Now ess ( )lim sE ss 0
="
( )
( )( )
lim limsG s
R ss sG s1
1s s0 0
=+
=+" "
or ess ( )%lim
sG s1 5
201
s 0= = =
"
Finite
But kv e1ss
= ( )lim sG s 20s 0
= ="
kv is finite for type 1 system having ramp input.
Sol. 57 Option (A) is correct.
Sol. 58 Option (C) is correct.Any point on real axis of s - is part of root locus if number of OL poles and zeros to right of that point is even. Thus (B) and (C) are possible option.The characteristics equation is
( ) ( )G s H s1 + 0=
or ( )( )
s sK s
13
1++-
0=
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or K s
s s1
32=
-+
For break away & break in point
dsdK ( )( )s s s s1 2 3 3 02= - + + + =
or s s2 32- + + 0=which gives s 3= , 1-Here 1- must be the break away point and 3 must be the break in point.
Sol. 59 Option (D) is correct.
( )G s ( )s s
e2
3 s2
= +-
or ( )G jw ( )j je
23 j2
w w= +w-
( )G jw 4
32w w
=+
Let at frequency gw the gain is 1. Thus
( )43
g g2w w +
1=
or 4 9g g4 2w w+ - 0=
or g2w .1 606=
or gw .1 26= rad/sec
Now ( )G j+ w 2 tan2 21w p w=- - - -
Let at frequency wf we have GH 180c+ =-
p- 2 tan2 21w p w=- - -f
f-
or 2 tan 21w w+f
f- 2p=
or 22 3
12
3w w w
+ -ff f
`c j m 2p=
or 2
524
3w w-f f
2p=
2
5wf 2
. p
or wf .0 63= rad
Sol. 60 Option (D) is correct.The gain at phase crossover frequency wf is
( )G j gw ( )43
2w w=
+f f
. ( . )0 63 0 63 43
2 2
1=+
or ( )G j gw .2 27= G.M. ( )log G j20 gw=- .log20 2 26- .7 08=- dB
Since G.M. is negative system is unstable.
The phase at gain cross over frequency is
( )G j g+ w 2 tan2 2gg1w p w=- - - -
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2 1.26 .tan2 21 261p
#=- - - -
or .4 65=- rad or .266 5c- PM ( )G j180 gc + w= + . .180 266 5 86 5c c c= - =-
Sol. 61 Option (D) is correct.The open loop transfer function is
( ) ( )G s H s ( )
ss2 1
2= +
Substituting s jw= we have
( ) ( )G j H jw w ( )j2 1
2ww=
-+
...(1)
( ) ( )G j H j+ w w 180 tan 1c w=- + -
The frequency at which phase becomes 180c- , is called phase crossover
frequency.
Thus 180- 180 tan 1c w=- + f-
or tan 1wf- 0=
or wf 0=The gain at 0w =f is
( ) ( )G j H jw w 2 12
23
ww= + =
Thus gain margin is 1 03
= = and in dB this is 3- .
Sol. 62 Option (C) is correct.Centroid is the point where all asymptotes intersects.
s No.of Open Loop Pole No.of Open Loop zero
Real of Open Loop Pole Real Part of Open Loop Pole= S S
S S-
-
3
1 3= - - .1 33=-
Sol. 63 Option (C) is correct.The given bode plot is shown below
At 1w = change in slope is +20 dB " 1 zero at 1w =At 10w = change in slope is 20- dB " 1 poles at 10w =At 100w = change in slope is 20- dB " 1 poles at 100w =
Thus ( )T s ( )( )
( )K s1 1
1s s10 100
=+ +
+
Now log K20 10 .K20 0 1"=- =
Thus ( )T s ( )( )
. ( )( )( )
( )ss s
s1 1
0 1 110 100
100 1s s10 100
=+ +
+ =+ +
+
Sol. 64 Option (C) is correct.
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We have ( )r t ( )u t10=
or ( )R s s10=
Now ( )H s s 2
1=+
( )C s ( ) ( )( )
H s R ss s s s2
1 102
10$ $= =+ +
or ( )C s s s5
25= -+
( )c t 5[1 ]e t2= - -
The steady state value of ( )c t is 5. It will reach 99% of steady state value
reaches at t , where
5[1 ]e t2- - .0 99 5#=or 1 e t2- - .0 99= e t2- .0 1=or t2- 0.1ln=or t . sec2 3=
Sol. 65 Option (A) is correct.Approximate (comparable to 90c) phase shift areDue to pole at .0 01 Hz 90" c-Due to pole at 80 Hz 90" c-Due to pole at 80 Hz 0"
Due to zero at 5 Hz 90" c
Due to zero at 100 Hz 0"
Due to zero at 200 Hz 0"
Thus approximate total 90c- phase shift is provided.
Sol. 66 Option (C) is correct.Mason Gain Formula
( )T s pk k
33S=
In given SFG there is only one forward path and 3 possible loop.
p1 abcd= 13 1=
13= - (sum of indivudual loops) - (Sum of two non touching loops)
( ) ( )L L L L L1 1 2 3 1 3= - + + +Non touching loop are L1 and L3 where
L L1 2 bedg=
Thus ( )( )
R sC s
( )be cf dg bedg
p1
1 13=- + + +
( )be cf dg bedg
abcd1
=- + + +
Sol. 67 Option (A) is correct.
We have A 21
23=
--= G
Characteristic equation is
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[ ]I Al - 0=
or 2
123
ll
+-
-+ 0=
or ( 2)( 3) 2+ +l l - 0=or 5 42l l+ + 0=Thus 1l 4=- and 12l =-Eigen values are 4- and 1- .
Eigen vectors for 41l =- ( )I A X1 1l - 0=
or xx
21
23
1
1
11
21
ll
+ -+= =G G 0=
xx
21
21
11
21
--
--= =G G 0=
or x x2 211 21- - 0=or x x11 21+ 0=We have only one independent equation x x11 21=- .
Let x K21 = , then x K11 =- , the Eigen vector will be
xx11
21= G
KK K
11=
-=
-= =G G
Now Eigen vector for 12l =- ( )I A X2 2l - 0=
or xx
21
23
2
2
12
22
ll
+-
-+= =G G 0=
or 11
22-
-= G
xx12
22= G 0=
We have only one independent equation x x212 22=Let x K22 = , then x K212 = . Thus Eigen vector will be
xx12
22= G
KK2
= = G K21= = G
Digonalizing matrix
M xx
xx
11
21
11
21
12
22= =
-= =G G
Now M 1- 31 1
121= -
---` j= G
Now Diagonal matrix of sinAt is D where
D ( )
( )sin
sint
t001
2
ll= = G
( )( )
sinsin
tt
40
02l
=-
= G
Now matrix B sinAt= MDM 1= -
( )
( )sin
sint
t31 1
121
40
0 11
21=-
- -- -
--` j= = =G G G
( 4 ) 2 ( )( ) ( )
( ) ( )( ) ( )
sin sinsin sin
sin sinsin sin
t tt t
t tt t3
14 2
2 4 22 4=-
- - - -- +
- - -- - - -` j= G
( ) ( )( ) ( )
( ) ( )( ) ( )
sin sinsin sin
sin sinsin sin
t tt t
t tt t3
1 4 24
2 4 22 4 2=-
- - - -- - -
- - -- - + -` j= G
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( ) ( )
( ( )( ) ( )( ) ( )
sin sinsin sin
sin sinsin sin
t tt t
t tt t s
31 4 2
42 4 22 4=
- + -- - + -
- - + -- + -` j= G
Sol. 68 Option (A) is correct.For ufb system the characteristic equation is
( )G s1 + 0=
( )( )s s s s
K12 2 3
( )G s
2
1+
+ + +
+ 0=
s s s s K4 5 64 3 2+ + + + 0=The routh table is shown below. For system to be stable,
K0 < and /
( )K0
2 721 4< -
This gives K0421< <
s4 1 5 K
s3 4 6 0
s227 K
s1/
K7 2
21 4- 0
s0 K
Sol. 69 Option (B) is correct.
We have ( )P s s s s s2 3 155 4 3= + + + +The routh table is shown below.
If 0"e + then 2 12e
e+ is positive and 2 1215 24 1442
ee e
+- - - is negative. Thus there are two
sign change in first column. Hence system has 2 root on RHS of plane.
s5 1 2 3
s4 1 2 15
s3 e 12- 0
s2 2 12e
e+ 15 0
s12 12
15 24 1442
ee e
+- - -
s0 0
Sol. 70 Option (D) is correct.
We have xx
1
2= G
xx u
32
10
10
1
2= +
- -= = =G G G
and Y [ ]1 0=xx u
12
1
2+= =G G
Here A 32
10=
- -= G, B
10= = G and C [ ]1 0=
The controllability matrix is
QC [ ]B AB=
10
32=
-= G
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detQC 0! Thus controllable
The observability matrix is
Q0 [ ]C A CT T T=
10
31 0!=
--= G
detQ0 0! Thus observable
Sol. 71 Option (B) is correct.
( )sI A- s
s00 1
001= -= =G G
ss
10
01=
--= G
( )sI A 1- - ( )
( )( )s
ss1
1 10
012=
-
--= G 0
0s
s
11
11= -
-> H
eAt [( )]L sI A1 1= -- -
e
e00t
t= = G
Sol. 72 Option (A) is correct.
Z P N= -N " Net encirclement of ( )j1 0- + 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.
Sol. 73 Option (C) is correct.PD Controller may accentuate noise at higher frequency. It does not effect the type of system and it increases the damping. It also reduce the maximum overshoot.
Sol. 74 Option (D) is correct.Mason Gain Formula
( )T s pk k
33S=
In given SFG there is only forward path and 3 possible loop.
p1 1=
13 s s s
s1 3 24 27= + + = +
L1 ,s
Ls
2 242= - = - and L
s3
3 = -
where L1 and L3 are non-touching
This ( )( )
R sC s
1 (
ploop gain) pair of non touching loops
1 13+
=- -
.1 s s s s s
ss
3 24 2 2 3
27
1 s s
ss
29 6
27
2
=- - - +
=- - -
+
+ +
+
^
^ ^
h
h h
( )
s ss s
29 627
2=+ +
+
Sol. 75 Option (D) is correct.We have
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( ) ( )G s H s1 + 0=
or ( )( )s s s
K12 3
++ +
0=
or K ( )s s s s5 62 2=- + +
dsdK ( )s s3 10 6 02=- + + =
which gives s 0.784, 2.5486
10 100 72!= - - =- -
The location of poles on s - plane is
Since breakpoint must lie on root locus so .s 0 748=- is possible.
Sol. 76 Option (A) is correct.The given bode plot is shown below
At .0 1w = change in slope is 60+ dB 3" zeroes at .0 1w =At 10w = change in slope is 40- dB 2" poles at 10w =At 100w = change in slope is 20- dB 1" poles at 100w =
Thus ( )T s ( ) ( )
( )K1 1
1.s s
s
102
100
0 13
=+ +
+
Now log K20 10 20=or K 10=
Thus ( )T s ( ) ( )
( )1 1
10 1.s s
s
102
100
0 13
=+ +
+
( ) ( )( . )
s ss
10 10010 0 1
2
8 3
=+ +
+
Sol. 77 Option (B) is correct.The characteristics equation is
s s4 42 + + 0=Comparing with
s 2 n n2 2xw w+ + 0=
we get 2 nxw 4= and 4n2w =
Thus x 1= Critically damped
ts 41 2
4 2n #xw
= = =
Sol. 78 Option (B) is correct.
Sol. 79 Option (C) is correct.We have
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xx1
2o
o= G
xx
11
01
1
2= = =G G and
( )( )
xx
00
10
1
2== =G G
A 11
01= = G
( )sI A- s
ss
s00 1
101
11
01= - =
-- -= = =G G G
( )sI A 1- - ( )
( )( )s
ss1
1 11
01
0
( )
s
s s2
11
11
11
2
=-
-+ - = -
-+
-> >H H
[( ) ]L sI A1 1-- - eete e
0Att
t t= = = G
( )x t [ ( )]e x tete e
ete
0 10
Att
t t
t
t0#= = == = =G G G
Sol. 80 Option (C) is correct.The characteristics equation is
ks s 62 + + 0=
or sK
sK
1 62 + + 0=
Comparing with s s2 n n2 2xw w+ + 0= we have
we get 2 nxw K1= and
K6
n2w =
or . K2 0 5 6# # w K1= Given .0 5x =
or K6
K1
2= & K 61=
Sol. 81 Option (B) is correct.Any point on real axis lies on the root locus if total number of poles and zeros to the right of that point is odd. Here .s 1 5=- does not lie on real axis because there are total two poles and zeros (0 and 1- ) to the right of .s 1 5=- .
Sol. 82 Option (D) is correct.From the expression of OLTF it may be easily see that the maximum 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 does not exist, the phase margin is infinite.
Sol. 83 Option (D) is correct.
We have ( )x to ( ) ( )x t u t2 2=- + ...(i)
Taking Laplace transform we get
( )sX s ( ) ( )X s U s2 2=- +or ( ) ( )s X s2+ ( )U s2=
or ( )X s ( )
( )sU s
22=+
Now ( )y t . ( )x t0 5= ( )Y s . ( )X s0 5=
or ( )Y s . ( )
sU s2
0 5 2#=+
or ( )( )
U sY s
( )s 2
1=+
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Sol. 84 Option (D) is correct.From Mason gain formula we can write transfer function as
( )( )
R sY s
( ) ( )s K
K1 3 3s s
KsK
3=- +
= - --
For system to be stable ( )K3 0<- i.e. K 3>
Sol. 85 Option (B) is correct.The characteristics equation is
( )( )s s1 100+ + 0= s s101 1002 + + 0=Comparing with s 2 n n
2 2xw w+ + 0= we get
2 nxw 101= and 100n2w =
Thus x 20101= Overdamped
For overdamped system settling time can be determined by the dominant pole
of the closed loop system. In given system dominant pole consideration is at
s 1=- . Thus
T1 1= and Ts sec
T4 4= =
Sol. 86 Option (B) is correct.Routh table is shown below. Here all element in 3rd row are zero, so system is marginal stable.
s5 2 4 2
s4 1 2 1
s3 0 0 0
s2
s1
s0
Sol. 87 Option (B) is correct.The open loop transfer function is
( ) ( )G s H s ( )s s s 1
12=+ +
Substituting s jw= we have
( ) ( )G j H jw w ( )j j 1
12w w w
=- + +
( ) ( )G j H j+ w w ( )
tan2 11
2p
ww=- --
-
The frequency at which phase becomes 180c- , is called phase crossover
frequency.
Thus 180- 90 tan1
12w
w=- -- f
f-
or 90- tan1
12w
w=-- f
f-
or 1 2w- f 0=
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wf 1= rad/sec
The gain margin at this frequency 1w =f is
GM ( ) ( )log G j H j20 10 w w=- f f
( ( )log20 1102 2 2w w w= - +f f f log20 1 0=- =
Sol. 88 Option (A) is correct.
Z P N= -N " Net encirclement of ( )j1 0- + by Nyquist plot,
P " Number of open loop poles in right had side of s - plane
Z " Number of closed loop poles in right hand side of s - plane
Here N 0= (1 encirclement in CW direction and other in CCW)
and P 0=Thus Z 0=Hence there are no roots on RH of s - plane.
Sol. 89 Option (D) is correct.Take off point is moved after G2 as shown below
Sol. 90 Option (D) is correct.If roots of characteristics equation lie on negative axis at different positions (i.e. unequal), then system response is over damped.From the root locus diagram we see that for K0 1< < , the roots are on imaginary axis and for K1 5< < roots are on complex plain. For K 5> roots are again on imaginary axis.Thus system is over damped for 0 1K <# and K 5> .
Sol. 91 Option (C) is correct.
The characteristics equation is
s s2 22 + + 0=Comparing with s 2 n n
2 2xw w+ + 0= we get
2 nxw 2= and 2n2w =
nw 2=
and x 2
1=
Since 1<x thus system is under damped
Sol. 92 Option (C) is correct.From SFG we have
( )I s1 ( ) ( )G V s HI si1 2= + ...(1)
( )I s2 ( )G I s2 1= ...(2)
( )V s0 ( )G I s3 2= ...(3)
Now applying KVL in given block diagram we have
( )V si ( ) ( ) [ ( ) ( )] ( )I s Z s I s I s Z s1 1 1 2 3= + - ...(4)
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0 [ ( ) ( )] ( ) ( ) ( ) ( ) ( )I s I s Z s I s Z s I s Z s2 1 3 2 2 2 4= - + + ...(5)
From (4) we have
or ( )V si ( )[ ( ) ( )] ( ) ( )I s Z s Z S I s Z S1 1 3 2 3= + -
or ( )I s1 ( ) ( ) ( ) ( )
( )V
Z s Z sI
Z s Z sZ s1
i1 3
21 3
3=+
++
...(6)
From (5) we have
( ) ( )I s Z S1 3 ( )[ ( ) ( ) ( )]I s Z s Z s Z s2 2 3 4= + + ...(7)
or ( )I ss ( ) ( ) ( )
( ) ( )Z s Z s Z s
I s Z s3 2 4
1 3=+ +
Comparing (2) and (7) we have
G2 ( ) ( ) ( )( )
Z s Z s Z sZ s
3 2 4
3=+ +
Comparing (1) and (6) we have
H ( ) ( )
( )Z s Z s
Z s1 3
3=+
Sol. 93 Option (B) is correct.For unity negative feedback system the closed loop transfer function is
CLTF ( )
( )G s
G ss s
s1 7 13
42=
+=
+ ++ , ( )G s OL" Gain
or ( )
( )G s
G s1 +
ss s
47 132
=+
+ +
or ( )G s1
ss s
47 13 1
2=
++ + -
ss s
46 92
=+
+ +
or ( )G s s s
s6 9
42=+ ++
For DC gain s 0= , thus
Thus ( )G 0 94=
Sol. 94 Option (C) is correct.From the Block diagram transfer function is
( )T s ( ) ( )( )
G s H sG s
1=
+
Where ( )G s ( )( )s
K s22=
+-
and ( )H s ( )s 2= -The Characteristic equation is
( ) ( )G s H s1 + 0=
( )
( )( )
sK s
s12
222+
+- - 0=
or ( ) ( )s K s2 22 2+ + - 0=or ( ) ( )K s K s K1 4 1 4 42+ + - + + 0=Routh Table is shown below. For System to be stable k1 0>+ , and k4 4 0>+
and k4 4 0>- . This gives K1 1< <-As per question for K0 1<#
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s2 k1 + k4 4+
s1 k4 4- 0
s0 k4 4+
Sol. 95 Option (B) is correct.It is stable at all frequencies because for resistive network feedback factor is always less than unity. Hence overall gain decreases.
Sol. 96 Option (B) is correct.The characteristics equation is s s ks 32 2a+ + + 0=The Routh Table is shown below
For system to be stable 0>a and K 3 0>a
a -
Thus 0>a and K 3>a
s3 1 K
s2 a 3
s1 K 3a
a - 0
s0 3
Sol. 97 Option (B) is correct.Closed loop transfer function is given as
( )T s s s4 9
92=+ +
by comparing with standard form we get natural freq.
A2w 9=
nw 3= 2 nxw 4=
Damping factor x /2 34 2 3#
= =
For second order system the setting time for 2-percent band is given by
ts 4
nxw= /3 2 3
424
#= = 2=
Sol. 98 Option (D) is correct.Given loop transfer function is
Sol. 102 Option (C) is correct.Closed-loop transfer function is given by
( )T s ...s a s a s a
a s an n
n n
n n
11
1
1=+ + + +
+-
-
-
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1
...
...
s a s a sa s a
s a s a sa s a
n nn
n n
n nn
n n
11
22
1
11
22
1
=+
+ ++
+ ++
--
-
--
-
Thus ( ) ( )G s H s ....s a s a s
a s an n
n
n n
11
22
1=+ +
+-
-
-
For unity feed back ( )H s 1=
Thus ( )G s ....s a s a s
a s an n
n
n n
11
22
1=+ +
+-
-
-
Steady state error is given by
( )E s ( )( ) ( )
limR sG s H s1
1s 0
= +"
for unity feed back ( )H s 1=
Here input ( )R s ( )unit Ramps12=
so ( )E s ( )
lims G s1
11
s 0 2= +"
....
....lims s a s a
s a s a s1s n n
n
n nn
0 21
11
12
2
=+ + +
+ + +" -
--
aa
n
n 2= -
Sol. 103 Option (B) is correct.
Sol. 104 Option (A) is correct.
Sol. 105 Option (A) is correct.Applying Routh’s criteria
s s s5 7 33 2+ + + 0=
s3 1 7
s2 5 3
s15
7 5 3532=# - 0
s0 3
There is no sign change in the first column. Thus there is no root lying in the left-half plane.
Sol. 106 Option (A) is correct.Techometer acts like a differentiator so its transfer function is of the form ks .
Sol. 107 Option (A) is correct.Open loop transfer function is
( )G s ( )s s
K1
= +Steady state error
( )E s ( ) ( )( )
limG s H ssR s
1s 0= +"
Where ( )R s ( )input H s 1= = (unity feedback)
( )R s s1=
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so ( )E s
( )
( )lim lim
s sK
sss s Ks s
11
11
0s s0 0 2=
+ +
=+ +
+ =" "
Sol. 108 Option (B) is correct.Fig given below shows a unit impulse input given to a zero-order hold circuit which holds the input signal for a duration T & therefore, the output is a unit step function till duration T .
( )h t ( ) ( )u t u t T= - -Taking Laplace transform we have
( )H s s s e1 1 sT= - - s e1 1 sT= - -6 @
Sol. 109 Option (C) is correct.
Phase margin 180 gc q= + where gq value= of phase at gain crossover
frequency.
Here gq 125c=-so P.M 180 125 55c c c= - =
Sol. 110 Option (B) is correct.Open loop transfer function is given by
( ) ( )G s H s ( )( )
( . )s s s
K s1 1 2
1 0 5= + ++
Close looped system is of type 1.
It must be noted that type of the system is defined as no. of poles lyinglies at origin
in OLTF.
Sol. 111 Option (D) is correct.Transfer function of the phase lead controller is
.T F ( )( )
sTs
T jT j
11 3
11 3
ww= +
+ = ++
Phase is
( )f w ( ) ( )tan tanT T31 1w w= -- -
( )f w tanT
T T1 331
2 2ww w=
+--
; E
( )f w tanT
T1 3
212 2w
w=+
-; E
For maximum value of phase
( )
dd
wf w
0=
or 1 T3 2 2w=
Tw 3
1=
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So maximum phase is
maxf tanT
T1 3
212 2w
w=+
-; E at T
31w =
30tan tan1 3
2
311
31
31
1
#c=
+= =- -
> ;H E
Sol. 112 Option (A) is correct.( ) ( )G j H jw w enclose the ( 1, 0)- point so here ( ) ( )G j H j 1>p pw w
pw Phase= cross over frequency
Gain Margin ( ) ( )
logG j H j
20 1p p
10 w w=
so gain margin will be less than zero.
Sol. 113 Option (B) is correct.The denominator of Transfer function is called the characteristic equation of the system. so here characteristic equation is
( 1) ( 2)s s 02+ + =
Sol. 114 Option (C) is correct.In synchro error detector, output voltage is proportional to [ ( )]tw , where ( )tw is the rotor velocity so here n 1=
Sol. 115 Option (C) is correct.By masson’s gain formulae
