1 

CONTROLSYSTEMSChapter 6 : Polar & Nyquist Plot

 

GATE Objective & Numerical Type Solutions

Question 2 [Practice Book] [GATE EC 1989 IIT-Kanpur : 10 Marks]

Nyquist plot consider a feedback system where the OLTF is

1 2 3

( )(1 )(1 )(1 )

KG s

s sT sT sT

Draw Nyquist plot. Find also the range of K in terms of the crossover frequency pc for stability.

Sol. Given : 1 2 3

( )(1 )(1 )(1 )

KG s

s sT sT sT

1 2 3

1( )

(1 )(1 )(1 )G j

j j T j T j T

Magnitude can be written as,

2 2 2 2 2 2

1 2 3

1( )

1 1 1G j

T T T

Phase angle can be written as, 0 1 1 1

1 2 3( ) 90 tan tan tanG j T T T Nyquist path for the given transfer function is shown below.

Section 1 : Polar plot

At 0 ( )G j ( )G j 090

At ( ) 0G j ( )G j 0360

j�

0 to� �� � �

�

j��

Section 1

to 0� �� � �

0j �

Section 3

to� �� � � �

Section 2

Section 4

0 to 0� �� �0j �

j�

j��

�

j��

0j �

0j �

Polar plot

Mirror imageof polar plot

0 0

,

90 90

jRe R

� �

� �

0 0

, 0

90 90

jre r

�

� �

j��

CWCW

Re

Img0270�

090�

0

0�� �

�� � �0180�

0360�

  2 

Section 3 : Mirror image of polar plot

At ( ) 0G j ( )G j 03 6 0

At 0 ( )G j ( )G j 09 0

Section 2 : Semi-circle with radius tending to infinite.

At ( ) 0G j ( )G j 0360

At ( ) 0G j ( )G j 0360

Section 4 : Semi-circle with radius tending to zero.

At 0 ( )G j ( )G j 090

At 0 ( )G j ( )G j 09 0

Re

Img0270�

090�

0

0�� �

�� � �0180�0360�

Re

Img0270�

090�

�� � �0180�0360�

�� � �

Re

Img090

090�

00

  3 

The Nyquist plot is shown below.

The frequency at which phase angle of ( ω) ( ω)G j H j is 0180 is called phase crossover frequency.

0 0 1 1 11 2 3180 90 tan ( ) tan ( ) tan ( )pc pc pcT T T

0 1 1 11 2 390 tan ( ) tan ( ) tan ( )pc pc pcT T T

1 232

1 20

1 232

1 2

1tan 90

11

pc pcpc

pc

pc pcpc

pc

T TT

TT

T TT

TT

1 232

1 2

1 232

1 2

11or

01

1

pc pcpc

pc

pc pcpc

pc

T TT

TT

T TT

TT

21 2 2 3 3 11 0pc TT T T TT

1 2 2 3 3 1

1pc

T T T T T T

2 2 2 2 2 2

1 2 3

( )( 1) ( 1) ( 1)

pc

pc pc pc pc

KG j

T T T

Gain margin can be defined as reciprocal of the magnitude of the ( )G j measured at phase crossover frequency. For the system to be stable, gain margin measured in dB should be positive or ( )G j measured at phase crossover frequency should be less than 1.

2 2 2 2 2 2

1 2 3

1( 1) ( 1) ( 1)pc pc pc pc

K

T T T

2 2 2 2 2 21 2 3( 1) ( 1) ( 1)pc pc pc pcK T T T

where

1 2 2 3 3 1

1pc

T T T T T T

Question 4 [Practice Book] [GATE EC 1992 IIT-Delhi : 8 Marks]

A unity feedback system has open-loop transfer function

1

( )(2 1)( 1)

G ss s s

Sketch Nyquist plot for the system and from there obtain the gain margin and the phase margin.

Re

Img

0

0�� �

0�� �

�� � ��� � �

  4 

Sol. Given : 1

( )(2 1)( 1)

G ss s s

Put ,s j 1

( )(2 1) ( 1)

G jj j j

Magnitude can be written as,

2 2

1( )

(4 1) ( 1)G j

Phase angle can be written as,

0 1 1( ) 90 tan 2 tanG j The Nyquist path for the transfer function is shown below.

Section 1 : Polar plot

At 0 ( )G j ( )G j 090

At ( ) 0G j ( )G j 0270

Section 3 : Mirror image of Polar plot

At ( ) 0G j ( )G j 02 7 0

At 0 ( )G j ( )G j 09 0

j�

0 to� �� � �

�

j��

Section 1

to 0� �� � �

0j �

Section 3

to� �� � � �

Section 2

Section 4

0 to 0� �� �0j �

j�

j��

�

j��

0j �

0j �

Polar plot

Mirror imageof polar plot

0 0

,

90 90

jRe R

� �

� �

0 0

, 0

90 90

jre r

�

� �

j��

CWCW

Re

Img0270�

090�

0

0180�0360�

0�� �

�� � �

  5 

Section 2 : Semi-circle with radius tending to infinite.

At ( ) 0G j ( )G j 0270

At ( ) 0G j ( )G j 02 7 0

Section 4 : Semi-circle with radius tending to zero.

At 0 ( )G j ( )G j 09 0

At 0 ( )G j ( )G j 090

The Nyquist plot is shown below.

Re

Img0270

090

0

018000

0�� �

�� � �

Re

Img0270�

090�

�� � �0180�0360�

�� � �

Re

Img090

090�

00

Re

Img

0�� �

0

0�� �

�� � ��� � �

  6 

The frequency at which phase angle of ( ω) ( ω)G j H j is 0180 is called phase crossover frequency.

0 0 1 1180 90 tan 2 tanpc pc

0 1 190 tan (2 ) tan ( )pc pc

0 1

2

290 tan

1 2pc pc

pc

2

31

0 1 2pc

pc

21 2 0pc

1

rad/sec2

pc

1 1( )

11 1 1 (3) (1.5)4 1 1 22 22

pcG j

( ) 0.667pcG j

Gain margin can be defined as reciprocal of the magnitude of the ( ) ( )G j H j measured at phase

crossover frequency.

1 1

G.M. 1.490.667( )pcG j

In dB, G.M. = 20log 1

20log1.49 3.46 dB( )pcG j

The gain crossover is a point on the ( ) ( )G j H j plot at which the magnitude of ( ) ( )G j H j is equal

to unity (1) and the corresponding frequency is known as gain crossover frequency gc .

( ) ( ) 1gc gcG j H j

2 21 (4 1) ( 1)gc gc gc

2 2 21 (4 1)( 1)gc gc gc

4 2 2(4 )( 1) 1gc gc gc

6 4 4 24 4 1gc gc gc gc

6 4 24 5 1gc gc gc

2 saygcx

3 24 5 1x x x

3 24 5 1 0x x x

1 2 30.326, 0.788 0.379, 0.788 0.379x x j x j

0.326gc

0.57 rad/sec gc

  7 

Phase margin is defined as,

0P.M. 180 ( )gcG j

0 0 1 1P.M. 180 90 tan (2 ) tan ( )gc gc

0 1 1P.M. 90 tan (2 0.57) tan (0.57)

0P.M. 11.57 Question 6 [Practice Book] [GATE EC 1998 IIT-Delhi : 5 Marks]

The loop transfer function of a single loop control system is given by,

100

( ) ( )(1 0.01 )

sTG s H s es s

Find the condition for the closed loop system to be stable.

Sol. Given :

100( ) ( )

(1 0.01 )sTG s H s e

s s

Put ,s j 100

( ) ( )(1 0.01 )

j TG j H j ej j

Magnitude can be written as,

2

100( ) ( )

1 (0.01 )G j H j

Phase angle can be written as,

1rad radin rad

( ) ( ) ( ) tan (0.01 )2

G j H j T

At  0   ( ) ( )G j H j   0( ) ( ) 90G j H j  

At   ( ) ( ) 0G j H j   0( ) ( ) 180G j H j  

The frequency at which magnitude of ( ) ( )G j H j is unity i.e. 1 is called gain crossover frequency.

( ) ( ) 1gc gcG j H j

100

1(1 0.01 )

gcj T

gc gc

e

j j

2

1001

1 (0.01 )gc gc

78.6 rad/sec.gc

Phase margin is defined as,

P.M. ( ) ( )gc gcG j H j

1P.M. tan (0.01 )2gc gcT

For system to be stable, P.M. > 0

1radP.M. 78.6 tan (0.01 78.6) 0

2T

P.M. 78.6 0.9 0T

  8 

T < 0.01 sec system is stable.

Polar plot is shown in below figure.

Alternatively : Routh-Hurwitz Concept

The characteristic equation is given by,

1 ( ) ( ) 0G s H s

100(1 )

1 0(1 0.01 )

sT

s s

(1 0.01 ) 100(1 ) 0s s sT

2 0.01 100 100 0s s sT

20.01 (1 100 ) 100 0s s T

Routh Tabulation :

For system to be stable, first column of Routh array must be positive.

1 100 0T

0.01T

Question 7 [Practice Book] [GATE IN 1999 IIT-Bombay : 5 Marks]

The closed loop transfer function of a system consisting of a process ( )G s and a proportional controller

K is ( )

( )1 ( )

KG sH s

KG s

. The closed contour obtained by plotting the imaginary part of

1

( )H s versus the

real of 1

( )H s evaluated along the curve indicated in figure encircles the origin N times.

Re

Img

0� �

2s

1s

0s

0.01

1 100T�

100

100

0

�

j�

s-plane

R �

�

  9 

(a) Derive an expression for N, terms of the number of poles of H(s) in the right half of the s-plane.

(b) From a plot of the imaginary part of 1

( )G s versus the real part of 1

( )G s evaluated along the curve in

figure determine the condition on K that will make the closed loop system stable.

Sol. (a) ( )

( )1 ( )

KG sH s

KG s

1 1 ( ) 1

1( ) ( ) ( )

KG s

H s KG s KG s

Poles of (1 ( ))KG s Poles of ( )KG s

According to mapping theorem, N = Number of CW encirclements of (1/H) locus around the origin = Number of CW encirclements

of 1/ ( )KG s locus around ( 1, 0)j point.

= Number of zeros 1/ ( )H s in the RH-plane – Number of poles 1/ ( )H s in the RH-plane

= Number of poles H(s) in the RH-plane – Number of zeros H(s) in the RH-plane = P – Z For minimum phase functions, the second term is zero, N = P (b) For closed-loop stability, 0,P N Z

For the special case of minimum phase systems, N = 0 i.e. (1 / )KG locus should not enclose (–1, j0) point.

1 1

1,( ) ( )pc pc

KKG j G j

Where pc is the phase crossover frequency of 1

( )G j.

Question 8 [Practice Book] [GATE EE 2000 IIT-Kharagpur : 5 Marks]

Open-loop transfer function of a unity-feedback system is

1( ) ( )( 1)( 2)

D

D

ss e

G s G s es s s

Given : 1( ) 1G j when 0.466

(a) Determine the phase margin when 0.D

(b) Comment in one sentence on the effect of dead-time on the stability of the system. (c) Determine the maximum value of dead time D for the closed-loop system to be stable.

Sol. Given : 1( ) ( )( 1)( 2)

D

D

ss e

G s G s es s s

Put ,s j 1( ) ( )( 1)( 2)

D

D

jj e

G j G j ej j j

(a) 1( ) 0.466gc

G s

i.e. gain crossover frequency 0.466gc

1

1( )

( 1)( 2)G j

j j j

  10 

Phase angle can be written as,

0 1 1

1( ) 90 tan tan2

G j

Phase margin is given by,

01P.M. 180 ( )gcG j

P.M. 0 0 1 1180 90 tan tan1 2gc gc

P.M. 0 0 0 0 0180 (90 25 13 ) 52

0P.M. 52 Ans.

(b) If dead time increases the stability of the system decreases. Ans.

(c) For marginal stability 0P .M . 0 G.M. 0 dB

Transportation delay does not affect the value of gc .

0 0P.M. 0 180 ( )gcG j

0 0 0 1 1180 ( 57.3 ) 90 tan tan 01 2gc gc

gc D

With 0.466,gc

00.466 57.3D = 51.898

1.94D

For 1.94D system is marginal stable.

1.94D stable

1.94D unstable

1.94D is maximum value of D for closed loop system to be stable. Ans.

Note : Using Routh-Hurwitz stability, 1.67D

Question 4 [Work Book - Polar] [GATE EC 2001 IIT-Kanpur : 1 Mark]

For a unity feedback system represented by ( )(0.2 1)(0.05 1)

KG s

s s s

. For K = 1 gain margin is 28 dB.

For gain margin 20 dB, the value of K is

(A) 5 (B) 4 (C) 2.5 (D) 2

Ans. (C)

Sol. Given : ( )(0.2 1)(0.05 1)

KG s

s s s

Put ,s j ( )(0.2 1)(0.05 1)

KG j

j j j

Magnitude can be written as,

2 2 2 2

( ) ( )(0.2 ) 1 (0.05 ) 1

KG j H j

  11 

1K for gain margin 28 dB

1

G.M. 20 log( ) ( )pc pcG j H j

1

28 20log( ) ( )pc pcG j H j

( ) ( ) 0.0398G j H j …(i)

2 2 2 2

( ) ( )(0.2 ) 1 (0.05 ) 1

pc pc

pc pc pc

KG j H j

2 2

10.0398

0.04 1 0.0025 1pc pc pc

10.02 rad/spc

For G.M . 20 dB

2 2

( ) ( )0.04 1 0.0025 1pc pc pc

KG j H j

2 2

0.110 0.04 10 1 0.0025 10 1

K

2.5K

Hence, the correct option is (C).

Alternatively :

G.M. mar

desired

K

K

For a system marK is fixed.

Given : 1desiredK

In dB, G.M. 20log G.M.

28 20log G.M.

G.M. 25.11 G.M. 25.11mar desiredK K

In dB, G.M. 20log G.M.

20 20log G.M.

G.M. 10

25.11

2.511G.M. 10

mardesired

KK

Question 7 [Work Book - Polar] [GATE EC 2002 IISc-Bangalore : 2 Marks]

The system with the open loop transfer function 2

1( ) ( )

( 1)G s H s

s s s

has a gain margin of

(A) – 6 dB (B) 0 dB (C) 3.5 dB (D) 6 dB Ans. (B)

Sol. Given : 2

1( ) ( )

( 1)G s H s

s s s

  12 

Put ,s j 2

1( ) ( )

(1 )G j H j

j j

….. (i)

Magnitude can be written as,

2 2 2

1( ) ( )

(1 )G j H j

Phase angle can be written as,

12

( ) ( ) tan2 1

G j H j

The frequency at which phase angle of ( ω) ( ω)G j H j is 0180 is called phase crossover frequency.

1

2( ) ( ) tan

2 1pc

pc pcpc

G j H j

0 0 12

180 90 tan1

pc

pc

21 0pc 1 rad/secpc

Gain margin can be defined as reciprocal of the magnitude of the ( ) ( )G j H j  measured at phase crossover frequency.

1

G.M.( ω ) ( ω )

pc pcG j H j

 

2 2 2

1( ) ( ) 1

1 (1 1 ) 1pc pcG j H j

1G.M. 1

1

In dB, G.M. 20 log(1) 0 dB

Hence, the correct option is (B).

Alternatively 1 :

G.M. mar

desired

K

K

Given 1desiredK

Characteristic equation is given by, 1 ( ) ( ) 0G s H s

2

11 0

( 1)s s s

3 2 0s s s K For third order system to be marginal stable, IP = OP

1marK

G.M. 1 In dB, G.M. 20log1 0 dB

Alternatively 2 :

To calculate pc , equating imaginary part of ( )G j to

zero.

3 2Img 0j j

3 0pc pc

1 rad/secpc

Note : This method is only valid when polar plot

crosses 0180 axis.

Re

Img

� � �

0� �

1

1pc

� �

0180�

  13 

Question 8 [Work Book - Polar] [GATE IN 2002 IISc-Bangalore : 2 Marks]

The loop transfer function of a system is given by 10

( ) ( )Lse

G s H ss

. The phase cross-over frequency is

5 rad/s. The value of the dead time L is

(A) 20

(B)

10

(C)

20

(D) 0

Ans. (B)

Sol. Given : 10

( ) ( )Lse

G s H ss

and

ω 5pc rad/sec

Put ,s j 10

( ) ( )Lje

G j H jj

Phase angle can be written as,

     π

( ω) ( ω) ω2

G j H j L  

The frequency at which phase angle of ( ω) ( ω)G j H j is 0180 is called phase crossover frequency.

π

ω π2pcL

π π

2ω 10pc

L

Hence, the correct option is (B). Question 10 [Practice Book] [GATE EC 2004 IIT-Delhi : 2 Marks]

A system has poles at 0.01 Hz, 1 Hz and 80 Hz; zeros at 5 Hz, 100 Hz and 200 Hz. The approximate phase of the system response at 20 Hz is

(A) 090 (B) 00 (C) 090 (D) 0180 Ans. (A) Sol. The expression for phase will be

1 1 1 1 1 1tan tan tan tan tan tan5 100 200 0.01 1 80

f f f f f f

At f = 20 Hz,

1 1 1 1 1 120 20 20 20 20 20tan tan tan tan tan tan

5 100 200 0.01 1 80

098.2

So, 090

Hence, the correct option is (A). Question 14 [Practice Book] [GATE IN 2004 IIT-Delhi : 2 Marks]

Figure shows the polar plot of a system. The transfer function of the system is

(A) 5(1 + 0.1s) (B) (1 + 0.5s) (C) 5(1 + 10s) (D) 5(1 + s)

0� �

10� �

045Re

Img

5

  14 

Ans. (A) Sol. Figure shows the polar plot of a system.

   From figure we can conclude that | ( ω) | 5GH j at ω 0

0( ω) 45GH j at ω 10 0[ tan 45 1]

Only option (A) is satisfying both conditions.

2| ( ω) | 5 1 (0.1ω)GH j

1( ω) tan (0.1ω)GH j

At ω 0 | ( ω) | 5GH j

At 10, 0( ) 45GH j

Hence, the correct option is (A). Question 11 [Work Book - Polar] [GATE EC 2005 IIT-Bombay : 2 Marks]

The polar diagram of a conditionally stable system for open loop grain K = 1 is shown in the figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for

(A) 5K and 1 1

2 8K (B)

1

8K and

15

2K

(C) 1

8K and 5 K (D)

1

8K and 5 K

Ans. (B) Sol. Given : Open loop transfer function is stable, number of right sided poles, P = 0. Nyquist stability criterion is given by, N Z P Where N = number of encirclements of the (–1, j0) point made by the ( ) ( )G s H s plot in CW direction. Z = number of zeros of 1 ( ) ( )G s H s that are inside the Nyquist path (i.e. the right-half s-plane); notice that the zeros of 1 ( ) ( )G s H s are the same as poles of closed-loop transfer function.

0� �

10� �

045Re

Img

5

Img

Re– 8 –2 – 0.2

GH-plane

  15 

P = number of poles of 1 ( ) ( )G s H s that are inside the Nyquist path (i.e. the right-half

s-plane); notice that the poles of 1 ( ) ( )G s H s are the same as those of ( ) ( )G s H s .

There are four cases : Case 1 : If critical point – 1 + j0 lies between 0 and – 0.2K

0.2 1K 5K

Number of encirclement of – 1+ j0, N = 2 2N Z P 2Z [Unstable system with two right poles]

Case 2 : If critical point – 1 + j0 lies between – 2K and – 0.2K then encirclements of – 1 + j0 are in opposite direction and net encirclement is zero.

Case 3 : If point lies between – 2K and – 8K

8 1 2K K

So 1

8K and 1

2K

Img

ReIV III II I

– 0.2K

N = 2(2 CW)

CW

CW– 2K– 8K

0

Img

ReIV III II I

– 0.2K

N = 0(1 CW + 1 CCW)

– 2K– 8K

0

CW

CCW

CW

Img

ReIV III II I

– 0.2K

N = 2(2 CW)

CW

CW– 2K– 8K

0

  16 

Number of encirclement of – 1+ j0, N = 2

2N Z P 2Z [Unstable system with two right hand poles]

Case 4 : If point – 8K > – 1 or 8K < 1 then there is no encirclements of – 1 + j0.

So two conditions are

2K > 1, 0.2K < 1

1, 5

2K K

1 15 and 8 1 or

2 8K K K

Hence, the correct option is (B).

Question 4 [Work Book – Nyquist] [GATE EE 2006 IIT-Kharagpur : 2 Marks]

Consider the following Nyquist plots of loop transfer functions over 0 to . Which of these plots represents a stable closed loop system?

1. 2.

3. 4.

(A) 1 only (B) all, except 1 (C) all, except 3 (D) 1 and 2 only

Ans. (D)

Sol. If open loop system is stable then no poles should lie in the right half of s-plane i.e. P = 0.

For option (A), For option (B),

N = 0 [No encirclement] Type 3 system as starts from 0270 and ends at 0270

Stable closed loop system N = Z – P = 0 0Z

Stable closed loop system

Re

Img

� � �

1�

�

Re

Img

� � �1�

Re

Img

� � �1�

�

Re

Img

� � �1�

�

– 1Re

Img

–1

Img

Re

N = 0(1 CW + 1 CCW)

CWCW CCW

CCW

  17 

For option (C), For option (D),

N = Z – P = 2 2Z Type 1 system as starts from 090 and ends at 090

Unstable closed loops system N = Z – P = 2 2Z

Unstable closed loop system

Hence, the correct option is (D).

.Statement For Linked Answer Questions 21 & 22.

Consider a unity feedback system with open loop transfer function 2

(1 6 )( )

(1 )(1 2 )

sG s

s s s

Question 21 [Practice Book] [GATE IN 2008 IISc-Bangalore : 2 Marks]

The phase crossover frequency of the system in radians per second is

(A) 0.125 (B) 0.25 (C) 0.5 (D) 1

Ans. (C)

Sol. Given : 2

(1 6 )( )

(1 ) (1 2 )

sG s

s s s

Put ,s j 2

(1 6 ω)( ω)

( ω) (1 ω) (1 2 ω)

jG j

j j j

Phase angle can be written as,

0 1 1 1( ω) 180 tan (6ω) tan (ω) tan (2ω)G j

The frequency at which phase angle of ( ω) ( ω)G j H j is 0180 is called phase crossover frequency.

0 0 1 1 1180 180 tan 6ω tan ω tan 2ωpc pc pc

….. (i)

ω 0 rad/secpc

Polar Plot :

At 0 ( )G j ( )G j 0180

At ( ) 0G j ( )G j 0270

Re

Img

� � �1�

0� �N = 2

(2 CW)

CW

CW

–1

Img

Re

N = 2(2 CW)

CW

CW

0180�pc

�pc

�

Img

Re

CCW

CW

  18 

This system consist two ω pc .

For second ω pc :

From equation (i),

1 1 1tan 6 tan tan 2pc pc pc

1 12

2tan 6 tan

1 2pc pc

pcpc

Apply 1 1 1tan tan tan

1

A BA B

AB

2

ω 2ω6ω

1 2ωpc pc

pcpc

2

12

1 2ω pc

ω 0.5 rad/secpc

Hence, the correct option is (C).

Important concept of polar plot : 1. (a) Direction of polar plot is given by,

1 2

Where 1 2( 0) ( )GH j GH j

(b) If ve the direction clockwise

(c) If ve the direction Anti clockwise or counter clockwise

2. If transfer function consist finite left hand poles and zeros then starting direction of polar plot is given by either finite pole or finite zero which is nearer to origin and ending direction is given by

1 2

3. If finite pole exist nearer to origin than the starting direction will be clockwise. 4. If finite zero exist nearer to origin then the starting direction will be counter-clockwise.

. Common Data Questions 5 & 6 .

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

Question 5 [Work Book – Nyquist] [GATE EC 2009 IIT-Roorkee : 2 Marks]

Which of the following statements is true ? (A) G(s) is an all-pass filter. (B) G(s) has a zero in the right-half plane. (C) G(s) is the impedance of a passive network. (D) G(s) is marginally stable.

Ans. (B) Sol. In option (A) it is given that G(s) is all pass filter, and we know that all pass filter has constant (fixed)

magnitude for all frequency so its Nyquist plot should be a circle of constant radius with center at origin. So option (A) is not correct.

Re

Img

j�

– 1.0

– 0.5

( )G s

  19 

In option (D) it is given that G(s) is marginally stable but in the question G(s) is stable. Therefore, option (D) cannot be correct.

In option (B) it is given that G(s) has a zero in the right half of s-plane so assuming a ( )G s

1

1

( )s z

G s Ks p

where 1 10 & 0z p

Put ,s j 1

1

( )j z

G j Kj p

Magnitude can be written as,

2 2

1

2 21

( )z

G j Kp

Phase angle is given by,

0 1 1

1 1

( ) ( ) 180 tan tanG jz p

At 0 1

1

( )z

G j Kp

( )G j 0180

At ( )G j K ( )G j 00

This will satisfy the Nyquist plot given in the question.

Comparing the table and Nyquist plot we have

1

1

0.5z

Kp

….. (v)

In option (C) it is given that ( )G s is impedance of a passive network. Assuming a passive network series

combination of R & C, its impedance will be

1 1

( ) , 0, 0RCs

G s R R CCs Cs

a� a Re

ja

ja�

Img

( ) ( )G s H s Plane

Re

Img

1�

j�

j

0.5�

0�� �� � � �

� � � �0�� �

  20 

1

( )RC s

RCG s

Cs

1R s

RCs

Put ,s j

1

( )R j

RCG j

j

Phase angle is given by,

1 1( ) ( ) tan tan1 0

G j

RC

0 1( ) 90 tan ( )RC

Magnitude can be calculated as,

22 1

( )RC

G j R

 

At 0 ( )G j ( )G j 090

At ( )G j R ( )G j 00

Here at 0 magnitude is but for given Nyquist plot all the frequencies have finite magnitude so

option (C) is not satisfying.

Note : Impedance of passive network represents minimum phase system i.e. left hand poles and zeros of s-plane. In minimum phase system, there is no possibility of negative sign of ( )G j at 0 or .

Hence, the correct option is (B).

Question 24 [Practice Book] [GATE IN 2009 IIT-Roorkee : 2 Marks]

A unity feedback control loop with an open loop transfer function of the form ( )

K

s s a has a gain

crossover frequency of 1 rad/sec and a phase margin of 060 . If an element having a transfer function

( 3)

( 3 )

s

s

is inserted into the loop, the phase margin will become :

(A) 0 (B) 30 (C) 45 (D) 60 Ans. (*)

Sol. Given : P.M. = 600 , ( ) , 1rad / sec.( ) gc

KG s

s s a

Transfer function 3

( )3

sT s

s

inserted in loop.

Phase angle can be written as,

0 1( ) 90 tanG ja

……. (i)

  21 

Phase margin is given by,

0P.M. 180 ( )gcG j

0 060 180 ( )gcG j

0( ) 120gcG j

From equation (i), 0 1 090 tan 120gc

a

1 0tan 30gc

a

1

3gc

a

1 1

3a

3a

Gain crossover frequency is the frequency at which ( ) ( ) 1gc gcG j H j

( ) ( ) 1gc gcG j H j

2

13gc gc

K

2K [ 1]gc

Modified transfer function can be written as,

2 ( 3)

'( ) '( )( 3) ( 3)

sG s H s

s s s

The new transfer function ( )T s inserted has magnitude 1. So ( )gc new well remain same.

0 1 1 0 1'( ) '( ) 90 tan tan 180 tan3 3 3

G j H j

0 0 0 0 0 0'( ) '( ) 90 30 30 180 30 0gc gcG j H j

Modified phase margin is given by,

0P.M.' 180 '( ) '( )gc gcG j H j

0P.M.' 180 Correct option is 0180 which is not given in the option.

Question 20 [Work Book – Polar] [GATE EE 2010 IIT-Guwahati : 2 Marks]

The frequency response of 1

( )( 1)( 2)

G ss s s

plotted in the complex ( )G j plane (for 0) is

(A)

(B)

Img

Re–3 / 4

0� �

Img

Re–3 / 4

0� �

  22 

(C)

(D)

Ans. (A)

Sol. Given : 1

( )( 1)( 2)

G ss s s

Put ,s j 1

( )(1 ) (2 )

G jj j j

2 2

(1 ) (2 )

(1 ) (4 )

j j

j

2

2 2

(2 ) 3( )

(1 ) (4 )

jG j

j

2

2 2 2 2

3 2

(1 ) (4 ) ( 1) ( 4)j

x jy

Let Re[ ( )], Im [ ( )]x G j y G j

2

2 2 2 2

3 2,

(1 ) (4 ) ( 1)( 4)x y

Phase angle can be written as,

0 1 1( ) 90 tan tan2

G j

At 0 3

4x y ( )G j 090

At 0x 0y ( )G j 0270

For positive frequency i.e. 0 Nyquist plot is termed as polar plot.

Hence, the correct option is (A).

. Common Data Question 26 & 27 .

The input output transfer function of a plant 2

100( )

( 10)

H s

s s. The plant is placed in a unity negative

feedback configuration as shown in the figure below.

Img

Re

–1 / 6

0� �Img

Re

–1 / 60� �

y

x–3 / 4

0� �

�2

100( )

( 10)H s

s s�

�r

�

� yu

plant

  23 

Question 26 [Practice Book] [GATE EC 2011 IIT-Madras : 2 Marks]

The gain margin of the system under closed loop unity negative feedback is

(A) 0 dB (B) 20 dB (C) 26 dB (D) 46 dB

Ans. (C)

Sol. The plant with unity feedback configuration is shown below.

2

100( )

( 10)

H s

s s

2

100( )

( 10)

H j

j j

Magnitude can be written as,

2 2

100( )

( 100)H j

Phase angle can be calculated as,

     0 1 ω

( ω) 90 2tan10

H j

The frequency at which phase angle of ( ω) ( ω)G j H j is 0180 is called phase crossover frequency.

0 0 1 ω180 90 2 tan

10

pc  

      0 1 ω45 tan ω 10 rad/sec

10pc

pc

 

  Gain margin can be defined as reciprocal of the magnitude of the ( ) ( )G j H j measured at phase

crossover frequency.

ω

1G.M.=

( ω) ( ω)pc

G j H j 

In dB, 1

G.M.= 20log( ω) ( ω)G j H j

 

     100 1

( ω)10(100 100) 20pc

H j

 

      G.M .= 20 log 20 26 dB  

Hence, the correct option is (C).

�2

100( )

( 10)H s

s s�

�r

�

� yu

plant

  24 

Alternatively 1 : Characteristic equation is given by, 1 ( ) 0H s

2

1001 0

( 10)s s

100desiredK

For

marK replace gain with K in

( )H s

2

1 0( 10)

K

s s

3 220 100 0s s s K For third order system to be marginal stable, IP = OP

20 100 marK

2000marK

2000

G.M. 20100

mar

desired

K

K

In dB, G.M.= 20log 20 26 dB

Alternatively 2 :

For calculating pc equating imaginary part of

( ) ( )G j H j to zero, we get

3 2Img 100 20 0j j

3 100 0pc pc

10 rad/secpc  

 Note : This method is only valid when polar plot

crosses 0180 axis.

Question 23 [Work Book – Polar] [GATE IN 2011 IIT-Madras : 2 Marks]

The value of K for the damping ratio to be 0.5, corresponding to the dominant closed-loop complex

conjugate pole pair is (A) 250 (B) 125 (C) 75 (D) 50

Ans. (B) Sol. Closed loop transfer function is given by,

( )T s ( )

1 ( ) ( )

G s

G s H s

( )T s3

3

3

( 5)( 5)1 1

( 5)

KKs

K s Ks

Characteristic equation is given by,

3( 5) 0s K

3 215 75 125 0s s s K ….. (i) Characteristic for the dominant conjugate pole pair,

2 2( 2 )( ) 0n ns s s

0.5 (is given)

2 2( )( ) 0n ns s s

3 2 2 2( ) ( ) 0n n n ns s s ….. (ii)

Comparing equation (i) and (ii), we get

15 15n n

Re

Img

� � �

0� �

1

20

10pc

� �

0180�

  25 

2 75n n

2(15 ) 75n n n

2 215 75n n n

15 75n

5rad/secn

and 15 5 10

10

2125 nK 210 (5) 250

125K

Hence, the correct option is (B).

Alternatively : Root Locus Concept

Given : 3

( ) ( )( 5)

KG s H s

s

Number of poles P = 3 Number of zero Z = 0 Number of branches terminating at infinite = P – Z = 3 Asymptotes is given by,

0(2 1)180

AP Z

0,1, ..... 1P Z

0(2 1)180

3A

0, 1, 2

0 0 060 , 180 , 300A

Centroid is given by,

5 5 5 0

53

Intersection with imaginary axis can be calculated using Routh array. Characteristics equation is given by,

3

1 0( 5)

K

s

3 2125 15 75 0s s s K

3 215 75 125 0s s s K

Routh Tabulation :

Row of zeros can be formed by equating 1000

015

K

  26 

i.e. mar 1000K

Auxiliary equation can be formed as,

215 125 0s K

215 1125 0s

2 75s

5 3s j

5 3 rad / secn

Root locus can be drawn as shown below.

We know that, cos

1 0cos 0.5 60

In ,COB 5 3

tan 35

060

So 0 0 0 0180 (60 60 ) 60BAO

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

product of length from zero to point

BK

B

5 5 51

AB AB ABK

125K

Question 7 [Work Book – Nyquist] [GATE EC 2014 (Set-01) IIT-Kharagpur : 1 Mark]

Consider the feedback system shown in the figure. The Nyquist plot of G(s) is also shown. Which one of the following conclusions is correct?

�

j�

�

��

K � �

0K �

5� B

A

1000K �

1000K �5 3j�

5 3j

K � �

K � �

  27 

(A) G(s) is an all-pass filter.

(B) G(s) is a strictly proper transfer function.

(C) G(s) is a stable and minimum–phase transfer function.

(D) The closed-loop system is unstable for sufficiently large and positive K.

Ans. (D)

Sol. In option (A) it is given that G(s) is all pass filter, and we know that all pass filter has constant (fixed) magnitude for all frequency so its Nyquist plot should be a circle of constant radius with center at origin.

So option (A) is not correct.

In figure (b), '( ) ( )G s G s

Let 1 0.5

( )0.25

sG s

s

2

2 2

1 (0.5 )( )

(0.25)

G j

1 0.5

'( )0.25

sG s K

s

Phase angle is given by,

1 1( ) tan 0.5 tan0.25

G j

At 0 ( ) 4.0G j 0( ) 0G j

At ( ) 0.5G j ( )G j 0180

Here the transfer function which we have designed is stable and non-minimum phase system.

The figure drawn above is for value of K = 1 but suppose value K is change that is 0.5K > 1 then K > 2 system stable.

So, system here is conditional stable system.

For above case 0.5K critical point then lie inside the figure, system become unstable. So when sufficiently large value is provided then system becomes unstable.

Hence, the correct option is (D).

K G s( )–1 +1

Im ( )G j�

Re ( )G j�

�

j�

– 0.25 2 – 0.5

4Re

Img

0

0.5

  28 

Question 9 [Work Book – Nyquist] [GATE EC 2016 (Set-02) IISc-Bangalore : 1 Mark]

The number and direction of encirelements around the point 1 0j in the complex plane by the Nyquist

plot of 1( )

4 2

sG s

s

is

(A) zero (B) one, anti-clockwise (C) one, clockwise, (D) two, clockwise

Ans. (A)

Sol. Method 1 :

Given : 1

( )4 2

sG s

s

And 1 1( ) tan tan2

G s

At 0,s 0

0

1( ) 0.25 and ( ) 0

4G s G s

At s, 0( ) 0.5 and ( ) 180G s G s

Hence Nyquist plot is

Hence number of encirclement of ( 1 0)j is 0.

Method 2 :

We know N P Z

Where P = Number of open loop poles lie in right half of s-plane

1( )

4 2

sG s

s

Hence 0P

Z = Number of closed loop poles lie in right half of s-plane

To find Z-apply R-H criteria

Characteristic equation is 1 ( ) ( ) 0G s H s

(1 )

1 0(4 2 )

s

s

4 2 1 0s s

5 0s 5s

Hence pole lie on left half of s-plane and Z = 0.

0 0N P Z

0N

And number of encirclement to (–1 + j0) is zero.

Im

Re0.25– 5– 1

  29 

IES Objective Solutions

Question 3 [Practice Book] [IES EC 1993]

The gain-phase plots of open-loop transfer function of four different systems are shown below. The correct sequence of the increasing order of stability of these four systems will be

(A) D, C, B, A (B) A, B, C, D (C) B, C, A, D (D) A, D, B, C Ans. (B)

Sol. Gain margin (G.M.) is defined as,

where is frequency at which

Phase margin (P.M.) is defined as,

where is frequency at which

Now, we define the stability of a system with respect to gain margin and phase margin as 1. For G.M. in dB and P.M. in degree both positive, system will stable. 2. For G.M. in dB and P.M. in degree both negative, system will unstable. 3. For G.M. = 0 dB and P.M. system will marginally stable. Thus, the system will be more stable for greater G.M. and P.M. Using the analysis, we compare the

stability of given system plots as Plot A :

System is unstable. Plot B :

System is marginally stable.

Plot C :

0 0 0P.M. 180 ( 112.5 ) 67.5

System is stable.

Plot D :

A

BC

D

–2700

–2700

–2250

–2250

–1350

–1350

–900

–900

– 450

– 450

– 20 dB– 20 dB

– 40 dB– 40 dB

10 dB10 dB

30 dB30 dB

� increasing

50 dB50 dB

1G.M. 20log

( )pcG j

pc 0( ) 180pcG j 0P.M. 180 ( )gcG j

gc ( ) 1 0 dBgcG j

00

G.M. 40 dB 40 dB 0 0 0P.M. 180 ( 270 ) 90

G.M. 0 dB0 0 0P.M. 180 ( 180 ) 0

G.M. 30 dB 30 dB

G.M. 50 dB 50 dB 0 0P.M. 180 ( 67.5 ) 112.5

A

BC

D

–2700

–2700

–2250

–2250

–1350

–1350

–900

–900

– 450

– 450

– 20 dB– 20 dB

– 40 dB– 40 dB

10 dB10 dB

30 dB30 dB

� increasing

50 dB50 dB

–1800

–1800

0 dB0 dB

– 30 dB– 30 dB

– 10 dB– 10 dB

20 dB20 dB

40 dB40 dB

– 50 dB– 50 dB

–112.50

–112.50

–67.50

–67.50

  30 

Plot D is more stable than plot C. Hence, stability sequence will be A < B < C < D Hence, the correct option is (B). Question 16 [Practice Book] [IES EE 2001]

The Nyquist plot of a servo system is shown in the below figure. The root loci for the system would be

(A) (B)

(C) (D) None of these

Ans. (B) Sol. From the Nyquist plot open loop transfer function is given by,

Root locus for the above open-loop transfer function can be drawn as shown in figure.

Hence, the correct option is (B).

Img

Re

Increasing�

j�

�

j�

�

j�

�Double poleat origin

1( ) ( )

( 1)G s H s

s sT

0(2 1)1800,1, ..... 1A P Z

P Z

0(2 1)180

0,12

A

0 090 and 270A j�

�1

T

� �

  31 

Question 45 [Practice Book] [IES EC 2007]

Which one of the following statement is correct for gain margin and phase margin of two closed–loop system having loop function G(s) H(s) and exp (–s) G(s) H(s) ? (A) Both gain and phase margins of the two systems will be identical. (B) Both gain and phase margins of G(s) H(s) will be more. (C) Gain margins of the two systems are the same but phase margin of G(s) H(s) will be more. (D) Phase margins of the two systems are the same but gain margin of G(s) H(s) will be less.

Ans. (*) Sol. Relative stability parameters without delay can be written as, Assume proper transfer function. 1. Phase crossover frequency,

2. Gain crossover frequency,

3. Gain margin,

In dB,

4. Phase margin,

Relative stability parameters with delay can be written as, 1. Phase crossover frequency,

2. Gain crossover frequency,

3. Gain margin,

In dB,

Gain margin expressed in dB, 4. Phase angle with delay becomes more negative than phase angle without delay. Phase margin,

Fig. Effect of delay upon frequency response

Hence, all options are wrong.

0( ) 180pcG j

( ) 1gcG j

1G.M.

( )pcG j

1G.M. 20 log

( )pcG j

0P.M. 180 ( )gcG j

'pc pc

'gc gc

1G.M.'

( ' )pcG j

1G.M.' 20 log

( ' )pcG j

G.M.' G.M.

0P.M.' 180 '( ' )gcG j P.M.' P.M.

M (dB)

0 dB

Gain plot

(rad/sec)�

Phase (deg)

1800 (rad/sec)�

'gc gc

� � � pc�

Phase plot without delay

Phase plot with delay

P.M.'

P.M.

G.M.'G.M.

'pc

�

  32 

Question 46 [Practice Book] [IES EC 2007]

Consider the following statement with reference of relative stability of a system. 1. Phase margin is related to effective damping of the system.

2. Gain margin gives better estimate of damping ratio than phase margin.

3. When expressed in dB, gain margin is negative for a stable system.

Which of the statements given above are correct ?

(A) 1 and 2 only (B) 1 and 3 only (C) 2 and 3 only (D) 1, 2 and 3

Ans. (*)

Sol. The P.M. is defined in terms of damping ratio as,

P.M.

Thus, the phase margin is related to effective damping of the system.

Gain margin must be positive when expressed in dB for a stable system. A negative G.M. means that the system is unstable. Phase margin gives a better estimate of damping ratio and therefore of the transient overshoot of the system than does the gain margin.

None of the option is matched.

1

2 4

2ξtan

2ξ 4ξ 1