Control Systems by Nagrath & Gopal Solution Manual

CHAPTER 2 DYNAMIC MODELS AND DYNAMIC RESPONSE

2.1 Revisiting Examples 2.4 and 2.5 will be helpful.

(a) R(s) = A; Y sKAs

( ) =+t 1

y(t) = KA e tt

t- / ; yss = 0

(b) R(s) = As

; Y(s) = KAs s( ) + 1

y(t) = KA( )/1 e t ; yss = KA

(c) R(s) = As2

; Y(s) = KAs s2 1( ) +

y(t) = KA( )/t e t + ; yss = KA( )t -t

(d) R(s) = As

w

w2 2+; Y(s) = KA

s s

( )( )+ +1 2 2

y(t) = KA e KA tt

2 2 2 21 1++

++ / sin( )

= tan ( ) 1

y tt

( )fi

= KA t

2 2 1++sin( )

2.2 (a) The following result follows from Eqns (2.57 - 2.58)

y(t) =11

2 12

12

+

FHG

IKJ

e tnt

d

sin tan

d = n 1 2The final value theorem is applicable: the function sY(s) does not have poleson the jw-axis and right half of s-plane.

yss = lim ( )s

sY sfi

=0

1

(b) The following result follows from Review Examples 2.2.

http://ebook29.blogspot.com/


SOLUTION MANUAL 3

y(t) = t e tn

t

dd

n

+ +

FHG

IKJ

2 2 112

sin tan

yss = tn

2

The final value theorem is not applicable; the function sY(s) has a pole onthe jw -axis.

2.3 Revisiting Review Example 2.1 will be helpful.

E sE s

RRCsi

04101

( )( )

/=

+ ; E s

si( ) =

102

e tR

t e RCt0 410 10( ) ( );/= - + =-t t tt

For the output to track the input with a steady-state delay 100 106 sec, itis necessary that

R10

14 = , and t = = -RC 100 10 6

This gives R = 10 kW ; C = 0.1 mF

Steady-state error = 10 t - (10 t 10t )= 0.001

2.4 Revisiting Review Example 2.2 will be helpful.

w z z wn n= = =100 3 2 6 100; ; / / sec = 60 msec

Steady-state error = 25 25 25 2t tn

- - FHG

IKJ

z

w

= 1.5

2.5 My t By t Ky t&&( ) &( ) ( )+ + = F(t) = 1000m (t)

Y(s) = 110 1002s s s( )+ +

n = 10, = 0.5, wd = 5 3Using Eqns (2.57)-(2.58), we obtain



4 CONTROL SYSTEMS: PRINCIPLES AND DESIGN

y(t) = 0.01 1 23

5 3 35 1- +LNMOQP

- -e tt sin( tan

2.6 Y sR s

G ss s s s

( )( ) ( ) ( )( )= = + + = + +

67 6

61 62

| ( )|G j = 2 = 35 2 ; G(j2) = 81.87

yss = 3

5 22 81 87sin( . )t

= 3

50 sin 2t 21

50 cos 2t

2.7 Y sR s

( )( ) = G(s) =

s

s s

+

+ +3

7 102 =

s

s s

++ +

32 5( )( )

(a) R(s) = 1 1s +

y(t) = 12

13

16

2 5e e e tt t t FH IK( )(b) (s2 + 7s + 10) Y(s) sy(0) &y (0) 7y(0)

= (s + 3) R(s) sr(0)Initial conditions before application of the input are

y(0) = 1, &y (0) = 12 , r(0) = 0

Y(s) = ss s

s

s s s

++ +

++

+ + +15 2

2 53

2 5 1/

( )( ) ( )( )( )

y(t) = 1232

2 5e e et t t- - -+ -

2.8 X = X x I I i+ = +;

M d X xdt

2

2( )+

= F X x I i Mg( , )+ +



SOLUTION MANUAL 5

Mx&& = F X I FX

xFI

i MgX I X I

( ), ,

+ +

FHG

IKJ +

FHG

IKJ

F X I( )+ = Mg = 8 4 10 9 83. . NewtonsFor this value of force, we get from Fig. 2.8b,

X I= =0 0 6.27 ; .cm amps

Again from Fig. P 2.8b,

K FX X I1

0 14=

=,

. Newtons/cm

K FI X I

2 0 4=

=

,

. Newtons/amp

&& ;( )( )

.

.

xKM

xKM

i X sI s s

= + =-

1 22

47 616 67

2.9 Mx Bx Kx K y x&& & ( )+ + = -1Gravitational effect has been eliminated by appropriate choice of the zeroposition.

X sY s

( )( ) = G s s s( )

.

( . )( . )= + +0 1667

0 0033 1 0 0217 1

w = 2 11 63p lv / .= rad /sec

| ( |.

G j =11 63 = 0.1615x(peak) = 7.5 0.1615 = 1.2113 cm

2.10 (a) Mx Bx Kx F t&& & ( )+ + =Gravitational effect has been eliminated by appropriate choice of the zeroposition.

X sF s

( )( ) =

12Ms Bs K+ +

Force transmitted to the ground

= K X(s) + Bs X(s)




=

BK

s F s

MK

sBK

s

+FH IK+ +

1

12

( )

F(t) = A sin w tPeak amplitude of the force transmitted to the ground at steady-state

=

A BK

MK

BK

1

1

2

2 2 2

+ FH IK

FHG

IKJ + FH IK

(b) Mx B x y K x y&& ( & &) ( )+ - + - = 0

X sY s

( )( ) =

Bs KMs Bs K

+

+ +2

Peak amplitude of machine vibration

=

A BK

v

MvK

BvK

1

1

2

2 2 2

+ FH IK-

FHG

IKJ + FH IK`

2.11 M y K y y B y y K y y B y y1 1 1 1 0 1 1 0 2 1 2 2 1 2&& ( ) ( & & ) ( ) ( & & )+ + + + = 0M y K y y B y y2 2 2 2 1 2 2 1&& ( ) ( & & )+ + = 0Gravitational effect has been eliminated by appropriate choice of the zeroposition.

2.12 The following result follows from Section 11.2 (Eqn (11.9)).

(a) E sE si0( )( ) =

1 21 2

22

1 22

1 22

1 22

+ +

+ + +

RC s R C C sR C C s R C C s( )

(b) E sE si0( )( ) =

1 21 2

1 1 22 2

1 2 1 22 2

+ +

+ + +

R Cs R R C sR R Cs R R C s( )

2.13 Refer Example 11.6

2.14 Refer Example 11.7



SOLUTION MANUAL 7

2.15 The following result follows from Example 12.9.

x x x z x z r F t1 2 3 4= = = = =f f, & , , &, ( )&x = Ax + br

A =

0 1 0 04 4537 0 0 0

0 0 0 10 0 0 0

.

.5809

L

N

MMMM

O

Q

PPPP; b =

00 3947

00

L

N

MMMM

O

Q

PPPP.

.9211

2.16 Mixing valve obeys the following equations:

( ) [ ( )] ( )Q q c Q Q q c Q c Qi i H i i C i i+ + - + = +r q r q r qqi = kv x

The perturbation equation is

K H Cv ( )q q- x(t) = Q tiq ( )or, x(t) = K t K Q Ki H Cq q q( ); / [ ( )]= -v

The tank obeys the equations

V c ddt

rq

= Q c t tid id i Dr q q q q t( ); ( ) ( )- = -

This gives

q

q

( )( )s

si=

e

s

s-

+

1 5

1

.

2.17 C ddt1

1= q t R

RUAm

( ) ;l q q- - =1 21

11

C ddt2

2q=

q q q q1 2

1

2

2

-+

-

R Ri ; C V c R Q c2 2

1= =r

r;

From these equations, we get

&1 = + +1 1 4 461 2.92 .92 . qm& 2 = 0 078 0 2 0 1251 2. . .q q q- + i




2.18 At steady-state

0 = q q q q1 21

2

2

-+

-

R Ri ; this gives 1 120= C

0 = Qm 1 2

1

R; this gives Qm = 17 26. / minkg

2.19 Energy balance on process fluid:

V c ddt2 2 2 2 2

( )+ = Q c UAi2 2 2 2 2 2 2 2 1 1 [ ( )] [ ] + +

At steady-state

0 = Q c UAi2 2 2 2 2 2 1r q q q q( ) ( );- - - this gives q1 40= CThe perturbation equation is

V c ddt

Q c UA2 2 2 2 2 2 2 2 2 1rq

r q q q- - -( )

This gives

544 5 02 2 1. .432+ + =ddtq

q q

Energy balance on the cooling water:

V c ddt1 1 1 1 1

r q q+d i = ( ) [ ( )]Q q c i1 1 1 1 1 1 1+ - +r q q q

+ UA[ ]q q q q2 2 1 1+ - -

At steady-state

0 = Q c UAi1 1 1 1 1 2 1r q q q q[ ] [ ];- + - this gives

Q1 = 5.28 103 m3/secThe perturbation equation is

V c ddt1 1 1

1rq

= ( ) ( ) i c q Q c UA1 1 1 1 1 1 1 1 1 2 1 +



SOLUTION MANUAL 9

This gives

184.55 ddtq

q1 1+ = 0.465 q1 1318.2 q1

Manipulation of the perturbation equation gives

q

q2

1

( )( )s

s=

557 6100 10 729 0 83 2

.

.5 . + +s s2.20 Tank 1:

C dpdt1

1= q1 q10 q11

q10 = flow through R0; q11 = flow through R1Ag

gdhdt

1 1r

r = qghR

g h hR1

1

0

1 2

1- -

-r r ( )

ordhdt

1= - +

FHG

IKJ + +

r rgA R R

h gA R

hA

q1 0 1

11 1

21

11 1 1

= 3h1 + 2h2 + r1Tank 2:

C dpdt2

2= q2 + q11 q20

q20 = flow through R2A

gg

dhdt

2 2r

r = r rg h h

RghR

q( )1 2

1

2

22

-- +

ordhdt

2=

r rgA R

h gA R R

hA

q2 1

12 1 2

22

21 1 1

- +FHG

IKJ +

= 4h1 5h2 + r2

2.21 Addt

H h( )+ = Q q Q q Q q1 1 2 2+ + + - -

At steady-state

0 = Q Q Q1 2+ - ; this gives Q = 30 litres/sec





A dhdt = q1 + q2 q

The turbulent flow is governed by the relation

Q(t) = K H t( ) = f(H)Linearizing about the operating point, we obtain

Q(t) = f H f HH

H HH H

( ) ( ) ( )+

FHG

IKJ -=

= Q KH

h t+2

( )

Therefore

q(t) = KH

h t2

( ) = K HH

h t2

( ) = QH

h t2

( )

dh tdt( )

= - + +1 1 1

1 2Aq t

Aq t

Aq t( ) ( ) ( )

= 0.01 h(t) + 0.133 q1(t) + 0.133 q2(t)Mass balance on salt in the tank:

A ddt

H h t C c t[( ( ) ( ( ))]+ + = C Q q t C Q q t1 1 1 2 2 2[ ( )] [ ( )]+ + +

+ +[ ( )][ ( )]C c t Q q tAt steady state,

0 = C Q C Q CQ1 1 2 2+ - ; this gives C = 15


AH dc tdt

AC dh tdt

( ) ( )+ = C q t C q t Cq t Qc t1 1 2 2( ) ( ) ( ) ( )+ - -

This gives

dc tdt( )

= 0.02 c(t) 0.004 q1(t) + 0.002 q2(t)



SOLUTION MANUAL 11

2.22 e Ds = 12 3! 4 5

2 2 3 3 4 4 5 5- + - + - +t

t t t tD

D D D Dss s s s

! ! !...

It is easy to calculate with long division that

1 21 2-

+

t

tD

D

s

s

//

= 12 4

2 2 3 3- + - +t

t tD

D Dss s

...

1 2 121 2 12

2 2

2 2- +

+ +

t t

t tD D

D D

s s

s s

/ // /

= 12 6

2 2 3 3- + -t

t tD

D Dss s

+ + -t tD Ds s

4 4 5 5

12 144...



CHAPTER 3 MODELS OF INDUSTRIAL CONTROL DEVICES ANDSYSTEMS

3.1

R G2G1YG +G3 4

H2

H G1 2G G + G2 3 4

+ + +

Y sR s

( )( ) =

G G G GG G G G H G H G

1 2 3 4

2 3 4 1 2 1 1 21( )

( )( )+

+ + + +

3.2

R YG1 G2

G4

+

+

+ +

H2

G3

(1 )HG1 1

Y sR s

( )( ) = G

G G GG G H G H G4

1 2 3

2 3 2 2 1 11 1+

+ + -( )

3.3Y sR s

( )( ) = ( ) /P P P P1 1 2 2 3 3 4 4D D D D D+ + +

P1 = G1G2; P2 = G1G3; P3 = G4H2G1G2;

P4 = G4H2G1G3 = 1 ( G1G3H1H2 G1G2H1H2)

1 = 2 = 3 = 4 = 1



SOLUTION MANUAL 13

Y sR s

( )( ) =

( )( )( )

11

4 2 2 3 1

1 1 2 2 3

+ +

+ +

G H G G GG H H G G

R 1 1 YG1

G2

G4

H2

H1

G3

3.4

RG2G1

YG3

1+G H3 2

H /G33

H1

+ +

Y sR s W

( )( )

=0= M(s) = G G G

G H G H G G G H1 2 3

2 3 3 2 1 2 3 11 + + +

W

G3

Y

H G23

(G1 1 2 2H G +H )+

++

Y sW s R

( )( )

= 0= MW(s) =

G H GH G G G H G H

3 3 2

3 2 3 1 1 2 2

11

( )( )

+

+ + +




3.5

Y1

Y2

G1 G2 G3

G4 G5 G6

1

1

1

1

11 1

R1

R2

H2H1

YR R

1

1 02 ==

P1 1DD

=

G G G GG G G G H G G H G G G

1 2 3 4

1 2 4 1 2 4 5 1 1 2 4

11

( )( )

+

- - - + +

YR R

1

2 01 ==

G G H G G G4 5 1 1 2 3D

YR R

2

1 02 ==

G G G G H1 4 5 6 2D

;

YR R

2

2 01= =

G G G G G4 5 6 1 21( )+D

3.6

+ +

er

et

KA Kgif ia

Ra

Kb

JsKT

1 1 w

Kt

w ( )( )s

E sr=

5010 375s + .

For Er(s) = 100/s,



SOLUTION MANUAL 15

w (s) =100 50

10 375

+s s( . )

w( )t = 481.8 (1 e10.375 t)3.7

+

qR qM qLKP KA KTsL + Rff Js + Bs21 1 1

50

q

qL

R

s

s

( )( ) =

10 1 1 0 2 1 1s s s( . )( . )+ + +

3.8 TM = K1ec + K M2 & = & && &J Beq M eq Mq q+

K2 = slope of the characteristic lines shown in Fig. P3.8b

=

3 00 300-

- = 0.01

K1 = Te

M

cM

& =0 =

3 230 20-

- = 0.1

Jeq = JM + N NN N

JL1 32 4

2FHG

IKJ = 0.0032

Beq = N NN N

BL1 32 4

2FHG

IKJ = 0.00001

qM

c

s

E s( )( ) =

100 32 1s s( . )+ ;

q L

c

s

E s( )

=

10 32 1s s( . )+

3.9 (i) Jeq = JM + JL &

&

q

qL

M

FHG

IKJ

2

= 0.6

Beq = BM + BL &

&

q

qL

M

FHG

IKJ

2

= 0.015

KT = Kb




q M

a

s

E s( )( ) =

Ks J s B sL R K K

T

eq eq a a T b[ )( ) ]+ + +

=

16 7100 19 22

.

( . )s s s+ +

(ii) A sketch of multi-loop configuration may be made on the lines ofExample 3.5.

(iii) It will behave as a speed control system.(iv) Relative stability and speed of response may become unsatisfactory.

3.10

J = JM = n2 JL = 1.5 10

5; n = 1

B = BM + n2BL = 1 10

5

M

R

s

s

( )( ) =

13 333 13 332

.

.5 .s s+ +

+ ++

KA K1

K2

1 1Ksec

Js +B

TMqR qM

Kt

s

qM

3.11

+ +

KA Kg

Kb

KTJs + B

1Ra

1 1Ks sL +Rf f

qR qMs

qM

M

R

s

s

( )( ) =

L

R

s

s

( )( ) =

15 1054 200 15 10

3

3 2 3

+ + + s s s

3.12 (a) ( )( )s

E sr=

K

s sn n

1 2 122

+ +; K = 17.2, z = 1.066, w n = 61.2

(b) Modification in the drive system may be made on the lines of Fig. 3.64.



SOLUTION MANUAL 17

+ +

KTJs + B

1

sL +Ra a

er wKA Kr

Kb

Kt

3.13

++

KP KA KT G s( )

N s( )

qL

TW

TMqR

TM = JM && ( & & ) ( )q q q q qM M L M LB K+ - + -

B KM L M L( & & ) ( )q q q q- + - = J TL L W&&q +(JMs2 + Bs + K) M(s) (Bs + K) L(s) = TM(s)(Bs + K) M (s) (JLs2 + Bs + K) L(s) = TW(s)qL s( ) = G(s) TM(s) N(s) TW(s)

q

qL

R

s

s

( )( ) =

K K K G sK K K G sP A T

P A T

( )( )1+

qL

W

s

T s( )( )- =

N sK K K G sP A T

( )( )1+

G(s) = Bs Ks J J s J J Bs J J KM L M L M L

+

+ + + +2 2[ ( ) ( ) ]




N(s) = J s Bs Ks J J s J J Bs J J K

M

M L M L M L

2

2 2+ +

+ + + +[ ( ) ( ) ]

Revisiting Example 2.7 will be helpful.

3.14 The required block diagram easily follows from Fig. 3.49.

K x Ay1 - & = K p2D

( )K p AK2 2

D= My&&

(r y) Kp KA K = xY sR s

( )( ) =

20 02 22s s+ +.

3.15 Y sY sr

( )( ) =

K K K AK KMs B A K s K K K K A K

A r

A

v

v

1 22 2

2 1 2

/( / ) /+ + + s

y y

Ms + B

+++

Kr

er

e0

ea x

yrKA

Ks

K1K2

Fw

A

A

1s1

Kv

3.16 Revisiting Example 3.7 will be helpful. From geometry of the linkage, weget

E(s) = ba b

X s aa b

Y s+

-+

( ) ( )

Y sY sr

( )( ) =

bK AK a b KMs B A K s K aAK a b K

r 1 22 2

2 1 2

/ ( )( / ) / ( )

+

+ + + + +



SOLUTION MANUAL 19

y

Ms + Bs + K2++

Kr

yr x eK1 K2

a + b

a

a + b

Ab

sA

1

3.17 Let z(t) = displacement of the power pistonY(s) = e s D- t Z(s) ; t D =

dv

sec

Y sE sr

( )( ) =

G sG s H s

( )( ) ( )1+

G(s) = ( / )( / )K K K A K es Ms B A K

As D

v 1 22

2

-

+ +

t

; H(s) = Ks

z

+ + +

Fw

K2 sMs + B

A

A

yestD

1

zxi

1K1

Ks

KAer

e0

Kv

3.18 Adhdt1

1=

q q ghRw1

1

1- -

r

A dhdt2

2=

r rghR

ghR

1

1

2

2-




+

+

R / g2 r

q1hr h2

Q sw( )

H s2 ( )Q s1( )

(t s + 1)1

t1 t2;= =

(t s + 1)2

er

e0

Ks Ka

rg rgA R1 1 A R2 2

Ke Kv

Ks

3.19 Cddttq

= h Ra

t

--q q

+

+

+

qa

q

q

1/Rt

Rth

er

ts+ 1t = R Ct t

Rf

R2Kt

R1

KhheA

R1

3.20 Heater:

C ddt1

1q = h R

--q q1

1



SOLUTION MANUAL 21

Air: V cddt

rq

=

q qr q q1

1

-- -

RQ c i( )

These equations give us

t = R1C1, Kg =1

Q cr , w nQ

R C V=

1 1

2zw n =R C Q c V c C

VR C c1 1 1

1 1

r r

r

+ +

++

+

qi

qr

q

q

h

ts+ 1

1+ 1s

2z

Kg

s2

wn2 wn

Kg

Ks KAeA

Khh

Ks

er

e0

3.21 Process:

V c ddt2 2 2 2 2

r q q( )+ = Q c UAi i2 2 2 2 2 2 2 1 1r q q q q q q( ) ( )+ - + - -

Q c2 2 2 2 2r q q( )+Linearized equation is

V c ddt2 2 2

2rq

= Q c UA Q ci2 2 2 2 2 1 2 2 2 2r q q q r q- - -( )




Rearranging this equation and taking Laplace transform yields

( ) ( )t q2 21s s+ = + K s K si1 1 2 2 ( ) ( )t 2 = 544 0 423 01 2.5; . ; .577 = =K K

Cooling water:

V c ddt1 1 1 1 1

r q q( )+ = ( ) ( ) (Q q c UAi i1 1 1 1 1 1 2 2+ + + +r q q q q- - - + +q q r q q1 1 1 1 1 1 1 1) ( ) ( )Q q c

Linearized equation is

V c ddt1 1 1

1rq

= Q c c q UAi i1 1 1 1 1 1 1 1 2 1r q q r q q+ + -( )

Q c c q1 1 1 1 1 1 1 1 From this equation, we obtain

( ) ( )t q1 11s s+ = + + K Q s K s K si3 1 4 2 5 1( ) ( ) ( )

t1 = 182 1000 4184

3550 5 4 1000 41841.

.

+ Q

K3 = ( )

.

- +

+

27 1000 41843550 5 4 1000 4184

1

1

q

Q

K4 = 3550 5 4

3550 5 4 1000 41841

+

.

. Q

K5 = Q

Q1

1

1000 41843550 5 4 1000 4184

+ .

q1 and Q1 are obtained from the following energy-balance equations atsteady state:

Q c UAi1 1 1 1 1 2 1r q q q q( ) ( )- + - = 0

Q c UAi2 2 2 2 2 2 1r q q q q( ) ( )- - - = 0These equations yield

q1 = 40C; Q1 = 5.28 103 m3/sec



SOLUTION MANUAL 23

+

+

+

+

+

+

qi1

q1

qr

q2

q2

qi2

K5

q1

q1

K2

K3

K4

e

1 1

t1s+ 1 t2s+ 1K1

K1

K1 K2 K3 K4

With these parameters,

t 1 = 184.55 sec, K3 = 1318.2, K4 = 0.465, K5 = 0.535

q

q2 ( )( )s

sr=

G sG s( )

( )1+ ; G(s) =( . )

. .

5 55 107 26 10 119 10

31 2 3 4

2 3 5

+ +

-

- -

K K K Ks s

Refer Fig. 3.57 for suitable hardware required to implement the proposedcontrol scheme.

3.22

+ +

q1

qr

q2

Controller I

Set-pointadjustment

Controller II

Sensor II

Sensor I

Process II Process I

Refer Fig. 3.57 for suitable hardware required to implement the proposedcontrol scheme.



CHAPTER 4 BASIC PRINCIPLES OF FEEDBACK CONTROL

4.1 SGM = 1

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

s s

( )( )

+

+ +

11 50

| ( )|S jGM w w=1 = 0.0289

SHM

=

-

+

G s H sG s H s( ) ( )( ) ( )1 =

-

+ +

501 50s s( )

| ( )|S jHM w w =1 = 1.02

4.2Y sR s

( )( ) = M(s) =

P1 1DD

P1 = K K

s s

3 12 1( ) ;+ D = 1

51 1 12

1 2 3 12- - +

-+

-+

FHG

IKJs s

K Ks

K Ks s( ) ( ) ; D1 = 1

M(s) = 51 5 5 5

12

1 1

Ks s K K( )+ + + +

SKM

1=

MK

KM1

1 =

s s K K ss s K K

21 1

2

21 1

1 5 5 51 5 5 5

( )( )+ + + -

+ + + +

| ( )|S jKM1 0w w= = 55 51K += 0.5

4.3 For G(s) = 20/(s + 1), and R(s) = 1/s,y t( )|open-loop = 20 (1 et)

y t( )|closed-loop = 2021

1 21( )- -e t

For G'(s) = 20/(s + 0.4), and R(s) = 1/s,y t( )|open-loop = 50 (1 e0.4t)

y t( )|closed-loop = 20

20 41 20 4

.

( ).- -e t

The transient response of the closed-loop system is less sensitive tovariations in plant parameters



SOLUTION MANUAL 25

4.4 For G(s) = 10/( ),t s + 1 and R(s) = 1/s,ess |open-loop = 0

ess |closed-loop = 1

1 10+ Kp= 0.0099

For G'(s) = 11/( )t s +1 , and R(s) = 1/s,ess |open-loop = 0.1

ess |closed-loop = 0.009The steady-state response of the closed-loop system is less sensitive tovariations in plant parameters.

4.5 t f |open-loop = LR =

250 = 0.04 sec

For the feedback system,

K e KiA f f( )- = Ldidt

R R if s f+ +( )

This gives

I sE s

f

f

( )( ) =

KsL R R K K

A

s A+ + +

t f |closed-loop = L

R R K Ks A+ + =

251 90+ K

= 0.004

This gives K = 4.99

4.6 The given block diagram is equivalent to a single-loop block diagram givenbelow,

+

0.4

10( + )K s

1W s( ) Y s( )

s + 1

s + 1

Y sW s

( )( ) =

0 411 10 1

.

s K+ +




For W(s) = 1/s,

yss = lim ( )s

sY s0

=

0 410 1

.

K + = 0.01

This gives K = 3.9

4.7 Case I:

Y sW s

( )( ) =

s s

s s K( )

( )t

t

+

+ +

11

for W(s) = 1/s, yss = 0Case II:

Y sW s

( )( ) =

Ks s K( )t + +1

For W(s) = 1/s, yss = 1The control scheme shown in the following figure will eliminate the error inCase II.

++

Kc1

1 +T sI

W s( )

Y s( )Plant

4.8 A unity-feedback configuration of the given system is shown below.

+

qrq

D s( )e 200 0.02

( + 1)s ( + 2)sG s( ) =

E ssr

( )( )q =

11+ D s G s( ) ( ) ;ess = lim ( )s sE s0 ; qr s( ) =

1s

(i) ess = 1/3(ii) ess = 0

(iii) ess = 1/3The integral term improves the steady-state performance and the derivativeterm has no effect on steady-state error.

4.9 (i) s2 + 1 = 0; oscillatory(ii) s2 + 2s + 1 = 0; stable

(iii) s3 + s + 2 = 0; unstable



SOLUTION MANUAL 27

The derivative term improves the relative stability; and the integral termhas the opposite effect.

4.10 (a) er = 50 0.03 = 1.5 volts

(b)V sE sr

( )( ) = M(s) =

0 651 5 1 0 0195

.

( )( ) .K

s s Ke

e+ + +

SKM

e=

MK

KMe

e=

( )( )( )( ) .

s s

s s

+ +

+ + +

1 5 11 5 1 9 75

(c) V(s) =325

1 5 1 9 7519 5 1

1 5 1 9 75( )( ) . ( ). ( )

( )( ) . ( )s s E ss

s sW sr

+ + +-

+

+ + +

Suppose that the engine stalls when constant percent gradient is x%. Thenfor Er(s) = 1.5/s and W(s) = x/s,

vss = lim ( )s sV sfi0 =325 1 5

10 7519 5

10 75

-.

.

.

.

x = 0

This gives x = 25

4.11 (a) The block diagram is shown in the figure below.

++

KA

KA

Kg

IL

Ra

eoer 1

sL + Rf f

(b)E sE sr IL

0

0

( )( )

==

102 1 10s K+ +

For Er(s) = 50/s,

eoss =lim ( )s

sE sfi0 0 =

5001 10+ K = 250

This gives K = 0.1

With K = 0.1, we have




Eo(s) =10

2 22 12 2s

E s ss

I sr L+

-+

+( ) ( )

When Er(s) = 50/s and IL(s) = 30/s, we get

eoss = 10 50

2302

- = 235 V

Let x V be the reference voltage required to restore the terminal voltage of250 V.

102

302

-

x= 250; this gives x = 53V

(c) Eo(s) =10

2 1sE s I sr L

+-( ) ( )

er = 25 V gives eoss = 250 V when IL = 0

er = 25 V gives eoss = 220 V when IL = 30 amps

(d) The effect of load disturbance on terminal voltage has been reduceddue to feedback action.

4.12

++

wr w0Kg KTRa JsKAKt

1 1

Kb

Kt

(a) ww

o

r

s

s

( )( )

= M(s) =

K Ks K K

A g

A g2 5( )+ +For wr(s)= 10/s,

wo(t) = 12513

1 130( )- -e t ; woss = 9.61 rad/sec

(b) When the feedback loop is opened,w

wo

r

s

s

( )( ) =

502 5

Ks

A

( )+



SOLUTION MANUAL 29

KA = 0.192 gives woss = 9.61 for wr = 10.

w o t( ) = 9.61 (1 e5t)Time-constant in the open-loop case is 1/5 sec, and in closed-loop caseis 1/130 sec; system dynamics becomes faster with feedback action

(c) SKMA =

MK

KMA

A =+

+ +

2 102 10

s

s K KA gKg = Kg, where K is a constant.

S MK

KM

MK

KM

s

s K KgM

g

g

g

g

g

g

A gw

w

w=

=

=

+

+ +

2 102 10

The open-loop transfer function of the system is

G sK K

s

A g( ) ( )= +2 5Therefore

S SKG G

A g= =1 w

S S S SKM

KG M G

A A g g<


S jKMg ( ) ~.

ww =

-0 1

44 10

(c) q Lw

s

T s( )( )- =

( ) /( ) ( )

s

s s s Kg+

+ + +

4 21 4 10

Wind gust torque on the load shaft = 100 N-m.

Therefore, on the motor shaft

Tw (s) = n 100/s = 50/s Lss = lim ( ) .

sLs s

fi=

00 1q rad

(d) Replace KA by D(s) = K T sc I1 1+FHG

IKJ to reduce the steady-state error

to zero value.

4.14 (a) qq

( )( )s

sr open loop- = G(s) = K K K K s

s s s

P A e ( )( . . )

25 10 25 0 02 12

+

+ +

q

q

( )( )s

sr closed loop-= M(s) = G s

G s( )

( )1 +

SKG = 1; SKM

= s s s

s s s K K K K sP A e

( . . )( . . ) ( )

0 25 0 02 10 25 0 02 1 25 1

2

2+ +

+ + + +

S SKM

KG 0.528

5.5 With s = $s 1, the characteristic equation becomes

$s 3 + 3K $s 2 + (K + 2) $s + 4 = 0

From the Routh array, we find that for K > 0.528, all the s-plane roots lie tothe left of the line at s = 1.

5.6 G(s) = 42 120 1

1 20 0 1K

s

s

s+

+

+ +

LNM

OQP

/ ( ).2 / ( )

H(s) = 0 054 1.

s +

1 + G(s) H(s) = 0 gives8s3 + 46s2 + 31s + 5 + 4K = 0

From the Routh array, we find that the closed-loop system is stable forK < 43.3.

5.7 (i) s3 + 5s2 + 9s + K = 0; K < 45(ii) s3 + 5s2 + (9 K) s + K = 0; K < 7.5

(iii) s4 + 7s3 + 19s2 + (18 K) s + 2K = 0; K < 10.15.8 (a) D(s) = s3 + 10s2 + (21 + K)s + 13K = 0

(i) K < 70 (ii) K = 70(iii) for K = 70, A(s) = s2 + 91 = 0D(s) = (s2 + 91) (s + 10) = (s + j9.54) (s j9.54) (s + 10)

(b) D(s) = s3 + 5s2 + (K 6)s + K = 0




(i) K > 7.5 (ii) K = 7.5(iii) For K = 7.5, A(s) = s2 + 1.5 = 0

D(s) = (s2 + 1.5) (s + 5) = (s + j1.223) (s j1.223) (s + 5)(c) D(s) = s4 + 7s3 + 15s2 + (25 + K)s + 2K = 0

(i) K < 28.1 (ii) K = 28.1(iii) For K = 28.1, A(s) = s2 + 7.58 = 0

D(s) = (s2 + 7.58) (s2 + 7s + 7.414)= (s + j2.75) (s j2.75) (s + 5.7) (s + 1.3)

5.9 (a) D(s) = s4 + 12s3 + 69s2 + 198s + 200 + K = 0From the Routh array we find that the system is stable for K < 666.25.For K = 666.25,

A(s) = 52.5s2 + 866.25 = 0This gives

s = j 4.06Frequency of oscillations is 4.06 rad/sec when K = 666.25.

(b) D(s) = s4 + 3s3 + 12s2 + (K 16) s + K = 0From the Routh array, we find that the system is stable for23.3 < K < 35.7.For K = 23.3,

A(s) = 9.57s2 + 23.3 = 0; s = j1.56For K = 35.7,

A(s) = 5.43s2 + 35.7 = 0; s = j2.56Frequency of oscillation is 1.56 rad/sec when K = 23.3; and 2.56 rad/sec when K =

35.7.

5.10 D(s) = s3 + as2 + (2 + K) s + 1 + K = 0From the Routh array we find that for the system to oscillate,

(2 + K)a = 1 + K

Oscillation frequency = 1+ Ka

= 2

These equations give a = 0.75, K = 2

5.11 D(s) = 0.02s3 + 0.3s2 + s + K = 0(a) K < 15(b) For K = 15, the auxiliary equation is A(s) = s2 + 50 = 0. The

oscillation frequency is 7.07 rad/sec at K = 15.(c) With K = 7.5 and s = $s 1,

D( $)s = 0.2 $s3 + 2.4 $s 2 + 4.6 $s + 67.8 = 0Two sign changes are there in the first column of the Routh array;therefore two roots have time-constant larger than 1 sec.

5.12 (a) Unstable for all Kc(b) For stability, TD > t.

5.13 (a) D(s) = s3 + 15s2 + 50s + 25K = 0; K < 30



SOLUTION MANUAL 41

(b) D(s) = ts3 + (1 + 5t)s2 + 5s + 50 = 0; t < 0.2

(c) D(s) = ts3 + (1 + 5t)s2 + 5s + 2.5K = 0; K < 2 10t+FH IK

5.14 D(s) = s4 + 5s3 + 6s2 + Kcs + Kca = 0; a = 1TI

From the Routh array, we find that K < (30 25a) results in systemstability. Larger the a (smaller the TI), lower is the limit on gain forstability.

5.15 D(s) = s3 + 15s2 + (50 + 100Kt)s + 100K = 0From the Routh array, we find that for stability K < (7.5 + 15Kt). Limit onK for stability increases with increasing value of Kt.



CHAPTER 6 THE PERFORMANCE OF FEEDBACK SYSTEMS

6.1 (a) Refer Section 6.3 for the derivation(b) The characteristic equation is

s2 + 10s + 100 = s2 + 2zwn s + w n2

= 0

(i) wn = 10 rad/sec; z = 0.5; wd = w n 1 2-z = 8.66 rad/sec

(ii) tr = p z

w- -cos 1

d = 0.242 sec; tp =

pw d

= 0.363 sec

Mp = e- -pz z/ 1 2

= 16.32%; ts = 4

zw n = 0.8 sec

(iii) Kp = limsfi0

G(s) = ; Kv = limsfi0

sG(s) = 10;

Ka = limsfi0

s2G(s) = 0

(iv) ess = 11+ K p = 0; ess =

1Kv

= 0.1; ess = 1

Ka =

6.2 Characteristic equation is

s2 + 10s + 10KA = s2 + 2zwn s + w n

2 = 0

(a) KA = 2.5 gives z = 1 which meets the response requirements.(b) Kp = , Kv = 2.5

ess = 5

11

++

K Kp v= 0.4

6.3 (a) Using Routh criterion, it can be checked that the close-loop system isstable.

(i) ess = 10Kv

= 10

= 0 ; (ii) ess = 10 0 2K Kav

+.

= 10 0

0 1+

.2.

= 2

(b) The closed-loop system is unstable.6.4 The closed-loop system is stable. This can easily be verified by Routh

criterion.

G(s) = 50 1 1 1 0 1( . )( )( .2 )s s s+ + +



SOLUTION MANUAL 43

ess|r = 10 = 101+ K p =

101 5+

Y sW s

( )( ) =

0 1 11 0 1 0 1 1 5

.

( )( .2 )( . )s

s s s

+

+ + + +

For W(s) = 4/s, yss = 4/6ess|total = 7/3

6.5 (a) The characteristic equation of the system is900s3 + 420s2 + 43s + 1 + 0.8 KA = 0

Using Routh criterion, we find that the system is stable for KA < 23.84.

(b) Rearrangement of the block diagram of Fig. P6.5 is shown below.

+

qr q50

1

0.016

30 1s +

10 1s +

3 1s +KA

G(s) = 0 016 503 1 30 1

.

( )( )

+ +

Ks s

A

H(s) = 110 1s +

; the dc gain of H(s) is unity.

Kp = limsfi0

G(s) = 0.8 KA ; ess = 101+ K p = 1

This gives KA = 11.25

6.6 (a) The system is stable for Kc < 9.Kp = lim

sfi0 D(s) G(s) = Kc; ess = 11+ Kc

ess= 0.1 (10%) is the minimum possible value for steady-stateerror. Therefore ess less than 2% is not possible with propor-tional compensator.




(b) Replace Kc by D(s) = 3 + Ks

I. The closed-loop system is stable for

0 < K1 < 3. Any value in this range satisfies the static accuracyrequirements.

6.7 (a) Kv= 1000

10Kc

= 1000

z = 10 10002 100+

KD = 0.5

These equations give Kc = 10 and KD = 0.09.

(b) The closed-loop poles have real part = zwn = 50. The zero ispresent at s = Kc/KD = 111.11. The zero will result in pronouncedearly peak. z is not an accurate estimate of Mp.

6.8 (a) Kv = KI = 10(b) With KI = 10, the characteristic equation becomes

D(s) = s3 + 10s2 + 100(1 + Kc)s + 1000 = 0With s = $s 1,

D( $s ) = $ $s s3 27+ + [100(1 + Kc) 17] $s + 1009 100 (1 + Kc) = 0

In Routh array formation, Kc = 0.41 results in auxiliary equation withpurely imaginary roots. Therefore Kc = 0.41 results in dominants-plane poles with real part = 1.

(c) Routh array formation for D( $s ) gives the following auxiliary equation.A( $s ) = 7 $s 2 + 868 = 0; $s = j11.14

The complex-conjugate s-plane roots are 1 j11.14.wd = wn 1

2-z = 11.14; zwn = 1

These equations give z = 0.09.

(d) D( $s ) = ( $s + 7) ( $s + j11.14) ( $s j11.14) = 0The third s-plane closed-loop pole is at s = 8. The dominancecondition in terms of third closed-loop pole is reasonably satisfied.The zero is at s = KI/Kc = 24.39. The zero will not give pronouncedearly peak. Therefore z approximately represents Mp.

6.9 (a) With TI = ,

D(s)G(s) = 80 18 802

( )++ +

T ss s

D



SOLUTION MANUAL 45

The characteristic polynomial becomes

D(s) = s2 + (8 + 80TD)s + 160TD = 0.216 gives z = 1

(b) The zero is at s = 1/0.216 = 4.63. Real part of the closed-loop pole= zwn = 12.65. Therefore, small overshoot will be observed.

(c) The characteristic equation becomes

s3 + 25.28s2 + 160s + 80TI

= 0

TI > 0.0198 for stability.

6.10 (a) G(s) = Ks s( )+1 ; Kv = limsfi0 sG(s) = K

K = 10 will give steady-state unit-ramp following error of 0.1.

D(s) = s2 + s + K = s2 + 2zwns + w n2

K = 10 gives z = 0.158.

(b) G(s) = 101 10s s Kt( )+ +

D(s) = s2 + (1 + 10Kt)s + 10 = s2 + 2zwns + w n2

Kt = 0.216 gives z = 0.5.

Kv = limsfi0

sG(s) = 101 10+ Kt

= 10

3 16.

ess = 0.316

(c) G(s) = 101 10

Ks s K

A

t( )+ +

Kv = 10

1 10K

KA

t+

D(s) = s2 + (1 + 10Kt)s + 10 KA = s2 + 2zwns + w n2

From these equations, we find that

KA = 10 and Kt = 0.9 give ess = 0.1 and z = 0.5.




6.11 (a) z2 = ( )( )In

In M

Mp

p

2

2 2+ pi

Mp = 0.1 gives z = 0.59

ts = 4

zw n = 0.05. Therefore, wn = 135.59

G(s) = 50 1 100

1

2

Ks s K( .2 )+ +

D(s) = 0.2s2 + s(1 + 100K2) + 5K1 = 0

Therefore s2 + 5(1 + 100K2)s + 25K1 = s2 + 2zwns + w n2

This equation gives K1 = 735.39 and K2 = 0.31

(b) Kp = , Kv = 5

1 1001

2

KK+

= 114.9, Ka = 0

6.12 q ( )( )s

T sw =

41 10 10s s K K st( )+ + +

For Tw(s) = 1/s, qss = 410K = 0.05

This gives K = 8

D(s) = s2 + (1 + 10Kt)s + 10K = s2 + 2zwns + w n2

Kt = 0.79 gives z = 0.5

6.13 Y sF s

( )( ) =

12Ms Bs K+ +

. For F(s) = 9/s, yss = 9K = 0.03

This gives K = 300 Newtons/m

e- -pz z/ 12

= 0 0030 03.

.

= 0.1

This gives z = 0.59

tp = p

w zn 12-

= p

w0 81. n = 2

This gives wn = 1.94 rad/sec

From the equation



SOLUTION MANUAL 47

s2 + BM

sKM

+ = s2 + 2zwns + w n2

we obtain

M = 79.71 kg, B = 182.47 Newtons per m/sec

6.14

qq

( )( )s

sR =

K J

sBJ

sKJ

T

T

/2 + +

= w

zw wn

n ns s

2

2 22+ +

+ +

qR q1

Js + Bs2KT

Tw

z2 = ( )

( )ln

ln M

Mp

p

2

2 2+ pi

For Mp = 0.25, z = 0.4

Steady-state error to unit-ramp input = 2zw n

= 0.04

This gives wn = 20

q( )( )s

T sw-=

12Js Bs KT+ +

qss = 10KT

= 0.01

From these equations, we obtain the following values of system parameters.

KT = 1000 N-m/rad

B = 40 N-m/(rad/sec)J = 2.5 kg-m2




6.15 J &w + Bw = KT(wr w)ww

( )( )s

sr=

45100 50s +

wr = 50 260p

rad/sec = step input

w(t) = 1.5p (1 e0.5t)wss = 1.5p rad/sec = 45 rpm; ess = 5 rpm

A control scheme employing gain adjustment with integral errorcompensation will remove the offset.

6.16 J&& &q q+ B = KT(qr q)q

q( )( )s

sr=

2400150 600 24002s s+ +

qr(s) = p3s ; q(s) = p3

164 162s s s( )+ +

LNM

OQP

q(t) = pz

w zzw

31

1 21-

-+

LNMM

OQPP

--e t

nt

dsin cos( )

= p p3

1 11547 3 463

2- +FH IKLNMOQP

-. sin .e tt

Mp = e- -pz z/ 1 2

= 16.3%

The following control schemes have the potential of eliminating overshoot:

(i) gain adjustment with derivative error compensation; and(ii) gain adjustment with derivative output compensation.

6.17

+ +

qR ece qLqMKs KA K1

sK2

1 1

50Js + Bs2



SOLUTION MANUAL 49

(b) G(s) = 4100K

s s

A

( .5)+Kv = 4KA/100.5

For a ramp input qR(t) = p t, the steady-state error = pKv

The specification is 5 deg or 5180

p rad.

Therefore

1004

.5pKA

= p

180 ; this gives KA = 904.5

The characteristic

equation is

s2 + 100.5s + 4 904.5 = 0

wn = 60.15; z = 0.835; Mp = 0.85%; ts = 0.0796 sec

(c) With PI control the system becomes type-2; and the steady-state errorto ramp inputs is zero provided the closed-loop system is stable.

G(s) = 4 1 1

100

Ks

s s

A +FH IK+( .5) ; KA = 904.5

The characteristic equation is

s3 + 100.5s2 + 4 904.5s + 4 904.5 = 0

It can easily be verified that the closed-loop system is stable.

(d) The characteristic equation can be expressed as(s + 1.0291) (s2 + 99.4508s + 3514.6332) = 0

or

(s + 1.0291) (s + 49.7254 + j32.2803) (s + 49.7254 j32.2803) = 0The dominance condition is satisfied because the real pole is very close tothe zero. Therefore, the transient response resembles that of a second-ordersystem with characteristic equation

s2 + 99.4508s + 3514.6332 = 0

wn = 59.28; z = 0.84; Mp = 0.772%; ts = 0.08 sec




6.18

qLqR

e

JL

LfKA

Rf

etIa

(const)

Ket

+ +

qR qM qLKA

sKKt

KP

KPe

KT1

J seq2

1

50

(c) The characteristic equation of the system is

s2 + 100 KKAs + 6KA = s2 + 2zwn s + w n2

= 0

This gives KA = 2.67; K = 0.024



SOLUTION MANUAL 51

6.19

++

qR qLKAKP KA

sKb

KTRa

1

Js + Bs21

10

(b) Open-loop transfer function G(s) = 204 202

Ks s

A

( )+

Kv = 20202

KA =

10 01.

; this gives KA = 1010.

The characteristic equation becomes

s2 + 50.5s + 5050 = 0 = s2 + 2zwn s + w n2

This gives z = 0.355; Mp = 30.33%

(c) G(s) = 204 202

( )( )K sK

s s

A D+

+

Kv = 20202

KA = 100; no effect on ess.

Characteristic equation of the system is

s2 + (50.5 + 5KD)s + 5050 = 0 = s2 + 2zwns + w n2

This gives KD = 6.95.

For z = 0.6, Mp = 9.5%

(d) System zero: s = -KK

A

D = 145.32

Real part of closed-loop poles = zwn = 42.636.

The zero will affect the overshoot. The peak overshoot will be slightlymore than 9.5%.

6.20 (a) G(s) = 30 3 1 1 2

Ks s s

A

( . )( )+ +




Kv = 3KA; ess = 0 033

.

KA =

0 01.KA

(b) Y sF sw

( )( ) =

0 3 1 20 3 1 1 2 3

. ( )( . )( )

+

+ + +

s

s s s KA

yss = 0 1.KA

(c) Characteristic equation is0.6s3 + 2.3s2 + s + 3KA = 0

For stability, KA < 1.28.

Minimum value of error in part (a) = 7.81 103

Minimum value of error in part (b) = 0.0786.21

Controller

Power amp

1

e

u i

100

x47.6

s2 16.67

(a) Characteristic equation of the system iss2 16.67s + 4760Kc = 0

Unstable or oscillatory for all Kc > 0.

(b) U(s) = Kc(1 + sTD) E(s)Characteristic equation becomes

s2 + 4760 KcTDs + 4760Kc 16.67 = 0

For stability, Kc > 3.5 103 and TD > 0.

(c) z2 = [ ( .2)][ ( .2)]ln

ln 0

0

2

2 2+ pi ; z = 0.456

ts = 4

zw n = 0.4; wn = 21.93



SOLUTION MANUAL 53

s2 + 4760 KcTDs + 4760 Kc 16.67 = s2 + 2zwn s + w n2

This gives Kc = 0.1045, TD = 0.04.

6.22 (a) System-type number = 2

(b) H sQ sw( )

( )- = 0 0 1 1

0 1 1 02.5 ( . )

( . ) .5( )s s

s s K sKc D+

+ + +

hss = -0

0.5

.5Kc ; Kc > 10


0.1s3 + s2 + 0.5KDs + 0.5Kc = 0

For stability, KD > 0.1Kc6.23

+ qqr K2 1+K1 s s + 1

Characteristic equation is

s2 + (1 + K1)s + K2 = 0 = s2 + 2zwn s + w n2

Mp = 20% fi z = 0.456; ts = 2 fi wn = 4.386

K2 = 19.237; K1 = 3



CHAPTER 7 COMPENSATOR DESIGN USING ROOT LOCUSPLOTS

7.1 (a) sA = 2 ; fA = 60, 180, 300 ; intersection with jw-axis at j5.042; angle of departure fp from 1 + j4 = 40.

(b) sA = 1.25 ; fA = 45, 135, 225, 315 ; intersection with jw-axis at j1. 1; angle of departure fp from 1 + j1 = 71.6; multiple roots ats = 2.3.

(c) Angle of arrival fz at 3 + j4 = 77.5 ; multiple roots at s = 0.45.(d) sA = 1.33 ; fA = 60; 180, 300; intersection with jw-axis at

j 5 ; angle of departure fp from 2 + j1 = 63.43; multiple rootsat s = 1, 1.67.

(e) sA = 1.5 ; fA = 90, 270(f) sA = 5.5 ; fA = 90, 270; multiple roots at s = 2.31, 5.18.(g) sA = 0.5 ; fA = 90, 270; intersection with jw-axis at j2.92.

7.2 (i) sA = 2; fA = 60, 180, 300; intersection with jw-axis at j 10 ;angle of departure fp from 3 + j1 = 71.5.Two root loci break away from s = 1.1835 at 90. Two root lociapproach s = 2.8165 at 90

jwj 10

s

3 + 1j

2

(i)



SOLUTION MANUAL 55

(ii) Intersection with jw-axis at j2 3 ; angle of departure fp from 3 + j 3 = 60.Three root loci approach the point s = 2, and then break away indirections 120 apart.

jw jw

j 3

ss

2

(ii) (iii)

9 4

3 +

60

j23

(iii) sA = 4 ; fA = 90, 270The characteristic equation has three roots at s = 3; the three rootloci originating from open-loop poles approach this point and thenbreakaway. Tangents to the three loci breaking away from s = 3 are120 apart.

(iv) sA = 1 ; fA = 45, 135, 45, 135; intersection with the jw axisat j 1.58; angle of departure fp from 1 + j2 = 90There are two roots at s = 1, two roots at s = 1 + j1.22, and tworoots at s = 1 j1.22. The break away directions are shown in thefigure.

(v) There are four roots at s = 1. The break away directions are shown inthe figure.




jw jw

s s

1 + 1j 1 + 2j

2 2

(v)(iv)

45

45

7.3 sA = 1 ; fA = 60, 180, 300; intersection with the jw-axis at j 2 ;multiple roots at s = 0.42.The z = 0.5 loci passes through the origin and makes an angle of q =cos1 z = 60 with the negative real axis. The point sd = 0.33 + j0.58 onthis line satisfies the angle criterion. By magnitude criterion, the value of Kat this point is found to be 1.04. Using this value of K, the third pole isfound ats = 2.33. Therefore,

M(s) = 1 040 33 0 0 33 0 2 33)

.

( . .58)( . .58)( .s j s j s+ + + - +7.4 sA = 3 ; fA = 45, 135, 225, 315; intersection with the jw-axis at

j3.25; departure angle fp from 4 + j4 = 225; multiple roots at s = 1.5.At the intersection of the z = 0.707 line with the root locus, the value of

K, by magnitude criterion is 130. The remaining pair of complex roots forK = 130 can be approximately located graphically. It turns out that real partof complex pair away from jw-axis is approximately four times as large asthat of the pair near jw-axis. Therefore, the transient response term due tothe pair away from the jw-axis will decay much more rapidly than thetransient response term due to the pair near jw-axis.

7.5 sA = 2.67 ; fA = 60, 180, 300 ; intersection with jw-axis at j 32 ;departure angle fp from 4 + j4 = 45.

The point sd = 2 + j3.4 on the z = 0.5 line satisfies the angle criterion.By magnitude criterion, the value of K at this point is found to be 65. Usingthis value of K, the third pole is found at s = 4; the dominance condition isnot satisfied.

7.6 sA = 2; fA = 60, 180, 300; intersection with the jw-axis at j 5 ;multiple roots at s = 0.473.



SOLUTION MANUAL 57

The least value of K to give an oscillatory response is the value of K atthe multiple roots (K = 1.128).

The greatest value of K that can be used before continuous oscillationsoccur is the value of K at the point of intersection with the jw-axis. Fromthe Routh array formulation it is found that the greatest value of K is 30 andat this gain continuous oscillations of frequency 5 rad/sec occur.

7.7 The proof given in Example 7.1.7.8 The proof follows from Example 7.1.

Draw a line from the origin tangential to the circular part of the locus. Thisline corresponds to damping ratio for maximum oscillatory response. Thetangential line corresponds to z = 0.82 and the value of K at the point oftangency is 2.1.

7.9 The proof given in Example 7.2

7.10 s = s + jw

tan1 ws

ws

++

--1 11tan = tan tan ( )- -++

+1 1

2180 2 1w

sw

sq

Taking tangents on both sides of this equation, and noting that

tan tan +

LNMOQP

1

2180

=

ws + 2

,

we obtain

ws

ws

ws

ws

++

-

-+FH IK

-FH IK

1 1

1 1 1 =

ws

ws

ws

ws

++

- FH IK +FH IK

21

2

Manipulation of this equation gives

s w-FH IK + -LNM

OQP

12

54

22

= 0

There exists a circular root locus with centre at s = , w = 0 and the radiusequal to 5 2 .

7.11 Use the result in Problem 7.7 for plotting the root locus. Complex-rootbranches form a circle.The z = 0.707 line intersects the root locus at two points; the values of K atthese points are 1 and 5. The point that corresponds to K = 5 results in lowervalue of ts; therefore we choose K = 5(sd = 3 + j3).




For z = 0.707, Mp = e- -pz z/ 12

= 4.3%

ts = 4

zw n =

43

sec

G(s) = K ss s

( )( )( )

+

+ -

42 1

E(s) = 11+ G s( ) R(s) =

( )( )( )( ) ( ) ( )

s s

s s K sR s+ +

+ - + +

2 12 1 4

For R(s) = 1/s (assuming selection of K that results in stable closed-looppoles)

ess = lim ( )s

sE sfi0

= -

- +

22 4K

For K = 5, ess = 1/9

Therefore, steady-state error is 11.11%.7.12 Breakaway points obtained from the solution of the equation dK/ds = 0,

are s = 0.634, 2.366. Two root loci break away from the real axis ats = 0.634, and break into the real axis at s = 2.366. By determining asufficient number of points that satisfy the angle criterion, it can be foundthat the root locus is a circle with centre at 1.5 that passes through thebreakaway points.

Both the breakaway points correspond to z = 1. For minimum steady-state error we should select larger value of K.

s = 0.634 corresponds to K = 0.0718

s = 2.366 corresponds to K = 14

7.13 The characteristic equation is

1 + 0 810 1 30 1 3 1

.

( )( )( )K

s s s+ + + = 0

or 1 +

+FH IK +FH IK +FH IKK

s s s1

101

3013

= 0 ; K' = 0.000888 K

sA = 0.155 ; fA = 60, 180, 300; intersection with the jw-axis at j 0.22; multiple roots at s = 0.063The point on the z = 0.707 line that satisfies the angle criterion is sd = 0.06 j0.06 ; the value of K at this point, obtained by magnitude criterion,is 0.00131. Therefore K = 1.475.



SOLUTION MANUAL 59

7.14 (a) Ideal feedback sensor:

1 + Ks s( . )0 1 1+ = 1 +

1010K

s s( )+ = 0

K = 10 gives z = 0.5.

(b) Sensor with appreciable time-constant:

1 + Ks s s( . )( . )0 1 1 0 02 1+ + = 0

or 1 + + +

Ks s s( )( )10 20 = 0 ; K = 500 K

Locate a root locus point on the real axis that gives K = 5000 (i.e., K= 10). Complex-conjugate roots can then be found by long division.The process gives z = 0.4.

7.15 Characteristic equation is given by

D(s) = s3 + 6s2 + 8s + 0.1K(s + 10) = 0which can be arranged as

1 + ++ +

K ss s s

( )( )( )

102 4

= 0 ; K = 0.1K

sA = 2 ; fA = 90, 90; intersection with the jw-axis at j 20 ; multipleroots at s = 0.8951.

The point on z = 0.5 line that satisfies the angle criterion is sd = 0.75 + j1.3. The value of K at this point, obtained by magnitude criterion,is 1.0154. Hence K = 10.154.

The third closed-loop pole is found at s = 4.5

7.16 Characteristic equation of the system is

s3 + s + 10Kts + 10 = 0

which can be rearranged as

1 + Kss s2 10+ +

= 0 ; K = 10Kt

Notice that a zero is located at the origin and open-loop poles are located ats = 0.5 j3.1225. As per the result of Problem 7.9, a circular root locusexists with the centre at zero and radius equal to 10 .

The point on the z = 0.7 line that satisfies the angle criterion is sd = 2.214 + j2.258.




The gain K corresponding to this point is 3.427. Hence the desired value ofvelocity feedback gain Kt is 0.3427.

7.17 Characteristic equation of the system is

D(s) = s(s + 1) (s + 4) + 20Kts + 20 = 0which may be rearranged as

1 + Kss j s j s( )( )( )+ - +2 2 5 = 0 ; K = 20Kt

sA = 2.5 ; fA = 90, 90; departure angle fp from s = j2 is 158.2.The following two points on z = 0.4 line satisfy the angle criterion:

s1 = 1.05 + j2.41, s2 = 2.16 + j4.97The value of K at s1 is 0.449, and at s2 is 1.4130.

The third pole corresponding to K = 0.449 is found at s = 2.9, and forK = 1.413 at 0.68.From the root locus plot, one may infer that the closed-loop pole at s = 0.68 is close to system zero at the origin. This is not the case. In fact theclosed-loop system does not have a zero at the origin:

Y sR s

( )( ) =

201 4 20 1s s s K st( )( ) ( )+ + + +

In the root locus plot, the zero at the origin was introduced because of theprocess of modifying the characteristic equation so that the adjustablevariable K = 20Kt appears as a multiplying factor.

Kt = 0.449 satisfies the dominance condition to a reasonable extent.For Kt = 1.413, complex-conjugate poles are not dominant.

7.18 The characteristic equation of the system is

s(s + 1) (s + 3) + 2s + 2a = 0which may be rearranged as

1 + Ks s s( )2 4 5+ + = 0 ; K = 2a

sA = 1.3333 ; fA = 60, 60, 180 ; intersection with the jw-axis at j 5 ; departure angle fp from complex pole in the upper half of s-plane = 63.43; multiple roots at s = 1, 1.666.

The point on z = 0.5 line that satisfies the angle criterion is sd = 0.63 + j1.09. The value of K at this point, obtained by magnitude criterion,is 4.32. Therefore a = 2.16.The third pole for K = 4.32 is at s = 2.75.



SOLUTION MANUAL 61


s3 + 9s2 + 18s + 10a = 0

which may be rearranged as

1 + Ks s s( )( )+ +3 6 = 0 ; K = 10a

sA = 3 ; fA = 60, 60, 180; intersection with the jw-axis at j3 2 ;multiple roots at s = 1.268.The point on z = 0.5 line that satisfies the angle criterion is sd = 1 + j1.732. The value of K at this point is 28. Therefore a = 2.8. The thirdpole for K = 28 is at s = 7.

7.20 (a) 1 + Ks s( )- 2 = 0

System is unstable for all K > 0.(b) In practice, the poles and zeros can never exactly cancel since they are

determined by two independent pieces of hardware whose numericalvalues are neither precisely known nor precisely fixed. We shouldtherefore never attempt to cancel poles in the right-half plane, sinceany inexact cancellation will result in an unstable closed-loop system.

(c) 1 + K ss s s

( )( )( )

+

+

12 8

= 0

From the Routh array, we find that the loci cross the jw-axis whenK = 19.2. For K > 19.2, the closed-loop poles are always in the left-half plane and the system is stable.

E sR s

( )( ) =

11+ G s( ) =

s s s

s s s Ks K( )( )

( )( )- +

- + + +

2 82 8

For R(s) = 1/s and K > 19.2 (required for stability),ess = lim

sfi0 sE(s) = 0

For R(s) = 1/s2 and K > 19.2,

ess = limsfi0

sE(s) = -16K

7.21 sA = 2 ; fA = 60, 180, 300; intersection with the jw-axis at j 5 ;multiple roots at s = 0.473.The value of K at the point of intersection of the root locus and z = 0.3 lineis 7.0.




Kv = limsfi0

sG(s) = K/5 = 1.4 ; ts = 4/zwn = 11.6 sec.

With the compensator, the characteristic equation becomes

1 + K ss s s s

( )( )( )( )

10 1100 1 1 5

+

+ + + = 0

or

1 + ++ + +

K ss s s s

( . )( . )( )( )

0 10 01 1 5

= 0 ; K = 0.1K

The value of K at the point of intersection of z = 0.3 line and the root locusis 6.0. Therefore K = 60.

Kv = limsfi0

sD(s) G(s) = K/5 = 12, ; ts = 4/zwn = 12.7 sec.

7.22 1 + K ss s s

( .5)( )( )

+

+ +

21 a

= 0

Desired closed-loop pole; sd = 1.6 + j4. Angle criterion at sd is satisfiedwhen a = 5.3. The value of K at sd is 23.2. The third pole is at s = 3.1.

7.23 (a) 1 + Ks s( )+ 2 = 0

z = 0.707 ; ts = 4

zw n = 2. Therefore, zwn = 2

Proportional control does not meet this requirement.

(b) 1 + K sKs s

c D+

+( )2 = 1 + K s K K

s s

D c D( / )( )+

+ 2

At the points of intersection of z = 0.707 line and zwn = 2 line, theangle criterion is satisfied when Kc/KD = 4.The root locus plot of

1 + K ss s

D ( )( )

+

+

42

= 0

has circular complex-root branches (refer Problem 7.7). At the pointof intersection of z = 0.707 line and zwn = 2 line, we find by magnitudecriterion that KD = 2. Therefore Kc = 8.

7.24 G(s) = K ss s s

( . )( . )

+

+ +

0 10 8 42

; K = 4000



SOLUTION MANUAL 63

D(s) = s ss s

2 0 8 40 384 10 42

+ +

+ +

.

( . )( . )D(s) improves transient response considerably, and there is no effect onsteady-state performance.

7.25 z = 0.45 (specified). Let us select the real part of the desired roots as zwn =4 for rapid settling.

The zero of the compensator is placed at s = z = 4, directly belowthe desired root location. For the angle criterion to be satisfied at the desiredroot location, the pole is placed at s = p = 10.6. The gain of thecompensated system, obtained by magnitude criterion, is 96.5. Thecompensated system is then

D(s)G(s) = 96.5 42 10 6

( )( )( . )

s

s s s

+

+ +

Kv = limsfi0

sD(s) G(s) = 18.2.

The Kv of the compensated system is less than the desired value of 20.Therefore, we must repeat the design procedure for a second choice of thedesired root.

7.26 Desired root location may be taken as

sd = 1.588 + j3.152

It meets the requirements on Mp and settling time. With D(s) = ss

+

+

110

,

angle criterion is met. Magnitude criterion gives K = 147. The compensatedsystem has Kv = 2.94.

7.27 z = 0.707; 4/zwn = 1.4 fi zwn = 2.85

wn = 4 ; w zn 12- = 2.85

sd = 2.85 + j2.85

With D(s) = ( .5)( .5)s

s

+

+

13

2

2 , angle criterion is satisfied at sd. The value of K,

found by magnitude criterion, is 37.

7.28 From the root locus plot of the uncompensated system, we find that for gainof 1.06, the dominant closed-loop poles are at 0.33 j 0.58. The value ofz is 0.5 and that of wn is 0.66. The velocity error constant is 0.53.

The desired dominant closed-loop poles are 0.33 j0.58 with a Kv of5. The lag compensator




D(s) = ss

+

+

0 10 001

.

.

gives Kv boost of the factor of 10. The angle contribution of thiscompensator at 0.33 + j0.58 is about seven degrees. Since the anglecontribution is not small, there will be a small change in the new root locusnear the desired dominant closed-loop poles. If the damping of the newdominant poles is kept at z = 0.5, then the dominant poles from the new rootlocus are found at 0.28 j 0.51. The undamped natural frequency reducesto 0.56. This implies that the transient response of the compensated systemis slower than the original.

The gain at 0.28 + j0.51 from the new root locus plot is 0.98.Therefore K = 0.98/1.06 = 0.925 and

D(s) = 0 0 10 01

.925( . ).

s

s

+

+

7.29 The dominant closed-loop poles of uncompensated system are located ats = 3.6 j4.8 with z = 0.6. The value of K is found as 820.Therefore Kv = lim

sfi0 sG(s) = 820/200 = 4.1.

Lag compensator D(s) = ss

+

+

00 025

.25.

gives a Kv boost of the factor of 10.

The angle contribution of this compensator at 3.6 + j4.8 is 1.8, which isacceptable in the present problem.

7.30 (a) 1 + Ks s( )( .5 )2 1 0 1+ + = 1 +

Ks s( .5)( )+ +0 2 = 0

From the root locus we find that no value of K will yield zwn = 0.75.

(b) D(s) = KT sc I

1 1+FHGIKJ = Kc + KI /s = Kc

s K Ks

I c+FH IK/

D(s)G(s) = K s K Ks s s

c I c( / )( .5)( )

+

+ +0 2

At the point corresponding to z = 0.6 and zwn = 0.75, the angle criterion issatisfied when KI /Kc = 0.75.

The gain at the desired point is Kc = 2.06. This yields KI = 1.545.7.31 Kv = 20 demands K = 2000. However, using Routh criterion, we find that

when K = 2000, the roots of the characteristic equation are at j10. Clearlythe roots of the system when Kv requirement is satisfied are a long wayfrom satisfying the z requirement. It will be difficult to bring the dominant



SOLUTION MANUAL 65

roots from the jw-axis to the z = 0.707 line by using a lead compensator.Therefore, we will attempt to satisfy the Kv and z requirements by a lagcompensator.

From the uncompensated root locus we find that corresponding toz = 0.707, the dominant roots are at 2.9 j2.9 and the value of K is 236.Therefore necessary ratio of zero to pole of the compensator is

| || |z

p =

2000236

= 8.5

We will choose z = 0.1 and p = 0.1/9. For this choice, we find that the anglecontribution of the compensator at 2.9 + j.2.9 is negligible.

7.32 D(s)G(s) = K s ss s s s s

( / )( / )( )( )( / )( / )

+ +

+ + + +

1 11 5 1

1 2

1 2

t tat a t

11t

= 0.05 ; 12t

= 1 ; 11at

= 0.005 ; at 2

= 10

D(s)G(s) = K ss s s s

( . )( . )( )( )

+

+ + +

0 050 005 5 10

For z = 0.45, we obtain

sd = 1.571 + j3.119, K = 146.3, Kv = 29.3, wn = 3.49

7.33 D(s) = Kc + KDs = Kc 1+FHG

IKJ

KK

sD

c

(a) D(s)G(s) = K K

Ks

s

cD

c

1

12

+FHG

IKJ

+

s2 + 1 + Kc + KDs = (s + 1 + j 3 ) (s + 1 j 3 )This equation gives Kc = 3 and KD = 2

Kp = limsfi0

D(s)G(s) = 3.

(b) With PID controller

D(s) = K s ss

( .2)( . )+ +0 1 3332




the angle criterion at sd = 1 + j 3 is satisfied. The gain K is 2.143.7.34 (a) Uncompensated system:

Mp = 20% fi z = 0.45

At the point of intersection of z = 0.45 line and the root locus, zwn =1.1 and K = 56

(b) ts = 4/zwn = 3.6 secKv = lim

sfi0 sG(s) = 2.07

(c) Compensated system:Mp = 15% fi z = 0.55

ts = 3 62.

.5 = 1.45 ; zwn = 4/1.45 = 2.75

wn = 2 750.

.55 = 5 ; sd = 2.75 + j4.18

Kv 20

Lead-lag compensated system:

D1(s) D2(s) G(s) = K s ss s s s s

( .9)( . )( )( . )

+ +

+ + + +

3)( 00 15 3)( 9 13 6 ; K = 506

7.35 G(s) = Ks s KKt( )+ +2

E sR s

( )( ) =

11+ G s( ) =

s s KKs s KK K

t

t

( )( )

+ +

+ + +

22

For R(s) = 1/s2,

ess = 2 0 35+ KK

Kt

.

z 0.707 ; ts 3 sec.

The characteristic equation of the system is

s2 + 2s + KKts + K = 0

which may be rearranged as



SOLUTION MANUAL 67

1 + KK s

s s Kt

2 2+ + = 0

The locus of the roots as K varies (set KKt = 0) is determined from thefollowing equation:

1 + Ks s( )+ 2 = 0

For K = 20, the roots are 1 j4.36.Then the effect of varying KKt is determined from the locus equation

1 + KK ss j s j

t

( . )( . )+ + + -1 4 36 1 4 36 = 0

Note that complex root branches follow a circular path.At the intersection of the root locus and z = 0.707 line, we obtain KKt =

4.3. The real part of the point of intersection is s = 3.15, and therefore thesettling time is 1.27 sec.


1 + KK s

s s Kt

2 2+ + = 0

or

1 + KK ss j K s j K

t

( ( )( ( )+ + - + - -1 1 1 1 = 0

The root contour plotted for various values of K with Kt = = KKt varyingfrom 0 to are shown in the figure below.




jw

s

1j

2j

1j

K =t 0

K =t

Kt

K =t 0

K =t 0

K = 2

K = 5

K =t 0

j2

5 2 1

7.37 G(s) = AKs s s KK st( )( )+ + +1 5


s(s + 1) (s + 5) + KKt (s + A/Kt) = 0which may be rearranged as

1 + KK s A Ks s s

t t( / )( )( )

+

+ +1 5 = 0

The angle criterion at s = 1 + j2 is satisfied when A/Kt = 2.5. Let us takeA = 2 and Kt = 0.8.

Kv = limsfi0

sG(s) = AKKKt5 +

With K = 10, Kv = 1.54

The closed-loop transfer function of the system is

Y sR s

( )( ) =

201 2 1 2 4( )( )( )s j s j s+ + + - +


s(s + 1) (s + 5) + KKt s + AK = 0



SOLUTION MANUAL 69

or

s(s + 1) (s + 5) + bs + a = 0 ; b = KKt ; a = AKThe root locus equation as a function of a with b = 0 is

1 + a

s s s( )( )+ +1 5 = 0

Sketch a root locus plot for a varying from 0 to . The points on these locibecome the open-loop poles for the root locus equation

1 + ba

s

s s s( )( )+ + +1 5 = 0

For some selected values of a, the loci for b are sketched (refer Fig. 7.44).

7.39 es = - -+

( )ss

22

1 K ss s

( )( )( )

-

+ +

21 2

= 0 = 1 F(s)

F(s) = K jj j( )

( )( )s w

s w s w+ -

+ + + +

21 2

= K j

j( )

( )( ) (s w

s s w w s- +

+ + - + +

21 2 2 3)2

F(s) = tan1 ws

w ss s w-

-+

+ + -

-

22 3)

1 21

2tan(

( )( )

= tan1

ws

w ss s w

ws

w ss s w

--

+

+ + -

+-

+

+ + -

LNM

OQP

RS||

T||

UV||

W||

( )22 3)

1 2

12

2 3)1 2

2

2

(( )(

( )( )

F(s) = 0 ifw

sw s

s s w--

+

+ + -22 3)

1 2 2(

( )( ) = 0

Manipulation of this equation gives

(s 2)2 + w2 = 12Centre = (2, 0) and radius = 12 = 3.464




It can easily be verified that at every point on this circle, F(s) = 0.Revisiting Example 7.14 will be helpful.

7.40 The proof runs parallel to that in Example 7.2, with a change in anglecriterion as K 0.

Centre at zero at s = 0

Radius = a b2 2+ = ( .5) ( .5)0 02 2+ = 0.707The root locus plot is shown in the figure below.

j0.5

0.5

Kfi Kfi

K = 0

K = 0

7.41 G(s) = 20 7 3)2 8

. (( )( )

s

s s s

+

+ +

(i) a = a0 + Da = 8 + DaThe characteristic equation is

s(s + 2) (s + 8) + Da s(s + 2) + 20.7 (s + 3) = 0When Da = 0, the roots are (may be determined by root-locus methodor the Newton-Raphson method)

l1,2 = 2.36 j2.48, l3 = 5.27The root locus for Da is determined using the root locus equation

1 + Dal l l

s s

s s s

( )( )( )( )

+

- - -

21 2 3

= 0



SOLUTION MANUAL 71

The angle of departure at s = l1, is 80.

Near s = l1, the locus may be approximated by a line drawn from 2.36 + j2.48 at an angle of 80. For a change of Dl1 = 0.2 80along the departure line, Da is determined by magnitude criterion:

Da = 0.48

Therefore the sensitivity at l1 is

Sal+1

= D

D

la a

1

0/ =

0 800 48 8.2

. / -

= 3.34 80

(ii) Sensitivity of the root s = l1 to a change in zero at s = 3 is determinedas follows.

b = b0 + Db = 3 + Db


s(s + 2) (s + 8) + 20.7 (s + 3 + Db) = 0or

1 + 20 71 2 3

.

( )( )( )Db

l l ls s s- - - = 0

The angle of departure of l1 is 50.

For a change of Dl1 = 0.2 + 50, we obtain Db = 0.21.Therefore

Sbl1

= D

D

lb b

1

0/ =

0 500 3

.2 .21 / +

= 2.84 + 50

The sensitivity of the system to the pole can be considered to be lessthan the sensitivity to the zero because of the direction of departurefrom the pole at s = l1.

7.42 (a) The characteristic equation is

1 + K ss s s

( )( )(

+

+ +

52 3) = 0

It can easily be verified from the root locus plot that for K = 8 theclosed-loop poles are

l1,2 = 0.5 j 3.12, l3 = 4(b) s(s + 2 + d) (s + 3) + 8(s + 5) = 0




or

1 + d s ss s s s

( )( )( ) ( )

+

+ + + +

32 3 8 5

= 0

or 1 + dl l l

s s

s s s s

( )( )( )( )

+

- - -

31 2 3

= 0

Case I: d > 0

From the root locus plot we find that the direction of departure fromthe pole at s = l1, is away from the jw-axis.Case II: d < 0

From the root locus plot we find that the direction of departure fromthe pole at s = l1 is towards the jw-axis. Therefore the variation d < 0is dangerous.



CHAPTER 8 THE NYQUIST STABILITY CRITERIONAND STABILITY MARGINS

8.1 (a) Revisiting Example 8.1 will be helpful.

(i) 1 0 (ii) 0 180 (iv) t tt t

1 2

1 290

+ -

(b) Revisiting Example 8.2 will be helpful.(i) t j (ii) 0 180

(c) Revisiting Example 8.3 will be helpful.(i) 180 (ii) 0 270

(d) (i) (t1 + t2) j (ii) 0 270

(iii) t tt t

1 2

1 2180

+ -

(e) (i) -180 (ii) 0 0 (iv) t t t tt t

1 2 1 2

1 290

( )

+ -

8.2 (a) Number of poles of G(s) in right half s-plane, P = 1Number of clockwise encirclements of the critical point, N = 1Z = number of zeros of 1 + G(s) in right half s-plane = N + P = 2The closed-loop system is unstable.

(b) P = 2Number of counterclockwise encirclements of the critical point = 2

Therefore N = number of clockwise encirclements = 2Z = N + P = 2 + 2 = 0

The closed-loop system is stable.(c) P = 0

N = (Number of clockwise encirclements of the critical point numberof counterclockwise encirclements of the critical point) = 0Z = N + P = 0; the closed-loop system is stable.

8.3 The polar plot of G(jw) for w = 0+ to w = + is given in Fig. P8.3b. Plot ofG(jw) for w = to w = 0 is the reflection of the given polar plot withrespect to the real axis. Since

G j( )w w=0 = 180,G(s) has double pole at the origin. The map of Nyquist contour semicircles = r ejf, r fi 0, f varying from 90 at w = 0 through 0 to + 90 at w =0+, into the Nyquist plot is given by 2f (an infinite semicircletraversed clockwise).




With this information, Nyquist plot for the given system can easily bedrawn.(i) P = 0, N = 0; Z = N + P = 0

The closed-loop system is stable(ii) P = 1, N = 0; Z = N + P = 1

The closed-loop system is unstable(iii) P = 0, N = 0; Z = N + P = 0

The closed-loop system is stable.8.4 (a) The key points to the polar plot are:

G j H j( ) ( )w ww =

= 18 0

G j H j( ) ( )w ww =

= 0 270

The intersections of the polar plot with the axes of G(s)H(s)-plane caneasily be ascertained by identifying the real and imaginary parts ofG(jw)H(jw). When we set Im [G(jw)H(jw)] to zero, we get w = 4.123and

G j H j( ) ( ).

w w w=4 123 = 1.428

Similarly, setting Re [G(jw)H(jw)] to zero, we get intersection withthe imaginary axis.

Based on this information, a rough sketch of the Nyquist plot caneasily be made. From the Nyquist plot, we find that N = 2. Since P = 0,we have Z = N + P = 2, i.e., two closed-loop poles in right half s-plane.

(b) The key points of the polar plot are (refer Problem 8.1d):G j H j( ) ( )w w

w =0 = 3 j

G j H j( ) ( )w ww =

= 0 270

Intersection with the real axis at w = 1/ 2 ;

G j H j( ) ( )w ww =

12

= -43

The map of Nyquist contour semicircle s = re jf, r fi 0, f varyingfrom 90 at w = 0 through 0 to + 90 at w = 0+, into the Nyquist plotis given by f (an infinite semicircle traversed clockwise).

Based on this information, a rough sketch of Nyquist plot can easilybe made.

P = 0, N = 2. Therefore Z = N + P = 2



SOLUTION MANUAL 75

(c) G j H j( ) ( )w w w=0 = 2 180G j H j( ) ( )w w w= = 0 90No more intersections with real/imaginary axis. From the Nyquist plot,we find that N = 1. Since P = 1, we have Z = N + P = 0; the closed-loop system is stable.

(d) G j H j( ) ( )w w w=0 = 0 90 ; G j H j( ) ( )w w w=1 = 2.6161 ;G j H j( ) ( )w w w=2 = 2.6 198 ; G j H j( ) ( )w w w=10 = 0.8 107

G j H j( ) ( )w w w= = 0 90The Nyquist plot is given in Fig. P8.2b.

P = 2, N = 2 ; Z = N + P = 0

(e) G j H j( ) ( )w ww =0 = 180

G j H j( ) ( )w ww =

= 0 90

No intersections with real/imaginary axis.The map of Nyquist contour semicircle s = re jf, r fi 0, f varyingfrom 90 at w = 0 through 0 to + 90 at w = 0+, into the Nyquist plotis given by 2f (an infinite semicircle traversed clockwise).With this information, Nyquist plot for the given system can easily bedrawn.

P = 0, N = 0 ; Z = N + P = 0

(f) G j H j( ) ( )w ww =0 =

1100

0

G j H j( ) ( )w ww=

-10 = 0

Consider the Nyquist contour shown in figure below. For the semi-circle s = j10 + rejf, r fi 0, f varying from 90 at w = 0 through 0to + 90 at w = 10+,

G(s)H(s) = limrfi0

110 1002( )j e j+ +r f = lim ( )r f fr rfi +0

120e j ej j

= limr

f

rfi-

0

1e

j

It is an infinite semicircle from w = 10 to w = 10+ traversed clockwise.

G j H j( ) ( )w ww =

= 0 180




The Nyquist plot is shown in the figure below.The Nyquist plot passes through 1 + j0 point. The Nyquist criterionis not applicable.

1 + 0j

10j

10j

s

j

j

Re

Im

w =

jw

w = 0+w = 10+

Nyquistplot

Nyquistcontour

100

1

(g) G j H j( ) ( )w ww =0 = 8 + j

G j H j( ) ( )w w w= = 0 90

G j H j( ) ( )w w w= 3 = 2 + j0No intersection with the imaginary axis.

The map of Nyquist contour semicircle s = rejf, r fi 0, f varyingfrom 90 at w = 0 through 0to + 90 w = 0+, into the Nyquist plot isgiven by f (an infinite semicircle traversed clockwise).

With this information, a rough sketch of Nyquist plot can easily bemade (refer Fig. P8.2a).P = 1, N = 1. Therefore Z = N + P = 0.

8.5 G(jw)H(jw) = - ++

+-

+

K j K( )( )

( )( )

11 1

21 2

2 212

1 22 2

12

w t t

w w t

w t t

w w t

(i) For t1 < t2, the polar plot of G(jw)H(jw) is entirely in the thirdquadrant as shown in figure below.



SOLUTION MANUAL 77

P = 0, N = 0. Therefore Z = N + P = 0; the closed-loop system is stable.(ii) For t1 > t2, the polar plot of G(jw)H(jw) is entirely in the second

quadrant, as shown in figure below.

P = 0, N = 2. Therefore Z = N + P = 2; the closed-loop system isunstable.

Re

(i) (ii)

Im

w = w = w =

w =

w = 0

w = 0

w = 0+

w = 0+

8.6 (a) G j( )ww=0 = 40 ; G j( )w w= = 0 270

Intersection with the real axis at 0.8.The critical point 1 + j0 will be encircled by the Nyquist plot ifK > 1/0.8. Since P = 0, we want net encirclements of the critical pointto be zero for stability. Therefore K must be less than 5/4.

(b) G(s) = 4 11 0 12( )

( . )+

+

s

s s = +

+

K ss s

( )( )t

t2

21

11

; t1 < t2

The Nyquist plot for this transfer function has already been given inthe Solution of Problem 8.5. From this plot we see that the closed-loopsystem is table for all K.

(c) G(s) = 4 1 0 112

( . )( )+

+

s

s s =

+

+

K ss s

( )( )t

t2

21

11

; t1 > t2

The Nyquist plot for this transfer function has already been given inthe Solution of Problem 8.5. From this plot we see that the closed-loopsystem is unstable for all K.




(d) G(jw) = ejj-

+

0 8

1

. w

w

= 1

1 2+w [(cos 0.8w w sin 0.8w) j(sin 0.8w + w cos 0.8w)]

The imaginary part is equal to zero if

sin 0.8w + w cos 0.8w = 0

This gives

w = tan 0.8w

Solving this equation for smallest positive value of w, we get w =2.4482.

G j( )ww=0 = 1 0 ; G j( ) .w w=2 4482 = 0.378 + j0

G j( )ww=

= 0

The polar plot will spiral into the w fi point at the origin (refer Fig.8.39).

The critical value of K is obtained by letting G(j 2.4482) equal 1.This gives K = 2.65. The closed-loop system is stable for K < 2.65.

8.7 The figure given below shows the Nyquist plot of G(s)H(s) for K = 1. Asgain K is varied, we can visualize the Nyquist plot in this figure expanding(increased gain) or shrinking (decreased gain) like a balloon.

Since P = 2, we require N = 2 for stability, i.e., the critical point mustbe encircled ccw two times. This is true if 1.33 K < 1 or K > 0.75.

Re

1.33

1

1

Im

w =

w = 0+w =11



SOLUTION MANUAL 79

8.8 10K

G j H j( ) ( )w ww=

= 1 + j0 ; 1K

G j H j( ) ( )w ww=

= 0.05 + j0

10 62K

G j H j( ) ( ).

w ww=

= 0.167 + j0

The Nyquist plot is shown in the figure below. If the critical point liesinside the larger loop, N = 1. Since P = 1, we have Z = N + P = 2 and theclosed-loop system is unstable.

If the critical point lies inside the smaller loop, N = 1, Z = N + P = 0and the closed-loop system is stable. Therefore, for stability 0.167K < 1 or K > 6.

Re

0.167

1

Im

w =

w = 0+

8.9 Eliminating the minor-loop we obtain forward-path transfer function of anequivalent single-loop system.

G(s) = K ss s

( .5)++ +

013 2

By Routh criterion, we find that the polynomial (s3 + s2 + 1) has two rootsin the right half s-plane. Therefore P = 2.

10K

G j( )ww=

= 0.50 ; 1K

G j( )ww=

= 0 180




11 4K

G j( ).

ww=

= 0.5 + j0

The polar plot intersects the positive imaginary axis; the point of

intersection can be found by setting real part of 1K

G j( )w , equal to zero.

With this information, Nyquist plot can easily be constructed. From theNyquist plot we find that the critical point is encircled twice in ccw if 0.5 K < 1 or K > 2. For this range of K, the closed loop system is stable.The figure given below illustrates this result.

Re 0.5 0.5

0.5j

1.5j

1j

Im

w = 1.4 w = 0+w =

1

KG j( ) plotw

8.10 Fig. P8.10a:

G(s) = Kes s s

s-

+ +

2

1 4 1( )( )

10K

G j( )ww=

= 90; 10 26K

G j( ).

ww=

= 2.55 + j0

1K

G j( )ww =

= 0



SOLUTION MANUAL 81

The polar plot will spiral into the w fi point at the origin (refer Fig.8.39).It is found that the maximum value of K for stability is 1/2.55 = 0.385

Fig. P8.10b:

D(s)G(s) = Kes s

s-

+

2

2 4 1( )

10K

D j G j( ) ( )w ww=

= 180 ; 1K

D j G j( ) ( )w ww =

= 0

The polar plot will spiral into the w fi point at the origin. A rough sketchof the Nyquist plot is shown in figure below. It is observed that the systemis unstable for all values of K.

Re

Im

w = 0+

8.11 Revisiting Example 8.15 will be helpful.

1 + G(s) = 0may be manipulated as

e

s s

s D-

+

t

( )1 = 1 or G1(s) = es Dt

; G1(s) = 1 1s s( )+

The polar plot of G1(jw) intersects the polar plot of e j Dw t at a pointcorresponding to w = 0.75 at the G(jw)-locus. At this point, the phase of thepolar plot of e j Dw t is found to be 52. Therefore,




0.75tD = 52(p/180)This gives

tD = 1.2 sec

8.12 G j( )ww =0 = 2.2. j ; G j( )w w= = 0 270

G j( ).

w w=15 9 = 0.182 + j0; G j( ) .w w=6 2 = 1 148.3GM = 1/0.182 = 5.5 ; FM = 31.7

wg = 6.2 rad/sec ; wf = 15.9 rad/sec.

8.13 (a) This result can easily proved using Routh criterion.(b) General shape of the Nyquist plot of given transfer function is as

shown in Fig. P8.2c. When K = 7, the polar plot intersects the negativereal axis at the point 1.4 + j0. The resulting Nyquist plot encirclesthe critical point once in clockwise direction and once in counter-clockwise direction. Therefore N = 0. Since P = 0, the system is stablefor K = 7. Reducing the gain by a factor of 1/1.4 will bring the systemto the verge of instability. Therefore GM = 0.7.

The phase margin is found as +10.

8.14 (a) The system is type-0; so the low-frequency asymptote has a slope of0 dB/decade, and is plotted at dB = 20 log 25 = 27.96.

Asymptote slope changes to 20 dB/decade at the first cornerfrequency wc1 = 1; then to 40 dB/decade at wc2 = 10, and to 60 dB/decade at wc3 = 20.

Asymptotic crossing of the 0 dB-axis is at wg = 16.(b) The system is type-1; so the low-frequency asymptote has a slope of

20 dB/decade and is drawn so that its extension would intersect the0 dB-axis at w = K = 50. The asymptote slope changes to 40 dB/decade at the first corner frequency at wc1 = 1; then to 20dB/decade at wc2 = 5, and back to 40 dB/decade at wc3 = 50. Theasymptotic gain crossover is at wg = 10.2.

(c) The system is type-2; the low-frequency asymptote is drawn with aslope of 40 dB/decade, and is located such that its extension wouldintersect the 0 dB-axis at w = K = 500 = 22.36. The asymptoteslope changes to 60 dB/decade at the first corner frequency wc1 = 1;then to 40 dB/decade at wc2 = 5, to 20 dB/decade at wc3 = 10 and to 40 dB/decade at wc4 = 50. The asymptotic gain crossover is atwg = 10.

(d) The system is type-1, and the low-frequency asymptote has a slope of 20 dB/decade and is drawn so that its extension would cross the



SOLUTION MANUAL 83

0 dB-axis at w = K = 50. The asymptote slope changes to40 dB/decade at wc1 = 10; then to 20 dB/decade at wc2 = 20; to40 dB/decade at wc3 = 50; and to 80 dB/decade at wc4 = 200 (notethat the corner frequency for the complex poles is wc4 = wn = 200).The asymptotic gain crossover is at wg = 26.

8.15 (a) The low-frequency asymptote has a slope of 20 dB/decade andpasses through the point (w = 1, dB = 20). The asymptote slopechanges to 40 dB/decade at wc = 10.Compensate the asymptotic plot by 3 dB at w = 10, by 1dB at w =5 and by 1 dB at w = 20. The compensated magnitude plot crossesthe 0 dB line at wg = 7.86.

The phase is computed from

G(jw) = 90 tan1 0.1wThe phase shift at wg = 7.86 is 128.2, and the resulting phase marginis 128.2 ( 180) = 51.8. Because the phase curve never reaches 180 line, the gain margin is infinity.

(b) Low-frequency asymptote has a slope of 20 dB/decade and passesthrough the point (w = 1, dB = 20)The asymptote slope changes to 60 dB/decade at the cornerfrequency wc = 10. Compensate the asymptotic plot by 6 dB at w =10, by 2 dB at w = 5 and 2 dB at w = 20. The compensatedmagnitude plot crosses the 0 dB line at wg = 6.8. The phase iscomputed from

G(jw) = 90 2 tan1 0.1wThe phase shift at wg = 6.8 is 158.6. Therefore, the phase margin is180 158.6 = 21.4. At a frequency of wf = 10, the phase shift is 180; the gain margin is 6 dB.

(c) The low-frequency asymptote has a slope of 20 dB/decade and passesthrough the point (w = 0.1, dB = 20log 200 = 46). The slope changes to 40 dB/decade at the first corner frequency wc1 = 2, and back to 20dB/decade at the second corner frequency wc2 = 5. Compensate theasymptotic plot by 3 dB at w = 2, by 1 dB at w = 1, by 1 dB at w= 4, by + 3 dB at

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Control Systems by Nagrath & Gopal Solution Manual