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 sKA

s( ) =

+t 1

y(t) =KA

e t

tt- / ; yss = 0

(b) R(s) =A

s; Y(s) = KA

s s( )τ + 1

y(t) = KA( )/1− −e t τ ; yss = KA

(c) R(s) =A

s2 ; Y(s) = KAs s2 1( )τ +

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

(d) R(s) =A

s

w

w2 2+; Y(s) = KA

s sω

τ ω( )( )+ +1 2 2

y(t) = KA e KA ttωττ ω τ ω

ω θτ2 2 2 21 1+

++

+− / sin( )

θ = tan ( )− −1 ωτ

y tt

( )®¥

= KA tτ ω

ω θ2 2 1+

+sin( )

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

y(t) =11

21

2

12

−−

+ −FHG

IKJ

−−e t

nt

d

ζω

ζω ζ

ζsin tan

ω d = ω ζn 1 2−

The 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 s®

=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

211

2ζω ω

ω ζζ

ζωsin tan

yss = tn

− 2ζω

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 s

E s

R

RCsi

0410

1

( )

( )

/=

+ ; 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 × 10–6 sec, itis necessary that

R10

14 = , andt = = ´ -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 25 252

t tn

- - ´FHG

IKJ

z

w

= 1.5

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

Y(s) =1

10 1002s s s( )+ +

ω n = 10, ζ = 0.5, wd = 5 3

Using Eqns (2.57)-(2.58), we obtain



4 CONTROL SYSTEMS: PRINCIPLES AND DESIGN

y(t) = 0.01 12

35 3 35 1- +

LNM

OQP

- -e tt sin( tan

2.6Y s

R sG s

s s s s

( )

( )( )

( )( )= =

+ +=

+ +

6

7 6

6

1 62

| ( )|G jω ω = 2 = 35 2

; ∠G(j2) = – 81.87º

yss = 35 2

2 81 87sin( . º )t −

= 3

50 sin 2t –

21

50 cos 2t

2.7 Y sR s

( )( )

= G(s) = ss s

++ +

37 102 = s

s s+

+ +3

2 5( )( )

(a) R(s) = 1

1s+

y(t) = 12

13

16

2 5e e e tt t t− − −− −FH IKµ( )

(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–) = 1

2, r (0–) = 0

Y(s) = ss s

ss s s

++ +

+ ++ + +

15 22 5

32 5 1

/( )( ) ( )( )( )

y(t) = 1

2

3

22 5e e et t t- - -+ -

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

Md X x

dt

2

2( )+

= F X x I i Mg( , )+ + −



SOLUTION MANUAL 5

Mx&& = F X I FX

x FI

i MgX I X I

( ), ,

+ + ∂∂

FHG

IKJ + ∂

∂FHG

IKJ −

F X I( )+ = Mg = × ×−8 4 10 9 83. . Newtons

For 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 I

1 0 14=¶

¶=

,. Newtons/cm

KF

I X I2 0 4=

¶

¶=

,

. Newtons/amp

&& ;( )

( )

.

.x

K

Mx

K

Mi

X s

I s s= + =

-1 2

247 6

16 67

2.9 Mx Bx Kx K y x&& & ( )+ + = -1

Gravitational effect has been eliminated by appropriate choice of the zeroposition.

X sY s

( )( )

= G ss s

( ) .( . )( . )

=+ +0 1667

0 0033 1 0 0217 1

w = 2 1163p lv / .= rad /sec

| ( | .G jω ω =11 63 = 0.1615

x(peak) = 7.5 × 0.1615 = 1.2113 cm

2.10 (a) Mx Bx Kx F t&& & ( )+ + =


X s

F s

( )

( ) = 1

2Ms Bs K+ +

Force transmitted to the ground

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




=

BK

s F s

MK

s BK

s

+FH IK+ +

1

12

( )

F(t) = A sin w t

Peak amplitude of the force transmitted to the ground at steady-state

=A B

K

MK

BK

1

1

2

2 2 2

+ FH IK

−FHG

IKJ + FH IK

ω

ω ω

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

X sY s

( )( )

= Bs KMs Bs K

++ +2

Peak amplitude of machine vibration

= A B

Kv

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&& ( ) ( & & ) ( ) ( & & )+ − + − + − + − = 0

M y K y y B y y2 2 2 2 1 2 2 1&& ( ) ( & & )+ − + − = 0


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

(a)E s

E si

0( )

( ) =1 2

1 22

21 2

2

1 22

1 22

+ +

+ + +

RC s R C C s

R C C s R C C s( )

(b)E s

E si

0( )

( ) =1 2

1 21 1 2

2 2

1 2 1 22 2

+ +

+ + +

R Cs R R C s

R 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 0

4 4537 0 0 0

0 0 0 1

0 0 0 0

.

.5809

L

N

MMMM

O

Q

PPPP; b =

0

0 3947

0

0

−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 q

qi = 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 cd

dtr

q= 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 Cd

dt11θ

= q tR

RUA

m( ) ;lq q

--

=1 2

11

1

Cd

dt22q =

q q q q1 2

1

2

2

-+

-

R Ri ; C V c R

Q c2 21

= =rr

;

From these equations, we get

&θ1 = − + +1 1 4 461 2.92 .92 .θ θ qm

&θ 2 = 0 078 0 2 0 1251 2. . .q q q- + i




2.18 At steady-state

0 =q q q q1 2

1

2

2

-+

-

R Ri ; this gives θ1 120= ºC

0 = Qmλ − θ θ1 2

1

−R

; this gives Qm =1726. / minkg

2.19 Energy balance on process fluid:

V c ddt2 2 2 2 2ρ θ θ( )+ = Q c UAi2 2 2 2 2 2 2 2 1 1ρ θ θ θ θ θ θ θ[ ( )] [ ]− + − + − −

At steady-state

0 = Q c UAi2 2 2 2 2 2 1r q q q q( ) ( );- - - this gives q1 40= ºC


V cd

dtQ c UA2 2 2

22 2 2 2 2 1r

qr q q q- - -( )

This gives

544 5 022 1. .432+ + =

d

dt

qq q

Energy balance on the cooling water:

V cd

dt1 1 1 1 1r 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 ×10–3 m3/sec


V cd

dt1 1 11r

q= ( ) ( )θ θ ρ ρ θ θ θi c q Q c UA1 1 1 1 1 1 1 1 1 2 1− − + −



SOLUTION MANUAL 9

This gives

184.55 d

dt

qq1

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 s

2.20 Tank 1:

Cdp

dt11 = q1 – q10 – q11

q10 = flow through R0; q11 = flow through R1

A

gg

dh

dt1 1

rr´ = q

gh

R

g h h

R11

0

1 2

1- -

-r r ( )

ordh

dt1 = - +

FHG

IKJ + +

r rg

A R Rh

g

A Rh

Aq

1 0 11

1 12

11

1 1 1

= – 3h1 + 2h2 + r1

Tank 2:

Cdp

dt22 = q2 + q11 – q20

q20 = flow through R2

A

gg

dh

dt2 2

rr´ =

r rg h h

R

gh

Rq

( )1 2

1

2

22

-- +

ordh

dt2 =

r rg

A Rh

g

A R Rh

Aq

2 11

2 1 22

22

1 1 1- +

FHG

IKJ +

= 4h1 – 5h2 + r2

2.21 Ad

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

At steady-state

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





Adh

dt= 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 Hf H

HH H

H H

( )( )

( )+¶

¶

FHG

IKJ -

=

= QK

Hh t+

2( )

Therefore

q(t) =K

Hh t

2( ) =

K H

Hh t

2( ) =

Q

Hh t

2( )

dh t

dt

( ) = - + +

1 1 11 2A

q tA

q tA

q 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 t

At steady state,

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


AHdc t

dtAC

dh tdt

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

This gives

dc t

dt

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



SOLUTION MANUAL 11

2.22 e Ds−τ = 12 3! 4 5

2 2 3 3 4 4 5 5

- + - + - +tt t t t

DD D D Ds

s s s s

! ! !...

It is easy to calculate with long division that

1 2

1 2

-

+

t

tD

D

s

s

/

/= 1

2 4

2 2 3 3

- + - +tt t

DD Ds

s s...

1 2 12

1 2 12

2 2

2 2

- +

+ +

t t

t tD D

D D

s s

s s

/ /

/ /= 1

2 6

2 2 3 3

- + -tt t

DD Ds

s s

+ + -t tD Ds s4 4 5 5

12 144...



CHAPTER 3 MODELS OF INDUSTRIAL CONTROL DEVICES ANDSYSTEMS

3.1

R G2G1

YG +G3 4

H2

H G1 2

G G + G2 3 4

+ + +

–– –

Y s

R s

( )

( ) =G G G G

G G G G H G H G1 2 3 4

2 3 4 1 2 1 1 21

( )

( )( )

+

+ + + +

3.2

R YG1 G2

G4

+

+

+ +

–

–

H2

G3

(1– )HG1 1

Y s

R s

( )

( ) = GG G G

G G H G H G41 2 3

2 3 2 2 1 11 1+

+ + -( )

3.3Y s

R 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 s

R s

( )

( ) = ( )( )

( )

1

14 2 2 3 1

1 1 2 2 3

+ +

+ +

G H G G G

G 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 s

R 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 s

W s R

( )

( )= 0

= MW(s) =G H G

H G G G H G H3 3 2

3 2 3 1 1 2 2

1

1

( )

( )

++ + +




3.5

Y1

Y2

G1 G2 G3

G4 G5 G6

1

–1

–1

1

11 1

R1

R2

H2H1

Y

RR

1

1 02==

P1 1D

D

=G G G G

G G G G H G G H G G G1 2 3 4

1 2 4 1 2 4 5 1 1 2 4

1

1

( )

( )

+

- - - + +

Y

R R

1

2 01==

G G H G G G4 5 1 1 2 3

D

Y

R R

2

1 02 ==

G G G G H1 4 5 6 2

D;

Y

RR

2

2 01= =

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

D

3.6

+ +

– –

er

et

KA Kg

if iaRa

Kb

JsKT

1 1 w

Kt

w ( )

( )

s

E sr=

50

10 375s+ .

For Er(s) = 100/s,



SOLUTION MANUAL 15

w (s) =100 50

10 375

´

+s s( . )

w( )t = 481.8 (1 – e–10.375 t)

3.7

+

–

qRqM qL

KP KA KTsL + Rff Js + Bs21 1 1

50

q

qL

R

s

s

( )

( ) =1

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

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

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

= 3 0

0 300

-

- = – 0.01

K1 = ∆∆T

eM

cM

&θ =0

= 3 2

30 20

-

- = 0.1

Jeq = JM + N N

N NJL

1 3

2 4

2FHG

IKJ = 0.0032

Beq = N N

N NBL

1 3

2 4

2FHG

IKJ = 0.00001

qM

c

s

E s

( )

( ) = 10

0 32 1s s( . )+; q L

c

s

E s

( ) =

1

0 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

( )

( )=

K

s J s B sL R K KT

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

= 16 7

100 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

θθ

M

R

s

s

( )

( )= 13 33

3 13 332.

.5 .s s+ +

+ ++

– ––

KA K1

K2

1 1Ksec

Js +B

TMqR qM

Kt

s

qM

3.11

+ +

––

KA Kg

Kb

KT

Js + B1

Ra

1 1Ks sL +Rf f

qRqM

s

qM

θθ

M

R

s

s

( )

( )=

θθ

L

R

s

s

( )

( )= 15 10

54 200 15 10

3

3 2 3×

+ + + ×s s s

3.12 (a)ω( )

( )s

E sr

= K

s sn n

1 212

2

ωζ

ω+ +

; 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

+ +

––

KT

Js + 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) θM(s) – (Bs + K) θL(s) = TM(s)

(Bs + K) θM (s) – (JLs2 + Bs + K) θL(s) = TW(s)

qL s( ) = G(s) TM(s) – N(s) TW(s)

q

qL

R

s

s

( )

( )=

K K K G s

K K K G sP A T

P A T

( )

( )1+

qL

W

s

T s

( )

( )-=

N s

K K K G sP A T

( )

( )1+

G(s) = Bs K

s J J s J J Bs J J KM L M L M L

+

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




N(s) = J s Bs K

s J J s J J Bs J J KM

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 pA

K22

D = My&&

(r – y) Kp KA K = x

Y s

R s

( )

( ) = 2

0 02 22s s+ +.

3.15Y s

Y sr

( )

( )=

K K K AK K

Ms B A K s K K K K A KA 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) = b

a bX s

a

a bY s

+-

+( ) ( )

Y s

Y sr

( )

( )=

bK AK a b K

Ms B A K s K aAK a b Kr 1 2

2 22 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 piston

Y(s) = e s D- t Z(s) ; t D = d

v sec

Y s

E sr

( )

( )=

G s

G s H s

( )

( ) ( )1+

G(s) = ( / )

( / )

K K K A K e

s Ms B A KA

s Dv 1 2

22

-

+ +

t

; H(s) = Ks

z

+ + +

– –

–

Fw

K2sMs + B

A

A

ye–st

D1

zxi

1K1

Ks

KAer

e0

Kv

3.18 Adh

dt11

= q qgh

Rw11

1

- -r

Adh

dt22

= r rgh

R

gh

R1

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 rg

A R1 1 A R2 2

Ke Kv

Ks

3.19 Cd

dttq

= hR

a

t

--q q

+

+

+

–

qa

q

q

1/Rt

Rth

er

ts+ 1

t = R Ct t

Rf

R2Kt–

R1

KhheA

R–1

3.20 Heater:

Cd

dt11q = h R

--q q1

1



SOLUTION MANUAL 21

Air: V cd

dtr

q= q q

r 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 KA

eAKh

h

Ks

er

e0

3.21 Process:

V cd

dt2 2 2 2 2r 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 cd

dt2 2 22r

q= 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 2θ θ( ) ( )

t 2 = 544 0 423 01 2.5; . ; .577′ = ′ =K K

Cooling water:

V cd

dt1 1 1 1 1r 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 cLinearized equation is

V cd

dt1 1 11r

q= 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 1ρ θ θ ρ

From this equation, we obtain

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

t1 = 182 1000 4184

3550 5 4 1000 41841

.

.

´ ´

´ + ´ ´Q

′K3 = ( )

.

- + ´ ´

´ + ´ ´

27 1000 4184

3550 5 4 1000 41841

1

q

Q

′K4 = 3550 5 4

3550 5 4 1000 41841

´

´ + ´ ´

.

. Q

′K5 = Q

Q1

1

1000 4184

3550 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( ) ( )- - - = 0

These equations yield

q1 = 40ºC; Q1 = 5.28 × 10–3 m3/sec



SOLUTION MANUAL 23

+

+

+

+

+

+

–

–

qi1

q1

qr

q2

q2

qi2

K¢5

q1

q1

K¢2

K¢3

K¢4

e

1 1

t1s+ 1 t2s+ 1K¢

1

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 s

G s

( )

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

. .

5 55 10

7 26 10 119 10

31 2 3 4

2 3 5

´

+ ´ + ´

-

- -

K K K K

s 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

( )

( )

+

+ +

1

1 50

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

SHM =

-

+

G s H s

G s H s

( ) ( )

( ) ( )1 =

-

+ +

50

1 50s s( )

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

4.2Y s

R s

( )

( )= M(s) =

P1 1D

D

P1 = K K

s s3 1

2 1( );

+ D = 1

5

1 1 121 2 3 1

2- -+

-+

-+

FHG

IKJs s

K K

s

K K

s s( ) ( ); D1 = 1

M(s) = 5

1 5 5 51

21 1

K

s s K K( )+ + + +

SKM

1= ¶

¶´

M

K

K

M1

1 = s s K K s

s s K K

21 1

2

21 1

1 5 5 5

1 5 5 5

( )

( )

+ + + -

+ + + +

| ( )|S jKM

1 0w w= = 5

5 51K += 0.5

4.3 For G(s) = 20/(s + 1), and R(s) = 1/s,

y t( )|open-loop = 20 (1 – e–t)

y t( )|closed-loop = 20

211 21( )- -e t

For G'(s) = 20/(s + 0.4), and R(s) = 1/s,

y t( )|open-loop = 50 (1 – e–0.4t)

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 = 0.009

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

4.5 t f |open-loop = L

R=

2

50= 0.04 sec

For the feedback system,

K e KiA f f( )- = Ldi

dtR R if

s f+ +( )

This gives

I s

E sf

f

( )

( ) = K

sL R R K KA

s A+ + +

t f |closed-loop = L

R R K Ks A+ + =

2

51 90+ K = 0.004

This givesK = 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 s

W s

( )

( )=

0 4

11 10 1

.

s K+ +




For W(s) = 1/s,

yss = lim ( )s

sY s→0

= 0 4

10 1

.

K + = 0.01

This gives K = 3.9

4.7 Case I:

Y s

W s

( )

( )=

s s

s s K

( )

( )

t

t

+

+ +

1

1

for W(s) = 1/s, yss = 0

Case II:

Y s

W s

( )

( ) = K

s s K( )t + +1

For W(s) = 1/s, yss = 1

The 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 s

sr

( )

( )q =

1

1+ D s G s( ) ( );ess = lim ( )

ssE s

→0; qr s( ) =

1

s

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

The 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 s

E sr

( )

( ) = M(s) =0 65

1 5 1 0 0195

.

( )( ) .

K

s s Ke

e+ + +

SKM

e=

¶

¶´

M

K

K

Me

e=

( )( )

( )( ) .

s s

s s

+ +

+ + +

1 5 1

1 5 1 9 75

(c) V(s) =325

1 5 1 9 75

195 1

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

. ( )

( )( ) .( )

s sE s

s

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 s®0

=325 1 5

10 75

19 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 s

E sr IL

0

0

( )

( )=

= 10

2 1 10s K+ +

For Er(s) = 50/s,

eoss = lim ( )s

sE s®0

0 =500

1 10+ K = 250

This gives K = 0.1

With K = 0.1, we have




Eo(s) =10

2 2

2 1

2 2sE s

s

sI sr L

+-

+

+( ) ( )

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

eoss = 10 50

2

30

2

´- = 235 V

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

10

2

30

2

´-

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

++

––

wrw0

Kg KTRa JsKAKt

1 1

Kb

Kt

(a) w

wo

r

s

s

( )

( )= M(s) =

K K

s K KA g

A g2 5( )+ +

For wr(s)= 10/s,

wo(t) = 125

131 130( )- -e t

; woss = 9.61 rad/sec

(b) When the feedback loop is opened,

w

wo

r

s

s

( )

( ) = 50

2 5

K

sA

( )+



SOLUTION MANUAL 29

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

w o t( ) = 9.61 (1 – e–5t)

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) SKM

A =

¶

¶´

M

K

K

MA

A =+

+ +

2 102 10

s

s K KA g

Kg = Kωg, where K is a constant.

SM

K

K

MMK

K

M

s

s K Kg

M

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

sA g

( )( )

=+2 5

Therefore

S SKG G

A g= =1 w

S S S SKM

KG M G

A A g g< <; w w

4.13

+ + +

– –

–

– KA

KP

KP Kg

sKb

KT

Tw

Ra J s2eq

qLqR

qM = qC

1

sL + Rf f

1 1 1

2

(b) q

qL

R

g

g

s

sM s

K

s s s K

( )

( )( )

( ) ( )= =

+ + +

5

1 4 10

S MK

K

M

s s s

s s sKM

g

g

g=¶

¶´ =

+ +

+ + + ´

( ) ( )

( ) ( )

1 4

1 4 10 100




S jKM

g( ) ~

.w

w =

-´0 1

44 10

(c)q L

w

s

T s

( )

( )- =

( ) /( ) ( )

ss 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

θLss = lim ( ) .s

Ls s®

=0

0 1q rad

(d) Replace KA by D(s) = KT sc

I

1 1+

FHG

IKJ to reduce the steady-state error

to zero value.

4.14 (a)q

q

( )( )ssr open loop-

= G(s) = K K K K s

s s sP A e ( )

( . . )

25 1

0 25 0 02 12

+

+ +

q

q

( )( )s

sr closed loop-

= M(s) = G s

G s( )

( )1+

SKG = 1; SK

M = s s s

s s s K K K K sP A e

( . . )

( . . ) ( )

0 25 0 02 1

0 25 0 02 1 25 1

2

2

+ +

+ + + +

S SKM

KG<

(b) A PD control scheme with proportional controller in the forward path

and derivative action realized by feeding back &q , will increase aircraftdamping.

4.15 (a)q ( )

( )s

T sw

= 1

1

/ B

s JB

ssK K

BR

K K

BRb T

a

T A

a

+FH IK + +

For Tw(s) = Kw/s,

θss= lims® 0

sθ (s) = K R

K Kw a

T A



SOLUTION MANUAL 31

(b)q

q

( )( )ssr

= M (s) = K K BR

sJB

ssK K

BR

K K

BR

T A a

b T

a

T A

a

/

+FH

IK + +1

SJM =

¶

¶´

M

JJM

= -

+FH IK + +

JB

s

sJB

ssK K

BR

K K

BRb T

a

T A

a

2

1

(c) Excessive large magnitudes of signals at various levels in a controlsystem can drive the devices into nonlinear region of operation. Therequirement of linear operation of devices under various operatingconditions imposes a constraint on the use of large values of KA.

4.16 (a)dV

dt = qi − q; V = C h(t), q = h(t)/R

(b)

+

–

hr qi qhR 1RRCs + 1

K

(c)H sH sr

( )( )

= M (s) = KRRCs KR+ +1

SRM = ¶

¶´ =

+ +

MR

RM RCs KR

11

SRM

s = 0 = 1

1+ KR

(d) For Hr(s) = 1/s, hss = KRKR

ess1+; = 1

1+ KR

Replacing K by KT sc

l

1 1+

FHG

IKJ reduces the steady-state error to zero; and the

system becomes insensitive to changes in R under steady dc conditions.

4.17 V ddt

C c Q q C c Q C ci i i i( ) ( ) ( ) ( )+ = + + − +

The perturbation dynamics is given by

tdc t

dt( )

+ c(t) = C

Qq t

Q

Qc t V

Qi

ii

i( ) ( );+ =t

Block diagram of the closed-loop system is shown below.




+

+

–

qi 1 c

ts + 1K C /Qi

ci

Q /Qi

C sC si

( )( )

= M sQ Q

s KC Qi

i

( )/

/=

+ +τ 1

For Ci(s) = A/s,

css = AQ

Q KCi

i+

With integral control, css becomes zero.

4.18 (b) Open-loop case:

(i)q( )

( )s

E sr

= G(s) = KR

R CsSK

G1

1 11

+=;

+K

+

+

–

q

qi

er K

1/R1

Kt

er

et

KR C1 s + 1

R1



SOLUTION MANUAL 33

(ii)q

q

( )( )ssi

= = =1

1 1R Cs

s si

i ss. ( ) / ,Forq q

(iii) τ = R1C

Closed-loop case:

(i)q( )

( )s

E sr

= M(s) = KR

R Cs KK Rt

1

1 11+ +; S

R Cs

R Cs KK RKM

t

=+

+ +1

1 1

1

1

(ii)q

q

( )( )ssi

= 1

11 1R Cs KK Rt+ +. For θ i(s) = 1/s, θss =

11 1+ KK Rt

(iii) τ = R C

KK Rt

1

11 +

(c) K(er − et ) = C ddt R R

RUA

R R C

R Ri aq q q q q

t+-

+-

= =+1 2

21 2

1 2

1; ,

+K

+

++

+

–

q

qaqi

er K

R1

Kt

erK

R2

( + ) (R R1 2 ts + 1)

R R1 2

1 1

q

q

( )( )ssa open loop-

= R

R R s1

1 2 1( ) ( )+ +t; q ss open loop- = R1 (R1 + R2)

θθ

( )( )ssa closed loop−

= 1

1 2 1 2( ) ( 1) t

R

R R s KK R Rt+ + +

θ ss closed loop− = R

R R KK R Rt

1

1 2 1 2+ +

4.19 (a)Y sW s

( )( )

= 0 1 1

2 3 11252. ( )

..

s

s s

+

+ + For W(s) = 1/s, yss = 0.089




(b) Replace Kc by KT sc

l

1 1+

FHG

IKJ

(c)Y sW s

s

s s

K K

sf c( )

( )

. ( )

.=

+

+ +-

+

LNM

OQP

0 1 1

2 3 11251

12

For W(s) = 1/s, yss = 0.089 (1 − Kf Kc)

yss = 0 when Kf = 1/Kc = 0.8

The PI control scheme increases the order of the system; this makes it lessstable.

The feedforward scheme does not affect the characteristic roots of thesystem. A difficulty with feedforward compensation is that it is an open-loop technique; it contains no self-correcting action. If the value of Kc is notaccurately known, the gain Kf will not cancel the disturbance completely.The usefulness of a method must be determined in the light of the specificperformance requirements. If the uncertainty is large, self-correcting actioncan be obtained by using PI control law in conjunction with the feedforwardcompensation.

4.20

++

––

efif w1

Js B+sL + R + 1f f

KAKtr

Kt

KT

Time-constant (without current feedback) = Lf /(Rf + 1)

I s

E sf

f

( )

( ) =

K

sL R KA

f f A+ + + 1

Time-constant (with current feedback) = L

R Kf

f A+ + 1

Position control system is shown below.



SOLUTION MANUAL 35

+ + +

– – –

w wr Ktif 1

s

Kt

KP

KP

KA

sL + R + 1f f Js B+

4.21 (a) s2 + Kc TDs + Kc = (s + 1 + j1.732) (s + 1 − j 1.732)

This gives Kc = 4, TD = 0.5

(b) Alternative control scheme is shown below.

+ +

– –

qqr q11ssK1

K2

q

q

( )( )

; ,ss

K

s K s KK K

r

=+ +

= =12

2 11 24 2

4.22 Fig. P4.22a:

q

q

( )( )

( )ss

M s Ks s Kr

= =+ +

255 252

S MK

KM

s s

s sKM =

¶

¶´ =

+

+ +

( )5

5 252

S jKM ( ) .w

w ==

51 41

Fig. P4.22b:

q

q

( )( )ssr

= M (s) = KK

s KK KKs

22

1 21+ + +( ) =

25

5 252

K

s s K+ +

This gives K2 = 25 and K1 = 4




M (s) = 25

1 4 252K

s K s K+ + +( )

SKM =

s ss s

( )+

+ +

15 252

| ( )|S jKM w w=5 ≈ 1

This shows the superiority of the two-loop system over a single-loopsystem.

—-



CHAPTER 5 CONCEPTS OF STABILITY AND THE ROUTHSTABILITY CRITERION

5.1 (a) All the elements in the first column of the Routh array are +ve.Therefore, all the roots are in the left-half plane.

(b) Two sign changes are found in the first column of the Routh array.Therefore, two roots are in the right-half plane and the rest in the left-half plane.

(c) s3-row of the Routh array has zero pivot element, but the entire row isnot all zeros. We replace the pivot element by e and then proceed withthe construction of the Routh array. As e ® 0, two sign changes arefound in the first column of the Routh array. Therefore, two roots arein right-half plane and the rest in the left-half plane.

(d) s1-row of the Routh array is an all-zero row. Auxiliary polynomialformed using the elements of s2-row is given by

A(s) = s2 + 1We replace the elements of s1-row with the coefficients of

dA sds( )

= 2s + 0

and proceed with the construction of the Routh array. There are nosign changes in the resulting Routh array; the characteristicpolynomial does not have any root in the right-half plane. The roots ofthe 2nd-order auxiliary polynomial are therefore purely imaginary.The given characteristic equation has two roots on the imaginary axisand the rest in the left-half plane.

(e) Since all the coefficients of the given characteristic polynomial arenot of the same sign, the system is unstable. The Routh arrayformation is required only if the number of roots in the right-half planeare to be determined.Only one sign change (the s0-row has – ve pivot element) is found inthe Routh array. Therefore one root is in the right-half plane and therest in the left-half plane.

(f) s3-row of the Routh array is an all-zero row. The auxillary polynomialis

A(s) = 9s4 + 0s2 + 36We replace elements of s3-row with the coefficients of

dA sds( )

= 36s3 + 0s

The resulting Routh array has s2-row with zero pivot element but theentire row is not all zeros. We replace the pivot element with e andproceed with the construction of the Routh array. Two sign changesare found in the first column; the characteristic polynomial D(s) has




two roots in the right-half plane. Three possibilities exist: (i) the 4th-order auxiliary polynomial has all the four purely imaginary roots andthe two right-half plane roots are contributed by the factor D(s)/A(s),(ii) the 4th-order auxiliary polynomial has complex roots withquadrantal symmetry (two roots in the right-half plane) and the factorD(s)/A(s) has all the roots is the left-half plane, and (iii) the 4th-orderauxiliary polynomial has two real-axis roots and two imaginary-axisroots with quadrantal symmetry and the factor D(s)/A(s) has one rootin the right-half plane. Examining A(s) we find that auxiliary-polynomial roots are s = – 1 ± j1, 1 ± j1.The given system, therefore, has two roots in the right-half plane andthe rest in the left-hand plane.

5.2 (a) s1-row of the Routh array is an all-zero row. The auxiliary polynomialisA(s) = s2 + 9

By long division

D(s)/A(s) = (s4 + 2s3 + 11s2 + 18s + 18)/A(s) = s2 + 2s + 2

Therefore

D(s) = (s2 + 9) (s2 + 2s + 2)

= (s + j3) (s – j3) (s + 1 + j1) (s + 1 – j1)(b) s3-row of the Routh array is an all-zero row. The auxiliary polynomial

is

A(s) = s4 + 24s2 – 25

D(s)/A(s) = (s5 + 2s4 + 24s3 + 48s2 – 25s – 50)/A(s)

= s + 2

D(s) = (s + 2) (s4 + 24s2 – 25)

= (s + 2) (s2 – 1) (s2 + 25)

= (s + 2) (s + 1) (s – 1) (s + j5) (s – j5)(c) s3-row of the Routh array is an all-zero row. The auxiliary polynomial

is

A(s) = s4 + 3s2 + 2

D(s)/A(s) = (s6 + 3s5 + 5s4 + 9s3 + 8s2 + 6s + 4)/A(s)

= s2 + 3s + 2

D(s) = (s2 + 3s + 2) (s4 + 3s2 + 2)

= (s2 + 3s + 2) (s2 + 1) (s2 + 2)

= (s + 1) (s + 2) (s + j1) (s - j1) (s + j 2 ) (s – j 2 )



SOLUTION MANUAL 39

5.3 With s = $s – 1, the characteristic equation becomes

$s3 + $s

2 + $s + 1 = 0

$s1-row in the Routh array is an all-zero row. The auxiliary polynomial is

A( $s) = $s2 + 1 = ($s + j1) ($s – j1)

We replace the elements of $s2-row with the coefficients of

dA sds( $)$

= 2$s + 0

and proceed with the construction of the Routh array. There are no signchanges in the resulting Routh array. The s-polynomial does not have rootsto the right of the line at s = –1, and there are two roots at s = – 1 ± j1. Thelargest time-constant is therefore 1 sec.

5.4 (a) Unstable for all values of K(b) Unstable for all values of K(c) K < 14/9(d) K > 0.528

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

$s3 + 3K $s2 + (K + 2)$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) =4

2 120 1

1 20 0 1K

ss

s+

+

+ ´ +

LNM

OQP

/ ( ).2 / ( )

H(s) =0 054 1

.s+

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

8s3 + 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 < 10.1

5.8 (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 = 0

From the Routh array we find that for the system to oscillate,

(2 + K)a = 1 + K

Oscillation frequency = 1+ K

a = 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 $s2 + 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 = 1

TI

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 = 0

From 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 = p

w d = 0.363 sec

Mp = e- -pz z/ 1 2

= 16.32%; ts = 4

zw n

= 0.8 sec

(iii) Kp = lims®0

G(s) = ¥ ; Kv = lims®0

sG(s) = 10;

Ka = lims®0

s2G(s) = 0

(iv) ess = 11+ K p

= 0; ess = 1

Kv = 0.1; ess = 1

Ka

= ¥

6.2 Characteristic equation is

s2 + 10s + 10KA = s2 + 2zwn s + w n2 = 0

(a) KA = 2.5 gives z = 1 which meets the response requirements.

(b) Kp = ¥ , Kv = 2.5

ess = 5

1

1

++

K Kp v= 0.4

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

(i) ess = 10

Kv =

10¥

= 0 ; (ii) ess = 10 0 2

K 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 Routhcriterion.

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



SOLUTION MANUAL 43

ess|r = 10 = 101+ Kp

= 101 5+

Y sW s

( )( )

= 0 1 11 0 1 0 1 1 5

.( )( .2 )( . )

ss s s

+

+ + + +

For W(s) = 4/s, yss = 4/6

ess|total = 7/3

6.5 (a) The characteristic equation of the system is

900s3 + 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 50

3 1 30 1

.

( )( )

´

+ +

K

s sA

H(s) = 110 1s+ ; the dc gain of H(s) is unity.

Kp = lims®0

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 = lims®0

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 + K

sI . The closed-loop system is stable for

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

6.7 (a) Kv= 1000

10

Kc = 1000

z = 10 1000

2 100

+

´

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 = 0

With 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( $s) = 7$s2 + 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) = 0

The 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 1

8 802

( )+

+ +

T s

s sD



SOLUTION MANUAL 45

The characteristic polynomial becomes

D(s) = s2 + (8 + 80TD)s + 160

TD = 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 = lims®0

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 = lims®0

sG(s) = 101 10+ Kt

= 10316.

ess = 0.316

(c) G(s) = 10

1 10

K

s s KA

t( )+ +

Kv = 10

1 10

K

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+ π

Mp = 0.1 gives z = 0.59

ts = 4zw n

= 0.05. Therefore, wn = 135.59

G(s) = 5

0 1 1001

2

K

s 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

K

K+ = 114.9, Ka = 0

6.12q ( )

( )s

T sw

= 4

1 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

6.13Y sF s

( )( )

= 12Ms Bs K+ +

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

= 0.03

This gives K = 300 Newtons/m

e- -pz z/ 1 2

= 0 0030 03..

= 0.1

This gives z = 0.59

tp = p

w zn 1 2- =

pw0 81. n

= 2

This gives wn = 1.94 rad/sec

From the equation



SOLUTION MANUAL 47

s2 + BM

s KM

+ = s2 + 2zwns + w n2

we obtain

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

6.14

qq

( )( )ssR

= K J

s BJ

sK

J

T

T

/

2 + + =

wzw w

n

n ns s

2

2 22+ +

+ +

–

–qR q1

Js + Bs2KT

Tw

z2 = ( )

( )

ln

ln

M

Mp

p

2

2 2+ π

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-= 1

2Js 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

( )( )ssr

= 45100 50s+

wr = 50 × 260p rad/sec = step input

w(t) = 1.5p (1 – e–0.5t)

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)

qq

( )

( )

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 2

1--

+LNMM

OQPP

--e t

nt

dsin cos( )

= p p3

1 11547 3 463

2- +FH IKLNM

OQP

-. 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) = 4

100

K

s sA

( .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

= p180

; 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


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) = 0

The 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

et

Ia(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

KT

Ra

1

Js + Bs2

1

10

(b) Open-loop transfer function G(s) = 20

4 202

K

s sA

( )+

Kv = 20

202

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) = 20

4 202

( )

( )

K sK

s sA D+

+

Kv = 20

202

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 = -K

KA

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) = 3

0 3 1 1 2

K

s s sA

( . )( )+ +




Kv = 3KA; ess = 0 033.KA

= 0 01.KA

(b)Y sF sw

( )( )

= 0 3 1 2

0 3 1 1 2 3. ( )

( . )( )+

+ + +

ss s s KA

yss = 0 1.KA

(c) Characteristic equation is

0.6s3 + 2.3s2 + s + 3KA = 0

For stability, KA < 1.28.

Minimum value of error in part (a) = 7.81 × 10–3

Minimum value of error in part (b) = 0.078

6.21

–Controller

Power amp

1

e

u i

100

x47.6

s2– 16.67

(a) Characteristic equation of the system is

s2 – 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 × 10–3 and TD > 0.

(c) z2 = [ ( .2)]

[ ( .2)]ln

ln 0

0

2

2 2+ π ; 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 = - 00

.5.5Kc

; Kc > 10


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

For stability, KD > 0.1Kc

6.23

–

+ qqr K2 1+K1 s s + 1

Characteristic equation is

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

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

K2 = 19.237; K1 = 3

——



CHAPTER 7 COMPENSATOR DESIGN USING ROOT LOCUSPLOTS

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

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

(c) Angle of arrival fz at – 3 + j4 = 77.5º ; multiple roots at s = – 0.45.

(d) – sA = –1.33 ; fA = 60º; 180º, 300º; intersection with jw-axis at

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

(e) – sA = – 1.5 ; fA = 90º, 270º

(f) – sA = – 5.5 ; fA = 90º, 270º; multiple roots at s = – 2.31, – 5.18.

(g) – sA = 0.5 ; fA = 90º, 270º; intersection with jw-axis at ± j2.92.

7.2 (i) – sA = – 2; fA = 60º, 180º, 300º; intersection with jw-axis at ± j 10 ;angle of departure fp from – 3 + j1 = – 71.5º.

Two root loci break away from s = – 1.1835 at ± 90º. Two root lociapproach s = – 2.8165 at ± 90º

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 = – 60º.Three root loci approach the point s = – 2, and then break away indirections 120º apart.

jw jw

j ÷3

ss

– 2

(ii) (iii)

– 9 – 4

– 3 +

60°

j2÷3

(iii) – sA = – 4 ; fA = 90º, 270ºThe 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 are120º apart.

(iv) – sA = – 1 ; fA = 45º, 135º, – 45º, – 135º; intersection with the jw axisat ± j 1.58; angle of departure fp from – 1 + j2 = – 90ºThere 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 = 60º, 180º, 300º; 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 =cos–1 z = 60º 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 = 45º, 135º, 225º, 315º; intersection with the jw-axis at± j3.25; departure angle fp from – 4 + j4 = 225º; multiple roots at s = – 1.5.

At the intersection of the z = 0.707 line with the root locus, the value ofK, 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 = 60º, 180º, 300º ; intersection with jw-axis at ± j 32 ;departure angle fp from – 4 + j4 = – 45º.

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 = 60º, 180º, 300º; 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 and

at 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

tan–1 ws

ws

++

--1 11tan = tan tan º ( )- -++

± +1 1

2180 2 1w

sw

sq

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

tan tan º−

+±L

NMOQP

1

2180

ωσ

= 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

2

12

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 = ½, w = 0 and the radius

equal 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/ 1 2

= 4.3%

ts = 4zw n

= 43

sec

G(s) = K s

s s( )

( )( )+

+ -

42 1

E(s) = 11+G s( )

R(s) = ( )( )

( )( ) ( )( )

s ss s K s

R s+ +

+ - + +

2 12 1 4

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

ess = lim ( )s

sE s®0

= -

- +

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

.( )( )( )

Ks s s+ + +

= 0

or 1 + ¢

+FH IK +FH IK +FH IKK

s s s1

10130

13

= 0 ; K' = 0.000888 K

– sA = – 0.155 ; fA = 60º, 180º, 300º; intersection with the jw-axis at± j 0.22; multiple roots at s = – 0.063

The 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) = 0

which can be arranged as

1 + ¢ +

+ +

K ss s s

( )( )( )

102 4

= 0 ; ¢K = 0.1K

– sA = 2 ; fA = 90º, – 90º; 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 locus

exists 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 = 0

which may be rearranged as

1 + Ks

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

– sA = – 2.5 ; fA = 90º, – 90º; departure angle fp from s = j2 is 158.2º.

The following two points on z = 0.4 line satisfy the angle criterion:

s1 = – 1.05 + j2.41, s2 = – 2.16 + j4.97

The 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 = 0


1 + Ks s s( )2 4 5+ +

= 0 ; K = 2a

– sA = – 1.3333 ; fA = 60º, – 60º, 180º ; intersection with the jw-axis at

± j 5 ; departure angle fp from complex pole in the upper half of s-plane =– 63.43º; 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


1 + K

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

– sA = – 3 ; fA = 60º, – 60º, 180º; 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 s

s s s( )

( )( )+

− +1

2 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 = lims®0

sE(s) = 0

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

ess = lims®0

sE(s) = -16K

7.21 – sA = – 2 ; fA = 60º, 180º, 300º; 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 = lims®0

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

With the compensator, the characteristic equation becomes

1 + K s

s s s s( )

( )( )( )10 1

100 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 = lims®0

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

7.22 1 + K s

s 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 = 4zw n

= 2. Therefore, zwn = 2

Proportional control does not meet this requirement.

(b) 1 + K sK

s sc D+

+( )2 = 1 +

K s K K

s sD 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 s

s sD ( )

( )

+

+

4

2 = 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 s

s 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 4

2 10 6( )

( )( . )s

s s s+

+ +

Kv = lims®0

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 ® zwn = 2.85

wn = 4 ; w zn 1 2- = 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 1

0 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 = lims®0

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 – 1.8º, 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+

FHG

IKJ = Kc + KI /s = Kc

s K K

sI c+F

HIK

/

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

s s sc 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

| || |zp

= 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 s

s s s s s

( / )( / )

( )( )( / )( / )

+ +

+ + + +

1 1

1 5 11 2

1 2

t tat a t

1

1t = 0.05 ; 1

2t = 1 ; 1

1at = 0.005 ; a

t 2

= 10

D(s)G(s) = K s

s 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

K

KsD

c

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

KK

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 = lims®0

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% ® 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 sec

Kv = lims®0

sG(s) = 2.07

(c) Compensated system:

Mp = 15% ® 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 ³ 20

Lead-lag compensated system:

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

s s s s s( .9)

( . )( )( . )+ +

+ + + +

3)( 00 15 3)( 9 13 6

; K = 506

7.35 G(s) = K

s s KKt( )+ +2

E sR s

( )( )

= 11+G s( )

= s s KK

s s KK Kt

t

( )

( )

+ +

+ + +

2

2

For R(s) = 1/s2,

ess = 2

0 35+

£KK

Kt .

z ³ 0.707 ; ts £ 3 sec.

The characteristic equation of the system is

s2 + 2s + KKts + K = 0




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 s

s j s jt

( . )( . )+ + + -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 s

s j K s j Kt

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

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




jw

s

– 1j

– 2j

1j

K =¢t 0

K =¢t •

• ¨ K¢t

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) = 0


1 + KK s A K

s s st 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 = lims®0

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 = AK

The 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 + b

as

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

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

7.39 e–s = --

+

( )ss

22

1 – K s

s s( )

( )( )-

+ +

21 2

= 0 = 1 – F(s)

F(s) = K jj j

( )( )( )

s ws w s w

+ -

+ + + +

21 2

= K j

j

( )( )( ) (

s ws s w w s

- +

+ + - + +

21 2 2 3)2

ÐF(s) = tan–1 ws

w ss s w-

-+

+ + -

-

22 3)

1 21

2tan(

( )( )

= tan–1

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

(( )(

( )( )

ÐF(s) = 0º if

ws

w ss s w-

-+

+ + -22 3)

1 2 2(

( )( ) = 0

Manipulation of this equation gives

(s – 2)2 + w2 = 12

Centre = (2, 0) and radius = 12 = 3.464




It can easily be verified that at every point on this circle, ÐF(s) = 0º.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+ = 0.707

The root locus plot is shown in the figure below.

j0.5

– 0.5

KÆ – •KÆ – •

K = 0

K = 0

ññ

7.41 G(s) = 20 7 3)

2 8. (

( )( )s

s s s+

+ +

(i) a = a0 + Da = 8 + Da


s(s + 2) (s + 8) + Da s(s + 2) + 20.7 (s + 3) = 0

When Da = 0, the roots are (may be determined by root-locus methodor the Newton-Raphson method)

l1,2 = – 2.36 ± j2.48, l3 = – 5.27

The root locus for Da is determined using the root locus equation

1 + Da

l l ls s

s s s

( )( )( )( )

+

- - -

2

1 2 3

= 0



SOLUTION MANUAL 71

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

Near s = l1, the locus may be approximated by a line drawn from– 2.36 + j2.48 at an angle of – 80º. For a change of Dl1 = 0.2 Ð – 80ºalong 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 80

0 48 8.2 º

. /Ð - = 3.34 Ð – 80º

(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) = 0

or

1 + 20 7

1 2 3

.

( )( )( )

Dbl l ls s s- - -

= 0

The angle of departure of l1 is 50º.

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

Therefore

Sbl1 =

D

D

lb b

1

0/ = 0 50

0 3.2 º

.21/Ð + = 2.84 Ð + 50º

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 s

s 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 s

s s s s

( )

( )( ) ( )

+

+ + + +

3

2 3 8 5 = 0

or 1 + d

l l ls s

s s s s

( )

( )( )( )

+

- - -

3

1 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) 1Ð 0º (ii) 0Ð – 180º (iv) t t

t t1 2

1 2

90+

Ð - º

(b) Revisiting Example 8.2 will be helpful.

(i) – t – j¥ (ii) 0Ð – 180º

(c) Revisiting Example 8.3 will be helpful.

(i) ¥Ð – 180º (ii) 0Ð – 270º

(d) (i) – (t1 + t2) – j¥ (ii) 0 Ð – 270º

(iii) t t

t t1 2

1 2

180+

Ð - º

(e) (i) ¥Ð -180º (ii) 0 Ð 0º (iv)t t t t

t t

1 2 1 2

1 2

90( )

º+

Ð -

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 = 2

Number 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 = ¥Ð – 180º,

G(s) has double pole at the origin. The map of Nyquist contour semicircles = rejf, r ® 0, f varying from – 90º at w = 0– through 0º to + 90º at w =

0+, into the Nyquist plot is given by ¥Ð – 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 = 0The closed-loop system is stable

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

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

8.4 (a) The key points to the polar plot are:

G j H j( ) ( )w ww =¥

= 18Ð 0º

G j H j( ) ( )w ww =¥

= 0Ð – 270º

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 ww =0

= – 3 – j ¥

G j H j( ) ( )w ww =¥

= 0Ð – 270º

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

G j H j( ) ( )w ww =

1

2= - 4

3

The map of Nyquist contour semicircle s = rejf, r ® 0, f varyingfrom – 90º at w = 0– through 0º to + 90º at w = 0+, into the Nyquist plotis given by ¥Ð – 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 = 2Ð – 180º

G j H j( ) ( )w w w=¥ = 0Ð – 90º

No 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 = 0Ð 90º ; G j H j( ) ( )w w w=1 = 2.6Ð161º ;

G j H j( ) ( )w w w=2 = 2.6 Ð 198º ; G j H j( ) ( )w w w=10 = 0.8Ð – 107º

G j H j( ) ( )w w w=¥ = 0Ð – 90º

The Nyquist plot is given in Fig. P8.2b.

P = 2, N = – 2 ; Z = N + P = 0

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

= ¥Ð – 180º

G j H j( ) ( )w ww =¥

= 0 Ð – 90º

No intersections with real/imaginary axis.The map of Nyquist contour semicircle s = rejf, r ® 0, f varyingfrom – 90º at w = 0– through 0º to + 90º at w = 0+, into the Nyquist plotis given by ¥Ð – 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

0Ð º

G j H j( ) ( )w ww=

-10 = ¥ Ð 0º

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

G(s)H(s) = limr®0

1

10 1002( )j e j+ +r f = lim

( )r f fr r® +0

120e j ej j

= limr

f

r®

-

0

1 e j

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

G j H j( ) ( )w ww =¥

= 0Ð – 180º




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=¥ = 0º Ð – 90º

G j H j( ) ( )w w w= 3 = – 2 + j0

No intersection with the imaginary axis.

The map of Nyquist contour semicircle s = rejf, r ® 0, f varyingfrom – 90º at w = 0– through 0°to + 90° w = 0+, into the Nyquist plot isgiven by ¥Ð – 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( )

( )

( )

( )

1

1 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

= 4Ð0º ; G j( )w w=¥ = 0 Ð – 270º

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 s

s s

( )

( )

t

t2

21

1

1; 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 1

12( . )

( )+

+

s

s s =

¢ +

+

K s

s s

( )

( )

t

t2

21

1

1 ; 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) = ej

j-

+

0 8

1

. w

w

= 11 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

= 1Ð 0º ; G j( ) .w w=2 4482 = – 0.378 + j0

G j( )ww=¥

= 0

The polar plot will spiral into the w ®¥ 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 1

0KG j H j( ) ( )w w

w= = – 1 + j0 ;

1K

G j H j( ) ( )w ww=¥

= 0.05 + j0

1

0 62KG j H j( ) ( )

.w w

w= = – 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.

1

0KG j( )w

w=

= 0.5Ð0º ; 1K

G j( )ww=¥

= 0Ð – 180º




1

1 4KG j( )

.w

w=

= – 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( )( )

1

0KG j( )w

w=

= ¥Ð – 90º; 1

0 26KG j( )

.w

w= = – 2.55 + j0

1

KG j( )w

w =¥

= 0



SOLUTION MANUAL 81

The polar plot will spiral into the w ® ¥ 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) = Ke

s s

s-

+

2

2 4 1( )

1

0KD j G j( ) ( )w w

w= = ¥Ð – 180º ;

1

KD j G j( ) ( )w w

w =¥

= 0

The polar plot will spiral into the w ®¥ 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) = 0

may be manipulated as

e

s s

sD

-

+

t

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

1s s( )+

The polar plot of G1(jw) intersects the polar plot of – ej Dw t at a pointcorresponding to w = 0.75 at the G(jw)-locus. At this point, the phase of the

polar plot of –ej Dw t is found to be 52º. 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=¥ = 0Ð – 270º

G j( ) .w w=15 9 = – 0.182 + j0; G j( ) .w w=6 2 = 1 Ð – 148.3º

GM = 1/0.182 = 5.5 ; FM = 31.7º

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 asshown 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 +10º.

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 would

intersect 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 to–40 dB/decade at wc1 = 10; then to –20 dB/decade at wc2 = 20; to–40 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

ÐG(jw) = – 90º – tan–1 0.1w

The phase shift at wg = 7.86 is –128.2º, and the resulting phase marginis – 128.2º – (– 180º) = 51.8º. Because the phase curve never reaches– 180º 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

ÐG(jw) = – 90º – 2 tan–1 0.1w

The phase shift at wg = 6.8 is – 158.6º. Therefore, the phase margin is180º – 158.6º = 21.4º. At a frequency of wf = 10, the phase shift is– 180º; 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 w = 5, by + 1 dB at w = 2.5 and by + 1 dB at w = 10.The compensated magnitude plot crosses the 0 dB line at wg = 9. Thephase shift at this frequency is – 106.6º. Therefore the phase margin is73.4º.

Phase never reaches – 180º line; the gain margin is infinity.




(d) Reconsider the bode plot for system of Problem 8.15c.The time-delay factor e–0.1s will not change the magnitude plot; it willchange only the phase characteristics. For the system with dead-time,

wg = 9 rad/sec; FM = 22º; wf = 13.65 rad/sec; GM = 4.2 dB

(e) G(s) = 402 4 5( )( )( )s s s+ + +

= 112

1 14

1 15

1s s s+FH IK +FH IK +FH IKThe low-frequency asymptote coincides with 0 dB line; its slopechanges to – 20 dB/decade at wc1 = 2, to – 40 dB/decade at wc2 = 4,and to – 60 dB/decade at wc3 = 5. Compensated magnitude plot caneasily be obtained by applying corrections.The phase plot crosses the – 180º line at wf = 7 rad/sec, and the gainmargin is 20 dB. Since the gain never reaches 0 dB (wg = 0), the phasemargin is infinity.

(f) From the Bode plots of G(jw) we find that gain crossover frequencywg = 3.16 rad/sec, and the phase margin is –33º. The phase plot isasymptotic to the –180º line in the low-frequency range; it neverreaches – 180ºTherefore the GM = – ¥

8.16 G(s) = K

s s s

/

( )

5

1 15

1+ +FH IK

From the Bode plot of the system sketched for K = 10, we find that FM =21º, GM = 8 dB.Increasing the gain from K = 10 to K = 100 shifts the 0 dB axis down by 20dB. The phase and gain margins for the system with K = 100 are

FM = – 30º, GM = – 12 dB

Thus the system is stable for K = 10 but unstable for K = 100.

8.17 (a) G(s) = K

s s s

/ 4012

114

115

1+FH

IK +FH

IK +FH

IK

Since G(s) has all poles in left half s-plane, the open-loop system isstable. Hence the closed-loop system will be stable if the frequencyresponse has a gain less than unity when the phase is 180º.When K = 40, the gain margin is 20 dB (refer Problem 8.15e).Therefore, an increase in gain of +20 dB (i.e., by a factor of 10) ispossible before the system becomes unstable. Hence K < 400 forstability.



SOLUTION MANUAL 85

(b) From the Bode plot of G(jw) with K = 1, we find that the phasecrossover frequency wf = 15.88 rad/sec and gain margin = 34.82 dB.This means that the gain can be increased by 34.82 dB (i.e., by a factorof 55) before the system becomes unstable. Hence K < 55 for stability.

(c) From the Bode plot of G(jw) with K = 1, we find that wf = 0.66 rad/secand GM = 4.5 dB. Thus the critical value of K for stability is 1.67, i.e.,K < 1.67 for the system stability.

8.18 From the Bode plot of G(jw), it can easily be determined that (a) when tD =0, the GM = 12 dB, and FM = 33º; and (b) when tD = 0.04 sec, the GM =2.5 dB and FM = 18º. The relative stability of the system reduces due to thepresence of dead-time.

(c) The value of tD for the system to be on verge of instability is obtainedby setting the phase margin equal to zero, i.e.,

G jg

1( )ww w=

– w t

p

g D ´180º = – 180º

where G1(jw) is the system without dead-time, and wg is the gaincrossover frequency. From the Bode plot of G1(jw), we get wg = 7.4rad/sec and ÐG1(jwg) = – 147º.

Therefore,

– 147º – 7 4 180. ºt

pD ´

= – 180º

or tD = 0.078 sec

8.19 Since the systems have minimum-phase characteristics, zeros (if any) arein left half s-plane.(a) The low-frequency asymptote has a slope of – 20 dB/decade and its

extension intersects the 0 dB-axis at w = 4. The system is thereforetype-1; 4/s is a factor of the transfer function.

The asymptote slope changes from – 20 dB/decade to 0 dB/decade atw = 2 (this corresponds to the factor (1 + s/2)), then to – 20 dB/decadeat w = 10 (this corresponds to the factor (1 + s/10)). The transferfunction

G(s) = 4 1 1

2

1 110

+FH IK+

FH IK

s

s s

(b) The low-frequency asymptote has a slope of + 20 dB/decade and it




intersects 0 dB-axis at w = 0.2. The system is therefore type-0; Ks is afactor of the transfer function with 20 log 0.2K = 0 or K = 5.The low-frequency asymptote has a magnitude of 20 dB at a frequencywc1, where

20 log (5wc1) = 20. This gives wc1 = 2.

The transfer function

G(s) = 5

1 12

1 110

1 130

s

s s s+FH IK +FH IK +FH IK(c) The low-frequency asymptote has a slope of – 6 dB/octave with a

magnitude of – 9 dB at w = 1. The system is therefore type-1; K/s is afactor of the transfer function with 20 log (K/w) = – 9 at w = 1. Thisgives K = 0.355.The transfer function

G(s) = 0 355 1 1 1

20

1 140

. ( )+ +FH IK

+FH IK

s s

s s

(d) The low-frequency asymptote has a slope of – 20dB/decade with amagnitude of 40 dB at w = 2.5. The system is therefore type-1; K/s is afactor of the transfer function with 20 log (K/2.5) = 40. This gives K =250.

G(s) = 250

1 12

1 140

s s s+FH IK +FH IK.5

(e) The low-frequency asymptote has a slope of +12 dB/octave; thesystem is therefore type-0 and Ks2 is a factor of the transfer function.The first corner frequency wc1 = 0.5. The slope of the asymptotechanges to +6 dB/decade at this corner frequency. Therefore1/(2s + 1) is a factor of the transfer function.The asymptotic plot of Ks2/(2s + 1) is shown in the figure below.



SOLUTION MANUAL 87

w

20 log Kw2

– 20 log 2w

20

10

–20

–10

0.1 1

0.5

0

32

dB

From this figure, we find that

(20 log Kw2 – 20 log 2w)|w = 1 = 32

This gives K = 79.6.

The transfer function

G(s) = 79 61 2 1 1 0

2.( )( )( .2 )

ss s s+ + +

8.20 The low-frequency asymptote has a slope of 20 dB/decade. Ks is a factor ofthe transfer function with 20logK = 30. This gives K = 31.623. The transferfunction

G(s) = 31 623

1 1 15

1 120

.

( )

s

s s s+ +FH IK +FH IK(i) 20 log (31.623wg1) = 0; this gives wg1 = 0.0316

(ii) 20 log (31.623wg2) – 20 log wg2 – 20 log 15 2w g

FH IK

– 20 log 120 2w g

FH IK = 0; this gives wg2 = 56.234

8.21 The low-frequency asymptote has a slope of – 6 dB/octave and has amagnitude of 11 dB at w = 3. K/s is therefore a factor of the transferfunction with 20 log (K/3) = 11. This gives K = 10.64.The transfer function




G(s) = 10 6413

1 18

1

.

s s s+FH IK +FH IK

ÐG(jw) = – 90º – tan–1 13

w – tan–1 18

w

ÐG(jw) = – 180º at w = 4.9

The gain obtained from asymptotic magnitude plot at w = 4.9 is

20 log 10 644

20 43

.

.9log .9

- = 2.5 dB

Therefore, GM = – 2.5 dB

8.22 Revisit Review Example 8.3.

G(s) = 5 1

110

112

10 650

12500

2

+FH

IK

+FH

IK + +LNM

OQP

s

s s s s.

8.23 From the asymptotic magnitude plot we find that the system is type-0 witha double pole at s = – 3. The transfer function obtained from the magnitudeplot is

G(s) = 0 1

13

12

.

s+FH IKÐG(jw) vs w may be compared with the given phase characteristics tocheck the accuracy of identification of corner frequency at wc = 3.

——



CHAPTER 9 FEEDBACK SYSTEM PERFORMANCEBASED ON THE FREQUENCYRESPONSE

9.1 For derivation of the result, refer Section 9.3; Eqn. (9.9).9.2 For derivation of the result, refer Section 9.3; Eqn. (9.18).9.3 For derivation of the result, refer Section 9.3; Eqns (9.15) – (9.16).9.4 The relations

z 4 – z2 + 14 2Mr

= 0 ; wr = w zn

1 2 2-

yield z = 0.6 and wn = 21.8

Note that Mr – z relation gives two values of z for Mr = 1.8; z = 0.6 andz = 0.8. We select z = 0.6 as damping ratio larger than 0.707 yields no peakabove zero frequency.

The characteristic equation of the given system is

s2 + as + K = s2 + 2zwns + w n2 = 0

This gives K = 475 and a = 26.2

ts = 4

zw n

= 0.305 sec

wb = wn 1 2 2 4 42 2 41 2

- + - +z z z/

= 25.1 rad/sec

9.5 From the response curve, we find

Mp = 0.135, tp = 0.185 sec

z2 =( )

( )

ln

ln

M

Mp

p

2

2 2+ p

gives z = 0.535 for Mp = 0.135

p

w zn 1 2-

= 0.185 gives wn = 20

The corresponding frequency response performance indices are:

Mr =1

2 1 2z z- = 1.11 ; wr = w zn 1 2 2

- = 13.25




wb = w z z zn 1 2 2 4 42 2 41 2

- + - +/

= 24.6

9.6 z » 0.01fm

The characteristic equation of the system is

t s2 + (1 + KKt)s + K = 0

This gives

wn = /K t ; z = 1

2

+ KK

Kt

t = 0.01fm

K = 1

2 10 2 10 104 2 2 4 2

KK K

tm t m m t´ - ± ´ -

- - -f t f t f t

9.7 The polar plot of G(jw) crosses the real axis at w = 1/ t t1 2 . Themagnitude |G(jw)| at this frequency is given by

|G(jw)| = Kt t

t t1 2

1 2+

Therefore

Gm × Kt t

t t1 2

1 2+ = 1

This gives

K = 1 1 1

1 2Gm t t+

FHG

IKJ

9.8 (a) It can be determined from the Bode plot that wg = 1 rad/sec andFM = 59.2º

(b) FM = tan–1 2

4 1 24 2

z

z z+ -

FM = 59.2º ® z = 0.6

wg = w z zn 4 1 24 2+ -

This gives wn = 1.4

Therefore,

Mp = e- -pz z/ 1 2

= 9.48%



SOLUTION MANUAL 91

ts = 4zw n

= 4.76 sec.

9.9 The low-frequency asymptote has a slope of – 20 dB/decade and intersectsthe 0 dB-axis at w = 8; 8/s is a factor of the transfer function. The forward-path transfer function can easily be identified as

G(s) = 8 1 1

21 1

4

1 18

1 124

1 136

+FH IK +FH IK+

FH IK +FH IK +FH IK

s s

s s s s

The phase curve can now be constructed.

From the phase curve and the asymptotic magnitude curve, we obtain

FM = 50º ; GM = 24 dB

Phase margin of 50º gives z = 0.48 which corresponds to Mp » 18%.

9.10 (a) From the Bode plot, we get FM = 12º

(b) For a phase margin of 50º, we require that G(jw) H(jw) = 1 Ð – 130ºfor some value of w. From the phase curve of G(jw)H(jw), we find thatÐG(jw)H(jw) ~- –130º at w = 0.5. The magnitude of G(jw)H(jw) atthis frequency is approximately 3.5. The gain must be reduced by afactor of 3.5 to achieve a phase margin of 50º.

(c) FM of 50º gives z = 0.48 which corresponds to Mp ~- 18%.

9.11 G(jw) = K j

j

( )( )

w

w

+ 22

ÐG(jw) = tan–1 w

2 – 180º = – 130º

This equation gives w = 2.3835

The magnitude of G(jw) must be unity at w = 2.3835.

K j

j

( )

( ) .

w

w w

+

=

22

2 3835 = 1

This equation gives K = 1.826.

Since the phase curve never reaches – 180º line, the gain margin = ¥ .

9.12 (a) From the open-loop frequency response table, we find that wf = 10

rad/sec and GM = 20 log 10 64.

= 3.88 dB; wg = 8 rad/sec and FM = 10º.




(b) For the desired gain margin of 20 dB, we must decrease the gain by(20 – 3.88) = 16.12 dB. It means that magnitude curve must be loweredby 16.12 dB, i.e., the gain must be changed by a factor of b, given by20 log b = – 16.12. This gives b = 0.156.

(c) To obtain a phase margin of 60º, we must first determine the frequencyat which the phase angle of G(jw) is – 120º, and then adjust the gain sothat |G(jw)| at this frequency is 0 dB. From the Bode plot, we find thatlowering the magnitude plot by 16.65 dB gives the desired phasemargin. It equivalently means that the gain should be changed by afactor of a where 20 log a = – 16.65. This gives a = 0.147.

9.13 A 9.48% overshoot implies z = 0.6. For this damping, required phasemargin is 59.2º.

G(s) = 10036 100

Ks s s( )( )+ +

= K

s s s

/ 36136

1 1100

1+FH IK +FH IKMake a Bode plot for say K = 3.6.

From the plot we find that at w = 14.8 rad/sec, ÐG(jw) = – 120.8º. Atw = 14.8, the gain is – 44.18 dB. The magnitude curve has to be raised to 0dB at w = 14.8 to yield the required phase margin. The gain should bechanged by a factor of a where 20 log a = 44.18. This gives a = 161.808.Therefore,

K = 3.6 × 161.808 = 582.51

wg = 14.8 = wn 4 1 24 2z z+ -

This gives

wn = 20.68

tp = p

w zn 1 2-

= 0.19 sec

9.14 (a) From the Bode plot, it can easily be determined that GM = 5 dB andFM = 30º. The gain crossover frequency wg = 0.83 rad/sec.

(b) The magnitude plot should be lowered by 5 dB to obtain a gain marginof 10 dB. This is achieved by reducing the gain by 5 dB.

(c) The gain at w = 1.27 is –4.8 dB. Therefore, the gain should beincreased by 4.8 dB to obtain a gain crossover frequency of 1.27rad/sec.

(d) Gain at the frequency which givesÐ G(jw) = – 135º, should be 0 dB.



SOLUTION MANUAL 93

From the Bode plot we find that this requires reducing the gain by 3.5dB.

9.15 The requirements on FM and wg are satisfied if we increase the gain to thelimit of zero GM with a phase margin ³ 45º. From the Bode plot we findthat K = 37.67 meets this requirement.

9.16 (a) From the Bode plot we find that 0 dB crossing occurs at a frequency of0.47 rad/sec with a phase angle of – 145º. Therefore, the phase marginis 35º. Assuming a second-order approximation, FM = 35º ® z =0.33, Mp = 33%.

(b) The zero dB crossing occurs with a phase angle of – 118º. Thereforethe phase margin is 62º. FM = 62º ® z = 0.64, MP = 7.3%.

9.17 The 0 dB crossing occurs at w = 0.8, with a phase angle of – 140º when tD= 0, and – 183º when tD = 1. Therefore phase margin of the system withoutdead-time is 40º; and with dead-time added, the phase margin becomes– 3º. We find that with the dead-time added, the system becomes unstable.Therefore, the system gain must be reduced in order to provide a reasonablephase margin.

We find from the Bode plot that in order to provide a phase margin of30º, the gain would have to be decreased by 5 dB, i.e., by a factor of bwhere 20 log b = 5. This gives b = 1.78.

We find that the dead-time necessitates the reduction in loop gain is orderto obtain a stable response. The cost of stability is the resulting increase inthe steady-state error of the system as the loop gain is reduced.

9.18 Make a polar plot of

G(jw) = 50 181 3)(1 6

/( / / )j j jw w w+ +

G j( )w w=0 = - - ¥2518

j ; G j( )w w=¥ = 0 Ð – 270º

The polar plot intersects the negative real axis at w = 4.24.

The polar plot is tangential to the M = 1.8 circle (refer Eqn. (9.22)).Therefore, the resonance peak Mr = 1.8.

The bandwidth of a system is defined as the frequency at which themagnitude of the closed-loop frequency response is 0.707 of its magnitudeat w = 0. For the system under consideration, closed-loop gain is unity at w= 0; therefore, bandwidth is given by the frequency at which the M = 0.707circle intersects the polar plot. This frequency is found to be 3.61 rad/sec.Therefore wb = 3.61




z4 – z2 + 14 2Mr

= 0 ; Mp = e- -´

pz z/ 1 2

100

wb = wn ( )1 2 4 2 22 4 2- + - +z z z

ts = 4/zwn

The Mr – z relation gives z = 0.29, 0.96 for Mr = 1.8. We select z = 0.29because damping ratio larger than 0.707 yields no peak above zerofrequency. z = 0.29 corresponds to Mp = 38.6%.

For z = 0.29 and wb = 3.61, we get wn = 2.4721. The setting time ts =5.58 sec.

9.19 The polar plot is tangential to M = 1.4 circle (refer Eqn. (9.22)) at afrequency w = 4 rad/sec. Therefore Mr = 1.4 and wr = 4.

z4 – z2 + 1

4 2Mr

= 0

wr = wn 1 2 2- z

Mr = 1.4 ® z = 0.39.

The corresponding Mp = 26.43%.

wr = 4 and z = 0.39 ® wn = 4.8. The settling time ts = 2.14 sec.

The Mr – z relationship has been derived for standard second-ordersystems with zero-frequency closed-loop gain equal to unity. The answer isbased on the assumption that zero-frequency closed-loop gain for thesystem under consideration is unity.

9.20 From the Bode plot, we find that

wg = 8.3 rad/sec ; wf = 14.14 rad/sec;

GM = 8.2 dB ; FM = 27.7º

For each value of w, the magnitude and phase of G(jw) are transferred fromthe Bode plot to the Nichols chart. The resulting dB vs phase curve istangential to M = 6.7 dB contour of the Nichols chart at w = 9 rad/sec.Therefore Mr = 6.7 dB and wr = 9. The dB vs phase curve intersects the– 3 dB contour of the Nichols chart at w = 13.6. Therefore bandwidth wb= 13.6 rad/sec.

The following points are worth noting in the use of Nichols chart.

(i) The frequency parameters (wr or wb) can be easily determined bytransferring the dB and/or phase data from the dB vs phase curve to theBode plot.



SOLUTION MANUAL 95

(ii) The open-loop dB vs phase plot may not be tangential to one of theconstant-M contours of the Nichols chart. One must do a littleinterpolation. In the present problem, Mr = 6.7 dB has been obtainedby interpolation.

9.21 Plot the given frequency response on a dB scale against phase angle. Fromthe dB vs phase curve, we obtain,

GM = 7 dB ; FM = 17º

The dB vs phase curve when transferred on Nichols chart gives Mr = 10 dB;wr = 2.75 rad/sec; wb = 4.2 rad/sec

Note that the –3dB bandwidth definition is applicable to systems havingunity closed-loop gain at w = 0. We have made this assumption for thesystem under consideration.

9.22 For parts (a) and (b) of this problem, refer Problem 9.10.

(c) Since the data for –3dB contour of the Nichols chart has beenprovided, we don’t require the Nichols chart for bandwidthdetermination. On a linear scale graph sheet, dB vs phase cure of theopen-loop frequency response is plotted with data coming from Bodeplot. On the same graph sheet, the – 3dB contour using the given datais plotted.

The intersection of the two curves occurs at w = 0.911 rad/sec.Therefore bandwidth wb = 0.911.

(d) FM = tan–1 2

4 1 24 2

z

z z+ -

wb = wn 1 2 2 4 42 2 4- + - +z z z

FM = 50º ® z = 0.48 ; Mp » 18%

wb = 0.911 and z = 0.47 ® wn = 0.7025 ; ts = 11.93 sec.

9.23 The frequency-response data from the Bode plot of G(jw) when transferredto the Nichols chart, gives the following result.

Mr = 1.4 ; wr = 6.9 rad/sec

From the relations

Mr = 1

2 1 2z z- ; wr = wn 1 2- z 2

we obtain

z = 39 ; wn = 8.27 rad/sec




Note that the given type-0 system has zero-frequency closed-loop gain= 54/(25 + 54) = 0.6835. The approximation of the transient response ofthe system by z = 0.39 and wn = 8.27 has large error since the correlations(Mr, wr) ® (z, wn) used above, have been derived from systems with unityzero-frequency closed-loop gain.

9.24 (a) Transfer the frequency-response data from the Bode plot of G(jw) tothe Nichols chart.

Shifting the dB vs phase curve down by 5.4 dB makes this curvetangent to the M = 20 log1.4contour. Therefore gain must be changedby a factor of b, where 20 log b = – 5.4. This gives b = 0.54.

(b) The gain-compensated system intersects the – 3 dB contour of theNichols chart at w = 9.2. Therefore bandwidth wb = 9.2 rad/sec.

Note that the – 3dB bandwidth definition given by Eqn. (9.2b) is applicableto systems having unity closed-loop gain at w = 0. The given system doesnot satisfy this requirement. The answer, therefore, is an approximation ofthe bandwidth.

9.25 (a) From the Bode plot we find the phase crossover frequency wf = 2.1and GM = 14 dB. For the gain margin to be 20 dB, the magnitude plotis to be brought down by 6 dB. The value of K corresponding to thiscondition is given by the relation

20 log K = – 6

which gives K = 0.5

(b) From the Bode plot it is observed that FM = 60º will be obtained ifgain crossover frequency wg is 0.4 rad/sec. This is achieved ifmagnitude plot is brought down by 7 dB. This condition correspondsto K = 0.446.

(c) From the dB-phase plot on the Nichols chart, we find that for this plotto be tangent to M = 1 dB contour, the dB-phase curve must be broughtdown by 4 dB. This condition corresponds to K = 0.63. Thecorresponding value of wr (read off from Bode plot) is 0.5 rad/sec.

The value of bandwidth (wb) is the frequency at which the dB-phaseplot intersects the – 3dB contour. It is found that wb = 1 rad/sec.

(d) The dB vs phase plot on the Nichols chart when raised by 2.4 dB,intersects the –3 dB contour at w = 1.5 rad/sec. This conditioncorresponds to K = 1.35.

9.26 Plot the given frequency response on a dB scale against phase angle. Fromthis dB vs phase curve, we obtain

GM = 16.5 dB ; FM = 59º



SOLUTION MANUAL 97

Transfer the dB vs phase curve on the Nichols chart. We find that this curvewhen raised by about 5 dB, touches the M = 20 log1.4 contour of theNichols chart. Therefore the gain should be increased by a factor of b,where 20 log b = 5. This gives b = 1.75.

The gain-compensated system has GM = 11.6 dB and FM = 42.5º.

9.27 The dB vs phase plot becomes tangential to 20 log1.4 dB contour onNichols chart if the plot is lowered by about 11 dB. This means that thegain K must be reduced by a factor of a where 20 log (1/a) = –11. Thisgives a = 3.5.

Phase margin of the gain-compensated system is 42.5º

Mr = 1.4 ® z = 0.387 ; FM = 42.5º ® z = 0.394

The following comments may be carefully noted.

(i) Zero-frequency closed-loop gain has been assumed to be unity.

(ii) One is usually safe if the lower of the two values of z is utilized foranalysis and design purposes.

9.28 SGM (jw) =

11+ G j( )w

= G j

G j

-

-+

1

11

( )

( )

w

w

Magnitude of SGM (jw) can be obtained by plotting G–1(jw) on the Nichols

chart.

The dB vs phase plot of G–1(jw) is tangent to M = 20 log 2.18 dB contour ofNichols chart. Therefore

|SGM (jw)|max = 2.18

The peak occurs at w = 7 rad/sec.



CHAPTER 10 COMPENSATOR DESIGN USINGBODE PLOTS

10.1 (a) From the Bode plot of G(jw)H(jw), we find that GM = 6 dB andFM = 17º

(b) Now the gain and phase of the compensator with Kc = 1 are added tothe Bode plot. From the new Bode plot we find that the gain must beraised by 1.5 dB to have a gain margin of 6 dB.20 log Kc = 1.5 gives Kc = 1.2

10.2 (a) From the Bode plot of G(jw)H(jw) we find that wg = 4.08 rad/sec; FM= 3.9º; GM = 1.6 dB

(b) Now the gain and phase of the compensator are added to the Bode plot.From the new Bode plot, we find thatwg = 5 rad/sec; FM = 37.6º; GM = 18 dBThe increase in phase and gain margins implies that lead compensationincreases margin of stability. The increase in wg implies that leadcompensation increases speed of response.

10.3 (a) Kv = lims® 0

sD(s)G(s) = K = 12

(b) From the Bode plot and Nichols chart analysis of the system withK=12, we find that FM = 15º; wb = 5.5 rad/sec; Mr = 12 dB; wr = 3.5rad/sec

(c) The phase margin of the uncompensated system is 15º. The phase leadrequired at the gain crossover frequency of the compensated magnitudecurve = 40º – 15º + 5º = 30º

a = 1 301 30-

+

sin ºsin º

= 0.334

The frequency at which the uncompensated system has a magnitude of

– 20 log (1/ a ) = – 4.8 dB is 4.6 rad/sec. Selecting this frequency asgain crossover frequency of the compensated magnitude curve, we set

1a t

= 4.6

The transfer function of the lead compensator becomes

D(s) = 12 0 376 10 128 1( . ).

ss+

+

(d) FM = 42º; wb = 9 rad/sec; Mr = 3 dB;

wr = 4.6 rad/sec



SOLUTION MANUAL 99

10.4 Lead compensator D(s) = 10 0 5 1

01 1

( . )

.

s

s

+

+meets the phase margin and steady-

state accuracy requirements. These requirements are also met by the lag

compensator D(s) = 10 10 1

100 1

( )s

s

+

+

In this particularly simple example, specifications could be met by eithercompensation. In more realistic situations, there are additional performancespecifications such as bandwidth and there are constraints on loop gain.Had there been additional specifications and constraints, it would haveinfluenced the choice of compensator (lead or lag).

105. The settling time and peak overshoot requirements on performance may betranslated to the following equivalent specifications:

z = 0.45 ; wn = 2.22

z is related to the phase margin by the relation

FM »z

0 01.= 45º

wn is related to the bandwidth wb by the relation

wb = wn 1 2 2 4 42 2 41 2

- + - +z z z/

For a closed-loop system with z = 0.45, we estimate from this relation

wb = 1.33 wn

Therefore, we require a closed-loop bandwidth wb ~-3. The gain K of thecompensator

D(s) =K s

s

( )tat

+

+

1

1

may be set a value given by K = w n2 . This gives K ~- 5. To provide a

suitable margin for settling time, we select K = 10.

The phase margin of the uncompensated system is 0º because thedouble integration results in a constant 180º phase lag. Therefore, we mustadd a 45º phase lead at the gain crossover frequency of the compensatedmagnitude curve. Evaluating the value of a, we have

a =1 451 45-

+

sin ºsin º

= 0.172

To provide a margin of safety, we select a = 1/6.





– 20 log (1/ a ) = – 7.78 dB is 4.95 rad/sec. Selecting this frequency asthe gain crossover frequency of the compensated magnitude curve, we set

1a t

= 4.95

The closed-loop system with the compensator

D(s) = 10 0 083 10 1

( . ).5

ss

+

+

satisfies the performance specifications.

10.6 Kv = Aess

= AA0 05.

= 20

K = 20 realizes this value of Kv.

From the Bode plot of

G(jw) =200 1j jw w( .5 )+

we find that the phase margin of the uncompensated system is 18º. Thephase lead required at the gain crossover frequency of the compensatedmagnitude curve = 45º – 18º + 3º = 30º.

a = 1 301 30-

+

sin ºsin º

= 0.334


– 20 log (1/ a ) = – 4.8 dB is 8.4 rad/sec. Selecting this frequency as thegain crossover frequency of the compensated magnitude curve, we set

1a t

= 8.4

Therefore the compensator

D(s) = 0 10 0668 1

.2.

ss+

+

The phase margin of the compensated system is found to be 43.7º. If wedesire to have exactly a 45º phase margin, we should repeat the steps with adecreased value of a.From the Nichols chart analysis of the compensated and uncompensatedsystems, we find that the lead compensator has increased the bandwidthfrom 9.5 rad/sec to 12 rad/sec.



SOLUTION MANUAL 101

10.7 From the Bode plot of

G(jw) = 200 1j jw w( .5 )+

we find that the phase margin of the uncompensated system is 18º. Torealize a phase margin of 45º, the gain crossover frequency should bemoved to ¢wg where the phase angle of the uncompensated system is:

– 180º + 45º + 5º = – 130º.From the Bode plot of uncompensated system, we find that ¢wg = 1.5. The

attenuation necessary to cause ¢wg to be the new gain crossover frequency

is 20 dB. The b parameter of the lag compensator can now be calculated.

20 log b = 20

This gives b = 10. Placing the upper corner frequency of the compensator adecade below ¢wg , we have 1/t = 0.15.

Therefore, the lag compensator

D(s) = 6.66 166.6 1

ss+

+

As a final check, we numerically evaluate the phase margin and thebandwidth of the compensated system. It is found that that FM = 45º andwb = 2.5 rad/sec.

10.8 From the steady-state requirement, we set K = 100. From the Bode plot of

G(jw) = 1000 1 0 1j j jw w w( . )( .2 )+

we find that the phase margin is – 40º, which means that the system isunstable. The rapid decrease of phase of G(jw) at the gain crossoverfrequency wg = 17 rad/sec, implies that single-stage lead compensation maybe ineffective for this system. (The reader should, in fact, try a single-stagelead compensator). For the present system, in which the desired Kv is 100,a phase lead of more than 85º is required. For phase leads greater than 60º,it is advisable to use two or more cascaded stages of lead compensation(refer Fig. 10.15). The design approach may be that of achieving a portionof the desired phase margin improvement by each compensator stage.

We first add a single-stage lead compensator that will provide a phaselead of about 42.5º, i.e., the compensator will improve the phase margin toabout 2.5º.




Since the phase curve of the Bode plot of the uncompensated system at thegain crossover frequency has a large negative slope, the value of a = 0.19given by

a = 1 421 42-

+

sin .5ºsin .5º

will not yield a phase margin of 2.5º. In fact, when we set a = 0.08 andt = 0.1, the single-stage compensator improves the phase margin to 3º.

Therefore D1(s) = 0 1 10 008 1

..

ss+

+

Notice that the single-stage lead compensator not only improves the phasemargin, but also reduces the slope of the phase curve of the gain crossoverfrequency. By adding another stage of the compensator with D2(s) = D1(s),we obtain

D2(s)D1(s)G(s) = 100 0 1 1

0 1 0 008 12( . )

( .2 )( . )s

s s s+

+ +

Note that pole and zero at s = – 10 cancel each other.

The final compensated system has a phase margin of 45º.

10.9 G(s) = 250025K

s s( )+

K = 1 satisfies the steady-state performance requirement. The phase marginof the uncompensated system, read at the gain crossover frequency wg = 47rad/sec, is 28º.

The phase lead required at the gain crossover frequency of thecompensated magnitude curve = 45º – 28º + 8º = 25º

a = 1 251 25-

+

sin ºsin º

= 0.405


– 20 log (1/ a ) = – 3.93 dB is 60 rad/sec. Selecting this frequency as gaincrossover frequency of the compensated magnitude curve, we set

1a t

= 60

The transfer function of the lead compensator is

D(s) = 0 026 10 01 1..

ss+

+



SOLUTION MANUAL 103

The phase margin of the compensated system is found to be 47.5º. ByNichols chart analysis, it is found that the lead compensator reduces theresonance peak from 2 to 1.25 and increases the bandwidth from 74 to 98rad/sec.

10.10The phase margin of the uncompensated system

G(s) = 250025K

s s( )+; K = 1

read at the gain crossover frequency wg = 47 rad/sec is 28º. To realize aphase margin of 45º, the gain crossover frequency should be moved to ¢w g

where the phase angle of the uncompensated system is:

– 180º + 45º = – 135º.

From the Bode plot of the uncompensated system we find that ¢wg = 25.

Since the lag compensator contributes a small negative phase when theupper corner frequency of the compensator is placed at 1/10 of the value of

¢wg , it is a safe measure to choose ¢wg at somewhat less than 25 rad/sec,

say, 20 rad/sec.The attenuation necessary to cause ¢w g = 20 to be the new gain crossover

frequency is 14 dB.

The b parameter of the lag compensator can now be calculated.

20 log b = 14. This gives b = 5

Placing the upper corner frequency of the compensator a decade below ¢wg,

we have 1/t = 2. Therefore the lag compensator

D(s) = 0 12 1.5.5

ss+

+

From the Bode plot–Nichols chart analysis we find that the compensatedsystem has

FM = 51º; Mr = 1.2 and wb = 27.5 rad/sec.

10.11Kv = lims®0

sG(s) = 5

From the Bode plot we find that the uncompensated system has a phasemargin of – 20º; the system is therefore unstable.

We attempt a lag compensator. This choice is based on the observationmade from Bode plot of uncompensated system that there is a rapiddecrease in phase of G(jw) near wg. Lead compensator will not be effectivefor this system.




To realize a phase margin of 40º, the gain crossover frequency should bemoved to ¢w g where the phase angle of the uncompensated system is:

– 180º + 40º + 12º = – 128º

From the Bode plot of the uncompensated system, it is found that ¢wg = 0.5

rad/sec. The attenuation necessary to cause ¢wg to be the new gain crossover

frequency is 20 dB. The b parameter of the lag compensator can now becalculated.

20 log b = 20. This gives b = 10

Since we have taken a large safety margin of 12°, we can place the uppercorner frequency of the compensator at 0.1 rad/sec, i.e., 1/t = 0.1. Thus thetransfer function of the lag compensator becomes

D(s) = 10 1100 1

ss+

+

From the Bode plot of compensated system we find that the phase margin isabout 40° and the gain margin is about 11 dB.

10.12It easily follows that K = 30 satisfies the specification on Kv. From theBode plot of

G(jw) = 300 1 1 0 1j j jw w w( . )( .2 )+ +

we find that gain crossover frequency is 11 rad/sec and phase margin is– 24º. Nichols chart analysis gives wb = 14 rad/sec.If lead compensation is employed, the system bandwidth will increase stillfurther, resulting in an undesirable system which will be sensitive to noise.If lag compensation is attempted, the bandwidth will decrease sufficientlyso as to fall short of the specified value of 12 rad/sec, resulting in a sluggishsystem. These facts can be verified by designing lead and lag compensa-tors. We thus find that there is need to go in for lag-lead compensation.Since the full lag compensator will reduce the system bandwidthexcessively, the lag section of the lag-lead compensator must be designedso as to provide partial compensation only. The lag section, therefore,should move the gain crossover frequency to a value higher than the gaincrossover frequency of the fully lag-compensated system. We make achoice of new gain crossover frequency as ¢wg = 3.5 rad/sec. The

attenuation necessary to cause ¢wg to be the new gain crossover frequency

is 18.5 dB. This gives the b parameter of the lag section as

20 log b = 18.5; b = 8.32, say, 10



SOLUTION MANUAL 105

Placing the upper corner frequency of the lag section at 1/t1 = 1, we get thetransfer function of the lag section as

D1(s) = t

bt1

1

1

1

s

s

+

+ = s

s+

+

110 1

It is found that the lag-section compensated system has a phase margin of24º.

We now proceed to design the lead section. The implementation of the lag-lead compensator is simpler if a and b parameters of the lead and lagcompensators, respectively, are related as a = 1/b. Let us first make thischoice. If our attempt fails, we will relax this constraint on a.

The frequency at which the lag-section compensated system has a

magnitude of – 20 log (1/a ) = – 10 dB is 7.5 rad/sec. Selecting thisfrequency as gain crossover frequency of the lag-lead compensatedmagnitude curve, we set

1

2a t= 7.5

The transfer function of the lead-section becomes

D2(s) =t

at2

2

1

1

s

s

+

+ = 0 422 1

0 0422 1..

ss+

+

The analysis of the lag-lead compensated system gives FM = 48º andwb = 13 rad/sec.

10.13z 2 = ( )

( )

ln

ln

M

Mp

p

2

2 2+ p

Mp = 0.2 ® z = 0.456

tp =p

w zn 1 2-

tp = 0.1 ® wn = 35.3

wb = wn [ ] /1 2 2 4 42 2 4 1 2- + - +z z z= 46.576

FM = tan–1 2

4 1 24 2

z

z z+ -

~- 48º




In order to meet the specification of Kv = 40, K must be set at 1440. Fromthe Bode plot of

G(jw) =144000

36 100j j jw w w( )( )+ +

we find that the gain crossover frequency is 29.7 rad/sec and phase marginis 34º.

The phase lead required at the gain crossover frequency of the compensatedmagnitude curve = 48º – 34º + 10º = 24º

a =1 24

1 24

-

+

sin º

sin º = 0.42


– 20 log ( / )1 a = – 3.77 dB is 39 rad/sec. Selecting this frequency as gaincrossover frequency of the compensated magnitude curve, we set

1

at= 39

The transfer function of the lead compensator becomes

D(s) =1440 0 04 1

0 0168 1

( . )

.

s

s

+

+

It can easily be verified by Nichols chart analysis that the bandwidth of thecompensated system exceeds the requirement. We assume the peak timespecification is met. This conclusion about the peak time is based on asecond-order approximation that should be checked via simulation.

10.14K = 2000 meets the requirement on Kv. We estimate a phase margin of 65ºto meet the requirement on z.(a) From the Bode plot we find that the uncompensated system with K =

2000 has a phase margin of zero degrees. From the Bode plot weobserve that there is a rapid decrease of phase at the gain crossoverfrequency. Since the requirement on phase lead is quite large, it is notadvisable to compensate this system by a single-stage leadcompensator (refer Fig. 10.15).

(b) Allowing 10º for the lag compensator, we locate the frequency atwhich the phase angle of the uncompensated system is:– 180º + 65º + 10º = – 105º. This frequency is equal to 1.5 rad/sec. Thegain crossover frequency ¢wg should be moved to this value. The

necessary attenuation is 23 dB. The b parameter of the lag com-pensator can now be calculated.

20 log b = 23. This gives b = 14.2



SOLUTION MANUAL 107

Placing the upper corner frequency of the compensator one decadebelow ¢wg, we have 1/t = 0.15. Therefore the lag compensator

D(s) =6 66 1

94 66 1

.

.

s

s

+

+

The phase margin of the compensated system is found to be 67º.

(c) The bandwidth of the compensated system is found to be wb = 2.08rad/sec.

10.15Approximate relation between fm and z is z » 0.01fm

z= 0.4 ® fm = 40º

Let us try lag compensation.

The realize a phase margin of 40º, the gain crossover frequency should bemoved to ¢wg where the phase angle of the uncompensated system is:

– 180º + 40º + 10º = – 130º. From the Bode plot of the uncompensated

system we find that, ¢wg = 6. The attenuation necessary to cause ¢wg to be

the new gain crossover frequency is 9 dB.

20 log b = 9. This gives b ~- 3

Placing the upper corner frequency of the lag compensator two octaves

below ¢wg, we have 1/t = 1.5.

The lag compensator

D(s) =0 67 1

2 1

. s

s

+

+

Nichols chart analysis of the compensated system gives wb = 11 rad/sec.Using second-order approximation

wb = wn 1 2 2 4 42 2 41 2

- + - +z z z/

we get wn = 10

Therefore ts = 4/zwn = 1 sec

Note that the zero-frequency closed-loop gain of the system is nonunity.Therefore, the use of – 3 dB bandwidth definition, and the second-ordersystem correlation between wb and z & wn may result in considerable error.Final design must be checked by simulation.

10.16(a) From the Bode plot of uncompensated system we find that FM = 0.63º




(b) Phase margin of the system with the second-order compensator is9.47º. There is no effect of the compensator on steady-stateperformance of the system.

(c) The lead compensator

D(s) = 1 0 0378

1 0 0012

+

+

.

.

s

s

meets the requirements on relative stability.

10.17 Kv = 4.8



CHAPTER 11 HARDWARE AND SOFTWAREIMPLEMENTATION OF COMMONCOMPENSATORS

11.1 (a)E s

Y s

( )

( ) = –

R

R C sD

D+1 / = –

T s

T sD

Da +1 ; a = R/RD ; TD = RD CD

(b)E s

Y s

( )

( ) = –

R

R R sR C1 2 21+ +/ ( )= –

K T s

T sc D

D

( )+

+

1

1a

Kc = R

R R1 2+ ; a =

R

R R1

1 2+ ; TD = R2C

11.2 Z1(s) = R2 + R1/(1 + R1Cs); Z2(s) = R1 + R2

E s

E s2

1

( )

( )= –

Z s

Z s2

1

1( )

( )[ ]-

=t

ats

s

+

+

1

1; t = R1C, a = R

R R2

1 2+ < 1

Refer Section 11.2 (Fig. 11.4) for the Bode plot and filtering properties ofthe lead compensator.

11.3 Z1(s) = R1 + R2 ; Z2(s) = R2 + R1/(1 + R1Cs)

E s

E s2

1

( )

( ) = –

Z s

Z s2

1

( )

( ) = –

tbt

s

s

+

+

1

1; t =

R R

R RC1 2

1 2+, b =

R R

R1 2

2

+ > 1

Add an inverting amplifier of gain unity. Refer Section 11.2 (Fig. 11.7) forthe Bode plot and filtering properties of the lag compensator.

11.4 Z1(s) = (R1C1s + 1)R3/[(R1 + R3)C1s + 1]

Z2(s) = (R2C2s + 1)R4/[(R2 + R4)C2s + 1]

E s

E s2

1

( )

( ) = – Z s

Z s

R

R2

1

6

5

( )

( )-LNM

OQP=

K s s

s s

c +FHG

IKJ +FHG

IKJ

+FHG

IKJ +FHG

IKJ

1 1

1 11 2

1 2

t t

at bt

t1 = (R1 + R3)C1 ; t2 = R2C2 ; a = R

R R1

1 3+ ; b =

R R

R2 4

2

+ ;




Kc = R R R

R R R

R R

R R2 4 6

1 3 5

1 3

2 4

+

+

FHG

IKJ

11.5 Z1(s) = R1/(1 + R1Cs); Z2(s) = R2 + 1/C2s

E s

E s2

1

( )

( ) = –

Z s

Z s

R

R2

1

4

3

( )

( )-LNM

OQP= Kc 1

1

1

+ +LNM

OQPT s

T sD

Kc = R R C R C

R R C4 1 1 2 2

3 1 2

( )+ = 39.42; TI = R1C1 + R2C2 = 3.077

TD = R R C C

R C R C1 2 1 2

1 1 2 2+ = 0.7692

These equations give

R1 = R2 = 153.85 kW; R4 = 197.1 kW

11.6 Z1(s) = R1/(1 + R1C1s); Z2(s) = R2/(1 + R2C2s)

E s

E s2

1

( )

( )= - -

LNM

OQP

Z s

Z s

R

R2

1

4

3

( )

( )

= R R

R R

R C s

R C s4 2

3 1

1 1

2 2

1

1

+

+

FHG

IKJ = 2.51

0 345 1

0185 1

.

.

s

s

+

+

FH

IK

This equation gives

R1 = 34.5 kW; R2 = 18.5 kW; R4 = 45.8 kW

11.7 By Trapezoidal rule for integration:

e t dtkT

( )0z = T

e e T e T e T e k T e kT( ) ( ) ( ) ( )...

( ) ( )0

2

2

2

1

2

++

++ +

- +LNM

OQP

= Te i T e iT

Ti

k ( ) ( )- +LNMM

OQPP=

å1

1

By backward-difference approximation for derivatives:

de t

dt t kT

( )

=

= e kT e k T

T

( ) (( ) )- -1

A difference equation model of the PID controller is, therefore, given by



SOLUTION MANUAL 111

u(k) = Kc e kT

T

e i e i T

Te k e k

I i

kD( )

( ) ( )[ ( ) ( )]+

- ++ - -

RSTUVW=

å1

21

1

11.8 Taking sampling frequency twenty times the closed-loop natural frequency(which is a measure of bandwidth), we have

ws = 20wn = 2pT

; this gives T = 0.5 sec

Settling time ts = 4/zwn = 13

1

10th of ts is 1.3 sec; T = 0.5 sec is therefore a safe choice.

U s

E s

( )

( ) =

2 2 0 1

0 01

. ( . )

.

s

s

+

+

&u(t) + 0.01 u(t) = 2.2 &e(t) + 0.22 e(t)

1

T[u(k) – u(k – 1)] + 0.01 u(k) =

2 2.

T [e(k) – e(k – 1)] + 0.22 e(k)

This gives

u(k) = 2

2 01. u(k – 1) +

4 62

2 01

.

. e(k) –

4 4

2 01

.

. e(k – 1)

Configuration of the digital control scheme is shown in Fig. 11.19.

11.9 (a) &u(t) + au(t) = K &e(t) + Kb e(t)

By backward-difference approximation:

u k u k

Tau k

( ) ( )( )

- -+

1 =

K

T [e(k) – e(k – 1)] + Kbe(k)

By backward-rectangular rule for integration:

u(t) = u(0) – a u d Ke t Ke Kb e dtt

( ) ( ) ( ) ( )t t t t+ - + zz 000

u(k) = u(k – 1) – aT u(k) + K e(k) – K e(k – 1) + Kb T e(k)

By both the approaches of discretization, we get the followingcomputer algorithm.

u(k) = 1

11

1

1 11

+- +

+

+-

+-

aTu k

K bT

aTe k

K

aTe k( )

( )( ) ( )




(b) u(k) = u(k– 1) – aT u k u k( ) ( )+ -LNM

OQP

1

2 + Ke(k) – Ke(k – 1)

+ KbT e k e k( ) ( )+ -LNM

OQP

1

2

= 1

2

12

-

+

aT

aT u(k – 1) + K

bT

aT

12

12

+FH

IK

+

e(k) – K

bT

aT

12

12

-FH

IK

+

e(k – 1)

11.10 For a choice of T = 0.015 sec, sampling frequency ws = 2p /T = 418.88rad/sec which is about 11 times the closed-loop bandwidth. This is a safechoice.

U s

E s

( )

( )=

s s

s

2

2

236

11236

1

7401

( )

.+ +

+FH

IK

1.826 × 10–6 &&u(t) + 2.7 × 10–3 &u(t) + u(t)

= 7.7 × 10–4 &&e(t) + 0.03 &e(t) + e(t)

Using the results of Eqns (11.36) – (11.37), we get

1.826 × 10–6 u k u k u k

T

( ) ( ) ( )- - + -LNM

OQP

2 1 22

+ 2.7 × 10–3 u k u k

T

( ) ( )- -LNM

OQP

1 + u(k)

= 7.7 × 10–4 e k e k e k

T

( ) ( ) ( )- - + -LNM

OQP

2 1 22

+ 0.03 e k e k

T

( ) ( )- -LNM

OQP

1 + e(k)

This gives

1.1881 u(k) – 0.1962 u(k – 1) + 0.0081 u(k – 2)

= 6.42 e(k) – 8.84 e(k – 1) + 3.42 e(k – 2)

By direct digital design, we will be able to achieve the desired closed-loopperformance using longer value of sampling interval.



SOLUTION MANUAL 113

11.11B s

Y s

( )

( ) =

K s

st

t +1

t &b(t) + b(t) = Kt &y(t)

b(t) = b(0) – 1

00

tq q

tb d

Ky t yt

t

( ) [ ( ) ( )]+ -zb(k) = b(k – 1) –

T b k b k Ky k y kt

t t( ) ( )

[ ( ) ( )]+ -L

NMOQP + - -

1

21

=1 2

1 2

-

+

T

T

/

/

tt

b(k – 1) + K

Ty k

K

Ty kt t/

/( )

/

/( )

tt

tt1 2 1 2

1+

-+

-

11.12 The set-point control, proportional action and derivative action are allincluded in the op amp circuit of Fig. 11.43. Only the integral action is tobe added. Connect the integral-action unit of Fig. 11.41 in cascade withthe proportional action unit of Fig. 11.43; the output e4 of the proportional-action unit of Fig. 11.43 becomes input to the integral-action unit (outpute3).

The resulting op amp circuit is governed by the following relation (referExamples 11.6 and 11.7);

E3(s) = Kc (1 + TDs) 1 1+

FHG

IKJT sI

[Y(s) – R(s)]

Kc =R

R2

2¢ ; TD = RDCD ; TI = RI CI

11.13 E2(s) = Y(s) – R(s) ; E3(s) = – TD sY(s) ; TD = RDCD

E4(s) = Kc [E3(s) – E2(s)] ; Kc = R

R2

2¢

= Kc [R(s) – (1 + TDs) Y(s)]

11.14 Kcu = 1.2, Tu = 4.5 min

From the tuning rules given in Table 11.1:

Kc = 0.45Kcu = 0.54 or 1

0 54. × 100 = 185% PB

TI = Tu/1.2 = 3.75 min or 1

3 75. = 0.266 repeats/min




(b) Kc: 110% PB ; TI: 0.355 repeats/min ; TD: 0.45 min

11.15 From the process reaction curve (refer Section 11.9) we obtain

tD = 6 sec, t = 12 sec, K = 1

From tuning rules given in Table 11.2;

Kc = 1.8, TI = 19.98 sec

11.16 The model obtained from process reaction curve (refer Section 11.9):

K = 1 ; t = 1.875 ; tD = 1.375

From tuning rules given in Table 11.2:

(a) Kc = 1.36

(b) Kc = 1.23 ; TI = 4.58 min

11.17 tND = 0.2 for process A, 0.5 for process B, and 0.5 for Process C. ProcessA is more controllable than processes B and C, which are equallycontrollable.

11.18 (a) From the tuning rules given in Table 11.2:

¢Kc = 3 ; ¢TI = 4 min ; ¢TD = 1 min

(b) tCD = tD + 1

2 T = 2 min +

1

2 × 8 sec = 2.067 min.

Replacing tD by tCD in the tuning formulas, we obtain

¢Kc = 2.9 ; ¢TI = 4.13 min ; ¢TD = 1.03 min

11.19 Replace tD in tuning formulas of Table 11.2 by tCD = tD + 1

2 T. Increase

in sampling time demands reduced Kc and increased TI and TD.

11.20 From tuning rules given in Table 11.1:

Kc = 2.25 ; TI = 28.33 sec

Correction to account for sampling is not required since the test hasdirectly been conducted on the digital loop.



CHAPTER 12 CONTROL SYSTEM ANALYSIS USINGSTATE VARIABLE METHODS

12.1

+

–

x4 x3 x2 x1

Kg

Kb

iaifu

KT1 1 11

sL + Rf f sJ + Be e sn

s a g( )L + L + R +Ra g

qM qLqM

Je = n2J = 0.4 ; Be = n2B = 0.01

&x1 = x2 ; 0.4&x2 + 0.01x2 = 1.2x3

0.1&x3 + 19x3 = 100 x4 – 1.2x2 ; 5 &x4 + 21x4 = 4

or &x = Ax + bu

A =

0 1 0 0

0 0 025 3 0

0 12 190 1000

0 0 0 4 2

-

- -

-

L

N

MMMM

O

Q

PPPP

.

.

; b =

0

0

0

0 2.

L

N

MMMM

O

Q

PPPP

y = qL = nx1 = 0.5x1

12.2

x3 x2 x1

ifu qqKA KTs f fL + R + 1

1 1

Js + B

1s

&x1 = x2 ; 0.5&x2 + 0.5x2 = 10x3

20 &x3 + 100x3 = 50u ; y = x1

A =

0 1 0

0 1 20

0 0 5

-

-

L

NMMM

O

QPPP ; b =

0

0

2 5.

L

NMMM

O

QPPP ; c = 1 0 0




12.3 x1 = qM ; x2 = &q M , x3 = ia ; y = qL

&x1 = x2

2 &x2 + x2 = 38 x3

2 &x3 + 21x3 = ea – 0.5 x2

ea = k1(qR – qL) – k2 &q M = k1qR – k1

20 x1 – k2x2

A =

0 1 0

0 0 5 19

40

0 5

2212

1 2

-

- -+

-

L

N

MMMMM

O

Q

PPPPP

.( . )k k

; b =

0

0

21k

L

N

MMMM

O

Q

PPPP ; c =

1

200 0L

NMOQP

12.4 x1 = w ; x2 = ia

J &w + Bw = KT ia

Raia + La di

dta = ea – Kbw

ea = Kcec = Kc [k1 (er – Kt w) – k2 ia}

A =-

- + - +

L

N

MMMM

O

Q

PPPP

B

J

K

Jk K K K

L

R k K

L

T

t c b

a

a c

a

( ) ( )1 2 ;

b =0

1k K

Lc

a

L

NMM

O

QPP ; c = 1 0

12.5 A = P–1 AP = -

-

LNM

OQP

11 6

15 8

b = P–1 b = 1

2LNMOQP ; c = cP = 2 1-

Y s

U s

( )

( )=

P1 1D

D =

s

s s

-

- -- - -

2

1 21 3 2( ) =

1

3 22s s+ +



SOLUTION MANUAL 117

X2X = Y1

s–1

–2

–3

11U

s–1

Y s

U s

( )

( )=

P P P P1 1 2 2 3 3 4 4D D D D

D

+ + +

= 2 1 8 15 1 2 1 11 2 6 2 1

1 11 8 15 6 11 8

1 1 2 1 1 1 1

1 1 1 1 1 1

s s s s s s s

s s s s s s

- - - - - - -

- - - - - -

- + + - + +

- - + + - + -

( ) ( ) ( )[ ] ( ) [ ]

[ ( )( )] [( )( )]

= 1

3 22s s+ +

X2

X1

Y

s–1

–1

6

2

–15–11

2

1

8

U

s–1

12.6 A = 0 1

0 0LNM

OQP ; b =

0

1LNMOQP

A = P–1 AP = 1 1

–1 –1LNM

OQP ; b= P–1 b =

0

1LNMOQP

|lI – A| = |lI – A | = l2

12.7 X(s) = (sI – A)–1 x0 + (sI – A)–1 b U(s)

= G(s)x0 + H(s) U(s)

G(s) = 1

D

s s s

s s s

s s

( )

( )

+ +

- +

- -

L

NMMM

O

QPPP

3 3 1

1 3

1 2

; H(s) = 1

1

2Ds

s

L

NMMM

O

QPPP




D = s3 + 3s2 + 1

12.8

–3

–1

u

x30 x2

0x10

x1x2x3

+

++

+

+

+

+

+

+

+

12.9 Taking outputs of integrators as state variables, we get (x1 being the outputof rightmost integrator),

&x1 = x2

&x2 = – 2x2 + x3

&x3 = – x3 – x2 – y + u

y = 2x1 – 2x2 + x3

A =

0 1 0

0 2 1

2 1 2

-

- -

L

NMMM

O

QPPP ; b =

0

0

1

L

NMMM

O

QPPP ; c = 2 2 1-

12.10 (a) G(s) = c (sI – A)–1 b

=s

s s

+

+ +

3

1 2( )( )

(b) G(s) =1

1 2( )( )s s+ +

12.11 &x1 = – 3x1 + 2x2 + [– 2x1 – 1.5x2 – 3.5x3]

&x2 = 4x1 – 5x2

&x3 = x2 – r



SOLUTION MANUAL 119

A =

- -

-

L

NMMM

O

QPPP

5 0 5 3 5

4 5 0

0 1 0

. .

; b =

0

0

1-

L

NMMM

O

QPPP ; c = 0 1 0

G(s) = c(sI – A)–1 b = 14

1 2 7( )( )( )s s s+ + +

12.12 (a) x1 = output of lag 1/(s + 2)

x2 = output of lag 1/(s + 1)

&x1 + 2x1 = x2 ; &x2 + x2 = – x1 + u

y = x2 + (– x1 + u)

A =-

- -

LNM

OQP

2 1

1 1 ; b =

0

1LNMOQP ; c = -1 1 ; d = 1

(b) x1 = output of lag 1/(s + 2)

x2 = output of lag 1/s

x3 = output of lag 1/(s + 1)

&x1 + 2x1 = y ; &x2 = – x1 + u

&x3 + x3 = – x1 + u ; y = x2 + x3

A =

-

-

- -

L

NMMM

O

QPPP

2 1 1

1 0 0

1 0 1

; b =

0

1

1

L

NMMM

O

QPPP ; c = 0 1 1

12.13(i) Y s

U s

( )

( )=

s

s s

+

+ +

3

1 2( )( ) = 2

1

1

2s s++

+

–

A =-

-

LNM

OQP

1 0

0 2 ; b =

1

1LNMOQP ; c = 2 1-




–1

2

–2

u +

+

+

+

+

+

x2

x1

y–1

(ii) Y s

U s

( )

( ) =

ba a a

3

s s s31

22 3+ + +

= 5

4 5 23 2s s s+ + +

From Eqns (12.46), the second companion form of the state model is givenbelow.

A =

0 0 2

1 0 5

0 1 4

-

-

-

L

NMMM

O

QPPP ; b =

5

0

0

L

NMMM

O

QPPP ; c = 0 0 1

2u + + +

– – –

x2x1 x3 = y

2 5 4

5

(iii) Y s

U s

( )

( )=

b b b ba a a

03

12

2 33

12

2 3

s s s

s s s

+ + +

+ + + =

s s s

s s s

3 2

3 2

8 17 8

6 11 6

+ + +

+ + +

From Eqns (12.44), the state model in second companion form is givenbelow.



SOLUTION MANUAL 121

A =

0 1 0

0 0 1

6 11 6- - -

L

NMMM

O

QPPP ; b =

0

0

1

L

NMMM

O

QPPP ; c = 2 6 2

6

8

y

81

6

17

11

u +

+ + +

+ + +

+ +++

–

x2 x1x3zz zz zz

12.14 (i)Y s

U s

( )

( )=

s

s s s

+

+ +

1

3 23 2 = b b

a a a2 3

31

22 3

s

s s s

+

+ + +

From Eqns (12.46):

A =

0 0 0

1 0 2

0 1 3

-

-

L

NMMM

O

QPPP ; b =

1

1

0

L

NMMM

O

QPPP ; c = 0 0 1

(ii)Y s

U s

( )

( )=

1

6 11 63 2s s s+ + + =

ba a a

33

12

2 3s s s+ + +

From Eqns (12.44):

A =

0 1 0

0 0 1

6 11 6- - -

L

NMMM

O

QPPP ; b =

0

0

1

L

NMMM

O

QPPP ; c = 1 0 0

(iii)Y s

U s

( )

( ) =

s s s

s s s

3 2

3 2

8 17 8

6 11 6

+ + +

+ + + = 1

1

1

2

2

1

3+

-

++

++

+s s s




A =

-

-

-

L

NMMM

O

QPPP

1 0 0

0 2 0

0 0 3

b =

1

1

1

L

NMMM

O

QPPP c = -1 2 1 ; d = 1

12.15 (a)Y s

U s

( )

( )=

1000 5000

52 1003 2

s

s s s

+

+ + =

b ba a a

2 33

12

2 3

s

s s s

+

+ + +

From Eqns (12.44):

A =

0 1 0

0 0 1

0 100 52- -

L

NMMM

O

QPPP ; b =

0

0

1

L

NMMM

O

QPPP ; c = 5000 1000 0

(b)Y s

U s

( )

( ) =

1000 5000

52 1003 2

s

s s s

+

+ + =

50 3125

2

18 75

50s s s+-

++-

+

. .

A =

0 0 0

0 2 0

0 0 50

-

-

L

NMMM

O

QPPP ; b =

50

3125

18 75

-

-

L

NMMM

O

QPPP

.

.

; c = 1 1 1

12.16 Y s

U s

( )

( ) =

2 6 5

1 2

2

2

s s

s s

+ +

+ +( ) ( ) =

1

1

1

1

1

22( )s s s++

++

+

From Eqns (12.55):

A =

-

-

-

L

NMMM

O

QPPP

1 1 0

0 1 0

0 0 2

; b =

0

1

1

L

NMMM

O

QPPP ; c = 1 1 1

x(t) =0

t

z eA(t – t) bu(t)dt

e–At = L –1 [(sI – A)–1] =

e te

e

e

t t

t

t

- -

-

-

L

N

MMM

O

Q

PPP

0

0 0

0 0 2



SOLUTION MANUAL 123

e u dtt

A b( ) ( )-z t t t0

=

( ) ( )

( )

( )

t e d

e d

e d

tt

tt

tt

-

L

N

MMMMMMMMM

O

Q

PPPPPPPPP

- -

- -

- -

zzz

t t

t

t

t

t

t

0

0

2

0

=

1

11

21 2

- -

-

-

L

N

MMMM

O

Q

PPPP

- -

-

-

e te

e

e

t t

t

t( )

y = x1 + x2 + x3 = 2.5 – 2e–t – te–t – 0.5 e–2t

u

y

++

+

+

+

+

+

–

x3

zz

zz

zz

1

1

1

–1

–2

–1

1

x2x1

12.17 F = eAT = 0 696 0 246

0123 0 572

. .

. .LNM

OQP

g = (e–AT – I ) A–1 b

=– . .

. .

0 304 0 246

0123 0 428-

LNM

OQP- -

- -

L

N

MMM

O

Q

PPP

-LNM

OQP

3

4

1

21

4

1

2

1

5 =

-LNM

OQP

0 021

0 747

.

.

c = 2 4- ; d = 6




12.18Y s

U s

( )

( )=

e

s

s-

+

0 4

1

.

& ( )x t1 = – x1(t) + u(t – 0.4)

tD = 0.4 ; therefore, N = 0, D = 0.4, m = 0.6

F = e–1 = 0.3679

g2 = e d-z s s0

0 6.

= 0.4512 ; g1 = e–0.6 e d-z q q0

0 4.

= 0.1809

x1(k + 1) = 0.3679 x1(k) + 0.1809 u(k – 1) + 0.4512u(k)

Introduce a new state x2(k) = u(k – 1)

x(k + 1) = Fx(k) = gu(k)

y(k) = cx(k)

F =0 3679 01809

0 0

. .LNM

OQP ; g =

0 4512

1

.LNM

OQP

c = 1 0

12.19

s–1

s–1

s–1

s–1

–2

x1x2

x20

x20

–2

1 1

1

1

u

G(s) =G s G s

G s G s11 12

21 22

( ) ( )

( ) ( )LNM

OQP ; H(s) =

H s

H s1

2

( )

( )LNM

OQP

X s

x1

10

( )= G11(s) =

s s

s s s s s

- -

- - - - -

+

- - - + + - -

1 1

1 1 2 1 1

1 2

1 2 2 2 2

( )

( ) ( )( )



SOLUTION MANUAL 125

=s s- -

+1 11 2( )

D =

1 2

1

1 2

3

/ /

s s++

+

X s

x1

20

( )= G12(s) =

s-2 1( )

D =

1 2

1

1 2

3

/ /

s s++-

+

X s

x2

10

( )= G21(s) =

s-2 1( )

D =

1 2

1

1 2

3

/ /

s s++-

+

X s

x2

20

( )= G22(s) =

s s- -+

1 11 2( )

D =

1 2

1

1 2

3

/ /

s s++

+

X s

U s1( )

( )= H1(s) =

s s s- - -+ +

2 1 11 1 2( ) ( )

D =

1

1s+

X s

U s2( )

( )= H2(s) =

1

1s+

Zero-input response:

x(t) = eAt x0 = L-1 [G(s)x0]

= 1

2

3 3

3 310

20

e e e e

e e e e

x

x

t t t t

t t t t

- - - -

- - - -

+ -

- +

LNM

OQPLNM

OQP

Zero-state response:

x(t) = e dtt

A b( ) ( )-z t u t t0

= L –1 [H(s)U(s)] = 1

1

-

-

LNM

OQP

-

-

e

e

t

t

12.20 A =0 1

2 3- -

LNM

OQP

; eigenvalues are l1 = – 1, l2 = – 2

Y(s) = c(sI – A)–1 x0 + c(sI – A)–1 b U(s)

=s

s s s s s

+

+ ++

+ +

4

1 2

1

1 2( )( ) ( )( )

=3

1

2

2

1 2 1

1

1 2

2s s s s s+-

++ -

++

+

/ /

y(t) =1

22

3

22

+ -- -e et t




12.21 &x1 = – 3x1 + 2x2 + [7r – 3x1 – 1.5 x2]

&x2 = 4x1 – 5x2

A =-

-

LNM

OQP

6 0 5

4 5

. ; b =

7

0LNMOQP ; c = 0 1

eAt = L –1 [(sI – A)–1]

=

1

3

2

3

0 5

3

0 5

34

3

4

3

2

3

1

3

4 7 4 7

4 7 4 7

e e e e

e e e e

t t t t

t t t t

- - - -

- - - -

+ -

- +

L

N

MMM

O

Q

PPP

. .

y(t) = x2(t) = 74

3

4

34 7

0

e e dt tt

- - - --

LNM

OQPz ( ) ( )t t t

=28

3

1

41

1

714 7( ) ( )- - -

LNM

OQP

- -e et t

12.22 eAt = e–t

+

- -

LNM

OQP

- -

- - -

te te

te e te

t t

t t t

x(t) = e tt tA x( ) ( )- 00

Given: x1(2)= 2 ; t0 = 1, t = 2

Manipulation of the equation gives

x1(2) = 2e–1 x1(1) + e–1 x2(1) = 2

If x2(1) = 2k, then x1(1) = e1 – k

Thus

e k

k

1

2

-LNM

OQP is a possible set of states

x

x1

2

1

1

( )

( )LNM

OQP for any k ¹ 0

12.23e

e

t

t

-

--

LNM

OQP

2

22 =

a b

c dLNM

OQP -

LNM

OQP

1

2;

e

e

t

t

-

--

LNM

OQP

= a b

c dLNM

OQP -

LNM

OQP

1

1

This gives

eAt =a b

c dLNM

OQP =

2

2 2 2

2 2

2 2

e e e e

e e e e

t t t t

t t t t

- - - -

- - - -

- -

- -

LNM

OQP



SOLUTION MANUAL 127

= L –1 [(sI – A)–1]

(sI – A)–1 =

2

1

1

2

1

1

1

22

2

2

1

2

2

1

1

s s s s

s s s s

+-

+ +-

+

+-

+ +-

+

L

N

MMM

O

Q

PPP

(sI – A) =s

s

-

+

LNM

OQP

1

2 3

A =0 1

2 3- -

LNM

OQP

12.24 V =

c

cA

cA

M

n-

L

N

MMMM

O

Q

PPPP1

is a triangular matrix with diagonal elements equal to unity;|V| = (– 1)n for all ai’s. This proves the result.

12.25 U = [b Ab ... An–1 b] is a triangular matrix with diagonal elements equalto unity; |U| = (– 1)n for all ai’s. This proves the result.

12.26 (i) U = [b Ab] = 1 2

0 1

-LNM

OQP ; r(U) = 2

Completely controllable

V =c

cALNM

OQP =

1 1

3 3

-

-

LNM

OQP ; r(V) = 1

Not completely observable.

(ii) The system is in Jordan canonical form; it is controllable but notobservable.

(iii) The system is in controllable companion form. The given system istherefore controllable. We have to test for observability property only.

r(V) = rc

cA

cA2

L

NMMM

O

QPPP= 3

The system is completely observable.




(iv) Given system is in observable companion form. We have to test forcontrollability property only.

r(U) = r[b Ab A2b] = 3

The system is completely controllable.

12.27 (i) Observable but not controllable

(ii) Controllable but not observable

(iii) Neither controllable nor observable.

Controllable and observable realization:

A = – 1 ; b = 1 ; c = 1

12.28 (i) G(s)= c(sI – A)–1 b = 1

2s+

Given state model is in observable companion form. Since there is apole-zero cancellation, the state model is uncontrollable.

(ii) G(s) =s

s s

+

+ +

4

2 3( )( )

The given state model is in controllable companion form. Since thereis a pole-zero cancellation, the model is unobservable.

12.29(a) |lI – A| = (l – 1) (l + 2) (l + 1)

The system is unstable.

(b) G(s) = c(sI – A)–1 b

=1

1 2( )( )s s+ +

The G(s) is stable

(c) The unstable mode et of the free response is hidden from the transferfunction representation

12.30 (a) G(s) =102s s+

= b

a a2

21 2s s+ +

A =0 1

0 1-LNM

OQP ; b =

0

1LNMOQP ; c = [10 0]

(b) Let us obtain controllable companion form realization of

G(s)=10 2

1 2

( )

( )( )

s

s s s

+

+ + =

10 20

3 23 2

s

s s s

+

+ + =

b ba a a

2 33

12

2 3

s

s s s

+

+ + +



SOLUTION MANUAL 129

A =

0 1 0

0 0 1

0 2 3- -

L

NMMM

O

QPPP ; b =

0

0

1

L

NMMM

O

QPPP ; c = 20 10 0

(c) Let us obtain observable companion form realization of

G(s) =10 2

1 2

( )

( )( )

s

s s s

+

+ +

A =

0 0 0

1 0 2

0 1 3

-

-

L

NMMM

O

QPPP ; b =

20

10

0

L

NMMM

O

QPPP ; c = 0 0 1



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Control Systems by Nagrath & Gopal Solution Manual