Control Systems – GATE batches

TYC
Basics and Laplace transforms
1.1
01. The Laplace transform of   t  2  is  0, t  0
f  t    3t then F(s) is
1 1 te , t  0
(a) (b)

Th
s2 s2 1 1

is
(a) (b)

PD
e 2s e 2s
s3  s  3 2

F
be
(c) (d) 1 1

lo
s s

ng
(c) (d)
s  s  3
s3 2

s
e 2 s

to
Given that G  s   , then g  t  is

Ab
02.
07. A function f(t) is defined as below:

hi
s 1

na
 0, t0
e   t  2

v
 t 2
 

Kr
(a) 
f t    

is
 then F(s) is

hn
e  t  2 
t
sin  2t  4  , t  0

a
(b)

M
  

te
e t   t  2 

fF
(c) s2 s2

R
(a) (b)

0
   

(a
e   t 
2
 t 2 2 s 4 s2  4

bh
(d)

in
av
03. The inverse Laplace transform of s2 s2

m
(c) (d)

kr
s2
s   

is
2
4 2 s2  4

hn
 s  2 2  1

a2
02
  2
It is given that F  s  

0@
08. ,
 s  1 s  2 
2
e t cos 2t et sin 2t

gm
(a) (b)

ai
l.c
(c) e2t cos t (d) e 2t sin t then f  t  is given by

om
,8
04. A linear time invariant system is initially
(a) f  t   2e  t  2te 2t  2e2t for t  0

54
at rest, when subjected to a unit step input,

72
(b) f  t   2e t  2te 2t  2e 2t for t  0

68
gives a response y  t   te  t ; t  0 . The

03
4)
transfer function of the system is (c) f  t   2e t  2te 2t  2e 2t for t  0

(a)
1
(b)
1 (d) f  t   2e t  2e 2t for t  0
 s  1 2
s  s  1
2
2s  1
09. The initial value of F  s   is 2
s 1 s  s 1
(c) (d)
 s  12 s  s  1 (a) 0 (b) 2
(c) ∞ (d) 0.5
05. A system is described as 10. A system’s output c, is required to the sys-
dc  t  tem’s input r, by the straight line relation-
2  c  t   r  t  t  2  where r(t) and
dt ship, c = 5r + 7. The system is
c(t) are input and output respectively. (a) Linear
C s (b) not linear because superposition prin-
Then is ciple fails
R s
(c) not linear because principle of homo-
1  2s 1  2s geneity fails
(a) (b)
s 1  2s 
2
s 1  2s  2 (d) not linear because both superposition
principle and principle of homogeneity
1  2s 1  2s
(c) (d) fails
s 1  2 s  s 1  2 s  11. Inverse Laplace transform of
06. A function f(t) is defined as below: 1
F s   2 2 is

s s 2 
1 Prepared by: BNSS Shankar, M.Tech (IITK) OHM Institute – Hyderabad GATE|ESE|PSUs
Control Systems TYC 1.1

(a) f t  
1  1 
t  sin t u t  (b) x  t   et  1u  t 
2 
    (c) x  t   e 2t  1u  t 
 1 
(b) f t   t  sin t  u t  (d) x  t   e  t  1u  t 
  
1 15. Obtain the solution of the differential
(c) f t   t  sin t ut  equation x  ax  A sin t , x  0   b
 2

Th
f t    2 t  sin t u t 

is
(d)

PD
Answer keys

F
s 1
Given that F s  

be
12. . Then
  TYC – 1.1

lo
s s  s 1
2

ng
s
f t  is

to
01 – c 02 – a 03 – c 04 – c 05 – a

Ab
hi
t
 3  06 – b 07 – d 08 – a 09 – b 10 – d

na
2
(a) 1  e cos t   for t  0

v
2
11 – a 12 – b 13 – a 14 – c

Kr
3  2 6

is
15:

hn
a
2 2t  3   A   at Aa

M
(b) 1  e cos  t   for t  0 x t    b  2 e  2 sin t

te
2

fF
3  2 6  a   a  2

R
0
A

(a
2  2t  3   cos t  t  0 

bh
(c) 1  e sin  t   for t  0  2

in
a  2

av
3  2 6

m
kr
is
2 2
t
 3 

hn
(d) 1  e sin  t   for t  0

a2
3  2 6

02
0@
13. Inverse Laplace transform of

gm
6s  3

ai
F  s   2 is

l.c
om
s

,8
f  t   6  3t , for t  0

54
(a)

72
68
(b) f  t   6t  3, for t  0

03
4)
(c) f  t   6  3t 2 , for t  0
3
(d) f  t   6  , for t  0
t
14. A differential equation is given as below:
x  2 x    t  , x 0  0 
Then, the solution is
(a) x  t   e 2t  1u  t 

2 Prepared by: BNSS Shankar, M.Tech (IITK) OHM Institute – Hyderabad GATE|ESE|PSUs
 TYC
Transfer functions – Electric Circuits
1.2
I s 04. Consider the following circuit. By put-
01. Consider the following circuit, is, L V s
Vi  s  ting, T1  and T2  CR then o is,
R Vi  s 

Th
C

is
i L

PD


F
be
i

lo
ng
Vin R Vin R

s
C vo

to
Ab
hi
na


v
Kr
is
hn
1 sC T1T2 1

a
(a) (b) (a) (b)

M
1  sRC 1  sRC s  s 1
2
s  s  T1T2 2

te
fF
C

R
(c) 1  sRC 1 1

0
(d) (c) (d)

(a
1  sRC

bh
T1T2 s  sT2  1
2
T1T2 s  sT1  1
2

in
V s

av
V s

m
02. Consider the following circuit, o is,

kr
Vi  s  05. Consider the following circuit, o is,

is
Vi  s 

hn
a2
C

02
1 1

0@
 

gm
i

ai
l.c
Vin vo
Vin R vo 1F 1F

om
,8
54
72
 

68
4 1

03
(a) 2 (b) 2

4)
sRC s  6s  4 s  3s  1
(a) (b) 1  sRC 1 3
1  sRC (c) 2 (d) 2
1 sC s  s 1 s  3s  3
(c) (d)
1  sRC 1  sRC V s
06. Consider the following circuit, o is,
03. Consider the following circuit. By put- Vi  s 
R2 V s 1F 0.5 F
ting, a  and CR1  T then o
R1  R2 Vi  s  
is, Vin 1 vo
C 1
i
 
2
s s
R1 (a) (b)
Vin R2 vo s  4s  2
2
s  s 1
2

s2 s
(c) 2 (d) 2
 s  4s  2 s  s 1

1  sT a 1  asT  07. Consider the following circuit, if R = C =

(a) (b)
a 1  sT  1  sT  V s
1 then o is,
a 1  sT  1  sT  Vi  s 
(c) (d)
1  asT 1  asT

3 Prepared by: BNSS Shankar, M.Tech (IITK) OHM Institute – Hyderabad GATE|ESE|PSUs
Control Systems TYC 1.2
R C s2
s
 (a) (b)
2s  s  2
2
s 2  3s  1
Vin R C vo 1 s2
(c) 2 (d) 2
2s  s  2 s  s 1
 V s
11. The transfer function G  s   o for
s s2 Vi  s 

Th
(a) 2 (b) 2

is
s  3s  1 s  3s  1

PD
the following network is

F
s2 1F

be
s 1
(c) 2 (d) 2

lo
 

ng
s  s 1 s  s 1

s
1

to
V s vi  t 

Ab
08. Consider the following circuit, o is, vo  t 

hi
Vi  s 

na
1F

v
Kr
L C  

is
hn
 s 1 s2

a
M
(a) (b)
2  s  2 2  s  1

te
fF
Vin R vo

R
0
s2 1

(a
(c) (d)

bh
 s 1 2

in
av
V s

m
The transfer function G  s   o

kr
sCR 12. for

is
Vi  s 

hn
(a)

a2
s LC  sRC  1
2

02
the following circuit is

0@
s
(b) 2 1H

gm
s LC  sRC  1

ai
l.c
sCR

om
1
(c) 2

,8
s RC  sLC  1 vi  t  


54
1 

72
sLR

68
(d) 2 1H vo  t 

03
s RC  sLR  1

4)
Vo  s  
09. Consider the following circuit, is,
Vi  s 
1 s
1 (a) (b) 2
 s  3s  1 2
s  3s  1
3s
(c) 2 (d) s 2  3s  1
vin  t  
 1H 1 vo  t  s  2s  2
13. In the following circuit, R1 = 360 kΩ, C1 =
 5.6 µF, R2 = 220 kΩ, C2 = 0.1 µF. The trans-
s 1
(a) (b) V s
s 1 s2 fer function G  s   o is
s s Vi  s 
(c) (d)
s2 2s  1 R2 C2
V s C1
10. Consider the following circuit, o is,
Vi  s 

1 1H vi  t  
 vo  t 
R1
vin  t  
 1 1F vo  t 
s 2  45.95s  22.55
(a) 1.232

s

4 Prepared by: BNSS Shankar, M.Tech (IITK) OHM Institute – Hyderabad GATE|ESE|PSUs
Control Systems TYC 1.2

s 2  45.95s  22.55

 s  20   s  10  
(b) 1.232 (c) (d)
s  s  10  s  s  20 
2 2
s
s  45.95s  22.55
2
V s
(c) 16. The transfer function G  s   o of the
s Vi  s 
 459.95 
(d) 1.232  s   22.55 following circuit is

Th
is
 s  100 k

PD
14. In the following circuit, R1 = 500 kΩ, C1

F
be
= 2 µF, R2 = 100 kΩ, C2 = 2 µF. The trans- 100 k 1μF

lo
ng
V s 

s
fer function G  s   o vi  t 

to
is  vo  t 

Ab
Vi  s 

hi
na
v
R2 C2

Kr
is
s  10 s  10

hn
(a) (b) 

a
C1 s2 s2

M
R1

te


fF
vi  t  s s

R
 vo  t  (c) (d) 

0
s  10 s  10

(a
bh
V s

in
av
17. The transfer function G  s   o of the

m
Vi  s 

kr
is
1  s  10  1 s2 

hn
(a)   (b)   following circuit is

a2
5  s  2  5  s  10 

02
R1

0@
gm
ai
1  s  5  s  5 R1

l.c
(c)   (d) 5 

om
5  s  1   s  1 


,8
54


72
68
15. In the following circuit, each resistor is vi  t 
R2 vo  t 

03
C

4)
100 kΩ and each capacitor is 1 µF, then
V s
the transfer function G  s   o is
Vi  s 

R2Cs  1 R Cs  1
(a)  (b)  1
R2Cs  1 R1Cs  1
 R2Cs  1 R1Cs  1
vi  t  
(c) (d)
vo  t  R2Cs  1 R1Cs  1
Answer keys
TYC – 1.2
s  s  20  s  s  10  01 – b 02 – a 03 – c 04 – d 05 – b
(a)  (b)  06 – c 07 – a 08 – a 09 – d 10 – c
 s  10   s  20 
2 2
11 – d 12 – b 13 – b 14 – c 15 – a
16 – d 17 – a

5 Prepared by: BNSS Shankar, M.Tech (IITK) OHM Institute – Hyderabad GATE|ESE|PSUs
 TYC

1.3 Block diagram reduction

01. The characteristic equation of the closed – 1 14

(c) (d)
loop system shown in the figure is s 2  14 s  44 s 2  44 s  14
R s C s
C

Th
3 04. For the block diagram shown in figure,

is
 4 R

PD
s 1

F
is given by

be
lo
ng
H2

s
10

to


Ab
s  10 R  C

hi
G1

na
G2


v
(a) s 2  11s  10  0 

Kr
is
hn
(b) s 2  11s  130  0

a
H1

M
te
(c) s  10 s  120  0
2

fF
G1G2

R
(a)

0
(a
(d) s  10 s  12  0
2
1  G1H1  G2 H 2

bh
in
av
C G1G2

m
02. For the block diagram shown in figure, (b)

kr
R 1  G1H 2  G2 H1

is
hn
a2
is given by G1G2

02
(c)

0@
1  G1H1  G2 H 2

gm
R C


ai
G1  G2 G1G2

l.c
(d)

om
1  G1H 2  G2 H1

,8
54
H1

72
C

68
05. For the block diagram shown in figure,

03
G1G2 R

4)
(a)
1  G1G2 H1  G2 H1  G1 is given by
G1G2 R K1 1 C

(b) 1 s s
1  G1G2 H1  G1 H 2  G2
G1G2
(c) K2
1  G1G2 H1  G2 H1  G1
G1G2 
(d)
1  G1 H1  G2 H1  G1
K1
C s (a)
03. for the block diagram shown in figure s  s 1  K1  K 2   K 2
2

R s
K1
is given by (b)
s  s 1  K1K 2   K1
2

R  s C s K2
 1 1 (c)

s4


s 8 s  s 1  K1 K 2   K 2
2

K2
(d)
s  s 1  K1  K 2   K 2
2
44 1
(a) (b)
s  14 s  44
2
s 2  14 s  14

6 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 1.3

C s C s
06. for the block diagram shown in figure 09. for the closed – loop system shown in
R s R s
is given by figure is given by
5
R  s   C s

Th
 R  s   C s

is
s s 1/ s 

PD
1/ s 2s


F
   

be
lo
ng
1/ s 2

s
to
Ab
hi
s

na
v
12 s 2s

Kr
s3  1 s3  1 (a) (b)

is
hn
(a) (b) 3s  5 3s  5

a
2s 3  s 2  2s  1 2s 4  s 2  2s  1

M
12 s 20 s

te
fF
s2 1 s3  1 (c) (d)

R
(c) (d) 6 s  13s  5
2
2s  1

0
(a
2s 4  s 2  2s 2s 4  s 2  2s

bh
C s

in
C s

av
10. for the closed – loop system shown in
R s

m
07. for the closed – loop system shown in

kr
R s

is
hn
figure is given by

a2
02
figure is given by

0@
s
R s  C s

gm
H1 

ai
l.c
R  s  C s  

om
 1/ s
G1 G2 G3

,8


54


72


68
s

03
 G1G3  1 G2  G2G3  1 G1

4)

(a) (b)
1  G1 H1 1  G1 H1 1/ s

(c)
 G1G2  1 G3 (d)
 G1G3  1 G2 s  s 2  1 s2  1
1  G1 H1 1  G2 H1 (a) (b) 4 2
s4  s2  1 s  s 1
C s s  s 2  1 s  s 2  1
08. for the closed – loop system shown in
R s (c)
s 4  s3  1
(d)
s4  s2 1
figure is given by C s
11. for the closed – loop system shown in
H1 R s

R  s
 figure is given by
G
 C s
 G1

R  s   C s
G2
H2  

G  H1 G  H2 H1
(a) (b) 
1  GH 2 1  GH1 
H2
G  H2 GH1
(c) (d)
1  GH1 1  GH 2

7 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 1.3

(a)
1  G1  G2 (d)
1
1  G2  H1  H 2  1  G2  H1  H 2 

(b)
1  G2  G1 Answer keys
TYC – 1.3
1  G2  H1  H 2 
G1G2 01 – b 02 – a 03 – c 04 – a 05 – b

Th
(c)
1  G2  H1  H 2 

is
06 – d 07 – c 08 – a 09 – b 10 – d

PD
11 – a

F
be
lo
ng
s
to
Ab
hi
na
v
Kr
is
hn
a
M
te
fF
R
0
(a
bh
in
av
m
kr
is
hn
a2
02
0@
gm
ai
l.c
om
,8
54
72
68
03
4)

8 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

 TYC

1.4 Signal Flow Graphs

C (c)
 s  27  (d)
 s  27 
01. for the signal flow graph shown in the
R s  29 s  48
2
s  29s  6
2

figure is given by C

Th
04. for the signal flow graph shown in the

is
R

PD
1

F
be
figure is given by

lo
1 G1 G2 1

ng
1
R

s
C 5

to
 H1

Ab
hi
H2

na
1 K 0.5 1/  s  1 1/ s 2 1

v
Kr
R C
G1G2

is
hn
(a)

a
1  G1G2 H 1  G2 H1  G1 s

M
te
fF
G1G2 1

R
(b)

0
(a
1  G1G2 H1  G1 H 2  G2

bh
0.5K

in
(a)

av
G1G2 s  3.5s 2  s  0.5K
3

m
(c)

kr
1  G1G2 H1  G2 H1  G1

is
K

hn
(b) 3

a2
G1G2 s  3.5s 2  s  K

02
(d)

0@
1  G1 H 1  G2 H1  G1 0.5K

gm
(c) 3

ai
s  3.5s 2  s  0.5K

l.c
C

om
02. for the signal flow graph shown in the

,8
R K

54
(d)

72
figure is given by 2s  3.5s 2  2s  K
3

68
03
05. Consider the following statements with re-

4)
5
spect to the following two signal flow
1 2 3 4 1
R C graphs.
1 a 1 b 1 c 1
1 1 1
24 18 e f
(a) (b) d
18 24 SFG-(i)
44 44 1 a b c 1
(c) (d)
18 23
C s d e f
03. for the signal flow graph shown in the SFG-(ii)
R s
1. SFG (i) and (ii) are equivalent
figure is given by
2. SFG (i) and (ii) are not equivalent
1 1/ s 6 1/ s
R s 3. Transfer function for SFG (ii) is
1 3 abc
2
4 1   ad  be  cf   adcf
C s
4. Transfer function for SFG (i) is
s  s  27  s  s  27 
(a) 2 (b) 2 abc
s  29 s  48 s  29s  6
1   ad  be  cf   adcf

9 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 1.4
Which of the above statements is/are cor- Y s
08. in Q07 is given by
rect? N  s  R s 0
 
(a) 1, 3 and 4 only
(b) 2 and 4 only s4 10  s  4 
(a) (b)
(c) 2, 3 and 4 only s  16s  20
2
s  16 s  20
2

s  s  1 10s  s  1

Th
(d) 2 and 3 only
(c) (d)

is
s  16 s  20 s 2  16 s  20

PD
2
C

F
06. for the signal flow graph shown in the

be
R Y5

lo
09. for the signal flow graph shown in the

ng
figure is given by Y2

s
to
Ab
figure is given by

hi
na
G1

v
G2 G3 G4 G4

Kr
R C

is
hn
1

a
1 1

M
1 1 G1 G2 G3 Y5 1

te
fF
Y1 Y5

R
Y2 Y3 Y4
G1G2 G3G4

0
(a
(a)

bh
2  G2G3G4  2G3G4  2G4  H1 H2

in
av
 H3

m
G1G2 G3G4

kr
(b)

is
hn
1  G2G3G4  G3G4  2G4

a2
02
G1G2 G3G4 G1G2  G4 1  G1G2G3

0@
(c) (a) (b)

gm
1  G2G3G4  G3G4  2G4 1  G3 H 2 G1G2  G4

ai
l.c
G1G2G3  G3G4

om
G1G2 G3G4 (c)
(d)

,8
1  G3 H 2

54
1  G2G3G4  G3G4  G4

72
68
07. In the closed – loop system shown below, (d) Cannot be determined

03
4)
10 10. Consider the closed – loop control system
consider G1  2; G2  s  2; G3 
s  s  1 shown in figure. If G p  s   Gc  s  then

Y s Y s
and H1  0.5s , then is given by is given by
R s N R s N
 s 0
 s  0

G1 N s
N s
  Y s R s   Y s
G2 G3 Gp  s 
R  s    
  

H1
Gc  s  H s

 
s4 10  s  4 
(a) 2 (b) 2 Gp  s 
s  16s  20 s  16 s  20 (a) G p  s  (b)
s  s  1 10s  s  1 H s
(c) (d)
s  16 s  20 s 2  16 s  20 (c) 1  G p  s  H  s  (d) G p  s  H  s 
2

10 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 1.4

Answer keys
TYC – 1.4

01 – a 02 – d 03 – b 04 – c 05 – d

Th
06 – a 07 – b 08 – c 09 – c 10 – a

is
PD
F
be
lo
ng
s
to
Ab
hi
na
v
Kr
is
hn
a
M
te
fF
R
0
(a
bh
in
av
m
kr
is
hn
a2
02
0@
gm
ai
l.c
om
,8
54
72
68
03
4)

11 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

 TYC

2.1 Transient response

01. Measurements conducted on a servomecha- 05. A certain feedback system is described by

nism show the system response to be the following transfer function:
c  t   1  0.2e60t  1.2e 10 t , when subjected 16
G s  2 ; H  s   Ks

Th
is
s  4s  16

PD
to a unit step input. The damping ratio and

F
The damping ratio of the system is 0.8. The

be
lo
undamped natural frequency (in rad/s) of

ng
percentage of peak overshoot in the step re-

s
the system, are respectively

to
sponse and K value, are respectively

Ab
hi
(a) 1.47 and 25.03 (b) 1.43 and 24.49

na
(a) 1.5 and 0.15 (b) 15 and 0.15

v
(c) 0.43 and 25.03 (d) 0.43 and 24.49

Kr
(c) 0.15 and 15 (d) 15 and 1.51

is
hn
02. In the system shown in figure, damping ra-

a
06. A feedback system employing output damp-

M
tio is 0.5 then the settling time (for 2% tol-

te
ing is shown in the figure. The second order

fF
R
erance) in unit step response is

0
closed – loop system has damping ratio of

(a
bh
C s

in
R s 
K 0.5 and frequency of damped oscillation is

av
s  s  10 

m


kr
9.5 rad/s. Then the values of K1 and K2 are

is
hn
respectively

a2
02
C s

0@
(a) 0.326 s (b) 1.326 s R s   1

gm
K1
(c) 0.8 s (d) 1.8 s s  s  1

ai
 

l.c
om
03. The open – loop transfer function of a unity

,8
K2 s

54
feedback system is given by

72
68
K

03
G  s  where K, T are constant

4)
s 1  Ts  (a) 120.34 and 9.97
having positive real values. The peak over- (b) 9.97 and 120.34
shoot of unit step response is required to re- (c) 99.4 and 9.97
duce from 75% to 25%, then the amplifier (d) 99.4 and 10.97
gain (K) 07. For a closed – loop system shown in the fig-
(a) should be reduced by a factor of 20 ure below, find the values of K and T such
(b) should be increased by a factor of 20 that the maximum overshoot in the unit step
(c) should be reduced by a factor of 36 response is 25% and peak time is 2 s.
(d) should be increased by a factor of 36 R  s  1 C s
K
04. A servo mechanism is represented by the  s 2

2
d y dy
equation 2
 4.8  144 E where
dt dt 1  Ts
E  C  0.5 y is the actuating signal. The (a) K  0.468 and T  2.924 s
damping ratio and damped natural fre- (b) K  2.924 and T  0.468s
quency (in rad/s) of oscillation in the re- (c) K  1.71and T  0.8s
sponse, are respectively
(d) K  0.8and T  1.71s
(a) 0.283 and 8.48 (b) 0.56 and 8.48
(c) 0.56 and 8.13 (d) 0.283 and 8.13

12 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 2.1

08. In the closed – loop system shown in figure C s

R s 
K
below, assume J = 1 kg – m2, the overshoot  s  s  5
in the unit step response is 25% and the peak
time is 2 s then the value of K1 and K2 are
Then value of K is given by
R  s   K1 1 C s

Th
(a) 0.591 (b) 1.79

is
  Js s

PD
(c) 4.23 (d) 17.9

F
be
12. vi  t  is a step voltage and L = 1 H in the cir-

lo
K2

ng
s
cuit shown in figure. The values of R and C

to
Ab
hi
to yield a 20% overshoot and 1 ms settling

na
v
(a) K1  12.95and K 2  0.471 time for vc  t  , are respectively

Kr
is
hn
(b) K1  1.72 and K 2  0.71

a
R L

M
te
(c) K1  2.924and K 2  0.468s 

fF
R
vi  t  

0
 vc  t 

(a
(d) K1  2.95and K 2  2.471

bh
in
av
09. Unit step response of the system shown be- 

m
kr
low is (a) 912 Ω and 1 µF

is
hn
a2
2s  1 C s (b) 8 kΩ and 13 µF

02
R s 

0@
 s 2
(c) 8 kΩ and 0.013 µF

gm
ai
(d) 912 Ω and 0.13 µF

l.c
om
13. vi  t  is a step voltage and L = 1 H and R1 =

,8
(a) c  t   1  e t  te t , t  0

54
72
1 MΩ in the circuit shown in figure. The

68
(b) c  t   2e t  te t , t  0

03
values of R2 and C to yield a 15% overshoot

4)
(c) c  t   2e t  te  t , t  0 and 1 ms settling time for vc  t  , are respec-
(d) c  t   1  e  te , t  0
t t
tively
L R2
10. Unit impulse response of the system shown

below is
C s vi  t  
 R1 vc  t 
 2s  1
R s
2
 s

(a) 912 Ω and 16 µF
(a) c  t   1  e t  te t , t  0 (b) 8 kΩ and 0.016 µF
(c) 8 kΩ and 0.013 µF
(b) c  t   2e t  te t , t  0
(d) 912 Ω and 0.13 µF
(c) c  t   2e t  te  t , t  0 14. For the control system shown in figure be-
(d) c  t   1  e  t  te t , t  0 low, peak time and settling time are respec-
tively
11. The step response has 10% overshoot in the
control system shown in the figure below.

13 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 2.1

R  s  C s
 20
10
   s  s  12 

10 0.2s

(a) 0.17 s and 0.5 s

Th
is
PD
(b) 0.5 s and 0.17 s

F
be
(c) 0.27 s and 0.375 s

lo
ng
(d) 0.07 s and 0.375 s

s
to
Ab
15. Transfer function of a closed – loop system

hi
na
is shown in figure below.

v
Kr
R  s

is
C s

hn
14.145

a
s 

M
2
 1.204 s  2.829  s  5 

te
fF
R
0
The approximate percentage of peak over-

(a
bh
in
shoot in the unit step response is

av
m
(a) 15% (b) 30%

kr
is
hn
(c) 10% (d) 20%

a2
02
0@
gm
ai
l.c
Answer keys

om
,8
TYC – 2.1

54
72
01 – b 02 – c 03 – a 04 – d 05 – a

68
03
06 – a 07 – b 08 – c 09 – d 10 – b

4)
11 – d 12 – c 13 – b 14 – a 15 – b

14 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

 TYC

2.2 Steady State Error

01. The forward transfer function of a unity 05. The open – loop transfer function of a unity
feedback type 1, second order system has a feedback control system is
K
pole at –2. The nature of gain K is so ad- G  s 

Th
. If the steady state
s  s  4  s  10 

is
PD
justed that damping ratio is 0.4. The above

F
be
system is subjected to input r  t   1  4t . error for a unit ramp input is 0.1 then K

lo
ng
value is

s
The steady state error is

to
Ab
(a) 400 (b) 40

hi
(a) 0 (b) ∞

na
(c) 4 (d) 0.4

v
(c) 1.28 (d) 0.32

Kr
is
06. A unity feedback control system has the fol-

hn
02. The open – loop transfer function of control

a
M
lowing forward transfer function
system is given by

te
fF
K1  2 s  1

R
  s G  s 

0
K
G s  o

(a
 s  4 s  1 s  1
2

bh
 e  s  s  s  4s  8
2

in
av
The input r  t   1  5t is applied to the sys-

m
kr
Consider step input having an error angle of

is
hn
a2
θe and assume K as the gain constant. Then tem. It is desired that the steady state value

02
0@
(a) output velocity is proportional to error of the error should be less than or equal to

gm
angle 0.1 for the given function. The minimum

ai
l.c
om
(b) output velocity is inversely proportional value of K1 required to satisfy the require-

,8
54
to error angle ment is

72
68
(c) output acceleration is proportional to er- (a) K1  50 (b) K1  5

03
4)
ror angle (c) K1  50
(d) output acceleration is inversely propor- (d) Such a value does not exist
tional to error angle
07. Consider a unity feedback control system
03. A feedback control system is described as with the closed – loop transfer function
50
G  s  ; H  s  1 C s Ks  b
s  s  2  s  5   . The steady state – error
R s s  as  b
2

Then, static error constants Kp, Kv and Ka

in the unit ramp response is
are respectively
ak ak
(a) 0, 0 and 5 (b) 0, 0 and 10 (a) (b)
b b
(c) ∞, ∞ and 10 (d) ∞, ∞ and 5
b b
04. A feedback control system is described as (c) (d)
ak ak
K
G  s  2 ; H  s  1 08. Consider the control system shown in fig-
s  s  20  s  30 
ure. It is required to eliminate steady state
It is required to limit the steady state error error due to unit ramp input. [ Note that
to 10 units with input r  t   1  10t  40t 2 , e  t   r t   c  t  ]
then the value of K is
(a) 120 (b) 1200
(c) 12 (d) 1.2
15 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs
Control Systems TYC 2.2
R s  n2 C s 12. In the system shown in figure, find the value
1  sK
 s  s  2n  of K so that for an input of 100t u  t  , there
will be an error of 0.01 in the steady – state.
R s  K C s
The value of K is
  s  s  1
 

Th
(a) (b)

is
n n2

PD
10s

F
K

be
4 2

lo
(c) (d)

ng
n n

s
to
(a) 11,000 (b) 1100

Ab
09. For the system shown inf figure, if the input

hi
(c) 110 (d) 110,000

na
is r  t   15t u  t  then the steady state error

v
Kr
13. The forward transfer function of a unity

is
hn
is feedback system is

a
M
K  s 2  3s  30 

te
fF
G s 

R
R s   C s . It is required to

0
1 2
s n  s  5

(a
bh
s s3

in
 

av
1

m
have error between an input of

kr
is
6000

hn
3

a2
10tu  t  and the output in the steady state.

02
0@
gm
(a) 2/9 (b) 67.5 The value of K and n are, respectively

ai
l.c
(c) 4.5 (d) ∞ (a) 10,000 and 1 (b) 1000 and 1

om
,8
10. For the system shown inf figure, if the input (c) 10,000 and 0 (d) 1000 and 0

54
72
is r  t   50 u  t  then the steady state error 14. A unity negative feedback system has the

68
03
forward transfer function

4)
is
K s  
R s  5 C s G  s  is to be designed to meet
s s   
 s  s  1 s  2 

the following requirements.
 Steady – state error for a unit ramp input
 s  3 = 0.1.
 the closed – loop poles should be at
(a) 0.75 (b) 0 1  j1. The value of  ,  and K are
(c) 37.5 (d) 0.33 (a)   1.11,   0.2, K  1.8
11. The system shown in the figure is (b)   0.2,   1.11, K  1.8
R s  100  s  2  C s
1000 (c)   1.11,   1.8, K  0.2
  s  s  5 s
(d)   1.8,   0.2, K  1.11
15. The system shown in the figure is required
10
to meet 10% overshoot and K v  100.
R s  K C s
(a) Type – 0 (b) Type – 1  ss  
(c) Type – 2 (d) Type – 3

16 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 2.2

The values of K and α are respectively R s  C s

K G s
(a) 1397 and 13970 (b) 1397 and 1397 
(c) 13970 and 139.7 (d) 13.97 and 1397
16. The steady – state error due to a unit step H

input and unit step disturbance in the system The sensitivity of overall transfer function

Th
is
shown in figure is with respect to changes in H is

PD
D  s s  s  1  s  1

F
be
(a) (b) 2

lo
ng
s  s  1000
2
s  s  1000

s
R  s   C s

to
1  100 1000 1000

Ab
(c) 2 (d) 2

hi
 s5 s2

na
s  s  1000 s  2s  1000

v
Kr
20. The sensitivity of steady – state error to

is
hn
a
changes in K for the system shown in figure

M
49 11

te
(a) (b)

fF
is

R
11 49

0
R s  C s

(a
K s  7

bh
11 49

in
(c) (d)

av
49 11  s 2  2s  10

m
kr
is
17. The system shown in the figure is

hn
a2
C s

02
R s  s 1 7 K 7K

0@
(a) (b)
 s  2

gm
2
 s 7 K  10 7 K  10

ai
l.c
K 7 K

om
(c) (d)

,8
7 K  10 7 K  10

54
K

72
21. Given the system shown in the figure. Find

68
03
the sensitivity of the steady – state error to

4)
(a) Type – 0 (b) Type – 1
(c) Type – 2 (d) Type – 3 parameter ‘a’. Assume a step input.
18. The system shown in given figure is R s  K C s
R s  K  s  1 C s  s  s  1 s  4 

 s 2
 s  2

s4 s  a
s3
(a) 1/  a  1 (b) a 1
(a) Type – 0 (b) Type – 1
(c) 1/  a  1 (d) 1/  a  1
(c) Type – 2 (d) Type – 3
19. A control system shown in the figure has Answer keys TYC – 2.2
01 – c 02 – a 03 – d 04 – b 05 – a
following parameters: K = 10 V/rad, H
06 – c 07 – a 08 – d 09 – b 10 – c
100
= 10 V/rad and G  s   11 – b 12 – d 13 – a 14 – a 15 – c
s  s  1 16 – d 17 – a 18 – a 19 – c 20 – a
21 – d

17 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

 TYC

3.1 Stability

01. A closed – loop system is shown in the fig- 05. The closed – loop transfer function of a con-
ure. The roots of closed – loop system are trol system is
located as s 3  2s 2  7 s  21

Th
T s  5

is
C s s  2s 4  3s 3  6s 2  2s  4

PD
R s  1000

F
be
 s  2  s  3 s  5 The closed – loop poles lie as follows:

lo


ng
(a) 1 in RHP and 4 in LHP

s
to
Ab
(b) 4 in RHP and 1 in LHP

hi
na
(a) 2 in RHP and 2 in LHP (c) 2 in RHP and 1 in LHP and 2 on jω axis

v
Kr
is
(b) 2 in RHP and 1 in LHP (d) 1 in RHP and 4 on jω axis

hn
a
(c) 1 in RHP and 2 in LHP

M
te
fF
(d) all 3 are in LHP 06. In the following control system,

R
0
(a
R s  K  s  20  C s

bh
in
s  s  2  s  3

av
02. Consider a polynomial: 

m
kr
P  s   3s  9s  6s  4s  7 s  8s  2s  6
7 6 5 4 3 2

is
hn
a2
02
Comment on the roots alignment.

0@
The range of K for the closed – loop system

gm
(a) 4 in RHP and 3 in LHP
to be stable is

ai
l.c
(b) 2 in RHP and 5 in LHP

om
(a) 0  K  2 (b) 0  K  2

,8
(c) 4 in RHP and 3 in LHP

54
(c) K  2 (d) K  2

72
(d) 5 in RHP and 2 in LHP

68
03
4)
07. In the following control system,
03. In the control system shown in figure, the
R s  K  s  4 C s
number of roots on imaginary axis is
s  s  1 s  2 
R s  200 C s 

 
s s3  6 s 2  11s  6 
The frequency of oscillations is

(a) 2 (b) 4 (a) 2 2 rad/s at K = 4

(c) 0 (d) 1 (b) 2 2 rad/s at K = 6
(c) 8 rad/s at K = 4
04. Control system shown in the figure is (d) 8 rad/s at K = 6
R  s  1 C s

  4 3
s 2 s  3s  2s  3s  2 2
 08. Open – loop transfer function of a unity neg-
ative feedback control system is
K  s  2
G  s  . The range of
(a) Stable s 2
 1  s  4  s  1
(b) Unstable with 2 roots in RHP
K to ensure closed – loop system to be stable
(c) Unstable with 4 roots in RHP
is
(d) Marginally stable
(a) K  12 (b) K  2
18 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs
Control Systems TYC 3.1

(c) K  33
(d) There is no value of K to satisfy stability

09. In the following control system, Answer keys TYC – 3.1

R s  C s 01 – b 02 – a 03 – c 04 – b 05 – d
 

Th
K s 2  2s  2 06 – a 07 – b 08 – d 09 – c 10 – c

is
PD

11 – d

F
be
lo
1

ng
s
2
s  2s  1

to
Ab
hi
the range of K for closed – loop system to

na
v
be stable is

Kr
is
hn
1 1

a
(a)   K  1 (b)  K 1

M
2 2

te
fF
R
1

0
(a
(c)   K  1 (d) K  1

bh
2

in
av
m
10. For the system shown in figure, the fre-

kr
is
hn
quency of sustained oscillations is

a2
02
R s  C  s

0@
 1
K

gm
  s  s  2 s  3

ai
l.c
om
,8
54
s

72
68
03
4)
(a) 7 rad/s at K = 5
(b) 7 rad/s at K = 35
(c) 7 rad/s at K = 35
(d) 7 rad/s at K = 5
11. Consider the following characteristic equa-
tion:
s 4  Ks 3  s 2  s  1  0
The range of K for stability is
(a) K  1 (b) K  1
(c) K  1
(d) There is no value of K

19 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

 TYC

3.2 Root Locus

(c) s1 only be located on root locus and cor-
01. Given that F s 
 s  1 then F(s) at responding K = 8.5
s  s  2
(d) s2 only be located on root locus and cor-

Th
s  3  j 4 is

is
responding K = 8.5

PD
(a) 0.217  114.3 (b) 0.217114.3

F
05. Consider a control system shown in figure

be
lo
(c) 0.17746.8 (d) 0.177  46.8

ng
(a) and its open loop pole locations is shown

s
 s  2  s  4 

to
in figure (b).
02. Given that F  s  

Ab
then

hi
s  s  3 s  6  R s  C s

na
v
G s

Kr
F(s) at s  7  j9 is

is


hn
a
(a) 0.06737.4 (b) 0.017  37.4

M
te
fF
(c) 0.096  110 (d) 0.096110

R
Fig  a 

0
(a
03. Consider a system given in the figure.

bh
Im

in
av
R s  C s

m
K

kr
is
s  s  10 

hn
 X

a2
X 0
Re

02
10

0@
Fig  b 

gm
ai
Which of the following statements is correct

l.c
Using root locus concept, the gain K for the

om
with respect to the given system?

,8
closed – loop system to have damping ratio

54
(a) The system is overdamped if K  25

72
1

68
of is

03
(b) The system is critically damped at K =

4)
2
25
1
(c) The system is underdamped if K  25 (a) (b) 100
100
(d) All of the above
1
04. Consider a control system shown in the fig- (c) 50 (d)
50
ure below:
06. Root locus diagram of a unity negative feed-
R s  K C s
back control system is shown below.
s  s  5 Im


Which of the following point(s) will be lo- X X X X Re

4 3 2 1 0
cated on the root locus diagram?
s1  3.5  j1.5 and s2  2.5  j1.5
(a) s1 only be located on root locus and cor-
responding K = 10.5
(b) s2 only be located on root locus and cor- The range of gain for closed loop stable op-
responding K = 10.5 eration is
(a) K  9.645

20 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 3.2

(b) 0  K  9.645 Im

(c) K  74.645
(d) 0  K  74.645
07. For the system shown in figure, given that
X X Re
K 3
G  s  2 1 0

Th
.
 s  2  s  4  s  6 

is
PD
F
R  s  C s

be
lo
G s

ng
s
 Fig  b 

to
Ab
hi
Consider the statements:

na
v
1. System is overdamped for small values

Kr
The value of K where root locus cross im-

is
hn
of gain ( 0  K  0.0718 )

a
aginary axis is

M
te
2. System is overdamped for large values

fF
(a) 10 (b) 48

R
of gain ( K  13.93 )

0
(a
(c) 96 (d) 480

bh
3. System is underdamped for medium

in
av
m
values of gain ( 0.0718  K  13.93 )

kr
08. For the system shown in figure, given that

is
hn
Which of the above statement(s) is/are cor-
K  s  2  s  4 

a2
G  s 

02
. rect?
s  6 s  25 

0@
2

gm
(a) 1 and 2 (b) 1 and 3

ai
C s

l.c
R  s  (c) 1, 2 and 3 (d) 2 and 3

om
G s

,8
11. The open – loop transfer function of a unity

54


72
feedback control system is given by

68
03
K  s  2

4)
The root locus has G  s  . The angle of departure
s 2  2s  2
(a) Break away point at s = 2.885
at the complex pole is
(b) Break away point at s = – 5.718
(a) 225° (b) – 135°
(c) Break in point at s = 2.885
(c) 45° (d) 135°
(d) Break in point at s = 3.56
12. A negative unity feedback system has the
09. The value of K at break-away/in point in the
forward transfer function
Q08 is
K  s  1
(a) 51.3 (b) 0.024 G s  . If K is set to 20, find the
s  s  2
(c) 31.04 (d) 32.8
10. A control system is shown in figure (a) and changes in closed – loop pole location for
its root locus is shown in figure (b). 5% change in K.
R s  C s 13. Root locus of a control system is shown in
s3
K  s  2 the figure. The closed – loop transfer func-
 s  s  1
tion of the system is

Fig  a 

21 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 3.2
Im The value of gain K where root locus
crosses jω axis is
(a) 56 (b) 1568
(c) 8 7 (d) 392
X X X
4 3 2 1 Re

Th
is
PD
Answer keys TYC – 3.2

F
be
lo
ng
01 – c 02 – a 03 – d 04 – d 05 – c

s
K  s  4

to
06 – b 07 – d 08 – c 09 – a 10 – c

Ab
(a)
s  s  1 s  2  s  3

hi
11 – d 12 - ** 13 – d 14 – b 15 – a

na
v
Kr
K  s  4

is
hn
(b) 12. s  0.9975 for closed – loop pole at
 s  1 s  2  s  3

a
M
s  21.05 and s  0.0025 for closed –

te
fF
K  s  4

R
loop pole at s  0.95

0
(c)

(a
 s  1 s  2  s  3  K  s  4 

bh
in
av
K  s  4

m
kr
(d)

is
 s  1 s  2  s  3  K  s  4 

hn
a2
02
0@
14. A negative unity feedback system has the

gm
K  s  1

ai
forward transfer function G  s  

l.c
.
s  s  1

om
,8
54
The root locus is

72
68
03
(a) a circle with radius of 2 units

4)
(b) a circle with radius of 2 units
(c) a straight line parallel to y – axis
(d) a square of each side is 2 units
15. The root locus of a control system is
shown in the figure below:
j

j 7

X X XX
5 3 1 0 

j 7

22 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

 TYC

4.1 Frequency response and Nyquist plots

10
01. Consider the unity – feedback system with (c) sin  4t  18.435 
the open – loop transfer function 2
10 (d) 15.81sin  4t  18.435 

Th
G s  . The steady – state output of

is
s 1

PD
04. The closed – loop bandwidth (in rad/s) re-

F
be
the system when it is subjected to input quired for 20% overshoot and 2 s settling

lo
ng
r  t   2cos  2t  45  is

s
time is

to
Ab
(a) Css  t   1.79 cos  2t  55.3  (a) 10.66 (b) 1.613

hi
na
v
(c) 5.79 (d) 2.29

Kr
(b) Css  t   8.94cos  2t  63.43

is
hn
05. The frequency response of the system

a
M
(c) Css  t   1.79sin  2t  53.3  e s d

te

G  s 

fF
at frequency   gives

R
(d) Css  t   8.94sin  2t  63.43 

0
s 2 d

(a
bh
in
magnitude and phase of

av
02. Consider the unity – feedback system with

m
2 d 2 d

kr
the open – loop transfer function

is
(a) and  90 (b) and  180

hn
 

a2
10
G s 

02
. The steady – state output of 2 d 

0@
s 1 (c) and 0 (d) and  180

gm
the system when it is subjected to input  2 d

ai
l.c
om
r  t   sin  t  30   2cos  2t  45 is 06. The sinusoidal transfer function of a feed-

,8
54
back control system is given by

72
(a)

68
n2

03
1.79cos  t  24.85   0.905sin  2t  53.3  G  j  

4)
where
(b)  n2   2  j 2n 
0.905sin  t  24.85   1.79cos  2t  55.3  0 
1
. If a sinusoidal excitation of
(c) 2

1.79cos  t  53.3  0.905sin  2t  24.85  cos  n 1  2 2  t is applied to the system,

 
(d)
the steady – state response magnitude is
1.79sin  t  53.3  0.905cos  2t  24.85 
1 1
(a) (b)
03. Consider the system with closed – loop
2 1   2 2
C s 2  0.5s  1
transfer function is  , if 1 1
R  s  0.25s  1 (c) (d)

 1  2 2
this system is subjected to an input of
07. Consider the following statements for an
10
r t   sin  4t  then the response is underdamped second – order system:
2
1. Peak overshoot in step – response re-
(a) 10sin  4t  18.435  duces as damping ratio is increased from 0.2
(b) 15.81sin  4t  18.435  to 0.6

23 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 4.1
2. Resonance peak in frequency response d j
reduces as damping ratio is increased from
0.2 to 0.6   0
5 2 
Which of these statement(s) is/are correct?
3
(a) 1 only (b) 2 only 1


Th
(c) 1 and 2 (d) neither 1 nor 2 2

is
PD
1 1 100
08. Polar plots of and 2 are respectively

F
10. Polar plot of G  s   is

be
s s s  10 s  100
2

lo
ng
(a) Negative real axis and negative imagi-

s
shown in the figure.

to
Ab
nary axis in s – plane j

hi
na
(b) Negative imaginary axis and negative

v
Kr
is
real axis in s – plane   0

hn
a


M
(c) Positive real axis and positive imagi-

te
fF
nary axis in s – plane

R
0
Y

(a
(d) Positive real axis and negative imagi-

bh
in
av
nary axis in s – plane At point Y, the magnitude of the phasor and

m
kr
frequency (in rad/s) are, respectively

is
10
09. The polar plot of G  s  

hn
is

a2
 s  1 s  2  (a) 1 and 100 (b) 0.5 and 10

02
0@
a  j (c) 1 and 10 (d) 0.5 and 100

gm
ai
11. Consider an open – loop system with the

l.c
om
  0 following transfer function,

,8
54
 s 1

72
G  s H  s  2
5 2

68
3
s  s  2

03
4)
 2 Comment on the stability of the system
 b j when the feedback path is closed
(a) feedback system is unstable with one
  0
pole in the right half of s – plane
10 2 
3
(b) feedback system is unstable with three
poles in the right half of s – plane
 2
(c) feedback system is stable
c j
(d) feedback system is unstable with two
  0
poles in the right half of s – plane

10 2 
3 12. Consider a system with open – loop transfer
K

1
function, G  s  H  s   .
2 s T1s  1T2 s  1
Using Nyquist stability criterion, the closed
loop system is
(a) stable for all values of K
(b) stable for large values of K and unstable
for small value of K
24 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs
Control Systems TYC 4.1

(c) stable for small value of K and unstable 16. With the value of K found in Q15, the phase
for large values of K margin in degree is
(d) unstable for all values of K (a) 189.7 (b) 9.7
13. Consider a system with open – loop transfer (c) 170.3 (d) 19.2
function, 17. The open loop transfer function of a feed-

Th
K  s  3

is
back system is
G s H s 

PD
;K 1
s  s  1 K 1  2 s 

F
be
G s H  s  .

lo
s 1  s  1  s  s 2 

ng
Using Nyquist stability criterion, which of

s
to
Ab
the following comment is true? The value of K for obtaining a gain margin

hi
na
(a) Both open loop and closed loop opera- of 3 dB is

v
Kr
is
tions are stable (a) 1.153 (b) 1.064

hn
a
(b) Both open loop and closed loop opera-

M
(c) 0.235 (d) 0.614

te
fF
tions are unstable 18. The open – loop transfer function of a feed-

R
0
(a
(c) open loop is stable and closed loop is back control system is

bh
in
unstable

av
s  0.25
G s H s  2

m
kr
(d) open loop is unstable and closed loop is s  s  1 s  0.5 

is
hn
a2
stable

02
The phase crossover frequency (in rad/s) is

0@
14. Consider a system with open – loop transfer
given by

gm
ai
function,

l.c
1 1

om
5 (a) (b)
G  s H s 

,8
8 2

54
s 1  s 

72
1 1

68
(c) (d)

03
Using Nyquist stability criterion, which of 2 2 2

4)
the following comment is true?
19. The open – loop transfer function of a unity
(a) Both open loop and closed loop opera-
negative feedback control system is
tions are stable
K
(b) Both open loop and closed loop opera- G s H s 
 s  2  s  4  s  6 
tions are unstable
If K = 1, the Nyquist plot is shown in the
(c) open loop is stable and closed loop is
figure. The Nyquist diagram crosses the real
unstable
Im
(d) open loop is unstable and closed loop is
stable
15. The open loop transfer function of a feed-
back system is
Re
K
G s H  s  .   44
s  s  1 s  2 
The value of K for obtaining a gain margin
of 3 dB is axis at   44 rad/s and at this frequency
(a) 4.25 (b) 1.064
1
(c) 0.235 (d) 6 G  j    . Now, if K is increased to
480
100, then the gain margin is
25 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs
Control Systems TYC 4.1
(a) 53.62 dB (b) 33.62 dB
(c) 13.62 dB (d) 27.3 dB
20. Frequency response of a system is given in
the following table.
ω (rad/s) Gain Phase (deg)
0.1 5 0

Th
is
1 1.5 – 90

PD
F
10 0.25 – 180

be
lo
100 0.1 – 235

ng
s
Gain margin is

to
Ab
(a) 0.25 (b) 4

hi
na
v
(c) 5 (d) 1.5

Kr
is
hn
a
M
te
fF
R
Answer keys TYC – 4.1

0
(a
bh
in
01 – a 02 – b 03 – d 04 – c 05 – b

av
m
kr
06 – a 07 – c 08 – b 09 – a 10 – c

is
hn
11 – d 12 – c 13 – d 14 – b 15 – a

a2
02
16 – b 17 – d 18 – c 19 – c 20 – b

0@
gm
ai
l.c
om
,8
54
72
68
03
4)

26 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

 TYC

4.2 Bode plots

1 100 1  s 
01. Consider G  j   . In the Bode (c)
 j 
n
10  s 100  s 
1  s 

Th
plot, the phase angle is

is
(d)

PD
(a) 90 n in the low frequency region 10  s 100  s 

F
be
lo
(b) 90 n in the high frequency region 04. The transfer function of the system whose

ng
s
(c) 90 n in the entire frequency region

to
Bode magnitude plot shown in the figure is

Ab
hi
(d) 90 n in the entire frequency region dB

na
v
Kr
02. Which of the following statement is/are true

is
0 dB/dec

hn
regarding the Bode – magnitude plot of

a
10 20 dB/dec

M
te
1
G s 

fF
by using asymptotic ap-

R
1  sT

0
20 dB/dec

(a
bh
proach?

in
av
m
1. The Maximum error occurs at the cor-

kr
1 100 

is
hn
ner frequency and is approximately equal to

a2
1

02
– 3 dB. (a)
 s  1 0.01s  1

0@
gm
2. The error at one octave above or below

ai
100

l.c
the corner frequency is approximately (b)

om
 s  1 s  100 

,8
– 1 dB.

54
72
3. The error at one decade above or below 3.16 s

68
(c)
 s  1 s  100 

03
the corner frequency is approximately

4)
– 0.04 dB. 3.16 s
(d)
Options:  s  1 0.01s  1
(a) 1 only (b) 1, 2 and 3 05. The transfer function of the system whose
(c) 1 and 2 (d) 2 and 3 Bode magnitude plot shown in the figure is
03. The transfer function of the system whose dB
Bode magnitude plot shown in the figure is 0 dB/dec
dB 20 dB/dec
0 dB/dec
0 dB/dec 20 dB/dec
0 0
20dB/dec
20 0.01 0.02 0.1 1 2 10 
20 dB/dec 1  50s 1  s 
0dB/dec (a)
1  10s 1  0.5s 
0.1 1 10 100  10 1  50s 1  s 
(b)
100 1  s  1  10s 1  0.5s 
(a)
1  0.1s 1  0.01s   0.02  s 1  s 
(c)
1  s   0.1  s  0.5  s 
(b)
1  0.1s 1  0.01s 
27 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs
Control Systems TYC 4.2

0.1 0.02  s 1  s  c db

(d)
 0.1  s  0.5  s  40 dB/dec

06. The transfer function of the system whose 60 dB/dec

Bode magnitude plot shown in the figure is
dB
log 

Th
10 1

is


PD
5

F
be
d

lo
db

ng
6 dB/oct

s
0.001 0.1 10 

to
Ab
1  0.1s 1  10s

hi
12 dB/oct

na
(a) (b)

v
1  10s 1  0.1s

Kr
is
hn
1 s 1  10s

a
(c) (d) log 

M
1  10s 1 s 1

te


fF
5

R
 1 

0
07. Consider G  s   5 1   . The magni-

(a
09. The transfer function of the system whose

bh
 2s 

in
av
Bode magnitude plot shown in the figure is

m
tude at break frequency, using asymptotic

kr
is
db

hn
approximation is

a2
02
(a) 20 dB (b) 23 dB 20 db/dec

0@
gm
(c) 14 dB (d) 17 dB

ai
50

l.c
 40 db/dec
08. Identify the correct Bode plot for

om
,8
K

54
G s  2 from the following op-

72
s  s  5 20 db/dec

68
0 db/dec

03
4)
tions.
a  db
2.5 10 50 
6 dB/oct 2
 s 
1  
12 dB/oct (a)  10 
 s  s 
s 1   1  
 2.5   50 
log 
 5
790  s  10 
2

(b)
b s  s  2.5  s  50 
db
2
40dB/dec  s 
987.5  1  
60dB/dec (c)  10 
 s  s 
s 1   1  
 2.5   50 
log 
987.5  s  10 
2
 5
(d)
s  s  2.5  s  50 
10. The approximate gain crossover frequency
for the Bode plot given in Q09 is
(a) 100 rad/s (b) 1000 rad/s

28 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 4.2

(c) 2000 rad/s (d) 5000 rad/s

Answer keys TYC – 4.2

01 – d 02 – b 03 – c 04 – d 05 – a

Th
is
06 – b 07 – c 08 – b 09 – d 10 – b

PD
F
be
lo
ng
s
to
Ab
hi
na
v
Kr
is
hn
a
M
te
fF
R
0
(a
bh
in
av
m
kr
is
hn
a2
02
0@
gm
ai
l.c
om
,8
54
72
68
03
4)

29 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

 TYC

5.1 Controllers and Compensators

01. The electrical network shown in the figure K

(b) ;0  a  1
is s  s  1 s  a 
2

Th
C1
(c) ;0  a 1

is
s  s  1 s  a 

PD
2

F
be
K s  a

lo
ng
(d) ;0  a  1
s 2  s  1

s
to
Ab
R1

hi
04. The electrical network shown in the figure

na
v
vi R2 vo

Kr
is

is
hn
R1

a
M
te
fF
R
(a) lag network (b) lead network

0
(a
R2

bh
(c) lead – lag (d) PID vi vo

in
av
m
02. The electrical network shown in the figure C2

kr
is
hn
is

a2
02
C1 (a) lag network (b) lead network

0@
gm
(c) lead – lag (d) PID

ai
l.c
3.5s  1.4
05. A transfer function Gc  s  

om
is an

,8
s2

54
72
example for

68
R1

03
R2 (a) lag network (b) lead network

4)
vi vo
(c) lead – lag (d) PID
C2
06. Consider a control system shown in the fig-
(a) lag network (b) lead network ure.
 C s
(c) lag - lead (d) PID R s Gc  s  G s
03. A control system with 

K
G s  2 ; H  s   1 is unstable for H s
s  s  1
U s  1 
all positive values of K. The system can be Where, Gc  s    4 1  
E s  2s 
stabilized with the following compensated
system. If e(t) is a unit step function, then u(t) versus
 C s t sketch is
R s Gc  s  G s
 a 
u t 

H s
4
Now, Gc  s  is
K s  a
(a) ;0  a 1 t
s 2  s  1 0

30 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 5.1

b required specifications are   0.6 and

u t 
damped natural frequency of 10 rad/s. Then
the values of Kp and Kd are respectively
R  s  1 C s
K p  Kd s
 s  s  1

Th
is
Controller

PD
t Process
0

F
be
c

lo
ng
u t 

s
(a) 156.25 and 14

to
Ab
(b) 14 and 156.25

hi
na
v
(c) 12.5 and 7

Kr
is
hn
4 (d) 7 and 12.5

a
M
09. Consider the following statements.

te
t

fF
0

R
1. The natural frequency (ωn) must be

0
(a
d

bh
large for good performance; however, band-
u t 

in
av
width (ωb) considerations impose a limit on

m
kr
is
increasing ωn

hn
a2
02
2. When design consideration imposes a

0@
2 limit on ωn, a PD controller with a filter is

gm
ai
l.c
t suitable.

om
0

,8
From these,

54
07. A control system is shown in the figure be-

72
(a) 1 and 2 are false

68
low.

03
(b) 1 is true and 2 is false

4)
D
(c) 2 is true and 1 is false
R0 E T 1 C (d) 1 and 2 are true

 Gc  s  
 s  Js  B  10. The lag – lead compensator has a transfer
 1  1 
s T  s T 
function Gc  K c  1
 2
,
Which of the following controller Gc  s  
s s 1 
 T1    T2 
can be used to eliminate the steady – state
error to the step disturbance torque? assume T1 = 2 s, Kc = 10, δ = β = 5 and
(a) Gc  s   K p T2  5T1 . Draw Bode plot and find the fre-
quency where the phase angle is 0°.
K
(b) Gc  s  
s
Answer keys TYC – 5.1
 1 
(c) Gc  s   K p 1   01 – b 02 – c 03 – d 04 – a 05 – b
 Ti s 
06 – c 07 – c 08 – a 09 – d
(d) either (a) or (d)
08. A type – 1 process is controlled by using PD
controller as shown in the figure below. The

31 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

 TYC

6.1 State space analysis

01. An electric circuit is shown in the figure.  1 1 1 

 
Consider state vector as iL vo  and vin(t)  RC1 RC1 RC1 
 1 1 

Th
as input variable. (a) A    0 

is
 L LC1

PD
L 

F
 1 1 

be

0

lo
 

ng
iL  t   RC2 RC2 
vi  t   vo  t 

s
 R

to
Ab
C
 1 1 1 

hi
  RC RC1 

na
C1

v
In the state space representation, A and B  1 

Kr
is
 1 

hn
matrices are (b) A    0 0 

a
L

M
 1   

te
1  1 1 

fF
0 L  0

R
 and B   L   RC2 RC2 

0
(a) A  

(a
  
1 1 

bh
 0

in
 C RC   1 1 

av
0

m
 RC RC1 

kr
is
 1   1 

hn
0 L   1 

a2
0 (c) A   0 0 

02
(b) A    and B   

0@
L
1 
1   L  

gm
 C 
RC   1 1 

ai
0

l.c
 RC2 RC2 

om
 1  

,8
0 1
C 

54
 1 1 1 
 and B   L 

72
(c) A    

68
1 1   
 RC1 C1 RC1 

03
 L  0

4)
RC   1 1 1 
(d) A   
 1   L C2 RC1 
0 C  0  1 1 
(d) A    and B     0 
1 1  L  RC2 RC2 
 L RC 
 0 1
02. An electric circuit is shown in the figure. 03. Given that x  t     x  t  . The eigen
 2 3
Consider state vector as  vc1 iL vc 2  ,
values of state matrix are
vin(t) as input variable and vc2(t) as output (a) – 1 and 2 (b) 1 and – 2
variable. (c) 2 and 3 (d) 1 and 2
C1 R 04. Consider a system defined by the state –
   space equations as follows:
vc1
vin   x1   5 1  x1   2 
 L iL C2 vc 2  x    3 1  x    5  u and
 2   2  

x 
In the state space representation, the state y  1 2  1 
 x2 
matrix A is

32 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 6.1

Y s 4 s 3  10 s 2  45s  105

The transfer function is (d)
U s  s 2  9  s 2  5
12s  59 1 08. Given
(a) (b) 2
s  6s  8
2
s  6s  8  x1  1 0   x1  1
12s  59 1  x   1 1   x   1 u  t  and

Th
(c) 2 (d) 2  2   2  

is
s  6s  5 s  6s  5

PD
1 

F
05. A system is represented by the state space x  0     then state transition matrix is

be
lo
0

ng
equations as follows.

s
to
 x1   2 1  x1   1   et 0  e t 0

Ab
(a)  t  (b)   t 
 x    3 5  x    2  u and

hi
te et  te e t 

na
 2   2  

v
Kr
 et 0  et 0 

is
x 

hn
y   3 2   1  . The poles of the system are (c)  t
(d)  t 

a
0 e  0 e 

M
 x2 

te
fF
09. The homogenous equation assuming the

R
(a) 5.8 and 1.21 (b) 5.8 and –1.21

0
(a
(c) –5.8 and –1.21 (d) –5.8 and 1.21 given initial state vector in Q08 is

bh
in
(a) x1  t   tet and x2  t   tet

av
06. A system is represented by the state space

m
kr
equations as follows.

is
(b) x1  t   et and x2  t   tet

hn
a2
0 2 3 0 

02
(c) x1  t   et and x2  t   et
x   0 6 5  x  1  u and
 

0@
gm
1 4 2  1  (d) x1  t   tet and x2  t   tet

ai
l.c
om
y  1 2 0 x , The poles of the system are  2 1 0 

,8
54
X   0 2 1  X then state

72
(a) 9.11, 0.534 and – 1.645 10. Given

68
03
(b) 9.11, – 0.534 and – 1.645  0 0 2 

4)
(c) – 9.11, 0.534 and – 1.645 transition matrix is
(d) – 9.11, 0.534 and 1.645  e 2 t te 2t t 2 e 2 t 
07. A system is represented by the state space  
(a)  0 0 te 2t 
equations as follows.  0 0 e2t 

 x1   1 2   x1  1
 x    3 1  x   1 sin 3t and  e 2 t 0 t 2 e 2t 
 2   2    
t 2 2 t 
x  2 (b)  0 e 2 t e
y  1 2   1  ; x  0     then Y  s  is  2 
 2 t 
 x2  1   0 0 e 
4 s 3  10 s 2  45s  105
(a)  2 t t 2 2 t 
 s 2  9  s 2  5 e e 2t
2
e 
 
4 s 3  10 s 2  45s  105 (c) e 2t 0 te 2t 
(b)  0
 s 2  9  s2  5 e 2t e 2 t 
 
 
4 s 3  10 s 2  45s  105
(c)
 s 2  9  s 2  5

33 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs

Control Systems TYC 6.1

 2t t 2 2 t  x 
e te2t
e  y  1 1  1  . If the input is unit step
 2   x2 
e 2t
(d)  0 te2t 
function, then the steady state error is____
 0
0 e 2t 
 
 

Th
11. Consider the following two systems: Answer keys TYC – 6.1

is
PD
System I:

F
be
 1 1 2  2 01 – a 02 – b 03 – d 04 – a 05 – c

lo
ng
06 – a 07 – c 08 – a 09 – b 10 – d
X   0 1 5  X  1  U
  

s
to
11 – c 12 - * 13 - *

Ab
 0 3 4 1 

hi
na
v
System II: 2

Kr
12 – No 13.

is
hn
 2 1 3  2 3

a
M
X   0 2 1  X  1  U and
  

te
fF
R
 7 8 9  2

0
(a
bh
Y   4 6 8 X

in
av
m
kr
Which of the following statements are true?

is
hn
a2
(a) System I is controllable and System II is

02
0@
not observable

gm
(b) System I is not controllable and System

ai
l.c
om
II is not observable

,8
(c) System I is controllable and System II is

54
72
observable

68
03
4)
(d) System I is not controllable and System
II is observable
12. Determine whether the system shown in fig-
ure is observable?
4

1 1
1 s 5
s
y
u x2 x1

21/ 4

5

13. A state space model is given below.

 x1   0 1   x1  0
 x    3 6   x   1  u and
 2   2  

34 Prepared by: BNSS Shankar, M.Tech(IITK) OHM Institute – Hyderabad GATE|ESE|PSUs