Principles of Control Systems Solution Manual

Solutions of Selected Unsolved Examples

3.3 Transfer Function

Q.8

Solution : The s-domain network is shown in the Fig. 3.1.

Applying KVL to the two loops,

� � � �R I (s) I (s) sL I (s) sL V(s)1 1 1 2 = 0

i.e. I (s) [R sL] I (s) sL1 1 2� � = V(s) ... (1)

� � � �I (s) R1sC

I (s) sL I (s) sL I (s)2 2 2 2 1 = 0

� I (s)1 =

I (s) R1sC

sL

sLI (s)

sR C 1 s LC

s LC

2 2

22

2

2

� ��

� �

� ��

��

�

... (2)

Using (2) in (1),

I (s)s LC sR C 1

s LC[R sL] I (s) sL2

22

2 1 2

� ��

��

�

� � = V(s)

� I (s)(s LC sR C 1) (R sL)

s LCsL2

22 1

2

� � ��

� �

��

��

��= V(s)

�I (s)

V(s)2 =

s LC

s LCR s R R C R s L C s L R C sL s L C

2

21 1 2 1

3 2 22

3 2� � � � � �

�I (s)

V(s)2 =

s LC

s LC (R R ) s (R R C L) R

2

21 2 1 2 1� � � �

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM(3 - 1)

3 Transfer Function and Impulse Response

�1(s)

R2R1

sL

+ –

+_

+ –+

–

–

+�2(s)

+

–

1sC

V(s)

Fig. 3.1

3.6 Laplace Transform of Electrical Network

Q.2

Solution : The s-domian network is shown in the Fig. 3.2 (a).

� T.F. =E (s)E (s)

ZZ Z

R1 sR C

R1 sR C

Ro

1

2

1 2

2

2 2

1

1 1

��

��

�� 2

2 21 sR C�

� =R (1 s R C )

(R R )+ sR R (C C )2 1 1

1 2 1 2 1 2

�

� �

��

Principles of Control Systems 3 - 2 Transfer Function and Impulse Response

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM

E (s)i

R1

R2

1–––sC1

1–––sC2

E (s)o

R || 1–––sC1

1

=R

–––––––1+sR C

1

1 1

R || 1–––sC2

2

=R

–––––––1+sR C

2

2 2

E (s)i

Z (s)1

Z (s)2 E (s)o

(a) (b)

Fig. 3.2


5.3 Rules for Block Diagram Reduction

Q.11

Solution : Shifting take off point after G 2 and seperating feedback paths we get,


5Block Diagram Representation of

Control Systems

G1 G2

H2

1G2

G3

H3

H1

– – –

R(s)

Parallel (1 + = )

C(s)

Minor loop

1 + G2

G2

1G2

Minor loop

+

G1

H1

–

R(s)C(s)

Minor loop

G2

21 + G H2

1 + G2

G2

G3

1 + G H3 3

R(s)C(s)

1 + G2

G2

G3

1 + G H3 3

G 21

2 2

G

1 + G H

G 21 1

2 2

G H

1 + G H1 +

G1 G

1 + G H + G G H2

2 2 1 2 1

G1 G (1 + G )

1 + G H + G H + G G H H + G G H + G G G H H

3 2

3 3 2 2 2 3 2 3 1 2 1 1 2 3 1 3

R(s)C(s)

Series

Q.12

Solution : No series, parallel combination and no minor feedback loop exists. So shifting

take off point before the block ofs

s + 10�

��

�

��.

Principles of Control Systems 5 - 2 Block Diagram Representation of Control Systems


–

–

+

+ C(s)R(s)

Parallel1

s

s

s + 10

s

s + 10

s

s – 3

s + 8

s – 4

3

–

– C(s)R(s)

Minorloop

1

s + 10

s + 8

s – 4

3

s

s + 10

s

s – 3+

–

– C(s)R(s)

s + 8

s – 4

3

s

s + 10

s

s – 3+

s + 10

s + 11

s (2s + 7)

(s + 10) (s + 3)

1

1

s + 101 +

After simplification After simplification

–

– C(s)R(s)

s + 8

s – 4

3

s + 10

s + 11

s (2s + 7)

(s + 10) (s + 3)�

�C(s)

R(s)=

s(2s + 7)

(s + 11) (s 3)

1 +s(2s + 7)

(s + 11) (s 3)

(s + 8)

(s 4)

3

=s (2s + 7) (s 4)

(s + 11) (s 3) (s 4) + s(2s + 7) (s + 8)3

�C(s)

R(s)=

s(2s s 28)

2s + 7s + s + 20s 9s + 132

2

5 4 3 2

5.4 Analysis of Multiple Input Multiple Output Systems

Q.4

Solution : As there are two inputs, consider each input separately. Consider R(s),assuming Y(s) = 0.

C(s)

R(s)=

G G

1 G G H1 2

1 2 1�i.e. C(s) =

R(s) G G

1 G G H1 2

1 2 1�

Now consider Y(s) acting with R(s) = 0.

Now sign of signal obtained from H1 is negative which must be carried forward, though

summing point at R(s) is removed, as R(s) = 0, so we get,

Key Point While finding equivalent G, trace forward path from input summing point to

output in direction of signal. While finding equivalent H, trace the feedback path from output

to input summing point in the direction of signal.

Now equivalent G = G 2 , tracing forward path from input summing point to output.

Equivalent H = G H1 1 tracing feedback path from output to input summing point.

While sign of the final feedback is positive at the input summing point.



–

+C(s)

�

R(s)G1 G2

H1

–

+ C(s)R(s)G G1 2

H1

Minor feedbackloop

�

++ C(s)

–1 G1 G2

H1

Y(s)To carry forwardnegative sign

++ C(s)

G2

–G H1 1

Y(s)

�C(s)

Y(s)=

G1 GH

=G

1 G ( G H )2

2 1 1 … H itself is negative

�C(s)

Y(s)=

G

1 G G H2

1 2 1�i.e. C(s) =

Y(s) G

1 G G H2

1 2 1�

Hence the net output C(s) is given by algebraically adding its two components,

C(s) =G G R(s) + G Y(s)

1 G G H1 2 2

1 2 1�

Q.5

Solution : For C/R1, consider R2 = 0, hence summing point at R2 can be removed.

�C

R1=

G1 GH1�

where G =(G G )G G

1 G H1 2 3 4

3 2

�

�

�C

R1=

(G G )G G

1+ G H

1(G G )G G H

1 G H

1 2 3 4

3 2

1 2 3 4 1

3 2

�

��

�

=(G G )G G

1 G H (G G )G G H1 2 3 4

3 2 1 2 3 4 1

�

� � �



–

G2

G1

+

G3 G4

H2

++

–

H1

C

ParallelG + G1 2

Minorloop

G

1+ G H3

3 2

R1

Fig. 5.1 (a)

G + G1 2

G

1+ G H3

3 2

G4

H2 H1

+

–

Series G

R1

C+R1

G C

–

Fig. 5.1 (b)

For C/R2 , consider R1 = 0 but negative sign of H1 must be considered while removing

summing point at R1. The parallel combination of G , G1 2 and minor loop of G 3 and H2

can be used directly.

G eq = G 4

Heq = �

�

(G G )G H

(1 G H )1 2 3 1

3 2

�C

R2=

G

1 G Heq

eq eq=

G

1 G(G G )G H

(1 G H )

4

41 2 3 1

3 2 �

�

�

��

��

�C

R2=

G (1 G H )

1 G H (G G )G G H4 3 2

3 2 1 2 3 4 1

�

� � �

Q.6

Solution : i) With N(s) = 0 block diagram becomes

Minor feedback loop =

10s (s + 1)

1 +10

s (s + 1)0.5s

=10

s + s + 5s2=

10

s + 6s2

Assume output of second summing points as X(s),

Hence E(s) = R(s) – C(s) ... (i)

C(s) = X(s)10 (s + 4)

s + 6s2... (ii)



G + G1 2

G

1+ G H3

3 2

G4

H1

+C– 1

+

R2G (Forward path)eq

H (Feedback path)eq

Fig. 5.1 (c)

Geq

+C

+

R2

Heq

Fig. 5.1 (d)

– –

+C(s)R(s) E(s)

3

s + 410

s(s + 1)

0.5 s

Minor loop

X(s) = E(s) +3

s + 4R(s) ... (iii)

Substituting value of X(s) and R(s) from (i) & (ii) in (iii) we get,

s + 6s

10 (s + 4)

2

C(s) = E(s) +3

s + 4E(s) +

3s + 4

C(s)

s + 6s

10(s + 4)3

(s + 4)

2

��

�

��

C(s) = 1 +3

s + 4�

��

�

�� E(s)

(s + 6s 30)

10 (s + 4)

2 C(s) =

(s + 7)

(s + 4)E(s)

� C(s)

E(s)=

10 (s + 7)

s + 6s 302 when N(s) = 0

ii) To findC(s)

R(s), we have to reduce block diagram solving minor feedback loop and

shifting summing point to the left as shown earlier in (i).

So referring to block diagram after these two steps i.e.



–

+C(s)R(s) E(s)

3

s + 4

10

s + 6s2X(s)

s + 4

–

+C(s)R(s) E(s)

3

s + 410

s + 6s2

–

+C(s)R(s) E(s)

3

s + 4

10

s + 6s2

s + 4

Exchanging two summing points using associative law,

� Block diagram becomes,

�C(s)

R(s)=

s + 7

s + 4

10 (s + 4)

s + 16s + 402

�

��

�

��

�

��

�

�� =

10(s + 7)

s + 16s + 402

iii) With R(s) = 0 block diagram becomes,

The block of ‘3’ will not exist as R(s) = 0. Similarly first summing point will also vanish

but student should note that negative sign of feedback must be considered as it is, though

summing point gets deleted.

In general while deleting

summing point, it is

necessary to consider the

signs of the different signals

at that summing points and

should not be disturbed. So

introducing block of ‘–1’ to

consider negative sign.



–

+C(s)R(s)

3

s + 4

10 (

s + 6s2

s + 4)

Parallel of '1' and

= 1 +

3––––s + 4

3––––s + 4

Minor loopG

––––––1 + GH

=

10 (s + 4)––––––––

s + 6s2

10 (s + 4)––––––––

s + 6s21 +

C(s)R(s) 3

s + 4

10 (

s + 16s + 402

s + 4)1 +

– –

++

+ C(s)

N(s)

s + 410

s(s + 1)

0.5 s

–

++

+ C(s)

N(s)

s + 4– 110

s(s + 1)

0.5 s

Two blocks are in parallel, adding them with signs.

Removing summing point, as sign is positive no need of adding a block.

�C(s)

N(s)=

1

110 (1.5s + 4)

s(s + 1)

��

��

=1

1 +15s + 40

s(s + 1)

�C(s)

N(s)=

s(s + 1)

s + 16s + 402

5.5 Block Diagram from System Equations

Q.2

Solution : The network can be redrawn in s-domain as,



–+

+

+ C(s)

N(s)

10

s(s + 1)

– (s + 4)

0.5 s

Parallel

+

+

+ C(s)

N(s)

10

s(s + 1)

– (s + 4) – 0.5 s

After simplification ( – 1.5 s – 4)

+

+ C(s)

N(s)

10

s(s + 1)

– (1.5 s + 4)

+

+

C(s)N(s)

– 10

s(s + 1)

(1.5 s + 4)

Minor loop with G = 1

I (s)1 =V (s) V (s)

Z1

ZV (s)

1Z

V (s)1 a

1 11

1a

� ... (1)

V (s)a = � �I I Z Z I (s) Z I (s)1 2 2 2 1 2 2 � ... (2)

I (s)2 =V (s) V (s)

Z1

ZV (s)

1Z

V (s)a 2

3 3a

32

� ... (3)

V (s)2 = Z I (s)4 2 ... (4)

Simulating each equation, the complete block diagram is,



V (s)1 V (s)2

Z (s)1 Z (s)3

Z (s)2 Z (s)4

A

V (s)a

I1(s) I2(s)

Z = R || ––– = –––––––––

Z = R + sL

Z = R || ––– = –––––––––

Z = R

1 1

2 2 2

3 3

4 4

1sC1

R

1 + sC R1

1 1

1sC3

R

1 + sR C3

3 3

1__Z1

Z2

+I1(s)

+ 1__Z3

Z4

1__Z1

–

Z2

1__Z3

V (s)2

V (s)a

– –

Shift

I2(s)

Shift

V (s)1

1__Z1

Z2

+ + 1__Z3

Z4

1__Z1

–

Z2

1__Z3

V (s)2– –

V (s)1

+

1__Z2

1__Z4

Minor feedback loopInterchange

�V (s)

V (s)2

1=

� � � �

� � � �

1Z

Z Z

Z Z

Z

Z Z

11

Z

Z Z

Z Z

Z

Z Z1

1 2

1 2

4

3 4

4

1 2

1 2

4

3 4

��

��

�

�

��

�

��

�

�

��

�

��

=� � � �

Z Z

Z Z Z Z Z Z2 4

1 2 3 4 1 2� � �

Substituting the values of Z1, Z2 , Z 3 , Z4

� V (s)

V (s)2

1=

� �R s L R

R

1 sC RR sL

R

1 sR CR

2 2 4

1

1 12 2

3

3 34

�

��

�

��

�

��

��

��

�

�

� ��

�

�

�

��

�

��

R R sL

1 sR C1 2 2

1 1

... Ans.

��



1__Z1

Z21__Z3

–

V (s)2–

V (s)1

1__Z1

1__Z4

Z Z______Z + Z

3 4

3 4

Z4

1 + –––Z

Z4

3

Minor feedback loop

1__Z1

–

V (s)2V (s)1

1__Z4

Z Z_________Z (Z + Z )

3 4

3 3 4

Z Z_______Z + Z

1 2

1 2

Minor feedback loop

Z2

1 + –––Z

Z2

1


6.5 Mason's Gain Formula

Q.11

Solution : Number of forward paths = K = 2

� T.F. =

TK KK 1

�

��

�2

using Mason's gain Formula

T1 = G G G G1 2 3 4

T2 = G G1 5

Individual feedback loops,

Loops L and L1 3 are non touching loops.� � = � � � �L + L + L + L + L L1 2 3 4 1 3

= 1 – [–G G H – G G H – G H G H H ] [G G H H ]1 2 3 2 3 2 4 1 5 1 2 1 2 4 1 3� � G

Consider T1 , all loops are touching � � �1 1

Consider T2 , all loops are touching � � �2 1


6Signal Flow Graph Representation of

Control Systems

–H2 –H1

G5

–H1

G4

G1 G2

–H3

G2

–H2

G3

L = –G1 1G H2 3 L = –2 G G H2 3 2

L = –3 G H4 1 L = +4 G H H5 1 2

� T.F. =T + T1 1 2 2� �

�=

G G G G 1 + G G 11 + G G H + G G H + G H G H H + G G G

1 2 3 4 1 5

1 2 3 2 3 2 4 1 5 1 2 1 2

4 1 3H H

C(s)

R(s)=

G G G G + G G

1 + G G H + G G H + G H G H H + G G G H H1 2 3 4 1 5

1 2 3 2 3 2 4 1 5 1 2 1 2 4 1 3

Q.12

Solution : Number of forward paths K = 2

Mason's gain formula,

T.F. =

TK KK 1

�

��

�2

=T T1 � � �

�1 2 2

T1 = G G G G1 2 3 4 , T2 = G G5 4

Individual feedback loops are,

L and L1 3 are two non touching loops.

� � = 1 [L + L + L ]+ [L L ]1 2 3 1 3

= 1 [ G H G G G G H G G H ]+ [G H G G H ]2 1 1 2 3 4 2 5 4 2 2 1 5 4 2

= 1 + G H + G G G G H + G G H + G G G H H2 1 1 2 3 4 2 5 4 2 2 5 4 1 2

For T1 all loops are touching

Principles of Control Systems 6 - 2 Signal Flow Graph Representation of Control Systems


G1 G2

G2

–H2

–H2

–H1

G3

G5

G4

G4

L = –G H1 2 1

L =2 � G G G G H1 2 3 4 2

L =3 � G G H5 4 2

G2

–H1

G5

G4L is non touching to T .1 2

� � 1 = 1 eliminating all loop gains and products from �

Consider T2 ,

� � 2 = 1 – [L ] 1 – (–G H ) 1 G H1 2 1 2 1� � �

�C(s)

R(s)=

T T1 2� � �

�1 2 =

G G G G 1 + G G (1 + G H )1 2 3 4 5 4 2 1

�

�C(s)

R(s)=

G G G G + G G (1 + G H )1 + G H + G G G G H + G G H + G G

1 2 3 4 4 5 2 1

2 1 1 2 3 4 2 5 4 2 2 5 G H H4 1 2

Q.13


� T.F. =T1 �

�1 .... Mason's gain formula

T1 = G G G G G1 2 3 4 5

Individual feedback loops,

Combinations of two non touching loops,

i) L and L1 2 ii) L and L1 5 iii) L and L1 4

iv) L and L2 5 v) L and L3 5

Combination of three non touching loops, is L , L and L1 2 5 .� � = 1 – [L L L L L ] [L L + L L L L L L L L ]– [L L1 2 3 4 5 1 2 1 5 1 4 2 5 3 5 1� � � � � � � � 2 5L ]

� � = 1 + G H + G H + G G G H + G H + G H1 1 3 3 1 2 3 2 4 4 5 5

� � � �G G H H G G H H G G H H G G1 3 1 3 1 4 1 4 1 5 1 5 3 5 H H3 5

� �G G G G H H G G G H H H1 2 3 5 2 5 1 3 5 1 3 5

Now considering T G G G G G1 1 2 3 4 5�

All loops are touching to this forward path hence,� 1 = 1

�C(s)

R(s)=

T 1 1�

�=

G G G G G 11 2 3 4 5

�



G1 G2 G3

–H2

–H4

G4

–H5

G5G1

–H1

G3

–H3

L = –G1 1H1 L = –G2 3H3 L = –G4 4H4 L = –G5 5H5L = –G3 1G G H2 3 2

�

C(s)

R(s)=

G G G G G

1 + G H + G H + G G G H + G H + G H

+ G G H H +

1 2 3 4 5

1 1 3 3 1 2 3 2 4 4 5 5

1 3 1 3 G G H H + G G H H

+ G G H H + G G G G H H

+ G G G H H

1 4 1 4 1 5 1 5

3 5 3 5 1 2 3 5 2 5

1 3 5 1 3 H5

Q.14

Solution : Number of forward paths K = 3

T1= G G G G2 4 5 6 , T2 = G G G G1 4 5 6 , T3 = G G G G3 4 5 6


L1 = G G H2 4 1 L2 = G G H1 4 1 L 3 = G G H3 4 1

� � = 1 [ ]L + L + L1 2 3

No combination of non touching loops

� = 1 � G G H + G G H + G G H2 4 1 1 4 1 3 4 1

Consider T1 = G G G G2 4 5 6 All loops are touching, � � �1 1

T2 = G G G G1 4 5 6 All loops are touching, � � �2 1

T3 = G G G G3 4 5 6 All loops are touching, � � �3 1

� T.F. =T + T + T1 1 2 2 3 3� � �

�

=G G G G 1 + G G G G 1 + G G G G 12 4 5 6 1 4 5 6 3 4 5 6

�

�C(s)

R(s)=

G G G G + G G G G + G G G G

1 + G G H + G G H + G G H2 4 5 6 1 4 5 6 3 4 5 6

2 4 1 1 4 1 3 4 1

Q.15


T1 = G G G G1 2 3 4 , T2 = G G G1 2 6 , T3 = G G G G1 2 3 5

Individual feedback loops



G1

G2

–H1 –H1 –H1

G4 G4

G4

G3

G1 G2 G3G3

–H1

–H3

G4

L1 = – G H3 3 L2 = – G G G G H1 2 3 4 1

No combination of nontouching loops

� � = � �1 L + L1 2 = 1 � G H + G G G G H3 3 1 2 3 4 1

Consider T1 = G G G G1 2 3 4 All loops are touching � 1 = 1

T2 = G G G1 2 6 All loops are touching � �2 1

T3 = G G G G1 2 3 5 All loops are touching � �3 1

� T.F. =T + T + T1 1 2 2 3 3� � �

�

=G G G G 1 + G G G 1 + G G G G 11 2 3 4 1 2 6 1 2 3 5

�

�C(s)

R(s)=

G G G G G G G G G G G

1 G H G1 2 3 4 1 2 6 1 2 3 5

3 3 1

� �

� � G G G H2 3 4 1

Q.16


� T1 = G G G G1 2 3 4

Individual feedback loops

L G H1 1 1� L G H2 3 4� L G H3 4 3�

L G G G H2 3 4 24 � L G5 5�

Combinations of two nontouching loops

i) L and L1 2 ii) L and L1 3 iii) L and L1 5 iv) L and L2 5

Combination of three nontouching loops.

i) L , L and L1 2 5

� � = � � � � � �1 L + L + L + L + L + L L + L L + L L + L L L L L1 2 3 4 5 1 2 1 3 1 5 2 5 1 2 5

� = 1 + G H + G H + G H + G G G H G + G G H H1 1 3 4 4 3 2 3 4 2 5 1 3 1 4

G G H H G H G G H G G G G H1 4 1 3 1 1 5 3 4 5 1 3 5 1� H4



–H3

G4G1

–H1

G3

–H4

G5

SELF LOOPG2 G3

–H2

G4

All loops are touching to forward path T1 hence � 1 = 1.

�C(s)

R(s)=

T1 �

�1 =

G G G G 11 2 3 4

�

�

C(s)

R(s)=

G G G G

1 G H G H G H G G G H1 2 3 4

1 1 3 4 4 3 2 3 4 2� � � � G G G H H

G G H H G H G G5 1 3 1 4

1 4 1 3 1 1 5 3

� �

� G H G G G H H5 4 1 3 5 1 4

Q.17

Solution :

T1 = G G1 2

Individual loops are,

L1 = – G H1 1 L2 = – G G1 2 L 3 = – G H2 2

L and L1 3 is combination of two non-touching loops

� � = � � � �1 L + L + L + L L1 2 3 1 3

= 1 + G H + G G + G H + G G H H1 1 1 2 2 2 1 2 1 2

�C(s)

R(s)=

T1 1�

�

Considering T1, all loops are touching � � 1 = 1



–

––

G2

C(s)R(s)G1

s1 s2

t2t1

s3

H1H2

–H1

–H2

G1s1 t2t1s2 s3R(s) C(s)11 11 G2

–1

–H1

G11

–H2

G2G1 1 G2

–1

�C(s)

R(s)=

G G

1 + G H + G G + G H + G G H H1 2

1 1 1 2 2 2 1 2 1 2

Q.18

Solution : The signal flow graph is show below representing each summing and take off

point by separate node.

Forward paths : T1 = G G1 2

Individual feedback loop gains are,

Nontouching loops : L L G G H H1 2 1 2 1 2�

� � = 1 [L L L ] [L L ]1 2 3 1 2 � � �

All loops are touching to forward path T1 hence � 1 1� .

�C(s)

R(s)=

T1 1�

�=

G G

1 G H G H G G H G G H H1 2

1 1 2 2 1 2 3 1 2 1 2� � � �

By block diagram reduction, eliminate the two minor loops,

�C(s)

R(s)=

G G

(1 G H ) (1 G H )

1G G

(1 G H ) (1 G H )H

1 2

1 1 1 2

1 2

1 1 2 23

� �

��

�

�G G

(1 G H ) (1 G H ) G G H2 2

2 1 2 2 1 2 3� � �

�C(s)

R(s)=

G G

1 G H G H G G H H G G G1 2

1 1 2 2 1 2 1 2 1 2 3� � � �



G2

–H3

G11

–H1 –H2

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

–H1

G1

L = –G H1 1 1

1

–H3

G2G11

L = –G3 1G H2 3

–H2

G2

L = –G H2 2 2

1

G

H1

11 + G1

G

H2

21 + G2

H3

R(s)C(s)

+

–

G G

( H )(1 + G H )1 2

1 2 21 + G1

H3

R(s)C(s)

+

–

6.7 Application of Mason's Gain Formula to Electrical Network

Q.1

Solution : Convert the given network in its laplace form and assume different loop

currents and node voltages as shown.

Writing down equations for I , V , I , V1 1 2 o we get,

I 1 =(V V )

1sC

i 1

1

= sC (V V )1 i o ... (I)

V1 = (I I ) R1 2 1 ... (II)

I 2 =(V V )

1sC

1 o

2

= sC (V V )2 1 o ... (III)

Vo = I R2 2 ... (IV)

Simulating above equations by signal flow graph

Combining we get signal flow graph for given network.



Vo(s)R1

V1

1sC1

1sC2

R2Vi(s)I1 I2

VoI2 R2

V1 VoI2sC2

sC2

V1 I2I1 R1

–R1

V1I1Vi

sC1

–sC1

For equation (I)

For equation (II)

For equation (III)

For equation (IV)

V1 Vo

–sC1 –sC2–R1

R2R1 I2I1 V1 sC2sC1

To find T.F. apply Mason's gain formula

T.F. =� �

�

TK 1 Number of forward path = K = 1

� T.F. =T1 1�

�

T1 = sC R sC R1 1 2 2 = s R R C C21 2 1 2

Individual loops :

L1 = – R C s1 1 , L2 = – sR C1 2 L 3 = – R sC2 2

Out of three, L and L1 3 are nontouching

� = 1 – � � � �L + L + L + L L1 2 3 1 3

= 1 – � � � � sR C sR C sR C + s R C R C1 1 1 2 2 22

1 1 2 2

= 1 + s � � � �R C + R C + R C + s R C R C1 1 1 2 2 22

1 1 2 2

�V (s)

V (s)o

i=

T1 1�

�

All loops are touching to T1, � � 1 = 1

�V (s)

V (s)o

i=

� � � �s R C R C

1 + s R C + R C + R C + s R C R C

21 1 2 2

1 1 1 2 2 22

1 1 2 2

Q.2

Solution : Laplace Transform of the given network is,

Equations for different currents and voltages are



Vi(s) Vo(s)

R1 V1R3

R2 R4I1(s) I2(s)

S.F.G. (I) I1 = �V V1Ri 1

1 � ... (I)

S.F.G. (II)V1 = (I I ) R1 2 2 ... (II)

S.F.G. (III) I 2 = �V V1

R1 o3

� ... (III)

S.F.G. (IV) Vo = I R2 4 ... (IV)

Total S.F.G. is as shown in the following figure.

Use Mason's gain formula to findV

Vo

i

V

Vo

i=

� �

�

TK K

Number of forward paths = K = 1 =T1 1�

�


� T1 =R R

R R2 4

1 3

L1 = R

R2

1, L2 = –

R

R2

3, L 3 = –

R

R4

3

L and L1 3 are nontouching

� = 1 – � � � �L + L + L + L L1 2 3 1 3

= 1 +R

R+

R

R+

R

R+

R R

R R2

1

2

3

4

3

2 4

1 3,

� � 1 = 1 … all loops are touching to T1

� =R R + R R + R R + R R + R R

R R1 3 2 3 1 2 1 4 2 4

1 3



V1 I2R2

–R2

I1

V1 VoI2

1R3

–1R3

VoI2 R4

V1Vi

1R1

–1R1

I1

1R3

1R1

Vi Vo

I2R2

–R2

R4I1 V1

–1R1

–1R3

V

Vo

i=

T1 1�

�

�V

Vo

i=

R R

R R + R R + R R + R R + R R2 4

2 4 1 4 1 2 2 3 1 3

Q.3

Solution : The Laplace domain representation of the given network is shown below.

The various branch currents are shown,

� I (s)1 =V (s) V (s)

1sC

i 1

��

��

= sC V (s) sC V (s)i 1 ... (1)

Then, I (s)2 =V (s) – V (s)

Ri o

=1R

V (s) –1R

V (s)i o ... (2)

V (s)1 = [I (s) I (s)]R1 2�

= R I (s) R I (s)1 2� ... (3)

and also, I (s)2 =V (s) V (s)

1sC

o 1

= sC V (s) sC V (s)o 1

From this obtain the equation for V (s)o as I (s)2 equation is already obtained.

Note : Write the separate equation for separate branch and each element must beconsidered atleast once.

� V (s)o = V (s)1sC

I (s)i 2� ... (4)



1sC

1sC

R

R

V (s)i V (s)o

V (s)oV (s)i

I2(s)

I1(s)

V (s)1

I2(s)

I2(s)

( + )I1 2I

( + )1 2I I

( + )1 2I I

Hence the signal flow graph is,

The forward path gains are,

T1 = sCR T2 =1

sCRT3 =

1R

R 1� � = 1

The various loop gains are,

L1 = – sCR L2 = 1

sCRL 3 = � �

1R

R 1 = – 1

The loops L and L1 2 are non-touching� L L1 2 = 1

Hence system determinant is,

� = 1 [L L L ] [L L ]1 2 3 1 2 � � �

= 1 sCR1

sCR1 1� � � � =

3sCR s C R 1sCR

2 2 2� �

For T1, � 1 = 1 all loops touching to T1

For T2 , � 2 = 1 – L1 as L1 is non touching to T2

= (1 + sCR)

For T3 , � 3 = 1 all loops touching to T3

According to Mason’s gain formula,V (s)

V (s)o

i=

T T T1 1 2 2 3 3� � � � �

�

=

sCR1

sCR(1 sCR) 1

3sCR s C R 1sCR

2 2 2

� ��

��

� ��

��

�

��

=

s C R 1 sCR sCRsCR

3sCR s C R 1sCR

2 2 2

2 2 2

� � ��

��

�

��

� ��

��

�

��

�V (s)

V (s)o

i=

s C R 2sCR 1

s C R 3sCR 1

2 2 2

2 2 2

� �

� �



R

R

–1R

1sCsC

1R

–sC

+1

V (s)1V (s)i V (s)oI2(s)I1(s)

Q.4

Solution : Laplace transform of the given network.

I(s) =V V

Ri o

1

... (1)

V (s)o = I(s) R +1sC2

�

��

��... (2)

S.F.G. for equation (1) : S.F.G. for equation (2) :

Complete S.F.G. is :

Use Mason's gain formula. Number of forward path = K = 1

T1 =1

RR +

1sC1

2 ��

�� =

1 + sR C

R sC2

1

Individual loop, L1 =R sC+ 1

sC1

R2

1

�

��

�

�� =

1 + R sC

R sC2

1

� � = 1 – [L1] = 1 +(1 + R sC)

R sC2

1

As L1 is touching to T ,1 1� = 1

�V (s)

V (s)o

i=

T1 1�

�=

1 + sR C

R sC

1 +(1 + R sC)

R sC

2

1

2

1

V (s)

V (s)o

i=

1 + sR C

1 + R sC+ R sC2

2 1=

1+ sR C

1+ sC(R + R )2

1 2

��



Vo(s)R2

R1

Vi(s)I(s) 1

sC

Vo

I

Vi

1R1

1R1

–

VoI

R +21

sC

Vo

I

Vi

1R1

1R1

–

R +21

sC

1– —–

R1

1R + —–

sC2


7.8 Analysis of TYPE 0, 1 and 2 Systems

Q.11

Solution : Express the given open loop transfer function in time constant form.

G(s)H(s) =10 (s 2) (s 3)

s (s 1) (s 5) (s 4)

10.2.3 1s

� �

� � ��

� �2

1s3

s 1 5 4 (1 s) 1s5

�

��

�

��

��

�

� � � � � ��

��

�

��

��

�

1

s4

=

3 1s2

1s3

s (1 s) 1s5

1s

� ��

��

�

��

��

�

� ��

��

�

�

4�

��

�

Now, Kp = Lim G(s)H(s)s 0�

��

Kv = Lim s G(s)H(s)s 0�

�3

Ka = Lim s G(s)H(s)s 0

2

�

� 0

Now input is, r(t) = 3 t t 3 t 2t2

22

� � � � �

The input is combination of three standard inputs.

A 1 = 3, step of 3

A 2 = 1, ramp of 1

A 3 = 2, parabolic input of 2

Note that parabolic input must be expressed asA2

t 2 .

a) For step of 3 the error is,

ess1 =A

1 K3

101

p��

� ��


(7 - 1)

7Time Domain Analysis

of Control Systems

b) For ramp of 1 the error is,

ess2 =A

K12

v�

3

c) For parabolic of 2, the error is

ess3 =A

K23

a� � �

0

Hence steady state error is,

ess = e e e1

ss1 ss2 ss3� � � � � �03

= �

Q.12

Solution : From the system shown we can write,

G(s) =K

s(s 1)�, H(s) = 1

The input is r(t) = 0.1 t i.e. ramp of magnitude 0.1. For ramp input Kv controls the error.

� Kv =Lim

s 0�sG(s)H(s) =

Lim

s 0�=

s.Ks(s 1)�

= K

� ess =A

Kv=

0.1K

Maximum ess allowed is 0.005

� 0.005 =0.1K

� K =0.1

0.005= 20

For any value of K greater than 20, ess will be less than 0.005. Hence the range of value of

K for ess � 0.005 is,

20 � K < �

Q.13

Solution : G(s) H(s) =K

s (s + 1) (1 + 0.4 s)

i) ess =Lim

s � 0

s R(s)

1 + G(s) H(s),

r(t) = 4t, R(s) =4

s2

Principles of Control Systems 7 - 2 Time Domain Analysis of Control Systems


=Lim

s � 0

s4

s

1 +K

s (s + 1) (1 + 0.4 s)

2

=Lim

s � 0

4

s +K s

s (1 + s) (1 + 0.4 s)

=4

0 + K=

4K

K = 2 given

� ess = 2

ii) Now for the same input

ess =4K

and it is desired to have ess = 0.2

� 0.2 =4K

� K = 20

iii) Now input is r(t) = 2 + 6t

R(s) =2s

+6

s2

� ess =Lim

s 0�

s2s

+6

s

1 +K

s (s + 1) (1 + 0.4s)

2

�

��

�

��

=Lim

s 0

2

1K

s (1 s) (1 0.4s)

Lim

s

6

sK s

s�

��

��

� 0

(1 s) (1 0.4s)� �

�

�

�

�

�

�

�

�

�

�

�

�

= 0 +6K

, K = 10 given

� ess =6

10= 0.6

Q.14

Solution : Solving internal feedback loop we get,

� G(s) =

10s (1 + 4s)

1 +10

s(1 + 4s)6s

=10

4s + s + 60s2=

10s[61 + 4s]



10

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

–10

s (1+4s)6s1 +

Fig. 7.1

G(s)H(s) =

1061

s 1 +461

s�

��

�

��

Comparing with standard form =K(1 + T s) (1 + T s) ......

s (1 + T s) (1 + T s) ........

1 2j

a b

where j = type of system.

i.e. j = 1, System is TYPE 1 system

Error coefficients :

Kp =Lim

s 0�G(s)H(s) =

Lim

s 0�

1061

s 1 +461

s

=�

��

�

��

�

Kv =Lim

s 0�s G(s)H(s) =

Lim

s 0�

1061

s

s 1 +461

s

=

�

��

�

��

1061

Ka =Lim

s 0�s G(s)H(s2 ) =

Lim

s 0�

s1061

s 1 +461

s

2 �

��

�

��

�

��

�

��

= 0

Q.15

Solution : G(s) = 1 +0.1s

�

��

�

200(s 1) (s 2)� �

�

�

��

=(s + 0.1)200

s(s 1) (s 2)� �, H(s) = 0.02

� G(s)H(s) =200(s + 0.1)

s(s 1) (s 2)� �� 0.02

=200 0.02 0.1� �

�1 2�

(1 + 10s)

s(1 s) (1 0.5s)� �=

0.2(1 + 10s)

s(1 s) (1 0.5s)� �

i) For unit step input

Kp =Lim

s 0�G(s)H(s) =

Lim

s 0�

0.2(1 + 10s)

s(1 s) (1 0.5s)� �= �

� ess =1

1 Kp�= 0

ii) For unit ramp input

Kv =Lim

s 0�sG(s)H(s) =

Lim

s 0�

0.2(1 + 10s)

s(1 s) (1 0.5s)� �= 0.2

� ess =1

Kv=

10 2.

= 5



Q.16

Solution : When system is not in the simple closed loop form then we can not apply the

error coefficients.

In such case, we have to use final value theorem

i.e. ess =Lim

t ��e(t) =

Lim

s 0�sE(s)

In the system given E(s) = – C(s) when R(s) = 0

Now let us find outC(s)

T(s), so that for unit step disturbance we can calculate C(s) and hence

E(s). When R = 0, summing point at R(s) can be removed and block of `–1' is to be addedto consider sign of the signal at that summing point.

Shifting summing point to right.

Combining the two summing points and redrawing the diagram.



100

s2

10

1 + 0.1 s

0.1s

10–1

–

C(s)

++

+

T

Fig. 7.2

100

s2

10

1 + 0.1 s

0.1s 10

1 + 0.1s

�

–10

–

C(s)

+

+

+

T

Fig. 7.3

100

s2

s

(1 + 0.1 s)

–100

(1 + 0.1 s)

–

C(s)T(s)

+

Fig. 7.4

Negative sign of�

�

�

��

�

1001 0.1s

can be taken out to change sign of the signal at the summing

point from positive to negative.

�C(s)

T(s)=

100

s

1100

s

(100 s)

(1 0.1s)

2

2�

�

�

=100 (1 + 0.1s)

(0.1s s 100s 10000)3 2� � �

For T(s) =1s

, C(s) =1s�

100 (1 + 0.1s)

(0.1s s 100s 10000)3 2� � �

Now E(s) = – C(s) = –100 (1 + 0.1s)

s(0.1s s 100s 10000)3 2� � �

� ess =Lim

s 0�s E(s)

� ess =Lim

s 0�s �

�

� � �

�

��

��

�

�

100 (1 + 0.1s)

s(0.1s s 100s 10000)3 2

�

��

= –100

10000

Steady state error = – 0.01

Q.17

Solution : G(s)H(s) =K(1 + 2s) (1 + 4s)

s (s 2s 10)2 2� �

Given K = 2

� G(s)H(s) =2(1 + 2s) (1 + 4s)

s 10 (1 0.2s 0.1s )2 2� � � �

=0.2(1 + 2s) (1 + 4s)

s (1 0.2s 0.1s )2 2� �

Type 2 system,



100

s2

s

1 + 0.1 s

100

(1 + 0.1 s)

––

C(s)T(s)

Fig. 7.5

100

s2

100 + s

(1 + 0.1 s)

–

C(s)T(s)

Fig. 7.6

Kp =Lim

s 0�G(s) H(s) = �

Kv =Lim

s 0�sG(s) H(s) = �

Ka =Lim

s 0�s2 G(s) H(s) = 0.2

Q.18

Solution : G(s) =K(1 2s)

s(1 s) (1 0.4s) 2

�

� �

The system is Type-1 system. For an input of r(t) = t, calculate Kv .

Kv = Lim G(s) Lims K(1 2s)

s(1s 0 s 0� �

��

� s) (1 0.4s)K

2�

�

� ess =A

K1Kv

� … A = 1 for r(t) = t

But ess = 10 % = 0.1 (given)

� 0.1 =1K

i.e. K = 10

7.17 Derivations of Time Domain Specifications

Q.9

Solution :C(s)

R(s)=

20(s + 1) (s + 4)

1 +20

(s + 1) (s + 4)

=20

s + 5s + 242

Key Point Now though T.F. is not in standard form, denominator always reflect 2� �n and

�n2 from middle term and the last term respectively.

� comparing s + 5s + 242 with s + 2 s +2n n

2� � �

� �n2 = 24 � �n = 4.8989 rad/sec.

2 n� � = 5 � � = 0.51031

�d = �n21 � � = 4.2129 rad/sec.

Now for c(t) we can use standard expression forC(s)

R(s)in standard form. So writing



C(s)

R(s)=

2024

24

s + 5s + 242

�

��

��

�

��

��

For the bracket term use standard expression, and then c(t) can be obtained by multiplying

this expression by constant2024

.

� c(t) =2024

1e

1sin ( t + )

n t

2d�

� �

�

�

�

�

�

�

�

�

� � �

�

= tan11

2�

� �

�radians = 1.03 radians

� c(t) = ! "2024

1 1.1628 e sin (4.2129 t + 1.03)2.5 t�

�

Q.10

Solution : System differential equation is,

d y

dt+ 4

dy

dt+ 8y

2

2= 8x

To find T.F.Y(s)

X(s), take Laplace transform from above equation and neglect initial

conditions.

s Y(s) + 4s Y(s) + 8 Y(s)2 = 8 X(s)

� Y(s) [ ]s + 4s + 82 = 8 X(s)

� T.F.Y(s)

X(s)=

8

s + 4s + 82

Comparing this with standard T.F. of second order system�

� �

n2

2n n

2s + 2 s +�

� �n2 = 8

� �n = 2.83 rad/sec.

2 n� � = 4 � � = 0.7067

� �d = �n21 � � = 2.83 1 (0.7067) 2

� = 2.002 rad/sec.

� Tp = Time for peak overshoot

=#

�

#

d=

2.002= 1.57 sec.



% Mp = e 1001 2� � � �

�# = e 0.706 1 (0.706)2 100� � � �# = 4.33%

Ts = Settling time =4

=4

0.7067 2.83n� ��= 2 sec.

c(t) = 1 –e

1sin ( t + )

n t

2d

� �

� �

�

�

where = tan11

2�

� �

�

�

�

�

��

�

= 45$ =#

4rad

� c(t) = 1 –e

1 (0.7067)sin 2t +

4

0.7067 2.87 t

2

� �

�

�

��

�

#

c(t) = 1 – 1.41 e sin 2t +4

2t� �

��

�

#

�n = 2.83 rad/sec. Tp = 1.57 sec.

�d = 2.002 rad/sec. % Mp = 4.33%

Ts = 2 sec. � = 0.7067

Q.11

Solution : Assume TL as zero for [I]

i) To calculate M ,p

G(s) =6

0.15 s 0.9s2�

H(s) = 1

�C(s)

R(s)=

G(s)

1 G(s)H(s)�=

6s ( 0.15 s 0.9)

16

s (0.15s 0.9)

�

��

=40

s 6 s 402� �

Comparing the characteristic equation with s 2 s2n n

2� �� ,

� �n2 = 40 i.e. �n = 6.3245

2��n = 6

� � = 0.4743

Now, Mp = e� ��

#� �/ 1 2100 = 18.4 %

ii) For unit ramp input,

Kv = Lims 0�

s G(s)H(s) = Lims 0�

s.6

s ( 0.15 s 0.9)�.1 = 6.67



� ess =1

Kv= 0.15 rad

II) For TL is 1 Nm, assume R(s) as zero. With R(s) zero, the system gets modified as

From the Fig. 7.7, we can write.

G(s) =1

0.15 s 0.9 s2�

and H(s) = – 6

�C(s)

T (s)L= –

G(s)

1 G(s) H(s)�

�

��

�

��

Negative sign as sign of T (s)L applied is negative and 1 – G(s)H(s) as sign of the feedback

is positive. A block of – 1 is to continue sign of C(s), though R(s) = 0.

�C(s)

T (s)L= –

1

( 0.15 s 0.9s)

11

( 0.15 s 0.9 s). ( 6)

2

2

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

= –1

0.15 s 0.9 s 62� �

�

��

�

��

T (s)L =1s

� C(s) =�

� �

1

s [0.15 s 0.9 s 6]2

Css = Lims 0�

sC(s) = Lims 0�

s.1

s [ 0.15 s 0.9 s 6]2

�

� �

= �1

6= – 0.166

Q.12

Solution : R(s) =1s

T(s) =C(s)

R(s)=

Ks + a

hence C(s) =K

s(s a)�

Finding partial fractions, we get

C(s) =A

s+

A

s + a1 2



– 1

0.15 s + 0.9 s26–1

C(s)

T (s)L

+E(s)

Fig. 7.7

where A 1 =Ka

and A 2 = �Ka

� C(s) =

Ka

s

Ka

s + a

�

��

�

�

�

��

�

Taking inverse Laplace transform,

� c(t) =Ka

Ka

e at�

�

The steady state of c(t) is 2

�Lim

t ��c(t) = 2

�Ka

= 2

While slope at t = 0 is2.02 0.

= 1

i.e.dc(t)

d tt = 0

= 1

� –Ka

( a) e at

t = 0� �

� = 1

� K = 1 and a = 1/2

Q.13

Solution : The response of first order system to the step input is given as

c(t) = A ( t e )t T�

�

Where A = Final response value

T = Time constant

Now c(t) = 0.98 A for t = 60 sec.

� 0.98 A = A (1 e )60 T�

�

� 0.98 = 1 – e 60 t�

� e T60� = 0.02

� �60T

= – 3.912

� T = 15.33 sec.

Now, i = Input temperature

0 = Output temperature



For first order system,

0

i

(s)

(s)=

11 + Ts

=1

1 + 15.33 s

� 0 = i1

1 + 15.33 s�

��

�

��

Now error = i 0(s) (s)�

� E( )s = i i(s) (s)1

1 + 15.33 s�

�

��

�

��

= i (s) 11

1 + 15.33 s�

�

��

�

��

=15.33 s (s)

(1 + 15.33 s)i

But input temperature is varying as a rate of 10º C/min i.e.16

ºC/sec.

� i (t) =16

t�

� i (s) =1

6 s2

� E(s) =1

6 s

15.33 s

(1 + 15.33 s)2�

� ess =Lim

s 0�s E(s) =

Lim

s 0�

1

6 s2�

�

15.33 s(1 15.33 s)

=15 33

6.

= 2.556

Q.14

Solution : Closed loop T.F. =G(s)

1 + G(s), H(s) = 1

=

Ks(1 + Ts)

1 +K

s(1 + Ts)

=K

s T + s + K=

KT

s +1T

s +KT

2 2

Comparing with standard form�

� �

n2

2n n

2s + 2 s +�

�n2 =

KT

�n =KT

2� �n =1T

� =1

2KT

T

=1

2 K T



Now for Mp = 0.6 , Let � = �1

� 0.6 = e 1 112

� � � �#

i.e. – 0.51 = –# �

� �

1

121

� Solving, �1 = 0.1602

For Mp = 0.2, Let � = �2

� 0.2 = e 1 122

� � � �#

i.e. – 1.6094 =� �

� �

# 2

221

� Solving, �2 = 0.4559

For �1 = 0.1602, K = %1 and �2 = 0.4559, K = K2

�1 =1

2 K T1

� 0.1602 =1

2 K T1

... (1)

�2 =1

2 K T2

� 0.4559 =1

2 K T2

... (2)

From equation (1) K T1 =1

2 0.1602�= 3.1210

� TK1 = 9.74 i.e. TK 11 � = 8.7412

From equation (2) K T2 =1

2 0.4559�= 1.09673

� TK2 = 1.2028 i.e. TK2 – 1 = 0.2028

�TK 1

TK 11

2

�

�=

8.74120.2028

= 43.3

Q.15

Solution : To prove that system is overdamped means to prove � > 1 and not dependent

on the values of R and C.

So first to find its C.L.T.F.V (s)

V (s)o

iuse signal

flow graph method.

� I (s)1 =V V

Ri 1�

V (s)1 = (I I )1sC1 2�

I (s)2 =V V

R1 o�

V (s)o = I1sC2



V (s)i

V (s)1

I1(s) I2(s)

Vo(s)

R R

1/sC 1/sC

Fig. 7.8

� Signal flow graph is shown in the following Fig. 7.9.

T1 =1

s C R2 2 2, L1 = L = L =

1sCR2 3�

L and L1 3 is combination of 2 nontouching loops.

� L L1 3 =1

s C R2 2 2

Using Mason's gain formula,

� & = 1– ! " ! "L + L + L + L L1 2 3 1 3 = 1 +3

sCR+

1

s C R2 2 2

&1 = 1 as all loops are touching to T,

�C(s)

R(s)=

T=

1

s C R

1 +3

sCR+

1

s C R

1 12 2 2

2 2 2

&

&

=1

s C R + 3sCR + 12 2 2=

1

C2 R2

s +3s

CR+

1

C R

22 2

Comparing with standard form,

�n2 =

1

C R2 2� �n =

1CR

rad/sec.

2 n� � =3

CR� � =

3CR

12

CR =32

= 1.5

As � > 1 system is overdamped and � is independent of R and C values.

For this, two resistances must be equal, and two capacitor values must be equal.

Q.16

Solution :C(s)

R(s)=

G(s)

1 + G(s)H(s)=

1s (Js K)

1 +1

s (Js K)(s + m)

�

�

=1

Js Ks + s + m2�

C(s)

R(s)=

1 J

s +K+ 1

Js +

mJ

2 ��

��

�

��



Vi VoV1

L1 L2 L3

I1 I21/R 1/R1

sC1

sC

–1/R –1/R–1sC

Fig. 7.9

Comparing denominator with s + 2 s +2n n

2� � � = 0

� �n2 =

mJ

� �n =mJ

2� �n =! "� K+ 1

Ji.e. � =

! " ! "� �K+ 1

2JJ

m=

K+ 1

2 J m

For Mp = 25% , we can calculate �.

25 = 100 �� e 1 2# i.e. ln 0.25 =

� �

� �

#

1 2

Solving, � = 0.4037

Now Tp =#

�d=

#

�n21 � �

i.e. 2 =#

�n21 (0.4037)�

� �n = 1.7169 rad/sec.

Now steady state of system is 5 units.

Now steady state is defined as,

Css =Lim

t ��c(t)

Using final value theorem we can say,

Css =Lim

t ��c(t) =

Lim

s 0�sC(s)

From C.L.T.F. C(s) = R(s) �

1J

s +1 K

Js +

mJ

2 �and R(s) = 1/s

� C(s) =

1J

s s +1 K

Js +

mJ

2 ��

��

�

��

Css =Lim

s 0�sC(s)

=Lim

s 0�s

1J

s s +1 K

Js +

mJ

2 ��

��

�

��

= 5 (given)

�1m

= 5 i.e. m = 0.2



Now �n =mJ

hence 1.7169 =0.2J

� J = 0.06784

and � =K + 1

2 J m

�i.e. 0.437 =

�

�

K+ 1

2 0.06784 0.2

K = + 0.9059

Q.17

Solution : From the system shown,

G(s) =K

s2and H(s) = 1 + Ts

�C(s)

R(s)=

G(s)

1 G(s)H(s)�=

K

s

1K

s(1 Ts)

2

2� �

=K

s KTs K2� �

Comparing denominator with s sn n22 2� � ��

�n2 = K � �n = K ... (1)

and 2��n = KT � � =12

K T ... (2)

Now Mp = 20%

� 0.2 = e� � � �# / 1 2

� ln (0.2) =�#

'�

�

�2

Solving for � , � = 0.4559 ... (3)

Tp = 2 sec.

� 2 =#

�d=

#

�n 1 2� �

=

( )

#

�n 1 0 4559 2� .

... (4)

� �n = 1.7648 rad/sec

Substituting (4) and (3) in (1) and (2)

K = 3.1147 and T = 0.5166

Q.18

Solution : The T.F. of the system is,

C(s)

R(s)=

2 1

12 1

2

2

s

ss

s

�

��

=2 1

2 12

s

s s

�

� �



Now as numerator has term of 's' standard expression of c(t) cannot be used, thoughsystem is second order. Hence use partial fraction method substituting R(s).

R(s) =1s, as unit step input

� C(s) =1s

2 1

2 12

s

s s

�

� �

=1s

2 1

1 2

s

s

�

�( )

=A

s

B

(s 1)

C

(s 1)2�

�

��

… Partial fraction

� A (s + 1)2 + Bs + Cs (s + 1) = 2s + 1

� As2 + 2As + A + Bs + Cs2 + Cs = 2s + 1

� A + C = 0, 2A + B + C = 2, A = 1

� C = – 1, B = 1

� C(s) =1

s

1

(s 1)

1

(s 1)2�

�

��

Taking inverse Laplace transform,

� c(t) = 1 – te– t – e– t

Q.19

Solution : G(s) =A

s(s P)�, H(s) = 1

�C(s)

R(s)=

As(s P)

1A

s(s P)

A

s Ps A2�

��

�

� � s 2 s

n2

2n n

2�

� �

�

��

� �n2 = A i.e. � �� n nA and 2 P i. e.

P

2 A� � �

% Mp = e 100 i. e. 10 e1 2 1 2� � � �

� � �#� � #� � 100

Solving for �, � = 0.5911

Ts =4

4 i. e.4

4 0.59111.6917 ra

nn

��

�� d sec

� A = � �n2 and P 2 A� � � �2.862 2



7.21 Sensitivity of a Control System

Q.6

Solution : From given signal flow graph,

T1 = U(s)Q(s)G(s), T2 = M(s)G(s)

L1 = – Q(s)G(s), &1 = 1, &2 = 1

� T(s) =( )

( )

Y s

R s=

( ) ( ) ( ) ( ) ( )

( ) ( )

U s Q s G s M s G s

1 Q s G s

�

�

� T(s) =( ) ( ) ( ) ( )! "

( ) ( )

G s U s Q s M s

1 Q s G s

�

�

Sensitivity of T(s) with respect to G(s) is,

SGT =

*

*

T T

G G=

GT

TG*

*

*

*

TG

=( ) ( )! " ( ) ( ) ( )! " ( ) ( ) ( )! " ( )! " ( )

( ) ( )! "

1 Q s G s U s Q s M s U s Q s M s Q s G s

1 Q s G s2

� � � �

�

=( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )! "

U s Q s U s Q s G s M s M s Q s G s U s Q s G s M s Q s G s

1 Q s G s

2 2

2

� � � � �

�

=( ) ( ) ( )! "

( ) ( )! "

U s Q s M s

1 Q s G s2

�

�

� SGT =

( )

( ) ( ) ( ) ( )! "

( ) ( )! "

( ) ( ) ( )! "

( ) ( )! "

G s

G s U s Q s M s

1 Q s G s

U s Q s M s

1 Q s G s2�

�

��

�

� SGT =

( ) ( )

1

1 Q s G s�... Required sensitivity

It does not depend on U(s) or M(s).

��




8.7 Special Cases of Routh's Criterion

Q.4

Solution :

s6 1 4 5 2

s5 3 6 3 0

s4 2 4 2 0

s3 0 0 0 0

Row of zeros

A(s) = 2s4 24 2� �s = 0 i.e. s4 22 1� �s = 0

dA(s)

ds= 4s 4s3 �

s6 1 4 5 2

s5 3 6 3 0

s4 2 4 2 0

s3 4 4 0 0

s2 2 2 0 0

s1 0 0 0 0

Row of zeros again

� A �(s) = 2s 22 � = 0

dA (s)

ds

�= 4s = 0


8 Stability Analysis of Control Systems

s6 1 4 5 2

s5 3 6 3 0

s4 2 4 2 0

s3 4 4 0 0

s2 2 2 0 0

s1 4 0 0 0

s0 2 0 0 0

No sign change, hence no root is located in R.H.S. of s-plane. As row of zeros occur,

system may be marginally stable or unstable. To examine that find the roots of first

auxiliary equation.

A(s) = s4 22 1� �s = 0 s2 =� � �2 4 4

2= – 1

s2 = – 1, s2 = – 1, s1 2, = � j, s 3 4, = � j

The roots of �A (s) = 0 are the roots of A(s) = 0. So do not

solve second auxiliary equation. Predict the stability from the

nature of roots of first auxiliary equation.

As there are repeated roots on imaginary axis, system is unstable.

8.9 Marginal K and Frequency of Sustained Oscillations

Q.16

Solution : H(s) = 1

� � � � �1+ G s H s =� � � �

1 02

��

K

s 2 s 6s 252

� (s 2) (s 6 s 25)2 2� � � � K = 0 i.e. � �s 10s 53s 124s K 1004 3 2� � � � � = 0

Routh's array is,

s4

1 53 K + 100 For stability, K + 100 > 0 ... from s0

i.e. K > – 100

� �5034.4 10 K 100 0� � ... from s1

� 4034.4 10Ki.e. K 403.44�

s3

10 124 0

s2

40.6 K + 100

s1 � �5034.4 10 K 100

40.6

� � 0

s0

K + 100

Principles of Control Systems 8 - 2 Stability Analysis of Control Systems


+j

–j

�0

j�

Fig. 8.1

For stability, –100 K 403.44� �

For sustained oscillations, K = Kmar for which row of s1 becomes row of zeros.

� Kmar = 403.44

For this Kmar, � �A s = � �40.6 s K 100 02 � �

� s2 =� � � �

–K 100

40.6

403.44 100

40.612.4

� �

� �

� s = � j 3.5213 = � j �

Hence the frequency of oscillations is � 3.5213 rad / sec.

Q.17

Solution : The characteristic equation is 1 + G(s)H(s) = 0

1 +K (s 10) (s 20)

s (s 2)2

� �

�= 0 i.e. s3 + s2 (K + 2) + 30Ks + 200K = 0

Routh's array is,

s3

1 30K From s2

row, K + 2 > 0 K > – 2

200 K > 0, K > 0

30 K (K + 2) – 200 K > 0

as K > 0

i.e. K > 4.667

s2

K + 2 200K From s0

row,

s1 � �30K K

+2

� �2 200K

K0 From s

1row,

i.e. 30 (K + 2) – 200 > 0

s0

200K � 30 K > 140

� For 4.667 < K < , system is stable.

Kmar = 4.667 which makes row of s1 as row of zeros.

A(s) � (K + 2) s2 + 200K = 0 i.e. s2 = –200

2

K

K( )�

� s2 = –200 4.667(4.667 2)

��

= – 140 i.e. s = � j11.8321

Frequency of oscillations = 11.8321 rad/sec.

Q.18

Solution : Refer example 8.9.1 for the procedure and verify the answers :

0 < K < 4.1667, Kmar = 4.1667, � = 2.0412 rad/sec.

��

Principles of Control Systems 8 - 3 Stability Analysis of Control Systems



10.4 Conceptual Approach to Frequency Response

Q.3

Solution : T(s) =10 2( )s

(s + 1) (s + 5)

�

� T(j )� =10 (2 + j )

(1 + j ) (5 + j )

�

� �

Now the steady state response is given by,

c(t) = A M sin ��t + )�

where A = amplitude , � = input frequency, � = phase angle of transfer functionM = Magnitude of transfer function

M = T (j )� =10 4

1 25

2

2 2

�

� � �

�

� �

and � = � T (j ) = tan tan1 1

��

��

2 51tan

A = 5 as x (t)i = 5 sin (� t + )

�Steady state response can be obtained as,

c(t) =5 10 4

1 25

2

2 2

� �

� �

� � ��

� �

� sin ( )t

=50 4

1 255

2

2 2

1 1�

� �

�

�

�

�

�

� �

� �

��

sin tant + + tan2

tan1�


10Introduction to Stability Analysis

in Frequency Domain

10.6 B. W. (Bandwidth)

Q.2

Solution : For the equivalent second order system Mr = 2 and �r = �n 1 2 2 � = 3

Now Mr =1

2 1 2� �

We are given Mr = 2

� 2 =1

2 1 2� �

i.e. 4 =1

4 12 2� �( )

16 � �2 21( ) = 1

16 � �2 416 = 1

16 � � �4 216 1 = 0

Hence �2 =

16 � 256 64

32=

1632

19232

� = 0.5 � 0.433

�2 = 0.0669 or 0.933 and � = 0.2588 or 0.966

Key Point A system with relative damping factor greater than 0.707 does not exhibit any

peak in the frequency response hence the acceptable answer is only �2 = 0.0669.

Also �r = �n 1 2 2 � i.e. 3 = �n 1 2 0 2588 2

( . )

� �n = 3.223 rad sec

The transfer function of equivalent second order system is,

M(s) =�

� �

n

n ns s

2

2 22� � �

=(3.223)

s 2 (0.2588) (3.223) s (3.223)

2

2 2� �

=10.387

s 1.668 s 10.3872� �

Rise time, Tr =�

�

dwhere = tan

�

�

�

�

�

�

�

�

121

radians

In this case, = tan

�

�

�

�

�

�

�

�

121 0 2588

0 2558

( . )

.= tan �

��

�

��

1 0 96590 2588..

= 1.3 radians

� Also �d = �n 1 2 � = 3.223 1 0 2588 2

( . ) = 3.11 rad/sec

Principles of Control Systems 10 - 2 Introduction to Stability Analysis in Frequency Domain


� Rise time, Tr =� 1.33.11

= 0.592 sec

Time to peak, Tp =�

�d=

�

3.11= 1.0101 sec

Settling time, Ts =4

n� �(For � 2 % error band) =

40.2588 3.223�

= 4.795 sec

Ts =3

n� �= 3.596 sec (for � 5 % error band)

Period of oscillations,

�osc. =2 �

�d=

2

3.11

�= 2.02 sec.

Number of oscillations during settling

N =T

Ts

osc.=

47952 02..

= 2.37

Peak overshoot, Mp = e � �� 1 2

Since�

�

�

1 2=

0.2588 �

1 0 2588 2 ( . )

= 0.8417

� Overshoot = e 0.8417 = 0.4309

Thus the system exhibits 43.1 % overshoot.

Sketches of frequency response and estimated time response are shown.

Q.3

Solution : Mp = 16.2 % and Tp =�

5 3



0 3 �

M

(rad/sec)

1

1

1 1

2 3 4 5 6 70

+ +0.37

c(t)

Total 2.37oscillationsbefore itsettles

t(sec)

M = 2r

Fig. 10.1 Frequency response Fig. 10.2 Step response (estimated)

Now Mp = e �

� � �1 2100

� 16.2 = e �

� � �1 2100

� ln(0.162) =

� �

�1 2

��

��

�

��

1 8201 2.

�=

�

�

�

1 2

i.e. �2 = 0.2513

� � = 0.5

And Tp =�

�d=

�

5 3

� �d = 5 3 = � �n 1 2

� �n = 10 rad/sec

For sinusoidal input,

i) �r = � �n 1 2 2 = 10 1 2 0 5 2

� � ( . ) = 7.071 rad/sec

ii) Mr =1

2 1 2� �

=1

2 0 5 1 0 5 2� � . ( . )

= 1.1547

Q.4

Solution : G(s) =K

s(s a)�, H(s) = 1

i)C(s)

R(s)=

G(s)

1 G(s)H(s)�=

Ks(s a)

1K

s(s a)

�

��

=K

s as K2� �

Comparing denominater with s s2n n

2� �2�� ,

�n2 = K, �n = K ....(1)

2��n = a, � �a

2 K....(2)

Now, Mr =1

2 1 2� �

= 1.04

Solving, � = 0.6021 ....� cannot be more than 0.707.

While �r = � �n 1 2 2 = 11.55



� �n = 22.027 rad/sec

From equation (1), K = 485.1861

From equation (2), a = 26.5248

ii) B.W. = � � � �n2 2 41 2 2 4 4 � �

= 22.027 1 2 0 6021 2 4 0 6021 4 0 60212 2 4 � � � �( . ) ( . ) ( . )

= 25.2308 rad/sec

Ts =4

n��=

40 6021 22 027. .�

= 0.3016 sec

Q.5

Solution : G(s) =K

s(1 + Ts), H(s) = 1

�C(s)

R(s)=

G(s)

1 + G(s) H(s)=

Ks(1 + Ts)

1 +K

s(1 + Ts)

=K

Ts + s + K2=

K T

s +1T

s +KT

2

Comparing this with�

� �

n2

2n n

2s + 2 s +�

,

�n2 =

KT

� �n =KT

� 2 n� � =1T

i.e. � =1

2 KT

Now Mp = 25 % for unit step input

25 = e 1001 2 � �

�� i.e. ln 0.25 = –

� �

�1 2

Solving, � = 0.4036

In frequency domain,

�r = 8 rad/sec = �n21 2 � and � = 0.4036

� 8 = �n21 2 (0.4036) �

� �n = 9.744 rad/sec

�n =KT

i.e. �n2 =

KT

... (1)



� =1

2 KTi.e. �

2 =1

4KT... (2)

� K = T n2

� From equation (1), substituting in equation (2),

�2 =

1

4 T Tn2

� � ��

� T2 = 0.0161 i.e. T = 0.1271

� K = 12.071

Mr =1

2 1

1

2 0.4036 1 (0.4036)2 2� �

�

�

= 1.354

Q.6

Solution : For the second order system,

Mp = 10 % i.e. 10 = e �

� � �1 2100

� �

�1 2= � �ln 01. i.e.

� �

�1 2

= 2.3025

Solving, � = 0.5911

Tr =�

�

dwhere = tan

121 �

�= 0.9383 rad

� 0.1 =�

� �

0.9383

1n2

i.e. �n =

� �

2.2032

0.1 1 0.5911 2�

� �n = 27.3158 rad/sec

From these values, Mr and B.W. can be calculated as,

Mr =1

2 1 2� �

=

� �

1

2 0.5911 1 0.5911 2� �

= 1.0487

B.W. = � � � �n2 2 41 2 2 4 4� � �

= � � � �27.3158 1 2 0.5911 2 4 0.5911 4 0.2 2� � � � � � �5911 4

= 31.686

��




11.7 Calculation of G.M. and P.M. from Bode Plot

Q.16

Solution : Step 1 : G(s) is in the time constant form.

Step 2 : In this example, K is unknown. Draw the magnitude plot without K. The factors are,

i) Pole at the origin1

(s).

The straight line of slop – 20 dB/dec passing through the intersection point of � = 1 and 0

dB.

ii) Simple pole,1

1 s� 0 5., T1 = 0.5, �C1 �

1

1T= 2.

The straight line of slop – 20 dB/dec for � � 2.

iii) Simple pole1

1 s� 0 2., T2 = 0.2, �C 2

2

1�

T= 5.

The straight line of slope – 20 dB/dec for � > 5.

Range of � Resultant slope

0 < � < �C1(2) – 20 dB/dec

2 < � < �C2 (5) – 20 – 20 = – 40 dB/dec

5 < � < � – 40 – 20 = – 60 dB/dec

Note : The effect of K is to shift the entire magnitude plot upwards or downwards by

20 Log K dB. So find the shift required in magnitude plot without K to match the required

specification.

Step 3 : Phase angle table,

G(j�) =K

j j j� � �( . ) ( . )1 0 5 1 0 2� �


11 Bode Plot

� 1j�

–tan–1 0.5� –tan–10.02� � R

0.2 – 90º – 5.71º – 2.29º – 98º

2 – 90º – 45º – 21.8º – 156.8º

5 – 90º – 68.19º – 45º – 203.19º

10 – 90º – 78.69º – 63.43º – 232.12º

� – 90º – 90º – 90º – 270º

Step 4 : Sketch the Bode plot. From the plot �pc = 3.3 rad/sec.

Steps to find K for given G.M. :

1) Draw the horizontal line below 0 dB at a distance of given G.M., till it intersects

vertical line of �pc. This is point �A .

2) Draw the vertical line from �pc, till it intersects magnitude plot without K. This is

point A.

3) The point A must be at �A to match given G.M. Hence A to �A is the shift

required i.e. contribution by K which is 20 Log K dB.

4) Upward shift must be taken positive and downward shift negative.

20 Log K = Shift (A �A )

In this example,

20 Log K = 6 dB ... G.M. = 6 dB

K = 2 ... G.M. = 6 dB

Steps to find K for given P.M. :

1) Draw the hoizontal line above –180º line at a distance of given P.M., till itintersects phase angle plot. This is point B.

2) Draw the vertical line from B, till it intersects the magnitude plot without K. Thisis point �C .

3) This �C must be on 0 dB line at C, as the vertical line through point B must be �gcline for given P.M.

4) Thus distance of �C from 0 dB line is the shift required to satisfy given P.M. Thismust be 20 Log K dB.

5) Upward shift ( �C C) must be taken positive while downward shift ( �C C) must betaken negative.

20 Log K = Shift ( �C C)

Principles of Control Systems 11 - 2 Bode Plot


In this example,

20 Log K = 6 dB ... for P.M. = 25º

K = 2 ... for P.M. = 25º

Note : For same value of K, both specifications are matched. But this may not be the

case, all the time. (See Fig. 11.1 on next page).



12

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45

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89

12

34

56

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23

45

67

89

12

34

56

78

91

SE

MI-

LO

GP

AP

ER

(5C

YC

LE

SX

1/1

0)

–27

0º

–24

0º

–21

0º

–18

0º

–15

0º

–12

0º

–90

º

–4

0º

�R

–20

º

–6

dB

0dB20

40

60

MR

0.1

1�

pc

3.3

10

10

01

00

0L

og

�Sca

le:O

nY

axis

1un

it=

20

dB

,3

0º

–20

dB

/dec

–60

dB

/dec

WithoutK

AA

'shiftfo

rG

.M.=

+6

dB

–40

dB

/dec

C'–

Cshift

for

P.M

.=

25º

A

(1/s

)

B2

5º

C

C'

A'

Fig. 11.1

Q.17

Solution : Step 1 : Arrange G(s)H(s) in time constant form

G(s)H(s) =K

(1 2s) (2) 1s2

(10) 1s

10� �

��

��

��

��

=

K20

(1 2s) (1 0.5s) (1 0.1s)� � �

Step 2 : Factors

i)K20

��

�� is unknown.

ii) Simple pole,1

1 2s�, T1 = 2, �C1

=1T1

= 0.5

iii) Simple pole,1

1 0.5s�, T2 = 0.5, �C2

=1

T2= 2

iv)Simple pole,1

1 0.1s�, T3 = 0.1, �C 3

=1

T3= 10

Step 3 : Magnitude plot analysis.

i) No pole at the origin,K20

��

�� is unknown so starting Bode plot is nothing but 0 dB line

which will change its slope at �C1= 0.5.

ii) At �C1= 0.5, simple pole occurs which contributes as –20 dB/dec line. This line will

continue upto �C2= 2.

iii) At �C2= 2, simple pole occurs which contributes as –20 dB/dec individually and

hence resultant slope will become –20 – 20 = –40 dB/dec from �C2= 2 onwards till

next corner frequency �C 3= 10.

iv)As �C 3= 10, simple pole occurs which contributes as –20 dB/decade individually

and hence resultant slope will become –40 – 20 = 60 dB/dec from �C 3= 10 onwards

till � as there is no other factor present.

Step 4 : Phase angle plot

G(j H(j� �) ) =

K20

j j j

��

��

� � �( ) ( . ) ( . )1 2 1 05 1 01� � �

Though ( )K20 is unknown, it is not going to contribute any phase angle. Hence phase angle

plot is independent of value of ( )K20 and hence �pc is true, resultant specification from

this phase angle plot which is independent of value of ( )K20 .

��1

1 2j�= � �tan 1 2� , �

�1

1 05. j�= � �tan .1 05� , �

�1

1 01. j�= � �tan .1 01� ,



� tan�12� � �tan 1 05. � � �tan 1 01. � �R

1 – 63.43 – 26.56 – 5.7 – 95.7º

2 – 75.96 – 45 – 11.3 – 132.26º

5 – 84.28 – 68.19 – 26.5 – 179.03º

10 – 87.13 – 78.69 – 45 – 210.82º

� – 90 – 90 – 90 – 270º

Step 5 : Sketch the Bode plot, as shown in the Fig. 11.2



+6

0

+4

0

+2

0

–2

0

–4

0

–9

0o

–1

20

o

–1

50

o

–1

80

o

–2

10

o

–2

40

o

0d

B

0.1

10

01

00

0�

C1

=0

.5�

C2

=2

�p

c=

4.9

Re

qu

ire

d�

gc

=3

.3fo

rP.M

.=

30

o

Sh

ifte

dm

ag

nitu

de

plo

tto

me

etP.M

.=

+3

0o

SO

LU

TIO

N

Fo

rP.M

.=

+3

0(K

/20

)=

7.9

4i.e

.K

=1

58

.86

o

AB

=U

pw

ard

sh

iftre

qu

ire

dto

me

et

P.M

.=

+3

0=

+1

8d

Bo

P.M

.=3

0o

Lo

g�

–6

0d

B/d

ec

–4

0d

B/d

ec

–2

0d

B/d

ec

AB

–4

0

–2

0

–6

0

�p

c

Ma

gn

itu

de

plo

tw

ith

ou

tK 20

11

0�

C3

12

34

56

78

91

23

45

67

89

12

34

56

78

91

23

45

67

89

12

Fig. 11.2

�pc = 4.9 rad/sec

For P.M. = +30º,

+30� = P.M. = 180� + � �G(j H(jgc

� � � �) )|

� �G(j H(jgc

� � � �) )| = – 150º

Without effect of ( )K20 , �gc cannot be determined. So now adjust effect of ( )K

20 i.e. shifting

plot upwards till it intersects 0 dB line at such a frequency where � �G(j H(jgc

� � � �) )| i.e.

P.M. = +30�.

To get `2.8’ as �gc, magnitude plot is required to be shifted upwards from A to B as

shown in plot which is the effect of ( )K20 i.e. 20 log ( )K

20 .

Shift AB = +18 dB upwards.

20 log ( )K20 = +18

( )K20 = 7.94

K = 158.86 to meet P.M. as 30º

Q.18

Solution : Step 1 : Arrange G(s)H(s) in the time constant form.

The factors of (s 30s 20)2 � � are (s + 29.31) (s + 0.682)

Key Point As the factors of quadratic are real, they must be treated as separate simple poles.

So quadratic factor no longer remains quadratic in nature.

G(s)H(s) =2000 0.3 (1 )

4(1 ) (s 29.31) (s 0

s0.3

s4

� � �

� � � .682)

=2000 0.3 (1 )

4 (1 ) 29.31 (1

s0.3

s4

s

� � �

� � � � 29.31s

0.682) 0.682 (1 )� � �

=7.5(1 3.33s)

(1 0.25s) (1 0.034s) (1 1.466s

�

� � � )

Step 2 : Factors

i) K = 7.5

ii) Simple zero, (1 + 3.33s) , T1 = 3.33, �C1=

1T1

= 0.3

iii) Simple pole,1

1 1 466� . s, T2 = 1.466, �C2

=1

T2= 0.682

iv) Simple pole,1

1 0.25s�, T3 = 0.25, �C 3

=1

0 25.= 4

v) Simple pole,1

1 0.034s�, T4 = 0.034, �C4

=1

T4= 29.3



Step 3 : Magnitude plot

i) Contribution by K = 7.5 is 20 log 7.5 = 17.5 dB. So starting magnitude plot as

� � 0 is straight line of slope 0 dB. This will continue till �C1= 0.3.

ii) At �C1= 0.3, simple zero occurs which contributes as +20 dB/dec individually

and hence resultant slope will become 0 + 20 = 20 dB/dec from �C1= 0.3

onwards till next corner frequency �C2= 0.682.

iii) At �C2= 0.682, simple pole occurs which contributes as –20 dB/decade

individually and hence resultant slope will become 20 – 20 = 0 dB/dec from

�C2= 0.682 onwards till next corner frequency �C 3

= 4.

iv) At �C 3= 4, simple pole occurs which contributes as –20 dB/dec individually and

hence resultant slope will become 0 – 20 = –20 dB/dec from �C 3= 4 onwards till

next corner frequency �C4= 29.3.

v) At �C4= 29.3 simple pole occurs which contributes as – 20 dB/dec individually

and hence resultant slope will become –20 – 20 = –40 dB/dec from �C4= 29.3

onwards till � � � as there is no other factor present.

Step 4 : Phase angle plot :

G(j H(j� �) ) =7.5(1 j

j j j

�

� � �

3 33

1 0 25 1 0 034 1 1 466

. )

( . ) ( . ) ( . )

�

� � �

� �7.5 j0 = 0� � �1 3 33. j� = +tan �1 3 33. �

��

11 0 25. j�

= � �tan 1 0 25. �, ��

11 0 034. j�

= � �tan .1 0 34 �,

��

11 1 466. j�

= – tan .�1 1 466 �

Phase angle table :

� + tan�1 3 33. � � �tan .1 0 25� � �tan .1 0 034� � �tan .1 1 466� �R

0.1 18.41 – 1.4320 – 0.1948 – 8.3401 + 8.45

1 73.28 – 14.036 – 1.9473 – 55.70 + 1.6

10 85.7 – 68.19 – 18.77 – 86.09 – 87.35

20 89.13 – 78.69 – 34.24 – 88.04 – 111.81

50 89.65 – 85.42 – 59.53 – 89.21 – 144.52



100 89.82 – 87.7 – 73.61 – 89.6 – 161.10

� + 90 – 90 – 90 – 90 – 180

Step 5 : Sketch the Bode plot

With K = 2000, P.M. = + 40º, �gc = 45 rad/sec, G.M. = + � dB and �pc = �

system stable.

For P.M. = + 30º, �gc should be 64 as shown in plot. So plot must be shifted upwards

from A to B to get P.M. = + 30º.

This shift AB = +7 dB upwards.

So effect of constant must be (17.5 due to �K + 7 dB) = 24.5 dB required

So 20 log �K = 24.5 i.e. �K = 16.78. But �K = K/266.67, by adjusting G(s)H(s) in time

constant form.

� New K = 16.78 � 266.67 = 4474.65 for P.M. = 30º. (See Fig. 11.3)



1 2 3 4 5 6 7 8 9

60

40

20

+30o

–30o

–60o

–90o

–120o

–150o

–180o

0o

–20

0 dB

0.1 100 1000�gc = 45

New = 64 required for P.M. = 30�gc

o

Log �

P.M. = +30o

For P.M. = +30º

+40 + P.M.o

A

B

�gc+20 dB/dec –20 dB/dec–40 dB/dec

For K = 7.5

Resultant

0 dB/dec

0 dB/dec

1 10

To get P.M. = 30additional shift required to beintroduced is A to Bi.e. +7 dB by increasing K.K = 4475.65 to get

P.M. = 30

o

o

SOLUTIONWith K = 2000

= 45 rad/sec

=

P.M. = +40

G.M. = +STABLE

�

� �

�

gc

pco

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

Fig. 11.3

Q.19

Solution : The open loop transfer function is

G(s) =K

s s s( . ) ( . )1 0 2 1 0 02� �

The various factors are :

i) Constant K is unknown

ii) One pole at the origin.

The magnitude plot is the straight line of slope – 20 dB/decade passing through

intersection point of � = 1 and 0 dB line.

iii) The simple pole,1

1 0 2( . )� s

�C1=

10 2.

= 5

Magnitude plot is straight line of 0 dB upto 5 and straight line of slope – 20 dB/decade

after 5.

iv) The simple pole,1

1 0 02( . )� s

�C2=

10 02.

= 50

Magnitude plot is straight line of 0 dB upto 50 and straight line of slope – 20 dB/decade

after 50.

The resultant slope table is,

Range of � Slope

start < � < 5 – 20 dB/decade

5 < � < 50 – 40 dB/decade

50 < � < � – 60 dB/decade

G(j�) =K

j j j� � �( . ) ( . )1 0 2 1 0 02� �

The resultant phase angle plot is,



��1 for s

– 90�

�2 for1

1 0 2� . s

– tan.� �

��

��1 0 2

1�

�3 for1

1 0 02� . s

– tan.� �

��

��1 0 02

1�

�R

0.5 – 90º – 5.7º – 0.57º – 96º

1 – 90º – 11.3º – 1.14º – 102º

5 – 90º – 45º – 5.7º – 141º

50 – 90º – 84º – 45º – 219º

25 – 90º – 78º – 26.5º – 194º

From the Bode plot shown in the Fig. 11.4. (See Fig. 11.4 on next page.)

�pc = 16 rad/sec

Now to find limiting value of K, �gc= �pc

So point A should at point B to get �gc= �pc. So shift required from A to B is the

contribution by K, which is allowed till system does not reach on the verge of instability.

20 log K = shift AB

20 log K = + 35 dB

K = 56.23

Hence range of K for stability is,

0 < K < 56.23

Q.20

Solution : G(s)H(s) =K

s(1 0.1s)(1 0.25s)(1 0.001s)� � �

The G(s)H(s) is in time constant form. The various factors are,

i) K is unknown.

ii) One pole at origin, straight line of slope – 20 dB /dec passing through intersection

of � = 1 and 0 dB .

iii)1

1 0.25s�, simple pole, T1 = 0.25, �C1 � �

10 25

4.

Straight line of slope – 20 dB/dec for � � 4.

iv)1

1 0.1 s�, simple pole, T2 � 01. , �C2 � �

101

10.

Straight line of slope – 20 dB/dec for � � 10.

v)1

1 0.001s�, simple pole , T3 � 0 001. , �C3 � �

10 001

1000.





12

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12

34

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78

91

23

45

67

89

12

34

56

78

91

�2

70�

0.1

11

01

00

10

00

Lo

g�

�2

40�

�2

10�

�1

80�

�1

50�

�1

20�

�9

0�

�R

�2

0

0d

B

+2

0

+4

0

+6

0�

20

dB

/de

c

�4

0d

B/d

ec

�p

c=

16

SH

IFT

A+

B=

35

dB

Sca

leO

nY

axis

:1

unit

=2

0d

B,30º

B A

-6

0d

B/d

ec

Fig. 11.4

Straight line of slope – 20 dB/dec for � � 1000 (See Fig. 11.5 on next page.)


0 4� �� C1( ) – 20 dB/dec

4 10� �� C2 ( ) – 20 – 20 = – 40 dB/dec

10 1000� �� C3 ( ) – 40 – 20 = – 60 dB/dec

1000 � � �� – 60 – 20 = – 80 dB/dec

Phase angle table : G(j�)H(j�) =K

j (1 0.25j )(1 0.1j )(1 0.001j )� � � ��

� 1j�

� �tan .1 0 25 � � �tan .1 0 1 � � �tan .1 0 001 � � R

0.4 – 90º – 5.71º – 2.29º – 0.022º – 98º

4 – 90º – 45º – 21.8º – 0.22º – 157.02º

10 – 90º – 68.19º – 45º – 0.57º – 203.76º

100 – 90º – 87.7º – 84.28º – 5.71º – 267.69º

� – 90º – 90º – 90º – 90º – 360º

� Draw the Bode plot without K.

� As P.M. = 40º, draw the horizontal line at 40º above –180º line till it intersects phase

angle plot at point C.

� Draw vertical line from C till it intersects magnitude plot at point D.

� The point D must be on 0 dB line to have P.M. = 40º as the frequency corresponding to

point D must be �gc. The shift required from D till 0 dB line is the contribution by K .

Shift DE = 20 Log K = + 8 dB (Upward hence positive)

K = 2.5118 ......For P.M. = 40º

� Then draw magnitude plot parallel to that without K and then find the point of

intersection of �pc line with new magnitude plot. This point is A. The corresponding

gain margin is A to B.

G.M. = + 12 dB

Q.21

Solution : G(s)H(s) =80000

s(s 2)(s 50)(s 200)� � �

G(s)H(s) =80000

s 2 1s2

50 1s

50200 1

s200

� � ��

��

��

��

��

��





12

34

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12

34

56

78

91

23

45

67

89

12

34

56

78

91

+6

0

+4

0

+2

0

0dB

–20

–2

70

º 0.1

11

01

00

10

00

Lo

g�

�p

c=

6.2

+8

0

M ind

BR

P.M

.=+

40º

–3

00

º

�g

c

2.5

EB

–2

0d

B/d

ec

(1/s

)

G.M

.=

+1

2d

BA

–4

0d

B/d

ec

–6

0d

B/d

ec

Ne

wm

ag

nitu

de

plo

t

–8

0d

B/d

ec W

ithout

K

DE

=U

pw

ard

sh

ift

togetP.M

.=40ºSca

le:

On

ya

xis

:1

unit

=2

0d

B,30º

�R

–2

40

º

–2

10

º

–1

80

º

–1

50

º

–1

20

º

–9

0º

D C

Fig. 11.5

=4

s(1 0.5s)(1 0.02s)(1 0.005s)� � �.... Time constant form

The various factors are,

i) K = 4 i.e. 20 log 4 = 12.04 dB, line parallel to log � axis.

ii)1s

, one pole at origin, straight line of slope –20 dB/dec passing through intersection

of � � 1 and 0 dB.

iii)1

1 0.5s�, simple pole, T1 � 0 5. , �C1

1

1T

2� �

Straight line at slope –20 dB/dec for � � 2.

iv)1

1 0.02s�, simple pole, T2 � 0 02. , �C2

2

1T

� � 50

Straight line of slope – 20 dB/dec for � � 50

v)1

1 0.005s�, simple pole, T3 0 005� . , �C3

3

1T

� � 200

Straight line of slope – 20 dB/dec for � � 200


0 2� �� C1( ) – 20 dB/ dec

2 50� �� C2( ) – 20 – 20 = – 40 dB / dec

50 200� �� C3( ) – 40 – 20 = – 60 dB / dec

200 � � �� – 60 – 20 = – 80 dB / dec.

Phase angle table : G(j�)H(j�) =4

j (1 0.5j )(1 0.02j )(1 0.005j )� � � ��

� 1j�

� �tan .1 0 5� � �tan .1 0 02� � �tan .1 0 005� �R

0.2 �90º �5 71. º �0 229. º �0 05. º �9619. º

2 �90º �45º �2 29. º �057. º �137 86. º

50 �90º �87 7. º �45º �14 03. º �23673. º

20 �90º �84 2. º �21 8. º �5 71. º �201 71. º

� �90º �90º �90º �90º � 360º

The Bode plot is shown in the Fig. 11.6. From Bode plot, (See Fig. 11.6 on next page.)



�gc = 3 rad/sec, �pc = 10 rad /sec.

G.M. = + 20 dB, P.M. = + 33º



12

34

56

78

91

23

45

67

89

12

34

56

78

91

23

45

67

89

12

34

56

78

91

SE

MI-

LO

GP

AP

ER

(5C

YC

LE

SX

1/1

0)

+6

0

+4

0

+2

0

0d

B 0.1

11

01

00

Lo

g�

�g

c=

3

+8

0

M ind

BR

–3

00

º

�p

cA

B

G.M

.=

+2

0d

B

–4

0d

B/d

ec

�p

c

10

�g

c=

3

–2

0d

B/d

ec

–8

0d

B/d

ec

10

00

Sc

ale

:O

ny

ax

is:

1u

nit

=2

0d

B,

30

º

K=

4,1

2d

B

P.M

.+

33

º

DC

(1/s

)

–6

0d

B/d

ec

–2

70

º

–2

40

º

–2

10

º

–1

80

º

–1

50

º

–1

20

º

–9

0º

–2

0

=3

rad/s

ec

=1

0ra

d/s

ec

G.M

.=

+2

0d

BP.M

.=

+3

3º

Syste

mS

table

� �

gc

pc

�R

Fig. 11.6

Q.22

Solution : G(s) =20 (1 0.2 s)

s (1 0.5 s)

�

�

Step 1 : G(s) is in time constant form.

Step 2 : Factors are,

i) K = 20 i.e. 20 log K = 26.02 dB, line parallel to log � axis.

ii) One pole at origin, straight line of slope – 20 dB/Dec passing through intersection

point of � = 1 and 0 dB.

iii) Simple pole,1

1 0.5 s�, T1 = 0.5, �C1

1

1T

� = 2 straight line of slope – 20 dB/dec for

�� 2.

iv) Simple zero, (1 + 0.2 s), T2 = 0.2, �C22

1T

� = 5 straight line of slope + 20 dB/dec for

� � 5.

Frequency range Resultant slope

0 (2)C1� �� – 20 dB/dec

2 (5)C2� �� – 20 – 20 = – 40 dB/dec

5 � � �� – 40 + 20 = – 20 dB/dec

Step 3 : Phase angle table

G(j )� =20 (1 0.2 j )

j (1 0.5 j )

�

�

�

� �

� 1j�

� �tan .1 05 � � �tan .1 0 2� �R

0.1 – 90º – 2.86º + 1.14º – 91.71º

2 – 90º – 45º + 21.8º – 113.19º

5 – 90º – 68.19º + 45º – 113.19º

20 – 90º – 84.28º + 75.96º – 98.31º

� – 90º – 90º + 90º – 90º

Step 4 : The bode plot is shown in the Fig. 11.7. (See Fig. 11.7 on next page.)

Step 5 : From the bode plot,

�gc = 9 rad/sec �pc = � rad/sec

G.M. = � � dB P.M. = + 70º

The system is absolutely stable in nature.





12

34

56

78

91

23

45

67

89

12

34

56

78

91

23

45

67

89

12

34

56

78

91

+6

0

+4

0

+2

0

0dB

–2

0

–4

0

–7

5º

–9

0º

–1

05

º

–1

20

º

–1

35

º

–1

50

º

–1

65

º

–1

80

º

�R

M ind

BR

Sca

leO

nY

axis

=1

unit

=2

0d

B,15º

K,2

6d

B

–2

0d

B/d

ec

–4

0d

B/d

ec

–2

0d

B/d

ec

�g

c=

9

(1/s

)

P.M

.+

70

º

=9

rad/s

ec

=

G.M

.=

+d

BP.M

.=

70

ºS

yste

ma

bso

lute

lysta

ble

gc

pc

� ��

�

0.1

11

01

00

10

00

Lo

g�

Fig. 11.7

11.10 Calculation of Transfer Function from Magnitude Plot

Q.1

Solution : Starting slope 0 dB so no pole or zero at origin. At each corner frequency slope

changes.

First change in slope is at � = 1, �C1 = 1.

Change in slope is second slope – initial slope = 20 – 0 = 20 dB/dec.

i.e. factor is simple zero with �C1 = 1

i.e. T1 =1

�C1= 1

Factor is (1 + T1 s) i.e. (1 + s)

Now next change in slope is at � 1

�C2 = � 1 Change in slope 0 – 20 = – 20 dB/dec so factor is simple pole. Now � 1 is

not known.

Let us write equation for that line,

y = m x + C i.e. magnitude in dB = m Log � + C

m = + 20 dB/dec

Now at � = 1, magnitude in dB = 0 dB

Substituting 0 = 20 Log 1 + C

C = 0

Equation is magnitude in dB = 20 Log �

Now at � = � 1,

Magnitude in dB = 15 dB given

Substituting 15 = 20 Log � 1

� 1 = 5.623 rad/sec = �C2 .

T2 =1

�C2=

15.623

= 0.177

Factor is1

1 2� T s=

11 0177� . s

Next change in slope is at �2 .

�C3 = �2 . change is – 20 – 0 = – 20 dB/dec

Factor is simple pole.

To find �2 write equation for that line having slope – 20 dB/dec.

mag in dB = m Log � + C

mag in dB = – 20 Log � + C.



At � = 1000, mag in dB = 0

Substituting 0 = – 20 Log 1000 + C

C = + 60 dB

Equation is, mag in dB = – 20 Log � + 60

At �2 , mag.in dB = 15 = – 20 Log �2 + 60

– 20 Log �2 = – 45

�2 = 177.82 = �C3

T3 =1

�C3= 0.0562

Factor is1

1 0.00562 s�

The last change in slope is at

� = 1000 = �C4

Change is 0 – (– 20) = + 20 dB/dec so factor is simple zero

�C4 = 1000

T4 =1

�C4= 0.001

Factor (1 T s)4� = (1 + 0.001s)

The total transfer function is product of all of them

G(s)H(s) =(1 s) (1 0.001 s)

(1 0.177 s) (1 0.00562 s)

� �

� �

Q.2

Solution : Assume that the magnitude at � = 40 is 0 dB.

Now the equation for the first straight line of slope – 40 dB/dec is,

y = – 40 log � + C1 ... (1)

At � = 40, y = 0 dB

0 = � 40 log 40 + C1

C1 = + 64.082

The equation of second straight line of slope – 20 dB/dec is,

y = – 20 log � + C2 ... (2)

At � = 40, y = 0 dB

0 = � 20 log 40 + C2

C2 = + 32.0411



i) To find �1, substitute y = – 20 in equation (2)

– 20 = � �20 1log � 32.0411

– 52.0411 = � log 1�

�1 = 400 rad/secii) To find A 1, substitute � � 10 in equation (1)

A 1 = � �40 10log 64.082

A 1 = 24.082 dBiii) Now the equation of third straight line of slope – 40 dB/dec is,

y = � 40 log + C 3�

At � = 1000, the y can obtained by substituting � in equation (2)

y = – 20 log 1000 + 32.0411

y = – 27.9589 dB

Thus A 2 = – 27.9589 dB

iv) The equation of third line now is,

y = � �40 3log � C ... (3)

At � = 1000, y = – 27.9589

– 27.9589 = � �40 1000 3log C

C 3 = 92.0411

So to find �2 , substitute y = – 40 dB in equation (3),

– 40 = � �40 2log � 92.0411

�2 = 2000 rad/sec

Q.3

Solution : The slope – 6 dB/octave is same as –20 dB/decade while the slope – 12 dB/octave

is same as – 40 dB/decade. Hence the starting slope is –20 dB/decade hence there is one

pole at the origin.

Now this line of slope –20 dB/decade would have intersected 0 dB line at � = 10. Hence

at � = 1 its magnitude is +20 dB/dec. This indicates that at � = 1 the magnitude plot has

experienced a shift of +20 dB. This is the contribution by K.

20 Log K = 20

Log K = 1

K = 10

The first change in slope is at �C1= rad/sec. The change is of +20 dB/dec due to which

resultants slope becomes dB/dec after �C1= 5. Hence the corresponding factor is simple

zero.



T1 =1

�C1

=15

= 0.2

Factor = (1 T s)1� = (1 + 0.2 s)

The next change in slope is at �C2= 30 rad/sec. The change is of – 40 dB/dec. Hence the

corresponding factor is simple pole with order 2. So repeated simple pole is the factor.

T2 =1

�C2=

130

= 0.033

Factor =1

(1 T s)22�

=1

(1 0.033 s) 2�

Hence the transfer function is the product of all the factors.

G(s)H(s) =10 (1 0.2 s)

s (1 0.033 s) 2

�

�

��




12.5 Stability Determination from Polar Plot

Q.4

Solution : For drawing the rough sketch of polar plot we note that ,

G(j�)H(j�) =100

� � � �� 2 2 2 2 2 22 4 8

and � = � � ��

��

�

��

�

�� 90

2 4 81 1 1tan tan tan

� � �

Giving gain and phase at � = 0.1, 1, 10 as given below

� 0.1 1 10

GH 15.62 1.39 0.00808

� – 95.01º – 137.7º – 288º

At � = 0.1, GH 100

0.1 2 4 8� � �

1006 4.

15.62

At � = 1, GH 100

1 5 4 8� � �=

100

71.55= 1.39

At � = 10, GH 100

10 10 10 164� � �

1

10 164= 0.00808

Similarly we calculate � and write in the

above table. It is clear that it is difficult to

plot the polar plot to scale, however a rough

sketch can be made. This is shown.

To determine stability we must find x. This

can be done provided, we know the

frequency at which � = � �180 . This can be

easily be achieved by trying more values of �

greater than 1.


12 Polar and Nyquist Plots

�=0

��x

–90º

Fig. 12.1 A rough sketch of polar plot

At � = 2, � = � � � � � � � �90 45 26.56 14 = – 175.5�

At � = 2.2, � = � � � � � �90 47 72 28 8 15 4. . . = – 181 92. �

Obviously the phase cross over frequency is very nearly = 2.2 rad/sec

Hence GH|� �� pc= x =

100

2 2 2 4 82 2 2 2 2 2. � � � � � �2.2 2.2 2.2

=100

2.2 2.973 4.565 8.296� � �=

100

247.7

= 0.403

Since x < 1, the system is stable and GM =1x

=1

0 403.= 2.477

Moreover GM in dB = 20 log 2.477 = 7.87 dB

Similarly by trial we find �gc , the gain cross over frequency.

For example at � = 1.2

GH =100

1 2 5 44 17 44 65 44. . . .� � �

=100

1 2 2 332 4176 8 08. . . .� � �

=10094 4.

= 1.05 1

Thus �gc is around 1.05 rad/sec

And � at �gc is,

� = � � � ��

��

�

�� 90

1 052

1 054

1 058

1 1 1tan.

tan.

tan.

�

��

– 138.5°

This gives PM = 180° + (– 138.5°) = 41.5°For the above problem

Gain cross over frequency = 1.05 rad/sec phase cross over frequency = 2.2 rad/sec

Gain Margin = GM = 2.477 or 7.87 dB

Phase Margin = 41.5°

The system is stable.

Verification of G.M. by ROUTH’S CRITERION.

We can find critical value of K by ROUTH’S criterion.

1 + G(s)H(s) = 0 gives

1��

Ks (s 2) (s 4) (s 8)

= 0 K = 1

Principles of Control Systems 12 - 2 Polar and Nyquist Plots


or s (s + 2) (s + 4) (s + 8) + K = 0

or s (s 6s 8) (s 8) K2 � � � � = 0

or s (s 14 s 56 s 64) K3 2� � � � = 0

s 14 s 56 s 64 s K4 3 2� � � � = 0

ROUTH'S TABLE

s4 1 56 K

s3 14 64

s2 51.4 K

s1 3289.6 14K

51.4�

s0 K

Thus for stability 3289.6 – 14 K > 0

or 14 K < 3289.6

or K < 234.97

critical K = 234.97

and GM =critical Kactual K

=234.97

100= 2.35 or 7.42 dB

The answer agrees very closely with what we determined by Polar Plot (� 2 dB is accepted

using Bode’s or M-� plot).

Incidentally when K = 234.47, s row in the Routh Table is all zero and the auxiliary

equation formed by s2 row is

51.4 s K2 � = 0

or s = � jK

51.4= � j

234.9751.4

= 2.13 rad/sec

This is close to �pc.

Q.5

Solution :

G(j�)H(j�) =K

j j j� � �( ) ( . )1 2 1 01� �

We can find �pc by trial. However for this problem there is a simpler method.

GH =K

� � �� 1 2 1 012 2( ) ( . )

= � � � �� 90 2 011 1tan ( ) tan ( . )� �



At � = �pc , � = – 180°

� – 180° = � � � �� 90 2 011 1tan ( ) tan ( . )� �

+ 90° = tan� ��1 12 01( ) tan ( . )� �

90° = � � �1 2

where tan �1 = 2� and tan �2 = 0.1 �

Hence tan 90° = tan (� � �1 2 ) =tan � � �

� � �1 2

1 21

tan

tan tan

The right hand side is infinite only if 1� � �tan tan1 2 = 0

This gives 1� �2 01� �. = 0

o 0 2 2. � = 1

or �2 = 5

or � = 5 = 2.236 rad/sec

Thus �pc = 2.236 rad/sec

Then GH|� �� pc= x =

K

2 236 1 2 2 336 1 0 22362 2. ( . ) ( . )� � � � �

x =K

2.236 4.582 1.0246� �

x =K

10 447.

The system oscillates if x = 1

or K = 10.497

Q.6

Solution : G (j )� =K

j a� �

Since G(j�) will be plotted as a vector, we find the x and y

components of the vector.

In this case G(j�) = x + jy, where x and y are obtained by

rationalisation.

G(j�) =K

a j

a j

a j��

�

��

�=

K ( )a j

a

�

�

�

�2 2

=Ka

aj

K

a2 ��

��

�

�2 2 2

Thus x =Ka

a 2 � �2... (i)

and � = ��

�

K

a 2 �2... (ii)

We now eliminate �



x

y

0

G(j )�

Fig. 12.2 (a)

Dividing equation (ii) by equation (i),y

x= �

�KKa

=�a

and � =ya

x... (Iii)

Substituting in equation (i) , we have

x =Ka

a ( y a x)2 2� �=

Ka

ay a

x

22 2

2�

�

�

��

cross multiplying,

a xy a

x2

2 2

� = Ka

Multiplying throughout by x

a x y a2 2 2 2� = Kax

Dividing by a2 ,

we have x y2 2� =Ka

x��

�� ... (iv)

We bring this to the form

(x x ) (y y )12

12� � � = R

2

which gives centre (x1 , y )1 and radius R

From equation (iv), xKa

x y2 2� � = 0

or x 2K2a

xK2a

K2a

y22 2

2� � � � ��

��

�

�� = 0

or xK2a

y2

2��

�� =

K2a

2��

�� ... (v)

This is the equation of the circle. With centreK2a

, 0��

��

and radiusK2a

��

��. Note that y is negative for all values

of � as indicated by equation (ii). Hence only the

lower part of the circle gives the polar plot.

For the given problem K = 40 and a = 2 hence the

centre is at40

2 2, 0

��

�� i.e. (10, 0) and radius is,

K2a

=40

410�



0(10,0) 10

10

Fig. 12.2 (b) Polar plot of

G(s) =K

s + afor K = 40 and a = 2

12.14 Steps to Solve Problems by Nyquist Criterion

Q.12

Solution : The open loop T.F. is

G(s)H(s) =K (s 10)

s

2

3

�

G(s)H(s) =K j j

j

( ) ( )

( )

� �

�

� �10 103

There is no pole on right hand side hence for the stability N = – P = 0. The Nyquist plot

should not encircle critical point – 1 + j0.

The Nyquist path is,

Section-I

s = + j � M = 0 � = – 90º

s = + j0 M = � � = – 270º

Rotation = – 270º – (– 90º)

= – 180º Clockwise

Section-II

s = + j0 � � – 270º

s = – j0 � � + 270º

Rotation = 270º – (– 270º)

= + 540� Anticlockwise

Section-III

Section-III is mirror image of section-I

To find intersection of Nyquist plot with negative real axis, rationalising G(j�)H(j�).

G(j�)H(j�) =K j j j

j j

( ) ( ) ( )

( ) ( )

� � �

� �

� �

� �

10 10 3

3 3

=� �j K j� � �

�

3 2

6

100 20� �

=�

��20 1004

6

3 2

6

K j K�

�

� �

�

( )

For intersection with negative real axis,

100 – �2 = 0

� �2 = 100

� �pc = 10



Section II

Section IV

Section I

Section IIIs - plane

+ j0

– j0

O

j�

– j�

R�

�

Fig. 12.3

� Intersection point Q =� �20 K 10

10

4

6

Point Q = –K5

Hence the Nyquist plot is :

i) For stability,

Q > 1

�K5

� > 1

� K > 5 for stability.

ii) For K = 7,

Q = –75

= – 1.4

� G.M. = 20 log1

OQ

� G.M. = 20 log1

1 4.= + 2.93 dB

Q.13

Solution :

� � � �G j H j� � =� � � �

� �

2 8

3

� �j j

j

� �

�

Step 1 : No pole in r.h.s. of s-plane, P = 0

Step 2 : For stability, N = – P = 0



–270º

–900º

–1+j0

Q

I

��

� �

��

N=0

O

Fig. 12.4

Step 3 : Nyquist path is,

Step 4 : Section I :

s = + j� 090 90

270�

º º

º= 0 90� � º – 270º – (–90º) = – 180º

Clockwise

s = + j0 ��0 0

270

º º

º= �� 270º

Section - II :

s = + j0 i.e. � = + 0 �� 270º 270º – (–270º) = 540º

Anticlockwises = – j0 i.e. � = – 0 �� 270º

Section - III is mirror image of section - I, about real axis.

Section - IV not required.

Step 5 : Calculate point Q which is intersection of Nyquist plot with negative real axis.

� � � �G j H j� � =� � � �

� �

2 8� �

�

j j

j

� �

�3

...1j

= – j

=� �j j16 10 2

3

� ��

�= � ��

� �10

162 3

2

� ��

j

� 16 2� � = 0 ... for making imaginary part zero.

� �pc = 4

� Point Q =�1016

= – 0.625

Step 6 : The Nyquist plot is shown in the Fig. 12.5 (b)



II

I

IV

+j�

– j�

�

+j0

– j0

III

Fig. 12.5 (a)

Step 7 : N = 2 from the Nyquist plot and must be zero for stability.

Thus there are 2 closed loop poles located in right half of s-plane, making systemunstable.

12.15 Behaviour of Right Half Pole

Q.3

Solution : G(s) =Ks

(s 1) (s 5)2� �

Step 1 : Two poles in R.H.S. of s-plane.� P = 2

Step 2 : According to Nyquist stability

criterion,

N = – P = – 2

So Nyquist plot must encircle – 1 + j0

twice in clockwise direction.

Step 3 : The Nyquist path is, as shown in

Fig. 12.6.

Step 4 : G(j�) =Kj

( 1+ ) ( 1+ ) (5+ )

�

� � �� j j j

Section I : s = + j� to s = + j0



II

I

III

IV

j�

– j�

j0

– j0

Section

Section

Section

Section

R��

R 0�

s - plane

Nyquist path

Fig. 12.6

–j�

�

–1+j0

+j�

+j0

–j0

–270º

– 90º

Q

–0.625

–180º

N = 2

Fig. 12.5 (b)

Starting point � � � 1 9090 90 90�

��

� � �= 0 � – 180� – 270º – (– 180º) = – 90º

Clockwise

Terminating point � � + 0 0 ��

� � �90

180 180 0= 0 270� � �

Section - II : s = + j0 to s = – j0

� � + 0 0 � – 270�– 450º – (– 270º) = – 180º

Clockwise� � – 0 0 ��

90180 180 0

= 0 450� � �

Key Point Pole in right half contributes + 180º for both � � + 0 and � � – 0.

Section III : Mirror image of Section I about real axis.

Section IV : Not required.

Step 5 : Intersection with negative real axis.

G(j�) =Kj j j

j j j j

� � �

� � � �

( ) ( )

( ) ( ) ( ) ( )

5 1

1 1 5 5

2

2 2

� � �

� � � � � �… Rationalizing

=Kj j j� � � �

� �

[ ]

( ) ( )

5 9 3

1 25

2 3

2 2 2

� � �

� �=

� � � �

� �

K Kj� � � �

� �

2 2 2

2 2 2

9 5 3

1 25

( ) ( )

( ) ( )

Equating imaginary part to zero, 5 – 3 �2 = 0

� �2 =53

= 1.667 i.e. �pc = 1.2909 rad/sec

Substituting in real part intersection with negative real axis is obtained,

� Q =� � � �

� �

K 1.667 (9 1.667)

(1 1.667) (25 1.667)2= – 0.09374 K

Step 6 : The Nyquist plot is as

shown in the Fig. 12.7.

Note that while approaching

from 0 � – 180º to � – 270º in

clockwise direction, Section I of

plot has to intersect negative

real axis at point Q. So from

origin, it will travel through

third quadrant and through

point Q will reach to 0 � – 270º.



– 270º

+j�

+j0

–1 + j0

–1 + j0 – j�

– j0

Q

N = 0unstable

N = – 2stable

– 180º

Section I

Section I

Fig. 12.7

Step 7 : For stability, N = – 2 which is possible only if,

|OQ| > 1 i.e. | – 0.09374 K | > 1

� K >1

0 09374.> 10.667

So range of K for stability is 10.667 < K < �.

Thus for K = 10, the system is unstable.

��




14.4 State Model of Electrical Systems

Q.4

Solution : Applying KVL to the two loops,

–di

dti –

di

dt– i u(t)1

12

2� � = 0 ... Loop 1

� �– i – i idi2dt1 2 2� � = 0 ... Loop 2

�di2dt

= i – 2 i1 2 ... (1)

Using (1) in equation for loop 1,

di

dt1 = � �–i i 2i – i u(t)1 1 2 2� � � = –2i i u(t)1 2� � ... (2)

using i1(t) = X1 and i2(t) = X2

� X1

�

= – 2 X X u(t)1 2� � ...(3)

X2

�

= X – 2 X1 2 ...(4)

and y(t) = i – i1 2 = X – X1 2 ...(5)

�X

X

1

2

�

�

�

�

�

=�

�

�

�

�

2 1

1 2

X

X1

2

�

�

� +

1

0

�

�

�

u(t) and y(t) = � �1 1–X

X1

2

�

�

�

�A =

�

�

�

�

�

2 1

1 2, B =

1

0

�

�

� , C = � �1 1– , D = [0]


14State Variable Models

and Analysis of Control Systems

Q.5

Solution : Select the two currents as shown

in the Fig. 14.1. And write the equations

using KVL and KCL.

i1(t) =U(t) (t)

1 10

26

�

�

v` … (1)

v2(t) =1C

(I I )1 2�� dt

i.e. i1 – i2 = Cd (t)

dt2v

… (2)

i2(t) =v v2 1

61 10

�

�

… (3)

And v1(t) =1C 2i� dt

� i2(t) = Cd (t)

dt1v

… (4)

Elliminate i1 and i2 from above equations and C = 1 × 10– 6 F.

Substituting (1) and (3) in (2) we get,

Cd

dt2v

=U(t) v (t)

1 10

(t) (t)

1 10

26

2 16

�

�

�

�

�

��

�

�

�

�

v v

�d

dt2v

=1C

U(t)

1 10

1C

(t)

1 10

1C

2 (t)

1 1061

62

6�

�

�

�

�

�

�

�

�

�

�

v v

�d

dt2v

= v1(t) – 2v2(t) + U(t) … (5)

Using equation (3) in equation (4),

Cd

dt1v

=v v2 1

61 10

�

�

�d

dt1v

= – v1(t) + v2(t) … (6)

Select X1(t) = v1(t) and X2(t) = v2(t)

Using selected state variables,

X1

�

= – X1 + X2, X2

�

= X1 – 2X2 + U(t) and Y(t) = v1(t) = X1(t)

Hence the state model is,

X�

(t) = AX(t) + BU(t) and Y(t) = CX(t) + DU(t)

whereA =

� �

�

�

�

�

1 1

1 2, B =

0

1

�

�

� , C = [1 0], D = 0

Principles of Control Systems 14 - 2 State Variable Models and Analysis of Control Systems


U(t)

U(t)

1 F� 1 F�

v (t)2 v (t)11 M�1 M�+

–

+ +– –

+ +

– –i1 i2

Fig. 14.1

Q.6

Solution : The currents and voltages

are shown in the Fig. 14.2. Applying

KVL to the two loops,

� � � � �i R Ldi

dtV e 01 1 1

1C 1

i.e.di

dt1 =

��

R

Li

1L

V1

Le1

11

1C

11 ... (1)

� � � �Ldi

dti R e V2

22 2 2 C = 0 i.e.

di

dt

R

Li

1L

V1

Le2 2

22

2C

22�

�� ... (2)

And i i1 2� = CdV

dtC i.e.

dV

dtC =

1C

i1C

i1 2� ... (3)

Using state variables as, X i1 1� , X i2 2� , X V3 C� and e U1 1� , e U2 2� and output

y = i2

� X1�

=�

� �R

LX

1L

X1

LU1

11

13

11, X2

�

=�

�R

LX

1L

X1

LU2

22

23

22–

X 3�

=1C

X1C

X1 2� , y = X2

� A =

��

�

�

�

�

R

L0

1L

0R

L1

L1C

1C

0

1

1 1

2

2 2, B =

1L

0

01

L0 0

1

2�

�

�

�

, C = � �0 1 0

14.8 State Space Representation using Phase Variables

Q.4

Solution : Divide N and D by s4.

��

� �

C s

R s=

3

12 3 2

3

12 3 2

4

3 4

4

3 4

s

s s s

s

s s s� � �

�

� � � ��

�

�

Comparing withT1 1�

�=

� �

T

L L L1 1

1 2 31

�

� � �

T1 =34s

, �1 1� , Ls12

��

, Ls

2 3

3� � , L

s3 4

2� �



+ – + – + – + –

+

–

+

–

+

–e1

R1 L1

i1

VC

C

L2R2

i2

e2+

–

Fig. 14.2

Hence the signal flow graph is as shown in the Fig. 14.3.

From the signal flow graph,

X 1

�

= � �2 1 2X X , X 2

�

= X 3 , X 3

�

= � �3 1 4X X ,

X 4

�

= � �� 2 1X U s , � �Y s = 3 1X

A =

�

�

�

�

�

�

�

�

�

�

2 1 0 0

0 0 1 0

3 0 0 1

2 0 0 0

0

0

0

1

, B � � � �, ,C D� �3 0 0 0 0

This is the required state model in the form X�

= AX + BU.

Q.5

Solution : Using direct decomposition for the denominator,

T(s) =s 3s 3

{ (s 2)s 3] s 1}

2� �

� � �



X4 X4 X31/s1

C(s)= Y(s)

X3 X2 X2 X1 X11 1/s 1 1/s 1 1/s 3

– 2

– 3

– 2

R(s)= (s)U

Fig. 14.3

–– –

1 / s 1 / s 1 / s

2

3

1

3

1

3

+

++

Y(s)UX1

X2X3

+ X3 X2 X1

Fig. 14.4

For numerator shift take off point once for 3s and twice for s2 hence total state diagram is,

Output of each integrator is a state variable.

� X�

1 = X2, X�

2 = X3, X�

3 = U – X1 – 3X2 – 2X3

and Y = 3X1 + 3X2 + X3

Thus the various matrices are,

A =

0 1 0

0 0 1

1 3 2� � �

�

�

�

, B =

0

0

1

�

�

�

, C = [3 3 1], D = [0]

Q.6

Solution : Select state variables as,

y = X1,dy

dt= X X1 2

�

� ,d y

dt

2

2= X X2 3

�

�

Henced y

dt

3

3= X

�

3 and use in given equation,

� X 3

�

� � �6X 11 X 6 X3 2 1 = u

� X 3

�

= � � � �6X 11 X 6 X u1 2 3

The block diagram is as shown in

the Fig. 14.5 (a).

Thus the state model is,

X AX Bu�

� � and y = CX with,

A =

0 1 0

0 0 1

6 11 6� � �

�

�

�

,

B =

0

0

1

�

�

�

, C = [1 0 0],

D = [0]

The signal flow graph is as shown

in the Fig. 14.5 (b).



–u y

+1/s 1/s 1/s

––

6

11

6

X3 X = X2 3 X = X1 2

X1

Fig. 14.5 (a)

–6

–11–6

u1 1/s 1/s 1/s 1

X3X = X3 2 X = X2 1

yX1

Fig. 14.5 (b)

14.9 State Space Representation using Canonical Variables

Q.3

Solution : The T.F. T(s) can be factorised as,

T(s) =(s 2)

s (s 5s 4)

2

2

�

� �

=(s 2)

s (s 1)(s 4)

2�

� �

� T(s) =As

Bs 1

Cs 4

��

��

…Partial fractions

Obtain A, B, C as A = 1, B = – 1/3, C = + 1/3

� T(s) =1 1 3

11 3

4s s s�

��

�

/ /

The state diagram is as shown in the Fig. 14.6.

Thus X�

1 = U, X�

2 = – X2 + U, X�

3 = – 4 X3 + U

Y = X1 –13

X2 +13

X3

� A =

0 0 0

0 1 0

0 0 4

�

�

�

�

�

, B =

1

1

1

�

�

�

, C = 113

13

��

�

�



–

–

1

4

1U

Y

+

+

+

X1

X2

X3

X2

X3

X11/s

1/s13

–

13

1/s

Fig. 14.6

14.17 Laplace Transform Method of Finding State Transition Matrix

Q.4

Solution :

[sI – A] =s 0

0 s

0 1

2 0

�

�

�

��

�

�

�

=s

2

–1

s + 3

�

�

�

Adj [ sI – A] =C

C

C

C11

21

12

22

T�

�

� =

s + 3

1

–2

s

T�

�

�

=s + 3

2

1

s–

�

�

�

|sI – A| = s + 3s + 22 = (s +1) (s + 2)

� eAt = L (sI – A)–1 –1[ ]

=

s + 3

(s + 1) (s + 2)

–2(s + 1) (s + 2)

1(s + 1) (s + 2)

s(s + 1) (s + 2)

�

�

�

= � (s)

Finding partial fractions,

eAt = L ( (s))–1� =

2s + 1

–1

s + 2

–2s + 1

+2

s + 2

1s + 1

–1

s + 2

–1s + 1

+2

s + 2

�

�

�

=2e – e

– 2e + 2e

e – e

– e + 2e

– t – 2 t

– t – 2 t

– t – 2t

– t – 2t

�

�

�

ZIR = e X(0)At = e1

0At �

�

�

=2e e

2e + 2e

t 2 t

t 2 t

– –

– –

–

–

�

�

�

To find ZSR = � �L (s) B U(s)–1�

U(t) = Unit step �U(s) = 1/s

� ZSR = L

s + 3

(s + 1) (s + 2)

–2(s + 1) (s + 2)

1(s + 1) (s + 2)

s(s + 1) (s + 2)

–1� �

�

�

�

�

�

�

�

�

��

�

�

�

�

�

��

�

�

�

0

51 / s

= L

5s (s + 1) (s + 2)

5(s + 1) (s + 2)

L

2.5s

–5

–1 –1

�

�

��

�

��

�

�

��

�

��

�s + 1

+2.5

s + 2

5s + 1

–5

s + 2

�

�

��

�

��

�

�

��

�

��



=2.5 5e + 2.5 e

5 e 5e

t 2 t

t 2 t

–

–

– –

– –

�

�

�

� X(t) = ZIR +ZSR =2.5 – 3e + 1.5 e

3 e – 3e

–t – 2 t

– t – 2 t

�

�

�

� Y(t) =0

–2

1

–3

�

�

�

X(t)

Y (t)

Y (t)1

2

�

�

� =

0

–2

1

–3

�

�

�

2.5 – 3e + 1.5 e

3 e – 3e

– t – 2 t

– t – 2 t

�

�

�

=3e 3e

5 + 6e 3e

t 2t

2t t

– –

– –

–

– –

�

�

�

� Y (t)1 = 3 (e – e )– t – 2t

Y (t)2 = – 5 + 3 (2e – e )–2t –t

These are the outputs for unit step input applied.

Q.5

Solution : The state transition matrix is,

eA t = � �L (sI A)

1 1� ��

[sI – A] = s1 0

0 1

�

�

�

–1 2

1 4

�

�

�

�

�

=s 1 2

1 s 4

� �

� �

�

�

�

� �s I A��1 =

Adj [sI A]

| sI A |

�

�

Adj [sI – A] =C C

C C

11 12

21 22

T�

�

� =

s + 4

2 s 1

T1

� �

�

�

�

=s + 4 2

1 s 1

�

�

�

�

�

|sI – A| = (s – 1) (s + 4) + 2 = s s 4 42

� � � �s 2

= s 3s 22

� � = (s – 0.561) (s + 3.561)

� �s I A��1 =

s 4

(s 0.561) (s 3.561)

2

(s 0.561) (s 3

�

� �

�

� � .561)

1

(s 0.561) (s 3.561)

(s 1)

(s 0.561)� �

�

� (s 3.561)�

�

�

�

Ls 4

(s 0.561) (s 3.561)

1� �

� �

�

�

�

= L1.106

(s 0.561)

0.106

(s 3.561)

1�

��

�

�

�

�

= 1.106 e 0.106 e0.561 t 3.561 t� �

�



L2

(s 0.561) (s 3.561)

1� �

� �

�

�

�

= L0.485

(s 0.561)

0.485

(s 3.561)

1��

��

�

�

�

�

= �� 0.485 e + 0.485 e

0.561 t 3.561 t

L(s 0.561) (s 3.561)

1�

� �

�

�

�

1= L

0.242

(s 0.561)

0.242

(s 3.561)

1�

��

�

�

�

�

= 0.242 e + 0.242 e0.561 t 3.561 t� �

Ls 1)

(s 0.561) (s 3.561)

1��

� �

�

�

�

(= L

0.106

(s 0.561)+

1.106

(s 3.561)

1��

� �

�

�

�

= �� 0.106 e + 1.106 e

0.561 t 3.561 t

� eA t =

1 0.106 e 0.106 e .485 e +0.561 t 3.561 t 0.561 t

� ��

0.485 e

.242 e 0.242 e

3.561 t

0.561 t 3.561 t

�

��0 �

�

�

�

�0.106 e + 0.106 e0.561 t 3.561 t

� X(t) = eA t X(0) = e

A t0 5

1

.�

�

�

Y(t) =0

0

.068 e + 0.432 e

.015 e +

0.561 t 3.561 t

0.561 t

�

0.985 e3.561 t�

�

�

�

Q.6

Solution : Let the solution is,

X(t) = e X(0)At

Let A =A A

A A1 2

3 4

�

�

�

Now X(t) =2e

e

4t

4t

�

�

�

�

�

X(t)�

=�

�

�

�

�

�

�

8 e

4 e

4t

4t

At t = 0, X(0) =2

1

�

�

�

X(0)�

=�

�

�

�

�

8

4

And X(t) =4e

e

t

2t

�

�

�

�

�

2

X(t)�

=�

�

�

�

�

�

�

8 e

e

2t

2t2

At t = 0 X(0) =4

1

�

�

�

X(0)�

=�

�

�

�

�

8

2

X(t)�

=A A

A A1 2

3 4

�

�

� X(t)



� X(0)�

=A A

A A1 2

3 4

�

�

� X(0)

��

�

�

�

�

8

4=

A A

A A1 2

3 4

�

�

�

2

1

�

�

�

� – 8 = 2A A1 2� ... (1)

– 4 = 2A A3 4� ... (2)

and�

�

�

�

�

8

2=

A A

A A1 2

3 4

�

�

�

4

1

�

�

�

� – 8 = 4A A1 2� ... (3)

– 2 = 4A A3 4� ... (4)

Solving (1), (3) and (2), (4) simultaneously we get,

A =0 8

1 6

�

�

�

�

�

State transition matrix = eA t = L [sI A]1 1� ��

� [sI – A] =s s

1 s 6� �

�

�

�

i.e. Adj [sI – A] =s+ 6 8

1 s

��

�

�

|sI – A| = s 6s 82� � = (s + 2) (s + 4)

� eAt = L

s 6 8

1 s

(s 2) (s 4)1�

� ��

�

�

� �

�

�

��

�

�

�

�

�

��

�

�

�

= L

s 6(s 2) (s 4)

8(s 2) (s 4)

1(s 2) (s 4)

s(s 2) (

1�

�

� �

�

� �

� � � s 4)�

�

�

�

= L

2s 2

1s 4

4s 2

4s 4

0.5s 2

0.5(s 4)

1(s 2)

2(s 4

1��

��

�

��

�

��

� ��

� )

�

�

�

eAt =2e e 4e 4e

0.5e 0.5e e 2e

2t 4t 2t 4t

2t 4t 2t 4t

� � � �

� � � �

� � �

� � �

�

�

�

Q.7

Solution : From the given model,

A =1 0

1 1

�

�

�



[sI – A] = s1 0

0 1

�

�

�

–1 0

1 1

�

�

�

=s 1 0

1 s 1

�

� �

�

�

�

Adj [sI – A] =s 1 1

s 1

T�

�

�

�

� 0

=s 1 0

1 s 1

�

�

�

�

�

|sI – A| = (s 1)2

�

� [sI A]1

�� =

Adj [sI A]

sI A

�

�=

s 1 0

1 s 1

(s 1)2

�

�

�

�

�

�

=

1

s 10

1

(s 1)

1

s 12

�

� �

�

�

�

eAt = L

1�[sI A]

1�

� = L1�

1

s 10

1

(s 1)

1

s 12

�

� �

�

�

�

=e 0

te e

t

t t

�

�

�

� X(t) = eAt X(0) = zero input response

=e 0

te e

t

t t

�

�

�

1

0

�

�

�

=e

t e

t

t

�

�

�

This is the required solution.

Q.8

Solution : Find eAt for A =0 1

2 3� �

�

�

�

thus [sI – A] =s 1

2 s 2

�

�

�

�

�

[sI – A]– 1 =Adj [sI A]

|sI A|

�

�=

s

s

s s

�

�

�

�

�

� �

3 1

2

3 22=

s

s

s s

�

�

�

�

�

� �

3 1

2

1 2( ) ( )

� � (s) = [sI – A]– 1 =

s(s 1) (s )

1(s 1) (s )

(s 1) (s 2)s

(s 1) (s 2

�

� � � �

�

� � � �

32 2

2)

�

�

�

=

2s 1

1s 2

1s 1

1s 2

2s 1

2s 2 s 1

2s 2

��

� ��

�

�

��

�

�

��

�

�

�

�

1

� eAt = L– 1 [� (s)] =2e e e e

e e e e

t t t t

t t t t

� � � �

� � � �

� �

� � � �

�

�

�

2 2

2 22 2 2



� ZIR = eAt X(0) = eAt 0

0

�

�

�

=0

0

�

�

�

Now ZSR = L– 1 {� (s) B U(s)}

U(s) =

1

12

s

s �

�

�

�

� ZSR = L– 1

s(s 1) (s 3) (s 1) (s 2)

2(s 1) (s 2)

s(s 1) (s 2

�

� � � �

�

� � � �

3 1

)

�

�

�

�

�

�

�

�

�

�

�

�

��

�

�

�

�

�

2 1

0 1

1

13

s

s

��

�

�

�

= L– 1

s(s 1) (s 3) (s 1) (s 2)

2(s 1) (s 2)

s(s 1) (s 2

�

� � � �

�

� � � �

3 1

)

�

�

�

��

�

�

�

�

�

�

��

�

�

�

�

�

��

�

�

�

2 13

13

s s

s

= L– 1

s(s 1) (s 3) (s 1) (s 2)

2(s 1) (s 2)

s(s 1) (s 2

�

� � � �

�

� � � �

3 1

)

�

�

�

�

�

�

�

�

�

�

�

��

�

�

�

�

�

3 2

3

13

( )

( )

s

s s

s

��

�

�

�

= L– 1

31

1

6

1 35

s(s ) (s 1) (s 2) ( + 3)

s(s ) (s ) (s 1) (s 2

��

� �

�

� ��

� �

s

) ( + 3)s

�

�

�

= L– 1

3 31

0 5

11

2

0 5

3

2 31

13

0 5

12

s s s s s

s s s s s

��

��

��

��

��

��

��

��

�

. .

.

2

1 5

3�

�

�

�

�

.

s

=3 2 5 0 5

2 2 5 2 2 5

2 3

2 3

� � �

� � � �

�

�

�

� � �

� � �

. .

. .

e e e

e e e

t t t

t t t

� X(t) =X (t)

X (t)1

2

�

�

� = ZIR + ZSR =

3 2 5 0 5

2 2 5 2 2 5

2 3

2 3

� � �

� � � �

�

�

�

� � �

� � �

. .

. .

e e e

e e e

t t t

t t t

�Y (t)

Y (t)1

2

�

�

� =

1 0

1 1

�

�

�

X (t)

X (t)1

2

�

�

�

� Y1(t) = X1(t) = 3 – 2.5 e– t – e– 2t + 0.5 e– 3t

and Y2(t) = X1(t) + X2(t) = 1 + e– 2t – 2 e– 3t



14.20 Observability

Q.5

Solution : From the given model,

A =

0 1 0

0 0 1

6 11 6– – –

�

�

�

, B =

1

0

1

�

�

�

, C = � �10 5 1

For controllability, Qc = � �B : AB : A B2 ..... n = 3

AB =

0 1 0

0 0 1

6 11 6– – –

�

�

�

1

0

1

�

�

�

=

0

1

12–

�

�

�

A2B = � �A AB =

0 1 0

0 0 1

6 11 6– – –

�

�

�

0

1

12–

�

�

�

=

1

12

61

–

�

�

�

� Qc =

1 0 1

0 1 12

1 12 61

–

–

�

�

�

Now Q c = – 84 � 0 hence rank of Q c = n = 3.

The system is completely controllable.

For observability, Qo = C : A C : A CT T T 2 TT�

�

�

A CT T =

0 0 6

1 0 11

0 1 6

–

–

–

�

�

�

10

5

1

�

�

�

=

–

–

–

6

1

1

�

�

�

A CT 2 T = � �A A CT T T =

0 0 6

1 0 11

0 1 6

–

–

–

�

�

�

–

–

–

6

1

1

�

�

�

=

6

5

5

�

�

�

� Qo =

10 6 6

5 1 5

1 1 5

–

–

–

�

�

�

Now Qo

� �96 0 hence rank of Qo = n = 3

The system is completely observable.



Q.6

Solution : For controllability, � �Q B : AB : A B62

� and n = 3

AB =

0 1 0

0 0 1

0 2 3

0

0

1� �

�

�

�

�

�

�

=

0

1

3�

�

�

�

, � �A B A AB

1

3

7

2� � �

�

�

�

� Q c =

0 0 1

0 1 3

1 3 7

�

�

�

�

�

, Q c = 1 � 0

Hence rank of Q 3 nc � � hence system is completely state controllable.

For observability, � �Q C : A C : A CoT T T T 2 T

��

�

�

A CT T =

0 0 0

1 0 2

0 1 3

10

0

0

0

10

0

�

�

�

�

�

�

�

�

�

�

�

�

, � � � �A C A A CT 2 T T T T� =

0

0

10

�

�

�

� Q o =

10 0 0

0 10 0

0 0 10

�

�

�

, Q 1000 0o � �

Hence the rank of Q 3 no � � hence system is completely observable.

��



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Principles of Control Systems Solution Manual