Practice Solutions :

SOL.1 (A)
d2y d 2x
Given :  x ( t  2) 
dt 2 dt 2
where x is the input and y is the output.
Using Laplace transform with initial conditions zero.
s 2Y ( s )  e  2 s X ( s )  s 2 X ( s )
s 2Y ( s )  (e 2 s  s 2 ) X ( s )
Y ( s ) e2 s  s 2 e 2 s
  1 
X ( s) s2 s2
Hence, the correct option is (A).
 Key Point
Whenever we find transfer function, initial conditions are assumed to be zero.
SOL.2 (B)
s4
Given : CLTF 
s  7 s  13
2

For unity negative feedback system the closed loop transfer function is given by,
G( s)
CLTF 
1  G( s)
where, G(s) = open loop transfer function
G( s) s4
 2
1  G( s) s  7s  13
1  G ( s ) s 2  7 s  13

G( s) s4
1 s 2  7 s  13 s 2  6s  9
 1 
G( s) s4 s4
s4
G( s) 
s  6s  9
2
 Control Systems 2 Basics of Control System
For DC gain, s = 0
4
G ( s ) s 0 
9
Hence, the correct option is (B).
SOL.3 (A)
1
Given : System 1 : G ( s )  ,
(2 s  1)
1
System 2 : G ( s ) 
(5s  1)
For first order system with time constant  ,
1
G (s)  , H (s)  1
s
Block diagram of first order system is shown below,
1
R( s) Y (s)
st

Fig. First order system

Closed loop transfer function of negative unity feedback system is given by,
Y (s) G (s)

R(s) 1  G (s)
Y ( s) 1/ s

R( s) 1  1
s
Standard representation of first order system is given by,
Y ( s) 1
G (s)  
R ( s ) s  1
For first order system, bandwidth (B.W.) is reciprocal of time constant.
For system 1 :
1
G ( s)  ,   2 sec
(2 s  1)
1 1
B.W.    0.5
 2
For system 2 :
1
G ( s)  ,   5sec
(5s  1)
1 1
B.W.    0.2
 5
Therefore, bandwidth of system 1 is greater than bandwidth of system 2.
Hence, the correct option is (A).
Control Systems 3 Basics of Control System
SOL.4 (C)
Given:
(i) Closed loop transfer function,
C (s) K (s  a)

R( s) (s  k )
(ii) Unit step input r (t )  u(t )
(iii) c (0  )  2
(iv) c()  10
 K ( s  a) 
C ( s )  R( s )  
 (s  k ) 
For step Input r (t )  u(t )
Taking Laplace transform of r (t )
1
R( s) 
s
k (s  a)
C ( s) 
s(s  k )
1. Initial value theorem:
c(0 )  lim sC ( s)
s 

sk ( s  a )
c (0  )  lim
s  s ( S  k )

 a
k 1  
2  lim 
s
s   k
1  
 s
2. Final value theorem:
c()  lim sC ( s )
s 0

sK ( s  a )
10  lim
s 0 s(s  k )
ka
10 
k
a  10
Hence, the correct option is (C)
SOL.5 (C)
Given:
k
(i) Open loop transfer function 
sT  1
Control Systems 4 Basics of Control System
(ii) Bandwidth = B
Location of pole of open loop transfer function is shown below,
jw

s
0
1
-
T

 Key Point
For first order system, bandwidth (B.W.) is reciprocal of time constant.
1
Hence, Bclosed loop B …(i)
T
Closed-loop transfer function for negative unity feedback is given by,
G (s)
T (s) 
1  G (s)
K
T ( s )  sT  1
K
1
sT  1
K
T (s) 
sT  1  K
Location of pole of closed loop transfer function is shown below,
jw

s
0
æ1+ k ö
-ç ÷
è T ø

From above figure,

1 k
Bclosed loop  …(ii)
T
From equation (i) and (ii),
Bclosed loop  (1  K ) B
Hence, the correct option is (C).
SOL.6 (C)
Given :
(i) Open-loop transfer function
k 2n
G( s) 
s 2  2 n s
 Control Systems 5 Basics of Control System

 Key Point
Sensitivity of closed loop system w.r.t. forward path parameter K is given by,
T / T 1
S KT  
K / K 1  GH
1
S KT 
KH n2
1 2
s  2  n s
Hence, the correct option is (C).
SOL.7 (C)
 Key Point
The sensitivity of closed loop system w.r.t feedback path parameter H is given by,
T / T GH
S HT  
H / H 1  GH
 K 2n H
s 2  2 n s
S HT 
KH 2n
1 2
s  2 n s
 K n2 H
S HT 
s 2  2  n s  KH n2
Hence, the correct option is (B).
SOL.8 (0)
Given : r (t )   3e 2 t u (t ),
s 2
H ( s) 
s3
and initial value of the output is  2.  
  . Method 1 :
Steady state
r (t ) = - 3e 2t u (t ) æ s-2ö output = y(t)
ç ÷
è s+3ø

Taking Laplace transform of r(t),

3
R( s) 
s 2
Transfer function of the system is,
Y ( s)
H ( s) 
R( s )
Y ( s )  R( s ) H ( s )
 3   s  2 
Y ( s)    
 s 2  s 3
Control Systems 6 Basics of Control System
3
Y ( s) 
s3
Using final value theorem steady state output is given by,
yss  lim y (t )  lim sY ( s )
t  s 0

 3 
y ss  lim s  
s0
 s  3
  3 0 
y ss  lim  0
s0
 03 
Hence, the value of the output at steady state is 0.
. Method 2 :
3
Y ( s) 
s3
Taking inverse Laplace transform,
y (t )   3e 3t
Steady state output is given by,
yss  lim y (t )
t 

yss  lim  3e 3t 

t 

y ss  0
Hence, the value of the output at steady state is 0.
SOL.9 (0.214)
Given : g (t )  e  2t (sin 5t  cos5t )
x(t )  (t )
Taking Laplace transform,
X (s)  1
g (t )  e  2t sin 5t  e 2t cos5t
Note : Laplace transform,
b
sin bt 
L.T.

s  b2
2

s
cos bt 
L.T.

s  b2
2

e  at x(t ) 
L.T.
 X ( s  a)
Taking Laplace transform of g (t ) ,
5 s2
G (s)  
( s  2)  5 ( s  2) 2  52
2 2
Control Systems 7 Basics of Control System

DC gain means G(s) s0  G(0)

5 2 7
G (0)   2 2  0.241
2 5 2 5
22
29
Hence, the DC gain of the control system is 0.241.
SOL.10 (1)
( s  2)
Given : G( s)  and
( s  1) ( s  3)
r (t )  u(t )  
r (t ) G (s) y (t )

. Method 1 :
1
R( s ) 
s
Laplace transform of the output is given by,
Y ( s )  G ( s ) R( s )
1 ( s  2)
Y ( s)  G( s)  
s s ( s  1) ( s  3)
y (0)  Initial value
Applying initial value theorem,
y (0)  lim sY ( s )
s 

( s  2)
y (0)  lim
s  ( s  1) ( s  3)

 2
1  
y (0)  lim  s
0
s   1  3
s 1   1  
 s  s 
 dy 
L    sY ( s )  y (0)
 dt 
 dy  s( s  2)
L    sY ( s ) 
 dt  s( s  1) ( s  3)
 dy  ( s  2)
L  
 dt  ( s  1) ( s  3)
dy  dy  s ( s  2)
 lim sL    lim
dt t 0 s  dt  s ( s  1) ( s  3)
 2
s 2 1  
dy
 lim  s
dt t 0 s  2  1  3
s 1   1  
 s  s 
Control Systems 8 Basics of Control System

 2
1  
dy
 lim  s
1
dt t 0 s   1  3
1   1  
 s  s 
dy
Hence, the value of at t  0 is 1.
dt
. Method 2 :

R( s) G (s) Y (s)

Y ( s)  R( s)G( s)
( s  2)
Y (s) 
s ( s  1) ( s  3)
2 / 3 3 / 2 5 / 6
Y ( s)   
s s 1 s  3
Taking inverse Laplace of Y ( s) ,
 2 3  t 5 3 t
y (t )   e  e
3 2 6
d 3 5
y ( t )  0  e  t  e  3t
dt 2 2
d 3 5
y (t )     1
dt t 0 2 2
dy
Hence, the value of at t  0 is 1.
dt
SOL.11 (2.5)
Given:
(i) Input = 5 V
(ii) Forward gain = 1
(iii) Feedback gain = 1
Block diagram representation is shown below,
5 + 1 V0
–

V0 1

5 11
1
V0  5 
2
V0  2.5 V
Control Systems 9 Basics of Control System
SOL.12 (C)
Given : K  1,   1 sec, Td  0.1 sec
Transfer function of first order system is given by,
K
Gc ( s ) 
1  s
where, K is static gain and  is the time constant of system.
Transfer function of transportation lag is given by,
Gd ( s )  e  sTd
where, Td  Transportation lag or delay.
The overall transfer function with delay is given by,
G ( s )  Gc ( s )  G d ( s )

Ke  sTd 1 e s (0.1) e 0.1s

G ( s)   
(1  s) 1  ( s 1) (1  s )
Hence, the correct option is (C).
SOL.13 (A)
Given : Closed loop transfer function is,
C (s) 2( s  1)

R( s ) ( s  2) ( s  1)
For step input, r (t )  u(t )
Taking Laplace transform of r (t ) ,
1
R( s) 
s
2( s  1)
C (s) 
s ( s  2)( s  1)
Applying partial fraction,
2( s  1) A B C
  
s ( s  2)( s  1) s ( s  2) ( s  1)
A  s C ( s ) s 0  1

B  ( s  2) C ( s ) s2  3

C  ( s  1) C ( s) s1  4
 1 3 4
Then, C (s)   
s s  2 s 1
Taking inverse Laplace transform of C ( s) ,
c (t )   1  3e 2 t  4e  t  u (t )

Hence, the correct option is (A).

Control Systems 10 Basics of Control System
SOL.14 (D)
0.8
Given : G ( s ) 
s s2
2

The characteristics equation is given by,

s2  s  2  0
( s  2) ( s  1)  0
s   2, 1
jw

s
-2 1

Due to right hand pole system is unstable (unbounded).

Hence, the correct option is (D).
 Key Point
Final value theorem is applicable only for absolute stable system (all poles should lie in left half side of
s-plane), hence always check the stability of the system before applying final value theorem.
SOL.15 (C)
Given : x(t )  u (t ), y (t )  te  t u (t )
Taking Laplace transform,
1
X ( s)  L u (t ) 
s
1
et u (t ) 
L.T.

s 1
d  1  1
te  t u (t ) 
L.T.
  
ds  s  1  ( s  1) 2
Property of Laplace transform
dn
t n x (t ) 
L.T.
( 1) n  X ( s)
ds n
1
Y ( s )  L te  t u (t )  
( s  1) 2
Transfer function of the system is,
Y (s) s
T (s)  
X ( s ) ( s  1) 2
Hence, the correct option is (C).
SOL.16 (B)
d 2 x 1 dx 1
Given : 2
  x  10  5e  4t  2e  5t
dt 2 dt 18
Control Systems 11 Basics of Control System

The R.H.S. equation 10  5e 4t  2e 5t represents the input.

d 2 x 1 dx 1
  x  r (t )
dt 2 2 dt 18
Applying Laplace transform to the left hand side of the equation,
 2 sX ( s ) X ( s ) 
 s X (s)     R( s)
 2 18 
 2 s 1
 s    X (s)  R(s)
 2 18 
Transfer function is given by,
X (s) 1
T (s)  
R(s) s 2  1 s  1
2 18
Characteristic equation is,
s 1
s2    0
2 18
 1  1
 s   s    0
 3  6
1 1
s ,
3 6
18
T ( s) 
(6 s  1) (3s  1)
jw

s
-1 / 3 -1 / 6

Reciprocal of magnitude of negative real roots represents time constants.

1 1
1   3 sec , 2   6 sec
1/ 3 1/ 6
Hence, the correct option is (B).
 Avoid This Mistake
1
5e 4t    sec
4
1
2e 5t    sec
5

 Key Point
(i) The forced response implies the response only due to input when initial conditions are zero.
(ii) The natural response implies the response only due to initial conditions when no input is applied.
Control Systems 12 Basics of Control System
SOL.17 (A)
d2y dy
Given : 2  3  2 y  x(t )
dt dt
Taking Laplace transform,
s 2Y ( s)  3sY ( s)  2Y ( s)  X ( s)
X ( s)
Y ( s)  …(i)
s  3s  2
2

Also, x(t )  2u(t )

From equation (i),
2
Y ( s) 
s( s  3s  2)
2

2
Y ( s) 
s( s  1)( s  2)
By partial fraction method,
1 2 1
Y ( s)   
s s 1 s  2
Taking inverse Laplace transform,
y (t )  u(t )  2et u(t )  e2t u(t )
y(t )  [1  2et  e2t ]u(t )
Hence, the correct option is (A).
SOL.18 (C)
The given figure is shown below,
Cascade connection

R( s) K G (s) C (s)

H (s)

Fig. (a)

R K G (s) C

H (s)

Fig. (b)
Transfer function is given by,
C (s) KG ( s )
T (s)  
R ( s ) 1  KG ( s ) H ( s )
Sensitivity of  with respect to  is given as
% change in   /   
S    
% change in   /   
 Control Systems 13 Basics of Control System
Therefore, sensitivity of T (s) with respect to G ( s) is given as,
G ( s ) T ( s )
      SGT (( ss ))   …(i)
T ( s ) G ( s )
G ( s ) 1  KG ( s ) H ( s )
      
T (s) K
T ( s) 1  KG ( s) H ( s)  K  KG ( s) KH ( s )
      
G ( s) 1  KG(s) H (s)
2

T ( s) K  K 2G ( s) H ( s)  K 2G ( s) H ( s )
      
G ( s) 1  KG(s) H (s)
2

T ( s ) K
      
G ( s ) 1  KG ( s ) H ( s ) 2
1  KG ( s ) H ( s ) K
      SGT (( ss ))  
1  KG ( s) H ( s)
2
K
1
      SGT (( ss ))  …(ii)
1  KG ( s ) H ( s )
H ( s ) T ( s )
Similarly, S HT ((ss))  
T ( s ) H ( s )
H ( s ) H ( s) 1  KG ( s) H ( s)
      
T ( s) KG ( s)
T ( s) 1  KG ( s) H ( s)  0  KG ( s)  KG ( s)
      
H ( s) 1  KG(s) H (s)
2

K 2 G ( s ) 
2
T ( s )
      
H ( s ) 1  KG ( s ) H ( s )2
Therefore,
H ( s )  1  KG ( s ) H ( s )  K 2 G ( s )
2
KG ( s ) H ( s )
S T (s)
  S HT ((ss))  …(iii)
1  KG ( s) H ( s)
H (s) 2
KG ( s ) 1  KG ( s ) H ( s )
From (ii) and (iii),
SGT ((ss))  SHT ((ss))
Hence, the transfer function is more sensitive to perturbations in H ( s) .
Also, for very high value of K,
SGT ((ss))  0 and SHT ((ss))  1
Hence, the correct option is (C).
 Key Point
(i) A good control system is less sensitive to forward path transfer function than feedback path transfer
function.
(ii) Lesser the sensitivity, better will be the performance of control system
Control Systems 14 Basics of Control System
SOL.19 (A)
Given :
1
(i) G ( s ) 
s  3s  2
2

(ii) Unit impulse applied at time t  1 will be r (t )  (t  1).

Taking Laplace transform of r (t ) ,
R( s)  e s
Output of the system is given by,
e s
C (s)  G (s) R(s) 
s 2  3s  2
The steady state value of the output is given by,
Css  lim c (t )  lim sC ( s )
t  s 0

se  s
Css  0
s 2  3s  2
Hence, the correct option is (A).
SOL.20 (C)
Given :
1
Input G1 G2 Output
R G3 C

Overall transfer function,

C ( s) 1 G1G2
G (s)   G1G2  
R( s) G3 G3
G1G2
Output C  R
G3
ln C  ln G1  ln G2  ln G3  ln R
Differentiating above equation,
1 1 1 1 1
dC  dG1  dG2  dG3  dR
C G1 G2 G3 R
As there is no error in R, so dR = 0.
dC dG1 dG2 dG3
  
C G1 G3 G3
  1   2  3
So, relative error in output  1   2  3
Hence, the correct option is (C).
SOL.21 (A)
d 2 x (t ) dx (t )
Given : M D  Kx (t )  f (t )
dt dt
Control Systems 15 Basics of Control System
Taking Laplace transform,
Ms 2 X ( s )  DsX ( s )  KX ( s )  F ( s )
( Ms 2  Ds  K ) X ( s )  F ( s )
X (s) 1

F ( s ) Ms  Ds  K
2

On putting the values of M , D and K ,

X (s) 1

F ( s ) 0.1s  2 s  10
2

X (s) 10
 2
F ( s ) s  20 s  100
Hence, the correct option is (A).
SOL.22 (C)
Given:
(i) Unit impulse response c (t )  4e  t  6e 2 t
 Key Point
Laplace transform of unit impulse response is known as transfer function.
Hence, H ( s )  L[ 4e  t  6e 2 t ]
4 6
H (s)  
S 1 S  2
C ( s ) 4 s  8  6 s  6

R(s) ( s  1)( s  2)
C (s) 2s  2

R ( s ) ( s  1)( s  2)
For unit step input r (t )  u(t )
Taking Laplace transform of r (t ) ,
1 2s  2
c(s)  
s ( s  1)( s  2)
Applying partial fraction,
2s  2 A B C
  
s( s  1)( s  2) s s  1 s  2
A  sC ( s ) s  0  1

B  ( s  1)C ( s ) s  1  4

C  ( s  2)C ( s ) s   2   3

1 4 3
C (s)   
s s 1 s  2
Control Systems 16 Basics of Control System
Taking inverse Laplace transform,
c (t )  4e  t  3e 2 t  1
Hence, the correct option is (C).
SOL.23 (D)
A closed loop system is shown in below figure,
R( s) + G (s) Output C(s)
–

Feedback H(s)

From above figure,

Closed loop control system is a system in which control action depends on the output due to feedback.
Hence, the correct option is (D)
SOL.24 (C)
Given:
(i) Unit-Impulse response h(t )  16e 2t  8e  t
 Key Point
Laplace transform of unit impulse response is known as transfer function.

Taking Laplace transform of h(t ),

16 8
H (s)  
s  2 s 1
C ( s) 16s  16  8s  16

R( s ) ( s  2)( s  1)
C ( s) 8s

R( s) ( s  2)( s  1)
For unit step input r (t )  u (t )
Taking Laplace transform of r (t ),
1
R( s) 
s
8s
C (s) 
s(s  1)( s  2)
8
C ( s) 
( s  1)( s  2)
Applying partial fraction,
8s A B
 
s( s  1)( s  2) ( s  1) ( s  2)
Control Systems 17 Basics of Control System

A  ( s  1) C ( s) s1  8
B  ( s  2) C ( s ) s 2  8
8 8
Then, C (s)  
( s  1) ( s  2)
Taking inverse Laplace transform of C ( s ) ,
c(t )  8[e t  e 2t ]
Hence, the correct option is (C)
SOL.25 (A)
Given:
(i) Impulse response h(t )  5u (t )
(ii) Input r (t )  et
 Key Point
Laplace transform of unit impulse response is known as transfer function.
Taking Laplace transform of h(t ) and r (t ) ,
C (s) 5
H (s)  
R( s ) s
1
R( s) 
s 1
5
C ( s) 
s( s  1)
Applying partial fraction,
5 A B
 
s(s  1) s ( s  1)
A  sC ( s ) s  0  5

B  ( s  1) C ( s) s1  5
5 5
Then, C ( s) 
s s 1
Taking inverse Laplace transform,
c(t )  5[1  e  t ]u (t )
Hence, the correct option is (A).
SOL.26 (C)
Given Pole zero plot is shown below
Im

Re
-2 -1

-j
Control Systems 18 Basics of Control System
From above figure, transfer function G ( s ) is given by,
k ( s  1)
G (s) 
[ s  (2  j )][ s  ( 2  j )]
k ( s  1)
G (s) 
[( s  2)  j ][ ( s  2)  j ]
k ( s  1) k ( s  1)
G (s)   2
s  4  4s  1 s  4s  5
2

Steady state gain is given by,

lim g (t )  lim G ( s ) (Since, there is no pole at origin)
t  s 0

k
2  k  10
5
10( s  1)
Hence, transfer function G ( s) 
s 2  4s  5
Hence, the correct option is (C).
SOL.27 (B)
Given figure is shown below,
R R

ei C C e0

Transform domain :
1
C
Cs
R Va R

ei ( s ) 1 1 e0 ( s )
Cs Cs

Applying KCL at node Va

Va  ei ( s ) Va Va  e0 ( s )
  0 …(i)
R 1 R
Cs
Applying voltage division rule in the outer loop,
1
e0 ( s )  Va  Cs
1
R
Cs
Va  e0 ( s )[1  RCs ] …(ii)
Control Systems 19 Basics of Control System
Using equation (i) and (ii)
e0 ( s )[1  RCs ]  ei ( s )  RCs[e0 ( s )(1  RCs )]  e0 ( s )[1  RCs ]  e0 ( s )  0
e0 ( s )[1  RCs  RCs  R 2C 2 s 2  1  RCs  1]  ei ( s )

ei ( s )  e0 ( s )[1  3RCs  R 2C 2 s 2 ]
e0 ( s ) 1
Transfer function   2 2 2
ei ( s ) R C s  3RCs  1
For R-C circuit,
Time constant, T =RC
e0 ( s ) 1
 2 2
ei ( s ) s T  3sT  1
Hence, the correct option is (B)


Practice Solutions :
SOL.1 (D)
The given block diagram is shown below,
X 2 ( s)

s 1
X 1 ( s) Y (s)
s +1 s

. Method 1 : SFG :
The signal flow graph representation of block diagram is shown below,
X 2 ( s)
s 1
1 s +1 s 1
X 1 ( s) Y (s)

-1

When X 1 ( s )  0 , the signal flow graph is reduced as shown below,

1
1 s 1
X 2 ( s) Y (s)
-s
s +1

1
Forward path gain : P1 
s
Individual loop gain :
s 1 1
L1   
s 1 s s 1
Number of two non-touching loops : 0
Determinant :
  1  L1
 1  s  2
  1  
 1  s  s  1
Control Systems 2 Block Diagram & Signal Flow Graph
Path factor :
1
1 s 1
X 2 (s) Y (s)
-s
s +1

The loop L1 touches the forward path P1 , hence

1  1  (isolated loop gain)
1  1  0  1
Using Mason’s gain formula, transfer function is given by,
C ( s) 1
  Pk  k
R( s )  k 1
So, the transfer function is,
1
1
Y ( s) P11 s s 1
  
X 2 ( s)  s  2 s ( s  2)
s 1
Hence, the correct option is (D).
. Method 2 : Block Diagram :
When X 1 ( s )  0, the block diagram is reduced as shown below,
1
X 2 (s) + Y (s)
- s

s
( s + 1)
The transfer function of negative feedback system is given by,
G( s)
T ( s) 
1  G( s) H ( s)
where, G ( s )  Open loop transfer function, H ( s )  Feedback factor
1
Y ( s) ( s  1)
s
 
X 2 ( s) 1 s  s( s  2)
1   
 s s 1
Hence, the correct option is (D).
SOL.2 (B)
Given signal flow graph is shown below.
X 1 (s) X 2 (s)

1 G1 G2
Y (s)

-1 -1
Control Systems 3 Block Diagram & Signal Flow Graph

When X 1 ( s )  0 , signal flow graph can be redrawn as,

X 2 (s)

G2
- G1 Y (s)

- G1

Forward Path gain : P1  G2

Individual Loop gain :
L1   G1G2 , L2   G1
Number of two non-touching loops : 0
Determinant :
  1   G1G2  G1   1  G1 (1  G2 )
Path Factor :
X 2 (s)

G2
- G1 Y (s)

- G1

All the loops touch forward paths.

1  1  (Isolated loop gain)
1  1  0  1
Using Mason’s gain formula, transfer function is given by,
C ( s) 1
  Pk  k
R( s )  k
So, the transfer function is,
Y ( s ) P11

X 2 ( s) 
Y ( s) G2

X 2 ( s ) 1  G1 (1  G2 )
Hence, the correct option is (B).
SOL.3 (C)
The block diagram representation is shown below,
( s + 3)
R(s) C(s)
s ( s + 2)

s +1
s+4
Control Systems 4 Block Diagram & Signal Flow Graph
The equivalent transfer function is given by,
C ( s) G(s)

R( s ) 1  GH ( s )
s3
C ( s) s ( s  2)

R ( s ) 1  ( s  3) (s  1)
s ( s  2) s  4
C (s) ( s  3)( s  4)

R ( s ) s ( s  2)( s  4)  ( s  3)( s  1)
C (s) s 2  7 s  12
 3
R ( s ) s  6 s 2  8s  s 2  4 s  3
C (s) s 2  7 s  12
 3
R ( s ) s  7 s 2  12 s  3
Hence, the correct option is (C).
SOL.4 (B)
The given block diagram is shown below,

1 1
R(s) Y(s)
s +1 s
-

The signal flow graph representation of block diagram is

-1

1 1
s +1 s
R( s) Y (s)
1 1 1

-1

Forward path gain :

1 1 1
P1  1 1  1 
s 1 s s ( s  1)
1 1
P2  1 1 1 
s s
Individual loop gain :
1 1 1
L1    1 
s 1 s s ( s  1)
Number of two non-touching loop : 0
1
Determinant :   1  L1  1 
s  ( s  1)
Control Systems 5 Block Diagram & Signal Flow Graph
Path factor :
-1

1 1
s +1 s
R( s) Y (s)
1 1 1

-1

All the loops touch forward paths.

1  1  (Isolated loop gain)
1  1  0  1
-1

1 1
s +1 s
R( s) Y (s)
1 1 1

-1

All the loops touch forward paths.

 2  1  (Isolated loop gain)
2  1  0  1
Using Mason’s gain formula, transfer function is given by,
C (s) 1
  PK  K
R(s)  K
So, the transfer function is
C ( s ) P11  P2  2

R( s) 
1 1
1  1
C ( s ) s ( s  1) s

R( s) 1
1
s ( s  1)
1 1  s
C (s) s ( s  1) s
  2
R(s) s ( s  1)  1 s  s 1
s ( s  1)
Hence, the correct option is (B)
SOL.5 (B)
The given block diagram is shown below,
G1 ( s )

X (s)

G2 ( s ) Y (s)
Control Systems 6 Block Diagram & Signal Flow Graph
From figure,
[ X ( s)  Y ( s )]G1 ( s )  X ( s) G2 ( s)  Y ( s )
G1 ( s )G2 ( s ) X ( s )  G1 ( s )G2 ( s )Y ( s )  G2 ( s ) X ( s )  Y ( s )
G2 ( s) X ( s)[1  G1 ( s)]  Y ( s)[1  G1 ( s)G2 ( s )]
Y ( s ) G2 ( s )[1  G1 ( s )]

X ( s ) 1  G1 ( s )G2 ( s )
Hence, the correct option is (B)
SOL.6 (C)
The given control system is shown below
N (s)

Ge ( s )

s+a c
X (s) Y (s)
s+b s(s + d )

Consider X ( s )  0 then response due to N(s) is given by,

sa C
Y ( s )  [ Y ( s )  N ( s )Ge ( s )]   N ( s)
s  b s(s  d )
Y ( s )( s  a )C N ( s )Ge ( s )C ( s  a )
Y (s)    N (s)
s ( s  b)( s  d ) s ( s  b)( s  d )
 s( s  b)( s  d )  C ( s  a)Ge ( s ) 
Y ( s)  N ( s)  
 s ( s  b)( s  d )  C ( s  a) 
Given Response due to N ( s )  0
Hence, Y ( s )  0
 s( s  b)( s  d )  C ( s  a)Ge ( s) 
N (s)  0
 s( s  b)( s  d )  C ( s  a) 
s( s  b)( s  d )  C ( s  a)Ge ( s)
s ( s  b)( s  d )
Ge ( s ) 
C (s  a)
Hence, the correct option is (C)
SOL.7 (A)
Given signal flow graph is shown below
1 3
s +1 s+4
R( s) Y (s)

-1
-1
Control Systems 7 Block Diagram & Signal Flow Graph
Forward path gain :
1 3 3
P1  1  1 
s 1 s  4 ( s  1)( s  4)
Individual loop gain :
1 3 3
L1    1 
s 1 s  4 ( s  1)( s  4)
1 1
L2   1 
s 1 s 1
Number of two non-touching loops : 0
 1 3 
Determinant :   1  [ L1  L2 ]  1    
 s  1 ( s  1)( s  4) 
1 3
  1 
s  1 ( s  1)( s  4)
Path factor
1 3
1 s +1 s+4 1
R( s) R( s)

-1 -1

All the loops touch forward path P1 ,

Hence 1  1  (Isolated loop gain)
1  1  0  1
Using Mason’s gain formula, transfer function is given by,
C (s) 1
  PK  K
R(s)  K
So, the transfer function is,
Y ( s ) P11

R(s) 
3
1
Y (s) ( s  1)( s  4)

R( s) 1  1  3
s  1 ( s  1)( s  4)
Y (s) 3

R ( s ) ( s  1)( s  4)  ( s  4)  3
Y (s) 3
 2
R ( s ) s  6 s  11
Hence, the correct option is (A)
Control Systems 8 Block Diagram & Signal Flow Graph
SOL.8 (A)
Given network is shown below,
Y (s)

Ei ( s ) I (s) Z (s) E0 ( s )

Method 1 :
E0 ( s)  Z ( s) I ( s) …(i)
Applying KVL,
I (s)
 Ei ( s )   Z ( s) I (s)  0
Y (s)
I (s)
 Ei ( s )   E0 ( s )  0
Y (s)
I ( s )  Y ( s) Ei ( s )  Y ( s) E0 ( s ) …(ii)
From equation (i) and (ii),
Y (s) I (s)
Ei ( s ) E0 ( s )
Z (s)

- Y (s)

Hence, the correct option is (A).

Method 2 :
Step 1 : Identify number of nodes.
Ei ( s ) E0 ( s )
I (s)

Step 2 : Mention admittance of series element and impedance of shunt element.

I (s)
Ei ( s ) E0 ( s )
Y (s) Z (s)

Step 3 : Draw one negative feedback path for every element except for the last element.
Y (s) I (s)
Ei ( s ) E0 ( s )
Z (s)

- Y (s)

Hence, the correct option is (A).

SOL.9 (D)
Given signal flow graph is shown below,
Control Systems 9 Block Diagram & Signal Flow Graph
x2

e g
x1 c
1 f 1
1 2

x3
h  x3  2 g …(i)
e  x1  h …(ii)
f  x2  e …(iii)
g f …(iv)
cg …(v)
From equation (iii) and (iv),
g  x2  e
From equation (iii),
g  x2  x1  h
From equation (ii),
g  x2  x1  x3  2 g
g   ( x1  x2  x3 )
From equation (v),
cg
c   x1  x2  x3
Hence, the correct option is (D).
SOL.10 (B)
Given block diagram representation is shown below,
G2 D

R G1 G3 G4 C

H2

H1

Put D  0 modified block diagram is given by,

G3
R +- G1 + G2 G4 C
1 + G3 H 2

H1
Control Systems 10 Block Diagram & Signal Flow Graph

G3G4 (G1 + G 2 )
R +- C
1 + G3 H 2

H1

G3G4G1  G3G4G2
C 1  G3 H 2

R 1  (G3G4G1  G3G4G2 ) H1
1  G2 H 2
C G3G4G1  G3G4G2

R 1  G3 H 2  G1G3G4 H1  G2G3G4 H1
Hence, the correct option is (B).
SOL.11 (C)
Modified block diagram representation with R  0 is given below,
D

G3
0 +- G1 + G2 +- G4 C
1 + G3 H 3

H1

0 +- G4 C

G3 H1 (G1 + G 2 )
1 + G3 H 2

C G4

D 1  G3G4 H1 (G1  G2 )
1  G3 H 2
C G4 (1  G3 H 2 )

D 1  G3 H 2  (G1  G2 )G3G4 H1
Hence, the correct option is (C).
SOL.12 (D)
The given block diagram is shown below
8

_
+ + +
r(t) _ ò 4
_ ò c(t)

5 10
Control Systems 11 Block Diagram & Signal Flow Graph
1
Replacing  block by s
and inter changing alternate summer blocks

R( s) +- +- 1
4 +-
1
C (s)
s s

10
5

Negative feedback Negative feedback

connection connection
8

+-
R( s) 1 1
4 C (s)
s+5 s + 10

Cascade connection
8

- 4
R( s) + C (s)
( s + 5)( s + 10)

4
C ( s) ( s  5)( s  10) 4
  2
R( s) 1  48 s  15s  82
( s  5)( s  10)
Hence, the correct option is (D)
SOL.13 (A)
The given control system is shown in below figure
N (s)

R( s) +- G1 +
+- G2 C (s)

H1

H2

Consider N ( s )  0 then the modified block diagram is shown below,

Negative feedback connection

R( s) +- G1 +- G2 C (s)

H1

H2
Control Systems 12 Block Diagram & Signal Flow Graph
Cascade connection

G2
R( s) +- G1 C (s)
1 + G2 H1

H2

G1G2
R( s) +- C (s)
1 + G2 H1

H2

G1G2
C ( s) 1  G2 H1 G1G2
 
R ( s ) 1  G1G2 H 2 1  G2 H1  G1G2 H 2
1  G2 H1
Hence, the correct option is (A)
SOL.14 (A)
Consider R ( s )  0 then modified block diagram is shown below,
N (s)

0 +- G1 +- G2 C (s)

H1

H2

G2
N (s) +- C (s)
1 + G2 H1

G1 H 2

G2
C ( s) 1  G2 H1

N ( s ) 1  G2G1 H 2
1  G2 H1

C ( s) G2

N ( s ) 1  G2 H1  G1G2 H 2
Hence, the correct option is (A)
Control Systems 13 Block Diagram & Signal Flow Graph
SOL.15 (5)
The given signal flow graph is shown below,

A B D E
R C

Forward paths :
P1  RABDEC , P2  RAEC , P3  RADEC , P4  RABDEC , P5  RABDEC
Hence the number of forward paths is 5
SOL.16 (6)
The given signal flow graph is shown below,

R C
A B D E

Individual loops :
L1  ABA, L2  BDB, L3  DED, L4  EE , L5  ABDA, L6  BDEB
Hence, the no. of closed loops is 6.
SOL.17 (C)
Given signal flow graph is shown below,
0.5

X1 X3
10 X2 1

Forward paths :
P1  10
Individual loops :
L1  0.5
Determinant :
  1  0.5  0.5
Path factor :
1  1
Using Mason’s gain formula transfer function can be written as,
1
T (s)   Pk  k
 k
X 3 P11 10 1
   20
X1  0.5
Hence, the correct option is (C).
Control Systems 14 Block Diagram & Signal Flow Graph
SOL.18 (100)
The given signal flow graph is shown below,
–1

10 2

4 2
R(s) C(s)
1 5 10
–1 –2
Forward path gain : P1  1 5  4  2 10  400
P2  110  2 10  200
Individual loop gain :
L1  4  1  4, L2  2  2  4, L3  1
No. of two non-touching loops :
L1 L3  1 4  4, L2 L3  4  1  4
Determinant :
  1  ( L1  L2  L 3 )  L1 L3  L2 L3
  1  ( 4  4  1)  4  4
  1  9  4  4  18
Path factor :
–1

10 2

4 2
R(s) C(s)
1 5 10
–1 –2
Only loops L1 and L3 touch forward path P1
1  1  (Isolated loop gain)
1  1  (1)  2
–1

10 2

4 2
R(s) C(s)
1 5 10
–1 –2
Only loops L2 and L3 touch forward path, hence
 2  1  (Isolated loop gain)
 2  1  (4)  5
Control Systems 15 Block Diagram & Signal Flow Graph
Using Mason’s gain formula transfer function is given by,
C (s) 1
  PK  K
R(s)  K
C ( s ) P11  P2  2

R( s) 
C ( s ) 400  2  200  5

R(s) 18
C ( s ) 1800
  100
R( s) 18
C ( s)
Hence, the value of is 100.
R( s)
SOL.19 (2)
Number of two non-touching loops : L1 L3 , L2 L3
Hence, Number of pairs of non – touching loops are 2
SOL.20 (16)
The given signal flow graph is shown below,

a b c d e
R C
f

Individual loops :
L1  a  b  a, L2  b  c  b, L3  a  c  b  a
L4  c  d  c, L5  d  e  d , L6  e  f  e
L7  d  f  e  d L8  f  f
Number of two non-touching loops :
L1  L1 L3 , L1 L4 , L1 L6 , L1 L7 , L1 L8
L2  L2 L5 , L2 L6 , L2 L7 , L2 L8
L3  L3 L5 , L3 L6 , L3 L7 , L3 L8
L4  L4 L7 , L4 L8
L5  L5 L7
Hence, total number of two non-touching loops = 5+4+4+2+1=16
SOL.21 (5)
Number of three Non-touching loops :
L1  L1 L4 L6 , L1 L 4 L8 , L1 L5 L8
L2  L2 L5 L8 ,
L3  L3 L5 L8 ,
Total number of three non-touching loops  3  1  1  5
Control Systems 16 Block Diagram & Signal Flow Graph
SOL.22 (B)
The given signal flow graph is shown below,
c

b d e
a h
x

l f
k
m g
y

n
Forward Path gain :
P1  abdfg , P2  ahfg , P3  aklfg , P4  akmg
Individual loop gain : L1  c, L2  de, L3  lfn, L4  mn
Hence, the correct option is (B).
SOL.23 (C)
The given electrical system and its signal-flow graph representations are shown in the given below figures
Z1 ( s ) Z 2 (s) V0 ( s )

Vi ( s ) Z3 (s) Z 4 (s)
I1 ( s ) I 2 (s)

G1 I1 ( s ) G2 I 2 ( s ) G3
Vi ( s ) V0 ( s )

H
From first figure,
Vi  ( Z1 ( s)  Z3 ( s )) I1  Z3 ( s) I 2 … (i)
( Z 2 ( s)  Z3 ( s )  Z 4 ( s )) I 2  Z3 ( s) I1  0 … (ii)
From second figure,
I 2 ( s )  G2 I1 ( s ) … (iii)
I1 ( s)  G1Vi ( s)  HI 2 ( s) … (iv)
Comparing eq. (ii) and (iii), we get
Z3 ( s)
G2 
Z 2 ( s)  Z3 ( s)  Z 4 ( s)
Comparing eq. (i) and (iv), we get
Z3 ( s)
H
Z1 ( s)  Z3 ( s)
Hence, the correct option is (C).
Control Systems 17 Block Diagram & Signal Flow Graph
SOL.24 (B)
. Method 1 :
The given block diagram is shown below,
9

u (t ) y (t )
Integrator 2 Integrator

3 12

Signal flow graph of the block diagram can be drawn as,

–9

1 1
1 s 2 s 1
U(s) Y (s )

–3 –12
1 1 2
Forward Path gain : P1   2   2
s s s
Individual loop gain :
1 3 1 12
L1   (3)  , L2   (12)  ,
s s s s
1 1 18
L3   2 9  2
s s s
Number of two Non-touching loops :
36
L 1L2  2
s
Determinant :
 3 12 18  36
  1  ( L1  L2  L3 )  ( L1L2 )  L1  1     2  2
 s s s  s
Path factor :
–9

1 1
1 s 2 s 1
U(s) Y(s)

–3 –12
All the loops touch forward paths.
1  1  (Isolated loop gain)
1  1  0  1
Using Mason’s gain formula, transfer function is given by,
C ( s) 1
  Pk  k
R( s)  k
Control Systems 18 Block Diagram & Signal Flow Graph
So, the transfer function is,
Y ( s ) P11

U ( s) 
2
Y (s) s2

U ( s ) 1  3  12  18  36
s s s2 s2
Y ( s) 2
 2
U ( s ) s  15s  54
Y ( s) 2

U ( s) ( s  6) ( s  9)
Y ( s) 1

U ( s)  s s
27 1   1  
 6 9
Hence, the correct option is (B).
. Method 2 :
The given block diagram is shown below.
9

u (t ) y (t )
Integrator 2 Integrator

3 12

1
Integrator can be replaced by block .
s
9

1 1
u (t ) 2 y (t )
s s

3 12

1 1
U (s) 2 Y (s)
s s

3 12

Negative feedback Negative feedback

1 1
U (s) 2 Y (s)
s+3 s + 12
Cascade connection
Control Systems 19 Block Diagram & Signal Flow Graph

2
U (s) Y (s)
( s + 3) ( s + 12)

The transfer function of the system is

Y ( s) G(s)

U (s) 1  G (s) H (s)
2
Y ( s) ( s  3) ( s  12)

U ( s) 1  29
( s  3) ( s  12)
Y (s) 2
 2
U ( s ) s  15s  54
Y ( s) 2

U ( s) ( s  6) ( s  9)
Y ( s) 1

U ( s)  s s
27 1   1  
 6 9
Hence, the correct option is (B).
SOL.25 (A)
The given signal flow graph is shown in figure below,
1 1
X(s)
k
-1
-k s
Y(s)
1 1
Forward path gain :
1 1
P1  11  1 1  , P2  1 1  k  1   k
s s
k
Individual loop gain : L1 
s
Number of two non-touching loops : 0
k
Determinant :   1  L1  1 
s
Path factor :
(i)
1 1
X(s)

k
-1
-k s
Y(s)
1 1
Control Systems 20 Block Diagram & Signal Flow Graph
The loop L1 touches forward path P1 , hence
1  1  (isolated loop gain)
1  1  0  1
(ii)
1 1
X(s)

k
-1
-k s
Y(s)
1 1
The loop L1 touches forward path P2 , hence
 2  1  (isolated loop gain)
2  1  0  1
Using Mason’s gain formula, transfer function is given by,
C ( s) 1
  Pk  k
R( s)  k
So, the transfer function is,
1
 k
Y ( s ) P11  P2  2
  s
X ( s)  1
k
s
Y ( s )  1  ks 

X ( s) s  k 
Hence, the correct option is (A).
SOL.26 (A)
The given signal flow graph is shown below.
1

1 s -1 s -1 1
U (s) Y (s)
-4

-2

1 1
Forward path gain : P1  2
, P2 
s s
Individual loop gain :
4 2 2
L1 
, L2   4, L3  2 , L4 
s s s
Number of two non-touching loos : 0
Determinant :
4 2 2 4 2 2
  1   L1  L2  L3  L4   1    4  2    1  4  2 
 s s s s s s
Control Systems 21 Block Diagram & Signal Flow Graph
Path factor :
1

1 s -1 s -1 1
U (s) Y (s)
-4

-2

All the loops touch forward paths.

1  1  (Isolated loop gain)
1  1  0  1
1

1 s -1 s -1 1
U (s) Y (s)
-4

-2

All the loops touch forward paths.

 2  1  (Isolated loop gain)
2  1  0  1
Using Mason’s gain formula, transfer function is given by,
C ( s) 1
  Pk  k
R( s)  k
So, the transfer function is
1 1
Y  s  P11  P2  2 
  s2 s
U s  4 2 2
1  4  2 
s s s
Y s 1 s
 2
U  s  s  4s  4s 2  2  2 s

Y s s 1
 2
U  s  5s  6 s  2
Hence, the correct option is (A).
SOL.27 (C)
Given block diagram is shown below,
R( s) G1 G2 C (s)

. Method 1 : SFG :
The block diagram can be converted into signal flow graph as shown below,
Control Systems 22 Block Diagram & Signal Flow Graph
1 G1 G2 1
R( s) C (s)

1 1

Fig. (a)
Alternate signal flow graph is shown below,
1 G1 G2 1
R( s) C (s)

1
Forward path gain :
P1  1  G1  G2  1  G1G2 ,
P2  G2  1  G2 , P3  1  1  1
Individual loop gain : There is no individual loops.
Number of two non-touching loops : 0
Determinant :   1  0  1
Path factor :
(i)
1 G1 G2 1
R( s) C (s)

1 1

Since, there is no loop, hence

1  1  (isolated loop gain)
1  1  0  1
(ii)
1 G1 G2 1
R( s) C (s)

1 1

Since, there is no loop, hence

 2  1  (isolated loop gain)
2  1  0  1
(iii)
1 G1 G2 1
R( s) C (s)

1 1
Control Systems 23 Block Diagram & Signal Flow Graph
Since, there is no loop, hence
 3  1  (isolated loop gain)
3  1  0  1
Using Mason’s gain formula, transfer function is given by,
C ( s) 1
  Pk  k
R( s )  k
So, the transfer function is,
C ( s ) P11  P2  2  P3 3

R( s) 
C ( s)
 G1G2  G2  1
R( s)
Hence, the correct option is (C).
. Method 2 : Block Diagram :
Given block diagram can be redrawn as shown below,
Parallel connection

G2

Cascade connection

Transfer function is given by,

C ( s)
 (1  G1 )G2  1
R( s)
C ( s)
 1  G2  G1G2
R( s)
Hence, the correct option is (C).
SOL.28 (C)
Given : h1  b1 , h0  b0  b1a1
The given signal flow graph is shown in figure below.
Control Systems 24 Block Diagram & Signal Flow Graph
h1

1 1
h0 1 s 1 s 1
U (s) C (s)

-a1

-a0

h0 h
Forward path gain : P1  2
, P2  1
s s
Individual loop gain :
a1 a
L1  , L2  20
s s
Number of two non-touching loops : 0
Determinant :
 a a 
  1  ( L1  L2 )  1   1  20 
 s s 
s 2  a1s  a0

s2
Path factor :
h1

1 1
h0 1 s 1 s 1
U (s) C (s)

-a1

-a0

All loop touches forward path.

1  1  0  1,
h1

1 1
h0 1 s 1 s 1
U (s) C (s)

-a1

-a0

Loop L1 does not touch forward path.

  a   a   s  a1 
2 1   1   1  1    
 s   s  s 
Using Mason’s gain formula, transfer function is given by,
C ( s) 1
  Pk  k
R( s )  k
Control Systems 25 Block Diagram & Signal Flow Graph
So, the input-output transfer function is,
C ( s ) P11  P2  2
G( s)  
U ( s) 
 h0   h1   s  a1 
 s 2  1   s    s 
G ( s)    2    
 s  a1s  a0 
 
 s2 
 h  h ( s  a1 ) 
G ( s)   02 1 
 s  a1s  a0 
Putting the value of h0 and h1 ,
(b0  b1a1 )  b1 ( s  a1 )
G( s) 
( s 2  a1s  a0 )
 b s  b0 
G( s)   2 1 
 s  a1s  a0 
Hence, the correct option is (C).
SOL.29 (C)
Given block diagram is shown below.
X - HY
G1 ( X - HY )
G1

X (s) Y ( s ) = G1 ( X - HY )
H
HY Y (s) + G2 ( X - HY )

G2
G2 ( X - HY )
X - HY

From the block diagram

Y  G1 ( X  HY )  G2 ( X  HY )
Y  X (G1  G2 )  HY (G1  G2 )
Y 1  H (G1  G2 )  X (G1  G2 )
Y G1  G2

X 1  H (G1  G2 )
Hence, the correct option is (C).
SOL.30 (A)
Given signal flow graph is shown below.
–R U2
–1
1 1
Ul 1/L 1/s kl 1/J 1/s Y

–k2
Control Systems 26 Block Diagram & Signal Flow Graph

Y ( s)
For transfer function signal flow graph is redrawn as shown below,
U1 ( s ) U
2 0

–R
1 1
Ul 1/L 1/s kl 1/J 1/s Y

–k2

Forward path gain :

1 1 1 1
P1  1    k1  
L s J s
k1
P1 
JL s 2
R k k
Individual loop gain : L1  , L2  1 22
Ls JL s
Number of two non-touching loos : 0
Determinant :
R k1k2
  1  ( L1  L2 )  1  
Ls JL s 2
Path factor :
–R
1 1
Ul 1/L 1/s kl 1/J 1/s Y

–k2

All the loops touch forward paths.

1  1  (Isolated loop gain)
1  1  0  1
Using Mason’s gain formula, transfer function is given by,
C ( s) 1
  Pk  k
R( s )  k
So, the transfer function is,
Y ( s) P11

U1 ( s ) 
k1
Y ( s) JL s 2

U1 ( s ) 1  R  k1k2
Ls JL s 2
Control Systems 27 Block Diagram & Signal Flow Graph
Y ( s) k1

U1 ( s ) JL s  JR s  k1k2
2

Hence, the correct option is (A).

SOL.31 (A)
The given block diagram is shown below,
G2 ( s )

+
+ + +
R(s) G1 ( s ) G3 ( s ) C(s)
– –

H1 ( s )

.. Method 1 : Block Diagram..

The given block diagram can be redrawn as shown below,
Parallel connection
G2 ( s )

+
+ + +
R(s) G1 ( s ) G3 ( s ) C(s)
– –

H1 ( s )

Fig. (a)
Negative feedback connection
+ +
R(s) G1 ( s ) + G2 ( s ) G3 ( s ) C(s)
– –

H1 ( s )

Fig. (b)
Cascade connection

+ G1 ( s ) + G2 ( s )
R(s) G3 ( s ) C(s)
1 + H1 ( s )[G1 ( s ) + G2 ( s )]
–

Fig. (c)
+ [G1 ( s ) + G2 ( s )]G3 ( s )
R(s) C(s)
1 + H1 ( s )[G1 ( s ) + G2 ( s )]
–

Fig. (d)
Control Systems 28 Block Diagram & Signal Flow Graph
The overall closed loop transfer function of a unity feedback is given by,
C (s) [G1 ( s )  G2 ( s )]G3 ( s )

R( s ) 1  H1 ( s)[G1 ( s)  G2 ( s )]  G3 ( s )[G1 ( s )  G2 ( s )]
C ( s) [G1 ( s )  G2 ( s )]G3 ( s )

R( s ) 1  [G1 ( s )  G2 ( s )][G3 ( s )  H1 ( s )]
Hence, the correct option is (A).
.. Method 2 : SFG..
Signal flow graph of the given block diagram can be drawn as,
G2 ( s )
1 1 1 1 G3 ( s )
R( s) C (s)
G1 ( s )

- H1 ( s )

-1

Forward path gain : P1  G1 ( s )G3 ( s ), P2  G2 ( s )G3 ( s )

Individual loop gain :
L1   G1 ( s ) H1 ( s ) , L2   G1 ( s )G3 ( s ) ,
L3   G2 ( s ) H1 ( s ) , L4   G2 ( s )G3 ( s )
Number of two non-touching loops : 0
Determinant :   1  [ L1  L2  L3  L4 ]
  1  G1 ( s )G3 ( s )  G2 ( s )G3 ( s )  G1 ( s ) H1 ( s )  G2 ( s ) H1 ( s )
Path factor :
(i)
G2 ( s )
1 1 1 1 G3 ( s )
R( s) C (s)
G1 ( s )

- H1 ( s )

-1

All the loops touch forward path P1 , hence

1  1  (isolated loop gain)
1  1  0  1
(ii)
G2 ( s )
1 1 1 1 G3 ( s )
R( s) C (s)
G1 ( s )

- H1 ( s )

-1
Control Systems 29 Block Diagram & Signal Flow Graph
All the loops touch forward path P2 , hence
 2  1  (isolated loop gain)
 2  1  0 1
Using Mason’s gain formula, transfer function is given by,
C ( s) 1
  Pk  k
R( s)  k
So, the transfer function is,
C ( s ) P11  P2  2

R( s) 
C (s) G1 ( s )G3 ( s )  G2 ( s )G3 ( s )

R( s) 1  G1 ( s)G3 ( s)  G2 ( s )G3 ( s )  G1 ( s ) H1 ( s )  G2 ( s ) H1 ( s )
C ( s) [G1 ( s)  G2 ( s )]G3 ( s )

R( s) 1  [G1 ( s)  G2 ( s)][G3 ( s )  H1 ( s )]
Hence, the correct option is (A).


Practice Solutions :
SOL.1 (B)
Given block diagram is shown below,
2
R( s) + Ka + C ( s)
- - s ( s + 2)

sK t

2Ka
R( s) C ( s)
s ( s + 2 K t + 2)

2Ka
C ( s) s  2 s ( K t  1)
2

R( s) 1  2Ka
s  2 s ( K t  1)
2

C (s) 2Ka
 2 … (i)
R ( s ) s  2 s ( K t  1)  2 K a
On comparing equation (i), with standard characteristic equation,
n  2 K a
2n  2( K t  1)
(2  0.5)n  2( K t  1)
n  2 K t  2

2 K a  2 Kt  2 … (ii)
For unit ramp input,
K v  lim sGH ( s )
s 0
Control Systems 2 Time Response Analysis

 2Ka 
K v  lim s  2 
s 0
 s  2 s ( K t  1) 
2Ka
Kv 
0  2 Kt  2
Ka
Kv 
Kt  1
Steady state error is given by,
1
ess 
Kv
Kt  1
ess 
Ka
Given ess  0.2 ,
Kt  1
0.2 
Ka
0.2 K a  K t  1 … (iii)
From equation (ii) and (iii),
2 K a  2(0.2 K a  1)  2
2 K a  (0.4 K a ) 2
2 K a  0.16 K a 2
1
Ka   12.5
0.08
From equation (iii),
K t  0.2 K a  1
K t  0.2  12.5  1  1.5
Hence, the correct option is (B).
SOL.2 (D)
Given :
16
(i) G ( s ) 
s ( s  4)
(ii) H ( s )  1  ks
(iii)   0.6
The characteristic equation is given by,
1  G (s) H (s)  0
16(1  ks )
1 0
s ( s  4)
s 2  4 s  16  16 ks  0
s 2  s (16k  4)  16  0 …(i)
Control Systems 3 Time Response Analysis
Standard characteristic equation for second order system is given by,
s 2  2n s  2n  0 …(ii)
On comparing equation (i) and equation (ii),
2n  16
n  4
2n  16k  4
2  0.6  4  16k  4
0.8
k  0.05
16
Hence, the correct option is (D).
SOL.3 (C)
Given : Damping ratio   0.5456
and natural frequency n  316 rad/sec
Maximum peak overshoot is given by,


1 2
% MPO  e  100
0.5456

1 (0.5456) 2
% MPO  e  100
% MPO  12.93
Hence, the correct option is (C).
SOL.4 (C)
Given : ξ = 0.7 and ωn  4 rad/sec
K
R( s) C (s)
s (s + 2 )

1 + as

K
G ( s)  , H ( s )  1  as,
s ( s  2)
The characteristic equation is given by,
1  G ( s) H ( s)  0
K
1 (1  as )  0
s ( s  2)
s 2  ( Ka  2) s  K  0 …(i)
Standard characteristic equation for second order system is given by,
s 2  2 n s  2n  0 …(ii)
On comparing equation (i) and (ii),
2n  K  K  16
Control Systems 4 Time Response Analysis

2ξ ωn  (2  Ka )
2  0.7  4  (2  Ka)
a  0.225
Hence, the correct option is (C).
SOL.5 (A)
K
Given : G ( s )    [Type-1 system]
s( s  1)
and   r ( t )  u (t )
For step input (position input), steady state error is given by,
1
ess 
1 Kp
where, K p is position error coefficient.
The position error coefficient is given by,
K p  lim G ( s )
s 0

K
K p  lim 
s 0 s( s  1)
1
ess  0
1 
Hence, the correct option is (A).
SOL.6 (D)
an 1s  an
Given : T ( s )  n n 1
s  a1s  .....  an 1s  an
and r( t ) = t u ( t )
To find steady state error open loop transfer function is required.
Closed-loop transfer function for negative unity feedback is given by,
G( s)
T ( s)  … (i)
1  G( s)
where, G ( s )  Open loop transfer function
an 1s  an
s  a1s n 1  .....  an 2 s 2
n
T ( s)  … (ii)
an 1s  an
1 n
s  a1s n 1  .....  an 2 s 2
Comparing equation (i) and (ii),
an 1s  an
G( s)  n
s  a1s n 1  .....  an 2 s 2
For ramp input (velocity input) steady state error is given by,
1
ess  … (iii)
Kv
 Control Systems 5 Time Response Analysis

where, K v is velocity error coefficient.

The velocity error coefficient is given by,
Kv  lim sG ( s )
s 0

s ( an 1s  an )
      K v  lim 
s 0 s  a1s n 1  .....  an 2 s 2
n

From equation (iii),

1
ess 
0

Hence, the correct option is (D).
SOL.7 (C)
Given :   0.5
s6
s6 K
G( s)  2 
Ks  s  6 s 2  s  6
K K
Characteristics equation is,
s 6
s2    0 …(i)
K K
The characteristics equation for second-order system is given by,
s 2  2 n s  2n  0 …(ii)
Comparing equation (i) and equation (ii),
1 6
2ξn  and 2n 
K K
1
2  0.5n 
K
1
n 
K
2
1 1 6
  2
2
n    
K K K
1
K
6
Hence, the correct option is (C).
SOL.8 (B)
Given :
100 100
      G( s)   2 …(i)
( s  1)( s  100) s  101s  100
The transfer function for second-order system is given by,
C ( s) 2n
 …(ii)
R( s ) s 2  2 n s  2n
Control Systems 6 Time Response Analysis
where,   damping ratio,
n  natural angular frequency
On comparing equation (i) and equation (ii),
2n  100  n  10
2n  101    5.05
As   1 , system is over-damped.
For 2% error band, settling time of over-damped system is given by,
4
ts 
n  n 2  1
4
ts   4sec
5.05  10  10 5.052  1
Hence, the correct option is (B).
 Key Point
(i) Settling time for 2% error band for under-damped system is given by,
4
ts  sec
n
(ii) Settling time for 2% error band for over-damped system is given by,
4
ts  sec  
n  n 2  1
(iii)Settling time for 2% error band for critical damped system is given by,
ts  5.8  
1 1
  Where,    
n n
(iii)Settling time for 2% error band for undamped system is given by,
ts    
SOL.9 (B)
Given : MPO = 16 % [By default 1st peak overshoot]
Percentage MPO is given by,


1 2
% MPO  e 100


12
16  e 100
 
 ln(0.16)
1  2
1
2 
3.939
  0.5
Control Systems 7 Time Response Analysis

OR
  cos  
e  cot   0.16    cot  
 1  
2 sin  
 ln (0.16)
cot    0.5833

tan   1.714
  cos   cos (59.74)  0.503
Damping ratio ()  0.5
The second peak overshoot is given by,
3 

12
% MPO (2 )  end
100
The decay ratio is given by,
3 
 2 
12 
e 12
Decay ratio  
e  0.0265

2
e 1
Hence, the correct option is (B).
 Key Point
Concept of decay ratio :
c(t)
st
1 peak overshoot
nd
cmax 2 peak overshoot

t
0 tp

2nd peak overshoot

Decay ratio 
1st peak overshoot
Decay ratio can also be obtained by
2nd peak undershoot
Decay ratio   
1st peak undershoot
SOL.10 (B)
100 π
Given : G ( s )  and t p  sec.
s( s  p) 8
Peak time is given by,

tp 
d
Control Systems 8 Time Response Analysis
 
    
8 d
ωd  8 rad/sec.
Closed-loop transfer function for negative feedback system is given by,
G (s)
T (s) 
1  G ( s) H (s)
100
s ( s  p)
T ( s) 
100
1
s ( s  p)
100
T ( s)  …(i)
s  sp  100
2

Transfer function for second-order system is given by,

C (s) 2n
 2 …(ii)
R ( s) s  2 n s  2n
where,   damping ratio,
n  natural angular frequency
On comparing equation (i) and (ii),
2n  100  n  10 rad/sec.
2  n  p
The damped frequency of oscillation is given by,
d  n 1   2

8  10 1  2
(0.8) 2  1   2
 2  0.36
ξ = 0.6
Hence, the correct option is (B).
SOL.11 (B)
We have 2  n  p
p  2  0.6 10  12
Hence, the correct option is (B).
SOL.12 (C)
1
Given : closed loop  open loop
100
where,  represents time constant.
Control Systems 9 Time Response Analysis
Va ( s )
10
R( s) +– Ka w( s )
1 + 10s

From figure,
( s ) 10
Open loop transfer function of DC motor  
Va ( s ) 1  10 s
Location of pole of open loop transfer function is shown below,
jw

s
-1
10

Time constant is defined as reciprocal of magnitude of negative real root.

Hence, open loop  10 sec
1
closed loop  open loop  0.1 sec …(i)
100
Closed-loop transfer function for negative unity feedback is given by,
G (s)
T (s) 
1  G( s)
 10 
Here, G( s)  Ka  
 1  10 s 
 10 
Ka  
( s )  1  10 s 
T ( s)  
R( s )  10 
1  Ka  
 1  10 s 
10 K a 10 K a
T ( s)  
1  10s  10 K a 10s  (10 K a  1)
Location of pole of closed loop transfer function is shown below,
jw

s
æ 1 + 10 K a ö
-ç ÷
è 10 ø

From above figure,

10
closed loop  …(ii)
10 K a  1
Control Systems 10 Time Response Analysis
From equation (i) and (ii),
10
 0.1 sec
10 K a  1
10 1

10 K a  1 10
10 K a  1  100
10 K a  99
K a  9.9  10
Hence, the correct option is (C).
SOL.13 (D)
Given : (i) Forward path transfer function
1
G (s) 
s  s 1
2

(ii) Feedback path transfer function H ( s )  Ks

(iii)Critical damped transient response
Characteristic equation is given by,
1  G (s) H (s)  0
Ks
1 0
s  s 1
2

s 2  s (1  K )  1  0 … (i)
Standard characteristic equation for second order system is given by,
s 2  2n s  2n  0 … (ii)
n  1 rad/sec
2n  k  1
2  k 1
k 1
Hence, the correct option is (D).
SOL.14 (C)
Given : Unit step response,
C (t )  1  0.2e 60t  1.2e 10t
Taking Laplace transform of C (t )
1 0.2 1.2
C (s)   
s s  60 s  10
( s  60)( s  10)  0.2 s ( s  10)  1.2 s ( s  60)
C (s) 
s ( s  60)( s  10)
s 2  70s  600  0.2 s 2  2s  1.2s 2  72s
C ( s) 
s ( s 2  70 s  600)
Control Systems 11 Time Response Analysis
600
C (s) 
s ( s  70 s  600)
2

For unit step input, r (t )  u (t )

Taking Laplace transform of r (t )
1
R(s) 
s
C (s)
Transfer function, H ( s ) 
R(s)
600
H ( s) 
s ( s  70s  600)
2

1
s
600
H( s )  …(i)
s  70 s  600
2

Transfer function for standard second-order system is given by,

C (s) 2n
 H (s)  2 …(ii)
R( s ) s  2n s  2n
On comparing equation (i) and (ii),
2n  600

n  10 6  24.5 rad/sec
2n  70
35
  1.43
24.5
Hence, the correct option is (C).
SOL.15 (B)
Given :
(i) Forward path transfer function
40
G (s) 
s ( s  2)( s 2  2 s  30)
5t 2
(ii) Input r (t ) 
2
t2
For unit ramp input, r (t ) 
2
1
Steady-state error, ess 
Ka

 t2 
But given input  5  
2
Control Systems 12 Time Response Analysis
5
Hence, ess 
Ka
K a  lim s 2G ( s )
s 0

s 2  40
K a  lim 0
s 0 s ( s  2)( s 2  2s  30)
5
Steady-state error, ess  
0
Hence, the correct option is (B).
SOL.16 (B)
Given : Closed loop transfer function
9
T (s)  … (i)
s  4s  9
2

Transfer function for standard second order system is given by,

9
T (s) 
s  4s  9
2

C (s) 2n
 … (ii)
R ( s ) s 2  2n s  2n
On comparing equation (i) and (ii),
2n  9  n  3 rad/sec
2n  4
4 2
 
23 3
The settling time for 2% band is given by,
4
ts 
n
4
ts   2 sec
2
3
3
Hence, the correct option is (B).
SOL.17 (D)
Given : Characteristic equation  s 2  0.75 s  0.25 …(i)
Standard characteristic equation for second order system is given by,
s 2  2n s  n2  0 …(ii)
On comparing equation (i) and (ii)
2n  0.25
n  0.5 rad/sec
2n  0.75
Control Systems 13 Time Response Analysis
0.75
  0.75
2  0.5
Since,  is negative
Hence unstable system
Hence, the correct option is (D).
SOL.18 (B)
s
Given : (i) Transfer function H ( s ) 
s  4s  3
2

(ii) Unit ramp input

C ( s) s
H (s)   2
R(s) s  4s  3
For unit ramp input, r (t )  t u (t )
Taking laplace transform of r (t )
1
R(s) 
s2
s R(s)
C (s) 
s  4s  3
2

1
2
s
C ( s)  2 s
s  4s  3
1
C (s) 
s ( s  4 s  3)
2

Steady state output is given by,

lim C (t )  lim s C ( s )
t  s 0

s 1
lim 
s0 s ( s  4 s  3) 3
2

Hence, the correct option is (B).

SOL.19 (A)
Given block diagram is shown below,
e(t )  x(t )  y (t ) … (i)
Taking Laplace transform of equation (i),
E (s)  X (s)  Y (s) … (ii)
From block diagram,
d2y
 e(t ) … (iii)
dt 2
Taking Laplace transform of equation (iii),
s 2Y ( s)   E ( s)
 E ( s)
Y (s)  … (iv)
s2
Control Systems 14 Time Response Analysis
From equation (ii) and (iv),
E ( s)
E (s)  X (s) 
s2
Since, x(t )  tu (t )
1
Hence X ( s ) 
s2
E ( s) 1
E ( s)   2
s2 s
1  s 2  1
E (s)  2   2
 s  s
1
E ( s) 
s 1 2

Taking inverse Laplace transform of above equation

e(t )  sin t
Hence, the correct option is (A).
SOL.20 (D)
2n
Given open loop transfer function, G ( s ) 
s ( s  2n )
Closed loop transfer function is given by,
G (s)
T (s) 
1  G (s)
2n
T (s) 
s ( s  2n )
2n
1
s ( s  2n )
2n
T (s)  …(i)
s 2  2n s  2n
Transfer function for standard second-order system is given by,
C ( s) 2n
 2 …(ii)
R( s) s  2n s  2n
Where,   damping ratio
n  natural angular frequency
On comparing equation (i) and (ii),
 1
i.e. system is critically damped
Hence, the correct option is (D).
SOL.21 (D)
2s
Given open loop transfer function, G ( s ) 
( s  1)( s  2)
Control Systems 15 Time Response Analysis
For unity negative feedback, closed loop transfer function is given by,
G(s) G (s)

R(s) 1  G (s)
2s
C (s) ( s  1)( s  2)

R(s) 1  2s
( s  1)( s  2)
C (s) 2s
 2
R ( s ) s  5s  2
For unit step input r (t )  u (t )
Taking laplace transform of r (t ),
1
R(s) 
s
2s 1
C (s)  
s  2s  2 s
2

2
C ( s) 
s  2s  2
2

The steady state value can be calculated as,

2s
C ()  lim sC ( s )  lim
s 0 s 0 s  2s  2
2

C ( )  0
Hence, the correct option is (D).
SOL.22 (B)
Given unity feedback system is shown below,
Steady state error for unit ramp input is given by,
1
ess 
KV
K K
Where, KV  lim sG ( s )  lim s 
s 0 s 0 s( s  a) a
1 a
ess  
K K
a
(i) Sensitivity with respect to parameter K :
ess
e K e
S Kess  ss   ss
K ess K
K
K a
S Kess    1
a K2
K
Control Systems 16 Time Response Analysis
(ii) Sensitivity with respect to parameter a :
ess
e a e
S aess  ss   ss
a ess a
a
a 1
S aess   1
a K
K
Hence, the correct option is (B).
SOL.23 (D)
Given RLC network is shown below
1W 1H
+ +
+
Vi ( s ) 1F V0 ( s )

Transform domain :
1 s
+ +
1
Vi ( s ) V0 ( s )
s

 1 
 
V0 ( s )  Vi ( s )  3  [By VDR ]
1
1  s  
 s
Vi ( s )
V0 ( s ) 
s  s 1
2

V0 ( s ) 1
 2
Vi ( s ) s  s  1
Characteristic equation is given by,
s2  s 1  0 … (i)
Standard characteristic equation for second order system is given by,
s 2  2n s  n2  0 … (ii)
On comparing equation (i) and (ii)
n  1 and 2n  1
1
  0.5
2 1
Hence, the correct option is (D).
Control Systems 17 Time Response Analysis
SOL.24 (D)
Given black diagram shown in below figure,
1 1
R( s) C (s)
s s

Modified block diagram is given by,

1 1
R( s) R( s) C (s)
s s

Negative feedback

Cascade
Connection
1 1
R( s) C (s)
s +1 s

1
R( s) C (s)
s ( s + 1)

1
C ( s) s ( s  1)
Transfer function 
R( s) 1  1
s ( s  1)
C (s) 1
 2
R(s) s  s  1
Hence, the correct option is (D).
SOL.25 (B)
C (s) 1
 2
R(s) s  s  1
For unit impulse input, r (t )  (t )
Taking laplace transform of r (t ),
R(s)  1
1
C (s)  1
s  s 1
2

1
C ( s)  2
1  3
2

s   
 2  2 
Control Systems 18 Time Response Analysis

3 2

2 3
C ( s)  2
1  3
2

s   
 2  2 
Taking inverse laplace transform of C ( s )

2 2t  3 
C (t )  e  sin  t 
3  2 
C (t )

t
0

Hence, the correct option is (B).

SOL.26 (C)
C (s) 1
Transfer function  2 … (i)
R(s) s  s  1
Transfer function for standard second order system is given by,
C (s) 2n
 2 … (ii)
R ( s ) s  2n s  2n
Where,   damping ratio
n  natural angular frequency
On comparing equation (i) and (ii),
n  1 rad/sec and 2n  1
1
  0.5
2 1
Maximum peak overshoot is given by,
 0.5
 
1 2
MPO  e e 1 0.52
 0.163
Hence, the correct option is (C).
SOL.27 (D)
K
(i) G ( s ) 
s ( s  5)
(ii) Peak overshoot  16%
Maximum peak overshoot is given by,
12
MPO  e  /  0.16    0.5
Control Systems 19 Time Response Analysis
Closed loop transfer function is given by,
G (s)
T (s) 
1  GH ( s )
K
s ( s  5)
T (s) 
K
1
s ( s  5)
K
T (s)  … (i)
s  5s  K
2

Transfer function for standard second-order system is given by,

C (s) 2n
 2 … (ii)
R ( s ) s  2n s  2n
Where   damping ratio,
n  natural angular frequency
On comparing equation (i) and (ii),
2n  K and 2n  5
2  0.5  n  5
n  5
K  52
K  25
Hence, the correct option is (D).
SOL.28 (38.16)
Given : n  40 rad/sec and   0.3
The damped natural frequency is given by,
d  n 1  2

d  40 1  (0.3) 2  38.16 rad/sec

Hence, the damped natural frequency is 38.16 rad/sec.
SOL.29 (0)
Given : The closed loop transfer function of a system is
4
T (s)  2
( s  0.4 s  4)
4
T ( s)  s  0.4 s
2

 4 
1  2 
 s  0.4 s 
4
s ( s  0.4)
T (s)  …(i)
 4 
1  s ( s  0.4) 
 
Control Systems 20 Time Response Analysis
Closed-loop transfer function for negative unity feedback system is given by,
G (s)
T (s)  …(ii)
1  G (s)
On comparing equation (i) and equation (ii),
Open loop transfer function,
4
G (s)  [Type 1 system]
s ( s  0.4)
For step input (position input) steady state error is given by,
1
ess 
1 K p
where, K p is the position error coefficient.
The position error coefficient is given by,
K p  lim G ( s )
s 0

K
K p  lim 
s 0 s ( s  0.4)
1
ess  0
1 
Hence, the steady state error is 0.
SOL.30 (400)
K
Given : G ( s )  ,   0.25
s  10s
2

The closed-loop transfer function for negative unity feedback is given by,
K
G( s)
 s  10s
2
T ( s) 
1  G( s) 1  K
s 2  10 s
K
T ( s)  …(i)
s  10s  K
2

Transfer function of second-order system is given by,

2n
T ( s)  …(ii)
s 2  2n s  2n
Comparing equation (i) and (ii),
K  2n  n  K
2n  10
10
  0.25
2 K
10
K   20
0.5
Control Systems 21 Time Response Analysis

K  (20)2  400
Hence, the gain K for the system is 400.
SOL.31 (2.872)
K
Given : G ( s ) 
s ( s  2)
% MPO = 10 %
The characteristics equation is given by,
1  G(s)  0
K
1 0
s ( s  2)
s 2  2s  K  0 …(i)
Standard second order characteristic equation is given by,
s 2  2n s  n2  0 …(ii)
From equation (i) and (ii),
2n  K                       n  K
2n  2
1 1 1
   K   …(iii)
n K 2
Peak overshoot is given by,
 
1 2
MPO  e
 
1 2
0.1  e
 
 ln(0.1)
1  2

 2.3
1  2
 2  2  (2.3) 2 (1   2 )
15.16 2  (2.3) 2
  0.59
From equation (i),
1 1
K   2.872
 2
(0.59) 2
Hence, the value of K is 2.872.
SOL.32 (B)
25
Given : G ( s ) 
s ( s  6)
Control Systems 22 Time Response Analysis
Closed-loop transfer function for unity negative feedback is given by,
G (s)
T (s) 
1  G (s)
where, G ( s )  Open loop transfer function
25
s ( s  6) 25
T ( s)   2 …(i)
25 s  6 s  25
1
s ( s  6)
Transfer function for standard second-order system is given by,
C ( s) 2n
 2 …(ii)
R ( s) s  2 n s  n2
where,   damping ratio,
n  natural angular frequency
On comparing equation (i) and equation (ii),
ωn  5 rad/sec and 2ξ ωn  6
ξ  0.6
Maximum peak overshoot is given by,

1 2
MPO  e
0.6

MPO  e 10.62
 0.0947
MPO  9.47 %  10%
Hence, the correct option is (B).
SOL.33 (A)
The given block diagram is shown in figure.
3 15
u (t ) y (t )
s + 15 s +1

The open loop transfer function is given by,

45
G(s)  [Type-0 system]
( s  1) ( s  15)
For step input (position input) steady state error is given by,
1
ess 
1 K p
where, K p is position error coefficient.
The position error coefficient is given by,
K p  lim G( s)
s 0
Control Systems 23 Time Response Analysis
45
K p  lim 3
s  0 ( s  1) ( s  15)

1
ess   0.25
1 3
% steady state error = 25% [Refer Table 3.1]
Hence, the correct option is (A).
SOL.34 (D)
The characteristic equation is given by,
1  G (s)  0
45
1 0
( s  1) ( s  15)
45
1 0
s  16 s  15
2

s 2  16s  60  0
( s  6) ( s  10)  0
s   6 ,  10
Hence, the correct option is (D).
SOL.35 (B)
Given :
Kp K
R( s) +– C (s)
1 + st

Kp K
G (s)  , H ( s)  1
1  sτ
1
For unit step input, R( s ) 
s
Steady state error is given by,
sR ( s )
ess  lim sE ( s )  lim
s 0 s 0 1  G ( s )

1
s
s 1
ess  lim 
s 0 K K 1 K pK
1 p
1  sτ
Hence, the correct option is (B).
SOL.36 (D)
Given : ωn  5 and ξ = 0.7
K
R( s) C (s)
s (s + 2 )

1 + sP
Control Systems 24 Time Response Analysis
K
G (s)  , H ( s )  1  sP
s ( s  2)
The characteristic equation is given by,
1  G ( s) H ( s)  0
K (1  sP)
1 0
s ( s  2)
s 2  2s  K  KP s  0
s 2  (2  KP ) s  K  0 …(i)
Standard characteristic equation for second order system is given by,
s 2  2 n s  n2  0 …(ii)
On comparing equation (i) and equation (ii),
ω 2n  K  25
2 ξ ωn  2  KP
2  ξ  5  2  25  P
2  0.7  5  2  25P
P  0.2
Hence, the correct option is (D).
SOL.37 (A)
Given : r (t )  u (t )
Taking Laplace transform of r (t ) ,
1
R( s) 
s
The given feedback system is shown below.
KI
1
s
Z(s)
0 R( s) w2
Kp C (s)
E (s) s + 2xws + w2
2

2
G ( s) 
s 2  2  s  2
Z (s) K I
 …(i)
E (s) s
E (s) æ KI ö
R (s) ç K p + s ÷ G (s) C (s)
è ø

 K 
G '( s )   K p  I  G ( s)
 s 
Control Systems 25 Time Response Analysis
Closed-loop transfer function for negative feedback is given by,
C ( s) G '( s )
T ( s)  
R( s ) 1  G '( s )
 K 
 K p  I  G ( s)
C (s)
 
s 
R( s)  K 
1   K p  I  G(s)
 s 
From the figure, the error signal is given by,
E (s)  R (s)  C (s)
  KI  
  Kp   G (s) 
E ( s )  R ( s ) 1  
s 

 1   K  K I  G (s) 
  p  
 s 
 
1 1 
E (s)   
s  K  
1  K p  I  G ( s) 
  s  
From equation (i),
KI K  s 
Z ( s)  E ( s )  2I  
s s  s  ( sK p  K I ) G ( s) 

 
 
K 1
Z ( s)  I  
s   2 
 s  ( sK p  K I )  2 2 
  s  2 s    

KI  s 2  2  s  2 
Z ( s)   2 
 s ( s  2   s   )   ( sK p  K I ) 
2 2
s

Steady state value can be calculated as,

zss  lim z (t )  lim s Z ( s )
t  s 0

2
zss  K I  1
2  K I
Hence, the correct option is (A).
SOL.38 (D)
Transfer function for second-order system is given by,
C ( s) 2n
 …(i)
R ( s ) s 2  2 n s  n2
where,   Damping ratio,
n  Natural angular frequency
Control Systems 26 Time Response Analysis
25
For P  …(ii)
s  25
2

Comparing equation (i) and (ii),

ξ  0 (undamped)
36
For Q  …(iii)
s  20 s  36
2

On comparing equation (i) and (iii),

n  6, 2n  20  ξ  1.67  1   (over damped)
36
For R  …(iv)
s  12 s  36
2

Comparing equation (i) and (iv),

n  6, 2n  12  ξ  1 (critically damped)
49
For S  …(v)
s  7 s  49
2

Comparing equation (i) and (v),

n  7, 2n  7  ξ  0.5  1 (under damped)
Hence, the correct option is (D).
SOL.39 (A)
Given : A two-loop position control system is shown below,
Motor
1
R( s) Y (s)
s ( s + 1)

ks
Tacho-generator

1
R( s) Y (s)
s + s (k + 1)
2

Y (s) 1
 2 …(i)
R ( s ) s  ( k  1) s  1
Transfer function for standard second-order system is given by,
C ( s) 2n
 2 …(ii)
R ( s) s  2 n s  2n
where,   damping ratio,
n  natural angular frequency
On comparing equation (i) and equation (ii),
ωn  1 and 2 ξ ωn  k  1
k 1
ξ …(iii)
2
 Control Systems 27 Time Response Analysis
Maximum peak overshoot is given by,

12
MPO  e
Peak overshoots depends on damping factor ξ .
From equation (iii), ξ is proportional to gain k.
Therefore, gain k of the Tacho-generator influences mainly by peak overshoot.
Hence, the correct option is (A).
SOL.40 (C)
4
Given : G ( s )  and H ( s )  1
s ( s  4)
The closed-loop transfer function for negative unity feedback system is given by,
C (s) G (s)
  
R(s) 1  G (s) H (s)
4
Y (s) s ( s  4)
      
U (s) 1  4
s ( s  4)
Y (s) 4
 2 …(i)
U (s) s  4s  4
Transfer function of second-order system is given by,
C ( s) 2n
 2 …(ii)
R( s) s  2 n s  2n
where,   Damping ratio,
n  Natural angular frequency
On comparing equation (i) and (ii),
2n  4  n  2 rad/sec.

Hence, the correct option is (C).

SOL.41 (B)
Given :
(i) The damping ratio   0.5 & undamped natural frequency n  10 rad/sec.
(ii) Steady state value of the output to a unit step input is 1.02.
Transfer function for standard second-order system is given by,
C (s) K n 2
T (s)   2
R( s) s  2n s  n2
where,   damping ratio,
n  natural angular frequency,
K  DC gain of the system
Control Systems 28 Time Response Analysis

n  0.5  10
n  5 …(i)
2n  100 …(ii)
For unit step input, output is given by,
1 K n 2
C (s)   2
s s  2n s  n 2
Steady state output is given by,
C ()  lim C (t )  lim sC ( s )
t  s 0

K 2n
C ()  lim  1.02
s 0 s 2  2n s  n2
K  1.02 …(iii)
From equation (i), (ii) and (iii), transfer function of the system is,
1.02  100 102
T ( s)   2
s  2  5s  100 s  10s  100
2

Hence the correct option is (B).

 Key Point
You can directly select the answer from options by checking from the given data.

SOL.42 (C)
Maximum peak overshoot is given by,


12
MPO  e …(i)
Standard characteristic equation of a second order system is given by,
s 2  2n s  n2  0 …(ii)
where,   damping ratio
n  natural angular frequency
For option (A) :
Characteristic equation is,
s 2  10s  100  0 …(iii)
Comparing equation (ii) and (iii),
n  10 and 2n  10
  0.5
For option (B) :
Characteristic equation is,
s 2  15s  100  0 …(iv)
Comparing equation (ii) and (iv),
n  10 and 2n  15
  0.75
Control Systems 29 Time Response Analysis
For option (C) :
Characteristic equation is,
s 2  5s  100  0 …(v)
Comparing equation (ii) and (v),
n  10 and 2n  5
  0.25
For option (D) :
Characteristic equation is,
s 2  20s  100  0 …(vi)
Comparing equation (ii) and (vi),
n  10 and 2n  20
 1
For critically damped the system, there is no peak overshoot.
Note : Lowest value of damping ratio will provide maximum peak value of overshoot.
From all the options minimum value of damping ratio is 0.25. Therefore, MPO is maximum for   0.25
Hence, the correct option is (C).
 Key Point
MPO
1
1
MPO µ
x

x
0
Fig. MPO Vs 

SOL.43 (0.3375)
Given :   0.8, n  4 rad/sec
+ k
R(s) C(s)
s ( s + 1)
–

(1 + k p s )

k
G ( s) 
s ( s  1)
H ( s )  (1  k p s )
Transfer function of a closed loop system is given by,
k
C ( s) s ( s  1)

R( s) k (1  k p s )
1
s ( s  1)
Control Systems 30 Time Response Analysis
C (s) k

R( s ) s ( s  1)  k (1  k p s )
C (s) k
 2 …(i)
R( s ) s  (1  kk p ) s  k
Transfer function of standard second order system is given by,
C ( s) 2n
 2 …(ii)
R( s) s  2n s  2n
On comparing equations (i) and (ii),
2n  k
k  (4) 2  16
2n  1  kk p
2  0.8  4  1  16 k p
5.4
kp   0.3375
16
Hence, the value of k p is 0.3375.
SOL.44 (0.67)
1
Given : G ( s ) H ( s )   H ( s)  1
( s  1)( s  2)
For unit step input (position input) steady state error is given by,
1
ess 
1 Kp
where, K p is position error coefficient
The position error coefficient is given by,
K p  lim G ( s ) H ( s )
s 0

1 1
K p  lim 
s 0 ( s  1)( s  2) 2
1 1
ess    0.67
1 1.5
1
2
Hence, steady state error in the output of the system for a unit-step input is 0.67.
SOL.45 (B)
Given differential equation,
d2 d
4 2
C (t )  8 C (t )  16C (t )  16u (t )
dt dt
Taking laplace transform of above equation,
4 s 2C ( s )  8C ( s )  16C ( s )  16/s
Control Systems 31 Time Response Analysis
16
C ( s )[4 s 2  8s  16] 
s
16
C (s) 
s ( s  2 s  4)
2

C (s)
Transfer function 
R(s)
For unit step input r (t )  u (t ),
Taking laplace transform of r (t ),
1
R(s) 
s
u
C ( s ) s ( s  2s  4)
2

R( s) 1
s
C (s) 4
 2 … (i)
R(s) s  2s  4
Transfer function for standard second-order system is given by,
C (s) 5n
 … (ii)
R ( s ) s 2  2n s  2n
On comparing equation (i) and (ii)
n  2 rad/sec and 2n  2
1
  0.5
2
Hence, the correct option is (B).
SOL.46 (B)
20
Given open loop transfer function 
s ( s  5)
Closed loop transfer function is given by,
20
s ( s  5)
T (s) 
20
1
s ( s  5)
20
T (s)  …(i)
s  5s  20
2

Transfer function for standard second-order system is given by,

C ( s) 2n
 …(ii)
R( s) s 2  2n s  2n
On comparing equation (i) and (ii)
2n  5
Control Systems 32 Time Response Analysis

n  2.5
For 2% error band settling time is given by,
4 4
ts    1.60 sec
n 2.5
Hence, the correct option is (B).
SOL.47 (D)
Note : The steady state error for type system is zero when it is excited by unit-step input. Hence there is
a mistake in the question.
Modified Question :
10
A control system has G ( s )  and H ( s)  K . What is the value of K error for unit for which the
s ( s  2)
steady-state error for unit ramp input is less than 50 %.
For unit ramp input steady state error is given by,
1
ess 
KV
Where KV  lim sG ( s ) H ( s )
s 0

10 K
KV  lim s   2K
s 0 s ( s  5)
1
ess 
2K
1 50

2 K 100
100
K
100
K 1
Hence, the correct option is (D).
SOL.48 (0.4)
Given : Closed loop transfer function
C (s) 2

R ( s ) ( s  10)( s  100)

Pole location : s1  10 (significant pole) and s2  100 (insignificant pole)

s2
Since,  10
s1
Hence, dominant pole concept is applicable
C ( s) 2 0.02
 
R ( s ) 100( s  10) s  10
Control Systems 33 Time Response Analysis
For first order system, time required to reach 2 % of steady-state value is given by,
t s  4

ts  4   0.4 sec
10
Hence, time required to reach 2 % of steady-state value is 0.4 sec.
SOL.49 (B)
Given block diagram is shown below
25
R( s) C (s)
s ( s + 6)

Closer loop transfer function is given by,

25
C ( s) s ( s  6)

R ( s ) 1  25
s ( s  6)
C (s) 25
 2 …(i)
R ( s ) s  6 s  25
C ( s) 2n
 …(ii)
R( s) s 2  2n s  2n
Where,   damping ratio
n  natural angular frequency
On comparing equation (i) and (ii)
2n  6
n  3
For 2% error band settling time is given by,
4
ts 
n
4
ts   1.33 sec
3
Hence, the correct option is (B).
SOL.50 (C)
1
G (s) 
( s  1) 2
C (s) G (s)
Closed loop transfer function 
R(s) 1  G (s)
1
C ( s) s 2  2s  1

R( s) 11
s  2s  2
2
Control Systems 34 Time Response Analysis

C ( s) 2n
 …(ii)
R ( s ) s 2  2n  2n
Comparing equation (i) and (ii)
n  2 rad/sec and 2n  2
1
  0.707
2
Hence, the correct option is (C).
SOL.51 (C)
Given block diagram is shown below,
25
R( s)
s ( s + 6)

Closed loop transfer function is given by,

15
C ( s) s ( s  6)

R ( s ) 1  25
s ( s  6)
C (s) 25
 2 … (i)
R ( s ) s  6 s  25
Comparing equation (i) and (ii)
n  5 rad/sec and 2n  6
3
  0.6
5

Peak time, tp 

 
tp    0.79 sec
n 1   2 5 1  0.62
Hence, the correct option is (C).
SOL.52 (C)
Given : C (t )  Ae 6t sin (8t  ) …(i)
Transient response of standard second order under damped system is given by,
C (t )  ke nt sin (d t  ) …(ii)
On comparing equation (i) and (ii)
n  6 and d  8

n 1   2  8
2n (1   2 )  64
Control Systems 35 Time Response Analysis

2n  n2  2  64
2n  62  64
2n  36  64  100
n  10 rad/sec
Hence, the correct option is (C).


Practice Solutions :
Sol.1 (C)
Given : Characteristic equation,
2 s 4  s 3  3s 2  5s  10  0
Routh Tabulation :
s4 2 3 10
s3 1 5 0
s2 -7 10 0
s1 6.428 0 0
s0 10 0 0
Since, there are two sign changes in the first column of Routh tabulation, then the equation has two
roots in the right-half of s-plane.
Hence, the correct option is (C).
Sol.2 (D)
The given signal flow graph for system is shown below,
1 K s -1 1
R( s) Y (s)

3
-1

Forward path gain : P1  Ks  1

Individual loop gain : L1  3s  1 , L2   Ks 1
Number of two non-touching loops : 0
Determinant :   1  ( L1  L2 )  1  3s 1  Ks 1
Path factor :
1 K s -1 1
R( s) Y (s)

3
-1

All the loops touch forward path.

1  1  (isolated loop gain)
Control Systems 2 Routh‐Hurwitz Stability

1  1  0  1
Using Mason’s gain formula, transfer function is given by,
C ( s) 1
  Pk  k
R( s)  k
So, the transfer function is
Y ( s ) P
 1 1
R( s) 
Y ( s) Ks 1 K
 1 1

R( s ) 1  3s  Ks s  3 K
Pole of this transfer function is,
s=3–K
For the system to be stable, all the roots must be in the left-half of s-plane
3 K  0  K  3
Hence, the correct option is (D).
Sol.3 (A)
Given : The open loop transfer function of a unity feedback system is,
K
G( s) 
s ( s  s  2) ( s  3)
2

The characteristic equation is given by,

1  G ( s)  0
K
1 0
s ( s  s  2) ( s  3)
2

s 4  4 s 3  5s 2  6 s  K  0
Routh Tabulation :
s4   1  5 K 
s3   4 6  0 
s2   3.5 K  0 
21  4 K
s1   0  0 
3.5  
s0   K  0  0 

For the system to be stable, all the roots must be in the left-half of s-plane, thus all the coefficients in
the first column of Routh's tabulation must have the same sign. Therefore, the coefficient of first
column of the Routh’s table should be positive.
This leads to the following conditions :
21  4 K
(i) 0
3.5
(ii) K  0
Therefore, the condition of K for the system to be stable is,
21
0 K 
4
Hence, the correct option is (A).
Control Systems 3 Routh‐Hurwitz Stability
Sol.4 (B)
Given : The characteristic equation is,
P( s)  s5  s 4  2s 3  2s 2  3s  15  0
Routh Tabulation :
s5   1  2 3 
s4   1 2  15 
s3   0 – 12  0 

Since, the first element of the s 3 row is zero, the all elements in the s 2 row would be infinite. To
overcome this difficulty, we replace the zero in the s 3 row with a small positive variable ε and then
proceed with the tabulation. Starting with the s 3 row, the results are as follows,
s3     – 12  0 
2  12
s2     15  0 

15
12 
1
s    2  12    0  0 
 
  
s0   15  0  0 

From the Routh table,

2  12
lim 
0 
 
 15 
and lim  12    12
0
  2  12  
 
    
Since, there are two sign changes in the first column of Routh's tabulation, then the equation has two
roots in the right-half of s-plane.
Hence, the correct option is (B).
Sol.5 (C)
K ( s  1)
Given : G ( s ) 
s (1  Ts ) (1  2 s )
Where K  0 , T  0
. Method 1 :
Characteristic equation for unity feedback is given by,
1  G (s)  0
K ( s  1)
1 0
s (1  Ts ) (1  2 s )
s (1  Ts ) (1  2 s )  K ( s  1)  0
s(1  2s  Ts  2Ts 2 )  K ( s  1)  0
2Ts3  (2  T )s 2  s  K (s  1)  0
2Ts3  (2  T )s 2  (1  K )s  K  0
Control Systems 4 Routh‐Hurwitz Stability
Routh Tabulation :
s3   2T 1 + K  0 
s2   2 + T K  0 
(2  T )(1  K )  2TK
s1     0  0 
2T
s0   K  0  0 
For the system to be stable, all the roots must be in the left-half of s-plane, thus all the coefficients in
the first column of Routh's tabulation must have the same sign. Therefore, the coefficient of first
column of the Routh’s table should be positive.
This leads to the following conditions :
(i) K  0 …(i)
(2  T )(1  K )  2TK
(ii) 0
2 T
(2  T )(1  K )  2TK
(1  K ) 2T

K 2 T
1 2T
 1
K 2T
1 2T  2  T

K 2T
1 T 2

K T 2
T 2
K …(ii)
T 2
On combining equation (i) and (ii),
T 2
0K 
T 2
Hence, the correct option is (C).
. Method 2 :
The characteristic equation is,
EP

2Ts 3 + (2 + T ) s 2 + (1 + K ) s + K = 0

IP
Internal product (IP) = (2  T )(1  K )
External product (EP) = 2T  K
EP will exist, if 2TK  0
For the system to be stable, IP > EP
Hence, (2  T )(1  K ) > 2TK
1 K 2T

K 2 T
1 2T
 1
K 2 T
 Control Systems 5 Routh‐Hurwitz Stability
1 T 2

K T 2
T 2
K
T 2
T 2
Hence, range of K for the system to be stable is 0  K  .
T 2
Hence, the correct option is (C).
Sol.6 (C)
Given : Characteristic equation,
s 4  2 s 3  3s 2  2 s  K  0
Routh tabulation :

s4   1  3  K 

s3   2  2  0 

s2   2  K  0 

s1   2‐K  0  0 

s0   K  0  0 

For oscillations, there must be at least one row of zeros except last row,
2 K  0
K 2
Since, a row of zeros appears, we form auxiliary equation using coefficient of s 2 row,
2s 2  K  0
 K 2
s2    1
2 2
s   j1
s   j   j1
  1 rad/sec
Hence, the correct option is (C).
Sol.7 (A)
K
Given : G ( s ) 
s (1  sT1 )(1  sT2 )
. Method 1 :
Characteristic equation for unity feedback is given by,
1  G (s)  0
K
1 0
s (1  sT1 )(1  sT2 )
s (1  sT1 )(1  sT2 )  K  0
s (1  s 2T1T2  ST1  ST2 )  K  0
T1T2 s 3  (T1  T2 ) s 2  s  K  0
Control Systems 6 Routh‐Hurwitz Stability
Routh tabulation :

s3   T1T2   1 

s2   T1  T2   K 

T1  T2  KTT1 2
s1   0 
T1  T2
s0   K  0 

For the system to be stable, all the roots must be in the left-half of s-plane, thus all the coefficients in
the first column of Routh's tabulation must have the same sign. Therefore, the coefficient of first
column of the Routh’s table should be positive.
This leads to the following conditions :
(i) K  a
T  T  KT1T2
(ii) 1 2 0
T1  T2
KTT 1 2
1
T1  T2
T1  T2
K
T1T2
Hence, the correct option is (A).
. Method 2 :
The characteristic equation is,
T1T2 s 3  (T1  T2 ) s 2  s  K  0
Internal product ( IP )  T1  T2
External product ( EP )  (T1T2 ) ( K )
For the system to be stable, IP  EP
Hence, T1  T2  T1T2 K
T1  T2
K
T1T2
Hence, the correct option is (A).
Sol.8 (D)
K
Given : G ( s )  2
s (s  a)
The characteristic equation for unity secondary is given by,
1  G (s)  0
K
1 0
s (s  a)
2

s 2 (s  a)  K  0
s 3  as 2  K  0
s 3  as 2  0  s  K  0
Control Systems 7 Routh‐Hurwitz Stability

Since, s1 term is missing in the characteristic equation hence closed loop system is unstable for all the
values of K.
Hence, the correct option is (D).
Sol.9 (B)
K ( s  3)
Given : Open loop transfer function, G ( s ) 
s ( s  1)
Pole-zero plot is given by,
jw

+
-3 0 1

Since, one pole in the right half of s-plane

Hence, open loop transfer is unstable
The characteristic equation for unity negative feedback is given by,
1  G (s)  0
K ( s  3)
1 0
s ( s  1)
s ( s  1)  K ( s  3)  0
s 2  s  sK  3K  0
s 2  ( K  1) s  3 K  0
Routh Tabulation:
s2   1  3K 

s1   K 1 0 

s0   3K   0 

Since given k>1 hence all the coefficient of first column of the Routh table are positive
Hence, closed loop system is stable
Sol.10 (A)
Given : Characteristic equation
s 6  2 s 5  3s 4  4 s 3  3s 2  2 s  1  0
Routh tabulation:
s6   1  3  3  1 

s5   2  4  2  0 

s4   1  2  1  0 

s3   0  0  0  0 

For oscillations, there must be at least one row of zeros except last Row,
For the auxiliary equation using the coefficients of s 4 row,
s 4  2s 2  1  0
Control Systems 8 Routh‐Hurwitz Stability

Let s2  X
Then, X 2  2X 1  0
( X  1) 2  0
X  1,  1
Hence, s 2  1
Putting s  j,
( j) 2  1  2  1
  1 radian/sec
Hence, the correct option is (A).
Sol.11 (C)
 Key Point
To determine absolute stability of a system we must know number of right hand poles. Since, Routh
Hurwitz criterion gives exact number of right hand poles.
Hence, Routh Hurwitz criterion is used to determine absolute stability of the system.
Hence, the correct option is (C).
Sol.12 (A)
Given : Characteristic equation,
2 s 3  3s 2  4 s  6  0
Routh tabulation :
s3 2 4
s2 3 6
s1   0 0

A( s)  3s 2  6  0
6
s2   2
3
sj 2
Roots of auxiliary equation are the roots of characteristic equation. Hence s   j 2 are the roots of
characteristic equation.
2
3s 2 + 6 2 s 3 + 3 s 2 + 4 s + 6 s +1
3
- 2
2s + 4s

3s 2 + 6
-
3s 2 + 6
0
2
s 1  0
3
3
s  1.5
2
Hence, the correct option is (A).
Control Systems 9 Routh‐Hurwitz Stability
Sol.13 (A)
Given : Characteristic equation,
s 6  2 s 5  s 4  2 s 2  8s  8  0
Routh tabulation :
s6   1  1   2   8 

s5   2   0  8  0 

s4   1 2  8  0 

s3   4 8   0  0 

s2   4  8  0  0 

s1   0  0  0  0 

Since, a row of zeros appears, we form auxiliary equation using coefficient of s 4 row,
A( s)  4s 2  8
The derivative of A( s ) with respect to s is,
dA( s)
 8s
ds
From which the coefficient 4 replace the zero in the s1 row of the original tabulation. The remaining
portion of the Routh's tabulation is,
s1   8  0  0  0 

s0   8  0  0  0 

Since, there are two sign changes in the first column of Routh's tabulation, then the equation has two
roots in the right-half of s-plane.
Hence, the correct option is (A).
Sol.14 (D)
Given : Characteristic equation,
s 3  ks 2  5s  10  0
Routh tabulation :
s3   1  5 

s2   K  10 
5K  10
s1     0 
K
s0   10  0 
Given system is oscillating at frequency of  rad/sec and for oscillation, there must be at least one
row of zeros except last row.
5 K  10
0
K
K 2
Auxiliary equation is given by,
Ks 2  10  0
2 s 2  10  0
Control Systems 10 Routh‐Hurwitz Stability

s2  5  0
sj 5
 j   j 5
  5 rad/sec
Hence, the correct option is (D).
Sol.15 (C)
Given : Characteristic equation,
s 4  s3  2s 2  2s  3  0
Routh tabulation :
s4   1  2  3 

s3   1  2  0 

s2     3  0 
2 3
s1     0  0 

s0   3  0  0 

From the Routh table,

2  3
lim  
0 
Since, there are two sign changes in the first column of Routh's tabulation, then the equation has two
roots in the right-half of s-plane.
Hence, the correct option is (C).
Sol.16 (0.618)
The given figure is shown below.
( s + a)
R( s) +– C (s)
s 3 + (1 + a) s 2 + (a - 1) s + (1 - a)

The characteristic equation is given by,

1  G ( s)  0
( s  α)
1 0
s  (1  α) s  (α  1) s  (1  α)
3 2

s3  (1  α) s 2  (α  1)s  (1  α)  (s  α)  0
s3  (1  α) s 2  αs  1  0
Routh Tabulation :
s3   1  α 
s2   1 α   1 
α(1+α)  1
s1     0 
1 α
s0   1  0 
Control Systems 11 Routh‐Hurwitz Stability
For the system to be stable, all the roots must be in the left-half s-plane, thus all the coefficients in the
first column of Routh's tabulation must have the same sign. Therefore, first column of the Routh’s
table should be positive.
This leads to the following conditions :
(i) 1  α  0  α  1
α(1+α)  1
(ii)  0  α  0.618, 1.618
1 α
On combining the above two condition, the minimum value of  for the system to be stable is 0.618.
Hence, the value of α is 0.618.
Sol.17 (5)
K
Given : G ( s ) 
s ( s  2)( s 2  2 s  2)
The characteristic equation is given by,
1  G ( s)  0
K
1 0
s ( s  2)( s 2  2 s  2)
s(s  2)( s 2  2s  2)  K  0
s 4  4s3  6s 2  4s  K  0
Routh Tabulation :
s4   1  6 K 
s3   4 4  0 
s2   5 K  0 
20  4 K
s1     0  0 
5
s0   K  0  0 
For the system to be stable, all the roots must be in the left-half of s-plane, thus all the coefficients in
the first column of Routh's tabulation must have the same sign. Therefore, the coefficient of first
column of the Routh’s table should be positive.
20  4 K
0  K 5
5
Hence, the value of K is 5.
Sol.18 (2)
Given : Characteristic equation,
s 5  s 4  2s3  2s 2  4s  6  0
Routh tabulation :
s5 1 2 4
s4 1 2 6
s3   () 2 0
2
s2   2 6 0

6 2
1
s   2  0 0
2  2
s0   6 0 0
Control Systems 12 Routh‐Hurwitz Stability
From Routh table,
2
lim 2  
0 
6 2
And lim  2   2
0 2  2
Since, there are two sign changes in the first column of Routh tabulation hence the number of roots of
the equation which lie in the right half of s -plane is 2.
Sol.19 (C)
Given : Characteristic equation is,
2s 4  s3  3s 2  5s  7  0
Routh Tabulation :
s4   2  3 7 
3
s   1 5  0 
s2   – 7 7  0 
s1   6 0  0 
0
s   7  0  0 
Since, there are two sign changes in the first column of Routh's tabulation, then the equation has two
roots in the right-half of s-plane.
Hence, the correct option is (C).
Sol.20 (2)
Given : Characteristic equation,
2 s 4  s 3  3s 2  5s  10  0
Routh Tabulation :
s4 2 3 10
s3 1 5 0
s2 -7 10 0
s1 6.428 0 0
s0 10 0 0
Since, there are two sign changes in the first column of Routh tabulation, then the equation has two
roots in the right-half of s-plane.
The number of roots of the equation which lie in the right half of the s-plane is 2.
Sol.21 (A)
Given : Frequency of oscillation = 2 rad/sec
i.e. s   j   j 2
The partial Routh array of the characteristic equation of a system is given by,
s4 1 a 8

s3 3 12

The complete Routh table is,

s4 1 a 8

s3 3 12 0
Control Systems 13 Routh‐Hurwitz Stability

s2 a–4 8 0

12 (a  4)  24
s1 0 0
a4
s0 8 0 0

Given system is oscillating at frequency of 2 rad/sec and for oscillation, there must be at least one row
of zeros except last row.
12 (a  4)  24
Therefore, 0
a4
12(a  4)  24  0
    a4 2
    a6
Hence, the correct option is (A).
 Key Point
(i) ROZ occurs only for odd power of s.
(ii) There is no significance of ROZ in last row.
(iii) Purely imaginary roots of auxiliary equation gives frequency of oscillation.
(iv) Roots of auxiliary equation are roots of closed loop transfer function.
(v) If ROZ exists, then poles are symmetrical about the origin.
(vi) The roots of auxiliary equation (AE) are always symmetrical about y-axis.
(vii) The number of sign changes below the row of auxiliary equation gives number of Right Hand
Pole (RHP) of auxiliary equation and by symmetry, same number of poles exist in Left Hand
Pole (LHP).
(viii) Number of poles on imaginary axis = Order of AE – 2  Number of sign changes below AE.

Sol.22 (B)
Given : Characteristic equation is,
s3  4s 2  s  6  0
Routh Tabulation :
s3   1 1 
s2   – 4 6 
1
s   2.5   0 
0
s   6  0 
Since, there are two sign changes in the first column of Routh's tabulation, then the equation has two
roots in the right-half of s-plane.
Total roots = 3
Total number of left half root =1
Sol.23 (D)
The given block diagram is shown below.
s -1
U1
s+2

1
U2
s -1
Control Systems 14 Routh‐Hurwitz Stability
Applying superposition theorem :
(i) When U 2 (s)  0 , the transfer function is,
Y1 ( s )
U1 ( s ) U
2 ( s )0

s -1
U1 ( s ) Y1 ( s )
s+2

1
s -1
s 1
Y1 ( s) s2 s 1 s 1
  
U1 ( s) 1  s  1  1 s  2 1 s  3
s  2 s 1
Pole s   3 is in left half of s-plane.
Hence, it is stable for U1 .
(ii) When U1 ( s)  0 , the transfer function is,
Y2 ( s )
U 2 ( s) U ( s )0
1

- Y2 ( s ) s -1
U1 = 0
s+2

1
U 2 (s)
Y2 ( s ) s -1
Fig. (a)
s -1
-1
s+2

1
U 2 (s)
Y2 ( s ) s -1
Fig. (b)

U 2 (s) 1
Y2 ( s )
s -1

- ( s - 1)
s+2
Fig. (c)
1
U 2 (s) Y2 ( s )
s -1

s -1
s+2
Fig. (d)
Control Systems 15 Routh‐Hurwitz Stability

1 1
Y2 ( s ) s 1 ( s  1)
 
U 2 ( s) 1  1  s  1 1  1
s 1 s  2 ( s  2)
Y2 ( s ) s2

U 2 ( s ) ( s  1) ( s  3)
Poles are 1 and – 3.
Since, one pole is in right half of s-plane.
Therefore, it is unstable for U 2 .
Hence, the correct option is (D).
Sol.24 (A)
 K  1
Given : G ( s )   K P  I 
 s  s ( s  2)
( sK P  K I )
G( s) 
(s3  2s2 )

æ K ö 1
r G (s) = ç K P + I ÷ y
è s ø s ( s + 2)

The characteristic equation is given by,

1  G( s)  0
s 3  2s 2  K p s  K I  0
Routh Tabulation :
s3   1 KP  
s2   2 KI  
(2 K P  K I )
s1     0 
2
s0   KI   0 

For the system to be stable, all the roots must be in the left-half of s-plane, thus all the coefficients in
the first column of Routh's tabulation must have the same sign. Therefore, the coefficient of first
column of the Routh’s table should be positive.
This leads to the following conditions :
(i) K I  0
 2K  KI 
(ii)  P 0
 2 
KI
KP 
2
On combining above two conditions,
K
KP  I  0
2
Hence, the correct option is (A).
Control Systems 16 Routh‐Hurwitz Stability
Sol.25 (2)
Given : The characteristic equation is,
s 5  2 s 4  3s 3  6 s 2  4 s  8  0
Routh Tabulation :
s5   1  3  – 4 
s4   2  6 – 8 
s3   0  0  0 

Since, a row of zeros appears, we form the auxiliary equation using the coefficients of s 4 row,
A( s)  2s 4  6s 2  8
The derivative of A(s) with respect to s is,
d
 2 s 4  6s 2  8  8s 3  12 s
ds
From which the coefficients 8 and 12 replace the zeros in the s 3 row of the original tabulation. The
remaining portion of the Routh's tabulation is,
s3   8 12  0 
s2   3 – 8  0 
s1   33.33  0  0 
s0   – 8  0  0 

Let x  s 2 , then from auxiliary equation,

2 x2  6x  8  0
x  1,  4
s 2  1,  4  s  1,  j 2
Number of roots lying in RHS is one (i.e. s  1) .
Number of roots lying in the imaginary axis = 2 (i.e. s   j 2 )
Thus, number of roots in LHS of s-plane
 5  (2  1)  2
Hence, the number of roots that lie strictly in the left half of s-plane is 2.
Sol.26 (D)
From the Routh table, Routh array can be written as,
Routh Tabulation :
s3 1 (2K + 3)
s2 2K 4

1 4K 2  6K  4
s 0
2K
s0 4 0

For the system to be stable, all the roots must be in the left-half of s-plane, thus all the coefficients in
the first column of Routh's tabulation must have the same sign. Therefore, the coefficient of first
column of the Routh’s table should be positive.
Control Systems 17 Routh‐Hurwitz Stability
This leads to the following conditions :
(i) 2 K  0  K > 0 …(i)
4 K 2  6K  4
(ii) 0
2K
2 K 2  3K  2  0
(2 K  1) ( K  2)  0
1
K or K   2 …(ii)
2
On combining equation (i) and (ii),
1
K
2
Thus, from options, 0.5  K   .
Hence, the correct option is (D).
Sol.27 (0.75)
Given : The open loop transfer function of unity feedback system is given by,
2( s  1)
G( s) 
s  Ks 2  2 s  1
3

. Method 1 :
The characteristic equation is given by,
1  G( s)  0
s 3  Ks 2  2 s  1  2 s  2  0
s 3  Ks 2  4 s  3  0
Routh tabulation :
s3 1 4
s2 K 3
4K  3
s1 0  Row of zeros
K
s0 3 0

For the system to be marginal stable, row of zeros should appears.

This leads to the following conditions :
4K  3 3
0  K  0.75
K 4
Hence, the value of K is 0.75.
. Method 2 :
The system is oscillating hence this is a marginally stable system.
For marginally stable system :
IP

s 3 + Ks 2 + 4 s + 3 = 0

EP
Control Systems 18 Routh‐Hurwitz Stability
Internal product = 4 K ,
External product = 3
Internal product = External product
3
4K  3  K   0.75
4
Hence, the value of K is 0.75.
 Key Point
For a third order system only,
IP > EP : system is stable.
IP < EP : system is unstable.
IP = EP : system is marginal stable.

Sol.28 (D)
Given : Characteristic equation is,
s3  Ks 2  ( K  2)s  3  0
. Method 1 :
Routh Tabulation :
s3   1 K + 2 
s2   K 3 
K ( K  2)  3
s1   0 
K
s0   3  0 
For the system to be stable, all the roots must be in the left-half of s-plane, thus all the coefficients in
the first column of Routh's tabulation must have the same sign. Therefore, the coefficient of first
column of the Routh’s table should be positive.
This leads to the following conditions :
(i) K  0 …(i)
(ii) K ( K  2)  3  0
K 2  2K  3  0
( K  3) ( K  1)  0
K   3 and K  1 …(ii)
On combining equation (i) and (ii),
K 1
Hence, the correct option is (D).
. Method 2 :
The characteristic equation is,
EP

s 3 + Ks 2 + ( K + 2) s + 3 = 0

IP
Internal product (IP) = K ( K  2)
External product (EP) = 3 1  3
Control Systems 19 Routh‐Hurwitz Stability
For the system to be stable, IP > EP
K ( K  2)  3
K 2  2K  3  0
( K  3) ( K  1)  0
K   3 and K  1
Therefore, range of K for stability is K  1 .
Hence, the correct option is (D).
Sol.29 (A)
Given : Characteristic equation is,
s 3  3s 2  2 s  K  0
. Method 1 :
Routh Tabulation :
s3   1 2 
s2   3 K 
6K
s1   0 
3  
s0   K  0 
For the system to be stable, all the roots must be in the left-half of s-plane, thus all the coefficients in
the first column of Routh's tabulation must have the same sign. Therefore, the coefficient of first
column of the Routh’s table should be positive.
This leads to the following conditions :
6 K
(i) K  0 (ii) 0
3
K 6
Therefore, the range of K for which all the roots in the left half of s-plane is,
0 K 6
Hence, the correct option is (A).
. Method 2 :
The characteristic equation is,
EP

s 3 + 3s 2 + 2 s + K = 0

IP
Internal product (IP) = 3  2  6
External product (EP) = K 1  K
EP will exist, if K  0
For the system to be stable, IP > EP
6K  K 6
Therefore, the range of K for which all the roots in the left half of s-plane is,
0 K 6
Hence, the correct option is (A).
 Control Systems 20 Routh‐Hurwitz Stability
Sol.30 (D)
Given : The given feedback system is shown below,
+ s+a
R(s) C(s)
s 3 + 2a s 2 + a s + 1
–

The characteristic equation is given by,

1  G (s) H (s)  0
( s  )
1  0 
s  2s 2  s  1
3

      s 3  2 s 2   s  1  ( s   )  0  
s 3  2 s 2  (  1) s    1  0  
Routh Tabulation :
s3 1 (  1)
s2 2 (  1)
2(  1)  (  1)
s1 0
2
s0 (  1) 0
For the system to be stable, all the roots must be in the left-half of s-plane, thus all the coefficients in
the first column of Routh's tabulation must have the same sign. Therefore, the coefficient of first
column of the Routh’s table should be positive.
This leads to the following conditions :
2(  1)  (  1)
(i) 0    0.5
2
(ii)   1  0    1
Hence, the correct option is (D).
Sol.31 (9)
1 s
Given : G ( s )  , H (s) 
( s  1)( s  2) s
. Method 1 :
The characteristic equation is given by,
1  G(s) H (s)  0
( s  )
1 0
s ( s  1) ( s  2)
s 3  3s 2  2 s  s    0
s 3  3s 2  3s    0
Routh Tabulation :
s3 1 3 0

s2 3  0
9
s1 0 0
3
s0  0 0
Control Systems 21 Routh‐Hurwitz Stability
For marginally stable system, there exist one row of zeros.
This leads to the following conditions :
9
0
3
9
Hence, the value of  is 9.

. Method 2 :
The system has poles on the imaginary axis, so system will oscillate hence this is marginally stable
system.
For marginally stable system :
EP

s 3 + 3s 2 + 3s + a = 0

IP
Internal product (IP) = 3  3  9
External product (EP) = 1   
Internal product = External product
9
Hence, the value of  is 9.
Sol.32 (D)
Given : Characteristic equation,
s 4  4 s 3  4 s 2  3s  K  0
Routh tabulation :
s4 1 4 K

s3   4 3 0

s2 3.25 K 0

9.75  4 K
s1 0 0
3.25
s0 K 0 0

For marginally stable system, there exist one row of zeros.

This leads to the following condition :
9.75  4 K
0
3.25
3 13
K
4 4
39
K
16
Hence, the correct option is (D).
Sol.33 (D)
Given : Characteristic equation,
s 4  8s 3  24 s 2  32 s  K  0
Control Systems 22 Routh‐Hurwitz Stability
Routh tabulation :
s4 1 24 K

s3   8 32 0

s2 20 K 0

640  8K
s1 0 0
20
s0 K 0 0

For the system to be stable, all the roots must be in the left-half of s-plane, thus all the coefficients in
the first column of Routh's tabulation must have the same sign. Therefore, the coefficient of first
column of the Routh’s table should be positive.
Therefore, for unstability,
640  8 K
0
20
8K
 32
20
K  80
Hence, the correct option is (D).
Sol.34 (B)
Given Routh table is shown below,
Row I II III IV V
Sign + + – + –
Since there are is sign change in the first column of Routh table
Hence, the system has three roots in the right-half of s-plane
Total number of roots  5
Number of roots in the left-half of s-plane  5  3  2
Since three roots in right-half of s-plane
Hence, the system is unstable
Hence, the correct option is (B).
Sol.35 (B)
Given polynomial equation,
2 s 4  s 3  2 s 2  5s  10  0
Routh Tabulation :
s4 2 2 10

s3   1 5 0

s2 8 10 0

s1 6.25 0 0

s0 10 0 0

Since, there are two sign changes in the first column of Routh's tabulation, then the equation has two
roots in the right-half of s-plane.
Hence, the correct option is (B).
Control Systems 23 Routh‐Hurwitz Stability
Sol.36 (D)
Given : Open-loop transfer function
10
G (s) H (s) 
s ( s  2)( s  K )
Characteristic equation is given by,
1  GH ( s )  0
10
1 0
s ( s  2)( s  K )
s (s 2  2K  2s  Ks)  10  0
s3 (2  K ) s 2  2 Ks  10  0
Routh tabulation :
s3   1 2K

s2 2 K 10
2 K (2  K )  10
s1 0
(2  K )
s0 10 0

For the system to be stable, all the roots must be in the left-half s-plane, thus all the coefficients in the
first column of Routh's tabulation must have the same sign. Therefore, first column of the Routh’s
table should be positive.
This leads to the following condition :
2 K (2  K )  10
0
2 K
5
K
2 K
5  2K  K 2
K 2  2K  5  0
2  4  20
K
2
2  24
K
2
K  1  6
K  1.45
Hence, the correct option is (D).
Sol.37 (A)
Given : Closed loop transfer function,
C (s) s2
 3
R ( s ) s  8s  19 s  12
2

Characteristic equation is given by,

s 3  8s 2  19s  12  0
Control Systems 24 Routh‐Hurwitz Stability
Routh tabulation :
s3   1 19

s2 8 12
35
s1 0
2
s0 12 0

Since, there are no sign changes in the first column of Routh's tabulation, then the equation has no
roots in the right-half of s-plane.
Therefore, system is stable.
Hence, the correct option is (A).
Sol.38 (C)
Given : Characteristic equation,
R( s)  s5  2s 4  2s3  4s 2  11s  10
Routh tabulation :
s5   1 2 11
4
s   2 4 10

s3  6 0
4  12
s2 10 0

s1 6 0 0

s0   10  0 0

4  12
lim  
0 
Since, there are two sign changes in the first column of Routh's tabulation, then the equation has two
roots in the right-half of s-plane.
Hence, the correct option is (C).
Sol.39 (C)
Given : Characteristic equation,
s 4  2 s 3  11s 2  18s  18  0
Routh tabulation :
s4   1 11 18

s3 2 18 0

s2 2 18 0

s1 0 0 0

Since, a row of zeros appears, we form the auxiliary equation using the coefficients of s 2 row,
A( s)  2s 2  18
The derivative of A(s) with respect to s is,
dA( s)
 4s
ds
Control Systems 25 Routh‐Hurwitz Stability

From which the coefficient 4 replace the zero in the s1 row of the original tabulation. The remaining
portion of the Routh's tabulation is,
s1 4 0

s0 18 0

The roots of auxiliary equation (AE) are always symmetrical about y(imaginary)-axis
Since in this question auxiliary equation is a polynomial of order 2
Thus, two poles lie symmetrically on the imaginary axis of the s-plane
Hence, the correct option is (C).

Practice Solutions :
Sol.1 (0)
Given : Open loop poles are at  2  j 2 and 0.
Single zero at  4  j 0
. Method 1 :
Figure to calculate angle of departure is shown below.
(- 2 + 2 j ) jw Poles
P2 Zeros
+2 j

f p1
f z1
s
-4 -2 0 P1
Z1
f p2
-2 j
P3
(- 2 - 2 j )

Angle of departure :
The angle of departure from an open loop complex pole is given by (for K  0)
d  1800  (  p    z )
where,   z : Sum of all the angles subtended by zeros.
  p : Sum of all the angles subtended by remaining poles.
At pole (2  2 j ) ,
2
 p1  1800  tan 1    1350
2
 p 2  900

2
 z1  tan 1    450
2
 p   p1   p 2  1350  900  2250
Control Systems 2 Root Locus

 z   z1  450
d  1800  (2250  450 )  00
At pole (2  2 j ) , d   00  00
Hence, the angle of departure is 00 .
. Method 2 :
Open loop transfer function is,
K ( s  4)
G (s) H (s) 
s ( s  2  j 2) ( s  2  j 2)
Angle of departure is given by,
d  1800  G ( s ) H ( s ) s  ve imaginary complex pole

K (2  j 2  4)
G ( s ) H ( s ) s 2 j 2 
(2  j 2)(2  j 2  2  j 2)( 2  j 2  2  j 2)
K (2  j 2) K (1  j )
G ( s ) H ( s ) s 2 j 2  
(2  j 2)( j 4) (1  j )( j 4)
G ( s ) H ( s ) s 2 j 2  tan 1 (1)  1800  tan 1 (1)  900

G ( s ) H ( s ) s 2 j 2  450  1800  450  900

G ( s ) H ( s ) s 2 j 2  1800

d  1800  1800  00
At pole (2  2 j ) , d   00  00
Hence, the angle of departure is 00 .
Sol.2 (D)
In a transfer function,
(i) Numerator gives zeros (m) of the system.
(ii) Denominator gives poles (n) of the system.
Difference between poles and zeros (n  m) gives number of asymptotes.
Hence, the correct option is (D).
Sol.3 (B)
Given : The root locus plot of the unity feedback system is shown in figure,
jw

s
–8 –4 0

From the general predictions about the break points :

(i) Whenever there are two adjacently placed poles, with the section of real axis between them as a part
of root locus, then there exists minimum one break away point between them.
 Control Systems 3 Root Locus
(ii) If there is a zero on real axis, with no poles or zeros to the left of that zero, with entire real axis to the
left of that zero as a part of root locus, then there exists minimum one break point to the left of that
zero.
From given root locus plot, there are break points between  4    0 and      8 , so according to
the above predictions, option (B) can have the root locus as given in the question.
Hence, checking the validity for option (B).
K ( s  8)
G(s) 
s ( s  4)
(i) Number of poles and zeros :
Number of poles = 2
Number of zero = 1
(ii) Location of poles and zeros :
Location of poles, s  0,  4  
    Location of zero, s   8
jw

s
-8 -4 0

(iii)Root locus branch on real axis :

Any section on real axis will be part of root locus branch, if the sum of number of open loop poles and
open loop zeros on real axis to right of this section will be odd.
Hence, root locus branch on real axis will lie between
 4    0 and      8 .
(iv) Number of branches (B) :
BP2 (P  Z )
(v) Number of asymptotes (A) :
A  P  Z  2 1  1 (P  Z )
(vi) Angle of asymptotes ( A ) :
Angle of asymptotes is given by,
(2  1)  1800
A 
PZ
where,   0, 1, 2,.... ( P  Z  1)
  0 , A  180  
0

As there is only one asymptote, so there is no need to calculate centroid which gives the intersection
point of asymptotes on real axis.
 Control Systems 4 Root Locus
(vii) Root locus diagram :
jw
s = -13.656
Break-in point

K =¥ K =¥ K =0 K =0
s
-8 -4 0
Virtual zero

s = - 2.344
Breakaway point

Hence, the correct option is (B).

 Key Point
(i) The intersection point of asymptotes on real axis is called centroid.
(ii) Centroid is also referred as center of gravity.
(iii)Centroid may be located anywhere on the real axis.
(iv) The origin of asymptotic line is centroid.
(v) Direction of root locus is always from open loop pole to open loop zero.
(vi) Root locus is always symmetrical about real axis.
(vii) Number of root locus branches (B) :
1. For P  Z (Proper transfer function)
B  P (Number of branches = Number of poles)
2. For Z  P (Improper transfer function)
B  Z (Number of branches = Number of zeros)
3. For Z  P (Improper transfer function)
BPZ
(viii) If the open loop transfer function is of the following form :
K  s  b
G ( s) H ( s) 
s s  a

Then root locus is described by a circle with center  b and radius b(b  a) .

Breakaway point in the form  b  b(b  a) .

Sol.4 (B)
K ( s  10)
Given : G ( s ) H ( s ) 
( s  2) ( s  5)
. Method 1 :
(i) Number of poles and zeros :
Number of poles = 2
Number of zero = 1
(ii) Location of poles and zeros :
Location of poles, s   2,  5  
    Location of zero, s  10
Control Systems 5 Root Locus
jw

s
-10 -5 -2

(iii)Root locus branch on real axis :

Any section on real axis will be part of root locus branch, if the sum of number of open loop poles and
open loop zeros on real axis to right of this section will be odd.
Hence, root locus branch on real axis will lie between
 5     2 and     10 .
jw

RL RL
s
-10 -5 -2

(iv) Number of branches (B) :

BP2 (P  Z )
(v) Number of asymptotes (A) :
A  P  Z  2 1  1 (P  Z )
(vi) Angle of asymptotes ( A ) :
Angle of asymptotes is given by,
(2  1)  1800
A 
PZ
where,   0, 1, 2,.... ( P  Z  1)
  0 , A  180  
0

As there is only one asymptote, so there is no need to calculate centroid which gives the intersection
point of asymptotes on real axis.
(vii) Break-away/Break-in point :
The characteristic equation is given by,
1  G( s ) H ( s )  0
K ( s  10)
1 0
( s  2) ( s  5)
(s  2) (s  5)  K (s  10)  0
 ( s 2  7 s  10)
K
( s  10)
dK
0
ds
Control Systems 6 Root Locus

d  ( s  10) (2s  7)  ( s 2  7 s  10) 

ds  ( s  10) 2   0 

2s 2  7s  20s  70  s 2  7s  10
s 2  20s  60  0
s  3.67, 16.32
Since, the point s   3.67 is lying between two adjacent poles on root locus branch of real axis, hence
it is a valid break-away point.
Since, the point s  16.32 is lying between two adjacent zeros (one real and one virtual zero) on root
locus branch of real axis, hence it is a valid break-in point.
s = – 16.32 jw
Break-in point

s
– 10 –5 –2 0

s = – 3.67
Break-away point

(viii) Root locus diagram :

jw

Virtual zero
K =¥ K =¥ K =0 K =0
s
– 10 –5 –2 0
Break-in point
s = – 16.32
Breakaway point
s = – 3.67

Hence, the correct option is (B).

. Method 2 :
If there are adjacently placed poles on the real axis and the real axis between them is a part of root locus
then there exists minimum one break-away point between them.
Therefore, a break-away point should lie between – 2 and – 5.
Hence, the correct option is (B).
 Key Point
In case of valid root locus branch, Break-away point exists between two adjacent poles and break-in
point exists between two adjacent zeros.
Sol.5 (D)
The given root locus diagram is shown below,
jw

K =¥ K =¥ K =0 K =0
Virtual s
zero -3 -2 -1
K =5 K =1

(i) At K = 1 and K = 5, breakpoint will exist, hence for these values of K system will be critically damped.
(ii) For system to be over-damped, poles should be real, negative and unequal.
Control Systems 7 Root Locus
From the figure, over-damped poles exists only for 5  K   and 0  K  1 .
Hence, the correct option is (D).

 Key Point
(i) Type of damping with the variation of K in given root locus diagram is given below,

(ii) Root locus represent the path or locus of closed loop poles for different values of gain of open loop
transfer function K.
Sol.6 (B)
Let the gain of controller be ‘K’. Figure shows the system with controller gain K.
2
R( s) K C (s)
s ( s + 1) ( s + 2)

The characteristic equation is given by,

1  G( s) H (s)  0
2K
1 0
s ( s  1) ( s  2)
s 3  3s 2  2s  2 K  0
The maximum possible controller gain, for which the unity feedback system is stable, will be marginal
value of K i.e. K mar .
Routh Tabulation :
s3 1 2

s2 3 2K

6  2K
s1 0
3
s0 2K 0

The intersection of root locus plot with imaginary axis is given by the value of K obtained by solving the
following equation.
6  2 K mar
0  K mar  3
3
Hence, the correct option is (B).
Control Systems 8 Root Locus
Sol.7 (D)
Given :
jw

j 2

s
–2 –1 0

– 0.423 -j 2

At ‘Breakaway point’ system consist critical oscillations (non-oscillatory response).

The characteristic equation,
s 3  3s 2  2s  2 K  0
 ( s 3  3s 2  2 s )
K …(i)
2
From the given figure, the breakaway point is s   0.423 .   
At s   0.423, from equation (i),
K  0.192  0.20    
Hence, the correct option is (D).
 Key Point
Break away points represents the point where the value of K is maximum for the poles to lie on real axis.
The response will be non oscillatory till the poles are on real axis.
Sol.8 (B)
Given : ( s 2  4) ( s  1)  K ( s  1)  0
K ( s  1)
1 0 …(i)
( s  4) ( s  1)
2

Characteristic equation is given by,

1  G (s) H (s)  0 …(ii)
On comparing equation (i) and (ii),
K ( s  1)
G (s) H (s) 
( s  4) ( s  1)
2

.Method 1 : Procedure Based.

(i) Number of poles and zeros :
Number of zero = 1
Number of poles = 3
(ii) Location of poles and zeros :
Location of zero, s  1
Location of poles, s   1,  2, 2
Control Systems 9 Root Locus
jw

s
–2 –1 1 2

(iii)Root locus branch on real axis :

Any section on real axis will be part of root locus branch, if the sum of the number of open loop poles
and open loop zeros on real axis to right of this section will be odd.
Hence, root locus branch on real axis will lie between 1    2 and  2     1 .
jw

RL RL
s
–2 –1 1 2

(iv) Number of branches (B) :

BP3 (P  Z )
(v) Number of asymptotes (A) :
A  P  Z  3 1  2
(vi) Angle of asymptotes (  A ) :
(2  1)  1800
A 
PZ
Where,   0, 1, 2,.....( P  Z  1)
  0, 1  
A  900 , 2700
(vii) Centroid (  ) :
The intersection point of asymptotes on the real axis is called centroid.
Centroid is given by,


 Re(Poles)   Re(Zeros)
PZ
1  1
  1
3 1
jw

0
270 900
s
s = -1
 Control Systems 10 Root Locus
(viii) Break-away/break-in point :
Characteristic equation is given by,
1  G (s) H (s)  0
K ( s  1)
1 0
( s  4)( s  1)
2

( s 2  4)( s  1)
K 
s 1
dK
0
ds
d  ( s 2  4)( s  1) 
 0
ds  s 1 
d  s3  s 2  4s  4 
ds  s 1 0

      (3s 2  2 s  4)( s  1)  ( s 3  s 2  4 s  4)  0
3s 3  2s 2  4s  3s 2  2s  4  s 3  s 2  4s  4  0  
      2s3  2s 2  2s  8  0
s   1.48, 1.24, 1.24
Since, point s  1.48 is lying between two adjacent poles on root locus branch of real axis, hence it
will be valid break-away point.
Since, point s  1.24 is not lie in root locus branch. Hence, it is invalid break point.
s = – 1.48 jw
Break-away
point

s
–2 –1 1 2

(ix) Root locus diagram :

Virtual zero

K =¥
jw

K =0 K =0 K =¥ K =0
s
–2 –1 1 2
s = - 1.48
Break-away
point K =¥
Virtual zero
Hence, the correct option is (B).
.Method 2 : Concept Based.
(i) Location of poles and zeros :
Location of zero, s = 1
Location of poles, s   2, 1, 2
Control Systems 11 Root Locus
Hence, either option (A) or option (B) is correct.
jw

s
–2 –1 1 2

(ii) Root locus branch on real axis :

Any section on real axis will be part of root locus branch, if the sum of the number of open loop poles
and open loop zeros on real axis to the right of this section will be odd.
Hence, root locus branch on real axis will lie between  2     1 and 1    2
jw

RL RL
s
–2 –1 1 2

Hence, the correct option is (B).

Note : For option (A)
jw
Invalid root
locus
K =0 K =0 K =¥ K =0
s
–2 –1 1 2

Invalid root
locus

Sol.9 (A)
 2
K s 
3
Given : G ( s )  2 , H (s)  1
s ( s  2)
 2
K s 
3
G ( s ) H ( s )  2  
s ( s  2)
. Method 1 :
(i) Number of poles and zeros :
Number of zero = 1
Number of poles = 3
(ii) Location of poles and zeros :
2
Location of zero, s  
3
Location of poles, s  0, 0,  2
Control Systems 12 Root Locus
jw

s
–2 – 2/3 0

(iii)Root locus branch on real axis :

Any section on real axis will be part of root locus branch, if the sum of the number of open loop poles
and open loop zeros on real axis to the right of this section will be odd.
2
Hence, root loci branch on real axis will lie between  2     .
3
jw

RL
s
–2 – 2/3 0

(iv) Number of branches (B) :

BP3 (P  Z )
(v) Number of asymptotes (A) :
A  P  Z  3  1  2 (P  Z )
(vi) Angle of asymptotes (  A ) :
(2  1)  1800
A 
PZ
Where,   0, 1, 2, ..... ( P  Z  1)
  0,1
A  900 , 2700
(vii) Centroid (  ) :
The intersection point of asymptotes on the real axis is called centroid.
Centroid is given by,


 Re(Poles)   Re(Zeros)
PZ
2
2
 3 2
3 1 3
jw

0
270
0 90
s = -2 / 3
s
0
Control Systems 13 Root Locus
(viii) Break-away/break-in point :
Characteristic equation is given by,
1  G (s) H (s)  0
 2
Ks 
1  2 3 0
s ( s  2)
s 2 ( s  2)
K 
2
s
3
dK
0
ds
d  s 2 ( s  2) 
 0
ds  s
2 

 3 
d  s3  2s 2 
 0 
ds  s  2 
 3 
 2
(3s 2  4 s )  s    s 3  2 s 2  0
 3
8
3s 3  4 s 2  2 s 2  s  s 3  2 s 2  0
3
8
2s3  4s 2  s  0
3
s (6 s 2  12 s  8)  0
s  0,  1  j 0.57
Since, point s  0 is a multiple pole, hence it is a valid break away point.
As points s  1  j 0.57 do not satisfied angle condition, hence these are invalid break away points.
jw s=0
Break-away
point

s
–2 – 2/3 0

(ix) Intersection with j axis :

The value of K at point where root locus branch crosses the imaginary axis is determined by applying
routh hurwitz criterion to the characteristic equation.
From characteristic equation,
1  G (s) H (s)  0
 2
Ks 
1  2 3 0
s ( s  2)
Control Systems 14 Root Locus
2K
s 3  2 s 2  Ks  0
3
Routh tabulation :
s3 1 K
2K
s2 2
3
2K
2K 
s1 3  2K 0  ROZ
2 3
2K
s0 0
3
From auxiliary equation, intersection point will be calculated and this equation will exist if all element
of s1 row will be zero.
2K
Hence, 0 K 0
3
Auxiliary equation is given by,
2K
A( s )  2 s 2  0
3
A( s )  2 s 2  0  0
Hence, root locus will intersect j axis at s  0 and at this point value of K is 0.
(x) Root locus diagram :
Virtual zero K =¥ jw

K =0 2700 900 K = 0, 0
s
-2 K =¥
-2
3

-2
Re(s ) =
3

Virtual zero K =¥
2
Since, all the three root loci branch terminate at Re( s )   , hence as the value of K tends to infinity,
3
2
the real value of all the three pole will approach to  , so all the three roots have nearly equal real
3
part, as K approaches to infinity.
Hence, the correct option is (A).
. Method 2 :
From Routh tabulation, there is no sign changes in the first column.
Hence, number of right half poles is zero.
 Control Systems 15 Root Locus
Therefore, option (B) and (D) are incorrect.
Since, root locus crosses the imaginary axis at K = 0 therefore option (C) is also incorrect.
Hence, the correct option is (A).
Sol.10 (C)
Given : s(s  3)  K (s  5)  0
K ( s  5)
1 0 …(i)
s ( s  3)
The characteristics equation is given by,
1  G( s ) H ( s )  0 …(ii)
On comparing equation (i) and (ii),
K ( s  5)
G ( s) H ( s) 
s ( s  3)
. Method 1 :
(i) Number of poles and zeros :
Number of poles = 2
Number of zero = 1
(ii) Location of poles and zeros :
Location of poles, s  0,  3  
    Location of zero, s   5
jw

s
-5 -3 0

(iii)Root locus branch on real axis :

Any section on real axis will be part of root locus branch, if the sum of number of open loop poles and
open loop zeros on real axis to right of this section will be odd.
Hence, root locus branch on real axis will lie between
 3    0 and      5.
jw

RL RL
s
-5 -3 0

(iv) Number of branches (B) :

BP2 (P  Z )
 Control Systems 16 Root Locus
(v) Number of asymptotes (A) :
A  P  Z  2 1  1 (P  Z )
(vi) Angle of asymptotes ( A ) :
Angle of asymptotes is given by,
(2  1)  1800
A 
PZ
where,   0, 1, 2,.... ( P  Z  1)
  0, A  1800  
As there is only one asymptote, so there is no need to calculate centroid which gives the intersection
point of asymptotes on real axis.
(vii) Break-away/break-in point :
The characteristic equation is given by,
1  G( s ) H ( s )  0  
K ( s  5)
      1  0 
s ( s  3)
s 2  3s
      K 
s5
dK
0
ds
d  s 2  3s 
 0
ds  s  5 
( s  5)(2 s  3)  ( s 2  3s )
0
( s  5) 2
2s 2  13s  15  s 2  3s  0
s 2  10s  15  0
10  100  4  15
s1,2 
2
s1,2   5  10   1.837,  8.162
Since, the point s  1.837 is lying between two adjacent poles on root locus branch of real axis,
hence it is a valid break-away point.
Since, the point s   8.162 is lying between two adjacent zeros (one real and one virtual zero) on root
locus branch of real axis, hence it is a valid break-in point.
jw
s = - 8.162
Break-in point

s
-5 -3 0
s = - 1.837
Break-away point
 Control Systems 17 Root Locus
(viii) Root locus diagram :
jw
s = - 8.162
Break-in point

K =¥ K =¥ K =0 K =0
s
-5 -3 0
Virtual zero

s = - 1.837
Break-away point  
Hence, the correct option is (C).
. Method 2 :
K  s  5
G( s) H ( s)   
s  s  3
For the given form of open loop transfer function breakaway point is given by,
BP   b  b(b  a)  
We have b  5, a  3  
      BP   5  5(5  3)  
BP   5  10
Hence, the correct option is (C).
 Key Point
Break points = Centre  radius
Transfer
function Centre Radius
K ( s  b)
( b, 0) b(b  a)
s(s  a)
K ( s  b)
( b, 0) (b  a1 ) (b  a2 )
( s  a1 )( s  a2 )

K ( s  b1 ) ( s  b2 )   b1b2  b1b2
 , 0
s2  b1  b2  b1  b2
Sol.11 (– 4)
K ( s  2)
Given : G (s) H (s) 
s 2 ( s  10)
(i) Number of poles and zeros :
Number of poles = 3
Number of zero = 1
(ii) Location of poles and zeros :
Location of poles, s  0,0, 10  
    Location of zero, s   2
Control Systems 18 Root Locus
jw

s
-10 -2 0

(iii)Root locus branch on real axis :

Any section on real axis will be part of root locus branch, if the sum of number of open loop poles and
open loop zeros on real axis to right of this section will be odd.
Hence, root locus branch on real axis will lie between 10     2 .
jw

RL
s
-10 -2 0

(iv) Number of branches (B) :

BP3 (P  Z )
(v) Number of asymptotes (A) :
A  P  Z  3 1  2 (P  Z )
(vi) Angle of asymptotes ( A ) :
Angle of asymptotes is given by,
(2  1)  1800
A 
PZ
where,   0, 1, 2,.... ( P  Z  1)
  0,1 , A  900 , 2700
(vii) Centroid (  ) :
The intersection point of asymptotes on the real axis is called centroid.
Centroid is given by,


 Re(Poles)   Re(Zeros)  
PZ
10  2
  4
3 1
Hence, the point of intersection of the asymptotes on the real axis is – 4.
Sol.12 (A)
Ks
Given : OLTF = G ( s ) 
( s  1)( s  4)
(i) Number of poles and zeros :
Number of zeros = 1
Number of poles = 2
Control Systems 19 Root Locus
(ii) Location of poles and zeros :
Location of zero, s  0
Location of poles, s  1, 4
jw

s
0 1 4

(iii)Root locus branch on real axis :

Any section on real axis will be part of root locus branch, if the sum of the number of open loop poles
and open loop zeros on real axis to the right of this section will be odd.
Hence, root locus branch on real axis will lie between 1    4 and   0
jw

RL RL
s
0 1 4

(iv) Break-away/break-in point :

Characteristic equation is given by,
1  G (s)  0
Ks
1 0
( s  1)( s  4)
( s  1)( s  4)
K  …(i)
s
dK
0
ds
d  s 2  5s  4 
 0
ds  s 
(2 s  5) s  s 2  5s  4  0
2 s 2  5s  s 2  5s  4  0
s2  4  0  s   2, 2
Since, point s  2 is lying between two adjacent poles on root locus branch of real axis, hence it will
be break-away point.
Since, point s   2 is lying between two adjacent zeros (one real and one virtual zero) on root locus
branch of real axis, hence it will be break-in point.
Control Systems 20 Root Locus
jw
s=–2
Break-in point

s
0 1 4
Virtual s=2
zero Break-away
point
From equation (i), gain at the break away point s  2 is given by,
(2  1)(2  4)
K s 2   1 
2
From equation (i), gain at the break-in point s  2 is given by,
( 2  1)( 2  4)
K s  2   9
2
(v) Root locus diagram :
jw

j2
K =5
K =9
K =1

K =¥ K =¥ K =0 K =0
s
0 1 2 4
s = -2
Virtual
zero - j2

Hence, the correct option is (A).

Note : You can directly find break-in point from step (viii). We are providing all steps to draw root locus
diagram.
 Key Point
(i) Type of damping with the variation of K in given root locus diagram is given below,
Location of
Range of K Damping closed loop Stability
poles
0 < K <1 – ve over Real and Unstable
damping unequal
K= 1 – ve critical Real and Unstable
Damping equal
1< K < 5 – ve under Complex Unstable
damping conjugate
K= 5 Undamped Imaginary Marginal
stable
5< K <9 Under Complex Stable
damping conjugate
K= 9 Critical Real and Stable
damping equal
9<K <¥ Over Real and Stable
damping unequal
(ii) Root locus represent the path or locus of closed loop poles for different values of K .
Control Systems 21 Root Locus
Sol.13 (B)
Given : s 2  6 Ks  2s  5  0
6 Ks
1 0 …(i)
s  2s  5
2

Characteristic equation is given by,

1  G( s) H ( s)  0 …(ii)
On comparing equation (i) and (ii),
6 Ks
G (s) H ( s) 
s  2s  5
2

(i) Number of poles and zeros :

Number of zero = 1
Number of poles = 2
(ii) Location of poles and zero :
Location of zero, s  0
Location of poles, s  1  j 2
jw

+ j2

s
-1 0

- j2

(iii)Root locus branch on real axis :

Any section on real axis will be part of root locus branch, if the sum of the number of open loop poles
and open loop zeros on real axis to the right of this section will be odd.
Hence, root locus branch on real axis will lie between     0 .
jw
+ j2

RL
s
-1 0

- j2

(iv) Number of branches (B) :

BP2 (P  Z )
(v) Number of asymptotes (A) :
A  P  Z  2  1  1 (P  Z )
(vi) Angle of asymptotes (  A ) :
(2  1)  1800
A 
PZ
Control Systems 22 Root Locus
where,   0, 1, 2, ..... ( P  Z  1)
0
A  1800
(vii) Centroid (  ) :
The intersection point of asymptotes on the real axis is called centroid.
Centroid is given by,

  Re(Poles)   Re(Zeros)
PZ
2 0
  2
2 1
(viii) Break-away/break-in point :
Characteristic equation is given by,
1  G( s) H ( s)  0
Ks
1 0
s  2s  5
2

s 2  2s  5
K 
s
dK
0
ds
d  s 2  2s  5 
 0
ds  s 
(2 s  2) s  ( s 2  2 s  5)  0
s2  5  0
s 5
Since, the point s   5 is lying between two adjacent zeros (one real and one virtual zero) hence it
will be break-in point.
jw
Virtual
zero + j2

s
-1 0

s=- 5 - j2
Break-in point

(ix) Angle of departure (  d ) :

Angle of departure is given by,
d  1800  ( P  Z )
 Control Systems 23 Root Locus
jw
+ j2

fZ
s
-1 0
fP
- j2

where, Z is sum of all the angles subtended by zeros to any particular pole.
P is sum of all the angles subtended by other poles to any particular pole.
P  900
Z  1800  tan 1 2  116.560
d  1800  (900  116.560 )
d  206.560
(x) Root locus diagram :
jw
K =0
j2

K =¥ fZ
s
- 5 -1 0 K =¥

fP
Virtual - j2
zero K =0

Therefore, root locus will enter into real axis at s   5 .

Hence, the correct option is (B).
 Key Point
(i) At break-away point root locus leaves the real axis.
(ii) At break-in point root locus enters into real axis.

Sol.14 (–1.1)
K ( s  6)
Given : G ( s ) H ( s ) 
s ( s  2)
. Method 1 :
(i) Number of poles and zeros :
Number of poles = 2
Number of zero = 1
(ii) Location of poles and zeros :
Location of poles, s  0,  2  
    Location of zero, s   6
Control Systems 24 Root Locus
jw

s
-6 -2 0

(iii)Root locus branch on real axis :

Any section on real axis will be part of root locus branch, if the sum of number of open loop poles and
open loop zeros on real axis to right of this section will be odd.
Hence, root locus branch on real axis will lie between
 2    0  and      6
jw

RL RL
s
-6 -2 0

(iv) Number of branches (B) :

BP2 (P  Z )
(v) Number of asymptotes (A) :
A  P  Z  2 1  1 (P  Z )
(vi) Angle of asymptotes ( A ) :
Angle of asymptotes is given by,
(2  1)  1800
A 
PZ
where,   0, 1, 2,.... ( P  Z  1)
  0 , A  180
0

(vii) Centroid (  ) :
The intersection point of asymptotes on the real axis is called centroid.
Centroid is given by,


 Re(Poles)   Re(Zeros)
PZ
026
  4 
2 1
(viii) Break-away/Break-in point :
The characteristic equation is given by,
1  G( s ) H ( s )  0
K ( s  6)
1 0 
s ( s  2)
 Control Systems 25 Root Locus

 ( s 2  2s)
      K  
( s  6)
dK
       0   
ds
d   ( s 2  2s ) 
      0 
ds  ( s  6) 
( s  6) (2 s  2)  ( s 2  2s )
      0 
( s  6) 2
      2 s 2  12 s  2 s  12  s 2  2 s  0  
      s 2  12s  12  0  
 12  144  48
      s1, 2   
2
      s1, 2   1.101,  10.895  
Since, the point s  1.101 is lying between two adjacent poles on root locus branch of real axis,
hence it is a valid break-away point.
Since, the point s  10.895 is lying between two adjacent zeros (one real and one virtual zero) on
root locus branch of real axis, hence it is a valid break-in point.
jw
s = – 10.895
break-in point

s
-6 -2 0
s = – 1.101
break-away point

(ix) Root locus diagram :

jw
s = – 10.895
break-in point

K =¥ K =¥ K =0 K =0
s
-6 -2 0
Virtual zero

s = – 1.101
break-away point  
Hence, the breakaway point of the root-loci will be – 1.10.
. Method 2 :
K ( s  6)
G (s) H (s)   
s ( s  2)
For the given form of open loop transfer function breakaway point is given by,
BP   b  b(b  a)  
We have b  6, a  2  
 Control Systems 26 Root Locus

      BP   6  6(6  2)  1.10
Hence, the breakaway point of the root-loci will be – 1.10.
Sol.15 (A)
The given root locus is shown below,

jw

s
-7 -6 -5 -4 -3 -2 -1

From above figure,

Root locus diagram starts at a pole s  1 and s  2, and terminates at a zero s  5 and at infinity
Hence, open loop transfer function is given by,
K ( s  5)
G (s) 
( s  1) ( s  2)
Hence, the correct option is (A).
Sol.16 (B)
4K
Given : (i) Open loop transfer function, G ( s ) 
( s  1) ( s  3)
(ii) Damping ratio,   0.707
Characteristic equation is given by,
1  G(s)  0
4K
1 0
( s  1)( s  3)
s 2  4s  3  4 K  0 ... (i)
Second-order standard characteristic equation is given by,
s 2  2n s  2n  0 ... (ii)
On comparing equation (i) and (ii)
n  3  4 K and 2n  4
2
n  2 2
0.707
2 2  3  4K
8  3  4K
3
K
4
Hence, the correct option is (B).
Control Systems 27 Root Locus
Sol.17 (B)
Given : Open loop transfer function
K ( s  4)
G (s) 
( s  1) ( s  2)
For above transfer function pole-zero plot is given as,
jw

s
–4 –2 –1

(iii) Root locus branch on real axis :

Any section on real axis will be part of root locus branch, if the sum of the number of open loop
poles and open loop zeros on real axis to the right of this section will be odd.

jw

s
–4 –2 –1

Hence root loci branch on real axis will lie between 2    1 and   4
Hence, the correct option is (B).
Sol.18 (B)
K ( s  2)
Given : G ( s)  2 , H (s)  1
s  2s  3
Pole location :
s 2  2s  3  0
2  4  12
s
2
Break-away/break in point :
1  G( s) H ( s)  0
K ( s  2)
1 0
s 2  2s  3
s 2  2s  3  K ( s  2)  0
( s 2  2s  3)
K
( s  2)
Control Systems 28 Root Locus
dK
0
ds
d   ( s 2  2s  3) 
ds  s2 0

( s  2)  (2s  2)  ( s 2  2s  3)  0
2s 2  6s  4  s 2  2s  3  0
s 2  4s  1  0
 4  16  4
s
2
42 3
s  2  3
2
s  3.732,  0.267

j 2

– 3.732 –2 –1 – 0.267

-j 2

Hence, valid break-away point is –3.732

Hence, the correct option is (B).
Sol.19 (B)
Given : Closed loop transfer function is given below,
K
T ( s)  2
s  (3  K ) s  1
K
 s  3s  1
K 2
T ( s)  2 …(i)
s  3s  Ks  1 1  Ks
s 2  3s  1
Transfer function for negative feedback system is given by,
G( s)
T ( s)  …(ii)
1  G ( s) H ( s)
Comparing equation (i) and (ii),
K
G( s)  2 , H ( s)   s
s  3s  1
Hence, open loop transfer function is,
 Ks
G( s) H ( s)  2
s  3s  1
Due to negative sign, it is based on inverse root locus (IRL) or complementary root locus.
Control Systems 29 Root Locus

. Method 1 : Procedure Based :

(i) Number of poles and zeros :
Number of zero = 1
Number of poles = 2
(ii) Location of poles and zeros :
Location of zero : s  0
Location of poles :
3  9  4  3  2.23
s 
2 2
s   2.615,  0.385
jw

s
–2.615 –0.385 0

(iii) Root locus branch on real axis :

Any section on real axis will be part of root locus, if sum of number of open loop poles and open loop
zeros to right of that section will be even (complimentary root locus). Hence,   0 and
 2.615     0.385 will be part of root locus.
jw

CRL CRL
s
–2.615 –0.385 0

Fig. Pole-zero diagram

(iv) Number of branches (B)
B=P=2 (P > Z)
(v) Number of asymptotes (A) :
A = P – Z = 2 – 1 = 1 (P > Z)
(vi) Angle of asymptotes ( A ) :
Angle of asymptotes is given by,
2  1800
A 
PZ
where,   0, 1, 2, ....( P  Z  1)
0
A  00
(vii) Centroid (  ) :
The intersection point of asymptotes on the real axis is called centroid.
Control Systems 30 Root Locus
Centroid is given by,


 Re (Poles)   Re (Zeros)
PZ
 2.615  0.385
  3
2 1
jw

s
s = -3

(viii) Break-away / break-in point :

Characteristics equation is given by,
1  G( s ) H ( s )  0
Ks
1 0
s  3s  1
2

s 2  3s  1
K
s
dK
0
ds
(2 s  3) s  ( s 2  3s  1)
Hence, 0
s2
2s 2  3s  s 2  3s  1  0
s2  1  0
s  1
Since, point s  1 is lying between one physical zero and one virtual zero on root locus branch of
real axis, hence s  1 will be break-in point.
Since, point s  1 is lying between two adjacent poles on root locus branch of real axis, hence s  1
will be break-away point.
jw
s = –1, valid
break-away point

s
–2.615 –0.385 0
s = 1, valid
break-in point
(ix) Intersection with imaginary axis :
The value of K at point where root locus branch crosses the imaginary axis is determined by applying
Routh Hurwitz criterion to the characteristic equation.
From characteristics equation,
Control Systems 31 Root Locus
1  G( s ) H ( s )  0
Ks
1 0
s  3s  1
2

s 2  (3  K ) s  1  0
Routh Tabulation :
s2 1 1

s1 3 K 0  Row of zeros

s0 1 0

The intersection of root locus plot with imaginary axis is given by the value of K obtained by solving
the following equation.
3 K  0  K 3
The auxiliary equation can be formed using the row above the ROZ.
A( s )  s 2  1  0
s j
Hence, root locus will intersect j axis at  j and at these point value of K is 3.
(x) Root locus diagram :
jw
K=3
j

K=1
K=0 K=0 K =¥ K =¥
s
–2.615 –1 –0.385 0 1
virtual zero
-j
K=3

Hence, the correct option is (B).

. Method 2 : Concept Based :
Ks
G( s) H ( s) 
s  3s  1
2

Pole-zero location of OLTF is shown below,

jw

s
–2.615 –0.385

Hence, from the given options, only option (B) will satisfy to pole-zero location shown above.
Hence, the correct option is (B).
Control Systems 32 Root Locus

 Key Point
(i) In case of valid root locus branch, Break-away point exists between two adjacent poles and break-in
point exists between two adjacent zeros.
(ii) The intersection point of asymptotes on real axis is called centroid.
(ii) Centroid is also referred as center of gravity.
(iv) Centroid may be located anywhere on the real axis.
(v) The origin of asymptotic line is centroid.
(vi) This question can be solved either by taking positive feedback or negative feedback (value of H(s) is
negative).
Sol.20 (A)
Given : Open loop transfer function is given by,
K (s  a)
OLTF = ;ba
s 2 ( s  b)
Let us consider two cases as explained below,
4
Case 1 : a  and b  12
3
(i) Number of poles and zeros :
Number of zero  1
Number of poles  3
(ii) Location are poles and zeros :
4
Location of zero, s  
3
Location of poles, s  0, 0, 12
jw

s
– 12 – 4/3 0

(iii)Root locus branch on real axis :

Any section on real axis will be part of root locus branch, if the sum of number of open loop poles and
open loop zeros on real axis to right of this section will be odd.
4
Hence, root locus branch on real axis will lie between 12    
3
jw

RL
s
– 12 – 4/3 0
Control Systems 33 Root Locus
(iv) Number of branches ( B ) :
BP3 (P  Z )
(v) Number of asymptotes ( A) :
A  P  Z  3 1  2 (P  Z )
(vi) Angle of asymptotes (A) :
Angle of asymptotes is given by,
(2  1)  1800
A 
PZ
Where,   0, 1, 2, ..... ( P  Z  1)
  0, 1
A  900 , 2700
(vii) Centroid (  ) :
The intersection point of asymptotes on the real axis is called centroid.
Centroid is given by,


 Re(Poles)   Re(Zeros)
PZ
4
12 
 3   5.33
3 1
jw

270
0
900
s = – 5.33
s

(viii) Break-away/break-in point :

Characteristics equation is given by,
1  G (s) H (s)  0
 4
Ks 
3
1 2 0
s ( s  12)
 s 2 ( s  12) ( s 3  12s 2 )
K 
 4  4
s  s 
 3  3
dK
0
ds
d    s 3  12s 2  
 0
ds   4 
s 
  3  
Control Systems 34 Root Locus

 4 2
 s    3s  24 s    s  12 s 
3 2


3
2
0
 4
s 
 3
3s 3  24 s 2  4 s 2  32 s  s 3  12 s 2  0
2 s 3  16 s 2  32 s  0
s ( s 2  8s  16)  0
s ( s  4) 2  0
s = 0, – 4, – 4
Since, point s   4 is lying on root locus branch of real axis, hence it is a valid break-away point.
Since, point s  0 is a multiple pole, hence it is a valid break away point.
s = -4 jw
s=0
Break-away Break-away
point point
s
– 12 –2 0

(ix) Root locus diagram :

Virtual zero
K =¥ jw

K =0 K =¥ K = 0, 0
s
– 12 – 4/3
s=0
s=–4 break-away point
break-away/break-in
point K =¥
Virtual zero

Case 2 : a  2 and b  4
(i) Number of poles and zeros :
Number of zero  1
Number of poles  3
(ii) Location are poles and zeros :
Location of zero, s   2
Location of poles, s  0, 0,  4
jw

s
–4 –2 0
Control Systems 35 Root Locus
(iii)Root locus branch on real axis :
Any section on real axis will be part of root locus branch, if the sum of number of open loop poles and
open loop zeros on real axis to right of this section will be odd.
Hence, root locus branch on real axis will lie between  4     2
jw

RL
s
–4 –2 0

(iv) Number of branches ( B ) :

BP3 (P  Z )
(v) Number of asymptotes ( A) :
A  P  Z  3 1  2 (P  Z )
(vi) Angle of asymptotes (A) :
Angle of asymptotes is given by,
(2  1)  1800
A 
PZ
Where,   0, 1, 2, ..... ( P  Z  1)
  0, 1
A  900 , 2700  
(vii) Centroid (  ) :
The intersection point of asymptotes on the real axis is called centroid.
Centroid is given by,


 Re(Poles)   Re(Zeros)
PZ
4 2
  1
2
jw

2700 900
s
s=–1

(viii) Break-away/break-in point :

Characteristics equation is given by,
1  G (s) H (s)  0
 s 2 ( s  4) ( s 3  4 s 2 )
K 
( s  2) ( s  2)
 Control Systems 36 Root Locus

d   ( s 3  4s 2 ) 
0
ds  ( s  2) 
( s  2) (3s 2  8s )  ( s 3  4 s 2 )
 0
( s  2) 2
3s 3  8 s 2  6 s 2  16 s  s 3  4 s 2  0
2 s 3  10 s 2  16 s  0
s (2 s 2  10 s  16)  0
s  0, s   2.5  j1.322  
Since, point s  0 is a multiple pole, hence it is a valid break away point.
At point s   2.5  j1.332 ,
 s 2 ( s  4)
K
( s  2)
 ( 2.5  j1.322) 2 ( 2.5  j1.322  4)
                  K
( 2.5  j1.322  2)
K  6.5  j 9.26
Since, point s   2.5  j1.332 gives imaginary value of K, hence, it is invalid break point.
(ix) Root locus diagram :
Virtual zero jw
K =¥

K =0 K =¥ K = 0, 0
s
–4 –2

s=0
K =¥ break-away point
Virtual zero
Hence, the correct options are (A) and (C).
Sol.21 (B)
K
Given : G ( s ) H ( s ) 
( s  1) 4
Existence of points in root locus can be calculated using angle condition.
Angle condition :
G ( s ) H ( s )  (2n  1)1800 n  0,1, 2,..........
(i) At s1  3  j 4 , G ( s1 ) H ( s1 ) is given by,
K K
      G ( s1 ) H ( s1 )  
( 3  j 4  1) 4
( 2  j 4) 4
K is some positive real number, hence K  00
( 2  j 4)  1800  tan 1 (2)
( 2  j 4)  4   4  1800  tan 1 (2) 
( 2  j 4)  4   466.260
 Control Systems 37 Root Locus
Phase of G ( s1 ) H ( s1 ) is given by,
G ( s1 ) H ( s1 )  K  ( 2  j 4)  4
G ( s1 ) H ( s1 )  00  466.260   466.260

Since, G ( s1 ) H ( s1 ) is not odd multiples of 1800 , hence it will not lie on the root locus.
(ii) At s2  3  j 2 , G ( s2 ) H ( s2 ) is given by,
K
G ( s 2 ) H ( s2 ) 
( 3  j 2  1) 4
K
G ( s2 ) H ( s2 ) 
( 2  j 2) 4
2
( 2  j 2)  1800  tan 1    2250
2
( 2  j 2) 4  4  2250  9000
Phase of G ( s2 ) H ( s2 ) is given by,
      G ( s2 ) H ( s2 )  K  ( 2  j 2) 4  00  9000
G ( s2 ) H ( s2 )   9000  5  ( 1800 )

Since, G ( s2 ) H ( s2 ) is odd integer multiples of 1800 , hence it will lie on the root locus.
The root locus diagram is shown in below figure,
jw
s1 = -3 + 4 j
j4
Virtual zero j3 Virtual zero

K =¥ j2 K =¥

j1
K =0
0
s
–5 –4 –3 –2 –1 1 2 3
- j1
s2 = -3 - 2 j - j2
K =¥ K =¥
Virtual zero - j3 Virtual zero
- j4

Hence, the correct option is (B).

Sol.22 (B)
K
Given : G ( s ) H ( s ) 
s( s  1) ( s  3)
. Method 1 : Procedure Based :
(i) Number of poles and zeros :
Number of open loop poles P = 3
Number of open loop zero Z = 0
Control Systems 38 Root Locus
(ii) Location of poles and zeros :
Location of poles, s = 0, – 1, – 3
jw

s
-3 -1 0

(iii)Root locus branch on real axis :

Any section on real axis will be part of root locus, if sum of number of open loop poles and open loop
zeros to right of that section will be odd.
So, 1    0 and    3 will be part of root locus branch on real axis.
jw

RL RL
s
-3 -1 0

Fig. Pole-zero diagram

(iv) Number of branches (B) :
B=P=3 (P > Z)
(v) Number of asymptotes (A) :
A  P  Z  3 (P > Z)
(vi) Angle of asymptotes ( A) :
(2  1)  1800
A 
PZ
where,   0, 1, .....( P  Z  1)
  0, 1, 2
Hence, A  600 , 1800 , 3000
(vii) Centroid (  ) :
The intersection point of asymptotes on the real axis is called centroid.
Centroid is given by,


 Re (Poles)   Re (Zeros)
PZ
(0  1  3)  0  4
   1.33
3 0 3
jw

3000 600
s
s = -1.33
Control Systems 39 Root Locus
(viii) Break-away / Break-in point :
dK
0
ds
d
  s( s  1) ( s  3)  0
ds
d 3
 s  4 s 2  3s   0
ds
3s 2  8 s  3  0
s   0.45,  2.21
Since, point s   0.45 is lying between two adjacent poles on root locus branch of real axis, hence it
will be a valid break-away point.
Since, point s   2.21 is not lying on root locus branch of real axis, hence it will be invalid break-
away point.
(s = – 0.45) valid jw
break-away point

s
-3 -1 0

(s = – 2.21) invalid
break-away point
(ix) Intersection with imaginary axis :
The value of K at point where root locus branch crosses the imaginary axis is determined by applying
Routh Hurwitz criterion to the characteristics equation.
Characteristic equation is given by,
1  G(s) H ( s)  0
K
1 0
s ( s  1) ( s  3)
s 3  4 s 2  3s  K  0
Routh Tabulation :
s3 1 3
s2 4 K
12  K
s1 0  Row of zeros
4
s0 K 0

Intersection of root locus plot with imaginary axis is given by the value of K, obtained by solving
following equation.
12  K
0
4
K  12  
Control Systems 40 Root Locus
The auxiliary equation can be formed by using row present above ROZ,
A( s )  4 s 2  K  0
4 s 2  12  0
sj 3
Hence, root locus will intersect j axis at  j 3 and at these point value of K is 12.
(x) Root locus diagram :
virtual zero

jw K =¥

virtual zero s = – 1.5 invalid j 3

K =0 K =0 K =0
s
-1.33
K =¥ -1 0
-j 3
s = – 3 valid
s = – ¥ valid K =¥
s = - j 3 valid
virtual zero
Hence, the correct option is (B).
. Method 2 : Concept Based :
K
G( s) 
s ( s  1) ( s  3)
(i) Pole zero location is shown below,
jw

s
-3 -1 0

(ii) Root locus branch on real axis is shown below,

jw

RL RL
s
-3 - 1.5 -1 0

From above figure, it is clear that s   1.5 will not lie on root locus.
Hence, the correct option is (B).
Sol.23 (A)
Given : The Routh-Hurwitz array is given for a third order characteristic equation,
s3 1 b 0
2
s a c 0
(6  K )
s1 0 0
3
s0 K 0 0
 Control Systems 41 Root Locus
jw

j 1.41

- j 1.41

To calculate the intersection with jω  axis there should be row of zeros in Routh array.
(6  K )
i.e., 0  K 6
3  
1
To obtain first element of s row,
(ab  c) (6  K )
  
a 3
On comparing both the side of above equation,
a3
b2
cK
Therefore, cK 6
Hence, the correct option is (A).
Sol.24 (C)
Given : Open loop transfer function
K ( s  1)
G (s) H (s) 
s ( s  2) ( s 2  s  1)
Number of poles P  4
Number of zeros Z  1
Number of zeros at infinity is equal to P  Z  4  1  3
Hence, the correct option is (C).
Sol.25 (D)
jw

The above given root locus diagram is incorrect because movement of Root locus points is from open loop
zeros to open loop poles.
Hence, the correct option is (D).
Sol.26 (C)
Given: Characteristic equation
s ( s  2)  K ( s  4)  0
 Control Systems 42 Root Locus
Hence open loop transfer function is given by,
K ( s  4)
G(s) H (s)  … (i)
s ( s  2)

 Key Point
Break points = Centre  radius
Transfer
function Centre Radius
K ( s  b)
( b, 0) b(b  a)
s(s  a)
K ( s  b)
( b, 0) (b  a1 ) (b  a2 )
( s  a1 )( s  a2 )
K ( s  b)
On comparing equation (i) with we get, b  4
s( s  a)
Hence centre of the circle is = ( b, 0) = ( 4, 0)
Hence, the correct option is (C).
Sol.27 (B)
K
Given : G ( s ) H ( s ) 
s( s  1) ( s  3)
. Method 1 : Procedure Based :
(i) Number of poles and zeros :
Number of open loop poles P = 3
Number of open loop zero Z = 0
(ii) Location of poles and zeros :
Location of poles, s = 0, – 1, – 3
jw

s
-3 -1 0

(iii)Root locus branch on real axis :

Any section on real axis will be part of root locus, if sum of number of open loop poles and open loop
zeros to right of that section will be odd.
So,  1    0 and    3 will be part of root locus branch on real axis.
jw

RL RL
s
-3 -1 0

Fig. Pole-zero diagram

Control Systems 43 Root Locus
(iv) Number of branches (B) :
B=P=3 (P > Z)
(v) Number of asymptotes (A) :
A  P  Z  3 (P > Z)
(vi) Angle of asymptotes ( A) :
(2  1)  1800
A 
PZ
where,   0, 1, .....( P  Z  1)
  0, 1, 2
Hence, A  600 , 1800 , 3000
(vii) Centroid (  ) :
The intersection point of asymptotes on the real axis is called centroid.
Centroid is given by,


 Re (Poles)   Re (Zeros)
PZ
(0  1  3)  0  4
   1.33
3 0 3
jw

3000 600
s
s = -1.33

(viii) Break-away / Break-in point :

dK
0
ds
d
  s( s  1) ( s  3)  0
ds
d 3
 s  4 s 2  3s   0
ds 
3s 2  8 s  3  0
s   0.45,  2.21
Since, point s   0.45 is lying between two adjacent poles on root locus branch of real axis, hence it
will be a valid break-away point.
Since, point s   2.21 is not lying on root locus branch of real axis, hence it will be invalid break-
away point.
(s = – 0.45) valid jw
break-away point

s
-3 -1 0

(s = – 2.21) invalid
break-away point
Control Systems 44 Root Locus
(ix) Intersection with imaginary axis :
The value of K at point where root locus branch crosses the imaginary axis is determined by applying
Routh Hurwitz criterion to the characteristics equation.
Characteristic equation is given by,
1  G(s) H ( s)  0
K
1 0
s ( s  1) ( s  3)
s 3  4 s 2  3s  K  0
Routh Tabulation :
s3 1 3
s2 4 K
12  K
s1 0  Row of zeros
4
s0 K 0

Intersection of root locus plot with imaginary axis is given by the value of K, obtained by solving
following equation.
12  K
0
4
K  12  
The auxiliary equation can be formed by using row present above ROZ,
A( s )  4 s 2  K  0
4 s 2  12  0
sj 3
Hence, root locus will intersect j axis at  j 3 and at these point value of K is 12.
(x) Root locus diagram :
virtual zero

jw K =¥

virtual zero s = – 1.5 invalid j 3

K =0 K =0 K =0
s
-1.33
K =¥ -1 0
-j 3
s = – 3 valid
s = – ¥ valid K =¥
s = - j 3 valid
virtual zero
Hence, the correct option is (B).
. Method 2 : Concept Based :
K
G( s) 
s ( s  1) ( s  3)
Control Systems 45 Root Locus
(i) Pole zero location is shown below,
jw

s
-3 -1 0

(ii) Root locus branch on real axis is shown below,

jw

RL RL
s
-3 - 1.5 -1 0

From above figure, it is clear that s   1.5 will not lie on root locus.
Hence, the correct option is (B).
Sol.28 (C)
If a pole is added to the forward path transfer function then the root locus shifts towards the imaginary
axis. Hence, the closed loop system becomes less stable.
Hence, the correct option is (C).
Sol.29 (D)
The root locus describes the path of closed loop transfer function for different values of a specific
parameter (usually, gain K) varied from zero to infinity
Hence, the correct option is (D).
Sol.30 (C)
The given root-locus plot is shown below,
Im

K1
2
Re
-1

From above figure,

n  1 and n 1   2  2
2n  n2  2  2
2n  1  2
n  3
1 1
   0.577
n 3
Hence, the correct option is (C).
Control Systems 46 Root Locus
Sol.31 (B)
The given control system is shown below,
2
R( s) C (s)
s (s + 1)

( s + 0.5)
(s + 3)
Since, positive feedback control system hence this is the case of complementary control system.
Open loop transfer function is given by,
2 ( s  0.5)
G ( s) H (s)  
s ( s  1) ( s  3)
2 ( s  0.5)
G ( s) H (s) 
s ( s  1) ( s  3)
P  3 and Z  1
For complementary Root locus,
(2) 180
Angle of asymptotes A 
PZ
Where   0, 1, 2  P  Z  1
Putting P  3 and Z  1
  0, 1
2  0  1800
For   0, A   00
2
2  1 1800
For   1, A   1800
2
Hence, the correct option is (B).
Sol.32 (D)
The given block diagram is shown below,
K ( s + 2) ( s + 1)
R( s) C (s)
( s 2 - 2 s + 2)

Characteristic equation is given by,

1  G(s)  0
1  K ( s  2) ( s  1)
( s 2  2s  2)
s 2  2s  2  K ( s 2  3s  2)  0
s 2 (1  K )  s (3K  2) s  2  2 K  0
Control Systems 47 Root Locus
Routh Tabulation
s2 1 K 2  2K
s1 3K  2 0
s0 2  2K 0

The intersection of root locus plot with imaginary axis is given by the value of K obtained by solving the
following equation.
3K  2  0
2
K  0.67
3
The auxiliary equation can be formed using the row above the ROZ.
(1  K ) s 2  2  2K  0
 2 2 2
1   s  2  2   0
 3 3
5 s 2  10  0
s 2  2
s   j 2   j1.414
Hence, the correct option is (D).
Sol.33 (D)
The given root locus is shown below,
jw

K=4

s
–1

Since pole at s  1 has multiplicity of 4 hence, open loop transfer function is given by,
K 4
G(s)  
( s  1) 4
( s  1) 4
Hence, the correct option is (D).
Sol.34 (A)
Given : Characteristic equation s 2  5Ks  10  0
5 Ks
1 0 … (i)
s  10
2

Standard C.E. 1  G( s) H ( s)  0 … (ii)

On comparing equation (i) and (ii)
5Ks
GH ( s)  2
s  10
Control Systems 48 Root Locus
Poles at s   j10 and zero at s  0
Root locus starts from open loop poles s   j10 ( K  0) and ends at open loop zero or virtual zero
( K  )
Hence, the correct option is (A).
Sol.35 (C)
Open loop transfer function,
K
G (s) H (s) 
s ( s  4) ( s  2)
Characteristic equation is given by,
1  G(s) H (s)  0
s ( s 2  6s  8)  K  0
s 3  6 s 2  8s  K  0
Routh Tabulation
s3 1 8
s2 6 K
8 K
s1 0
6
s0 K 0

The intersection of root locus plot with imaginary axis is given by the value of K obtained by solving the
following equation.
8 K
0
6
K  48
The auxiliary equation can be formed using the row above the ROZ.
6s 2  K  0
6s 2  48  0
s 2  8
s   j 2.83
Hence, the correct option is (C).
Sol.36 (A)
K ( s  3)
Given : Open loop transfer function G ( s ) H ( s ) 
s ( s  2)
Any point (  j) on root locus satisfies the angle criteria for transfer function.
K (  j  3)
So, G ( s ) s  j 
(  j) (  j  2)

G(  j)  1800

Control Systems 49 Root Locus

   1    1   
00  tan 1    tan    tan    180
0

 3  2

   
  
       2   1800
tan 1    tan 1

 3  2 
1   (  2) 
 
   2  
 2
  3   2  2
2  2  2  2  2  2  3    3  0
2  2  6  6  0
(  3)2  (  0)2  3

 3
2
(  3) 2  (  0) 2 

Above equation represents a circle having center (–3, 0) and radius 3.

Hence, the correct option is (A).
Sol.37 (B)
k ( s  1)
Given : Open loop transfer function G ( s) 
( s  0.4s  0.4)
2

Pole location:
s 2  0.4s  0.4  0
 0.4  0.16  1.6
s
2
 0.4  j12
s   0.2  j 0.6
2

j 0.6

RL
–1 –0.2

- j 0.6

Break-away/break in point :
Characteristic equation is given by,
1  G(s)  0
K ( s  1)
1
( s  0.4 s  0.4)
2
Control Systems 50 Root Locus

 ( s 2  0.4s  0.4)
K
( s  1)
dK
0
ds
d  (2  1)(2s  0.4)  ( s 2  0.4s  0.4) 
ds  ( s  1)2 0

2s 2  0.4s  2s  0.4  s 2  0.4s  0.4  0
s 2  2s  0
s ( s  2)  0
s  0,  2
jw

j 0.6
Virtual zero

s
–1 –0.2

s = -2
Break-in point - j 0.6

Since, the point s  2 is lying between two adjacent zeros (one real and one virtual zero) hence it will be
break-in-point.
Hence, the correct option is (B).
Sol.38 (C)
s 2  0.4s  0.4
K
s 1 s 2

4  0.4  2  0.4
K  3.6
2  1
Hence, the correct option is (C).
Sol.39 (C)
Root locus diagram is shown below,
jw

+ j 0.6
3.6
=

Virtual zero
K

s
–2 –1 –0.2

- j 0.6
Control Systems 51 Root Locus
After s  2 (break-in point) response is over-damped break in point is known as minima point hence
minimum value of K on real axis is 3.6
Hence for K  3.6 system response is over-damped.
Hence, the correct option is (C).
Sol.40 (12)
Given : The open loop transfer function of a unity negative feedback system is,
K
G( s) 
s ( s  1)( s  3)
. Method 1 :
The value of K at the point where root locus branch crosses the imaginary axis is determined by applying
Routh Hurwitz criterion to the characteristic equation.
The characteristics equation is given by,
1  G(s) H ( s)  0
K
1 0
s ( s  1) ( s  3)
s 3  4 s 2  3s  K  0
Routh Tabulation :
s3 1 3
s2 4 K
12  K
s1 0  Row of zeros
4
s0 K 0

The intersection of root locus plot with imaginary axis is given by the value of K obtained by solving the
following equation.
12  K
0  K  12
4
Hence, the value of gain K (  0) at which the root locus crosses the imaginary axis is 12.
. Method 2 :
The characteristics equation is given by,
1  G(s) H ( s)  0
K
1 0
s ( s  1) ( s  3)
s 3  4 s 2  3s  K  0
For third order system to be marginally stable,
IP
s + 4 s + 3s + K = 0
3 2

EP
Control Systems 52 Root Locus
Internal product (IP) = External product (EP)
4  3  1 K
K  12
Hence, the value of gain K (  0) at which the root locus crosses the imaginary axis is 12.
 Key Point
For 3rd order system only,
IP > EP : system is stable.
IP < EP : system is unstable.
IP = EP : system is marginal stable
Sol.41 (1.25)
Given : The open-loop transfer function of a unity-feedback control system is,
K
G( s) 
s  5s  5
2

The characteristics equation is,

1  G (s)  0
K
1 0
s  5s  5
2

K   s 2  5s  5
Break-away / Break-in point :
dK
0
ds
dK
  2s  5  0
ds
s   2.5
Since, point s   2.5 is lying between two adjacent poles on the root locus of real axis, hence it is a
break-away point.
s = - 2.5 jw
Breakaway
point

s
-3.615 -1.385

Since, Break-away point will lie on root locus. Hence, from magnitude condition,
G( s) s 2.5  1 … (i)

Magnitude of G ( s ) at s   2.5 is given by,

K
G ( s ) s  2.5 
( 2.5)  5  ( 2.5)  5
2
Control Systems 53 Root Locus
K
G ( s ) s  2.5  … (ii)
1.25
From equation (i) and (ii),
K
1
1.25
K  1.25
Hence, the value of K at the break-away point is 1.25.
Sol.42 (8.6  103 )
The given root locus diagram is shown below,
jw

s
-2 -1 3 5

From above figure open loop transfer function is given by,

K ( s  3) ( s  5)
GH ( s ) 
( s  1) ( s  2)
Calculation of breakaway point :
Characteristic equation is given by,
1  GH ( s)  0
K ( s  3) ( s  5)
1 0
( s  1) ( s  2)

 ( s  1) ( s  2)  ( s 2  3s  2)
K  2
( s  3) ( s  5) ( s  8s  15)
dK
0
ds
d   ( s 2  3s  2) 
0
ds  ( s 2  8s  15) 

( s 2  8s  15) (2s  3)  ( s 2  3s  2) (2s  8)  0

11s 2  26s  61  0
26  262  4 11 61
s
2 11
and s  1.45
Since break-away point lies between 2  s  1 hence valid break away point is s  1.45
 Control Systems 54 Root Locus
Calculation of gain K :
 ( s  1) ( s  2)
K
( s  3) ( s  5) s 1.45

 (1.45  1) (1.45  2)
K  8.6 103
(1.45  3) (1.45  5)
Sol.43 (0.28)
Given : Open loop transfer function,
K ( s  1)( s  2)
G (s) 
( s  0.1) ( s  1)
 Key Point
Break-away point is known as maxima point because at this point point the value of K is maximum on the
real axis
Calculation of maxima point (Breakaway point) :
Characteristic equation is given by
1  G(s)  0
K ( s  1) ( s  2)
1 0
( s  0.1) ( s  1)
 ( s 2  0.9 s  0.1)
K
s 2  3s  2
dK
0
ds
d   ( s 2  0.9s  0.1) 
0
ds  s 2  3s  2 
( s 2  3s  2) (2s  0.9)  ( s 2  0.9s  0.1) (2s  3)  0
3.9s 2  4.2s  1.5  0
 4.2  4.22  4 1.5  3.9
s
2  3.9
s  0.28 and s  1.35
Pole-zero plot of G( s) is given by,
jw

RL –1.35 RL
s
–2 –1 –0.1 0.28 1

Hence, the valid breakaway point is s  0.28 .


Practice Solutions :
SOL.1
Given : Open loop transfer function is,
1
G ( s)  …(i)
s ( s  2) ( s  4)
Put s  j in equation (i),
1
G ( j) 
j( j  2) ( j  4)
Magnitude of open loop transfer function G ( j) H ( j) is given by,
1
      G ( j) H ( j)  …(ii)
   4 2  16
2

Phase angle of open loop transfer function G ( j) H ( j) is given by,
   
      G ( j) H ( j)  900  tan 1    tan 1   …(iii)
2 4
Calculation of  pc :
Phase angle at phase crossover frequency is given by,
G ( j pc ) H ( j pc )  1800 …(iv)
From equation (iii) and (iv),
  pc  1 
 pc 
      900  tan 1    tan    180
0

 2   4 
  pc  1 
 pc 
tan 1    tan    90
0

 2   4 
  pc  pc 
  
tan 
1 2 4   900
  2

 1
pc

 8 
 Control Systems 2 Polar Plot
3 pc
4 1
2pc 0
1
8
2pc
1 0
8
 pc  2 2 rad/sec
From equation (ii), gain at phase crossover frequency is,
1
G ( j pc ) H ( j pc ) 
2 2 8  4 8  16
1
G ( j pc ) H ( j pc ) 
2 2 12 24
1 1
G ( j pc ) H ( j pc )  
8  6 48
Gain margin is given by,
1
GM 
G ( j pc ) H ( j pc )
1
GM   48
1/ 48
Gain margin in dB,
GM dB  20 log10 48  33.62 dB
Hence, the gain margin is 33.62 dB.
SOL.2
1
Given : G ( j) H ( j) 
( j  1)3  
Magnitude of open loop transfer function G ( j) is given by,
1
G ( j) H ( j)    …(i) 
 
3
2  1

Magnitude at gain crossover frequency is given by,

G ( jgc ) H ( jgc )  1
From equation (i),
  1
1
 
3
       gc  1
2

ω gc  0 rad/sec  
Phase angle of open loop transfer function G ( j) is given by,
ω
       G ( jω) H ( jω)   3 tan 1             …(ii) 
1
 Control Systems 3 Polar Plot
Phase margin is given by,
PM  1800   G ( jω gc ) H ( jω gc )    
From equation (ii),   
      PM  1800  3 tan 1 (0)  1800 or π rad
Hence, the correct option is (B).
SOL.3
1
Given : G ( s ) 
( s  1)3
Put s  j in above equation,
1
G ( j) 
( j  1)3
Magnitude of open loop transfer function G ( j) is given by,
1
G ( j)  3
…(i)
(  1)
2 2

Phase angle of open loop transfer function G ( j) is given by,

ω
       G ( jω)   3 tan 1       …(ii)
1
. Method 1 :
Phase angle at phase crossover frequency is given by,
G ( j pc )  1800  
From equation (ii),
 3 tan 1 (ω pc )  1800

ω pc  tan 600  3 rad/sec

Gain margin is given by,
1
GM   
G ( jω pc ) H ( jω pc )
From equation (i),
1 1
G ( j pc )  3

8
(3  1) 2

1
GM  8
1/ 8
Hence, the correct option is (B).
. Method 2 :
To calculate  pc , equating imaginary part of G ( j) to zero.

Img   j3  1  32  j 3  0

 3pc  3 pc  0
Control Systems 4 Polar Plot

 pc  3 rad/sec
Img

-1800 w=¥ w=0

Re
0

w pc = 3

1 1
G ( j pc )  3

8
(3  1) 2

1
GM  8
1/ 8
Note : This method is only valid when polar plot crosses  1800 axis.
 Key Point
Whenever our OLTF has no S terms in numerator, then there is no need to rationalize we can directly
equate imaginary part of denominator to zero for finding  pc .

SOL.4
K
Given : G ( s ) 
s (0.2 s  1)(0.05s  1)
For K  1 , GM = 28 dB
In dB, GM  20 log GM
28  20 log GM
GM  25.11
. Method 1 :
Put s  j in the given equation,
K
G ( j) 
j(0.2 j  1)(0.05 j  1)
Magnitude of open loop transfer function G ( j) is given by,
K
G ( j) 
 (0.2) 2  1 (0.05) 2  1
Gain margin is given by,
1
GM 
G ( j pc )
1
25.11 
G ( j pc )

G ( j pc )  0.0398 …(i)
 Control Systems 5 Polar Plot
K
      G ( j pc ) 
 pc (0.2 pc ) 2  1 (0.05 pc ) 2  1

1
      0.0398 
 pc 0.04  1 0.00252pc  1
2
pc

1
      0.0398 
0.2  0.05   pc 2pc  25 2pc  400

100
0.0398 
 pc   25 2pc  400
2
pc

2
 100 
2pc (2pc  25) (2pc  400)     6312971.97
 0.0398 
Let 2pc  x , hence
x( x  25) ( x  400)  6312971.97
x3  425 x 2  104 x  6312971.97  0
By using calculator,
x  100.5,  341.6,  183.85
Thus,  pc  10.02, j18.48, j13.55 rad/sec
Since,  pc is a positive real frequency,
Therefore,  pc  10.02  10 rad/sec
For GM  20 dB ,
In dB, GM  20 log GM
20  20 log GM
GM  10
Gain margin is given by,
1
GM 
G ( j pc )
1
10 
G ( j pc )

G ( j pc )  0.1
K
      G ( j pc ) 
 pc 0.04  1 0.00252pc  1
2
pc

K
      0.1 
10 0.04  10  1 0.0025  102  1
2

K  2.5
Hence, the correct option is (C).
Control Systems 6 Polar Plot

. Method 2 :
K
G(s) 
s (0.2 s  1) (0.05s  1)
K
G ( s) 
 s  s 
s   1   1
 5   20 
100 K
G ( s)  …(i)
s ( s  5) ( s  20)
100 K
G ( s) 
s  25s 2  100 s
3

Put s  j in equation (i),

100 K
G ( j)    …(ii)
j( j  5) ( j  20)
100 K
G ( j) 
 j  252  j100
3

Magnitude of G ( j) is given by,

100 K
G ( j) 
 2  25 2  400
- 2700
Img

w=¥
-1800 00
Re
w pc

w=0
- 900

Fig. Type 1, order 3

From equation (i),
Img G ( j)   0

Img   j3  j100  252   0

3pc  100 pc  0
 pc  10 rad/sec
Gain margin is given by,
1
GM  …(ii)
G ( j pc )
From equation (i),
K
G ( j pc )  …(iii)
25
Control Systems 7 Polar Plot
GM  20 dB [Given]
In dB, 20  20 log GM
GM  10
From equation (ii) and (iii),
25
10 
K
K  2.5
Hence, the correct option is (C).
 Key Point
If we change the value of gain (K) of open loop transfer function, then
(i) Phase crossover frequency will not change.
(ii) Gain crossover frequency, gain margin and phase margin will change.

. Method 3 :
K mar
GM 
K desired
For a system, K mar is fixed.
K desired  1 [Given]
K mar  GM  K desired  25.11
In dB, GM (dB)  20 log GM
20  20 log GM [Given]
GM  10
K mar 25.11
K desired    2.511
GM 10
Hence, the correct option is (C).
SOL.5
(1  s )
Given : G ( s ) H ( s )  …(i)
(1  s )(2  s )
Put s  j in equation (i),
(1  j)
G ( j) H ( j) 
(1  j) (2  j)
Magnitude of open loop transfer function G ( j) H ( j) is given by,

1  2
G ( j) H ( j) 
1  2 4  2
1
G ( j) H ( j)  …(ii)
4  2
 Control Systems 8 Polar Plot

  G ( j) H ( j)  
0  0.5  
  0 

Magnitude at gain crossover frequency is given by,

G ( jgc ) H ( jgc )  1 …(iii)
From equation (ii) and (iii),
1
1
4  2gc

2gc   3

gc   j 3
Since, gc is a positive real frequency.
Therefore, gc does not exist.
Phase margin is given by,
PM  1800  G ( jgc ) H ( jgc )
Since, gc does not exist, hence phase margin does not exist or we can consider as  .
Hence, the correct option is (D).
SOL.6
1
Given : G ( s ) H ( s )  …(i)
s ( s  s  1)
2

. Method 1 :
Put s  j in equation (i),
1
G ( j) H ( j) 
j (1  2  j)
Magnitude of open loop transfer function G ( j) H ( j) is given by,
1
G ( j) H ( j)            …(ii)
 (1  2 ) 2  2
Phase angle of open loop transfer function G ( j) H ( j) is given by,
  
      G ( j) H ( j)   900  tan 1  2 
 1  
Calculation of  pc :
Phase angle at phase crossover frequency is given by,
G ( j pc ) H ( j pc )  1800

  pc 
 900  tan 1    180
0
 1  2
 pc 
 Control Systems 9 Polar Plot

  pc 
tan 1    90
0
 1  2
 pc 
1  2pc  0   pc   1 rad/sec

Since,  pc is a positive real frequency.

Therefore,  pc  1 rad/sec
From equation (ii),
1
      G ( j pc ) H ( j pc )  1
1 (1  12 ) 2  12
Gain margin is given by,
1 1
GM   1
G ( jω pc ) H ( jω pc ) 1
In dB, GM  20 log GM
GM (dB)  20 log(1)  0 dB
Hence, the correct option is (B).
. Method 2 :
1
G ( j) H ( j) 
 j  2  j
3

1
G ( j) H ( j)  …(i)
   j (  3 )
2

- 2700
Img

w=¥
-1800 00
Re
w pc

w=0
- 900

Fig. Type 1, order 3

At phase crossover frequency  pc , imaginary part of G ( j) H ( j) will be zero.
From equation (i),
Img G ( j) H ( j)   0

Img    j (1  2 )  0

1  2pc  0

 pc  1 rad/sec
Control Systems 10 Polar Plot
From equation (i),
G ( j pc ) H ( j pc )  1
In dB, gain margin is given by,
1
GM  20 log10
G ( j pc ) H ( j pc )
GM  20 log10 1  0 dB
Hence, the correct option is (B).
. Method 3 :
1
G (s) H (s)   
s ( s  s  1)
2

K
If we consider, G ( s ) H ( s ) 
s ( s  s  1)
2

From above open loop transfer function,

K desired  1
Gain margin is given by,
K
GM  mar …(i)
K desired
The characteristic equation is given by,
1  G ( s) H ( s)  0
K
1 0
s ( s  s  1)
2

s3  s 2  s  K  0
For third order system to be marginal stable,
Internal product = External product
1 1  1 K
K  K mar  1
From equation (i),
1
GM   1
1
In dB, GM  20 log1  0 dB
Hence, the correct option is (B).
SOL.7
10e  Ls
Given : G ( s) H ( s) 
s
and ω pc  5 rad/sec
Put s  j,
10e  Lj
G ( j) H ( j) 
j
 Control Systems 11 Polar Plot
Phase angle of open loop transfer function G ( j) is given by,
π
       G ( jω) H ( jω)   Lω   
2
Phase angle at phase crossover frequency is given by,
G ( j pc ) H ( j pc )  1800
π
 Lω pc   π
2
π π
L 
2ω pc 10
Hence, the correct option is (B).
SOL.8
s
Given : G (s) H (s) 
( s  100)3
Put s  j in above equation,
j
G ( j) H ( j) 
( j  100)3
Phase angle of open loop transfer function G ( j) H ( j) is given by,
  
G ( j) H ( j)  900  3 tan 1  
 100 
Magnitude of open loop transfer function G ( j) H ( j) is given by,

G ( j) H ( j)  3
[(100)   ]
2 2 2

Calculation of  pc :
Phase angle at phase crossover frequency is given by,
G ( j pc ) H ( j pc )  1800
  pc 
900  3 tan 1    180
0

 100 
  pc 
3 tan 1    270
0

 100 
  pc 
tan 1    90
0

 100 
  pc 
   tan 90
0
  pc  
 100 
OR

From above table,  pc  

Control Systems 12 Polar Plot
From table gain at phase crossover frequency is,
G ( j pc ) H ( j pc )  0
Gain margin is given by,
1 1
GM   
G ( jω pc ) H ( jω pc ) 0
Calculation of phase margin :
. Method 1 :

Hence the polar plot of open loop transfer function G ( j) H ( j) is shown below,
Img 900
0 + j1

3.7 ´ 10-5

-1 + j 0 w=0 1 + j0
w=¥ w1 Re
-1800 00

2.16 ´ 10-5
w2
Unit circle
0 - j1
- 90 0

Fig. Polar plot

From above figure, since polar plot is not intersecting unit circle, hence gain crossover frequency gc
does not exist and therefore phase margin does not exist or we can consider as  .
Hence, the correct option is (B).
. Method 2 :
Magnitude at gain crossover frequency is given by,
G ( jgc ) H ( jgc )  1

gc
1
(10  2gc )3/ 2
4

2gc  (104  2gc )3

Since, above equation will not be satisfied for any value of gc , hence gc does not exist and therefore
phase margin does not exist or we can consider as  .
Hence, the correct option is (B).
Control Systems 13 Polar Plot

. Method 3 :
For GM   option (B) and option (C) is satisfying.
Option (C) : In option (C), PM  00
Phase margin is given by,
PM  1800  G ( jgc ) H ( jgc )

00  1800  G ( jgc ) H ( jgc )

G ( jgc ) H ( jgc )  1800

From the table explained in method 1, at    G ( jgc ) H ( jgc ) is 1800 , hence    may be the
gain crossover frequency.    , will be gain crossover frequency if at this frequency G ( jgc ) H ( jgc )

will be one. But from the table G ( jgc ) H ( jgc ) is zero at    , hence    will not be gain crossover
frequency and therefore PM  00 is not correct.
Thus, the option (C) is wrong.
Hence, the correct option is (B).
SOL.9
Given : The open loop transfer function is,
2(1  s )
G(s) H (s) 
s2
Put s  j in above equation,
2(1  j)
G ( j) H ( j) 
( j) 2
Magnitude of open loop transfer function G ( j) H ( j) is given by,

2 1  2
G ( j) H ( j) 
2
Phase angle of open loop transfer function G ( j) H ( j) is given by,
G ( j) H ( j)   1800  tan 1
  G ( j) H ( j)   G ( j) H ( j)  

0     1800  
  0   900  

From above table,  pc  0 rad/sec

and G ( j pc ) H ( j pc )  
Gain margin is given by,
1 1
GM   0
G ( j pc ) H ( j pc ) 
Control Systems 14 Polar Plot

- 2700
Img

-1800 w=¥
00
w=0 Re

- 900

Fig. Type 2, order 2

Hence, the correct option is (B).
SOL.10
K
Given : G ( s ) 
( s  1) ( s  2)
Put s  j in above equation,
K
G ( jω) 
( jω  1) ( jω  2)
Magnitude of open loop transfer function G ( j) is given by,
K
G ( jω) 
(ω2  1) (ω2  4)
Phase angle of open loop transfer function G ( j) is given by,

G ( j)   tan 1   tan 1
2
  G ( j)   G ( j )  
K
0    00  
2
  0  1800  
Polar plot for open loop transfer function can be drawn as shown in figure,
-2700
Img

w=¥ K /2
-1800 00
w=0 Re

-900

Fig. Type 0, Order 2

From table, ω pc  

G ( jω pc )  0
Control Systems 15 Polar Plot
Gain margin is given by,
1 1
GM   
G ( jω pc ) 0
In dB, GM  20 log ()
1
GM  20 log    (20 log1  20 log 0)
0
GM  0  ( )   dB
Hence, the correct option is (D).
 Key Point
In case of system having only finite poles,
(i)  pc of second order system is always infinite.
(ii) Gain margin of second order system is always infinite.
SOL.11

Given : Phase margin 
4
The open-loop transfer function of a unity feedback system is,
as  1
G ( s)  2
s
Put s  j in above equation,
aj  1
G ( j) 
( j) 2
Magnitude of open loop transfer function G ( j) is given by,

a 2 ω2  1
G ( jω) 
ω2
Phase angle of open loop transfer function G ( j) is given by,
G ( j)  1800  tan 1 (aω)
Magnitude at gain crossover frequency is given by,
G ( jω gc )  1

1  2gc a 2
1
2gc
ω4gc  a 2 ω2gc  1  0 …(i)
Phase margin is given by,
PM  1800  G ( jgc )
1800  ( 1800  tan 1 aω gc )  450
tan 1 (aω gc )  450
aω gc  1 …(ii)
Control Systems 16 Polar Plot
From equation (i) and (ii),
ω4gc  1  1  0  ω4gc  2

ω gc  21/4  1.189 rad/sec

From equation (ii),
1 1
a   0.841
ω gc 1.189
Hence, the correct option is (C).
SOL.12
as + 1
d(t ) y (t )
s2
(Impulse
response)
as  1 as  1
Y (s)  2
 L{(t )} 
s s2
Taking inverse Laplace transform of Y ( s ) ,
 as  1 
y (t )  L1{Y ( s )}  L1  2 
 s 
 as  1
y (t )  L1  2   L1  2 
s  s 
a  1
y (t )  L1    L1  2 
s s 
y (t )  au (t )  tu (t )
At t  1 second, y (t ) is given by,
y (t  1)  a  1
y (t  1)  0.84  1  1.84
Hence, the correct option is (C).
SOL.13
K
Given : G ( s )  and H ( s )  1
( s  5)3
Put s  j,
K
G ( j) 
( j  5)3
Magnitude of open loop transfer function G ( j) is given by,
K
G ( j)  …(i)
(  25)3
2

Phase angle of open loop transfer function G ( j) is given by,

 
 G ( jω)   3 tan 1       …(ii) 
5
 Control Systems 17 Polar Plot
Phase margin is given by,
PM  1800   G ( jω gc )  
From equation (ii),
 ω gc 
450  1800  3 tan 1   
 5 
 gc 
      tan 1    45  
0

 5 
gc
       tan 450  
5
      ω gc  5 rad/sec  
Magnitude at gain crossover frequency is given by,
G ( jgc )  1
From equation (i),
K
1
 
3
2gc  25

K  250 2
Hence, the correct option is (B).
SOL.14
Given :   0.5
. Method 1 :
Closed loop transfer function is given by,
G(s)
T (s) 
1  G ( s) H ( s)
K
( s  5)3 K
T (s)  
1
K
 1 ( s  5)3  K
( s  5)3
The characteristic equation is given by,
( s  5)3  K  0
s 3  15s 2  75s  K  125  0 …(i)
The characteristic equation for the dominant conjugate pole pair is given by,
( s 2  2n s  n2 )( s  )  0
( s 2  n s  2n )( s  )  0
s 3  (  n ) s 2  (n  n2 ) s  2n  0 …(ii)
Comparing equation (i) and (ii),
  n  15 …(iii)
Control Systems 18 Polar Plot

K  125  2n …(iv)

n  n2  75
(  n ) n  75
From equation (iii),
15n  75
n  5 rad/sec
  15  5  10
From equation (iv),
K  125  10  (5) 2  250
K  125
Hence, the correct option is (B).
. Method 2 :
Root Locus Concept
K
Given : G ( s ) H ( s ) 
( s  5)3
Number of poles P = 3
Number of zero Z = 0
Number of branches terminating at infinite
=B=P=3
Angle of asymptotes is given by,
(2  1)1800
A    0,1, ..... P  Z  1
PZ
(2  1)1800
A 
3
  0, 1, 2
A  600 , 1800 , 3000
Centroid is given by,
5  5  5  0
  5
3
Intersection with imaginary axis can be calculated using Routh array.
Characteristics equation is given by,
K
1 0
( s  5)3
s 3  125  15s 2  75s  K  0
s 3  15s 2  75s  K  125  0
Control Systems 19 Polar Plot
Routh table :
s3 1 75

s2 15 K  125
1125  K  125
s1 0
15
s0 K  125 0
Row of zeros can be formed by equating
1125  K  125
0
15
i.e. K mar  1000
Auxiliary equation can be formed as,
15s 2  125  K  0
15s 2  1125  0
s 2  75
s   j5 3
n  5 3 rad/sec
Root locus can be drawn as shown below.
Virtual zero
jw
K =¥
K = 1000
j5 3
A
Virtual zero
K =0 f q s
K =¥ -5 B 0

- j5 3
K = 1000
K =¥

Virtual zero
We know that, cos   
  cos 1 0.5  600
In  AOB,
5 3
tan    3
5
  600
So, BAO  1800  (600  600 )  600
Since all the angles are equal, sides of triangle will also be same. So, it is a equilateral triangle.
Graphical calculation of K :
Product of length from pole to point B
K
Product of length from zero to point B
 Control Systems 20 Polar Plot
AB  AB  AB
K  5 5 5
1
K  125
Hence, the correct option is (B).
SOL.15
150
Given : G ( s )  , H (s)  1
s ( s  9)( s  25)
Magnitude of open loop transfer function G ( j) is given by,
150
      G ( jω) H ( jω) 
ω 2  92 2  252
Phase angle of open loop transfer function G ( j) is given by,
  
      G ( j) H ( j)   900  tan 1    tan 1  
9  25 
Phase angle at phase crossover frequency is given by,
      G ( j pc )  1800

  pc  1 
 pc 
      900  tan 1    tan    180
0

 9   25 
 pc  pc
      tan 1  tan 1  900
9 25
  pc  pc 
  
1  9 25 
      tan  900
  pc 2

1  
 225 

 A B 
Apply tan 1 A  tan 1 B  tan 1  
 1  AB 
  pc  pc 
  
       9 25 
 tan 900  
  pc  2

 1  
 225 
 pc2
      1  0   pc  15 rad/s
225
OR
For calculating  pc equating imaginary part of G ( j) to zero,

Img  j pc (9  j pc ) (25  j pc )   0

 pc (225  2pc )  0
 pc  15 rad/sec
Note : This method is only valid when polar plot crosses  1800 axis.
 Control Systems 21 Polar Plot
Gain margin is given by,
1
      GM   
G ( jω pc ) H ( jω pc )

150
      G ( jω pc ) H ( jω pc ) 
ω pc 2pc  92 2pc  252
 
150
      G ( jω pc ) H ( jω pc )   
15 152  92 152  252
      G ( jω pc ) H ( jω pc )  0.0196  
1
In dB, GM  20 log  34.1 dB  
0.0196
Hence, the correct option is (C).
SOL.16
100
Given : G(s)  …(i)
( s  1)3
Put s  j in equation (i),
100
G ( j) 
( j  1)3
Magnitude of G ( j) is given by,
100
G ( j)  …(ii)
( 1  2 )3
Phase angle of G ( j) is given by,
G ( j)   3 tan 1          …(iii)
Calculation of  pc :
Phase angle at phase crossover frequency is given by,
G ( j pc )   1800         …(iv)
From equation (iii) and (iv),
3 tan 1  pc   1800  

       pc  tan 600  3 rad/sec

Hence, the correct option is (A).
SOL.17
s
Given : Open loop transfer function G ( s ) 
1 s
Put s  j in the above equation,
j
G ( j) 
j  1
Control Systems 22 Polar Plot
Phase angle of G ( j) is given by,
G ( j)  900  tan 1 ()
Magnitude of G ( j) is given by,

G ( j) 
2  1
 G ( j) G ( j)
0 0 900
 1 00
Polar plot of G ( j) is given below,
Img

w=0 w=¥
Re

Hence, the correct option is (B).

SOL.18
The given polar plot is shown below,
Im
j1
0.25

–1 Re
300

- j1

1
Gain margin  4
0.25
Phase margin  1800  GH ( jgc )
From above figure GH ( jgc ) = 300
PM  1800  300  1500
Hence, the correct option is (A).
SOL.19
The given polar plot is shown below,
Img

w=¥ w=0
Re
0 1
– 1.18
Control Systems 23 Polar Plot
From above figure
G ( j) 0  1

Only in option (C) G ( j) 0  1

Hence, the correct option is (C).
SOL.20
Given : (i) Critical value of gain  K m arg inal  40
(ii) K desired  20
K mar 40
GM   2
K desired 20
GM(dB)  20 log 2  6 dB
Hence, the correct option is (C).
SOL.21
Given : Open loop transfer function
K
G(s) H (s) 
s  6 s  11s  6
3 2

Put s  j in the above equation

K
G ( j) H ( j) 
( j)  6 ( j) 2  11( j)  6
3

K
G ( j) H ( j) 
 j  6 2  11 j  6
3

At    pc , Img[G ( j pc ) H ( j pc )]  0

  pc 3  11  pc  0

 pc  11 rad/sec
Hence, the correct option is (B).
SOL.22
Key Point :
Phase crossover frequency and gain margin for standard third order transfer function :
K
G ( s) H ( s)  ,
s ( sT1  1) ( sT2  1)
1 T T
 pc  , G.M.  1 2
T1T2 KT1T2
1
P. G ( s)  … (i)
s ( s  1)( s  2)
2
G ( s)  … (ii)
 s
s (1  s ) 1  
 2
Control Systems 24 Polar Plot
On comparing equation (i) and (ii)
1
T1  1 and T2 
2
1 1
 pc    2 rad/sec
T1T2 1
1
2
1
Q. G ( s) 
s ( s  5)( s  10)
1
G(s)  50 … (iii)
 s  s 
s 1   1  
 5   10 
On comparing equation (i) and (iii)
1 1
T1  and T2 
5 10
1
 pc   50 rad/sec
1 1

5 10
1
R. G ( s)  … (iv)
s (1  0.1s ) (1  0.01s )
On comparing equation (i) and (iv)
T1  0.1 and T2  0.01
1
 pc   100 rad/sec
0.1 0.01
1
S. G(s) 
s (1  0.1s ) ( s  2)
1
G ( s)  2 … (v)
s (1  0.1s ) (1  0.5s )
On comparing equation (i) and (v)
T1  0.1 and T2  0.5
1
 pc   20 rad/sec
0.1 0.5
Hence, the correct option is (C).
SOL.23
The given polar plot is shown below,
Im GH ( s )
- 0.75

-1 Re GH ( s )

- 0.375
300
Control Systems 25 Polar Plot
1
Since, GM 
GH ( j pc )
From given figure,
1 4
GM  
0.75 3

Since, PM  1800  GH ( jgc )

PM  1800  (300  900 )

PM  600
Hence, the correct option is (A).
SOL.24
The given polar plot is show below,

GM in dB is given as,
1
GM  20 log10
GH ( j pc )

1
GM  20 log10  20 log10 2  6 dB
0.5

PM  1800  GH ( jgc )

PM  1800  600  1200

Hence, the correct option is (A).
SOL.25
The given Nyquist plot is shown below,
Control Systems 26 Polar Plot
Hence, Transfer function is a type ‘0’ and order ‘2’ system
Thus, only option (A) and (B) can be correct
1
From figure Gain at   0 is
4
1 1 1
(A)  GH ( j) 0  
( s  2) 2
(0  2) 2
4
1 1 1
(B) 2  GH ( j) 0  
s  3s  2 0  3 0  2 2
Hence, the correct option is (A).
SOL.26
Given nyquist plot is shown below,

Nyquist stability criteria is given by,

N  PZ
For stable OLTF P  0
From Nyquist plot,
N   2 (Two clockwise incirclement)
2  0 Z
Z 2
Hence, CLTF has two right hand poles
Thus closed loop system will be unstable.
For stable OLTF and unstable CLTF PM  00 and GM  0 dB
Only in figure (B) PM  00 and GM  0 dB
Hence, the correct option is (B).
Key point : P.M. and G.M. of stable, marginally stable and unstable system is shown below,
wgc < w pc wgc = w pc wgc > w pc
Stable system Marginally stable system Unstable system
G.M. > 1 and P.M. > 00 Img G.M. = 1 and P.M. = 00 Img G.M. < 1 and P.M. < 00 Img

1/G.M. Unit radius 1/G.M. Unit radius 1/G.M. Unit radius

circle circle w = wgc circle
w = w pc P
w = wgc w = w pc
Q w=¥ w=¥ P.M. w=¥
P.M.= 00 < 00
-180 0
O
Re -1800
O
Re -180 0
O
Re
P.M. P
-1 + j 0 > 00
-1 + j 0
-1 + j 0

P
ÐG ( jw) H ( jw) ÐG ( jw) H ( jw) ÐG ( jw) H ( jw)
w = wgc
w=0 at w = wgc w=0 at w = wgc w=0 at w = wgc
 Control Systems 27 Polar Plot
SOL.27
Given : Open loop transfer function
20
G(s) H (s) 
 1  1 
( s  1)  s    s  
 2  3
20
G ( j) H ( j) 
1  1 
(1  j)   j    j 
2  3 
Phase angle of open loop transfer function G ( j) H ( j) is given by,
G ( j) H ( j)   tan 1 ()  tan 1 (2)  tan 1 (3) … (i)
Calculation of  pc :
Phase angle at phase crossover frequency is given by,
G ( j pc ) H ( j pc )  1800 … (ii)
From equation (i) and (ii)
 tan 1 ( pc )  tan 1 (2 pc )  tan 1 (3 pc )  1800
   2 pc 
tan 1  pc 1
  tan (3 pc )  180
0
 1  22
 pc 
 ab 
Since, tan 1 (a )  tan 1 (b)  tan 1  
 1  ab 
  pc  2 pc 
  3 pc 
 
1  22pc 
tan    1800
 
1  3   pc  2 pc  
pc 
  1  2 pc
2
 


3 pc
 3 pc
1  22pc
 tan(1800 )  0
9 2

1 pc

1  2 2
pc

3 pc
 3 pc  0
1  22pc
1  1  22pc  0
2pc  1   pc  1rad/ sec
Hence, the correct option is (A)
SOL.28
1
Given : G ( s )  , H (s)  1
(1  s ) 2
Control Systems 28 Polar Plot
1
Open loop transfer function G ( s ) H ( s ) 
(1  s ) 2
Phase margin is given by,
PM  1800  GH ( jgc )
1
G ( j) H ( j) 
(1  j) 2
Magnitude at gain crossover frequency is given by,
G ( jgc ) H ( jgc )  1
1
1
 
2
1  2
gc

1  2gc  1  gc  0 rad/sec

Phase angle G ( j) H ( j)  2 tan 1 () 0  00
PM  1800  00  1800   rad
Hence, the correct option is (D)
SOL.29
Given :
(i) Open loop transfer function
5
G (s) H ( s) 
( s  1)(2 s  1)(3s  1)
(ii) f c  0.16 Hz
 pc  2 0.16  1rad/sec
5
G ( j) H ( j) 
( j  1)(2 j  1)(3 j  1)
Gain of the open loop transfer function is given by,
5
G ( j) H ( j) 
(  1)(42  1)(92  1)
2

1
GM  20 log 5
(12  1)(4 12  1)(9  12  1)
2  5  10 10
GM  20 log10  20 log
5 5
GM  20 log 2  6 dB
Hence, the correct option is (A).
SOL.30
If open loop transfer function is unstable then for an unstable closed loop system gain margin and phase
margin can be,
(i) Positive, negative
(ii) Negative, positive
 Control Systems 29 Polar Plot
(iii)Positive, negative
Hence, the correct option us (D).
SOL.31
If open loop transfer function is unstable then for an unstable closed loop system phase margin can be
positive and negative.
Hence, the correct option us (D).
SOL.32
1
Given : G ( s ) H ( s ) 
s ( s  1)(9 s  1)
. Method 1 :
Put s  j in above equation,
1
      G ( j) H ( j) 
j ( j  1)( j 9  1)
Phase angle of open loop transfer function G ( j) H ( j) is given by,
ω  9ω 
      G ( jω) H ( jω)   900  tan 1    tan 1   
1  1 
Phase angle at phase crossover frequency is given by,
      G ( j pc )  1800

 ω pc  1 
9ω pc 
       1800   900  tan 1    tan   
 1   1 
 ω pc  1 
9ω pc 
      900  tan 1    tan   
 1   1 
 ω + 9ω pc 
      900  tan 1  2 
 
 1  9ω pc 
      1  9ω2pc  0  
1
      ω pc  rad/sec  
3
Hence, the phase crossover frequency is 0.33 rad/sec.
. Method 2 :
For calculating  pc equating imaginary part of G ( j) H ( j) to zero, we get

Img  j pc ( j pc  1) (9 j pc  1)   0

 pc ( 92pc  1)  0
1
 pc  rad/sec
3
Control Systems 30 Polar Plot
Img
1

-1800 w=¥
Re

1
w pc =
3
w=0

Note : This method is only valid when polar plot crosses  1800 axis.
Hence, the phase crossover frequency is 0.33 rad/sec.
. Method 3 :
1
G(s)  …(i)
s ( s  1) (9s  1)
Transfer function is in the form below,
K
G ( s)  …(ii)
s ( sT1  1) ( sT2  1)
On comparing equation (i) and (ii),
T1  1, T2  9
Phase crossover frequency is given by,
1 1 1
 pc    rad/sec
T1T2 1 9 3
Hence, the phase crossover frequency is 0.33 rad/sec.
SOL.33
10
Given : G ( s ) 
s ( s  10)
Put s  j in above equation,
10
G ( j) 
j( j  10)
Magnitude of open loop transfer function G ( j) is given by,
10
G( j) 
 2  100
Phase angle of open loop transfer function G ( j) is given by,

G ( j)   900  tan 1
10
Magnitude at gain crossover frequency is given by,
G ( jgc ) H ( jgc )  1

10
1
gc 2gc  100
Control Systems 31 Polar Plot

2gc (2gc  100)  100

4gc  100 2gc  100  0

Put gc  x in above equation,

2

x 2  100 x  100  0
x  0.99,  100
Since, gc is a positive real frequency,

Therefore, gc  0.99

2

gc  0.994 rad/sec

 0.99 
G ( jgc )   900  tan 1  
 10 
G ( jgc )   95.650
Phase margin is defined as,
PM  1800  G ( jgc )

PM  1800  95.650  84.350

Hence, the phase margin is 84.350 .
SOL.34
10
Given : Open loop transfer function G ( s ) H ( s ) 
( s  5)3
10
G ( j) H ( j) 
( j  5)3
Phase angle of open loop transfer function is given by,
 
G ( j) H ( j)  3 tan 1   …(i)
5
At    pc , G ( j pc ) H ( j pc )  1800 …(ii)
From equation (i) and (ii),
  pc 
3 tan 1    180
0

 5 
 pc
 tan(600 )  3
5
 pc  5 3 rad/sec
Gain of open loop transfer function is given by,
10
G ( j) H ( j) 
 
3
2  52
Control Systems 32 Polar Plot
1
Gain margin  20 log
G ( j pc ) H ( j pc )
3
 
5 3 
2
 5 
2

GM  20 log10  
10
1000
GM  20 log10  20 log10 100
10
GM  20  2  40 dB
Hence, the gain margin will be 40 dB.
SOL.35
1
Given : Open loop transfer function GH ( s ) 
s ( s  1)
Put s  j in the above equation,
1
GH ( j)  …(i)
j ( j  1)

Magnitude of transfer function is given by,

1
GH ( j) 
 2  1
Phase angle of transfer function is given by,
GH ( j)  900  tan 1 () …(ii)
Calculation of  gc :

At   gc , GH ( j)  1 …(iii)

From equation (i) and (iii),
1
1
gc 2gc  1

2gc (2gc  1)  1

4gc  2gc  1  0

Let 2gc  X

X 2  X 1  0
1  1  4
X  0.618
2
2gc  0.618
gc  0.7861rad/sec

GH ( jgc )  900  tan 1 (0.7861)  128.17 0

 Control Systems 33 Polar Plot

PM  1800  G ( jgc )
PM  1800  128.170  51.820
Hence, the phase margin will be 51.82.
SOL.36
Given :
K
(i) Open-loop transfer function G ( s ) 
s ( s  5)
(ii) PM  450
Put s  j in the above equation,
K
GH ( j) 
j ( j  5)
Magnitude of open loop transfer function GH ( j) is given by,
K
GH ( j)  …(i)
 2  25
Phase angle of open loop transfer function GH ( j) is given by,
 
GH ( j)  900  tan 1   …(ii)
5
Phase margin is given by,
PM  1800  GH ( jgc ) …(iii)
From equation (ii) and (iii),
 gc 
450  1800  900  tan 1  
 5 
 gc 
tan 1    45
0

 5 
gc  5 rad/sec
Magnitude at gain crossover frequency is given given by,
GH ( jgc )  1 …(iv)
From equation (i) and (iv),
K
1
gc 2gc  25

K  5 52  25  35
Hence, the value of gain K is 35.
SOL.37
Im

3 1
+j j1
2 2
w=¥
Re

w=0
- j1
Control Systems 34 Polar Plot

 1 
3 1  
  j  1800  tan 1  2   1800  300  1500
2 2  3 
 
 2 
  1800  1500  300
PM      300
Hence, the value of phase margin is -30
SOL.38
Given : (i)  pc  gc  3 rad/sec
K
(ii) Open loop transfer function G ( s ) 
( s  1)3
K
G ( j) 
(1  j)3
Magnitude of loop transfer function G ( j) is given by,
K
G ( j)  …(i)
 
3
 12

Magnitude at gain crossover frequency is given by,

G ( jgc )  1 …(ii)

From equation (i) and (ii),

K
1
 
3
 1
2
gc

 
3
2 2
K 3 1 8

Hence, the value of phase margin is -30

SOL.39
Given : The open loop transfer function
1
G (s) H (s) 
(1  s )3
Put s  j in the above equation,
1
G ( j) H ( j) 
1  j
3

Phase angle of G ( j) H ( j) is given by,

G ( j ) H ( j)  3 tan 1 (3) ……….(i)

At    pc , G ( j pc ) H ( j pc )  1800 …(ii)
Control Systems 35 Polar Plot
From equation (i) and (ii),
3 tan 1 ( pc )  1800
 pc  tan(600 )  1.732 rad/ sec
Hence, the value of phase crossover frequency is 1.732 rad/ sec
SOL.40
1
GM  20 log
G ( j pc ) H ( j pc )
3
 
 3
2
GM  20 log  1  
 
GM  18.06 dB
Hence, the value of gain margin is 18.06 dB
SOL.41
Magnitude of G ( j) H ( j) is given by,
1
G ( j) H ( j) 
 
3
1  2

1
G ( j) H ( j) 0  1
 
3
1 0
Since, the magnitude of G ( j) H ( j) at   0 is 1 hence the gain crossover frequency is 0 rad/sec.
Hence, the value of gain crossover frequency is 0 rad/sec.
SOL.42
Phase margin is given by,
PM  1800  G ( jgc ) H ( jgc )
G ( j gc ) H ( jgc )  3 tan 1 (0)  00

PM  1800  0  1800  3.14 rad

Hence, the value of phase margin is 3.14 rad
SOL.43
Given : The open-loop transfer function
1000
G ( s) H ( s) 
s ( s  10) 2
Put s  j is the above equation,
1000
G ( j) H ( j) 
j ( j  10) 2
Phase-angle of G ( j) H ( j) is given by,

G ( j) H ( j)  900  2 tan 1   ….(i)
 10 
   pc , G ( j pc ) H ( j pc )  1800 ….(ii)
 Control Systems 36 Polar Plot
From equation (i) and (ii),
1800  900  2 tan 1 (0.1  pc )

tan 1 (0.1 pc )  450

 pc  10 rad/sec
Hence, the value of phase crossover frequency is 10 rad/sec.
SOL.44
10
Given : G ( s )  …(i)
s ( s  1) 2
Put s  j in equation (i),
10 10
      G ( j)  
j( j  1) 2
 j  j  22
3

Magnitude of open loop transfer function G ( j) is given by,

10
G ( j) 
(2  1)
Phase angle of open loop transfer function G ( j) is given by,
G ( j)   900  2 tan 1 ()
  G ( j ) H ( j ) G ( j  ) H ( j )
0     900  
  0   2700  

- 2700
Img
5

w=¥
-1800 00
Re
w = w0

w=0
- 900

Fig. Type 1, order 3

 Key Point
Phase crossover frequency is the frequency where polar plot or Nyquist plot crosses or touches the negative
real axis. Hence, 0   pc

. Method 1 :
Calculation of  pc :
Phase angle at phase crossover frequency is given by,
G ( j pc )  1800
 Control Systems 37 Polar Plot

 1800   900  2 tan 1 ( pc )

900  2 tan 1  pc
450  tan 1  pc
 pc  0  1 rad/sec
. Method 2 :
To calculate  pc , equating imaginary part of G ( j) to zero.

Img   j3  j  22   0

 3pc   pc  0

 pc 1  2pc   0
 pc  0
and (1  2pc )  0
2pc  1
 pc   1 rad/sec
Since,  pc is a positive real frequency.
Therefore,  pc  1 rad/sec
Note : This method is valid only when polar plot crosses the negative real axis.
Magnitude at phase crossover frequency is given by,
10 10
G ( j pc )   5
 pc ( pc  1) 2
2

At    pc , the polar plot crosses the negative real axis at – 5.

Hence, the correct option is (C).
SOL.45
e  0.1s
Given : Transfer function G ( s ) 
s
Put s  j ,
e  0.1 j
G ( j) 
j
Phase angle of loop transfer function G ( j) is given by,
π
G ( jω)   0.1ω 
2
Phase angle at phase crossover frequency is given by,
G ( j pc )  1800
π
       0.1ω pc   π
2
Control Systems 38 Polar Plot

 0.1 pc    
2
π
ω pc  rad/sec
0.2
Hence, the correct option is (C).

SOL.46
Given : The polar plot of a system is shown in below figure,
Img

w = 10

450 w=0
Re
5  
From the given plot,
| G ( jω) H ( jω) |  5 at ω  0
G ( jω) H ( jω)  450 at ω  10              [ tan 450  1]
From option (A) :
G ( s ) H ( s )  5 (1  0.1 s )

| G ( jω) H ( jω) |  5 1  (0.1ω) 2

G ( jω) H ( jω)  tan 1 (0.1ω)
At ω  0, | G ( jω) H ( jω) |  5
At   10, G ( jω) H ( j)  450
From option (B) :
G ( s ) H ( s )  (1  0.5 s )

| G ( jω) H ( jω) |  1  (0.5ω) 2

G ( jω) H ( jω)  tan 1 (0.5ω)
At ω  0,   | G ( jω) H ( jω) |  1
At   10, G ( jω) H ( j)  78.690
From option (C) :
G ( s ) H ( s )  5(1  10 s )

| G ( jω) H ( jω) |  5 1  (10 ω) 2

G ( jω) H ( jω)  tan 1 (10 ω)
At ω  0,   | G ( jω) H ( jω) |  5
At   10, G ( jω) H ( j)  89.420
 Control Systems 39 Polar Plot
From option (D) :
G ( s ) H ( s )  5(1  s )

| G ( jω) H ( jω) |  5 1  ω2
G ( jω) H ( jω)  tan 1 ω
At ω  0, | G ( jω) H ( jω) |  5
At   10, G ( jω) H ( j)  84.280  
Hence, the correct option is (A).
SOL.47
s2  4
Given : T ( s ) 
( s  1) ( s  4)
Put s  j in above equation,
2  4
      T ( j)   
 j  1  j  4 
Magnitude of T ( j) is given by,
2  4
T ( j) 
1  2 16  2
To find the frequency at which system output is zero, take T ( j)  0

2  4
      0  
1  2 16  2
       2  4  0  
        2 rad/sec  
Hence, the correct option is (C).
SOL.48
Given : The open loop transfer function is,
( s  1)
G ( s) 
s2
Put s  j in above equation,
( jω  1)
G ( jω) 
( jω) 2
Magnitude of G ( j) is given by,

2  1
G ( j)  …(i)
2
Phase angle of G ( j) is given by,
ω
       G ( jω) H ( jω)  1800  tan 1   …(ii)
1
 Control Systems 40 Polar Plot
Phase angle at phase crossover frequency is given by,
G ( j pc ) H ( j pc )  1800 …(iii)
From equation (ii) and (iii),
 ω pc 
1800  1800  tan 1  
       1 
ω pc  0
Gain margin is given by,
1
G.M.=
G ( jω pc ) H ( jω pc )
From equation (i),
G ( jω pc )  

1 1
G.M.   0
| G ( jω pc ) | 
Hence, the correct option is (A).
SOL.49
e  0.1s
Given : G ( s ) 
s
Put . s  j . in above equation
e  j 0.1
G ( j) 
j
Magnitude of G ( j) is given by,
1
G ( jω)  …(i)
ω
Phase angle of G ( j) is given by,
1800
G ( jω)   0.1ω   900  
π
Phase angle at phase crossover frequency is given by,
G ( jω pc )   1800

1800
       1800   0.1ω pc   900
π
1800
900  0.1 pc 


0.1 pc 
2
 pc  5
ω pc  15.7 rad/sec
Control Systems 41 Polar Plot
From equation (i),
1 1
G ( jω pc )  
ω pc 15.7
Gain margin is given by,
1
GM =  
G ( jω pc )
In dB, GM  20 log (15.7)  23.92 dB
Hence, the correct option is (D).
SOL.50
Given : G ( j)  5  j
For positive frequency i.e. 0  ω   , Nyquist plot is termed as polar plot.
Magnitude of transfer function is given by,
G ( j)  25  2
Phase angle of transfer function is given by,
 
G ( j)  tan 1  
5
  G ( j)   G ( j)  
0  5  00  
    900  

Img 900 w=¥

00
Re
5
w=0

Hence, the correct option is (A).

SOL.51
10( s  1)
Given : G ( s ) 
s  10
Put s  j in above equation,
10( j  1)
G ( j) 
( j  10)
Magnitude of open loop transfer function G ( j) is given by,

2  1
G ( j)  10
2  100
Phase angle of open loop transfer function G ( j) is given by,

G ( j)  tan 1   tan 1
10
Control Systems 42 Polar Plot

  G ( j)   G ( j)  
0  1  00  
10   7.106   39.280  
  10  00  

Img First quadrant

900

w=0 w=¥ 00
Re
0 1 10

Hence, the correct option is (A).

SOL.52
Given : Nyquist plots of two functions G1 ( s ) and G2 ( s ) are shown below,
- 2700
Img

w=¥
-1800 00
Re
G1 ( s )

w=0
- 900

Fig. Type 1, order 1 (one pole at origin)

- 2700
Img
w=¥

G2 ( s )

w=0
-1800 00
Re

- 900

Fig. One zero at origin

From figure,
1
G1 ( s )  …(i)
s
G2 ( s )  s …(ii)
Control Systems 43 Polar Plot
Put s  j in equation (i),
1
G1 ( j) 
j
Magnitude of G1 ( j) is given by,
1
G1 ( j) 

Phase angle of G1 ( j) is given by,
G1 ( j)   900
 G1 ( j) G1 ( j)
0   900
 0  900
Put s  j in equation (ii),
G2 ( j)  j
Magnitude of G2 ( j) is given by,
G2 ( j)  
Phase angle of G2 ( j) is given by,
G2 ( j)  900
 G2 ( j  )  G2 ( j  )
0 0 900
  900
Product of G1 ( s ) and G2 ( s ) is given by,
G1 ( s ) G2 ( s )

1
G1 ( s )G2 ( s )   s  1
s
G ( s)  1
Put s  j in above equation,
G ( j)  1
Magnitude of G ( j) is given by,
G ( j)  1
Phase angle of G ( j) is given by,
G ( j)  00
 G ( j) G ( j  )
0 1 00
 1 00
Control Systems 44 Polar Plot

- 2700
Img

-1800 00
1 Re

- 900

Hence, the correct option is (B).

SOL.53
Given : Phase margin  300
K (e - s )
U (s) Y (s)
s

Open loop transfer function is,

Ke  s
G ( s) H ( s)  …(i)
s
Put s  j in equation (i),
Ke  j
G ( j) H ( j)  …(ii)
j
Magnitude of open loop transfer function G ( j) H ( j) is given by
K
G ( j) H ( j)  …(iii)

Phase angle of open loop transfer function G ( j) H ( j) is given by
1800
G ( j) H ( j)  900          …(iv)

Phase margin is given by,
PM  1800  G ( jgc ) H ( jgc )

300  1800  G ( jgc ) H ( jgc )       …(v)

From equation (iv) and (v),
gc  1800
150  90 
0 0


gc  1800
600 

gc  1.047 rad/sec …(vi)
Magnitude at gain crossover frequency is given by,
G ( jgc ) H ( jgc )  1 …(vii)
Control Systems 45 Polar Plot
From equation (iii), (vi) and (vii),
K
1
gc
K  1.047 rad / sec
Hence, the value of K is 1.047 rad/sec.
SOL.54
1
Given : Open loop transfer function G ( s ) H ( s ) 
s ( s  1)( s  0.5)
Put s  j in the above equation
1
G ( j) H ( j) 
j ( j  1) ( j  0.5)
Phase angle of G ( j) H ( j) is given by,
G ( j) H ( j)  900  tan 1 ()  tan 1 (2) …(i)
At    pc , G ( j) H ( j pc )  1800 …(ii)
From equation (i) and (ii),
900  tan 1 ( pc )  tan 1 (2 pc )  1800
tan 1 ( pc )  tan 1 (2 pc )  900
   2 pc 
tan 1  pc   90
0

 1  2 pc 
2

1  22pc  0
1
 pc   0.707 rad/ sec
2
Hence, the correct option is (B)
SOL.55
Given : The given gain-phase plot is shown below,
G ( jw) , dB
w=2
2 dB

0 ÐG ( jw)
w = 10

-2 dB w = 100

-2700 -1800 -1400 -900

From figure,
At   100, G ( j)  1800
Hence,  pc  100 rad/sec
Control Systems 46 Polar Plot

At  pc  100 G ( j pc )  2 dB

20 log G ( j pc )  2

G ( j pc )  0.7940
1
GM  20 log
G ( j pc )
1
GM  20 log  2 dB
0.7940
(ii) Calculation of phase-margin
At   10, G ( j) dB  0 dB
Hence, gc  10 rad/ sec

At gc  10 rad/ sec, G ( j pc )  1400

PM  1800  G ( jgc )

PM  1800  1400  400

Hence, the phase-margin is 400
Hence, the correct option is (B)
SOL.56
1
Given : OLTF G ( s ) H ( s ) 
s (1  sT1 ) 1  sT2 
Put s  j in the above equation,
1
G ( j) H ( j) 
j (1  jT1 ) 1  jT2 
1
G ( j) H ( j)  … (i)
 1  2T12 1  2T22 

G ( j) H ( j)  900  tan 1  T1   tan 1  T2  … (ii)

Calculation of  pc :-

At    pc , G ( j pc ) H ( j pc )  1800 … (iii)
From equation (ii) and (iii),
900  tan 1 ( pcT1 )  tan 1 ( pcT2 )  900

tan 1 ( pcT1 )  tan 1 ( pcT2 )  900

 T   T 
tan 1  pc 1 2 pc 2   900
 1   pcT1T2 
1
1  2pcT1T2  0  pc 
T1T2
Control Systems 47 Polar Plot
1
G ( j pc ) H ( j pc ) 
 pc 1  2
pc 1
2
T 1  
2
pc 2
2
T 
T1T2 T1T2
G ( j pc ) H ( j pc )  
 T12   T22   T2T1   T1T2 
1   1     
 T1T2   T1T2   T2   T1 
T1T2  T1T2
G ( j pc ) H ( j pc ) 
T1  T2 
2

T1T2
G ( j pc ) H ( j pc ) 
T1  T2
1
GM 
G ( j pc ) H ( j pc )
T1  T2
GM 
T1T2
Hence, the correct option is (A)
SOL.57
K
Given : OLTF G ( s ) H ( s ) 
(1  s )(1  2 s )(1  3s )
Put s  j in the above equation
K
G ( j) H ( j) 
(1  j)(1  2 j)(1  3 j)
Phase angle of G ( j) H ( j) is given by,
G ( j) H ( j)   tan 1 ()  tan 1 (2)  tan 1 (3) …(i)
Calculation of  pc :

At    pc , G ( j pc ) H ( j pc )  1800 …(ii)
From equation (i) and (ii),
 tan 1 ( pc )  tan 1 (2 pc )  tan 1 (3 pc )  1800

 3 pc 
tan 1  1
  tan (3 pc )  180
0
 1  22
 pc 
 3 pc 
  3 pc 
1  2 pc
2

tan 1    1800
 92pc 
 1 
 1  2 pc 
2

3 pc
0
1  22 pc
Control Systems 48 Polar Plot

1  1  22pc  0
2pc  1   pc  1rad/se c
Hence, the correct option is (B)
SOL.58
Given : The given Nyquist plot is shown below,
Img

Re

Increasing
w

From above figure,

 GH ( j) 0  900  one pole at origin
GH ( j)   1800  second order system
Hence, OLTF is a type 1 and order 2 system
K
Consider G ( s ) 
s (s  a)
Only option (B) will satisfy the above open –loop transfer function GH G H ( s )
Hence, the correct option is (B).
SOL.59
Given : The given Nyquist plot is shown below,
Img

1.0
Re
w=¥ w=0

From above figure,

GH ( j) 0  00 , GH ( j) 0  1
And GH ( j)   1800
Only option (B) will satisfy this conditions
Hence, the correct option is (B)
SOL.60
Given : Gain margin is given by,
1
GM 
Gain of open loop system
Hence is gain of open loop system is doubled then the gain margin of the system is halved.
Hence, the correct option is (C).
 Control Systems 49 Polar Plot
SOL.61
Given : Open loop transfer function
2
G ( s) H ( s) 
s ( s  1)(2s  1)
Key point :
Hence T1  1, T2  2 and K  2
1  T1  T2 
GM   
K  T1T2 
1  1 2  3
GM   
2  1 2  4
Hence, the correct option is (A)
SOL.62
Given : Open loop transfer function
1
G ( j) 
( j) (1  jT )
2

Magnitude of G ( j) is given by,

1
G ( j) 
2 1  2T 2
Phase angle of G ( j) is given by,
G ( j)  1800  tan 1 (T )

 G ( j) G ( j)
0  1800
 0 2700
Polar plot is given by,
Img

w=0
w=¥
Re

Hence, the correct option is (C).

SOL.63
Given : Open loop transfer function
2
G(s) H (s) 
s ( s  1)
 Control Systems 50 Polar Plot
Put s  j in the above equation,
2
G ( j) H ( j) 
j ( j  1)
Phase angle of G ( j) H ( j) is given by,
G ( j) H ( j)  900  tan 1 ()
G ( j) H ( j)   900  tan  ()  1800
Hence  pc  
Since,  pc   , hence GM  
Hence the correct option is (D)
SOL.64
Given : Open loop transfer function
2
G ( s) H ( s) 
s ( s  1)
Put s  j in the above equation,
2
G ( j) H ( j) 
j ( j  1)
Phase angle of G ( j) H ( j) is given by,
G ( j) H ( j)  900  tan 1 ()
G ( j) H ( j)   900  tan  ()  1800
Hence  pc  
Since,  pc   , hence GM  
Hence the correct option is (D).
SOL.65
The plant with unity feedback configuration is shown below,
+ u 100
r S H (s) = y
s ( s + 10) 2
-
plant

. Method 1 :
100
OLTF, H (s) 
s ( s  10) 2
Put s  j in above equation,
100
H ( j) 
j( j  10) 2
100
H ( j) 
j (   100  j 2)
2
 Control Systems 51 Polar Plot
100
H ( j)  …(i)
 j  j100  22
3

Magnitude of open loop transfer function H ( j) is given by,

100
H ( j)  …(ii)
 (2  100) 2
Phase angle of open loop transfer function H ( j) is given by,
ω
 H ( jω)   900  2 tan 1  
 10 
Calculation of  pc :
Phase angle at phase crossover frequency is given by,
H ( j pc )  1800

 ω pc 
1800   900  2 tan 1  
 10 
 ω pc 
      450  tan 1    ω pc  10 rad/sec
 10 
OR
Equating imaginary part of H ( j) to zero,
From equation (i),
Img   j3  j100  202   0

 3pc  100 pc  0
 pc  10 rad/sec
- 2700
Img
1
20

-1800 w=¥
00
Re

w pc = 10
w=0
- 900

Fig. Type 1, order 3

Note : This method is only valid when polar plot crosses  1800 axis.
From equation (ii),
100
H ( j pc ) 
 pc (2pc  100) 2
100
H ( jω pc ) 
10 (102  102 ) 2
Control Systems 52 Polar Plot
100 1
H ( jω pc )  
10  (100  100) 20
Gain margin is given by,
1
GM   20
H ( jω pc )
In dB, GM (dB)  20 log 20  26 dB
Hence, the correct option is (C).
. Method 2 :
The characteristic equation is given by,
1  H ( s)  0
100
1 0
s ( s  10) 2
K desired  100
For K mar , replace gain with K in H ( s ) .
K
1 0
s ( s  10) 2
s 3  20 s 2  100 s  K  0
For third order system to be marginal stable,
Internal product (IP) = External product (EP)
20  100  K mar
K mar  2000
Gain margin is given by,
K mar
GM 
K desired
2000
GM   20
100
In dB, GM = 20 log 20  26 dB
Hence, the correct option is (C).
SOL.66
(i) Consider signal flow graph given in option (A),
1 1/s 1/s 1/s 100
u y

– 10 – 10
100
Forward path gain : P1 
s3
Individual loop gain :
10 10
L1  and L2 
s s
Control Systems 53 Polar Plot
Number of two non-touching loops :
100
L1 L2  2
s
Determinant :
  1  ( L1  L2 )  L1 L2
10 10 100
  1   2
s s s
20 100
  1  2
s s
Path factor : All the loop touches forward path.
1 1/s 1/s 1/s 100
u y

– 10 – 10
1  1  0  1
Using Mason’s gain formula, transfer function is given by,
Y ( s) 1
   k Pk
U ( s )  k i
100
1 3
Y ( s) 1 s
 (1 P1 ) 
U ( s)  20 100
1  2
s s
Y ( s) 100

U ( s ) s ( s  20 s  100)
2

Y ( s) 100
  H (s) [Given]
U ( s ) s ( s  10) 2
(ii) Consider signal flow graph given in option (B),
–100

1/s 1/s 1/s 100

u y

– 20
100
Forward path gain : P1 
s3
Individual loop gain :
20 100
L1  and L2  2
s s
Number of two non-touching loops = 0
Determinant :   1  ( L1  L2 )
 20 100 
  1   2 
 s s 
20 100
  1  2
s s
 Control Systems 54 Polar Plot
Path factor : All the loops touches forward path.
–100

1/s 1/s 1/s 100

u y

– 20
1  1  0  1
Using Mason’s gain formula, transfer function is given by,
Y ( s) 1
   k Pk
U ( s)  k
Y ( s) 1
 (1 P1 )
U ( s) 
100
1
Y ( s) s3 100
        3
U ( s ) 1  20  100 s  20 s 2  100 s
s s2
Y ( s) 100
  H (s) [Given]
U ( s ) s ( s  10) 2
(iii) Consider signal flow graph given in option (C),
–100

1/s 1/s 1/s 100

u y

– 20
100
Forward path gain : P1 
s3
Individual loop gain :
20 100
L1  and L2  2
s s
Number of two non-touching loops = 0
Determinant :   1  ( L1  L2 )
 20 100 
  1   2 
 s s 
20 100
  1  2
s s
Path factor : All the loops touches forward path.
–100

1/s 1/s 1/s 100

u y

– 20
1  1  0  1
Control Systems 55 Polar Plot
Using Mason’s gain formula, transfer function is given by,
Y ( s) 1
   k Pk
U ( s)  k
100
1 3
Y ( s) 1 s
 (1 P1 ) 
U ( s)  20 100
1  2
s s
Y ( s) 100
 3
U ( s ) s  20 s 2  100 s
Y ( s) 100

U ( s ) s ( s  20 s  100)
2

Y ( s) 100
  H ( s ) [Given]
U ( s ) s ( s  10) 2
(iv) Consider signal flow graph given in option (D),
–100

1/s 1/s 1/s 100

u y
100
Forward path gain : P1 
s3
100
Individual loop gain: L1  2
s
Number of two non-touching loops = 0
Determinant :
 100  s  100
2
  1  L1  1   2  
 s  s2
Path factor : All the loops touches forward path.
–100

1/s 1/s 1/s 100

u y
1  1  0  1
Using Mason’s gain formula, transfer function is given by,
Y ( s) 1 P
  Pk  k  1 1
U ( s)  k 
Y ( s) s2 100
 2  3
U ( s) s  100 s
Y ( s) 100
  H ( s ) [Given]
U ( s ) s ( s  100)
2

Hence, the correct option is (D).


Practice Solutions :
Sol.1
Given : G(s) has no poles in the right-half of s-plane i.e. P = 0.
Img

G(s) plane
CCW

CW

–1 Re

N=0
(1 CW + 1 CCW)

Nyquist stability criterion is given by,

N  PZ …(i)
where,
N = Number of encirclement about critical point ( 1  j 0) in counter clockwise (anti
clockwise) direction.
P = Number of right hand poles of open loop transfer function i.e. G ( s ) H ( s ) .
Z = Number of right hand poles of closed loop transfer function.
From figure, number of encirclement about critical point ( 1  j 0) is one clockwise and one
anticlockwise.
Thus, N=1–1=0
Hence, from equation (i),
0 0Z
Z  0 [stable]
Thus, no pole of closed transfer function will lie in RHS and therefore system is stable.
Hence, the correct option is (A).
Control Systems 2 Nyquist Stability Criterion
Sol.2
Given Nyquist plot is shown below,
Img

GH-plane
CCW
(–1, 0) w=¥
Re
w=0

N =1
Nyquist stability criterion is given by,
N= P–Z … (i)
where,
N = Number of encirclement about critical point ( 1  j 0) in counter clockwise (anti
clockwise) direction.
P = Number of right hand poles of open loop transfer function i.e. G ( s ) H ( s ) .
Z = Number of right hand poles of closed loop transfer function.
From figure, number of encirclement about critical point ( 1  j 0) is one anticlockwise.
Thus, N=1
Since, number of right hand pole of OLTF G ( s ) H ( s ) , P = 1 [Given]
Hence, from equation (i),
1  1 Z
Z 0
Thus, no pole of closed transfer function will lie in RHS and therefore system is stable.
Hence, the correct option is (A).
Sol.3
Given : Stable closed loop system.
Nyquist stability criterion is given by,
N= P–Z …(i)
where,
N = Number of encirclement about critical point ( 1  j 0) in counter clockwise (anti
clockwise) direction.
P = Number of right hand poles of open loop transfer function i.e. G ( s ) H ( s ) .
Z = Number of right hand poles of closed loop transfer function.
For stability of closed loop system, number of right hand poles of CLTF, Z  0
Hence, from equation (i),
Thus, N  P
N is positive for counter clockwise encirclement.
Hence, closed-loop control system is stable if the Nyquist plot of the corresponding open-loop transfer
function encircles the critical point ( 1  j 0) in the counter clockwise direction as many times as the
number of right-half s-plane poles.
Hence, the correct option is (A).
Control Systems 3 Nyquist Stability Criterion
Sol.4
Given: The s-plane contours enclose 3-zeros and 2-poles
N= P–Z
N= 2–3=–1
Hence Nyquist contour will encircle the origin of q(s) plane once in clockwise direction.
Hence, the correct option is (A).
Sol.5
Key Point :
Figure shows P.M. and G.M. of stable, marginally stable and unstable system.
wgc < w pc wgc = w pc wgc > w pc
Stable system Marginally stable system Unstable system
G.M. > 1 and P.M. > 00 Img G.M. = 1 and P.M. = 00 Img G.M. < 1 and P.M. < 00 Img

1/G.M. Unit radius 1/G.M. Unit radius 1/G.M. Unit radius

circle circle w = wgc circle
w = w pc P
w = wgc w = w pc
Q w=¥ w=¥ P.M. w=¥
P.M.= 00 < 00
-180 0
O
Re -180 0
O
Re -180 0
O
Re
P.M. P
-1 + j 0 > 00
-1 + j 0
-1 + j 0

P
ÐG ( jw) H ( jw) ÐG ( jw) H ( jw) ÐG ( jw) H ( jw)
w = wgc
w=0 at w = wgc w=0 at w = wgc w=0 at w = wgc

From above figure,

If Nyquist plot encloses the point ( 1, j 0) then the gain margin of the system in dB is less than zero.
Hence, the correct option is (B).
Sol.6
Given Nyquist plot is shown below,
Img
w=0
w<0

IV III II I w=¥
Re
R 0
Q P

w>0

w = 0+
Nyquist stability criterion is given by,
N  PZ …(i)
where,
N = Number of encirclement about critical point ( 1  j 0) in counter clockwise (anti
clockwise) direction.
P = Number of right hand poles of open loop transfer function i.e. G ( s ) H ( s ) .
Z = Number of right hand poles of closed loop transfer function.
Control Systems 4 Nyquist Stability Criterion
Consider stable OLTF hence P  0
Case 1 : Consider Region I (OP)
From figure, number of encirclement about critical point ( 1  j 0) is two clockwise.
Hence, N  2
N  PZ
2  0  Z
Z 2
Case 2 : Consider Region II ( PQ)
From figure, number of encirclement about critical point ( 1  j 0) is one clockwise and one
anticlockwise.
Hence, N  1 1  0
N  PZ
0  0Z
Z 0
Since no RHS pole
Hence stable CLTF
Case 3 : Consider Region III (QR )
From figure, number of encirclement about critical point ( 1  j 0) is two clockwise.
N  2
N  PZ
2  0  Z
Z 2
Since no RHS pole
Hence stable CLTF
Hence, the correct options is (B) and (D).
Sol.7
Given Nyquist plot is shown below,
Img

II I
Re

Given : (i) OLTF has no pole in right half of s-plane P  0

Nyquist stability criterion is given by,
N  PZ …(i)
where,
N = Number of encirclement about critical point ( 1  j 0) in counter clockwise (anti
clockwise) direction.
 Control Systems 5 Nyquist Stability Criterion
P = Number of right hand poles of open loop transfer function i.e. G ( s ) H ( s ) .
Z = Number of right hand poles of closed loop transfer function.
Case 1 : Consider region I :
From figure, number of encirclement about critical point ( 1  j 0) is one clockwise and one
anticlockwise.
N  1 1  0
N  PZ
0  0Z  Z  0
Since zero RHS poles hence stable system
Case 2 : Consider region II:
From figure, number of encirclement about critical point ( 1  j 0) is two clockwise.
N  2
N  PZ
2  0  Z  Z  2
Since two RHS poles hence unstable system
Hence, the correct option is (D).
Sol.8
s 2  0.2 s  100 1
Given : G ( s )  , H s 
s  0.2 s  100
2
2
The open loop transfer function is given by,
1  s 2  0.2 s  100 
G ( s) H (s)   2 
2  s  0.2 s  100 
Put s  j in above equation,
1 100  2  j 0.2 
G ( j) H ( j) 
2 100  2  j 0.2 
Magnitude of open loop transfer function G ( j) H ( j) is given by,

1  (100  2 ) 2  (0.2) 2 1
      G ( j) H ( j)   
2  (100   )  ( 0.2)
2 2 2 2

Phase angle of open loop transfer function G ( j) H ( j) is given by,
 0.2    0.2   0.2 
      G ( j ) H ( j)  tan 1  2 
 tan 1  2 
G ( j) H ( j)  2 tan 1  2 
 100     100     100   
 G ( j ) H ( j  ) G ( j  ) H ( j  )
1
0 00
2
1
100 1800
2
1
 00  3600
2
Control Systems 6 Nyquist Stability Criterion
Hence, Nyquist plot can be drawn as,
Im

Re
(-1 + j 0) 1
- 1
2 2

From the Nyquist plot shown, we can see that critical point (  1  j 0) lies outside the closed contour.
Hence, number of encirclement of (  1  j 0) point is zero i.e. N  0 .


Practice Solutions :
Sol.1 Given asymptotic Bode magnitude plot is shown below,
G ( jw), dB
0 dB/dec
160
140 20 dB/dec

60 dB/dec
20

w (rad/ sec )
0.1 10 100

For the given Bode magnitude plot, there are three corner frequencies ω1  0.1, ω2  10 and
ω3  100 rad/sec.
The initial slope is 0 dB/dec and line is constant and this corresponds to no pole or zero at origin in the
transfer function.
3
 s 
(i) At ω1  0.1 : The slope changes by  60 dB/dec and this is due to the factor  1   in the
 0.1 
numerator of the transfer function.
(ii) At ω2  10 : The slope changes by  40 dB/dec and resultant slope will be  20 dB/dec. This is due
2
 s 
to the factor  1   in the denominator of the transfer function.
 10 
(iii) At ω3  100 : The slope changes by  20 dB/dec and resultant slope will be 0 dB/dec. This is due to
 s 
the factor  1   in the denominator of the transfer function.
 100 
Calculation of K :
20log10 K  20 [From figure]
K  10
The overall transfer function can be written as,

 
Control Systems 2 Bode Plot
3
 s   s 
3
K 1   10 1  
 ω1   0.1 
G( s)  2
 2
 s   s  1  s  1  s 
 1  ω   1  ω   10   100 
 2   3 

108 ( s  0.1)3
G( s) 
( s  10)2 ( s  100)
Hence, the correct answer is (A).
Sol.2 Given Bode Plot is Shown below,
G ( jw)

100 dB
– 60 dB/dec

40 dB/dec

w (rad/sec)
10
For the given Bode magnitude plot there are one corner frequency : 1  10
The initial slope is  40 dB/sec and this corresponds to a factor s 2 in the numerator of transfer function.
At 1  10 rad/sec, the slope changes by  100 dB/sec
5
 s 
So resultant slope will be  60 dB/sec and this is due to the factor 1   in the denominator of the
 10 
transfer function.
Calculation of K :
100  20log K  40log10
20 log K  60
K  (10)3  1000
The overall transfer function can be written as,
Ks 2
G(s) H (s)  5
 s 
1  
 10 
1000 s 2
G(s) H (s) 
(1  0.1 s) 2
K
Sol.3 Given : Transfer function G ( j) 
j( jT  1)
Put j  s in the above equation,
K
G(s) 
 
 s 
s 1  
1
 
 T 

 
Control Systems 3 Bode Plot
(i) Since, type 1 system hence initial slope is  20 dB/dec
1
(ii) Corner frequency at  
T
(iii)Since type 1 system hence  pc  K
Hence, the correct option is (C).
Sol.4 Given : (i) Number of Zeros  2  Z
(ii) Number of Poles  6  P
High frequency asymptotes slope is given by,
Slope  20  Z  20  P dB/decade
Slope  20  2  20  6   80 dB/decade
10
Sol.5 Given : Transfer function G ( s ) 
0.66 s  2.33s  1
2

10
G(s) 
0.66 s  0.33s  2s  1
2

10
G( s) 
(2s  1) (1  0.33s )
10
G( s)  … (i)
 s  s 
1  1  
 0.5  3 
From equation (i),
Corner frequency is 1  0.5 and 2  3
Hence, the correct option is (D).
Sol.6 Given : (i) Slope  40 dB/decade   0.1 rad/sec
(ii) Slope  20 dB/decade , 0.1    10 rad/sec
(iii) Slope 0, 0.1    10 rad/sec
Since, Initial slope is  40 dB/dec and this corresponds to a factor s 2 in the denominator of the transfer
function i.e. two pole at origin
(i) At 1  0.1 : The slope changes by  20 dB/dec and resultant slope will be  20 dB/dec. This
 s 
corresponds to a factor 1   in the numerator of the transfer function
 0.1 
(ii) At   10 : The slope changes by  20 dB/dec and resultant slope will be 0 dB/dec. This is due to the
 s 
factor 1   in the numerator of the transfer function.
 10 
The overall transfer function can be written as,
 s  s 
K 1  1  
T (s)  
0.1  10 
s2
Hence, the correct option is (B).

 
 Control Systems 4 Bode Plot
Sol.7 Given magnitude plot is shown below,
Mag dB

20 dB

1 10 100 1000
0 dB –2 w rad/sec
c 0d
/de B/d
dB ec
20
– 20 dB

Initial slope is  20dB/dec and this corresponds to a factor s in the numerator of transfer function.
For the give Bode magnitude plot, there are one corner frequency : 1  10 rad/sec
At 1  10 rad/sec, the slope changes by  40 dB/dec So resultant slope will be  20 dB/dec and this is
2
 s 
due to the factor 1  
 10 
Calculation of K :
 20  20 log K  20 log10
K  0.01
The overall transfer function is given by,
0.01s 10 s
T (s)  
10  s 
2 2
 s 
 1  
 10 
Hence, the correct option is (C).
K
Sol.8 Given : OLTF G ( s )  2
s
K
Put s  j in the above equation, G ( j) 
( j) 2
K
Magnitude of G ( j) is given by, G ( j)  2

At   gc , G ( jgc )  1
K
1
2gc
gc  K rad/sec
Hence, the correct option is (B).
Sol.9

 
Control Systems 5 Bode Plot
From above figure,
(i) At   10 rad/sec
GH ( j)  1800 hence  pc  10 rad/sec
Also GH ( j pc )  4 dB hence GM  4 dB
(ii) At   100 rad/sec
GH ( j100)  0 dB
Hence gc  100 rad/sec
Also GH ( j100)   2100
PM  1800  GH ( j100)  1800  2100   300
Hence, the correct option is (A).
Sol.10
Key Point :
Error in Error in phase
G ( j) at
T(s) Initial slope magnitude (degree)

(dB)
1
– 20 dB/decade  90 0 – 3 dB  45 0
1  sT
1
 40 dB/decade 1800 – 6 dB 90 0
(1  sT ) 2
1
 60 dB/decade  270 0 – 9 dB  135 0
(1  sT ) 3
    
1
 20 N dB/decade  360 0 – 3N dB  45 0 N
(1  sT ) N
Hence, the correct option is (C).
Sol.11 Given asymptotic Bode magnitude plot is shown below,

The initial slope is – 20 dB/dec and this corresponds to a factor s in the denominator of the transfer
function i.e. one pole at origin.
(i) At ω1  3 : The slope changes by  20 dB/dec and resultant slope will be - 40 dB/dec, this is due to
 s
the factor 1   in the denominator of the transfer function.
 3

 
 Control Systems 6 Bode Plot

(ii) At ω2  8 : The slope changes by  20 dB/dec and resultant slope will be -60 dB/dec, this is due to
 s
the factor 1   in the denominator of the transfer function.
 8
The overall transfer function can be written as,
K K
H (s)  
 s  s   s  s 
1    1   1  3  1  8 
 ω2  ω3 
K
H ( s) 
s (0.33s  1) (0.125s  1)
Calculation of K :
11  20 log K  20 log 3 [From figure]
20 log K  20.542  
                     K  10.64
The overall transfer function can be written as,
10.64
                       H ( s) 
s (0.33s  1) (0.125s  1)
Hence, the correct answer is (A).
Sol.12 Given Bode plot is shown below,
dB

Y – 40 dB/dec

20 dB

0 dB w (rad / sec)
0.1 0.5 1 6 20
– 60 dB/dec
– 60 dB/dec
– 40 dB/dec

For the given Bode magnitude plot, there are three corner frequencies : ω1  0.5, ω2  6 and ω3  20.
The initial slope is  40 dB/dec and this corresponds to a factor s 2 in the denominator of the transfer
function.
At ω 1  0.5 rad/sec, the slope changes by  20 dB/dec so resultant slope will be  60 dB/dec and this is
 s 
due to the factor  1   in the denominator of the transfer function.
 0.5 
At ω 2  6 rad/sec, the slope changes by  20 dB/dec so resultant slope will be  40 dB/dec and this is
 s
due to the factor 1   in the numerator of the transfer function.
 6
At ω 3  20 rad/sec, the slope changes by  20 dB/dec so resultant slope will be  60 dB/dec and this is
 s 
due to the factor 1   in the denominator of the transfer function.
 20 

 
Control Systems 7 Bode Plot

Transfer function must have 2 poles at origin i.e. term 1/ s 2

20  20 log K  40 log1.0
K  10
Y  20log10  40log 0.1
Y  20  40  60dB
Hence, the correct option is (B).
Sol.13 Overall transfer function can be written as,
 s 
K 1  
 2 
T ( s) 
 s  s 
s 2 1   1  
 1  3 
 s
10  1  
 6
T ( s) 
 s 
s2 1  2s   1  
 20 
Hence, the correct option is (D).
Sol.14 Given Bode magnitude plot is shown below,

Since, 6dB/oct  20 dB/dec

And  12 dB/oct   40 dB/dec
3
 s 
At corner frequency 1 slope changes by 18 dB/oct or  60 dB/dec, this is due to a factor 1   in
 1 
the denominator of transfer function.
From the slope  6 dB/oct or  20 dB/dec
14  0
20 
log 1  log10
log 1  1.7
1  101.7  50 rad/sec
Calculation of K :
14  20 log K  20 log 50
20 log K   20
K  (10)1  0.1

 
Control Systems 8 Bode Plot
The overall transfer function is given by,
Ks 0.1 s
G( s) H (s)  3  3
 s   s 
1   1  
 1   50 
Hence, the correct option is (C).
Sol.15 The given magnitude plot is given by,
T ( jw)

x = 0.5 80 dB /dec

w (rad / sec )
0 dB 10
2
 2s  s  2 
At   10 rad/sec change in slope is  80 dB/dec and this is due to the factor 1    
 n  n  

Put   0.5 and n  10 in the above equation

2
 2  0.5s  s 2 
T ( s )  1    
 10  10  
2
 s  s  
2

T ( s )  1     
 10  10  
Hence, the correct option is (A).
Sol.16 Given magnitude plot is shown below,
T ( jw)
ec
/d

– 20 dB/dec
dB
20

w (rad / sec )
w = 0.1 4
x = 0.2
-10 dB

Initial slope is  20dB/dec and this corresponds to a factor s in the numerator of transfer function.
For the give Bode magnitude plot, there are one corner frequency : 1  4 rad/sec
At 1  4 rad/sec, the slope changes by  40 dB/dec

 2s  s 2 
So resultant slope will be  20 dB/dec and this is due to the factor 1    
 n  n  

Put   0.2 and n  4 rad/sec in the above equation,

2 2
2  0.2s  s  s
 1     1  0.1s   
4 4 4

 
Control Systems 9 Bode Plot
Calculation of K :
10  20 log K  20 log 4
K  3.162
The overall transfer function is given by,
Ks 3.162
T (s)  2
 2
s s
1  0.1s    1  0.1s   
4 4
Hence, the correct option is (D).
Sol.17 Given asymptotic Bode magnitude plot is shown below,
G ( jw) , dB

B C
40 dB – 40 dB/dec

40 dB/dec
A D
0 dB Freq. (Hz)
fL 300 900 fH

Calculation of fL :
For line AB,
40  0
40 
log10 (300)  log10 ( f L )
 300 
log10   1
 fL 
300  10 f L  f L  30 Hz …(i)
Calculation of fH :
0  40
For line CD,  40 
log10 f H  log10 (900)
 f 
log10  H   1
 900 
f H  900  10  9000 Hz …(ii)
f H  f L  9000  30  8970 Hz
Hence, the value of f H  f L is 8970 Hz.
K
Sol.18 Given : G ( s )        …(i)
( s  0.1) ( s  10) ( s  P1 )
Phase, f

0.01 0.1 1 10 100

0
0
w rad/sec
0
– 45

0
– 135

0
– 225
0
– 270

 
Control Systems 10 Bode Plot
From the given graph,
(a) 0.01    0.1
 450  00
Slope =   450 /dec …(ii)
 0.1 
log  
 0.01 
(b) 0.1    10
 2250  450
Slope =   900 /dec …(iii)
 10 
log  
 0.1 
(c) 10    100
 2700  2250
Slope=   450 /dec …(iv)
 100 
log  
 10 
From equation (i),
K
G ( s) 
( s  0.1) ( s  10) ( s  P1 )
K
G( s) 
 s  s  s
P1  1   1   1  
 0.1   10   P1 
Corner frequencies of G ( s ) are 0.1, 10 and P1 rad/sec.

Concept of Asymptotic Bode Phase Plot :

1. Consider the simple transfer function for real zero,
 s 
(a) G( s)  1  
 0 
 
G ( j)   1  j 
 0 

 
G ( j)  tan 1  
 0 
Phase of simple real zero is given by,
(i) At low frequency,   0 , the phase is approximately 00 .

(ii) At high frequency,   0 , the phase is 900 .

(iii) At corner frequency,   0 , the phase is 450 .

The rule for drawing the bode phase plot for real zero :
The phase plot is 00 until one tenth the corner frequency and then increase linearly to 900  at ten times
the corner frequency.

 
Control Systems 11 Bode Plot
ÐG ( jw) approximate plot (slope 450/decade)
exact plot
900

450

w (rad / sec)
0.1 w0 w0 10 w0
´10 ´10

2
 s 
(b) For G ( s )   1  
 0 
 
G ( j)  2 tan 1  
 0 
The phase plot is 00 until one tenth the corner frequency and then increase linearly to 1800  at ten times
the corner frequency.
ÐG ( jw) approximate plot (slope 900/decade)
exact plot
1800

900

w (rad / sec)
0.1 w0 w0 10 w0
´10 ´10

Note : An nth order zero rises to (90n)0 .

2. Consider the simple transfer function for real pole,
 
 1 
(a) G (s)   
1 s 
 0 
 
 
 1 
G ( j)   
1 j  
 0 
 
 
G ( j)   tan 1  
 0 
Phase of simple real pole is given by,
(i) At low frequency,   0 , the phase is approximately 00 .

(ii) At high frequency,   0 , the phase is 900 .

(iii) At corner frequency,   0 , the phase is  450 .

 
Control Systems 12 Bode Plot

The rule for drawing the bode phase plot for real pole :
The phase plot is 00 until one tenth the corner frequency and then decrease linearly 900  at ten times
the corner frequency.
ÐG ( jw)
´10 ´10
0.1 w0 w0 10 w0
00 w (rad / sec)

- 450

- 900
exact plot
approximate plot (slope – 450/decade)

1
(b) For G ( s )  2
 s 
1   
 0 

 
G ( j)   2 tan 1  
 0 
The phase plot is 00 until one tenth the corner frequency and then decrease linearly to 1800   at ten
times the corner frequency.
ÐG ( jw)
´10 ´10
0.1 w0 w0 10 w0
00 w (rad / sec)

- 900

-1800
exact plot 0
approximate plot (slope – 90 /decade)

Note : An nth order pole drops to (  90n)0 .

3. Effect of constant on phase :
A positive constant, K  0 , has no effect on phase. A negative constant, K  0 , will setup a phase
shift of 1800.
4. Effect of zeros at the origin on phase :
Zero at the origin s , cause a constant 900 phase shift for each zero.
Example : G ( s )  s
G( j)  900
ÐG ( jw)

900

w (rad / sec)

 
Control Systems 13 Bode Plot

5. Effect of poles at the origin on phase :

1
Pole at the origin , cause a constant 900 phase shift for each pole.
s
1
Example : G ( s ) 
s
G( j)  900
ÐG ( jw)

w (rad / sec)
00

- 900

Example 1 :
Draw asymptotic Bode phase plot for the following transfer function,
1
G( s) 
2 s  100
Solution :
 1 
1  
G( s)    100 
2 s  100  1  s 
 50 
Corner frequency 0  50 rad/sec
0 50
So the asymptotic Bode phase plot is 00 until   5 rad/sec and then drops linearly with slope
10 10
of  450 /dec until 100  10  50  500 rad/sec
ÐG ( jw)

5 50 500
00 w (rad / sec)

Slope = - 450 /decade

- 450

- 900

Example 2 :
Draw asymptotic Bode phase plot for the transfer function
20 s
G( s) 
 s  s 
1   1  
 5   500 
Solution :
Let, G( s)  K  G1 ( s )  G2 ( s)  G3 ( s)
Since there is zero at origin so it will provide constant phase shift of 900 .

 
Control Systems 14 Bode Plot

Let phase shift due to s will be G1 ( j) .

ÐG1 ( jw)

900

w (rad / sec)

Corner frequency 1  5 rad/sec so slope of asymptotic Bode phase plot will change in
1 1
 0.5 rad/sec and 10 1  50 rad/sec . Let phase shift due to will be G2 ( j) .
10 1
s
5
ÐG2 ( jw)

0.5 5 50
00 w (rad / sec)

Slope = - 450 /decade

- 450

- 900

Corner frequency 2  500 rad/sec so slope of asymptotic Bode phase plot will change in
2 1
 50 rad/sec and 10 2  5000 rad/sec . Let phase shift due to will be G3 ( j) .
10 1
s
500
ÐG3 ( jw)

50 500 5000
00 w (rad / sec)

Slope = - 450 /decade

- 450

- 900

So, asymptotic Bode phase plot for G ( s ) is

ÐG ( jw) = ÐG1 ( jw) + ÐG2 ( jw)

900

450

50 500 5000
00 w (rad / sec)
0.5 5

Slope = - 450 /decade

- 450

- 900

 
Control Systems 15 Bode Plot

From the given transfer function

K
G ( s) 
( s  0.1) ( s  10) ( s  P1 )
Let G( s)  K  G1 ( s )  G2 ( s)  G3 ( s)
1 10
Where, G1 ( s )   ,
s  0.1 1  s
0.1
1 0.1
G2 ( s )   ,
s  10 1  s
10
1
1 P1
G3 ( s )  
s  P1 1  s
P1
10
For G1 ( s )  , asymptotic Bode phase plot is shown below,
s
1
0.1
ÐG1 ( jw)

0.01 0.1 1
00 0
w (rad / sec)
0 /dec

- 450 /decade
- 450

00 /decade
- 900

0.1
For G2 ( s )  , asymptotic Bode phase plot is shown below,
s
1
10
ÐG2 ( jw)

1 10 100
00 w (rad / sec)
0
0 /dec

- 450 /decade
- 450

00 /decade
- 900

1
P1
For G3 ( s )  , asymptotic Bode phase plot is shown below,
s
1
P1

 
Control Systems 16 Bode Plot
ÐG3 ( jw)
P1
10 P1 10P1
00 w (rad / sec)
00 /dec

- 450 /decade
- 450

00 /decade
- 900

ÐG1 ( jw) × G2 ( jw)

0.01 0.1 1.0 10 100

00 w (rad / sec)
0
0 /dec
- 450
- 450 /decade
- 900

-1350

00 /decade
-1800
Fig. (a)
Phase, f
- 450 /decade
0.01 0.1 1 10 100
0
0
w rad/sec
00 /dec
0
– 45
- 900 /decade
0
– 135
- 450 /decade

0
– 225
0
– 270
00 /decade
Fig. (b)
Comparing both plots in figure (a) and (b), it is clear that between   0.1 and   10 , the slope is
different which means corner frequency P1 lies somewhere between 0.1 and 10.
Hence, let us assume, P1  1 ,
1 1
G3 ( s )  
s s
1 1
P1 1
ÐG3 ( jw)

0.1 1.0 10
00 w (rad / sec)
00 /dec
- 450 - 450 /decade

00 /decade
- 900

 
Control Systems 17 Bode Plot
ÐG3 ( jw)
0.01 0.1 1.0 10 100
00 w (rad / sec)
0
0 /decade
- 450 - 450 /decade

- 900
00 /decade
ÐG2 ( jw)
0.01 0.1 1.0 10 100
00 w (rad / sec)
00 /decade
- 450 - 450 /decade

- 900
00 /decade
ÐG1 ( jw)
0.01 0.1 1.0 10 100
00 0
w (rad / sec)
0 /dec
- 450 - 450 /decade
00 /decade
- 900

ÐG1 ( jw) + ÐG2 ( jw) + ÐG3 ( jw)

0.01 0.1 1.0 10 100
00 w (rad / sec)
- 45 /decade
0 0
0 /dec
- 450

-1350 - 900 /decade

- 2250 - 450 /decade

- 2700
00 /decade

Above drawn figure matches with the asymptotic bode phase plot given in the question. Therefore, our
assumption is correct.
Hence, the value of P1 is 1.
10
Sol.19 Given : Transfer function T ( s ) 
1 s
10
T ( j) 
1  j
Magnitude of T ( j) is given by,
10
T ( j) 
1  2

 
Control Systems 18 Bode Plot
10
T ( j) 0.1   10
1  0.12
T ( j) 0.1  20 log10  20 dB
Hence, the value of magnitude is 20 dB .
Sol.20 Given : Transfer function T ( s)  1  0.25 s  0.005 s 2
0.005 s 2  0.25 s  1  0
s 2  50 s  200  0 … (i)
For second-order system transfer function is given by,
s 2  2n s  n2  0 … (ii)
On comparing equation (i) and (ii)
n  10 2 and 2n  50
25
  1.7677
10 2
Error (dB)   20 log 2   20 log 2 1.7677  10.97
Error  10.97 dB
Hence, the value of error is 10.97 dB .
Sol.21 The given Bode Plot is shown below,

At gain crossover frequency (gc ) magnitude of G ( j) H ( j)  0 dB

From slope of 12 dB/octave or  40 dB/dec,
0  12
 40 
 
log  gc 
 2 
 gc 
log    0.3
 2 
gc
 (10)0.3  2
2
gc  2  2  4 rad/sec
Hence, the value of gain cross over frequency is  4 rad/sec .

 
Control Systems 19 Bode Plot
Sol.22 The given Bode Plot is shown below,

 6 dB/octave   20 dB/dec
12 dB/octave   40 dB/dec
(i) 20  20 log K
K  10
12  20
(ii)  20 
 2 
log  
 1 
 2 
log    0.4
 1 
1  0.796 rad/sec
The overall transfer function is given by,
10
G(s) H (s) 
 s  s 
1  1  
 0.296  2 
Since order 2 system hence  pc   and Gain margin  
Hence, the value of gain margin is   .
Sol.23 Transfer function is given by,
10
G ( j) H ( j) 
 j  j 
1  1  
 0.796  2 
Phase angle of G ( j) H ( j) is given by,
   1   
G ( j) H ( j)   tan 1    tan  
 0.796  2
Since gain crossover frequency gc  4 rad/sec

 4  1  4 
G ( jgc ) H ( jgc )   tan 1    tan    142 :18
0

 0.796  2
PM  1800  GH ( pc )  1800  142.180  37.820  
Hence, the value of phase margin is 37.820 .

 
Control Systems 20 Bode Plot
Sol.24 Given asymptotic Bode magnitude plot is shown in figure,
G ( jw) , dB

40 - 20 dB/decade

- 40 dB/decade

w (rad/sec )
w1 = 10 w2 = 250

From the given Bode magnitude plot, there are two corner frequencies :
ω1  10 and ω2  250 rad/sec
The initial slope is 0 dB/dec and this corresponds to no pole or zero at origin in the transfer function.
(i) At ω1  10 : The slope changes by  20 dB/dec and resultant slope will be – 20 dB/dec. This is due
 s 
to the factor 1   in the denominator of the transfer function.
 10 
(ii) At ω 2  250 : The slope changes by  20 dB/dec and resultant slope will be
 s 
– 40 dB/dec. This is due to the factor 1   in the denominator of the transfer function.
 250 
Calculation of K :
40  20 log10 K
K  102  100
The overall transfer function can be written as,
K 100
G ( s)  
 s  s  1  s  1  s 
1   1     
 ω1   ω2   10   250 
Hence, the correct option is (A).
Sol.25 Given asymptotic Bode magnitude plot is shown below,
dB
- 20 dB/decade
20 dB

w (rad / sec)
0.1 2 20
From the given Bode magnitude plot, there are two corner frequencies :
ω1  2 and ω2  20 rad/sec
The initial slope is  20 dB/dec and this corresponds to a factor s in the denominator of the transfer
function i.e. one pole at origin.
(i) At ω1  2 : The slope changes by  20 dB/dec and resultant slope will be 0 dB/dec. This is due to the
 s
factor 1   in the numerator of the transfer function.
 2

 
Control Systems 21 Bode Plot

(ii) At ω 2  20 : The slope changes by  20 dB/dec and resultant slope will be –20 dB/dec. This is due to
 s 
the factor 1   in the denominator of the transfer function.
 20 
Calculation of K :
20  20 log K  20 log 2
K  20
The overall transfer function can be written as,
 s 
K 1   20  1  s 
ω1 
G ( s)    
2
 s  s 1  s 
s 1    
 ω2   20 
Hence, the correct option is (B).
Sol.26 Given asymptotic Bode magnitude plot is shown below,
G ( jw) (dB)

60 - 20 dB/dec
40
– 40 dB/dec
20
0 20 w (rad / sec)
0.1 1 10
- 60 dB/dec

From the given Bode magnitude plot, there are two corner frequencies ω1  1 rad/sec and ω2  20
rad/sec .
The initial slope is  20 dB/dec and this corresponds to a factor s in the denominator of the transfer function
i.e. one pole at origin.
(i) At ω1  1 : The slope changes by  20 dB/dec and resultant slope will be  40 dB/dec. This is due to
 s
the factor  1   in the denominator of the transfer function.
 1
(ii) At ω2  20 : The slope changes by  20 dB/dec and resultant slope will be  60 dB/dec. This is due
 s 
to the factor  1   in the denominator of the transfer function.
 20 
Calculation of K :
60= 20log10 K  20log10 0.1   [From figure]
60  20log10 K  20
20log10 K  40
log10 K  2
K =100

 
Control Systems 22 Bode Plot
The overall transfer function can be written as,
K 100
G( s)  
 s  s   s  s 
s 1    1   s 1   1  
 ω1   ω2   1   20 
100
G ( s) 
s(1  s )(1  0.05s )
Hence, the correct option is (D).
Sol.27 The asymptotic Bode magnitude plot is shown below,
G ( jw)
(dB) – 40 dB/dec
20
- 20 dB/dec
w (rad / sec)
0 0.1

– 20
0 dB/dec

Initial slope is – 40 dB/dec it means that there are two poles at the origin.
(i) At ω1 : The slope changes by  20 dB/dec and resultant slope will be – 20 dB/dec, this is due to the
 s 
factor  1   in the numerator of the transfer function.
 1 
(ii) At ω2 : The slope changes by  20 dB/dec and resultant slope will be 0 dB/dec, this is due to the
 s 
factor 1   in the numerator of the transfer function.
 2 
Therefore, transfer function has two poles and two zeros.
The overall transfer function can be written as,
 s  s 
K  1  1  
1  2 
G(s)  
s2
Hence, the correct option is (C).
Sol.28 Given asymptotic Bode magnitude plot is shown below,
40 - 0
Slope = = 20 dB/dec
æ 10 ö
log ç ÷
dB è 0.1 ø
40

0
w (rad / sec)
0.001 0.1 10

For the given Bode magnitude plot, there are two corner frequencies ω1  0.1 rad/sec and ω2  10
rad/sec .

 
 Control Systems 23 Bode Plot
The initial slope is 0 dB/dec and this corresponds to no pole or zero at origin in the transfer function.
(i) At ω1  0.1 : The slope changes by  20 dB/dec and resultant slope will be + 20 dB/dec. This is due
 s 
to the factor  1   in the numerator of the transfer function.
 0.1 
(ii) At ω2  10 : The slope changes by  20 dB/dec and resultant slope will be 0 dB/dec. This is due to
 s 
the factor  1   in the denominator of the transfer function.
 10 
Calculation of K :
20log10 K  0 [From figure]
K =1
The overall transfer function can be written as,
 s 
K  1    1  s 
1   0.1  1  10s
      G( s)    
 s   s  1  0.1s
1    1  
 2 
 10 
Hence, the correct option is (A).

Sol.29 Given : G  s  
 s  1
 s  2  s  3
j


–3 –2 1

Stable Non-minimum Phase

system (NMPS)
Poles are in the left half of s-plane so the system is stable.
One zero is in the right half of s-plane. Therefore, the system is non-minimum phase type.
Hence, the correct option is (B).
 Key Point
Minimum phase system : All poles and zeros lie on the left half of s-plane and it is always stable system.
1
Example : T ( s ) 
( s  2)( s  3)
jw

s
-3 -2

Non-minimum phase system : When one or more zeros, poles or both lies on the right half of s-plane
and it may or may not be stable.

 
Control Systems 24 Bode Plot

( s  1)( s  2)
Example : T ( s ) 
( s  3)( s  4)
jw

s
-4 -3 -1 2

Sol.30 Given : Flat magnitude response of unity shows that it is an all pass system.
All pass System : All the poles lie in the left half of the s-plane and all the zeros lie in the right half of s-
plane and vice versa is also possible.
In all pass system, poles and zeros are symmetrical about j-axis.
Magnitude response is shown in figure.
H( f )
1

jw
-2+3j +2 + 3 j

-2-3j +2 - 3 j

Hence, the correct option is (D).

Sol.31 Given Bode magnitude plot is shown below,
G ( jw) , dB

40 -20 dB/decade

w (rad/sec)

Calculation of K :
40  20 log K [From figure]
K  100
For unit step input steady state error is given by,
1
ess 
1 K

 
Control Systems 25 Bode Plot
1 1
ess  
1  100 101
Hence, the correct option is (A).
Sol.32 The given Bode magnitude plot is shown below,
Gain (dB)

+ 20 dB/dec
26 dB - 20 dB/dec
20 dB

w1 w2 w (rad/sec)
0 dB 10
From slope of  20 dB/dec,
26  20
20 
log 1  log10
log 1  1.3  1  (10)1.3  20 rad/sec
From slope of  20 dB/dec,
0  26
 20 
log 2  log 1
log 2  log 20  1.3  2  (10)2.6  400 rad/sec
Hence, the correct option is (C).
1 s
Sol.33 Given : Transfer function T ( s ) 
1 s
For above transfer function pole-zero plot is given by,
jw

s
–1 0 1

Since one zero in the right half of s-plane

Hence the system is Non-minimum phase system.
Hence, the correct option is (A).
Sol.34 Since,  gc   pc system is unstable.
Gain (dB)
P.M.
20 dB w=0

w pc G.M.
Phase (Degrees)
0 0
- 270 0 wgc -180 0
- 900

– 20 dB
w=¥

 
Control Systems 26 Bode Plot
At  gc , G.M.  0

At gc , P.M.  1800  G ( jgc ) H ( jgc )

G ( jgc ) H ( jgc )  1800

P.M.  0
Hence, the correct option is (D).
Sol.35 Given : (i) Low frequency slope   60 dB/decade
(ii) High frequency slope   40 dB/decade
Since, low frequency slope is  60 dB/decade
Hence 3 poles at origin.
Hence, the correct option is (D).
1
Sol.36 In the asymptotic Bode magnitude plot of the term   the error in magnitude at the corner frequency 
(1  sT ) N
is – 3N dB
Hence, the correct option is (C).
Sol.37 Given : G ( jgc )  1250
Phase margin is given by,
PM  1800  G ( jgc )

PM  1800  1250  550

Hence, the correct option is (C).
5( s 2  10 s  100)
Sol.38 Given : Transfer function G ( s ) 
s 2 ( s 2  15s  1)
The transfer function in Bode Plot can be written as,
 2 s2 
K 1  s 2 
 n n1 
G ( s)   1

 2 s2 
s 2 1  s 2 
 n n2 
 2

Where n1 and n2 are corner frequency

 s s2 
K 1   2 
10 10 
G (s)  
 15s s 2 
s 2 1   
 1 12 
For the given transfer function, there are two corner frequencies :
n1  10 and n2  1 rad/sec
Hence, the correct option is (C).

 
Control Systems 27 Bode Plot
1
Sol.39 Given : Transfer function G ( s ) 
2s  1
1
G ( j) 
2 j  1
1
G ( j) 
1  42
1
DC gain G ( j) 0  1
1  02
1
High frequency gain  G ( j)   0
1  2
Hence, the correct option is (C).
Sol.40 Given Bode Plot is Shown below,

G ( jw)

100 dB
– 60 dB/dec

40 dB/dec

w (rad/sec)
10
For the given Bode magnitude plot there are one corner frequency : 1  10
The initial slope is  40 dB/sec and this corresponds to a factor s 2 in the numerator of transfer function.
At 1  10 rad/sec, the slope changes by  100 dB/sec
5
 s 
So resultant slope will be  60 dB/sec and this is due to the factor 1   in the denominator of the
 10 
transfer function.
The overall transfer function can be written as,
Ks 2
G(s) H (s)  5
 s 
1  
 10 
Hence, the correct option is (C).
Sol.41 Given asymptotic Bode magnitude plot is shown below,
G ( jw) H ( jw) dB
– 40 dB/dec

– 60 dB/dec

10
w (r/s)
5

– 40 dB/dec

 
Control Systems 28 Bode Plot

From the given Bode magnitude plot, there are two corner frequencies ω1  5 rad/sec and ω2  10 rad/sec
The initial slope is – 40 dB/dec and this corresponds to a factor s2 in the denominator of the transfer
function i.e. two poles at origin..
(i) At ω1  5 : The slope changes by  20 dB/dec and resultant slope will be  60 dB/dec. This is due
 s
to the factor 1   in the denominator of the transfer function.
 5
(ii) At ω2  10 : The slope changes by 20 dB/dec and resultant slope will be  40 dB/dec. This is due
 s 
to the factor 1   in the numerator of the transfer function.
 10 
The overall transfer function can be written as,
 s 
K 1   K 1  s 
ω2  10 
G ( s)    
 s   s
s 1   s 2 1  
 ω1   5
K ( s  10)
G( s) 
s 2 ( s  5)
Hence, the correct option is (C).



 
Practice Solutions :
Sol.1 The second order transfer function
4
T ( s)  …(i)
s  2s  4
2

Transfer function for second-order system is given by,

C ( s) 2n
 2 …(ii)
R( s) s  2n s  2n
On comparing equation (i) and (ii),
n  2 and 2n  2
1
  0.5
2
Maximum resonance peak is given by,
1
Mr 
2 1   2
1 1
Mr  
2  0.5 1  0.52 1
1
4
2
Mr 
3
Hence, the correct option is (D).
Sol.2 Given :
100
Transfer function M ( j)  …(i)
100  2  10 2 j
Transfer function for second order system is given by,
2n
T ( j)  …(ii)
2  2 jn  n2
 Control Systems 2 Frequency Response of Second Order System
On comparing equation (i) and (ii),
n  10 and 2n  10 2
1
  0.707
2
Maximum peak is given by,
1
Mp 
2 1   2
1
Mp 
2
1  1 
2 1  
2  2
1
Mp  1
1
2  1
2
Hence, the correct option is (B).
Sol.3 Given :
(i) Natural frequency of oscillations n  3 rad/sec
(ii) Damping ratio n  3
For standard second order system resonant frequency is given by,
r  n 1  22

r  3 1  2(0.5) 2  2.1 rad/sec

For standard second order system resonant peak is given by,
1
Mr 
2 1   2
1
Mr   1.16
2  0.5 1  0.52
Hence, the correct option is (D).
Sol.4 For a standard second order system phase margin (PM) is given by,
 2 
PM  tan 1  
 2  4  1 
2 4

From above equation PM  

When   0 , PM  tan 1 (00 )  0
Hence, the correct option is (D).
Sol.5 Given :
Resonant peak ( M r )  1
1
(M r )  1
2 1   2
 Control Systems 3 Frequency Response of Second Order System

42 (1  2 )  1
42  42  1  0
(22  1)2  0
22  1  0
1
2 
2
1

2
Hence, the correct option is (B).
Sol.6 Given :
Resonant peak ( M r )  2
For standard second order system resonant peak is given by,
1
Mr  2
2 1   2
1
42 (1  2 ) 
4
1
4  2  4 2  0
4
Assume   X ,
1
4X 2  4X  0
4
1
4  42  4  4 
X 4
2 4
4  16  4
X
8
X  2  0.933,0.06698
  0.256, 0.965
Resonant peak is valid only for   0.707
Thus,   0.256
Hence, the correct option is (B).
Sol.7 Given :
K
(i) OLTF G ( s)  , H (s)  1
s (1  Ts )
Closed loop transfer function is given by,
G(s)
T ( s) 
1  G ( s) H ( s)
 Control Systems 4 Frequency Response of Second Order System
K
s (1  Ts )
T (s) 
K
1
s ( HTs )
K
T (s) 
Ts  s  K
2

K /T
T ( s)  …(i)
s K
s  
2

T T
Transfer function for standard second order system is given by,
C ( s) 2n
 2 …(ii)
R( s) s  2n s  2n
On comparing equation (i) and (ii),
K 1
n  and 2n 
T T
1 1
 
K 2 KT
2T
T

% MPO   100%  25%
e 1  2
  0.40

Resonant frequency r  n 1  22

8  n 1  2  (0.4) 2
n  9.70 rad/sec
K
 9.70
T
K  94.11T …(i)
1
  0.40 
2 KT
KT  1.5625 …(ii)
From equation (i) and (ii),
94.11T 2  1.5625
T  0.127 …(iii)
Hence, the correct option is (A).
Sol.8 From equation (ii) and (iii),
K  0.127  1.5625
K  12.04
Hence, the correct option is (C).
 Control Systems 5 Frequency Response of Second Order System
Sol.9 Gain crossover frequency is given by,

gc  n 2 2  4 4  1

Put n  9.70 rad/sec and   0.40 in the above equation,

gc  9.70 2(0.40) 2  4(0.40) 4  1

gc  8.306 rad/sec

Hence, the correct option is (D).
Sol.10 Phase margin is given by,
 2 
PM  tan 1  
 2  4  1 
2 4

Put   0.40 in the above equation

 2  0.4 
PM  tan 1  
 2(0.4)  4(0.4)  1 
2 4

PM  43.40
Hence, the correct option is (A).
1
Sol.11 Mr 
2 1   2
1
Mr 
2  0.4 1  0.42
M r  1.355
Hence, the correct option is (A).
Sol.12 For a standard second order system 3-dB bandwidth is given by,

b  n 1  2 2  4 4  2  4 2

Put n  9.70 and   0.4 in the above equation,

b  9.70 1  2(0.4) 2  4(0.4) 4  4(0.4) 2  2

b  13.36
Hence, the correct option is (B).
Sol.13 Given :
K
(i) OLTF G ( s)  , H (s)  1
s(1  sT )
(ii) r  6 rad/sec
(iii) M p  20%  0.2
 Control Systems 6 Frequency Response of Second Order System
Closed loop transfer function is given by,
K
s (1  sT )
T (s) 
K
1
s (1  sT )
K
T ( s)  T …(i)
s K
s2  
T T
Transfer function for second order system is given by,
C ( s) 2n
 2 …(ii)
R( s) s  2n s  2n
On comparing equation (i) and (ii),
K 1
n  and 2n 
T T
1

2 KT


12
MPO  e  100  20 %
  0.456
r  n 1  2 2  6

n 1  0.4562  2  6
K
n  7.85 rad / sec 
T
K  61.62 T …(i)
1
  0.456 
2 KT
KT  1.20 …(ii)
Using equation (i) and (ii),
T  0.139
K  0.139  1.20
K  8.6
Hence, the value of gain K is 8.6.
1
Sol.14 Resonant peak M r 
2 1   2
Put   0.456 in the above equation,
1
Mr 
2  0.456 1  0.4562
M r  1.232
Hence, the value of resonant peak is 1.232.
 Control Systems 7 Frequency Response of Second Order System
Sol.15 Given :
(i) Maximum peak overshoot (MPO)  50%
(ii) Period of oscillation = 0.2 sec
For a second order system maximum peak overshoot is given by,


12
% MPO  e 100%


12
50%  e 100%
  0.215
1
Resonant peak ( M r ) 
2 1   2
1
(M r ) 
2  0.215 1  0.2152
M r  2.38
Hence, the value of resonant peak is 2.38.
Sol.16 Time period of oscillation is given by,
2
T0 
d
2
 0.2
d
d  31.415 rad/sec

d  n 1   2

31.415  n 1  0.2152
n  32.1621 rad/sec
For standard second order system resonant bandwidth is given by,

b  n 1  2 2  4 4  4 2  1

b  32.1621 1  2(0.215) 2  4(0.215) 4  4(0.215) 2  2

b  48.2 rad/sec
Hence, the value of resonant bandwidth is 48.32.


Practice Solutions :
Sol.1 Given :
 x1   1 0   x1  0
 x    0 2  x   1  u …(i)
 2    2  
x 
and y  1 2  1  …(ii)
 x2 
. Method 1 :
State equation is given by,
x  Ax  Bu …(iii)
Output state equation is given by,
y  Cx  Du …(iv)
On comparing equation (i), (ii) with equation (iii) and (iv),
 1 0  0
A   , B    and C  1 2
 0 2  1 
D0
The controllability matrix is defined as,
QC   B : AB : A2 B..... : An 1 B 
n n

where, n  number of state variable

 1 0  0  0 
AB       
 0 2 1  2
QC   B : AB
0 0 
QC   
1 2
QC  0
Thus, the system is not controllable.
Control Systems 2 State Space Analysis
The observability matrix is defined as,
QO  C T : AT C T : .... : ( An 1 )T C T 
n n

where, n  number of state variable

1  1 0 
CT    AT   
2  0 2 
 1 0  1   1 
AT C T       
 0 2   2    4 
QO  C T : AT C T 
1 1 
QO   
2  4
QO   2
QO  0
Thus, system is observable.
Hence, correct options are (B) and (C).
. Method 2 :
Diagonal Canonical Form (DCF) :
If poles of a transfer function are real and different, then the state variable representation can be arranged
in DCF form.
Example :
Y ( s) 1
Let,  2
U ( s ) s  3s  2
Y ( s) 1 K1 K2
  
U ( s ) ( s  1) ( s  2) ( s  1) ( s  2)
K1U ( s ) K 2U ( s )
Y ( s)   … (i)
( s  1) ( s  2)
K1U ( s ) K U (s)
Let, X 1( s)  , X 2 ( s)  2
( s  1) ( s  2)
sX1 ( s)   X 1 ( s)  K1U ( s) and
sX 2 ( s)  2 X 2 ( s)  K2U ( s)
Taking inverse Laplace transform,
Then, x1   x1  K1u … (ii)
x2   2 x2  K2u … (iii)
So, state variable representation in DCF form can be represented as,
Partial fraction coefficients

é x&1 ù é -1 0 ù é x1 ù é K1 ù
ê x& ú = ê 0 -2 ú ê x ú + ê K ú [u ]
ë 2û ë û ë 2û ë 2û

Poles/Eigen values
Control Systems 3 State Space Analysis

 x1 
and  y   1 1  
 x2 
OR
é x&1 ù é -1 0 ù é x1 ù é1ù
ê x& ú = ê 0 -2 ú ê x ú + ê1ú [u ]
ë 2û ë û ë 2û ë û

Poles/Eigen values

éx ù
[y ] = [K1 K2 ] ê 1 ú
ë x2 û

Partial fraction
coefficients
Properties of DCF :
(i) The principal diagonal elements of system matrix ‘A’ are roots of characteristic equation or eigen
values of the system.
(ii) Either input matrix ‘B’ or output matrix ‘C’ will contain coefficients of partial fraction.
If input matrix B will contain coefficient of partial fraction, then all the element of matrix C will be
1. If input matrix C will contain coefficient of partial fraction, then all the element of matrix B will
be 1.
Method to identify controllability and observability.
A system with distinct eigen values and a diagonal system matrix is controllable if the input matrix B
does not have any rows that are zero.
Distinct eigen values

é x&1 ù é -1 0ù é x1 ù é0ù
ê x& ú = ê 0 -2ú ê x ú + ê1ú u
Zero hence uncontrollable
ë 2û ë û ë 2û ë û
A system with distinct eigen values and diagonal system matrix is observable, if the output matrix C does
not have any column that are zero.
Distinct eigen values

é x&1 ù é -1 0ù é x1 ù é0ù
ê x& ú = ê 0 -2ú ê x ú + ê1 ú u
ë 2û ë û ë 2û ë û

éx ù
y = [1 2 ]ê 1ú
ë x2 û
Non-zero hence observable
Hence, the correct options are (B) and (C).
 Key Point
(i) If QO  0, then the system is observable.

If QO  0, then the system is not observable.

 Control Systems 4 State Space Analysis

(ii) Gilbert’s test :

It is applicable for diagonal canonical form (DCF) or Jordan canonical form (JCF).
(iii) Parallel realization of any system will always provide matrix A in the form of diagonal canonical
form (DCF).
(iv) If in any matrix, diagonal elements are non-zero and different, elements below and above diagonal
are zero, then that matrix is known as diagonal canonical form (DCF).
  P1 0 0 0  0 
 0  P2 0 0  0 

 0 0  P3 0  0 
 
 0  0  P4  0 
     
 
 0  0 0   Pn  nn
(v) If in any matrix, elements below the diagonal are zero, some element in diagonal are repeatitive and
element just above repeatitive diagonal element are one, then this type of matrix is known as Jordan
canonical form (JCF).
  P1 1 0 0  0 
 0  P1 1 0  0 
 
 0 0  P1 0  0 
 
 0  0  P4  0 
     
 
 0  0 0   Pn  nn

Sol.2 Given : x1 (t )  x2 (t )
x  Ax  Bu …(i)
 x1   a b   x1 (t )   e 
Let  x    c d   x (t )    f  u(t ) …(ii)
 2   2   
On comparing equation (i) and (ii),
a b  e
A   , B 
c d  f
 a b   e   ae  bf 
AB       
 c d   f   ce  df 
If x1 (t )  x2 (t ) then x1 (t )  x2 (t )
e  f and a  b  c  d
The controllability matrix is defined as,
QC  [ B : AB : A2 B..... : An 1 B ]nn
where, n  number of state variable
QC   B : AB
Control Systems 5 State Space Analysis

e ae  bf
QC 
f ce  df
QC  ce2  def  aef  bf 2

QC  ce 2  de 2  ae 2  be 2 [e  f ]

QC  e 2  c  d  ( a  b) a  b  c  d 
QC  0
Thus, the system is not completely controllable.
Hence, the correct option is (B).
 Key Point
(i) If QO  0, then the system is observable.

(ii) If QO  0, then the system is not observable.

(iii) If QC  0, then the system is controllable.

(iv) If QC  0, then the system is not controllable.

Sol.3 Given : x  Ax  Bu, y  Cx  Du

The transfer function of state space model is given by,
T(s) = C [ sI  A]1 B  D
Hence, the correct option is (A).
 x  1  1  x1  0
Sol.4 Given :  1         u
 x2  0 1   x2  1 
x 
y  1 1  1 
 x2 
and x1 (0)  1, x2 (0)  1, u(0)  0
. Method 1 :
From given state equation,
x1  x1  x2 …(i)
x2  x2  u …(ii)
y  x1  x2 …(iii)
From equation (i),
x1 (0)  x1 (0)  x2 (0)
x1 (0)  1  ( 1)  2
From equation (ii),
x2 (0)  x2 (0)  u(0)  1  0  1
Control Systems 6 State Space Analysis
From equation (iii),
dy dx1 dx2
   x1  x2
dt dt dt
dy
 x1 (0)  x2 (0)  2  1  1
dt t  0
Hence, the correct option is (A).
. Method 2 :
State transition matrix is given by,
(t )  L1  ( sI  A) 1 

 s 0 1 1  s  1 1 
 sI  A    
0 s  0 1   0 s  1
 s  1 1 
adj sI  A  
 0 s  1
sI  A  ( s  1) 2
Adj sI  A 1  s  1 1 
 sI  A1  
sI  A ( s  1)2  0 s  1
 1 1 
 s  1 ( s  1) 2 
 sI  A1   
 0 1 
 ( s  1) 

Taking inverse Laplace transform,
 et tet 
(t )  L1 ( sI  A) 1    
0 et 
 et tet   1 
x (t )  (t ) x (0)   
0 et   1
 et  tet 
x (t )   t 
 e 
Output is given by,
x 
y  1 1  1 
 x2 
 et  tet 
y (t )  1 112  t 
 tet 
  e  21
11

dy
Therefore,  tet  et
dt
dy
1
dt t 0
Hence, the correct option is (A).
 Control Systems 7 State Space Analysis
Sol.5 Given : x (t )   2 x ( t )  2u ( t ) …(i)
y (t )  0.5 x ( t ) …(ii)
. Method 1 :
Taking Laplace transform of equation (i),
sX ( s )   2 X ( s )  2U ( s )
X ( s )[ s  2]  2U ( s )
2U ( s )
X ( s)  …(iii)
( s  2)
Taking Laplace transform of equation (ii),
Y ( s )  0.5 X ( s ) …(iv)
From equation (iii) and (iv),
0.5  2U ( s )
Y ( s) 
s2
Y ( s) 1

U ( s ) ( s  2)
Hence, the correct option is (D).
. Method 2 :
The transfer function is given by,
Y ( s)
 C  sI  A  B 
1
…(v)
U ( s)
Comparing equation (i) and (ii) with
x  Ax  Bu and y  Cx ,
 A11    2 ,  B 11   2 , C 11  0.5
Put all these values in equation (v),
Y ( s)
 0.5   s  2   2
1

U ( s)
Y ( s) 1

U ( s) s  2
Hence, the correct option is (D).
 x  1 0  x1 
Sol.6 Given :  1      …(i)
 x2  1 1  x2 
 x1 (0)  1 
 x (0)   0 …(ii)
 2   
State equation is given by,
x  Ax  Bu …(iii)
On comparing equation (i) and (iii),
1 0 0
A   B 
1 1 0
Control Systems 8 State Space Analysis

 s 0 1 0
[ sI  A]    
0 s  1 1
s  1 0 
[ sI  A]   
 1 s  1
Adj[ sI  A]
[ sI  A]1 
sI  A

s  1 0 
Adj[ sI  A]  
 1 s  1
sI  A  ( s  1) 2  0  ( s  1) 2
Resolvent matrix of A is,
 s 1 
 ( s  1) 2 0 
( s )  [ sI  A]1   
 1 s 1 
 ( s  1) 2 ( s  1) 2 

 1 
 ( s  1) 0 
( s )   
 1 1 
 ( s  1) 2 ( s  1)  22

Taking inverse Laplace transform,
 et 0
(t )  L1[( s )]   t 
te e t  2 2
If input is zero, we consider zero input response.
ZIR = f (initial value)
 x (t ) ZIR 21  (t )22  x (0)21
 et 0  1   et 
 x(t ) ZIR 21   t   
te et  0 tet  21
Hence, the correct option is (C).

 2  e 2 t 0 
Sol.7 Given : x(0)    and (t )   
3  0 et 
Without any external input implies zero input. If input is zero we consider zero input response then,
ZIR  f (initial value)
ZIR  x ( t )  ( t ) x (0)
 e 2 t 0  2
 x(t )21    3 
0 e  t  22   21
Control Systems 9 State Space Analysis

 2e 2 t 
 x(t )21   t 
3e  21
0.271 
At t  1, ZIR   
1. 104
Hence, the correct option is (A).
3( s  2)
Sol.8 Given : H (s)  3
s  4s 2  2s  1
3s  6
H (s)  3
s  4s 2  2s  1
Y ( s) X ( s)
Let H (s) 
X ( s) U ( s)
X ( s) 1
where  3 …(i)
U ( s ) s  4s  2s  1
2

Y ( s)
 3s  6 ... (ii)
U ( s)
Using equation (i), we have
( s 3  4 s 2  2 s  1) X ( s )  U ( s )
s 3 X ( s )  4 s 2 X ( s )  2 sX ( s )  X ( s )  U ( s )
Taking inverse Laplace transform,
d 3x d 2x dx
3
4 2 2  x u …(iii)
dt dt dt
x  x1 …(iv)
dx
 x2  x1 …(v)
dt
d 2x
 x3  x2 …(vi)
dt 2
From equation (iii), (iv), (v) and (vi), we have
d 3x
  x1  2 x2  4 x3  u  x3 …(vii)
dt 3
Using equation (v), (vi) and (vii) we have,
 x1   0 1 0   x1  0 
 x    0 0 1   x   0  u
 2   2  
 x3   1 2 4  x3  1 
x  AX  Bu …(viii)
0 1 0  0 
When A   0 0 1  and B   0 
 1 2  4  1 
Control Systems 10 State Space Analysis
3( s  2)
Note : In original Gate question H ( s )  . Here order of characteristics equation is two. Order
4s 2  2s  1
of characteristics equation gives number of state variables, so accordingly there must be only two state
variables. But from given option it is observed that order of system matrix is three which as number of
state variables . So given function H ( s) has been corrected by taking characteristics equation as
s 3  4 s 2  2 s  1 instead of 4 s 2  2 s  1 .
Sol.9 Given : x  Ax  Bu , y  Cx
For zero initial conditions, x (0)  0
. Method 1 :
The transfer function is given by,
Y (s)
 C [ sI  A]1 B
U (s)
For impulse input, U ( s )  1
Y ( s )  C [ sI  A]1 B
State transition matrix is given by,
(t )  exp( At )  L1  sI  A
1

Y ( s )  C exp( At ) B
Hence, the correct option is (C).
. Method 2 :
Given x&  Ax  Bu and y  Cx
d
x(t )  Ax(t )  Bu (t )
dt
d
x(t )  Ax(t )  Bu (t )
dt
Pre-multiplying both sides by e  At
d
e At x(t )  Ae At x(t )  e At Bu (t )
dt
d  At
[e  x(t )]  e At .Bu (t )
dt
Taking initial time as t  0  and integrating from 0  to ‘t’ :
t
[e  At  x(t )]t0   e  A Bu () d 
0
t
e  At
x(t )  x(0 )   e  A Bu ()d 


0

Pre-multiplying both sides by e At

t

e
 At  A
e e At
x(t )  e x(0)  e
At At
Bu ()d 
0
 Control Systems 11 State Space Analysis
t

e
 A ( t  )
x (t )  e1424
At
x (03)  Bu ( ) d 
0
Used to find zeroinput response 1442443
Used to find zero state response

For zero initial conditions, x (0  )  0

t
x(t )   e A(t  ) Bu () d 
0

Given that response

y  Cx
t
y  C  e A(t  ) Bu () d 
0

For impulse response,

u (t )  (t ) u ( )   (  )
t
y  C  e A(t  ) B() d   C  e At  B
0

(using product property of impulse function)

Hence, the correct option is (C).
0 1 1 
Sol.10 Given : x    x  0  u, y  1 0 x …(i)
0 3   
State equation is given by,
x  Ax  Bu …(ii)
Output state equation is given by,
y  Cx  Du …(iii)
On comparing equation (i) with (ii) and (iii),
0 1  0 
A  , B    , C  1 0
 0  3 1 
and D  [0]
The transfer function is given by,
T(s) = C[ sI  A]1 B …(iv)
 s 0  0 1   s 1 
[ sI  A]     
0 s  0 3 0 s  3
 s  3 1
Adj[ sI  A]  
 0 s 
sI  A  s ( s  3)
Adj[ sI  A]
[ sI  A]1 
sI  A
Control Systems 12 State Space Analysis
1
1 s 1 
[ sI  A]   
0 s  3
1  s  3 1
[ sI  A]1 
s( s  3)  0 s 
From equation (iv),
1 1 
s s ( s  3)  1 
T ( s )  [1 0]12  
 1   0  21
0 ( s  3)  22

1
T ( s )  1 0  s 
 
0
Y (s) 1
T (s)    
U (s) s
Hence, the correct option is (A).
1 0  p
Sol.11 Given : A    , B  q
0 1  
The controllability matrix is defined as,
QC   B : AB : A2 B..... : An 1 B 
n n

where, n  number of state variable

1 0  p   p  0   p 
AB        
0 1  q   0  q   q 
QC   B : AB
p p
So, QC  
q q 
QC  0
Thus, the system is not controllable for all values of p and q.
Hence, the correct option is (C).
Sol.12 Given : The state and output equation is,
dx1 (t )
  3 x1 (t )  x2 (t )  2u (t ) …(i)
dt
dx2 (t )
  2 x2 ( t )  u ( t ) …(ii)
dt
y(t )  x1 (t ) …(iii)
From equation (i), (ii) and (iii),
  3 1 2
x    x  1  u …(iv)
0 2   
Control Systems 13 State Space Analysis

y  1 0 x …(v)
State equation is given by,
x  Ax  Bu …(vi)
Output state equation is given by,
y  Cx  Du …(vii)
On comparing equation (iv), (v) with (vi) and (vii),
 3 1   2
A   , B    , C  1 0
 0  2 1
D=0
Transfer function is given by,
T ( s)  C [sI  A]1 B …(viii)
 s 0  3 1 
[ sI  A]22     
0 s  22  0 2  22
s  3 1 
[ sI  A]22  
 0 s  2  22

sI  A  ( s  3)( s  2)  s 2  5s  6

s  2 1 
Adj[ sI  A]  
 0 s  3 22
Adj[ sI  A]
[ sI  A]1 
sI  A

 s2 1 
 s 2  5s  6 s  5s  6 
2
[ sI  A]212  
 s3 
0
 s 2  5s  6  22
From equation (vii), we get
 s2 1 
 s 2  5s  6 s  5s  6   2 
2
T ( s )  1 0  
 s  3   1 
0
 s 2  5s  6 

 2s  5 
 2 
T ( s )  1 012  s  5s  6 
 s3 
 s 2  5s  6  21

 2s  5 
T ( s)   2
 s  5s  6 11
Hence, the correct option is (C).
Control Systems 14 State Space Analysis
Sol.13
Method 1 :
The state transition matrix is given by
e At  (t )  L1[sI  A]1
Adj[ sI  A]
[ sI  A]1 
sI  A
1 s  2 1 
 sI  A1  
s  5s  6  0
2
s  3
1 s  2 1 
 sI  A1  
( s  2) ( s  3)  0 s  3
 1 1 
s  3  s  2  s  3 
( s )  
 1 
 0 s2 
2 2

 1 1 1 
s  3 
s  2 s  3
( s )   
 0 1 
 s2  22
Taking inverse Laplace transform,
 e  3t e  2 t  e  3t 
(t )  L 1
 sI  A 1
 
 0 e 2 t  22
Hence, the correct option is (B).
. Method 2 :
Check by options :
(i) From property of state transition matrix,
(0)  I
For option (A),
 e  3t 0 
(t )    2t  3t 
e  e e 2t 
 1 0
(0)   I
 2 1
This does not satisfy the property of STM.
For option (B),
 e  3t e  2t  e  3t 
(t )   
 0 e  2t 
1 0 
(0)   I
0 1 
This satisfies the property of STM.
Control Systems 15 State Space Analysis
For option (C),
 e  3t e  2 t  e  3t 
(t )   
 0 e 2 t 
1 2 
(0)   I
0 1 
This does not satisfy the property of STM.
For option (D),
 e3t e  2 t  e  3t 
(t )   
0 e  2t 
1 0 
(0)   I
0 1 
This satisfies the property of STM.
Option (A) and (C) are eliminated as does not satisfy property of STM.
So we have to check another property of STM for option (B) and (D).
(ii) From property of state transition matrix,
 '(0)  A
For option (B),
  3e  3t 2e  2 t  3e  3t 
 '(t )   
 0 2e  2t 
 3 1 
 '(0)   A
 0 2
This satisfies the property of STM.
For option (D),
3e3t 2e  2 t  3e  3t 
 '(t )   
 0 2 e  2 t 
3 1 
 '(0)   A
 0 2 
This does not satisfy the property of STM.
Only option (B) is satisfying both the property of STM.
Hence, the correct option is (B).
 x   4 1  x1  1
Sol.14 Given :  1          u,
 x2   3 1  x2  1
x 
y  1 0  1  …(i)
 x2 
State equation is given by,
x  Ax  Bu …(ii)
 Control Systems 16 State Space Analysis
Output state equation is given by,
y  Cx  Du …(iii)
On comparing equation (i) with (ii) and (iii),
  4 1 1
A  , B    , C  1 0
 3 1 1
The transfer function of state space equation is given by,
T ( s )  C  ( sI  A) 1  B  …(iv)

 s 0  4 1
 sI  A    
0 s   3 1
s  4 1 
 sI  A  
 3 s  1
 s  1 1 
Adj sI  A   
  3 s  4
sI  A  ( s  1)( s  4)  3  s 2  5s  1

 s 1 1 
Adj[ sI  A]  2 s  5s  1 
 sI  A   s  5s  1
2
1
[ sI  A]1  
sI  A  3 s4 
 s  5s  1
2
s 2  5s  1 
From equation (iv),
 s 1 1 
 s 2  5s  1 s  5s  1 
2
1
T ( s )  1 012   1
 3 s4    22
 s  5s  1
2
s 2  5s  1  22
 s 11   s 
 s 2  5s  1   s 2  5s  1 
T ( s )  1 0    1 0  
 3  s  4   s 1 
 s 2  5s  1   s 2  5s  1 
s
T (s) 
s  5s  1
2

Hence, the correct option is (A).

 x1   0 a1 0   x1  0
Sol.15 Given :  x2    0 0 a2   x2   0 u …(i)
      
 x3   a3 0 0   x3  1
 x1 
y  1 0 0  x2  …(ii)
 
 x3 
Control Systems 17 State Space Analysis
State equation is given by,
x  Ax  Bu …(iii)
On comparing equation (i) and (iii),
 0 a1 0  0
A   0 0 a2  and B  0
   
 a3 0 0  1
The controllability matrix is defined as,
QC   B : AB : A2 B..... : An 1 B 
n n

where, n  number of state variable

 0 a1 0  0  0 
AB   0 0 a2  0   a2 
    
 a3 0 0  1   0 

 0 a1 0   0 
A B   A  AB    0 0 a2   a2 
2
  
 a3 0 0   0 

 a1a2 
A2 B   0 
 
 0 

0 0 a1a2 
QC   B : AB : A2 B   0 a2 0 
33  
1 0 0  33

QC  (0  a1a22 )
Condition for controllability is,
QC  0
Hence,  a1a22  0
From above condition,
a1  0 and a2  0
Since, QC is independent of a3 , hence for controllability a3 may or may not be zero. Therefore for
controllability of given system, a1  0, a2  0 and a3 may or may not be zero.
Hence, the correct option is (D).
 x   0 1   x1 
Sol.16 Given :  1      …(i)
 x2   0 0   x2 
. Method 1 :
State equation is given by,
x  Ax  Bu …(ii)
Control Systems 18 State Space Analysis
Comparing equation (i) and (ii),
0 1 
A  B0
0 0 
 s 0 0 1   s 1
[ sI  A]     0 0  0 s 
 0 s     
Adj[sI  A]
[sI  A]1 
sI  A
 s 1
Adj[ sI  A]   
0 s 
sI  A  s2
The state transition matrix is given by,
e At  (t )  L1  ( s ) 
Resolvent matrix of A is,
1 1
1  s 1  s s2 
( s )   sI  A  2 
1
  
s 0 s   1
0
 s 
1 1
s s2 
Now, e At  (t )  L1  ( s )  L1  
0 1
 s 
1 t 
(t )   
0 1
Hence, the correct option is (D).
. Method 2 :
Check by options :
(i) From property of state transition matrix,
(0)  I
For option (A) :
t 1
(t )   
1 0
0 1 
(0)   I
1 0 
This does not satisfy the property of STM.
For option (B) :
1 0
(t )   
t 1
Control Systems 19 State Space Analysis

1 0 
(0)   I
0 1 
This satisfies the property of STM.
For option (C) :
0 1
(t )   
1 t 
0 1 
(0)   I
1 0 
This does not satisfy the property of STM.
For option (D) :
1 t 
(t )   
0 1
1 0 
(0)   I
0 1 
This satisfies the property of STM.
Option (A) and (C) are eliminated as they do not satisfy property of STM.
So we have to check another property of STM for option (B) and (D).
(ii) From property of state transition matrix,
 '(0)  A
For option (B) :
1 0
(t )   
t 1
0 0
 '(t )   
1 0 
0 0
 '(0)   A
 1 0 
This does not match with given matrix A.
For option (D) :
1 t 
(t )   
0 1
0 1 
 '(t )   
0 0
0 1 
 '(0)   A
0 0
This matches with given matrix A.
Hence, the correct option is (D).
Control Systems 20 State Space Analysis

 x  1 0   x1  1
Sol.17 Given :  1         u …(i)
 x2  1 1   x2  1
State equation is given by,
x  Ax  Bu …(ii)
y  Cx  Du …(iii)
Compare equation (i) and (ii),
1 0 1
A   , B  1
1 1 
. Method 1 :
The state transition matrix is given by,
e At  ( t )  L1  sI  A
1

 s 0  1 0   s  1 0 
[ sI  A]     
0 s  1 1   1 s  1
Adj[ sI  A]
[ sI  A]1 
sI  A
s  1 0 
Adj[ sI  A]  
 1 s  1 22
sI  A  ( s  1) 2

 s 1   1 
 ( s  1) 2 0   ( s  1) 0 
[ sI  A]  
1
 
 1 s 1   1 1 
 ( s  1) 2 ( s  1) 2   ( s  1) 2 ( s  1) 

Taking inverse Laplace transform,
 et 0
L1 ( sI  A) 1   (t )  e At   t 
te et 
Hence, the correct option is (C).
. Method 2 :
Check by options :
(i) From property of state transition matrix,
(0)  I
For option (A),
 et 0
(t )   t 
e et 
1 0
(0)   I
1 1 
This does not satisfies the property of STM.
Control Systems 21 State Space Analysis
For option (B),
 et 0
(t )   2 t 
t e et 
1 0 
(0)   I
0 1 
This satisfies the property of STM.
For option (C),
 et 0
(t )   t 
te et 
1 0 
(0)   I
0 1 
This satisfies the property of STM.
For option (D),
 et tet 
(t )   
0 et 
1 0 
(0)   I
0 1 
This satisfies the property of STM.
Only options (B), (C) and (D) are satisfying property of STM.
So we have to check another property of STM.
(ii) From property of state transition matrix,
 '(0)  A
For option (B),
 et 0
 '(t )   2 t 
t e  2te
t
et 
1 0 
 '(0)   A
0 1
This does not satisfy the property of STM.
For option (C),
 et 0
 '(t )   t t 
te  e et 
1 0
 '(0)    A
1 1
This satisfy the property of STM.
For option (D),
 et tet  et 
 '(t )   
0 et 
Control Systems 22 State Space Analysis

1 1
 '(0)    A
0 1
This does not satisfy the property of STM.
Only option (C) is satisfying both the property of STM.
Hence, the correct option is (C).
 1 2 1
Sol.18 Given : A    , B 
 6  1
The controllability matrix is defined as,
QC  [ B : AB : A2 B...... An 1 B ]
where, n  number of state variable
 1 2  1  3 
AB      
  6  1   6 
QC 22   B : AB 22

1 3 
QC   
1 α  6
The condition for the system to be uncontrollable is given by,
QC  0
1 3
QC  0
1 6
 63  0
  3
Hence, the value of  is – 3.
Sol.19 Given :
 x1 (t )  1 2   x1 (t )  1 
 x (t )    2 0  x (t )    2  u(t ) …(i)
 2    2   
 x (t ) 
y (t )  1 0  1  …(ii)
 x2 (t ) 
State equation is given by,
x  Ax  Bu …(iii)
and y ( t )  Cx  Du …(iv)
Comparing equation (i), (ii), (iii) and (iv),
1 2 1
A   , B    , C  1 0
 2 0 2
D=0
Transfer function is given by,
T ( s )  C   sI  A  B 
1

Control Systems 23 State Space Analysis

 s 0  1 2   s  1  2 
 sI  A    
0 s   2 0    2 s 
sI  A  s( s  1)  4  s 2  s  4
Note : From characteristic equation we can see that only option (D) is correct.
s 2 
Adj sI  A   
2 s  1
Adj sI  A 1 s 2 
 sI  A1   2
sI  A s  s  4  2 s  1
1 s 2  1 
 sI  A1  B 
s  s  4 2 s  1 2
2 

1  s  4
 sI  A1  B 
s  s  4  2s 
2 

 s4 
 2 
C  sI  A1  B   1 0  s  s  4 
 2s 
 s 2  s  4 
s4
T ( s )  C  sI  A B 
1

s s4
2

s4
T ( s) 
s s4
2

Hence, the correct option is (D).

Sol.20 Given :
 x1 (t )  0 1   x1 (t )   0 
 x (t )   0  2   x (t )   1  u (t ) …(i)
 2    2   
 x (t ) 
y (t )  [1 0]  1  …(ii)
 x2 (t ) 
 x1 (0)  1 
and  x (0)   0
 2   
State equation is given by,
x  Ax  Bu …(iii)
and y  Cx  Du …(iv)
Comparing equation (i), (ii), (iii) and (iv),
0 1  0
A  , B    and C  [1 0]
0  2  1 
D=0
Output matrix is given by,
 y (t )11  C 12  x(t )21 …(v)
Control Systems 24 State Space Analysis

where  x(t )22   x(t )ZIR   x(t )ZSR …(vi)

The state transition matrix is given by,
(t )  L1  ( s ) 
where, ( s )  [ sI  A]1
 s 1 
[ sI  A]22   
 0 s  2  22
sI  A  s ( s  2)
 s  2 1
Adj sI  A  
 0 s  22
Adj[ sI  A] 1  s  2 1
[ sI  A]1  
[ sI  A] s ( s  2)  0 s  22
1 1 
s s ( s  2) 
[ sI  A]1   
 1 
 0 s  2  22
1 11 1 
   
[ sI  A]1   s 2  s s  2 

 1 
 0 s2  22

1 11 1 
s   
2  s s  2 
( s )   
 1 
 0 s2  22
Taking inverse Laplace transform,
 1 
1 (1  e  2t ) 
(t )   2  STM
  2t 
0 e  22
Calculation of x ( t ) ZIR :
 1 
1 (1  e  2 t )  1 
[(t )][ x (0)]   2
   0  21
0 e  2t  22
1 
[(t )][ x(0)]    …(i)
0
Calculation of x(t ) ZSR :
1 11 1 
s   
2  s s  2  0  1
[( s )][ B ][U ( s )]    1  s
 1   
 0 s2 
Control Systems 25 State Space Analysis

 1 1 1 
 2 s  s  s  2  
[( s )][ B ][U ( s )]   
 1 
 s ( s  2) 
 
1  1 1 
 2  s 2  s ( s  2)  
[( s )][ B ][U ( s )]    
 1 
 
 s ( s  2) 
 1 1 1 
 2s 2  4s  4( s  2) 
(s)  B  U (s)   
 1 1 
 2s  2( s  2) 
 
Taking inverse Laplace transform,
 1 1 1  2 t  
 2 t  4  4 e  
L  ( s )   B  U ( s )   
1

  1 1  2t  
  2  2 e  
 
From equation (v) and (vi),
 1 1 1  2t 
t  e 
  2 4 4
1
[ x(t )]      
0   1  1 e  2t 
 2 2 
 1 3 1  2t  
 2 t  4  4 e  
[ x (t )]   
  1 1  2t  
  2  2 e  

From equation (v),

 x (t ) 
y (t )  [1 0]12  1 
 x2 (t )  21
 1 3 1  2 t  
 2 t  4  4 e  
y (t )  [1 0]12  
  1 1  2t  
  2  2 e  
21

1 3 1 
y (t )   t   e  2 t 
2 4 4 11
1 3 1 2
y (t ) t 1 sec    e  1.283
2 4 4
Hence, the value of output y(t) at t = 1 sec is 1.283.
Control Systems 26 State Space Analysis

  4 1.5 2
Sol.21 Given : x    x   u …(i)
4 0   0
y  1.5 0.625 x …(ii)
State space representation of the system is given by,
x  Ax  Bu …(iii)
y  Cx  Du …(iv)
Comparing equation (i), (ii) with (iii) and (iv),
  4 1.5 2
A  , B 
4 0  0
C  1.5 0.625 , D0
The transfer function of state space equation is given by,
T ( s )  C ( sI  A) 1  B  …(v)

 s 0  4 1.5
 sI  A   
0 s   4 0 
 s  4 1.5
 sI  A  
 4 s 
sI  A  s ( s  4)  (  4)  1.5  s 2  4 s  6

 s 1.5 
Adj  sI  A   
4 s  4
Adj  sI  A
 sI  A1 
sI  A

 s 1.5 
 s  4s  6 
 sI  A1   s  4 s  6
2 2

 4 s4 
 s  4 s  6
2
s 2  4 s  6 
 s 1.5 
 s  4s  6  2
 sI  A1  B   s  4 s  6
2 2
 0
 4 s4   
 s  4 s  6
2
s 2  4 s  6 
 2s 
 s  4s  6 
2
 sI  A  B  
1

 8 
 s 2  4 s  6 

 2s 
 s2  4s  6 
T ( s )  1.5 0.625  
 8 
 s 2  4 s  6 
Control Systems 27 State Space Analysis

 3s 5 
T ( s)   2  2
 s  4 s  6 s  4 s  6 
3s  5 
T ( s)11  
 s  4 s  6 11
2

Hence, the correct option is (A).

 1 0 
Sol.22 Given : x    x, …(i)
 0 2
1
with initial condition x(0)    .
1
State equation is given by,
x  Ax  Bu …(ii)
On comparing equation (i) and (ii),
 1 0 
A  , B   0
 0 2
The solution of homogeneous equation is given by,
x (t )  (t )  x (0)
. Method 1 :
State transition matrix is given by,
(t )  L1[( sI  A) 1 ]
 s 0  1 0   s  1 0 
[ sI  A]      0 2   0 s  2
 0 s     
s  2 0 
Adj[ sI  A]  
 0 s  1
sI  A  ( s  1)( s  2)  0  ( s  2)( s  1)
Adj[ sI  A]
[ sI  A]1 
sI  A
 1 
 s 1 0 
[ sI  A]1   
 0 1 
 s  2 
Taking inverse Laplace transform,
e t 0 
(t )  L1[ sI  A]1   
0 e 2t 
From equation (i),
et 0  1
x(t )  (t )  x(0)    
0 e2t  1
Hence, the correct option is (D).
Control Systems 28 State Space Analysis

. Method 2 :
Checking from the options
All the given options are of the form
x (t )  (t )  x (0)
State transition matrix (t ) must satisfy following three properties :
(i) (t ) t 0   I nn

(ii)  '(t ) t 0   Ann

(iii) ( t )   (t ) 

1

Only option (D) satisfies all three above mentioned properties.

Hence, the correct option is (D).
 3 0 0  6 0
Sol.23 Given : A   0  s 0  , B  0 6 
 0 0 1 0 0 

0 0 4 
C 
1 0 4 
Matrix QC   B AB A2 B 

 3 0 0  6 0   18 0 
AB   0  s 0  0 6    0 30 
   
 0 0 1 0 0   0 0 

 3 0 0   18 0  54 0 
A B  A( AB)   0  s 0 
2  0 30    0 150 
   
 0 0 1  0 0   0 0 

6 0 18 0 54 0 
QC  0 6 0 30 0 150 
0 0 0 0 0 0 
Since last column always zero hence if we take any 3  3 matrix determinant always zero hence
uncontrollable
Matrix QC  C T AT C T ( AT )2 C T 

 3 0 0  0 1  0 3
A C   0  s 0 
T T 0 0   0 0 
   
 0 0 1  4 4   4 4 

 3 0 0   0 3  0 9 
( A ) C   0  s 0 
T 2 T  0 0   0 0 
   
 0 0 1  4 4   4 4 
Control Systems 29 State Space Analysis

 0 1 0 3 0 9 
Q0   0 0 0 0 0 0 
 4 4 4 4 4 4 
Let take any 3  3 matrix
0 1 0
Q0'   0 0 0 
 4 4 4 

Q0'  0
Hence not completely observable
Hence, the correct option is (D).
Sol.24 Given,
The given transfer function of certain system is,
Y ( s) 1
 4
U ( s ) s  5s  7 s 2  6 s  3
3

Y ( s) X ( s)
Let, H (s)  
X ( s) U ( s)
Where,
X ( s) 1
 4 … (i)
U ( s ) s  5s  7 s 2  6 s  3
3

Y (s)
1 … (ii)
X ( s)
Using equation (i), we have
( s 4  5s 3  7 s 2  6s  3) X ( s)  U ( s)
s 4 X ( s )  5s 3 X ( s )  7 s 2 X ( s )  6sX ( s )  3 X ( s )  U ( s )
Taking inverse Laplace transform,
d 4x d 3x d 2x dx
4
 5 3
 7 2
 6  3x  u
dt dt dt dt
Let x  x1
dx
 x2  x1
dt
d 2x
 x3  x2  
x1 … (iii)
dt 2
d 3x
 x4  x3  
x2  
x1
dt 3
d 4x
 x5  x4  
x3  
x2  
x1
dt 4
x 4  u  3 x1  6 x2  7 x3  5 x4 … (iv)
 Control Systems 30 State Space Analysis
Using equation (iii) and (iv),
 x1   0 1 0 0  x1  0 
 x   0 0 1 0   x  0 
 2     2   u … (v)
 x3   0 0 0 1   x3  0 
       
 x4   3 6 7 5  x4  1 
The state equation is,
x  Ax  Bu … (vi)
On comparing equation (v) and (vi),
0 1 0 0 0 
0 0 1 0  
A  and B  0 
0 0 0 1 0 
   
 3 6 7 5 1 
Hence, the correct option is (A).
Sol.25 Given : x (t )  Ax (t )
 1 1 0 0 0
 0 1 1 0 0 

A   0 0 1 0 0
 
 0 0 0 3 4 
 0 0 0 4 3

Eigen values can be obtained as,

A  I  0
 1 1 0 0 0   0 0 0 0
 0 1 1 0 0   0  0 0 0 

A   I   0 0 1 0 0   0 0  0 0  0
   
 0 0 0 3 4   0 0 0  0
 0 0 0 4 3  0 0 0 0  

 1   1 0 0 0 
 0 1   1 0 0 
 
A  I   0 0 1   0 0 0
 
 0 0 0 3   4 
 0 0 0 4 3   

A I  (1 )[(1 )(1 )  (3 )(3 )  (16)  0

A I  (1)3[9  3  3  2 16]  0

A  I  [1  ]3[2  6  25]  0

 6  36 100 
A  I  [1  ]3  0
 2 
Control Systems 31 State Space Analysis

 6  j8 
A  I  [1  ]3  0
 2 
The Eigen values of the system are   1,  1,  1 and 3  j 4 .
Hence, the correct option is (D).
Sol.26 From the given figure,
1
y1  (u1  y2  y2 )
s 1
1 1
y1  u1 and y2  u2
s 1 s 1
 1 
0 
  1   s 1
y
y    1 

 2  0
 s  1 
Hence, the correct option is (C).
Sol.27 The given state diagram is shown below,
1

1 1/s 1 1/s x1 1 1
ul C2
x2 C1 C2

–2 –3
From above figure,
x1  x2  3x1
x2  2 x2  u
C2  x1  x2
 3 1  0 
C  1 1 , A    ,B   
 0 2  1 
Controllability matrix is given by,
QC   B : AB 

0 1 
QC   
1 2 
QC  0  1  0 Hence controllable matrix
Observability matrix is given by,
Q0  C T AT C T 

1 3
Q0   
1 2 
Q0  2  3  0
Hence completely observable matrix.
Hence, the correct option is (A).
Control Systems 32 State Space Analysis
Sol.28 Given :
0 1 0
x    [ x]    u  … (i)
 1 1 K 
x 
y  x1  x2 and x   1 
 x2 
x 
y  1 1  1  … (ii)
 x2 
State equation is,
x  Ax  Bu … (iii)
Output equation is,
y  Cx … (iv)
On comparing equation (iii), (iv) with (i) and (ii)
0 1 0
A  , B    , C  1 1 , D  0
 1 1 K 
Transfer function of state space model is,
T ( s )  C [ sI  A]1 B  D
the matrix [ sI  A] is,
 s 0  0 1 
[ sI  A]    
0 s   1 1
 s 1 
[ sI  A]   
1 s  1
 s  1 1  s  1  s  1
 1 s   1 s 
1
[ sI  A]     
s( s  1)  (1)(1) s2  s  1
 s  1 1
 1 s 
[ sI  A]   2
1 
s  s 1
 s  1 1 0
1 1   K 
 1 s    1 K
T ( s )  [C ] [ sI  A]1 [ B ]  1 1  
s2  s  1 s  s 1
2
 sK 
K  sK
T ( s) 
s2  s  1
y K ( s  1)
T ( s)   2
x s  s 1
Sol.29 Given : State variable model
0 1 1 
x    x   u
1 2  2
Control Systems 33 State Space Analysis
Characteristic equation is given by,
sI  A  0
s 1
0
1 s  2
s ( s  2)  1  0
s 2  2s  1  0 …(i)
For a second order system characteristic equation is given by,
s 2  2n s  2n  0 …(ii)
On comparing equation (i) and (ii),
n  1 rad/sec and 2n  2
 1
Hence, the correct option is (B).
0 1
Sol.30 Given : A   
 4 5
Characteristic equation is given by,
sI  A  0
s 1
0
4 s5
s ( s  5)  4  0
s 2  5s  4  0
( s  1)( s  4)  0
s  1,  4
Hence, the correct option is (A).
Sol.31 Given :
x1 (t )  2 x1 (t )  4 x2 (t )
x2 (t )  2 x1 (t )  x2 (t )  u(t )
 x1 (t )   2 4   x1 (t )  0
 x (t )    2  1  x (t )   1   u(t ) 11 …(i)
 2  21   2 2  2  21   21
State equation is given by,
x  Ax  Bu … (ii)
On comparing equation (i) and (ii),
 2 4  0
A   B 
 2 1 1 
The controllability matrix is defined as,
QC   B : AB : A2 B..... : An 1B 
n n
Control Systems 34 State Space Analysis
where, n  number of state variable
 2 4  0  4 
AB       
 2 1 1   1
0 4 
QC   B : AB    
1  1
QC  (0  1)  (1  4)   4  0
Thus, the system is controllable.
 s 0  2 4   s  2  4 
 sI  A     
0 s   2 1  2 s  1
The characteristic equation is given by,
sI  A  0
s2 4
sI  A   s 2  3s  6  0
2 s 1
Therefore, the roots of characteristic equation are,
s  1.37,  4.37
Img(jw)

Real (s)
– 4.37 1.37

One pole of the closed-loop system lie in right half of s-plane. Thus, the system is unstable.
Hence, the correct option is (B).
 Key Point
If QC  0 , then the system is controllable.
If QC  0 , then the system is uncontrollable.

Sol.32 Given :
1
R( s) 3 C (s)
s

. Method 1 :
1
R( s) 3 C (s)
s +1

Transfer function of given system is,

3
T ( s) 
s 1
C ( s) 3

R( s) s  1
Control Systems 35 State Space Analysis
sC ( s )  C ( s )  3R ( s )
Taking inverse Laplace transform,
dc(t )
 c(t )  3r (t )
dt
Taking x1  c(t )
dc(t )
So, x1 
dt
x1  3r(t )  x1
 x1    1 x1  3 r(t ) …(i)
The state equation of system is given by,
x  Ax  Bu …(ii)
On comparing equation (i) and (ii),
A   1
Therefore, the system matrix is [–1].
Hence, the correct option is (C).
. Method 2 :
Signal flow graph for given block diagram is shown below.
1
x&1 s x1 1
r (t ) c(t )
3

-1
c(t )  x1
x1  3r(t )  x1
 x1    1 x1  3 r(t ) …(i)
The state equation of system is given by,
x  Ax  Bu …(ii)
On comparing equation (i) and (ii),
A   1
Hence, the correct option is (C).
 x   2 0   x1  1
Sol.33 Given :  1          [u ],
 x2   0 4   x2  1
x 
y   4 0  1  …(i)
 x2 
Input, u  (t )
State equation is given by,
x  Ax  Bu …(ii)
Control Systems 36 State Space Analysis
Output state equation is given by,
y  Cx  Du …(iii)
From equation (i), (ii) and (iii),
2 0 1
A   , B    and C   4 0
 0 4 1
D=0
Taking Laplace transform of input,
U ( s)  1
As the initial conditions are zero. Output is only due to input i.e. ZSR.
Output of the system is given by,
T ( s)  C [ sI  A]1 B + D …(iv)
 s 0  2 0 
[ sI  A]    
0 s   0 4 
s  2 0 
[ sI  A]  
 0 s  4
Adj sI  A
 sI  A1 
sI  A

s  4 0 
Adj sI  A  
 0 s  2
sI  A  ( s  4)( s  2)

 s4 
 ( s  2)( s  4) 0 
 sI  A1   
 s2 
 0
 ( s  2)( s  4) 

 1 
 0 
 sI  A1   s  2 
 0 1 
 s  4 
From equation (iv),
 1 
s  2 0 
1
T ( s )   4 0    
 0 1  1
 s  4 
 1 
s  2
T ( s )   4 0  
 1 
 s  4 
Control Systems 37 State Space Analysis
4 Y ( s)
T ( s)  
s  2 U ( s)
4
Y ( s)  ; where U ( s )  1
s2
Taking inverse Laplace transform,
y (t )  4 e 2 t
Hence, the correct option is (B).
Y (s) s6
Sol.34 Given :  2 …(i)
U ( s ) s  5s  6
0 1 0
A  , B  1
  6  5  
The transfer function of state space equation is given by,
T ( s )  C  ( sI  A) 1  B 

 s 0   0 1   s 1 
[ sI  A]       6 5  6 s  5
 0 s     
Adj[ sI  A]
[ sI  A]1 
sI  A
 s  5 1
Adj[ sI  A]   
  6 s
sI  A  s ( s  5)  6

1  s  5 1 1  s  5 1
[ sI  A]1   2
s( s  5)  6   6 s  s  5s  6   6 s 
 

Let C   a b  , from equation (ii),

 s5 1 
 s 2  5s  6 s  5s  6 
2
0
T ( s)   a b   1 
 6 s   
 s  5s  6
2
s 2  5s  6 
 1 
 s  5s  6 
2
T ( s )   a b 12  
 s 
 s 2  5s  6  21
 a bs 
T (s)   2  2
 s  5s  6 s  5s  6 
a  bs
T (s)  …(iii)
s  5s  6
2

On comparing equations (i) and (iii),

a  6 and b  1
Control Systems 38 State Space Analysis

Hence, C   6 1
Hence, the correct option is (B).
 Key Point
In direct decomposition form,
(i) The last row of A contains the negative values of the coefficients of the homogeneous part of the
differential equation in ascending order, except for the coefficient of the highest-order term, which
is unity.
(ii) B is a column matrix with the last row equal to 1 and the rest of the elements are all zeros.
(iii) C is a row matrix with coefficients of numerator in ascending order of transfer function.
Note : Numerator terms of transfer function gives the C matrix.

Sol.35 From observation of options, we are taking x1 , x2 , x3 as state variables u1 and u2 as inputs.
-a

1 x&1 1/s x1
Input u1
a
b x&3
-b Output y
g 1/s
Input u2 1
x&2 1/s x2

-g

Given signal flow graph represents multiple input single output (MISO) system.
From above signal flow graph,
x1    x1  x2  u1
x2   x1  x2  u2
x3   x1  x2
y  x3
 x1     0  x1  1 0
 x      0  x    0 1   u1 
 2    2    u2 
 x3  31    0 33  x3  31  0 0 32 21

 x1 
 y 11  0 0 113  x2 
 x3 31
Hence, the correct option is (C).
Sol.36 Given :
 x1   2 0   x1  1
 x    0 1  x   1 u , x1 (0)  0,
 2   2  
x 
   x2 (0)  0 and y  1 0  1  …(i)
 x2 
Control Systems 39 State Space Analysis

. Method 1 :
State equation is given by,
x  Ax  Bu …(ii)
Output state equation is given by,
y  Cx  Du …(iii)
On comparing equation (i), (ii) and (iii), we get
 2 0  1
A   , B 
 0 1 1
C  1 0 and D = 0
The controllability matrix is defined as,
QC  [ B : AB : A2 B...... : An 1 B ]
Where n  number of state variable
 2 0  1  2
AB       
 0 1 1  1
QC   B : AB
1 2 
QC   
1 1 22
QC  1  ( 2)  1  0
Thus, the system is controllable.
The observability matrix is defined as,
QO  C T : AT C T : .... : ( An 1 )T C T 
where, n  number of state variable
1 
CT   
 0
 2 0  1  2
AT C T       
 0 1 0  0 
QO  C T : AT C T 

1 2
QO   
0 0 
QO  0
Thus, the system is not observable.
Hence, the correct option is (A).
. Method 2 :
A system with distinct eigen values and a diagonal system matrix is controllable if the input coupling
matrix B does not have any rows that are zero.
Control Systems 40 State Space Analysis
Distinct eigen values

é x&1 ù é -2 0 ù é x1 ù é1ù
ê x& ú = ê 0 -1ú ê x ú + ê1ú u Non-zero hence controllable
ë 2û ë ûë 2û ë û
For systems represented in parallel form with distinct eigenvalues, if any column of the output coupling
matrix C is zero, the diagonal system is not observable.
Distinct eigen values
Zero hence unobservable
é x&1 ù é -2 0 ù é x1 ù é1ù éx ù
ê x& ú = ê 0 -1ú ê x ú + ê1ú u y = [1 0]ê 1 ú
ë 2û ë ûë 2û ë û ë x2 û
Hence, the correct option is (A).

 Key Point
(i) Gilbert’s test :
It is applicable for diagonal canonical form (DCF) or Jordan canonical form (JCF).
(ii) Parallel realization of any system will always provide matrix A in the form of diagonal canonical
form (DCF).
(iii) If in any matrix, diagonal elements are non-zero and different, elements below and above diagonal
are zero, then that matrix is known as diagonal canonical form (DCF).
  P1 0 0 0  0 
 0  P2 0 0  0 
 
 0 0  P3 0  0 
 
 0  0  P4  0 
     
 
 0  0 0   Pn  nn
(iv) If in any matrix, elements below the diagonal are zero, some element in diagonal are repeatitive and
element just above repeatitive diagonal element are one, then this type of matrix is known as Jordan
canonical form (JCF).
  P1 1 0 0  0 
 0  P1 1 0  0 
 
 0 0  P1 0  0 
 
 0  0  P4  0 
     
 
 0  0 0   Pn  nn

Sol.37 The transfer function of state space model is given by,

T(s) = C[ sI  A]1 B …(i)
 s 0  2 0 
[ sI  A]    
0 s   0 1
s  2 0 
[ sI  A]  
 0 s  1
Control Systems 41 State Space Analysis
Adj[ sI  A]
[ sI  A]1 
sI  A
s  1 0 
Adj[ sI  A]  
 0 s  2
 ( s  1) 
 ( s  2)( s  1) 0 
sI  A  ( s  2)( s  1) [ sI  A]1   
 ( s  2) 
 0
 ( s  2)( s  1) 

 1 
 0 
[ sI  A]1   s  2 
 0 1 
 s  1 
From equation (i),
 1   1 
s  2 0  s  2
 
1
T ( s )  1 0      1 0  
 0 1  1  1 
 s  1   s  1 
Y ( s) 1
T ( s)  
X ( s) s  2
1
For unit step input X ( s )  ,
s
1 1 1 1 
Y ( s)    
s( s  2) 2  s s  2 
Taking inverse Laplace transform,
1 1 1
y (t )  (1  e 2 t )   e 2 t
2 2 2
Hence, the correct option is (A).
Sol.38 . Method 1 :
Given :
 x1   1 1 0   x1   0
 x    0 1 0   x    4  u, …(i)
 2   2  
 x3   0 0 2   x3   0 

 x1 
y  1 1 1  x2  …(ii)
 
 x3 
State equation is given by,
x  Ax  Bu …(iii)
Output equation is given by,
y  Cx  Du …(iv)
Control Systems 42 State Space Analysis
On comparing equation (i), (ii) with (iii) and (iv),
 1 1 0  0
A   0 1 0  , B   4  , C  [1 1 1]
   
 0 0 2   0 
D0
The controllability matrix is defined as,
QC   B : AB : A2 B..... : An 1 B 
n n

where, n  number of state variable

 1 1 0   0   4 
AB   0 1 0   4    4 
    
 0 0 2   0   0 
 1 1 0   4   8
A B  [ A][ AB ]   0 1 0   4    4 
2
    
 0 0 2   0   0 
 0 4 8
QC   B : AB : A B 
2
  4 4 4 
33  
 0 0 0  33
QC  0
Thus, system is not completely state controllable.
The observability matrix is defined as,
QO  C T : AT C T : .... : ( An 1 )T C T 
n n

where, n  number of state variable

1
C  1
T
 
1
 1 0 0  1  1
AT C T   1 1 0  1   0 
    
 0 0 2  1  2 
A2T C T  AT [ AT C T ]
 1 0 0   1  1 
A C   1 1 0   0    1
2T T
    
 0 0 2   2   4 
QO  C T : AT C T : ( A2 )T C T 
33

1 1 1 
QO  1 0 1
 
1 2 4  33
Control Systems 43 State Space Analysis

QO  1  0
Thus, system is completely state observable.
Hence, the correct option is (B).
Concept of Jordan Canonical Form (JCF) :
In a transfer function, if the denominator is in factored form with one or more repeatitive roots, then state
variable representation can be arranged in JCF form.
Example :
Y ( s) 1

U ( s ) ( s  1) ( s  2)
2

Y ( s) K1 K2 K3
  
U ( s ) ( s  1) 2
( s  1) ( s  2)
K1U ( s ) K 2U ( s) K 3U ( s )
Y ( s)    …(i)
( s  1) 2 ( s  1) ( s  2)
U ( s) X ( s)
Let, X 1 ( s)   2
( s  1) 2
( s  1)
U ( s)
X 2 ( s) 
( s  1)
U ( s)
X 3 ( s) 
( s  2)
Then, x1   x1  x2 …(ii)
x2   x2  u …(iii)
x3   2 x3  u …(iv)
From equation (i),
Y ( s)  K1 X 1 ( s)  K2 X 2 ( s)  K3 X 3 ( s)
y  K1 x1  K2 x2  K3 x3 …(v)
So, state variable representation in JCF form can be represented as,
Poles/Eigen values

é x&1 ù é -1 1 0ù é x1 ù é0 ù
ê x& ú = ê 0 -1 0 ú ê x ú + ê1 ú u
ê 2ú ê ú ê 2ú ê ú [ ]
êë x&3 úû êë 0 0 - 2 úû êë x3 úû êë1 úû

Jordan block

é x1 ù
y = [K1 K2 K 3 ] êê x2 úú
êë x3 úû
Partial fraction
coefficient
Control Systems 44 State Space Analysis

. Method 2 :
Method for identifying controllability and observability :
Since, the given state variable representation is in JCF, so we can apply Gilbert’s test for determining the
system controllability and observability.
(i) If A is in diagonal canonical form (DCF) or Jordan canonical form (JCF), then the pair (A, B) is
completely controllable if all the elements in the rows of B that correspond to the last row of each
Jordan block are nonzero.
Last row of
Jordan block-1
é x&1 ù é -1 1 0 ù é x1 ù é 0 ù Non-zero
ê x& ú = ê 0 -1 0 ú ê x ú + ê 4 ú u
ê 2ú ê ú ê 2 ú ê ú Zero hence uncontrollable
êë x&3 úû êë 0 0 -2 úû êë x3 úû êë 0 úû
Last row of
Jordan block-2

(ii) If A is in diagonal canonical form (DCF) or Jordan canonical form (JCF), then the pair (A, C) is
completely observable if all the elements in the columns of C that correspond to the first row of each
Jordan block are nonzero.
first row of Jordan block 1

é x&1 ù é -1 1 0 ù é x1 ù é 0 ù
ê x& ú = ê 0 -1 0 ú ê x ú + ê 4 ú u
ê 2ú ê úê 2ú ê ú
êë x&3 úû êë 0 0 -2 úû êë x3 úû êë 0 úû
first row of Jordan block 2
é x1 ù
y =[1 1 1 ] êê x2 úú
êë x3 úû

Non-zero hence observable

Hence, the correct option is (B).
dx1
Sol.39 Given :  2 x1  x2  u …(i)
dt
dx2
  2 x1  u …(ii)
dt
y  3x1 …(iii)
From equation (i), (ii) and (iii),
 x1   2 1   x1  1
 x     2 0   x   1 u  …(iv)
 2   2  
x 
y   3 0  1  …(v)
 x2 
State equation is given by,
x  Ax  Bu …(vi)
Control Systems 45 State Space Analysis
Output state equation is given by,
y  Cx  Du …(vii)
On comparing equation (iv), (v), (vi) and (vii),
 2 1 1
A   , B  1 and C   3 0
  2 0 
D0
Transfer function is given by,
T ( s)  C[sI  A]1 B …(viii)
 s 0  2 1  s  2 1
[ sI  A]    
 0 s    2 0  2 s 
Adj  sI  A
 sI  A
1

sI  A

s 1 
Adj[ sI  A]   
 2 s  2
sI  A  s( s  2)  2  s 2  2s  2

 s 1 
 s  2s  2
2
s  2s  2 
2
[ sI  A]1   
 2 s2 
 s  2 s  2
2
s 2  2 s  2 
From equation (viii),
 s 1 
 s2  2s  2 s  2 s  2  1
2
T ( s )   3 0  
 2 s  2  1
 s  2 s  2
2
s 2  2 s  2 
 s 1 
 s2  2s  2 
T ( s )   3 0  
 s4 
 s 2  2 s  2 
3( s  1)
T ( s) 
s  2s  2
2

Hence, the correct option is (A).

d
Sol.40 Given : x1 (t )  x2 (t )  0 …(i)
dt
d
x2 (t )  2 x1 (t )  3x2 (t )  r (t ) …(ii)
dt
From equation (i),
x1  x2  0
x1  x2 …(iii)
Control Systems 46 State Space Analysis
From equation (ii),
x2  2 x1  3x2  r(t )
x2   2 x1  3x2  r(t ) …(iv)
From equation (iii) and (iv), state space model is given by,
 x1   0 1   x1  0
 x    2 3  x   1  r (t ) …(v)
 2   2  
State equation is given by,
x  Ax  Bu …(vi)
Comparing equation (v) and (vi),
0 1 0
A  , B 
 2 3 22 1  21
The characteristic equation is given by,
sI  A  0

 s 0  0 1 
0 s    2 3  0
   
s 1
0
2 s3
s 2  3s  2  0 …(vii)
( s  2)( s  1)  0
s  2, s  1
The pole zero diagram shown below,
Img(jw)

Real (s)
–2 –1

Negative unequal real roots represent over-damped system (   1 ).

OR
For a second order system, the characteristics equation is given by,
s 2  2n s  2n  0 ... (viii)
On comparing equation (vii) and (viii),
2n  3 and n  2 rad/sec
2  2  3
  1.06 ξ > 1
Thus, the system is over-damped.
Hence, the correct option is (D).
Control Systems 47 State Space Analysis
2s  1
Sol.41 Transfer function G ( s ) 
s  7s  9
2

Y (s) X (s)
Let, H (s)  
X ( s) U ( s)
X (s) 1
Where,  2 …(i)
U ( s) s  7 s  9
Y (s)
 2s  1 …(ii)
X ( s)
Using equation (i), we have
( s 2  7 s  9) X ( s)  U ( s )
s 2 X ( s )  7 sX ( s)  9 X ( s)  U ( s )
Taking inverse Laplace transform,
d 2x dx
2
 7  9x  u
dt dt
Let x  x1
dx
 x2  x&1 …(iii)
dt
d 2x
 x&2 …(iv)
dt 2
x&2  7 x2  9 x1  u …(v)
Using equations (iii), (iv) and (v)
 x&1   0 1   x1  0 
 x&    9 7   x   1  u
 2   2  
From equation (ii),
Y (s)
 2s  1
X ( s)
(2 s  1) X ( s )  Y ( s )
Taking inverse Laplace transform,
dx
2 x y
dt
dx
x  x1 and  x2
dt
y  x1  2 x2

x 
y  [1 2]  1 
 x2 
Hence, the correct option is (A).
Control Systems 48 State Space Analysis
Sol.42 Given :
1 2  0 
x&    x    u and y  b 0 x
0 1  1 
1 2  0 
A  , B    , C  b 0
0 1  1 
The observability matrix is defined as,
Q0  C T AT C T 
 1 0  b   b 
AT C T       
 2 1  0   2b 
b b 
Q0   
0 2b 
Q0  2b 2
2b 2  0
b0
Hence, system will be observable for all non-zero values of b .
Hence, the correct option is (C).
 4 0 1 
Sol.43 Given: x    x    u and y =[1 1] x …(i)
 2 2  0
State equation is given by,
x  Ax  Bu …(ii)
y  Cx  Du
On comparing equation (i) and (ii)
 4 0 1 
A  , B    , C  1 1
 2 2  0
The controllability matrix is defined as,
QC  [ B : AB : A2 B...: An 1 B ]nn
Where, n number of state variable
 4 0  1   4 
AB       
 2 2  0   2 
1 4 
QC  [ B : AB]   
0 2 
QC  (2  0)  2
QC  0
Thus, the system is completely controllable.
The observability matrix is defined as,
Q0  [C T : AT C T ...( An 1 )T C T ]nn
Control Systems 49 State Space Analysis
Where, n = number of state variable,
 4 2  1  2 
AT C T       
 0 2  1  2 
Q0  [C T : AT C T ]
1 2 
Q0   
1 2 
Q0  2  2  0

Q0  0
Thus, the system is not observable
Hence, the correct option is (D)
Sol.44 Given: State space equation
 4 0 1 
x    x  0  u and y  1 1 x …(i)
 2 2   
State space equation is given by,
x  Ax  Bu …(ii)
y  Cx  Du
On comparing equation (i) and (ii)
 4 0 1 
A  , B    , C  1 1
 2 2  0
The state transition matrix (STM) e At is given by
(t )  e At  L1 [ sI  A]1

 s 0  4 0   s  4 0 
[ sI  A]       
0 s   2 2   2 s  2 
Adj[ sI  A] 1 s  2 0 
[ sI  A]1  
sI  A ( s  2)( s  4)  2 s  4 


 1 
 0 
s4
[sI  A]1   
 2 1 
 ( s  2)( s  4) s  2 
Taking inverse Laplace transform,
 1 
 0 
s4
(t )  e At  L1  
 1  1 1 
 s  2 ( s  4) s  2 
 e4t 0
(t )   2t 4t 
e  e e2t 
Control Systems 50 State Space Analysis

 e 4t 0
STM  (t )  e At   2t 4t 
e  e e 2t 
Hence, the correct option is (A)
Sol.45 Given:
 4 0 1 
(i) x    x   u
 2 2  0 
(ii) y  [1 1] x
State space equation is given by,
x  Ax  Bu
y  Cx  Du
 4 0 1
A  , B    and C  [11]
 2 2  0
 s 0  4 0 
[ sI  A]    
0 s   2 2 
s  4 0 
[ sI  A]  
 2 s  2 
Adj[ sI  A]
[ sI  A]1 
sI  A
s  2 0 
Adj [ sI  A]   
 2 s  4 
sI  A  ( s  4)( s  2)
1 s  2 0 
[ sI  A]1 
( s  4)( s  2)  2 s  4 


 1 
 0 
s4
[ sI  A]1   
 2 1 
 ( s  4)( s  2) s  2 
Transfer function is given by,
T ( s)  C[sI  A]1 B  D
 1 
 0 
s4 1 
T ( s )  [1 1]   0
 2 1  0
 ( s  4)( s  2) s  2 
 1 
 s4 
T ( s )  [1 1]  
 2 
 ( s  4)( s  2) 
Control Systems 51 State Space Analysis
1 2
T ( s)  
s  4 ( s  4)( s  2)
s22
T ( s) 
( s  4)( s  2)
s4 1
T ( s)  
( s  4)( s  2) s  2
Hence, the correct option is (D)
Sol.46 Given :
The given signal flow graph is,
1 s -1
1
K
1 -1
s -1 1
1
-2
Assume state vector and state variable in above figure as shown in below figure.
State State
vector variable
x1 x1
1 s -1
1
K y
u -1
x2 x 1
1 s -1 2
1
-2
From above figure,
x1  u  uK  x1  u (1  K )  x1
x2  u  u  2 x2  2u  2 x2
y  x1  x2
State equation will be
 x1   1 0   x1  1  K 
 x    0 2   x    2  [u ] … (i)
 2    2  
Standard equation of state space is,
x  Ax  Bu … (ii)
On comparing equation (i) and (ii)
 1 0  1  K 
A  and B   
 0 2   2 
The controllable matrix is,
QC   B : AB 
 Control Systems 52 State Space Analysis

 1 0  1  K 
Where, AB    
 0 2   2 
  (1  K ) 
AB   
 4 
1  K  (1  K ) 
QC  
 2 4 
For uncontrollable matrix QC  0
1 K 1  K
 QC  0
2 4
(1  K ) ( 4)   2 ( 1  K )   0
4  4 K  2  2 K  0
2  2 K  0
1  K  0
K  1
Hence, the correct option is (D).
Sol.47 Given: x1 (t )  x2 (t )
x2 (t )  x1 (t )  x2 (t )  u (t )
y (t )  x1 (t )  3x2 (t )

 x1 (t )   0 1  x1  0 
 x (t )   1 1  x   1  u (t ) 11 …(i)
 2  21   22  2  21   21
x 
 x(t )11  1 312  x1  …(ii)
 2  21
State equation is given by
x  Ax  Bu …(iii)
y  Cx  Du …(iv)
On comparing equation (i) and (iii)
0 1 0 
A  ,B   
1 1 1 
On comparing equation is given by.
C  [1 3], D  0
Transfer function.
T ( s)  C ( sI  A)1 B  D
 s 0 0 1 s 1 
( sI  A)         
 0 s  22 1 1 22  1 s  1 22
Control Systems 53 State Space Analysis

 s  1 1
Adj( sI  A)  
 1 s  22

sI  A  s( s  1)  1

sI  A  s 2  s  1

 s 1 1 
Adj( s I  A) 1  s  1 1  s2  s 1 s  s  1
2
[ sI  A]1   2   
sI  A s  s 1  1 2  22  1 s 
 s  s  1
2
s 2  s  1  22
 s 1 1   1 
 s2  s  1 s  s 1
2   0   s  s  1
2
[ s I  A]1 B       
 1 s   1  21  s 
 s 2  s  1 2

s  s  1  2 2 
 s  s  1  21
2

 1 
 s  s  1
2
C[ s I  A]1 B  [1 3]12  
 s 
 s 2  s  1  21
1 3s
T ( s )  C[ sI  A]1 B  D   2
s  s 1 s  s 1
2

(3s  1)
T (s) 
( s 2  s  1)
Hence, the correct option is (D)
Sol.48 The given state diagram is shown below,
1 1

a s -1 z2 1 s -1 z1 1
R(s) C(s)

- p2 - p1
X 2 (s) X 1 (s)

From above figure,

x2  aR( s )  P2 X 2 ( s )
and x1  Z 2 X 2 ( s )  P2 X 2 ( s )  P1 X 1 ( s )  aR( s)
 x1    P1 Z 2  P2   x1   a 
   P2   x    a  r (t )
 x2   0  2  
Hence, the correct option is (A).
1
Sol.49 Given : Transfer function G( s) 
s2
X ( s) 1

U (s) s 2
Control Systems 54 State Space Analysis

s 2 X (s)  U (s)
d 2 x(t )
u
dt 2
x2  u
 x1  0 1   x1  0 
 x    0 0   x   1  u
 2    2  
0 1 
A 
0 0
State transition matrix  L1[ sI  A]1
1
 s 1
1
[ sI  A]   
0 s 
1  s 1
[ sI  A]1  0 s 
s2  
1 1
s s2 
[ sI  A]1   
0 1
 s 
1 t 
STM  L1[ sI  A]1   
0 1
Hence, the correct option is (A).

Sol.50 Given:
State variable representation
0 1  0 
x    X    u , y  [11]x
0 1 1 
Resolvant matrix is given by,
1
 s 1 
1
( s )  [ sI  A]   
0 s  1
Adj[ sI  A]
[ sI  A]1 
sI  A

 s  1 1
 0 s 
( s )  [ sI  A]1  
s( s  1)
1  s  1 1
( s ) 
s( s  1)  0 s 
T ( s)  C[ sI  A]1 B [Since D  0]
Control Systems 55 State Space Analysis

1  s  1 1  0 
T ( s )  [1 1]
s( s  1)  0 s  1 
1 1
T ( s )  [1 1]
s( s  1)  s 

 1 
 s( s  1) 
T (s)  [1 1]  
 1 
 s  1 
1 1
T ( s)  
s( s  1) s  1
s 1 1
T ( s)  
s( s  1) s
Hence, the correct option is (B)
Sol.51 Given :
x1  v0 , x2  i
The given network is shown below,
L R

+ +
v(t ) i (t ) C v0 (t )
- -

Applying KVL in the above loop,

Ldi (t)
v (t )   i (t) R  v0 (t)
dt
di (t) v(t) i (t) R v0 (t)
  
dt L L L
v (t ) R x
x&2   x2  1 …(i)
L L L
dv0
iC
dt
x2  C x&1
x2
x&1 
C
 1 
0 0
 x&1   C   x1   
    1 v
 R   x2   
 x&2   1
 L L
L 
Hence, the correct option is (C).
Control Systems 56 State Space Analysis
Sol.52 The given state block diagram is shown below
1

x&1
2 ò
-2
U Y
4

x&2
ò
-5

From above figure,

x&1  x1  2 x2  2u
x&2  4 x1  5 x2  u
y  x1  x2

 x&1  1 2   x1   2 
Hence,  x&    4 5  x   1  u
 2   2  
x 
y  1 1  1 
 x2 
Transfer function  C  sI  A B
1

s 1 2 
 sI  A   
 4 s  5
1  s  5 2 
 sI  A1 
( s  1)( s  5)  8  4 s  1
1  s  5 2 
 sI  A1 
( s  1)( s  3)  4 s  1
1  s  5 2   2 
Transfer function  1 1
( s  1)( s  3)  4 s  1 1 
 2s  8 
 ( s  1)( s  3) 
T ( s )  1 1  
 s7 
 ( s  1)( s  3) 
 
2s  8  s  7
T ( s) 
( s  1)( s  3)
3( s  5)
T ( s) 
( s  1)( s  3)
Hence, the correct option is (C).
Control Systems 57 State Space Analysis
Sol.53 Given:
The state space equation is:
 x1   1 0   x1  1 
 x    0 3   x    0  u
 1   2  
x 
y  [1 1]  1 
 x2 
The characteristic equation is given by,
sI  A  0
s 1 0
0
0 s3
( s  1)( s  3)  0  0
s  1, 3
Roots of characteristic equation are poles of the system which also represent Eigen values of system
matrix A.
So Eigen values of A are 1  1, 2  3
Hence, the correct option is (B)
Sol.54 The characteristic equation is given by,
sI  A  0
s 1 0
0
0 s3
( s  1)( s  3)  0
s2  4s  3  0
Hence, the correct option is (B)
Sol.55 Transfer function is given by,
T ( s)  C[ sI  A]1 B [Since D  0]
s  1 0 
[ sI  A]  
 0 s  3
1 s  3 0 
[ sI  A]1 
( s  1)( s  3)  0 s  1
 1 
 s 1 0 
[ sI  A]1   
 0 1 
 s  3 
 1 
s 1 0 
1 
Transfer function T (s)  [1 1]   
 0 1  0
 s  3 
Control Systems 58 State Space Analysis

 1 
T (s)  [1 1]  s  1 
 
 0 
1
T (s) 
s 1
Hence, the correct option is (C)
Sol.56 The state transition matrix is given by,
(t )  L1[ sI  A]1
 1 
s 1 0 
(t )  L1  
 0 1 
 s  3 
et 0 
 (t )   3 t 
0 e 
Hence, the correct option is (A)
Sol.57 Given :
 1 0  1 
A  B    and C  [1 1]
 0 3 22 0
The controllability matrix is defined as,
Qc  [ B : AB : A2 B........... Ah 1 B ]nn
For n = 2,
Qc  [ B : AB]
 1 0  1   1
AB       
 0 3 0   0 
1 1
Qc   
0 0 
Qc  0  0  0
Thus, system is not completely state controller. The observability matrix is defined as,
Q0  [C T : AT C T .........( Am )T C T ]nn
For n = 2,
Q0  [C T : AT C T ]
 1 0  1  1
AT C t       
 0 3 1  3
1 1
Q0   
1 3
Q0  3  1  2  0
Thus, system is completely state observable.
Hence, the correct option is (B)
Control Systems 59 State Space Analysis
Sol.58 The characteristic equation is given by,
s2  4s  3  0 …(i)
For a standard second order system the characteristic equation is given by,
s 2  2n s  n2  0 …(ii)
On comparing equation (i) and (ii)
n  3 and 2n  4
2
  1.154
3
Hence, the correct option is (D)
Sol.59 Number of differential equations is same as the number of state equations.
Hence, the correct option is (C).
Sol.60 Given :
 2 1  1 
A  B  …(i)
 0 1 0
 0 1 0 
A  B  …(ii)
 1 0  1 
Controllability matrix is given by,
Qc  [ B : AB : A2 B......... An 1 B ]
For n = 2,
Qc  [ B : AB]
(i) Consider equation 1 :
 2 1  1   2 
AB       
 0 1 0   0 
1 2 
Qc   
0 0 
Qc  0
Thus, equation 1 is not completely state controllable.
(ii) Consider equation 2:
 0 1   0  1 
AB       
 1 0  1  0 
0 1 
Qc   
1 0 
Qc  1  0
Thus, equation 2 is completely state controllable
Hence, the correct option is (D)
Control Systems 60 State Space Analysis

 x1   1 0   x1 
Sol.61 Given :  x    0 2   x  …(i)
 2   2
Initial conditions are x1 (0)  1 and x2 (0)  1.
. Method 1 :
State equation is given by,
x  Ax  Bu …(ii)
On comparing equation (i) and (ii),
 1 0 
A  , input matrix B = 0
 0 2 
Solution of state equation is given by,
x ( t )  ZIR  ZSR
If input is not given then we consider zero input response.
x (t )  ZIR
 x(t )n1   (t )nn  x(0)n1
x (t )  L1[sI  A]1 x(0) …(iii)
Calculation of ZIR :
 s 0  1 0 
 sI  A    
0 s   0 2 
s  1 0 
 sI  A  
 0 s  2 
Adj[sI  A]
[ sI  A]1 
sI  A

s  2 0 
Adj[ sI  A]  
 0 s  1
sI  A  ( s  1)( s  2)

 s2 
 ( s  1)( s  2) 0 
[ sI  A]1   
 s 1 
 0
 ( s  1)( s  2) 

 1 
s 1 0 
[ sI  A]1   
 0 1 
 s  2 
Taking inverse Laplace transform,
e t 0 
(t )  L1[ sI  A]1   
0 e 2 t 
Control Systems 61 State Space Analysis
From equation (iii),
et 0  1
 21 
x ( t )  
0 e 2 t  22  1 21

 x1 (t )   et 
 x (t )    2 t 
 2   e 
Hence, the correct option is (C).
. Method 2 :
Check initial condition :
For option (A) :
x1 (t )  1, x2 (t )  2
Since, x1 (t ) and x2 (t ) are constant for all the time, hence cannot be initial value.
x1 (0)  1, x2 (0)  2
This does not match with given initial condition.
For option (B) :
x1 (t )   e  t , x2 (t )  2e  t
x1 (0)   e 0  1, x2 (0)  2e 0  2
This does not match with given initial condition.
For option (C) :
x1 (t )  e  t , x2 ( t )   e  2 t
x1 (0)  e 0  1, x2 (0)   e  20  1
This matches with given initial conditions.
For option (D) :
x1 (t )  e  t , x2 (t )   2e  t
x1 (0)  e 0  1, x2 (0)   2e0   2
This does not match with given initial conditions.
Hence, the correct option is (C).
1 0 
Sol.62 Given : x (t )    x (t ) …(i)
0 2
State equation for homogenous system is given by,
x (t )  A x (t ) …(ii)
On comparing equation (i) and (ii),
1 0 
A  
0 2 
Now, y (t )  C T x (t )
x(t )  (t ) x(0)  e At x(0)
Control Systems 62 State Space Analysis

where, e At  L1 [( sI  A)1 ] …(iii)

s  1 0 
sI  A  
 0 s  2
sI  A  ( s  1) ( s  2)

s  2 0 
Adj( sI  A)  
 0 s  1
Adj sI  A
( sI  A) 1 
sI  A

 1 
 s 1 0 
( sI  A) 1   
 0 1 
 s  2 
From equation (iii),
 1 
 s 1 0 
et 0
(t )  e At  L1   
 0 1   0 
e2t 
 s  2 
 et 0  1  et 
and x (t )     
0 e 2 t  1  e 2t 
 et 
y (t )  1 1  2 t    et  e 2t 
e 
y (t ) t log 2   eloge 2  e 2 loge 2 
e

y (t ) t log 2   e loge 2  eloge 4 

e

y (t ) t log 2  [2  4]  6
e

Hence, the value of y (t ) for t  loge 2 is 6.

 x1 (t )  0 0   x1 (t )   0 
Sol.63 Given :  x (t )   0 9   x (t )    45 u(t ) …(i)
 2    2   
 x1 (0)  0
and  x (0)   0
 2   
. Method 1 :
State equation is given by,
x  Ax  Bu …(ii)
On comparing equation (i) and (ii),
0 0  0
A  , B 
0 9   45
Control Systems 63 State Space Analysis
Solution of above state equation in time domain is given by,
x ( t )  ZIR  ZSR
where, ZIR  f [ x (0)] = zero input response
ZSR  f ( I /P ) x (0)0 = zero state response
In case of zero initial condition, we consider zero state response.
Thus, x (t )  ZSR
ZSR in transform domain is given by,
X ( s )  ( s ) BU ( s ) …(iii)
s 0 
[ sI  A]   
0 s  9
Resolvent matrix of A is,
( s)  [ sI  A]1
1 
0 
Adj sI  A  s
( s )   
sI  A 0 1 
 s  9 
0
 BU ( s)21    U ( s )11
 45 21
0
0  1
 BU ( s )21         45 
 45 21  s 11
 s  21
From equation (iii),
X ( s )  ( s ) BU ( s )
1   0 
 0  0
 X ( s)21   s   45    45 
0 1     
 s  
 s( s  9)  21
 s  9  22 21

Taking partial fraction,

 0 
 X ( s)21   5 5 


 s s  9  21
Taking inverse Laplace transform,
 0 
 x(t )21   9 t 
5  5e  21
 x1 (t )   0 
 x (t )   5  5e 9 t 
 2  21   21
x1 (t )  0 , x2 (t )  5  5e 9 t
x1 (  )  0, x2 (  )  5  5( e 9 )  5
Control Systems 64 State Space Analysis
The value of

lim x12 (t )  x22 (t )  x12 ( )  x22 ( )

t 

 0  (5) 2  5

Hence, the value of lim x12 (t )  x22 (t ) is 5.

t 

. Method 2 :
 x1 (t )  0 0   x1 (t )   0 
 x (t )   0 9   x (t )    45 u(t )
 2    2   
x1 (t )  x1 (0) …(i)
x2 (t )   9 x2 (t )  45x(t ) …(ii)
Taking Laplace transform on both sides of equation (i),
sX 1 ( s)  x1 (0)
x1 (0)
X 1 ( s)  0  x1 (0)  0
s
So, x1 (t )  0
Applying final value theorem,
lim x1 (t )  x1 ()  0
t 

Taking Laplace transform on both sides of equation (ii),

45
sX 2 ( s )  x2 (0)   9 X 2 ( s ) 
s
45
X 2 ( s)   x2 (0)  0
s ( s  9)
Applying final value theorem,
45
lim x2 (t )  x2 ( )  lim sX 2 ( s )  5
t  s 0 9
Required value  lim x12 (t )  x22 (t )  x12 (  )  x22 (  )  02  52
t 

Required value = 5
Hence, the value of lim x12 (t )  x22 (t ) is 5.
t 


Practice Solutions :
(0.5s  1)
Sol.1 Given : G ( s )        …(i) 
(0.05s  1)
Transfer function of lead compensator is given by,
(1  s)
G (s)  …(ii)
      (1  s )
where,   1
Comparing equation (i) and (ii),
  0.5 ,   0.05
  0.1
Maximum phase lead of compensator is given by,
 1  
m  sin 1  
 1  
9
m  sin 1    550
 11 
The frequency at which maximum phase occur is given by,
1
m   6.32rad/s
 
Hence, the correct option is (D).
Sol.2 The proportional band (PB) of the controller is given as,
100
KP 
PB
where, K P is proportional gain of the controller.
Hence, the correct option is (B).
Sol.3 Given : closed loop control system is shown below,
1
R( s) KC C (s)
s ( s + 3)
Control Systems 2 Controllers & Compensators
KC
G ( s)  , H (s)  1
s ( s  3)
Closed loop system transfer function is given by,
G ( s)
T (s) 
1  G (s) H ( s)
Y (s) KC
 2
R( s ) s  3s  K C
. Method 1 :
The characteristic equation is given by,
s 2  3s  K C  0
3  9  4 K C
Poles 
2
Given condition is,
Poles  1
3  9  4 KC
 1
2
 9  4KC  1
9  4KC  1
4KC  8
KC  2
Hence, the correct option is (C).
. Method 2 :
jw Im ( z )
Mapping from Stable
Stable
s-plane to z-plane

s Re ( z )
s = -1 z=0

s-plane z-plane

Put s 1  z
s  z 1
The characteristic equation is given by,
1  G(s) H (s)  0
KC
1 0
s ( s  3)
s 2  3s  K C  0
( z  1) 2  3( z  1)  K C  0
z 2  2 z  1  3z  3  KC  0
z 2  z  KC  2  0
Control Systems 3 Controllers & Compensators
Routh Tabulation :
z2 1 KC  2

z1 1 0

z0 KC  2 0

For the system to be stable, all the roots must be in the left-half of z-plane, thus all the coefficients in the
first column of Routh's tabulation must have the same sign. Therefore, the coefficient of first column of
the Routh’s table should be positive.
This leads to the following conditions :
KC  2  0
KC  2
Hence, the correct option is (C).
2
Sol.4 Given : Transfer function of plant  ,
s ( s  3)
K P'  1 and K I  0
Transfer function of PI controller is given by,
KI
GC ( s )  K P' 
s
 K  2 
Forward gain of the system   K P'  I   
 s   s ( s  3) 
2( K I  sK P' ) 2( K I  s )
G (s)   2      K P'  1
s 2 ( s  3) s ( s  3)
The steady state error for unit step input is given by,
1
ess  …(i)
1 KP
The position error coefficient is given by,
K P  lim G ( s ) H ( s)
s 0

2( s  K I )
K P  lim
s  0 s 2 ( s  3)

KP  
From equation (i),
1 1
ess   0
1 KP 1 
The steady state error is zero for any value of K I  0 so the lowest value of K I is zero.
Hence, the correct option is (A).
Sol.5 Option (A) :
For phase lag compensator, pole is more nearer to origin than zero in left half of s-plane.
 Control Systems 4 Controllers & Compensators
jw

Hence, above pole-zero location represents phase lag compensator.

Option (B) :
For phase lead compensator, zero is more nearer to origin than pole in left half of s-plane.
jw

Hence, above pole-zero location represents phase lead compensator.

Option (C) :
Lead compensator jw

Lag compensator

Hence, above pole-zero location represents phase lag-lead compensator.

Option (D) :
Lag compensator jw

Lead compensator
Hence, above pole-zero location represents phase lead-lag compensator.
Hence, the correct option is (A).
Sol.6 Lag compensator :
R1

R2
Vi V0
C

V0 ( s ) (1  s)

Vi ( s ) 1  s
R1  R2
where,   1,   ,   R2C
R2
   is lag compensator coefficient.
(i) Pole zero location of lag compensator is shown below,
Control Systems 5 Controllers & Compensators
jw

s
-1 -1
t bt

(ii) The maximum phase lag max occurs at mid frequency m between upper and lower corner
frequencies.
1
m 
 
1 
max  sin 1  
1  
(iii)In lag compensator, pole is much nearer to origin than zero in left half of s-plane.
(iv) It acts as low pass filter.
(v) It may increase the order of system.
Example :
Block diagram of uncompensated system is shown below,
R( s) G (s) Y (s)

1
Let, G ( s) 
s ( s  2) ( s  4)
Type-1, order - 3
Block diagram of compensated system is shown below,
R( s) GC ( s ) G (s) Y (s)

s4
(a) If GC ( s )   Lag compensator
s 1
1
G ( s )GC ( s ) 
s( s  1) ( s  2)
Type-1, order - 3
s5
(b) If GC ( s )   Lag compensator
s3
s5
G ( s )GC ( s ) 
s( s  4) ( s  2) ( s  3)
Type-1, order - 4
 Key Point
By the same way as explained above, we can perform analysis for lead compensator.
Lead compensator may increase order of system.
Control Systems 6 Controllers & Compensators
(vi) It increases the gain of system.
(vii) It decreases the steady state error of system i.e. it improves steady state response of system.
(viii) It decreases the bandwidth of system.
(ix) It decreases gain crossover gc frequency.
(x) It decreases the speed of response of system.
(xi) It increases the settling time of system.
(xii) It increases the peak overshoot of system.
(xiii) It decreases damping factor  .
(xiv) It decreases natural frequency n .
(xv) Lag compensator may stabilize an unstable system.
Example 1 :
Block diagram of uncompensated system is shown below,
R( s) G (s) Y (s)

1
Let, G ( s) 
( s  2) ( s  1)
OLTF - Unstable
G( s) 1
T ( s)   2
1  G( s) s  s  1
CLTF - Unstable
Block diagram of compensated system is shown below,
R( s) GC ( s ) G (s) Y (s)

s4
If GC ( s )   Lag compensator
s3
G ( s )GC ( s )
TC ( s )  … (i)
1  G ( s )GC ( s )
s4
TC ( s ) 
( s  3) ( s  s  1)  s  4
2

s4
TC ( s ) 
s  4 s 2  3s  1
3

Characteristic equation is given by,

s 3  4 s 2  3s  1
Routh Tabulation :
s3   1  3 
s2   4  1
s1   1/4  0
s0   1  0 
Control Systems 7 Controllers & Compensators
Since, there is no sign changes in the first column of Routh table.
Hence, compensated CLTF is stable
Example 2 :
1
Let, G ( s) 
( s  2) ( s  1)
OLTF - Unstable
G( s) 1
T ( s)   2
1  G ( s ) s  3s  3
CLTF - Unstable
From equation (i),
s4
TC ( s ) 
( s  3) ( s  3s  2)  s  4
2

s4
TC ( s ) 
s  6 s  10
3

Characteristic equation is given by,

s 3  6s  10
Routh Tabulation :
s3   1  – 6 
s2   0  10 

Since, the first element of the s 2 row is zero, the all elements in the s1 row would be infinite. To overcome
this difficulty, we replace the zero in the s 2 row with a small positive variable ε and then proceed with
the tabulation. Starting with the s 2 row, the results are as follows,
s2     10 
 6  10
s1   0 

s0   10  0 
From the Routh table,
 6ε  10  10 
lim  lim   6     
ε0   0
  
Since, there are two sign changes in the first column of Routh's tabulation, then the equation has two roots
in the right-half of s-plane.
Hence, compensated CLTF is unstable.
Therefore, Lag compensator may or may not stabilizes an unstable system
Hence, the correct option is (D).
Sol.7 Compensator transfer function is given below,
 s  s 
1  1  
 0.1  100 
C ( s) 
 s 
(1  s ) 1  
 10 
Control Systems 8 Controllers & Compensators

. Method 1 :
Concept of Asymptotic Bode Phase Plot :
Corner frequencies are 0.1, 1, 10 and 100 rad/sec.
Let C ( s )  H1 ( s )  H 2 ( s )  H 3 ( s )  H 4 ( s )
 s 
For H 1 ( s )   1 
 0.1 
 j 
H1 ( j)  1  
 0.1 
 
and H1 ( j)  tan 1  
 0.1 
At low frequency, H1 (0)  00
At high frequency, H1 ( )  900
At corner frequency, H 1 (0.1)  450
The phase plot is 00 until one tenth of the corner frequency then increases linearly to 900 at ten times the
corner frequency.
ÐH1 ( jw)

+ 900
ec
0 /d
5
+ 450 e4
lop
S
w (rad/sec)
0.01 0.1 1

 1 
For H 2 ( s )   
 1 s 
 1 
H 2 ( j)   
 1  j 
and H 2 ( j)   tan 1 ()
At low frequency, H 2 (0)  00
At high frequency, H 2 ( )   900
At corner frequency, H 2 (1)   450
The phase plot is 00 until one tenth of the corner frequency then decreases linearly to  900 at ten times
the corner frequency.
ÐH 2 ( jw)

0.1 1 10
w (rad/sec)
Sl
op

- 450
e–
45
0
/d
ec

- 900
Control Systems 9 Controllers & Compensators

1
For H 3 ( s ) 
 s 
 1  10 
 
1
H 3 ( j) 
 j 
1  
 10 

and H 3 ( j)   tan 1  
 10 
At low frequency, H 3 (0)  00
At high frequency, H 3 ( )   900
At corner frequency, H 3 (10)   450
ÐH 3 ( jw)

1 10 100
w (rad/sec)
Sl
op

- 450
e–
45
0
/d
ec

- 900

 s 
For H 4 ( s )   1 
 100 
 j 
H 4 ( j)  1  
 100 
  
and H 4 ( j)  tan 1  
 100 
At low frequency, H 4 (0)  00
At high frequency, H 4 ( )   900
At corner frequency, H 4 (100)   450

ÐH 4 ( jw)

+ 900
ec
0 /d
4 5
+ 450
ope
Sl
w (rad/sec)
10 100 1000
Control Systems 10 Controllers & Compensators
ÐH1 ( jw)

+ 900
ec
0 /d
+ 450 e 45
op
Sl
w (rad/sec)
0.01 0.1 1 10 100

ÐH 2 ( jw)

0.01 0.1 1 10 100

w (rad/sec)

Sl
op
- 450

e–
45
0
/d
ec
- 900

ÐH 3 ( jw)

0.01 0.1 1 10 100

w (rad/sec)
Sl
op
- 450 e–
45
0
/d
ec
- 900

ÐH 4 ( jw)

+ 900
ec
0 /d
+ 45 0
e 45
op
Sl
w (rad/sec)
0.01 0.1 1 10 100

ÐC ( jw)
+ 90 0
–9

+ 450
ec
0 /d
0

0 /d
45 1000
ec

w (rad/sec)
0.01 0.1 1 10 100 ec
0 /d
- 450 45

- 900

So, maximum phase lead occur between 0.1    1 .

Hence, the correct option is (A).
. Method 2 :
Pole-zero location of transfer function C ( s ) is shown below,
jw

Lag Lead
Control Systems 11 Controllers & Compensators
This is a representation of standard lead-lag compensator.
In a lead-lag compensator, lead occurs for initial frequency.
Therefore, for 0.1    1 , lead occurs.
Similarly, for later frequencies lag occurs. i.e. for 10    100 .
Hence, the correct option is (A).
Sol.8 Given :   1 and   0
In a lag compensator pole is nearer to origin.
Hence, transfer function of a phase lag compensator is given by,
 1
 s 
1  
T (s) 
 1 
s 
  
Hence, the correct option is (A).
1  0.5s
Sol.9 Given : Transfer function Gc ( s )  … (i)
1  0.05s
Transfer function of a standard lead compensator is given by,
 (1  s)
Gc ( s )  … (ii)
1  s
On comparing equation (i) and (ii),
  0.5
  0.1 and   0.05
The frequency at which phase of the compensator for is maximum is given by,
1
m 
 
1
m   6.32 rad/sec
0.5 0.1
The maximum phase lead max is given by,
 1   1  1  0.1  1  0.9 
max  sin 1    sin    sin  
1    1  0.1   1.1 
max  sin 1 (0.82)
Hence, the correct option is (D).
Sol.10 Given : Pole-zero plot is shown below,
jw

s
p z z p

GH

log w

+ 20 dB/dec
– 20 dB/dec
Control Systems 12 Controllers & Compensators
Since, first pair is pole-zero (lag compensator) then zero-pole (lead compensator) that means given pole-
zero plot is of lag-lead compensator.
Hence, the correct option is (B).
Sol.11 A phase lead network improves bandwidth.
Hence, the correct option is (A).
Sol.12 The integral control action reduces the steady state error to zero.
Hence, the correct option is (C).
0.4
Sol.13 Given : T ( s )  2  …(i)
s
For PI controller transfer function is given by,
KP
T ( s)  K P  …(ii)
TI s
On comparing equations (i) and (ii),
KP  2
KP
 0.4
TI
Integral time, TI  5
Proportional band is given by,
100
PB 
Gain
100 100
PB    50 %
KP 2
Hence, the correct option is (D).
Sol.14 Given : PB  50, Ti  0.2
Proportional band is given by,
100
PB 
Gain
100
PB 
KP
100
50 
KP
KP  2
For PI controller transfer function can be written as,
KP 2 10
T (s)  K P   2  2
Ti s 0.2 s s
Hence, the correct option is (C).
Control Systems 13 Controllers & Compensators
Sol.15 Given : Closed loop control system is shown below,
Negative feedback connection

KI 1
R( s) KP + C (s)
s ( s + 2) ( s + 10)

( sK P  K I )
G ( s)  , H (s)  1
s ( s  2) ( s  10)
The characteristics equation is given by,
1  G(s) H (s)  0
s ( s  2) ( s  10)  ( sK P  K I )  0
s 3  12s 2  20s  sK P  K I  0
s 3  12 s 2  (20  K P ) s  K I  0
Routh Tabulation :
s3 1 20  K P

s2 12 KI
12(20  K P )  K I
s1 0
12
s0 KI 0

For the system to be stable, all the roots must be in the left-half of s-plane, thus all the coefficients in the
first column of Routh's tabulation must have the same sign. Therefore, the coefficient of first column of
the Routh’s table should be positive.
This leads to the following conditions :
12(20  K P )  K I KI
(i) 0  KP   20
12 12
(ii) K I  0
Hence, the correct option is (D).
Sol.16 Given : K v  1000 ,   0.5
Control system with a PD controller is shown in the figure,
100
r KP + KD s c
s ( s + 10)

From above block diagram, open loop transfer function is given by,
100( K P  K D s )
G( s) 
s ( s  10)
Since there is one pole at origin hence, the type of system is one.
Velocity error coefficient is given by,
K v  lim sG ( s )
s 0
Control Systems 14 Controllers & Compensators
100( K P  K D s )
1000  lim s 
s 0 s( s  10)
100
1000  KP
10
K P  100
Closed loop transfer function of negative unity feedback is given by,
G ( s)
T ( s) 
1  G( s)
100( K P  K D s )
T ( s)      …(i)
s  (100 K D  10) s  100 K P
2

The transfer function of second order system is given by,

K 2n
T ( s)  … (ii)
s 2  2n s  2n
On comparing equation (i) and (ii),
2n  100 K P  100  100     n  100
2n  100K D  10
2  0.5  100  100 K D  10
K D  0.9
Hence, the correct option is (B).
Sol.17 Given :
e- s
R(s) KC C (s)
s +1

Loop transfer function of system is,

KC e s
G (s) H (s) 
1 s
K C e  j
Put s  j , G ( j) H ( j) 
1  j
Phase angle of open loop transfer function G ( j) H ( j) is given by,
1800
G ( j) H ( j)     tan 1 

At   1 , G ( j) H ( j)  1800

1800
 1  tan 1 1  1800

Using calculator,
1  2.028
Hence, the correct option is (D).
 Control Systems 15 Controllers & Compensators
Sol.18 Given : K C  1
e s
G (s) H (s) 
1 s
For unit step input (position input), steady state error is given by,
1
ess  …(i)
1 K p
where, K P is position error coefficient.
The position error coefficient is given by,
K P  lim G ( s ) H ( s )
s 0

e s
K P  lim 1
s 0 1  s

From equation (i),

1 1 1
ess   
1 K p 11 2
Hence, the correct option is (B).
Sol.19 Given : R2C2  R1C1
Given circuit is shown below,
Z
C1
Vi
R1 V0

From above circuit,

V0 Z  Z ( sC1R1  1)
    …(i)
Vi R1 R1
sC1R1  1
(i) For Q :  
 
R2 C2

( sC2 R2  1)
        Z  
sC2
From equation (i),
V0  ( sC2 R2  1) ( sC1R1  1)
 
Vi sC2 R1
V0  s 2C1C2 R1R2  s(C1R1  C2 R2 )  1

Vi sC2 R1
 Control Systems 16 Controllers & Compensators
V0  (C1R1  C2 R2 ) 1
   sC1R2
Vi C2 R1 sC2 R1
V0 K
= K1 + 2 + K 3 s
Vi s
Proportional

Integral
Derivative
Above equation represents gain of proportional integral derivative controller.
(ii) For R :
C2

R2

1 R2
Z  R2 || 
sC2 ( sC2 R2  1)
From equation (i),
V0 R2 ( sC1R1  1)
 
Vi ( sC2 R2  1) R1
 C1  1 
 s
V0 C2  R1C1 

Vi  1 
s  R C 
 2 2 

Since R2C2  R1C1 ,

1 1

R1C1 R2C2
Pole-zero location is shown below,
jw

s
-1 -1
R1C1 R2 C2

Since, pole is nearer to origin than zero in left half of s-plane, hence it is a lag compensator.
Hence, the correct option is (B).
k
Sol.20 Given : G ( s ) 
(1  s)
k
R( s) kp C (s)
1 + st
Control Systems 17 Controllers & Compensators
kk p
R( s) C (s)
1 + st

For step input steady state error is given by,

1
ess 
1  k p'
Where k p'  lim G ( s )
s 0

kk p
k p'  lim  kk p
s 0 1  s
1
ess 
1  kk p
Hence steady state error decreases.
Closed loop transfer function
C (s) G(s)

R( s) 1  G ( s)
kk p
C ( s ) (1  s) kk p
 
R( s) kk p 1  s  kk p
1
1  s
C (s) kk p

R(s)  1  kk p 
 s 
  
Pole-zero plot is given below,
jw

s
0
-(1 + kk p )
t

1 
Time constant  
1  kk p 1  kk p

Hence, time constant is also decreases.
Hence, the correct option is (B).
Sol.21 Given : Control system is shown below
Control Systems 18 Controllers & Compensators
1
R( s) 100 C (s)
s (1 + 4 s )

k0 s

100
R( s) C (s)
4s 2 + s + k0 s

C (s) 100 25
 2  … (i)
R ( s ) 4 s  s (1  k0 )  100 s 2  s (1  k0 )  25
4
For second-order standard transfer function is given by,
C ( s) 2n
 2 … (ii)
R( s) s  2n s  2n
On comparing equation (i) and (ii),
1  k0
n  5 and 2n 
4
 k0
2  0.5  5 
4
k0  19
Hence, the correct option is (B).
Sol.22 Given : The uncompensated system is,
900
G ( s) 
s ( s  1)( s  9)
Response of compensated system is,
900
G '( s)  Gc ( s )
s ( s  1)( s  9)
where, GC ( s )  Response of compensator
900
R( s) G (s) = C (s)
s ( s + 1) ( s + 9)

Fig. (a) Uncompensated system

R( s) GC ( s ) G (s) C (s)

Fig. (b) Compensated system

Control Systems 19 Controllers & Compensators
Given that gain-crossover frequency of compensated system is same as phase crossover frequency of un-
compensated system i.e.
gc  fig. (b)    pc  fig. (a) …(i)
Phase angle of uncompensated system at phase crossover frequency is given by,
G( j pc )  1800
 
900  tan 1 ( pc )  tan 1  pc   1800
 9 
  pc 
  pc  
tan 1  9   900
  pc
2

 1 
 9 
2pc
1 0
9
 pc  3 rad/sec
Hence, from equation (i),
( gc ) compensated  3 rad/sec
Magnitude at gain crossover frequency is given by,
G '( jgc )  1
900
GC ( jgc )  1
gc (  1) (2gc  81)
2
gc

900
GC ( jgc )  1
3 10  90
900
GC ( jgc )  1
3  30
1
GC ( jgc )   0.1
10
GC ( jgc )   20 dB (attenuation)
dB

Phase margin of compensated system is given by,

PM  1800  G '( jgc )
450  1800  900  tan 1 (gc )
 
 tan 1  gc   GC ( jgc )
 9 
1
3
450  900  tan 1 3  G ( j ) 450  900  900  G ( j )
1
C gc C gc
1  3 
3
GC ( jgc )  450
Hence, the correct option is (D).
Control Systems 20 Controllers & Compensators
Sol.23 Given : Phase lead network is shown below,
1F

1W
V1 0.5W V0

Transfer function
V0 ( s) 0.5

V1 ( s) 1 
 s 1 
0.5  
1 
 1 
s 
V0 ( s ) 0.5

V1 ( s ) 0.5  1
s 1
V0 ( s ) s 1
 … (i)
V1 ( s )  s
3 1  
 3
Transfer function of phase lead compensator is given by,
 (1  s )
Gc ( s )  … (ii)
1  s
Comparing equation (i) and (ii),
1
  1 and  
3
1

3
The maximum phase shift is given by,
 1
1  1    1
 1 3 
max  sin    sin  1 
 1    1 
 3
max  300
Hence, the correct option is (B).
