Solution Manual - Control Systems by Gopal

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CHAPTER 2 2.1

DYNAMIC MODELS AND DYNAMIC RESPONSE

Revisiting Examples 2.4 and 2.5 will be helpful. (a)

(b)

R(s) = A;

Y ( s) =

y(t) =

KA - t / t e ; t

R(s) =

A ; s

KA t s +1 yss = 0

Y(s) =

KA s (τs + 1)

y(t) = KA (1 − e − t / τ ) ; (c)

R(s) =

yss = KA

A KA ; Y(s) = 2 s2 s (τ s + 1)

y(t) = KA (t − τ + τe − t / τ ) ; (d)

KAω (τs + 1)( s 2 + ω 2 )

R(s) =

Aw ; s2 + w 2

y(t) =

KAωτ − t / τ KA sin(ω t + θ ) e + 2 2 2 τ ω +1 τ ω2 +1

θ y( t ) 2.2

yss = KA (t - t )

Y(s) =

= tan −1 ( −ωτ ) t®¥

=

KA

τ ω2 +1 2

sin(ω t + θ )

(a) The following result follows from Eqns (2.57 - 2.58) y(t) = 1 −

e −ζω n t 1−ζ2

F GH

sin ω d t + 2 tan −1

1−ζ2 ζ

I JK

ωd = ωn 1−ζ2 The final value theorem is applicable: the function sY(s) does not have poles on the jw-axis and right half of s-plane. yss = lim sY ( s ) = 1 s® 0

(b)

The following result follows from Review Examples 2.2.



SOLUTION MANUAL

y(t) = t −

yss = t −

F GH

2ζ e −ζω n t 1−ζ2 sin ω d t + 2 tan −1 + ωn ωd ζ

3

I JK

2ζ ωn

The final value theorem is not applicable; the function sY(s) has a pole on the jw -axis. 2.3

Revisiting Review Example 2.1 will be helpful. E0 (s ) R / 10 4 10 = ; Ei ( s ) = 2 Ei ( s) RCs + 1 s

e0 (t ) = 10 ´

R - t /t ( ); t = RC 4 t -t +te 10

For the output to track the input with a steady-state delay 100 × 10–6 sec, it is necessary that

R = 1, and t = RC = 100 ´ 10-6 10 4 This gives

R = 10 k W ;

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

FG H

Steady-state error = 25t - 25t - 25 ´

2z wn

IJ K

= 1.5 2.5

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

ωn

1 s( s + 10 s + 100 ) 2

= 10, ζ = 0.5, wd = 5 3

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



4

CONTROL SYSTEMS: PRINCIPLES AND DESIGN

LM N

2 -5 t e sin(5 3 t + tan -1 3 3

y(t) = 0.01 1 -

2.6

Y ( s) 6 6 = G( s ) = 2 = R( s) s + 7s + 6 ( s + 1)(s + 6) 3

| G( jω )|ω = 2 =

5 2 3

yss =

5 2

; ∠ G(j2) = – 81.87º sin( 2t − 81.87º )

3 21 sin 2t – cos 2t 50 50

=

2.7

OP Q

Y (s) s+3 s+3 = G(s) = 2 = R( s ) ( s + 2 )( s + 5) s + 7s + 10 (a)

R(s) =

1 s +1

y(t) =

FH 1 e 2

−t

IK

1 1 − e −2 t − e −5 t µ ( t ) 3 6

(s2 + 7s + 10) Y(s) – sy(0) – y& (0) – 7y(0)

(b)

= (s + 3) R(s) – sr(0) Initial conditions before application of the input are y(0–) = 1, y& (0–) =

2.8

1 , r(0–) = 0 2

Y(s) =

s + 15 / 2 s+3 + ( s + 2 )( s + 5) ( s + 2 )( s + 5)( s + 1)

y(t) =

1 - t 3 -2t e + e - e -5t 2 2

X = X + x; I = I + i

M

d 2 (X + x) = F ( X + x , I + i ) − Mg dt 2



SOLUTION MANUAL

FG H

∂F Mx&& = F ( X + I ) + ∂X

X ,I

5

IJ x + FG ∂F IJ i − Mg K H ∂I K X ,I

−3 F ( X + I ) = Mg = 8.4 × 10 × 9.8 Newtons

For this value of force, we get from Fig. 2.8b,

X = 0.27cm ; I = 0.6 amps Again from Fig. P 2.8b,

K1 =

¶F ¶X

X ,I

K2 =

¶F ¶I

X ,I

x&& =

2.9

= 0.14 Newtons/cm

= 0.4 Newtons/amp

K1 K X (s ) 47.6 x + 2 i; = 2 M M I (s ) s - 16.67

Mx&& + Bx& + Kx = K1 ( y - x )

Gravitational effect has been eliminated by appropriate choice of the zero position.

X (s) 0.1667 = G( s) = Y (s ) ( 0.0033 s + 1)( 0.0217 s + 1) w = 2p v / l = 11.63 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 )

Gravitational effect has been eliminated by appropriate choice of the zero position. X ( s) 1 F( s) = Ms 2 + Bs + K

Force transmitted to the ground = K X(s) + Bs X(s)



6


=

FH B s + 1IK F(s) K M 2 B s + s +1 K K A sin wt

F(t) =

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

FH B ω IK K FG1 − Mω IJ + F Bω I H K K HKK 2

A 1+

=

2

2

2

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

X (s) Bs + K = 2 Y (s ) Ms + Bs + K Peak amplitude of machine vibration

FH B vIK K FG1 - Mv IJ + F Bv I H K K H `K K 2

A 1+

=

2

2

2

2.11 M1 && y1 + K1 ( y1 − y0 ) + B1 ( y&1 − y&0 ) + K 2 ( y1 − y2 ) + B2 ( y&1 − y&2 ) = 0 M2 && y2 + K 2 ( y2 − y1 ) + B2 ( y&2 − y&1 ) = 0

Gravitational effect has been eliminated by appropriate choice of the zero position. 2.12 The following result follows from Section 11.2 (Eqn (11.9)). (a)

E0 (s ) 1 + 2 RC2 s + R2 C1C2 s 2 = Ei ( s) 1 + R(C1 + 2C2 )s + R2C1C2 s2

(b)

E0 (s ) 1 + 2 R1Cs + R1 R2C 2 s 2 = Ei ( s) 1 + (2 R1 + R2 )Cs + R1 R2 C 2 s 2

2.13 Refer Example 11.6 2.14 Refer Example 11.7



SOLUTION MANUAL

2.15 The following result follows from Example 12.9.

x1 = f, x 2 = f& , x 3 = z, x 4 = z&, r = F(t ) x& = Ax + br

A =

LM 0 MM4.4537 0 MN0.5809

OP LM 0 OP 0 P1P ; b = MM−0.3947 P 0 P MN 0.9211 PQ 0 PQ

1 0 0 0 0 0 0 0 0

2.16 Mixing valve obeys the following equations: (Qi + qi ) r cq H + [Q - (Qi + qi )] r cq C = Qr c(Qi + q i )

q i = kv x The perturbation equation is K v (q H - q C ) x(t) = Q q i (t )

x(t) = Kq i (t ); K = Q / [ K v (q H - q C )]

or,

The tank obeys the equations Vr c

dq = Qr c(q id - q ); q id (t ) = q i ( t - t D ) dt

This gives q (s ) e -1. 5s = q i ( s) s +1

2.17 C1 C2

q1 - q 2 1 dθ 1 = qm (t )l - R ; R1 = UA 1 dt

dq 2 dt

=

q1 - q 2 q i - q 2 + ; R1 R2

C2 = Vrc; R2 =

1 Qrc

From these equations, we get

θ& 1 = −1.92 θ 1 + 1.92 θ 2 + 4.46 q m θ& 2 = 0.078 q 1 - 0.2 q 2 + 0.125 q i


7


8


2.18 At steady-state 0 =

q1 - q 2 q i - q 2 ; this gives θ 1 = 120 º C + R1 R2

0 = Qm λ −

θ1 − θ 2 ; this gives Qm = 17.26 kg / min R1

2.19 Energy balance on process fluid:

V2 ρ 2 c2

d (θ 2 + θ 2 ) = Q2 ρ 2 c2 [θ i 2 − (θ 2 + θ 2 )] − UA[θ 2 + θ 2 − θ 1 − θ 1 ] dt

At steady-state 0 = Q2 r 2 c2 (q i 2 - q 2 ) - UA(q 2 - q 1 ); this gives q1 = 40 º C The perturbation equation is V2r 2c2

dq 2 - Q2 r 2c2q 2 - UA(q 2 - q 1 ) dt

544.5 +

dq 2 + q 2 = 0.432 q1 dt

This gives

Energy balance on the cooling water: V1r1c1

d q1 + q 1 = (Q1 + q1 )r 1c1 [q i1 - (q 1 + q 1 )] dt

d

i

+ UA[q 2 + q 2 - q1 - q 1 ] At steady-state 0 = Q1r1c1[q i1 - q1] + UA[q 2 - q1 ]; this gives –3 3 Q1 = 5.28 ×10 m /sec

The perturbation equation is V1r1c1

dq 1 = (θ i1 − θ 1 ) ρ1c1q1 − Q1 ρ1c1θ 1 + UA (θ 2 − θ 1 ) dt



SOLUTION MANUAL

This gives 184.55

dq1 + q1 = 0.465 q 1 – 1318.2 q1 dt

Manipulation of the perturbation equation gives q 2 (s ) 557.6 =– 3 2 q 1 ( s) 100.5 × 10 s + 729s + 0.8

2.20 Tank 1: C1

dp1 = q1 – q10 – q11 dt

q10 = flow through R0; q11 = flow through R1 A1 dh rgh1 rg(h1 - h2 ) ´ rg 1 = q1 R0 R1 dt rg

F GH

I JK

1 1 rg 1 rg dh1 + h1 + h2 + q1 =A1 R0 R1 A1 R1 A1 dt

or

= – 3h1 + 2h2 + r1 Tank 2: C2

dp2 = q2 + q11 – q20 dt

q20 = flow through R2 A2 dh rg(h1 - h2 ) rgh2 ´ rg 2 = + q2 dt rg R1 R2

FG H

IJ K

1 1 rg rg 1 dh2 + h1 h2 + q2 = A R A R R A dt 2 1 2 1 2 2

or

= 4h1 – 5h2 + r2 2.21 A

d ( H + h) = Q1 + q1 + Q2 + q2 - Q - q dt

At steady-state 0 = Q1 + Q2 - Q ; this gives Q = 30 litres/sec The perturbation equation is


9


10


A

dh = q1 + q 2 – q dt

The turbulent flow is governed by the relation Q(t) = K H(t ) = f(H) Linearizing about the operating point, we obtain Q(t) = f ( H ) +

= Q+

FG ¶f ( H) H ¶H

K

H= H

IJ (H - H ) K

h(t )

2 H

Therefore q(t) =

K 2 H

h(t ) =

K H Q h (t ) h(t ) = 2H 2H

1 1 1 dh(t ) = - q(t ) + q1(t ) + q2 (t ) dt A A A

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

A

d [( H + h( t ) ( C + c ( t ))] = C1[Q1 + q1(t )] + C2 [Q2 + q2 (t )] dt − [ C + c ( t )][ Q + q ( t )]

At steady state, 0 = C1Q1 + C2Q2 - CQ ; this gives C = 15 The perturbation equation is AH

dc( t ) dh( t ) = C1q1(t ) + C2q2 (t ) - Cq(t ) - Q c(t ) + AC dt dt

This gives dc(t ) = – 0.02 c(t) – 0.004 q1(t) + 0.002 q2(t) dt



SOLUTION MANUAL

2.22 e − τ D s = 1 - t D s +

t 2D s 2 t 3D s 3 t 4D s 4 t 5D s 5 + + ... 2! 3! 4! 5!

It is easy to calculate with long division that

1- t Ds / 2 t 2 s2 t 3 s3 = 1 - t D s + D - D + ... 2 4 1+ tDs / 2 1 - t D s / 2 + t D2 s 2 / 12 t D2 s 2 t D3 s 3 = 1 t + s D 2 6 1 + t D s / 2 + t D2 s 2 / 12

+

t D4 s 4 t D5 s 5 + - ... 12 144


11


CHAPTER 3

MODELS OF INDUSTRIAL CONTROL DEVICES AND SYSTEMS

3.1

R +

+ –

–

Y

G2 G3 +G4

+

G1

– H2 H1G2 G2G3 + G4

Y (s ) G1 (G2 G3 + G4 ) = R( s ) 1 + (G2 G3 + G4 )(G1 + H2 ) + G1 H1G2

3.2 H2 – R

G1

+

+

G2

G3

+

– (1–G1)H1 G4

G1G2G3 Y ( s) G4 + = R( s ) 1 + G2 G3 H2 + G2 H1 (1 - G1 )

3.3

+

Y ( s) ( P D + P D + P3 D 3 + P4 D 4 ) / D R( s ) = 1 1 2 2

P1 = G1G2; P2 = G1G3; P3 = G4H2G1G2; P4 = G4H2G1G3 ∆ = 1 – (– G1G3H1H2 – G1G2H1H2) ∆ 1 = ∆2 = ∆3 = ∆4 = 1


Y


SOLUTION MANUAL

13

Y ( s) (1 + G4 H2 )(G2 + G3 )G1 = 1 + G H H (G + G ) R( s ) 1 1 2 2 3

G2 R

G3

G1

1

Y

1

H1

–H2 – G4

3.4 H3 /G 3 R

+

G1

–

–

+

G3

G2

Y

1+G3 H 2

H1

Y (s) R(s ) W =0

= M(s) =

G1G2 G3 1 + G2 H3 + G3 H2 + G1G2 G3 H1 H3 G 2

– (G 1 H1 G 2+H 2)

Y (s ) W ( s)

+

= MW (s) = R=0

–

+

W +

G3 (1 + H 3G2 ) 1 + H3G2 + G3 ( G1 H1G2 + H 2 )


G3

Y


14


3.5 –1 R1

G1

1

G2

1

1

Y1

G6

1

Y2

1 Js

w

H1

H2 R2

G3

G4

1

G5

–1

Y1 R1

R2 = 0

=

P1D1 D

G1G2 G3 (1 + G4 )

= 1 - (-G G - G + G H G G H ) + G G G 1 2 4 1 2 4 5 1 1 2 4

Y1 R2 Y2 R1

=

G4G5 H1G1G2G3 D

=

G1G4G5G6 H2 Y2 ; R 2 D

R1 = 0

R2 = 0

R1 = 0

=

G4G5G6 (1 + G1G2 ) D

3.6 er

+

KA

if

+

Kg

–

– e t

1 Ra

ia

Kb Kt

w (s ) 50 = Er ( s) s + 10.375

For Er(s) = 100/s,


KT


SOLUTION MANUAL

100 ´ 50 w (s) = s( s + 10.375) –10.375 t ) w(t ) = 481.8 (1 – e

3.7 qR + –

KP

1 sLf + Rf

KA

KT

Js2

1 + Bs

qM

1 qL 50

q L (s) 1 q R (s ) = s( 0.1s + 1)(0.2 s + 1) + 1

3.8

TM = K1 ec + K 2θ& M

=

J&eq q&& M + Beq q& M

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

K1 =

3- 0 = – 0.01 0 - 300

∆TM ∆ec

J eq = JM +

Beq

θ& M =0

FG N N IJ HN N K

2

1 3

2

FNN I =G H N N JK

J L = 0.0032

4

2

1 3

2

3- 2 = 0.1 30 - 20

=

BL = 0.00001

4

q M (s) 10 1 q L (s ) = ; = s(0.32 s + 1) s(0.32 s + 1) Ec (s ) Es c

F q& I +J G H q& JK F q& I +B G& J Hq K

2

3.9 (i)

J eq = JM

L

= 0.6

L

M

Beq = BM

L

L

2

= 0.015

M

KT = Kb


15


16


KT q M (s) = s[ J eq s + Beq )( sLa + Ra ) + KT K b ] Ea ( s ) 16.7

= s(s 2 + 100 s + 19.2) (ii) A sketch of multi-loop configuration may be made on the lines of Example 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

θ M (s) 13.33 = 2 θ R (s) s + 3.5s + 13.33 qR +

Ks

–

+

KA

–

ec

+

K1

TM

1 Js +B

–

qM

1 s

K2 Kt

3.11 qR +

Ks

–

KA

Kg + –

1 sLf +Rf

qM 1 Js + B

KT Ra

1 s

qM

Kb

θ M (s) θ (s) = L θ R (s) θ R (s ) 3.12 (a)

ω (s) = Er ( s )

=

15 × 10 3 s + 54 s + 200 s + 15 × 10 3 3

2

K ; K = 17.2, z = 1.066, w n = 61.2 1 2 2ζ + + s s 1 ωn ω 2n

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


qM


SOLUTION MANUAL

er +

KA

Kr

–

+ –

w

KT Js + B

1 sLa + Ra

Kb

Kt

3.13 TW

N(s) qR +

KT

KA

KP

TM

G(s)

+

–

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

B(q& M - q& L ) + K (q M - q L ) = JL q&&L + TW (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)

K P K A K T G( s ) qL ( s ) = 1 + K P K A K T G( s ) qR ( s) N ( s) qL ( s ) = 1 + K K K G( s ) P A T - TW ( s) G(s) =

Bs + K s [ JM JL s + ( JM + JL ) Bs + ( JM + JL ) K ] 2

2


–

17

qL


18


N(s) =

JM s 2 + Bs + K s 2 [ JM JL s 2 + ( JM + JL ) Bs + ( JM + JL ) K ]

Revisiting Example 2.7 will be helpful. 3.14 The required block diagram easily follows from Fig. 3.49. K1 x - Ay& = K 2 Dp

( K 2 D p)

A && K 2 = My

(r – y) Kp KA K = x

Y ( s) 2 = 2 R( s) s + 0.02 s + 2 K A K v K1 AK r / K2 Y ( s) = 2 Yr ( s) Ms + ( B + A2 / K2 )s + K A K v K1 K s A / K 2

3.15

Fw

er yr

Kr

ea

+

KA

–

x Kv

K1

+ –

A K2

–

+

y 1 Ms + B

1 s

y

e0 A

Ks

3.16 Revisiting Example 3.7 will be helpful. From geometry of the linkage, we get b a X ( s) Y (s ) a+b a+b bK r AK1 / ( a + b) K2 Y ( s) = 2 Ms + ( B + A 2 / K 2 )s + K + aAK1 / ( a + b) K2 Yr ( s)

E(s) =



SOLUTION MANUAL

yr

x

Kr

b a+b

e

+ –

K1

+

1 Ms2 + Bs + K

A K2

–

a a+b

19

y

sA

3.17 Let z(t) = displacement of the power piston d Y(s) = e - st D Z(s) ; t D = sec v Y ( s) G( s ) = Er ( s ) 1 + G( s ) H ( s ) ( K A K v K1 A / K 2 )e - t D ; H(s) = Ks s( Ms + B + A 2 / K 2 ) s

G(s) =

i er +

KA

–

Fw

x Kv

K1

+ –

A K2

+

–

z 1 Ms + B

e0 A

Ks

3.18

A1

rgh1 dh1 = q1 - qw - R dt 1

A2

rgh1 rgh2 dh2 = R - R dt 1 2


z 1 s

e–stD

y


20


Qw(s) Q1(s) +

– R2/rg (t 1s + 1) (t 2s + 1)

H2 (s)

t 1 = A1R1 ; t 2 = A2R2 rg rg hr

er +

Ks

Ka

–

Ke

Kv

q1

h2

h

q

e0 Ks

3.19 Ct

dq dt

= h-

q -qa Rt

qa

1/Rt h

+

+

Rt ts+ 1

q

t = Rt Ct er

+ –

Rf R

–1

–

eA

R2 R1

Kh

Kt

3.20 Heater:

C1

q1 - q dq 1 = hR1 dt



SOLUTION MANUAL

Air:

Vrc

q1 - q dq - Qrc(q - q i ) = R1 dt

These equations give us

1 t = R1C1, Kg = Qrc , w n = 2zw n =

Q R1C1V

R1C1Qrc + Vrc + C1 VR1C1 rc

qi ts + 1 Kg +

h

Kg

+ 1 wn2

qr

Ks

er

+

KA

–

eA

s2

2z s+1 wn

q

Kh

h

q

e0 Ks

3.21 Process: V2 r 2 c2

d (q 2 + q 2 ) = Q2 r 2 c2 (q i 2 + q i 2 ) - UA(q 2 + q 2 - q 1 - q 1 ) dt

– Q2 r 2 c2 (q 2 + q 2 ) Linearized equation is

V2 r 2 c2

dq 2 = Q2 r 2 c2q i 2 - UA(q 2 - q 1 ) - Q2 r 2 c2q 2 dt


21


22


Rearranging this equation and taking Laplace transform yields (t 2 s + 1)q 2 ( s) = K1′θ 1 ( s ) + K 2′ θ i 2 ( s ) t 2 = 544.5; K1′ = 0.423; K 2′ = 0 .577

Cooling water: V1 r1c1

d (q 1 + q 1 ) = (Q + q ) r c (q i + q i ) + UA(q + q 1 1 1 1 1 1 2 2 dt - q 1 - q 1 ) - (Q1 + q1 )r 1c1 (q 1 + q 1 )

Linearized equation is

V1 r1c1

dq 1 = Q1r 1c1q i1 + q i1 r1c1q1 + UA(q 2 - q 1 ) dt − Q1 ρ 1c1θ 1 − θ 1ρ 1c1q1

From this equation, we obtain (t 1s + 1)q 1 ( s) = − K3′ Q1 ( s ) + K 4′ θ 2 ( s ) + K 5′θ i1 ( s )

1.82 ´ 1000 ´ 4184 t 1 = 3550 ´ 5.4 + Q ´ 1000 ´ 4184 1 K3′ =

( -27 + q 1 ) ´ 1000 ´ 4184 3550 ´ 5.4 + Q1 ´ 1000 ´ 4184

3550 ´ 5.4 K 4′ = 3550 ´ 5.4 + Q ´ 1000 ´ 4184 1 K 5′ =

Q1 ´ 1000 ´ 4184 3550 ´ 5.4 + Q1 ´ 1000 ´ 4184

q 1 and Q1 are obtained from the following energy-balance equations at steady state:

Q1r 1c1 (q i1 - q 1 ) + UA(q 2 - q 1 ) = 0 Q2 r 2 c2 (q i 2 - q 2 ) - UA(q 2 - q 1 ) = 0 These equations yield –3 3 q 1 = 40ºC; Q1 = 5.28 × 10 m /sec



SOLUTION MANUAL

qi1

qi2

K¢5

K¢2

+ q1

–

K¢3

+

q1 1 t1 s + 1

+

+

K¢1

23

q2 1 t2 s + 1

+

K¢4 qr

+

K1

e

K2

–

K3

K4

q2

q1

K1

With these parameters, t 1 = 184.55 sec, K3¢ = 1318.2, K 4¢ = 0.465, K 5¢ = 0.535 q 2 ( s) q r ( s)

=

(5.55 ´ 10 -3 ) K1 K 2 K 3 K 4 s 2 + 7.26 ´ 10 -3 s + 119 . ´ 10 -5

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

Refer Fig. 3.57 for suitable hardware required to implement the proposed control scheme. 3.22 q2

q1 qr

+ –

Controller I

+ –

Controller II

Process II

Process I

Sensor II

Set-point adjustment Sensor I

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



CHAPTER 4

BASIC PRINCIPLES OF FEEDBACK CONTROL

4.1

SGM =

1 s( s + 1) = 1 + G( s) H (s) s( s + 1) + 50

| SGM ( jw )|w =1 = 0.0289 SHM =

-50 - G( s ) H ( s ) = 1 + G( s) H ( s) s( s + 1) + 50

| S HM ( jw )|w =1 = 1.02

Y ( s) P1D 1 = M(s) = R( s) D

4.2

P1 =

FG H

M(s) =

SKM1 = | SKM1 ( jw )|w = 0 = 4.3

IJ K

K 3 K1 KK K K 5 ; D = 1- - 1 2 - 2 3 1 ; D1 = 1 2 s 2 ( s + 1) s ( s + 1) s + 1 s ( s + 1)

5K1 s ( s + 1 + 5K1 ) + 5 K1 + 5 2

s 2 ( s + 1 + 5 K1 ) + 5 - 5 K1s 2 ¶M K1 = 2 ´ s ( s + 1 + 5 K1 ) + 5 K1 + 5 ¶K1 M

5 5K1 + 5

= 0.5

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

y ( t )|open-loop = 20 (1 – e–t) y ( t )|closed-loop =

20 (1 - e -21t ) 21

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 (1 - e -20.4 t ) 20.4

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



SOLUTION MANUAL

4.4 For

G(s) = 10/ (t s + 1), and R(s)

25

= 1/s,

ess |open-loop = 0 1 ess |closed-loop = 1 + 10 K p = For

0.0099

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 to variations in plant parameters. 4.5

L 2 = = 0.04 sec R 50

t f |open-loop =

For the feedback system,

K A (e f - Ki f ) = L

di f dt

+ ( R + Rs )i f

This gives I f ( s)

KA = E f (s ) sL + R + Rs + K A K t f |closed-loop =

L 2 = = 0.004 R + Rs + K A K 51 + 90 K

This gives K = 4.99 4.6

The given block diagram is equivalent to a single-loop block diagram given below, W (s)

0.4 s+1

+ –

1

Y(s)

10(K + s) s+1

Y ( s) 0.4 = W ( s) 11s + 10 K + 1



26


For

W(s) = 1/s, yss = lim sY ( s ) = s →0

0.4 = 0.01 10 K + 1

This gives K = 3.9 4.7

Case I:

Y ( s) s(t s + 1) = W ( s) s(t s + 1) + K for W(s) = 1/s, yss = 0 Case II:

Y ( s) K = W ( s) s(t s + 1) + K For W(s) = 1/s, yss = 1 The control scheme shown in the following figure will eliminate the error in Case II. W (s) Kc 1 + 1 TIs

–

4.8

+

+ Plant

Y(s)

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

+ –

e

D(s)

G (s) =

200 ¥ 0.02 (s + 1) (s + 2)

q

1 1 E(s ) = ;ess = lim sE ( s ) ; qr ( s) = s s →0 qr ( s ) 1 + D( s )G( s) (i) ess = 1/3 (ii) ess = 0 (iii) ess = 1/3

4.9

The integral term improves the steady-state performance and the derivative term has no effect on steady-state error. (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 term has the opposite effect. er = 50 × 0.03 = 1.5 volts

4.10 (a)

V (s ) Er ( s )

(b)

= M(s) =

SKMe = (c)

V(s) =

0.65K e ( s + 1)(5s + 1) + 0.0195 K e

¶M Ke ( s + 1)(5s + 1) ´ = ¶K e M ( s + 1)(5s + 1) + 9.75

325 19.5( s + 1) Er ( s ) W ( s) ( s + 1)(5s + 1) + 9.75 ( s + 1)(5s + 1) + 9.75

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

sV (s) = vss = lim s ®0

325 ´ 1.5 19.5 ´ x =0 10.75 10.75

This gives x = 25 4.11 (a)

The block diagram is shown in the figure below. IL

Ra er +

KA

–

1 sLf + Rf

Kg

KA

(b)

E0 ( s ) Er ( s )

IL =0

=

10 2 s + 1 + 10 K

For Er(s) = 50/s,

sE0 ( s) = eoss = lim s ®0

500 = 250 1 + 10 K

This gives K = 0.1 With

K = 0.1, we have


+

–

eo


28


Eo(s) = When

10 2s + 1 Er ( s ) I L (s ) 2s + 2 2s + 2

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

10 ´ 50 30 = 235 V 2 2

Let x V be the reference voltage required to restore the terminal voltage of 250 V. 10 ´ x 30 = 250; this gives x = 53V 2 2

(c)

Eo(s) =

10 Er ( s ) - I L ( s ) 2s + 1

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 reduced due to feedback action. 4.12 wr

Kt

+

KA

–

Kg

+ –

1 Ra

Kb Kt

K A Kg (a) w o ( s) = M(s) = 2( s + 5) + K A K g w r ( s) For wr(s) = 10/s,

wo(t)

=

125 (1 - e -130t ) ; woss = 9.61 rad/sec 13

(b) When the feedback loop is opened,

w o ( s) 50 K A = w r ( s) 2( s + 5)


KT

1 Js

w0


SOLUTION MANUAL

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 case is 1/130 sec; system dynamics becomes faster with feedback action (c) SKMA =

2 s + 10 ¶M K A = ´ 2 s + 10 + K A K g ¶K A M Kg = Kωg, where K is a constant.

SwMg =

Kg ¶K g wg ¶M 2 s + 10 ¶M ´ ´ = ´ = M M 2 s + 10 + K A K g ¶K g ¶w g ¶K g

The open-loop transfer function of the system is

G (s) =

K A Kg 2 ( s + 5)

Therefore

SKGA = 1 = SwGg S KMA < S KGA ; SwMg < SwGg

4.13 Tw qR

KP

+– –

KA

1 sLf + Rf

Kg

+ –

1 Ra

KT

+

–

qM = qC 1 Jeqs2

sKb KP

(b)

5Kg q L ( s) = M (s) = q R ( s) s ( s + 1) ( s + 4 ) + 10 K g SKMg =

Kg s ( s + 1) ( s + 4 ) ¶M ´ = ¶K g M s ( s + 1) ( s + 4 ) + 10 ´ 100


1 2

qL

29


30


SKMg ( jw ) (c)

w = 0 .1

~ 4 ´ 10

-4

q L (s) (s + 4) / 2 = s ( s ) ( s + 4) + 10 K g + 1 - Tw ( s )

Wind gust torque on the load shaft = 100 N-m. Therefore, on the motor shaft Tw (s) = n × 100/s = 50/s θLss = lim sq L ( s ) = 0.1 rad s®0

(d)

FG H

Replace KA by D(s) = K c 1 +

1 TI s

IJ to reduce the steady-state error K

to zero value. 4.14 (a)

q (s) q r (s) q (s) q r (s)

= G(s) =

K P K A Ke K ( 25s + 1) s( 0.25s 2 + 0.02s + 1)

= M(s) =

G( s ) 1 + G( s )

open - loop

closed - loop

s( 0.25s 2 + 0.02 s + 1) s( 0.25s + 0.02 s + 1) + K P K A K e K (25s + 1)

M SKG = 1; SK =

2

SKM < SKG (b) A PD control scheme with proportional controller in the forward path and derivative action realized by feeding back q& , will increase aircraft damping. 4.15

(a)

q (s) Tw ( s )

=

1/ B sK b K T K K J s s +1 + + T A B BRa BRa

FH

IK

For Tw(s) = Kw/s,

θss = lim sθ (s) = s®0

K w Ra KT K A



SOLUTION MANUAL

(b)

q (s) = M (s) = q r (s)

SJM

31

K T K A / BRa sK K K K J s s +1 + b T + T A B BRa BRa

F H

I K

J - s2 ¶M J B ´ = = sK b K T K K J M ¶J s + T A s +1 + B BRa BRa

FH

IK

(c) Excessive large magnitudes of signals at various levels in a control system can drive the devices into nonlinear region of operation. The requirement of linear operation of devices under various operating conditions imposes a constraint on the use of large values of KA. 4.16 (a)

(b)

(c)

dV = qi − q; V = C h(t), q = h(t)/R dt hr + –

qi

K

R RCs + 1

h

1 R

q

H (s) KR = M (s) = RCs + 1 + KR Hr ( s ) S RM =

SRM

s=0

=

R 1 ¶M ´ = ¶R M RCs + 1 + KR 1 1 + KR

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

FG H

KR 1 ; ess = 1 + KR 1 + KR

IJ K

1 reduces the steady-state error to zero; and the Tl s system becomes insensitive to changes in R under steady dc conditions. Replacing K by K c 1 +

4.17 V

d (C + c ) = (Qi + qi ) ( Ci + ci ) − Q ( C + c ) dt

The perturbation dynamics is given by t

C Q dc ( t ) V + c(t) = i qi ( t ) + i ci (t );t = dt Q Q Q

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



32


ci

Qi /Q

qi

K

–

+

Ci /Q

+

1

c

ts + 1

Qi / Q C( s) = M(s) = Ci ( s ) τs + 1 + KCi / Q For Ci(s) = A/s, css =

AQi Q + KCi

With integral control, css becomes zero. 4.18 (b)

Open-loop case:

KR1 q( s ) ; SKG = 1 = G(s) = R1Cs + 1 Er ( s )

(i)

qi

er +

1/R1

er K K

–

+

+

R1 R1C s + 1

et Kt


q


SOLUTION MANUAL

(ii)

1 q (s) . For q i ( s) = 1 / s, q ss = 1 = Ri Cs q i (s)

(iii) τ = R1C Closed-loop case: (i)

KR1 R1Cs + 1 q( s ) = M(s) = ; SKM = R1Cs + 1 + KKt R1 R1Cs + 1 + KKt R1 Er ( s )

(ii)

q (s) 1 1 = . For θ i(s) = 1/s, θss = R1Cs + 1 + KKt R1 1 + KK t R1 q i (s)

(iii) τ =

(c)

R1C 1 + KK t R1

K(er − et ) = C

1 R1

+

er K K

+

qi

er + –

q - qi q - qa RRC 1 dq ; R2 = ,t = 1 2 + + dt R1 R2 UA R1 + R2 +

+

1 R2

qa

R1R2 (R1 + R2 ) (ts + 1)

q

Kt

q (s) q a (s)

= open - loop

R1 ; q ss ( R1 + R2 ) (t s + 1)

open - loop =

R1 (R1 + R2)

R1 θ (s) = ( R1 + R2 ) (t s + 1) + KK t R1R2 θ a ( s ) closed − loop

θ ss

4.19 (a)

closed − loop

=

R1 R1 + R2 + KK t R1 R2

0.1 ( s + 1) Y (s) = . For W(s) = 1/s, yss = 0.089 W (s) 2 s 2 + 3s + 1125 .


33


34


(b)

(c)

FG H

IJ K 0.1 ( s + 1) Y (s) LM1 - K K OP = W ( s ) 2 s + 3s + 1125 . N s +1 Q

Replace Kc by K c 1 +

1 Tl s

f

c

2

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 less stable. The feedforward scheme does not affect the characteristic roots of the system. A difficulty with feedforward compensation is that it is an openloop technique; it contains no self-correcting action. If the value of Kc is not accurately known, the gain Kf will not cancel the disturbance completely. The usefulness of a method must be determined in the light of the specific performance requirements. If the uncertainty is large, self-correcting action can be obtained by using PI control law in conjunction with the feedforward compensation. 4.20 r

Kt

+

ef +

–

–

KA

1 sLf + Rf + 1

Kt

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

I f (s ) E f (s)

=

KA sL f + R f + K A + 1

Time-constant (with current feedback) =

Lf Rf + KA + 1

Position control system is shown below.


if

w KT Js + B


SOLUTION MANUAL

r

KP

+

+

+

–

–

–

if

KA sLf + Rf + 1

Kt

w

Js + B

1 s

Kt KP

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

4.21 (a)

This gives Kc = 4, TD = 0.5 (b) qr

Alternative control scheme is shown below. +

K1

–

+

1 s

–

q

1 s

K2

K1 q (s) = 2 ; K1 = 4, K 2 = 2 q r ( s ) s + K 2 s + K1 4.22 Fig. P4.22a:

q (s ) 25 K = M (s ) = 2 q r (s) s + 5s + 25 K SKM =

s( s + 5) K ¶M ´ = 2 M ¶K s + 5s + 25

S KM ( jw ) w = 5 = 1.41

Fig. P4.22b: KK 2 25 K q (s) = M (s) = 2 = 2 q r (s) s + 5s + 25 K s + (1 + KK1 ) s + KK 2

This gives K2 = 25 and K1 = 4


q

35

w


36


M (s) =

SKM =

25 K s + (1 + 4 K ) s + 25 K 2

s( s + 1) s + 5s + 25 2

| SKM ( jw )|w =5 ≈ 1 This shows the superiority of the two-loop system over a single-loop system. —-



CHAPTER 5

5.1

CONCEPTS OF STABILITY AND THE ROUTH STABILITY CRITERION

(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 lefthalf plane. (c) s3-row of the Routh array has zero pivot element, but the entire row is not all zeros. We replace the pivot element by e and then proceed with the construction of the Routh array. As e ® 0, two sign changes are found in the first column of the Routh array. Therefore, two roots are in right-half plane and the rest in the left-half plane. (d) s1-row of the Routh array is an all-zero row. Auxiliary polynomial formed using the elements of s2-row is given by A(s) = s2 + 1 We replace the elements of s1-row with the coefficients of

dA( s ) = 2s + 0 ds and proceed with the construction of the Routh array. There are no sign changes in the resulting Routh array; the characteristic polynomial does not have any root in the right-half plane. The roots of the 2nd-order auxiliary polynomial are therefore purely imaginary. The given characteristic equation has two roots on the imaginary axis and the rest in the left-half plane. (e) Since all the coefficients of the given characteristic polynomial are not of the same sign, the system is unstable. The Routh array formation is required only if the number of roots in the right-half plane are to be determined. Only one sign change (the s0-row has – ve pivot element) is found in the Routh array. Therefore one root is in the right-half plane and the rest in the left-half plane. (f) s3-row of the Routh array is an all-zero row. The auxillary polynomial is A(s) = 9s4 + 0s2 + 36 We replace elements of s -row with the coefficients of 3

dA( s ) = 36s3 + 0s ds The resulting Routh array has s2-row with zero pivot element but the entire row is not all zeros. We replace the pivot element with e and proceed with the construction of the Routh array. Two sign changes are found in the first column; the characteristic polynomial D(s) has



38

5.2


two roots in the right-half plane. Three possibilities exist: (i) the 4thorder auxiliary polynomial has all the four purely imaginary roots and the two right-half plane roots are contributed by the factor D(s)/A(s), (ii) the 4th-order auxiliary polynomial has complex roots with quadrantal symmetry (two roots in the right-half plane) and the factor D(s)/A(s) has all the roots is the left-half plane, and (iii) the 4th-order auxiliary polynomial has two real-axis roots and two imaginary-axis roots with quadrantal symmetry and the factor D(s)/A(s) has one root in the right-half plane. Examining A(s) we find that auxiliarypolynomial roots are s = – 1 ± j1, 1 ± j1. The given system, therefore, has two roots in the right-half plane and the rest in the left-hand plane. (a) s1-row of the Routh array is an all-zero row. The auxiliary polynomial is A(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

5.3

39

With s = s$ – 1, the characteristic equation becomes 3 2 s$ + s$ + s$ + 1 = 0 1 s$ -row in the Routh array is an all-zero row. The auxiliary polynomial is

A( s$ ) = s$ 2 + 1 =

( s$ + j1) ( s$ – j1)

2

We replace the elements of s$ -row with the coefficients of dA( s$ ) ds$

5.4

5.5

= 2 s$ + 0

and proceed with the construction of the Routh array. There are no sign changes in the resulting Routh array. The s-polynomial does not have roots to the right of the line at s = –1, and there are two roots at s = – 1 ± j1. The largest time-constant is therefore 1 sec. (a) Unstable for all values of K (b) Unstable for all values of K (c) K < 14/9 (d) K > 0.528 With s = s$ – 1, the characteristic equation becomes

s$ 3 + 3K s$ 2 + (K + 2) s$ + 4 = 0 From the Routh array, we find that for K > 0.528, all the s-plane roots lie to the left of the line at s = –1. 5.6

LM N

G(s) =

20 / ( s + 1) 4K 2 s + 1 1 + 20 ´ 0 .2 / ( s + 1)

H(s) =

0.05 4s + 1

OP Q

1 + G(s) H(s) = 0 gives 8s3 + 46s2 + 31s + 5 + 4K = 0

5.7

5.8

From the Routh array, we find that the closed-loop system is stable for K < 43.3. (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 (a) D(s) = s3 + 10s2 + (21 + K)s + 13K = 0 (i) K < 70 (ii) K = 70 (iii) for K = 70, A(s) = s2 + 91 = 0 D(s) = (s2 + 91) (s + 10) = (s + j9.54) (s – j9.54) (s + 10) (b) D(s) = s3 + 5s2 + (K – 6)s + K = 0



40

5.9


(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) (a) D(s) = s4 + 12s3 + 69s2 + 198s + 200 + K = 0 From the Routh array we find that the system is stable for K < 666.25. For K = 666.25, A(s) = 52.5s2 + 866.25 = 0 This gives s = ± j 4.06 Frequency of oscillations is 4.06 rad/sec when K = 666.25. (b) D(s) = s4 + 3s3 + 12s2 + (K – 16) s + K = 0 From the Routh array, we find that the system is stable for 23.3 < K < 35.7. For K = 23.3, A(s) = 9.57s2 + 23.3 = 0; s = ± j1.56 For K = 35.7, A(s) = 5.43s2 + 35.7 = 0; s = ± j2.56 Frequency 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 =2 a

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 s$ 3 + 2.4 s$ 2 + 4.6 s$ + 67.8 = 0 Two 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) = t s3 + (1 + 5t)s2 + 5s + 50 = 0; t < 0.2 (c) D(s) = t s3 + (1 + 5t)s2 + 5s + 2.5K = 0; K < 5.14 D(s) = s4 + 5s3 + 6s2 + Kcs + Kca = 0; a =

FH 2 + 10IK t

1 TI

From the Routh array, we find that K < (30 – 25a) results in system stability. Larger the a (smaller the TI), lower is the limit on gain for stability. 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 on K for stability increases with increasing value of Kt.



CHAPTER 6 6.1

THE PERFORMANCE OF FEEDBACK SYSTEMS

(a) Refer Section 6.3 for the derivation (b) The characteristic equation is s2 + 10s + 100 = s2 + 2zwn s + w 2n = 0 (i) wn = 10 rad/sec; z = 0.5; wd = w n 1 - z 2 = 8.66 rad/sec (ii) tr =

p p - cos -1 z = 0.242 sec; tp = = 0.363 sec wd wd

Mp = e - pz /

1-z 2

= 16.32%; ts =

4 = 0.8 sec zw n

(iii) Kp = lim G(s) = ¥ ; Kv = lim sG(s) = 10; s®0

s®0

Ka = lim s2G(s) = 0 s®0

(iv) ess = 6.2

1 1 1 = 0; ess = = 0.1; ess = =¥ Ka Kv 1+ Kp

Characteristic equation is s2 + 10s + 10KA = s2 + 2zwn s + w 2n = 0 (a) KA = 2.5 gives z = 1 which meets the response requirements. (b) Kp = ¥ , Kv = 2.5 ess =

5 1 + = 0.4 1 + K p Kv

6.3 (a) Using Routh criterion, it can be checked that the close-loop system is stable. (i) ess =

10 10 = =0; ¥ Kv

(ii) ess =

10 0.2 10 0.2 + + = =2 ¥ 0.1 Kv K a

(b) The closed-loop system is unstable. 6.4

The closed-loop system is stable. This can easily be verified by Routh criterion. G(s) =

5 ( 0.1s + 1)( s + 1)( 0.2 s + 1)



SOLUTION MANUAL

ess|r = 10 =

Y (s) W (s)

=

43

10 10 = 1+ 5 1+ Kp 0.1s + 1 ( s + 1)( 0.2 s + 1)( 0.1s + 1) + 5

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

+

KA

–

0.016 3s + 1

50 30s + 1

q

1 10s + 1

G(s) =

0.016 ´ 50 K A ( 3s + 1)(30 s + 1)

H(s) =

1 ; the dc gain of H(s) is unity. 10 s + 1

Kp = lim G(s) = 0.8 KA ; ess = s®0

10 =1 1+ Kp

This gives KA = 11.25 6.6

(a) The system is stable for Kc < 9.

1 1 + Kc ess= 0.1 (10%) is the minimum possible value for steady-state error. Therefore ess less than 2% is not possible with proportional compensator.

Kp = lim D(s) G(s) = Kc; ess = s®0



44


KI . The closed-loop system is stable for s 0 < K1 < 3. Any value in this range satisfies the static accuracy requirements. 1000 Kc = 1000 (a) Kv = 10

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

6.7

z =

10 + 1000 KD = 0.5 2 ´ 100

These equations give Kc = 10 and KD = 0.09. (b) The closed-loop poles have real part = – zwn = – 50. The zero is present at s = – Kc/KD = – 111.11. The zero will result in pronounced early 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$ 3 + 7 s$ 2 + [100(1 + Kc) – 17] s$ + 1009 – 100 (1 + Kc) = 0 In Routh array formation, Kc = 0.41 results in auxiliary equation with purely imaginary roots. Therefore Kc = 0.41 results in dominant s-plane poles with real part = – 1. (c) Routh array formation for D( s$ ) gives the following auxiliary equation. A( s$ ) = 7 s$ 2 + 868 = 0; s$ = ± j11.14 The complex-conjugate s-plane roots are – 1 ± j11.14. wd = wn 1 - z 2 = 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 dominance condition in terms of third closed-loop pole is reasonably satisfied. The zero is at s = – KI/Kc = – 24.39. The zero will not give pronounced early peak. Therefore z approximately represents Mp.

6.9

(a) With TI = ¥ , D(s)G(s) =

80 (1 + TD s ) s 2 + 8 s + 80



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 +

80 =0 TI

TI > 0.0198 for stability. 6.10 (a) G(s) =

K ; Kv = lim sG(s) = K s®0 s( s + 1)

K = 10 will give steady-state unit-ramp following error of 0.1. D(s) = s2 + s + K = s2 + 2zwns + w 2n K = 10 gives z = 0.158. (b) G(s) =

10 s( s + 1 + 10 Kt )

D(s) = s2 + (1 + 10Kt)s + 10 = s2 + 2zwns + w 2n Kt = 0.216 gives z = 0.5. Kv = lim sG(s) = s®0

10 10 = 3.16 1 + 10 K t

ess = 0.316 (c) G(s) =

Kv =

10 K A s( s + 1 + 10 Kt ) 10 K A 1 + 10 Kt

D(s) = s2 + (1 + 10Kt)s + 10 KA = s2 + 2zwns + w 2n From these equations, we find that KA = 10 and Kt = 0.9 give ess = 0.1 and z = 0.5.



46


6.11 (a) z =

( In M p ) 2

2

( In M p ) 2 + π 2 M p = 0.1 gives z = 0.59 ts =

G(s) =

4 = 0.05. Therefore, wn = 135.59 zw n

5 K1 s( 0.2 s + 1 + 100 K2 )

D(s) = 0.2s2 + s(1 + 100K2) + 5K1 = 0 Therefore s2 + 5(1 + 100K2)s + 25K1 = s2 + 2zwns + w 2n This equation gives K1 = 735.39 and K2 = 0.31 5 K1 = 114.9, Ka = 0 1 + 100 K 2 q (s) 4 = 6.12 s( s + 1) + 10 K + 10 Kt s Tw ( s ) 4 For Tw(s) = 1/s, qss = = 0.05 10K

(b) Kp = ¥ , Kv =

This gives K = 8 D(s) = s2 + (1 + 10Kt)s + 10K = s2 + 2zwns + w 2n Kt = 0.79 gives z = 0.5 6.13

Y (s) 1 9 = . For F(s) = 9/s, yss = = 0.03 K Ms 2 + Bs + K F( s ) This gives K = 300 Newtons/m e - pz /

1-z 2

=

0.003 = 0.1 0.03

This gives z = 0.59 tp =

p wn 1-z

2

=

p =2 0.81w n

This gives wn = 1.94 rad/sec From the equation



SOLUTION MANUAL

s2 +

47

K B s+ = s2 + 2zwns + w 2n M M

we obtain M = 79.71 kg, B = 182.47 Newtons per m/sec 6.14

KT / J w 2n q (s) = = 2 K B q R (s) s + 2zw n s + w 2n s2 + s + T J J Tw

qR +

KT

–

z2 =

+

– Js2

1 + Bs

q

( ln M p ) 2 ( ln M p ) 2 + π 2

For Mp = 0.25, z = 0.4 Steady-state error to unit-ramp input =

2z = 0.04 wn

This gives wn = 20

q( s ) 1 = 2 - Tw ( s ) Js + Bs + K T qss =

10 = 0.01 KT

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



48


6.15 J w& + Bw = KT(wr – w)

w (s) 45 = 100 s + 50 w r (s ) wr = 50 ×

2p rad/sec = step input 60

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 error compensation will remove the offset. 6.16 Jq&& + Bq& = KT(qr – q)

q ( s) 2400 = 150 s 2 + 600 s + 2400 qr ( s ) qr(s) =

q(t) =

=

LM N

p 16 p ; q(s) = 2 3s 3 s ( s + 4 s + 16 )

LM MN p L 1 - 1.1547e 3 MN

OP Q

OP PQ p sin FH 3.46 t + IK O 3 PQ

e -zw nt p 1sin (w d t + cos -1z ) 2 3 1-z

M p = e - pz /

1-z 2

-2 t

= 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 + –

Ks

e

KA

ec

K1

+ –

Js2

qM 1 + Bs

sK2


1 50

qL


SOLUTION MANUAL

(b) G(s) = Kv

49

4KA s( s + 100.5)

= 4KA/100.5

For a ramp input qR(t) = p t, the steady-state error = The specification is 5 deg or

p Kv

5p rad. 180

Therefore

100.5p p = ; this gives KA = 904.5 180 4KA 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 error to ramp inputs is zero provided the closed-loop system is stable.

FH

IK

1 s ; K = 904.5 G(s) = A s( s + 100.5) 4KA 1 +

The characteristic equation is s3 + 100.5s2 + 4 × 904.5s + 4 × 904.5 = 0 It can easily be verified that the closed-loop system is stable. (d) The characteristic equation can be expressed as (s + 1.0291) (s2 + 99.4508s + 3514.6332) = 0 or (s + 1.0291) (s + 49.7254 + j32.2803) (s + 49.7254 – j32.2803) = 0 The dominance condition is satisfied because the real pole is very close to the zero. Therefore, the transient response resembles that of a second-order system with characteristic equation s2 + 99.4508s + 3514.6332 = 0 wn = 59.28; z = 0.84; Mp = 0.772%; ts = 0.08 sec



50


6.18

Ket

et

Ia

(const)

Rf Lf

KA

e

qR

JL

qL

qR

KP

+

e+

–

–

KA

KT

sKKt KP

(c) The characteristic equation of the system is s2 + 100 KKAs + 6KA = s2 + 2zwn s + w 2n = 0 This gives KA = 2.67; K = 0.024


1 Jeqs2

qM

1 qL 50


SOLUTION MANUAL

51

6.19 qR + –

KA

KP

+ –

1

KT Ra

Js2

+ Bs

1 10

qL

sKb

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

Kv =

20 K A s( 4 s + 202 )

20 K A 1 = ; this gives KA = 1010. 0.01 202

The characteristic equation becomes s2 + 50.5s + 5050 = 0 = s2 + 2zwn s + w 2n This gives z = 0.355; Mp = 30.33% (c) G(s) =

20( K A + sK D ) s( 4 s + 202) Kv =

20 K A = 100; no effect on ess. 202

Characteristic equation of the system is s2 + (50.5 + 5KD)s + 5050 = 0 = s2 + 2zwns + w 2n This gives KD = 6.95. For z = 0.6, Mp = 9.5% (d) System zero: s =

- KA = – 145.32 KD

Real part of closed-loop poles = – zwn = – 42.636. The zero will affect the overshoot. The peak overshoot will be slightly more than 9.5%. 6.20 (a) G(s) =

3KA s( 0.3s + 1)(1 + 2 s )



52


Kv = 3KA; ess =

0.03 0.01 = 3K A KA

Y (s) 0.3(1 + 2 s ) = Fw ( s ) s( 0.3s + 1)(1 + 2 s ) + 3 K A

(b)

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

–

u

1

i

47.6 16.67

s 2–

Power amp

e

100

(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 =

[ ln ( 0.2)]2 ; z = 0.456 [ ln ( 0.2)]2 + π 2 ts =

4 = 0.4; wn = 21.93 zw n


x


SOLUTION MANUAL

s2 + 4760 KcTDs + 4760 Kc – 16.67 = s2 + 2zwn s + w 2n This gives Kc = 0.1045, TD = 0.04. 6.22 (a) System-type number = 2 (b)

H (s) 0.5s( 0.1s + 1) = 2 - Qw ( s ) s ( 0.1s + 1) + 0.5( Kc + sK D ) hss = -

0.5 ; Kc > 10 0.5Kc

The characteristic equation is 0.1s3 + s2 + 0.5KDs + 0.5Kc = 0 For stability, KD > 0.1Kc 6.23 qr + –

K1 +

K2 s

1 s+1

Characteristic equation is s2 + (1 + K1)s + K2 = 0 = s2 + 2zwn s + w 2n Mp = 20% ® z = 0.456; ts = 2 ® wn = 4.386 K2 = 19.237; K1 = 3 ——


q

53


CHAPTER 7

7.1

COMPENSATOR DESIGN USING ROOT LOCUS PLOTS

(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 at s = – 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 roots at 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 loci approach s = – 2.8165 at ± 90º

jw j ÷10

– 3 + j1

s

–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 in directions 120º apart. jw

jw – 3 + j ÷3

j2÷3

60° s –2

(ii)

–9

s

–4

(iii)

(iii) – sA = – 4 ; fA = 90º, 270º The characteristic equation has three roots at s = – 3; the three root loci originating from open-loop poles approach this point and then breakaway. Tangents to the three loci breaking away from s = – 3 are 120º apart. (iv) – sA = – 1 ; fA = 45º, 135º, – 45º, – 135º; intersection with the jw axis at ± j 1.58; angle of departure fp from – 1 + j2 = – 90º There are two roots at s = – 1, two roots at s = – 1 + j1.22, and two roots at s = – 1 – j1.22. The break away directions are shown in the figure. (v) There are four roots at s = – 1. The break away directions are shown in the figure.



56


– 1 + j2

jw

– 1 + j1

jw

45° s

–2

(iv)

7.3

–2

45°

s

(v)

– 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 on this line satisfies the angle criterion. By magnitude criterion, the value of K at this point is found to be 1.04. Using this value of K, the third pole is found at s = – 2.33. Therefore, M(s) =

1.04 ( s + 0.33 + j 0.58)( s + 0.33 - j 0.58)( s + 2.33)

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 of K, by magnitude criterion is 130. The remaining pair of complex roots for K = 130 can be approximately located graphically. It turns out that real part of complex pair away from jw-axis is approximately four times as large as that of the pair near jw-axis. Therefore, the transient response term due to the pair away from the jw-axis will decay much more rapidly than the transient 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. Using this value of K, the third pole is found at s = – 4; the dominance condition is not 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 at the multiple roots (K = 1.128). The greatest value of K that can be used before continuous oscillations occur is the value of K at the point of intersection with the jw-axis. From the Routh array formulation it is found that the greatest value of K is 30 and 7.7 7.8

7.9

at this gain continuous oscillations of frequency 5 rad/sec occur. The proof given in Example 7.1. The proof follows from Example 7.1. Draw a line from the origin tangential to the circular part of the locus. This line corresponds to damping ratio for maximum oscillatory response. The tangential line corresponds to z = 0.82 and the value of K at the point of tangency is 2.1. The proof given in Example 7.2

7.10 s = s + jw tan–1

w +1 w -1 + tan -1 s s

= tan -1

w w + tan -1 ± 180 º (2 q + 1) s s +2

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

LM N

tan tan −1

w ω , ± 180 º = s +2 σ +2

OP Q

we obtain

w +1 w - 1 + s s w +1 w -1 1s s

FH

IK FH

IK

w w + s s +2 = w w 1s s +2

FH IK FH

IK

Manipulation of this equation gives

LMF s - 1 I NH 2 K

2

+w2 -

OP Q

5 =0 4

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-root branches form a circle. The z = 0.707 line intersects the root locus at two points; the values of K at these points are 1 and 5. The point that corresponds to K = 5 results in lower value of ts; therefore we choose K = 5(sd = – 3 + j3).



58


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

1-z 2

= 4.3%

4 4 = sec 3 zw n

G(s) =

K (s + 4) ( s + 2)( s - 1)

E(s) =

( s + 2 )( s + 1) 1 R( s ) R(s) = ( s + 2 )( s - 1) + K ( s + 4 ) 1 + G( s)

For R(s) = 1/s (assuming selection of K that results in stable closed-loop poles) ess = lim sE ( s ) = s ®0

-2 -2 + 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 at s = – 0.634, and break into the real axis at s = – 2.366. By determining a sufficient number of points that satisfy the angle criterion, it can be found that the root locus is a circle with centre at – 1.5 that passes through the breakaway points. Both the breakaway points correspond to z = 1. For minimum steadystate 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+

or

1+

0.8 K =0 (10 s + 1)(30 s + 1)( 3s + 1) K¢

FH s + 1 IK FH s + 1 IK FH s + 1 IK 10 30 3

= 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+

K 10 K =1+ =0 s( 0.1s + 1) s ( s + 10 )

K = 10 gives z = 0.5. (b) Sensor with appreciable time-constant:

or

1+

K =0 s( 0.1s + 1)( 0 .02 s + 1)

1+

K¢ = 0 ; K ¢ = 500 K s( s + 10 )( s + 20)

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 ¢( s + 10) = 0 ; K ¢ = 0.1K s( s + 2 )( s + 4)

– sA = 2 ; fA = 90º, – 90º; intersection with the jw-axis at ± j 20 ; multiple roots 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+

Ks = 0 ; K = 10Kt s 2 + s + 10

Notice that a zero is located at the origin and open-loop poles are located at s = – 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.



60


The gain K corresponding to this point is 3.427. Hence the desired value of velocity 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 = 0 ; K = 20Kt ( s + j 2)( s - j 2)(s + 5)

– 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 for K = 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 the closed-loop system does not have a zero at the origin:

Y (s) 20 = s( s + 1)( s + 4 ) + 20(1 + Kt s ) R( s ) In the root locus plot, the zero at the origin was introduced because of the process of modifying the characteristic equation so that the adjustable variable 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 which may be rearranged as 1+

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

– 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

7.19 The characteristic equation of the system is s3 + 9s2 + 18s + 10a = 0 which may be rearranged as 1+

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

– 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 third pole for K = 28 is at s = – 7. 7.20 (a) 1 +

K =0 s( s - 2 )

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 numerical values are neither precisely known nor precisely fixed. We should therefore never attempt to cancel poles in the right-half plane, since any inexact cancellation will result in an unstable closed-loop system. (c)

1+

K ( s + 1) =0 s( s − 2 )( s + 8)

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

E( s ) s ( s - 2 )( s + 8) 1 = = s( s - 2)( s + 8) + Ks + K R( s ) 1 + G( s) For R(s) = 1/s and K > 19.2 (required for stability), ess = lim sE(s) = 0 s®0

2

For R(s) = 1/s and K > 19.2, ess = lim sE(s) = s®0

-16 K

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 line is 7.0.



62


Kv = lim sG(s) = K/5 = 1.4 ; ts = 4/zwn = 11.6 sec. s®0

With the compensator, the characteristic equation becomes 1+

K (10 s + 1) =0 s(100 s + 1)( s + 1)( s + 5)

1+

K ¢( s + 0.1) = 0 ; K¢ = 0.1K s( s + 0.01)(s + 1)( s + 5)

or

The value of K¢ at the point of intersection of z = 0.3 line and the root locus is 6.0. Therefore K = 60. Kv = lim sD(s) G(s) = K/5 = 12, ; ts = 4/zwn = 12.7 sec. s®0

7.22 1 +

K ( s + 2.5) =0 s( s + 1)( s + a )

Desired closed-loop pole; sd = – 1.6 + j4. Angle criterion at sd is satisfied when a = 5.3. The value of K at sd is 23.2. The third pole is at s = – 3.1. 7.23 (a) 1 +

K =0 s( s + 2 )

z = 0.707 ; ts =

4 = 2. Therefore, zwn = 2 zw n

Proportional control does not meet this requirement. (b) 1 +

K c + sK D K ( s + Kc / K D ) =1+ D s(s + 2) s( s + 2)

At the points of intersection of z = 0.707 line and zwn = 2 line, the angle criterion is satisfied when Kc/KD = 4. The root locus plot of 1+

K D (s + 4) =0 s(s + 2)

has circular complex-root branches (refer Problem 7.7). At the point of intersection of z = 0.707 line and zwn = 2 line, we find by magnitude criterion that KD = 2. Therefore Kc = 8. 7.24 G(s) =

K ( s + 0.1) ; K = 4000 s( s 2 + 0.8s + 4)



SOLUTION MANUAL

D(s) =

63

s 2 + 0.8s + 4 ( s + 0.384)( s + 10.42)

D(s) improves transient response considerably, and there is no effect on steady-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 below the desired root location. For the angle criterion to be satisfied at the desired root location, the pole is placed at s = – p = – 10.6. The gain of the compensated system, obtained by magnitude criterion, is 96.5. The compensated system is then D(s)G(s) =

96.5( s + 4 ) s( s + 2 )( s + 10.6 )

Kv = lim sD(s) G(s) = 18.2. s®0

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 the desired root. 7.26 Desired root location may be taken as sd = – 1.588 + j3.152

s +1 , s + 10 angle criterion is met. Magnitude criterion gives K = 147. The compensated system has Kv = 2.94. It meets the requirements on Mp and settling time. With D(s) =

7.27 z = 0.707; 4/zwn = 1.4 ® zwn = 2.85 wn = 4 ; w n 1 - z 2 = 2.85 sd = – 2.85 + j2.85

( s + 1.5) 2 , angle criterion is satisfied at sd. The value of K, ( s + 3.5) 2 found by magnitude criterion, is 37.

With D(s) =

7.28 From the root locus plot of the uncompensated system, we find that for gain of 1.06, the dominant closed-loop poles are at – 0.33 ± j 0.58. The value of z 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 of 5. The lag compensator



64


s + 0.1 s + 0.001

D¢(s) =

gives Kv boost of the factor of 10. The angle contribution of this compensator at – 0.33 + j0.58 is about seven degrees. Since the angle contribution is not small, there will be a small change in the new root locus near the desired dominant closed-loop poles. If the damping of the new dominant poles is kept at z = 0.5, then the dominant poles from the new root locus are found at – 0.28 ± j 0.51. The undamped natural frequency reduces to 0.56. This implies that the transient response of the compensated system is 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

0.925( s + 0.1) s + 0.01

D(s) =

7.29 The dominant closed-loop poles of uncompensated system are located at s = –3.6 ± j4.8 with z = 0.6. The value of K is found as 820. Therefore Kv = lim sG(s) = 820/200 = 4.1. s®0

s + 0.25 gives a Kv boost of the factor of 10. s + 0.025 The angle contribution of this compensator at – 3.6 + j4.8 is – 1.8º, which is acceptable in the present problem.

Lag compensator D(s) =

7.30 (a) 1 +

K K =1+ =0 ( 2 s + 1)( 0.5s + 1) ( s + 0.5)( s + 2 )

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

FG H

(b) D(s) = K c 1 +

D(s)G(s) =

1 TI s

IJ = K + K /s = K F s + K / K I H s K K I

c

I

c

c

Kc (s + K I / Kc ) s( s + 0.5)( s + 2)

At the point corresponding to z = 0.6 and zwn = 0.75, the angle criterion is satisfied 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. Clearly the roots of the system when Kv requirement is satisfied are a long way from 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 lag compensator. From the uncompensated root locus we find that corresponding to z = 0.707, the dominant roots are at – 2.9 ± j2.9 and the value of K is 236. Therefore necessary ratio of zero to pole of the compensator is

| z| 2000 = = 8.5 236 | p| We will choose z = 0.1 and p = 0.1/9. For this choice, we find that the angle contribution of the compensator at – 2.9 + j.2.9 is negligible. 7.32 D(s)G(s) =

K ( s + 1 / t 1 )(s + 1/t 2 ) s( s + 1)(s + 5)(s + 1/ at 1 )(s + a /t 2 )

1 1 1 a = 0.05 ; =1; = 0.005 ; = 10 t2 t1 t2 at 1 D(s)G(s) =

K ( s + 0.05) s( s + 0.005)( s + 5)( s + 10)

For z = 0.45, we obtain sd = – 1.571 + j3.119, K = 146.3, Kv = 29.3, wn = 3.49

FG H

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

FG H

KD s Kc s2 + 1

Kc 1 +

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

KD s Kc

IJ K

IJ K

s2 + 1 + Kc + KDs = (s + 1 + j 3 ) (s + 1 – j 3 ) This equation gives Kc = 3 and KD = 2 Kp = lim D(s)G(s) = 3. s®0

(b) With PID controller D(s) =

K ( s + 0.2)( s + 1.3332 ) s



66


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 = lim sG(s) = 2.07 s®0

(c) Compensated system: M p = 15% ® z = 0.55 ts = wn =

3.6 = 1.45 ; zwn = 4/1.45 = 2.75 2.5 2.75 = 5 ; sd = – 2.75 + j4.18 0.55

³ 20

Kv

Lead-lag compensated system: D1(s) D2(s) G(s) = 7.35 G(s) =

K ( s + 3)( s + 0.9) ; K = 506 s( s + 0.15)(s + 3)( s + 9)(s + 13.6 )

K s( s + 2 + KKt ) E( s ) s( s + 2 + KK t ) 1 = = R( s ) s( s + 2 + KK t ) + K 1 + G( s)

For R(s) = 1/s2, ess =

2 + KKt £ 0.35 K

z ³ 0.707 ; ts £ 3 sec. The characteristic equation of the system is s2 + 2s + KKts + K = 0 which may be rearranged as



SOLUTION MANUAL

1+

67

KK t s =0 s + 2s + K 2

The locus of the roots as K varies (set KKt = 0) is determined from the following equation: 1+ For

K =0 s( s + 2 )

K = 20, the roots are – 1 ± j4.36.

Then the effect of varying KKt is determined from the locus equation 1+

KKt s =0 ( s + 1 + j 4.36)( s + 1 - j 4.36)

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 the settling time is 1.27 sec. 7.36 The characteristic equation of the system is 1+

KK t s =0 s + 2s + K

1+

KK t s =0 ( s + 1 + j ( K - 1 )( s + 1 - j ( K - 1)

2

or

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



68


jw K¢t = 0

K=5 K=2

K¢t = 0

j2

j1

• ¨ K¢t – ÷5

– ÷2

K¢t = •

–1

K¢t = 0

K¢t = 0

7.37 G(s) =

s

– j1

– j2

AK s( s + 1)( s + 5) + KK t s

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

KKt ( s + A / K t ) =0 s( s + 1)( s + 5)

The angle criterion at s = – 1 + j2 is satisfied when A/Kt = 2.5. Let us take A = 2 and Kt = 0.8. Kv = lim sG(s) = s®0

With

AK 5 + KK t

K = 10, Kv = 1.54

The closed-loop transfer function of the system is

Y (s) 20 = ( s + 1 + j 2 )( s + 1 - j 2 )( s + 4) R( s ) 7.38 The characteristic equation of the system is 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 =0 s ( s + 1)( s + 5 )

Sketch a root locus plot for a varying from 0 to ¥ . The points on these loci become the open-loop poles for the root locus equation 1+

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

For some selected values of a, the loci for b are sketched (refer Fig. 7.44). e–s = -

7.39

1–

(s - 2 ) s+2

K (s - 2) = 0 = 1 – F(s) ( s + 1)( s + 2 ) F(s) =

=

K (s + jw - 2 ) (s + jw + 1)(s + jw + 2 ) K (s - 2 + jw ) (s + 1)(s + 2 ) - w 2 + jw ( 2s + 3)

Ð F(s) = tan–1

= tan–1

w (2s + 3) w - tan -1 s -2 (s + 1)(s + 2 ) - w 2

R| w - w (2s + 3) S| s -w2 L(s + 1w)((s2s+ +2)3)- w |T 1 + s - 2 MN (s + 1)(s + 2) - w

Ð F(s) = 0º if

w ( 2s + 3) w =0 s - 2 (s + 1)(s + 2 ) - w 2 Manipulation of this equation gives (s – 2)2 + w2 = 12 Centre = (2, 0) and radius = 12 = 3.464


2

2

U| OP V| Q |W


70


It can easily be verified that at every point on this circle, Ð F(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 angle criterion as K £ 0. Centre at zero at s = 0 Radius = a 2 + b 2 = ( 0.5) 2 + ( 0.5) 2 = 0.707 The root locus plot is shown in the figure below.

j0.5

K=0

KÆ –ñ •

KÆ –ñ •

– 0.5

K=0

7.41 G(s) =

20.7( s + 3) s( s + 2 )( s + 8)

(i) a = a0 + Da = 8 + Da The characteristic equation is 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 method or 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 s( s + 2 ) =0 ( s - l 1 )( s - l 2 )( s - l 3 )



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 – 80º along the departure line, Da is determined by magnitude criterion: Da = 0.48 Therefore the sensitivity at l1 is Sal +1 =

Dl 1 0.2 Ð - 80 º = = 3.34 Ð – 80º Da / a 0 0. 48 / 8

(ii) Sensitivity of the root s = l1 to a change in zero at s = – 3 is determined as follows. b = b0 + Db = 3 + Db The characteristic equation is s(s + 2) (s + 8) + 20.7 (s + 3 + Db) = 0 or 1+

20.7 Db =0 ( s - l 1 )( s - l 2 )( s - l 3 )

The angle of departure of l1 is 50º. For a change of Dl1 = 0.2 Ð + 50º, we obtain Db = 0.21. Therefore

Sbl1 =

Dl 1 0.2 Ð + 50 º = = 2.84 Ð + 50º Db / b 0 0.21 / 3

The sensitivity of the system to the pole can be considered to be less than the sensitivity to the zero because of the direction of departure from the pole at s = l1. 7.42 (a) The characteristic equation is 1+

K ( s + 5) =0 s( s + 2 )( s + 3)

It can easily be verified from the root locus plot that for K = 8 the closed-loop poles are l1,2 = – 0.5 ± j 3.12, l3 = – 4 (b) s(s + 2 + d) (s + 3) + 8(s + 5) = 0



72


or 1+

or

1+

d s( s + 3) =0 s( s + 2)( s + 3) + 8(s + 5)

d s( s + 3) =0 s( s - l 1 )(s - l 2 )(s - l 3 )

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



CHAPTER 8

8.1

THE NYQUIST STABILITY CRITERION AND STABILITY MARGINS

(a) Revisiting Example 8.1 will be helpful. (i) 1 Ð 0º

(ii) 0 Ð – 180º

(iv)

t 1t 2 Ð - 90 º t1 + t 2

(b) Revisiting Example 8.2 will be helpful. (i) – t – j¥

(ii) 0 Ð – 180º

(c) Revisiting Example 8.3 will be helpful. (i) ¥Ð – 180º (d) (i) – (t1 + t2) – j ¥ (iii)

(ii) 0 Ð – 270º

t 1t 2 Ð - 180 º t1 +t 2

(e) (i) ¥Ð -180º 8.2

(ii) 0 Ð – 270º

(ii) 0 Ð 0º (iv)

t 1t 2 (t 1t 2 ) t1 + t 2

Ð - 90 º

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

8.3

Therefore N = number of clockwise encirclements = – 2 Z=N+P=–2+2=0 The closed-loop system is stable. (c) P = 0 N = (Number of clockwise encirclements of the critical point – number of counterclockwise encirclements of the critical point) = 0 Z = N + P = 0; the closed-loop system is stable. The polar plot of G(jw) for w = 0+ to w = + ¥ is given in Fig. P8.3b. Plot of G(jw) for w = – ¥ to w = 0– is the reflection of the given polar plot with respect to the real axis. Since

G( jw ) w =0 = ¥Ð – 180º, G(s) has double pole at the origin. The map of Nyquist contour semicircle s = r ejf, 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 semicircle traversed clockwise).



74 CONTROL SYSTEMS: PRINCIPLES AND DESIGN

With this information, Nyquist plot for the given system can easily be drawn. (i) P = 0, N = 0; Z = N + P = 0 The closed-loop system is stable (ii) P = 1, N = 0; Z = N + P = 1 The closed-loop system is unstable (iii) P = 0, N = 0; Z = N + P = 0 The closed-loop system is stable. 8.4

(a) The key points to the polar plot are: G( jw ) H ( jw ) w =¥ = 18 Ð 0º G( jw ) H ( jw ) w =¥ = 0 Ð – 270º

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

G( jw ) H ( jw ) w =4 .123 = – 1.428 Similarly, setting Re [G(jw)H(jw)] to zero, we get intersection with the imaginary axis. Based on this information, a rough sketch of the Nyquist plot can easily 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( jw ) H ( jw ) w =0 = – 3 – j ¥ G( jw ) H ( jw ) w =¥ = 0 Ð – 270º

Intersection with the real axis at w = 1/ 2 ; G ( jw ) H ( j w ) w =

1 2

=-

4 3

The map of Nyquist contour semicircle s = re jf, r ® 0, f varying from – 90º at w = 0– through 0º to + 90º at w = 0+, into the Nyquist plot is given by ¥Ð – f (an infinite semicircle traversed clockwise). Based on this information, a rough sketch of Nyquist plot can easily be made. P = 0, N = 2. Therefore Z = N + P = 2



SOLUTION MANUAL

75

(c) G( jw ) H ( jw ) w =0 = 2 Ð – 180º

G( jw ) H ( jw ) w =¥ = 0 Ð – 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 closedloop system is stable. (d) G( jw ) H ( jw ) w =0 = 0 Ð 90º ; G( jw ) H ( jw ) w =1 = 2.6 Ð 161º ;

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

G( jw ) H ( jw ) w =¥ = 0 Ð – 90º The Nyquist plot is given in Fig. P8.2b. P = 2, N = – 2 ; Z = N + P = 0 (e) G( jw ) H ( jw ) w =0 = ¥Ð – 180º G( jw ) H ( jw ) w =¥ = 0 Ð – 90º

No intersections with real/imaginary axis. The map of Nyquist contour semicircle s = re jf, 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 semicircle traversed clockwise). With this information, Nyquist plot for the given system can easily be drawn. P = 0, N = 0 ; Z = N + P = 0 (f) G( jw ) H ( jw ) w =0 =

1 Ð0 º 100

G( jw ) H ( jw ) w =10 - = ¥ Ð 0º

Consider the Nyquist contour shown in figure below. For the semicircle s = j10 + re jf, r ® 0, f varying from – 90º at w = 0– through 0º to + 90º at w = 10+, G(s)H(s) = lim

r®0

= lim

r ®0

1 1 = lim jf 2 jf ( j10 + re ) + 100 r ®0 re ( j 20 + re jf ) 1 - jf e r

It is an infinite semicircle from w = 10– to w = 10+ traversed clockwise. G( jw ) H ( jw ) w =¥ = 0 Ð – 180º




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

Im

j• 1 100

j10 s

Re w=

– j10

10+

– 1 + j0 w = •

w = 0+

– j• Nyquist plot

Nyquist contour

(g) G( jw ) H ( jw ) w =0 = – 8 + j ¥

G( jw ) H ( jw ) w =¥ = 0º Ð – 90º

G( jw ) H ( jw ) w =

3

= – 2 + j0

No intersection with the imaginary axis. The map of Nyquist contour semicircle s = rejf, r ® 0, f varying from – 90º at w = 0– through 0°to + 90° w = 0+, into the Nyquist plot is given by ¥Ð – f (an infinite semicircle traversed clockwise). With this information, a rough sketch of Nyquist plot can easily be made (refer Fig. P8.2a). P = 1, N = – 1. Therefore Z = N + P = 0. 8.5

G(jw)H(jw) =

- K (1 + w 2t 1t 2 ) jwK (t 1 - t 2 ) + 2 w 2 (1 + w 2t 12 ) w (1 + w 2t 12 )

(i) For t1 < t2, the polar plot of G (jw)H(jw) is entirely in the third quadrant 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 is unstable. Im

w = 0–

w = 0+ w=–• w=•

w=

0+

w=• w=–•

Re w = 0–

(i)

8.6

(ii)

(a) G( jw ) w =0 = 4 Ð 0º ; G( jw ) w =¥ = 0 Ð – 270º Intersection with the real axis at – 0.8. The critical point –1 + j0 will be encircled by the Nyquist plot if K > 1/0.8. Since P = 0, we want net encirclements of the critical point to be zero for stability. Therefore K must be less than 5/4. (b) G(s) =

K ¢(t s + 1) 4(1 + s ) =¢ 2 2 ; t1 < t2 s (1 + 0.1s ) s (t 1s + 1) 2

The Nyquist plot for this transfer function has already been given in the Solution of Problem 8.5. From this plot we see that the closed-loop system is table for all K. (c) G(s) =

K ¢(t s + 1) 4(1 + 0.1s ) = 2 2 ; t1 > t2 s 2 (1 + s ) s (t 1s + 1)

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




(d) G(jw) =

=

e -0 . 8 j w jw + 1

1 [(cos 0.8w – w sin 0.8w) – j(sin 0.8w + w cos 0.8w)] 1+w2

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( jw ) w =0 = 1 Ð 0º ; G( jw ) w =2 . 4482 = – 0.378 + j0 G( jw ) w =¥ = 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. As gain 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 must be encircled ccw two times. This is true if – 1.33 K < – 1 or K > 0.75. Im

w = ÷11

w = 0+ –1

w=•

– 1.33 –1


Re


SOLUTION MANUAL

8.8

79

1 1 G( jw ) H ( jw ) G( jw ) H ( jw ) = – 1 + j0 ; = 0.05 + j0 K K w =0 w =¥ 1 G( jw ) H ( jw ) = – 0.167 + j0 K w = 0 . 62 The Nyquist plot is shown in the figure below. If the critical point lies inside the larger loop, N = 1. Since P = 1, we have Z = N + P = 2 and the closed-loop system is unstable. If the critical point lies inside the smaller loop, N = – 1, Z = N + P = 0 and the closed-loop system is stable. Therefore, for stability – 0.167K < – 1 or K > 6. Im

– 0.167 w=

0+ w = • Re

–1

8.9

Eliminating the minor-loop we obtain forward-path transfer function of an equivalent single-loop system. G(s) =

K ( s + 0.5) s3 + s 2 + 1

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

1 1 G( jw ) G( jw ) = 0.5 Ð 0º ; = 0 Ð – 180º K K w =0 w =¥




1 G( jw ) = – 0.5 + j0 K w =1. 4 The polar plot intersects the positive imaginary axis; the point of 1 G( jw ) , equal to zero. intersection can be found by setting real part of K With this information, Nyquist plot can easily be constructed. From the Nyquist 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. Im j1.5

j1

1 G ( jw) – plot K

j0.5 w = 1.4 – 0.5

w=•

w = 0+ 0.5

Re

8.10 Fig. P8.10a: G(s) =

Ke -2 s s( s + 1)( 4 s + 1)

1 1 G( jw ) G ( jw ) = ¥Ð – 90º; = – 2.55 + j0 K K w =0 w = 0 . 26 1 G( jw ) =0 K w =¥



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 -2 s s ( 4 s + 1) 2

1 1 D( jw ) G( jw ) = ¥Ð – 180º ; D( jw )G( jw ) =0 K K w =0 w =¥ The polar plot will spiral into the w ® ¥ point at the origin. A rough sketch of the Nyquist plot is shown in figure below. It is observed that the system is unstable for all values of K. Im

w = 0+

Re

8.11 Revisiting Example 8.15 will be helpful. 1 + G(s) = 0 may be manipulated as - st

e D 1 = – 1 or G1(s) = – e st D ; G1(s) = s( s + 1) s( s + 1)

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




0.75tD = 52 (p/180) This gives tD = 1.2 sec 8.12 G( jw ) w =0 = – 2.2. – j ¥ ; G( jw ) w =¥ = 0 Ð – 270º

G( jw ) w =15.9 = – 0.182 + j0; G( jw ) 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 as shown in Fig. P8.2c. When K = 7, the polar plot intersects the negative real axis at the point – 1.4 + j0. The resulting Nyquist plot encircles the critical point once in clockwise direction and once in counterclockwise direction. Therefore N = 0. Since P = 0, the system is stable for K = 7. Reducing the gain by a factor of 1/1.4 will bring the system to 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 of 0 dB/decade, and is plotted at dB = 20 log 25 = 27.96. Asymptote slope changes to –20 dB/decade at the first corner frequency 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 the 0 dB-axis at w = K = 50. The asymptote slope changes to – 40 dB/decade at the first corner frequency at wc1 = 1; then to – 20 dB/decade at wc2 = 5, and back to – 40 dB/decade at wc3 = 50. The asymptotic gain crossover is at wg = 10.2. (c) The system is type-2; the low-frequency asymptote is drawn with a slope of – 40 dB/decade, and is located such that its extension would intersect the 0 dB-axis at w = K = 500 = 22.36. The asymptote slope 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 at wg = 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 (note that 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 and passes through the point (w = 1, dB = 20). The asymptote slope changes 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 crosses the 0 dB line at wg = 7.86. The phase is computed from Ð G(jw) = – 90º – tan–1 0.1w

The phase shift at wg = 7.86 is –128.2º, and the resulting phase margin is – 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 passes through the point (w = 1, dB = 20) The asymptote slope changes to – 60 dB/decade at the corner frequency 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 compensated magnitude plot crosses the 0 dB line at wg = 6.8. The phase is computed from Ð G(jw) = – 90º – 2 tan–1 0.1w

The phase shift at wg = 6.8 is – 158.6º. Therefore, the phase margin is 180º – 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 passes through 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 – 20 dB/decade at the second corner frequency wc2 = 5. Compensate the asymptotic 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. The phase shift at this frequency is – 106.6º. Therefore the phase margin is 73.4º. 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 will change 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) =

40 = ( s + 2 )( s + 4 )( s + 5 )

1

FH 1 s + 1IK FH 1 s + 1IK FH 1 s + 1IK 2 4 5

The low-frequency asymptote coincides with 0 dB line; its slope changes 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 can easily be obtained by applying corrections. The phase plot crosses the – 180º line at wf = 7 rad/sec, and the gain margin is 20 dB. Since the gain never reaches 0 dB (wg = 0), the phase margin is infinity. (f) From the Bode plots of G(jw) we find that gain crossover frequency wg = 3.16 rad/sec, and the phase margin is –33º. The phase plot is asymptotic to the –180º line in the low-frequency range; it never reaches – 180º Therefore the GM = – ¥ 8.16

G(s) =

K /5 1 s ( s + 1) s +1 5

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 20 dB. 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) =

F 1 s + 1I F H 2 KH

K / 40 1 1 s +1 s +1 4 5

IF KH

I K

Since G(s) has all poles in left half s-plane, the open-loop system is stable. Hence the closed-loop system will be stable if the frequency response 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) is possible before the system becomes unstable. Hence K < 400 for stability.



SOLUTION MANUAL

85

(b) From the Bode plot of G(jw) with K = 1, we find that the phase crossover 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 factor of 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/sec and 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 the presence of dead-time. (c) The value of tD for the system to be on verge of instability is obtained by setting the phase margin equal to zero, i.e.,

G1 ( jw ) w =w –

w gt D ´ 180º p

g

= – 180º

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

7.4t D ´ 180 º = – 180º p

or tD = 0.078 sec 8.19 Since the systems have minimum-phase characteristics, zeros (if any) are in 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 therefore type-1; 4/s is a factor of the transfer function. The asymptote slope changes from – 20 dB/decade to 0 dB/decade at w = 2 (this corresponds to the factor (1 + s/2)), then to – 20 dB/decade at w = 10 (this corresponds to the factor (1 + s/10)). The transfer function

FH 1 sIK 2 G(s) = FsH1 + 1 sIK 4 1+

10

(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 a factor 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 frequency wc1, where 20 log (5wc1) = 20. This gives wc1 = 2. The transfer function G(s) =

FH

5s 1 1 1 s 1+ s 1+ s 1+ 30 10 2

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 a factor of the transfer function with 20 log (K/w) = – 9 at w = 1. This gives K = 0.355. The transfer function

FH

0.355(1 + s ) 1 + G(s) =

FH

s 1+

1 s 40

IK

1 s 20

IK

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

250 1 1 s 1+ s 1+ s 40 2.5

FH

IK FH

IK

(e) The low-frequency asymptote has a slope of +12 dB/octave; the system is therefore type-0 and Ks2 is a factor of the transfer function. The first corner frequency wc1 = 0.5. The slope of the asymptote changes to +6 dB/decade at this corner frequency. Therefore 1/(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

dB

32 20

20 log Kw 2

10 0

0.1

w

1

–10

0.5 – 20 log 2w

–20

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) =

79.6 s 2 (1 + 2 s )(1 + s )(1 + 0.2 s )

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

31.623s 1 1 (1 + s ) 1 + s 1 + s 20 5

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 – 20 log

FH 1 w IK = 0; this gives w 20 g2

g2

FH 1 w IK 5 g2

= 56.234

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




10.64 1 1 s s +1 s +1 8 3

G(s) =

FH

IK FH

Ð G(jw) = – 90º – tan–1

IK

1 1 w – tan–1 w 3 8

Ð G(jw) = – 180º at w = 4.9

The gain obtained from asymptotic magnitude plot at w = 4.9 is 20 log

4.9 10.64 - 20 log = 2.5 dB 3 4.9

Therefore, GM = – 2.5 dB 8.22 Revisit Review Example 8.3.

F H

I K

1 s 10 G(s) = 1 0.6 1 2 s 1+ s 1+ s+ s 2 50 2500 5 1+

F H

I LM KN

OP Q

8.23 From the asymptotic magnitude plot we find that the system is type-0 with a double pole at s = – 3. The transfer function obtained from the magnitude plot is G(s) =

FH

0.1 1 s +1 3

IK

2

Ð G(jw) vs w may be compared with the given phase characteristics to check the accuracy of identification of corner frequency at wc = 3.

——



CHAPTER 9

9.1 9.2 9.3 9.4

FEEDBACK SYSTEM PERFORMANCE BASED ON THE FREQUENCY RESPONSE

For derivation of the result, refer Section 9.3; Eqn. (9.9). For derivation of the result, refer Section 9.3; Eqn. (9.18). For derivation of the result, refer Section 9.3; Eqns (9.15) – (9.16). The relations z4 – z2 +

1 2 = 0 ; wr = w n 1 - 2z 2 4 Mr

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 and z = 0.8. We select z = 0.6 as damping ratio larger than 0.707 yields no peak above zero frequency. The characteristic equation of the given system is s2 + as + K = s2 + 2zwns + w 2n = 0 This gives K = 475 and a = 26.2 ts =

4 = 0.305 sec zw n

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

1/ 2

= 25.1 rad/sec 9.5

From the response curve, we find M p = 0.135, tp = 0.185 sec z2 =

p wn 1-z2

(ln M p ) 2 (ln M p ) 2 + p 2

gives z = 0.535 for Mp = 0.135

= 0.185 gives wn = 20

The corresponding frequency response performance indices are: Mr =

1 2z 1 - z

2

= 1.11 ; wr = w n 1 - 2z 2 = 13.25




wb = w n 1 - 2z 2 + 2 - 4z 2 + 4z 4 9.6

1/ 2

= 24.6

z » 0.01fm The characteristic equation of the system is t s2 + (1 + KKt)s + K = 0 This gives K /t ; z =

wn = K= 9.7

1 + KK t = 0.01fm 2 Kt

1 2 ´ 10 -4 f 2mt - Kt ± 2 ´ 10 -2 f m t 10 -4 f 2mt - Kt Kt

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

Kt 1t 2 t1 + t 2

|G(jw)| = Therefore Gm ×

Kt 1t 2 =1 t1 + t 2

This gives

9.8

FG H

IJ K

1 1 1 + Gm t 1 t 2 (a) It can be determined from the Bode plot that wg = 1 rad/sec and FM = 59.2º K=

2z

(b) FM = tan–1

4z + 1 - 2z 2 4

FM = 59.2º ® z = 0.6 wg

= wn

4z 4 + 1 - 2z 2

This gives wn = 1.4 Therefore, M p = e - pz /

1-z 2

= 9.48%



SOLUTION MANUAL

ts = 9.9

91

4 = 4.76 sec. zw n

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

FH

IK FH

IK

1 1 s 1+ s 2 4 G(s) = 1 1 1 s 1+ s 1+ s 1+ s 8 24 36 8 1+

FH

IK FH

IK FH

IK

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 – 130º for some value of w. From the phase curve of G(jw)H(jw), we find that Ð G(jw)H(jw) ~ - –130º at w = 0.5. The magnitude of G(jw)H(jw) at this frequency is approximately 3.5. The gain must be reduced by a factor 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 ( jw + 2 ) ( jw ) 2

Ð G(jw) = tan–1

w – 180º = – 130º 2

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

K ( jw + 2 ) ( jw ) 2

=1 w =2 . 3835

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 1 = 3.88 dB; wg = 8 rad/sec and FM = 10º. rad/sec and GM = 20 log 0.64




(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 lowered by 16.12 dB, i.e., the gain must be changed by a factor of b, given by 20 log b = – 16.12. This gives b = 0.156. (c) To obtain a phase margin of 60º, we must first determine the frequency at which the phase angle of G(jw) is – 120º, and then adjust the gain so that |G(jw)| at this frequency is 0 dB. From the Bode plot, we find that lowering the magnitude plot by 16.65 dB gives the desired phase margin. It equivalently means that the gain should be changed by a factor 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 phase margin is 59.2º. G(s) =

=

100 K s ( s + 36 )( s + 100 )

K / 36

1 1 sF H 36 s + 1IK FH 100 s + 1IK

Make a Bode plot for say K = 3.6. From the plot we find that at w = 14.8 rad/sec, Ð G(jw) = – 120.8º. At w = 14.8, the gain is – 44.18 dB. The magnitude curve has to be raised to 0 dB at w = 14.8 to yield the required phase margin. The gain should be changed 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

4z 4 + 1 - 2z 2

This gives wn = 20.68 tp =

p

= 0.19 sec wn 1-z2 9.14 (a) From the Bode plot, it can easily be determined that GM = 5 dB and FM = 30º. The gain crossover frequency wg = 0.83 rad/sec. (b) The magnitude plot should be lowered by 5 dB to obtain a gain margin of 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 be increased by 4.8 dB to obtain a gain crossover frequency of 1.27 rad/sec. (d) Gain at the frequency which gives Ð G(jw) = – 135º, should be 0 dB.



SOLUTION MANUAL

93

From the Bode plot we find that this requires reducing the gain by 3.5 dB. 9.15 The requirements on FM and wg are satisfied if we increase the gain to the limit of zero GM with a phase margin ³ 45º. From the Bode plot we find that K = 37.67 meets this requirement. 9.16 (a) From the Bode plot we find that 0 dB crossing occurs at a frequency of 0.47 rad/sec with a phase angle of – 145º. Therefore, the phase margin is 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. Therefore the 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 without dead-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 reasonable phase margin. We find from the Bode plot that in order to provide a phase margin of 30º, the gain would have to be decreased by 5 dB, i.e., by a factor of b where 20 log b = 5. This gives b = 1.78. We find that the dead-time necessitates the reduction in loop gain is order to obtain a stable response. The cost of stability is the resulting increase in the steady-state error of the system as the loop gain is reduced. 9.18 Make a polar plot of G(jw) =

50 / 18 jw (1 + jw / 3)(1 + jw / 6 )

G( jw ) w =0 = -

25 - j¥ ; G( jw ) w =¥ = 0 Ð – 270º 18

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 the magnitude of the closed-loop frequency response is 0.707 of its magnitude at 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 = 0.707 circle intersects the polar plot. This frequency is found to be 3.61 rad/sec. Therefore wb = 3.61




z 4 – z2 +

1 = 0 ; Mp = e - pz / 4 Mr2 wb = wn

1-z 2

´ 100

(1 - 2z 2 ) + 4z 4 - 2z 2 + 2

ts = 4/zwn The Mr – z relation gives z = 0.29, 0.96 for Mr = 1.8. We select z = 0.29 because damping ratio larger than 0.707 yields no peak above zero frequency. 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 a frequency w = 4 rad/sec. Therefore Mr = 1.4 and wr = 4. z4 – z2 +

1 =0 4 M r2

wr = wn 1 - 2z 2 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-order systems with zero-frequency closed-loop gain equal to unity. The answer is based on the assumption that zero-frequency closed-loop gain for the system 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 from the Bode plot to the Nichols chart. The resulting dB vs phase curve is tangential 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 by transferring the dB and/or phase data from the dB vs phase curve to the Bode plot.



SOLUTION MANUAL

95

(ii) The open-loop dB vs phase plot may not be tangential to one of the constant-M contours of the Nichols chart. One must do a little interpolation. In the present problem, Mr = 6.7 dB has been obtained by interpolation. 9.21 Plot the given frequency response on a dB scale against phase angle. From the 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 having unity closed-loop gain at w = 0. We have made this assumption for the system 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 been provided, we don’t require the Nichols chart for bandwidth determination. On a linear scale graph sheet, dB vs phase cure of the open-loop frequency response is plotted with data coming from Bode plot. On the same graph sheet, the – 3dB contour using the given data is plotted. The intersection of the two curves occurs at w = 0.911 rad/sec. Therefore bandwidth wb = 0.911. FM = tan–1

(d)

wb = wn

2z 4z 4 + 1 - 2z 2 1 - 2z 2 + 2 - 4z 2 + 4z 4

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 transferred to the Nichols chart, gives the following result. Mr = 1.4 ; wr = 6.9 rad/sec From the relations Mr =

1 2z 1 - z

2

; wr = wn 1 - 2z 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 of the 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 unity zero-frequency closed-loop gain. 9.24 (a) Transfer the frequency-response data from the Bode plot of G(jw) to the Nichols chart. Shifting the dB vs phase curve down by 5.4 dB makes this curve tangent to the M = 20 log 1.4contour. Therefore gain must be changed by 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 the Nichols chart at w = 9.2. Therefore bandwidth wb = 9.2 rad/sec. Note that the – 3dB bandwidth definition given by Eqn. (9.2b) is applicable to systems having unity closed-loop gain at w = 0. The given system does not satisfy this requirement. The answer, therefore, is an approximation of the bandwidth. 9.25 (a) From the Bode plot we find the phase crossover frequency wf = 2.1 and GM = 14 dB. For the gain margin to be 20 dB, the magnitude plot is to be brought down by 6 dB. The value of K corresponding to this condition 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 if gain crossover frequency wg is 0.4 rad/sec. This is achieved if magnitude plot is brought down by 7 dB. This condition corresponds to K = 0.446. (c) From the dB-phase plot on the Nichols chart, we find that for this plot to be tangent to M = 1 dB contour, the dB-phase curve must be brought down by 4 dB. This condition corresponds to K = 0.63. The corresponding 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-phase plot 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 condition corresponds to K = 1.35. 9.26 Plot the given frequency response on a dB scale against phase angle. From this 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 curve when raised by about 5 dB, touches the M = 20 log1.4 contour of the Nichols 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 on Nichols chart if the plot is lowered by about 11 dB. This means that the gain K must be reduced by a factor of a where 20 log (1/a) = –11. This gives 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 for analysis and design purposes. 9.28 SGM (jw) =

G - ( jw ) 1 = 1 + G( jw ) 1 + G -1 ( jw ) 1

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 of Nichols chart. Therefore

| SGM (jw)|max = 2.18 The peak occurs at w = 7 rad/sec.



CHAPTER 10

COMPENSATOR DESIGN USING BODE PLOTS

10.1 (a) From the Bode plot of G(jw)H(jw), we find that GM = 6 dB and FM = 17º (b) Now the gain and phase of the compensator with Kc = 1 are added to the Bode plot. From the new Bode plot we find that the gain must be raised 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 that wg = 5 rad/sec; FM = 37.6º; GM = 18 dB The increase in phase and gain margins implies that lead compensation increases margin of stability. The increase in wg implies that lead compensation increases speed of response. 10.3 (a) Kv = lim sD(s)G(s) = K = 12 s® 0

(b) From the Bode plot and Nichols chart analysis of the system with K=12, we find that FM = 15º; wb = 5.5 rad/sec; Mr = 12 dB; wr = 3.5 rad/sec (c) The phase margin of the uncompensated system is 15º. The phase lead required at the gain crossover frequency of the compensated magnitude curve = 40º – 15º + 5º = 30º a =

1 - sin 30 º = 0.334 1 + sin 30 º

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 as gain crossover frequency of the compensated magnitude curve, we set 1 = 4.6 at The transfer function of the lead compensator becomes 12 ( 0 .376 s + 1) D(s) = 0 .128 s + 1 (d) FM = 42º; wb = 9 rad/sec; Mr = 3 dB; wr = 4.6 rad/sec



SOLUTION MANUAL

99

10 ( 0.5s + 1) meets the phase margin and steady0.1s + 1 state accuracy requirements. These requirements are also met by the lag

10.4 Lead compensator D(s) =

compensator D(s) =

10 (10 s + 1) 100s + 1

In this particularly simple example, specifications could be met by either compensation. In more realistic situations, there are additional performance specifications such as bandwidth and there are constraints on loop gain. Had there been additional specifications and constraints, it would have influenced the choice of compensator (lead or lag). 105. The settling time and peak overshoot requirements on performance may be translated to the following equivalent specifications: z = 0.45 ; wn = 2.22 z is related to the phase margin by the relation FM »

z = 45º 0.01

wn is related to the bandwidth wb by the relation wb = wn 1 - 2z 2 + 2 - 4z 2 + 4z 4

1/ 2

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 the compensator D(s) =

K (t s + 1) at s + 1

may be set a value given by K = w 2n . 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 the double integration results in a constant 180º phase lag. Therefore, we must add a 45º phase lead at the gain crossover frequency of the compensated magnitude curve. Evaluating the value of a, we have a =

1 - sin 45º = 0.172 1 + sin 45º

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



100


The frequency at which the uncompensated system has a magnitude of – 20 log (1/ a ) = – 7.78 dB is 4.95 rad/sec. Selecting this frequency as the gain crossover frequency of the compensated magnitude curve, we set

1 at

= 4.95

The closed-loop system with the compensator D(s) =

10 ( 0 .083s + 1) 0.5 s + 1

satisfies the performance specifications. 10.6 Kv =

A A = = 20 0.05 A ess

K = 20 realizes this value of Kv. From the Bode plot of G(jw) =

20 jw ( j 0.5w + 1)

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

1 - sin 30 º = 0.334 1 + sin 30 º

The frequency at which the uncompensated system has a magnitude of – 20 log (1/ a ) = – 4.8 dB is 8.4 rad/sec. Selecting this frequency as the gain crossover frequency of the compensated magnitude curve, we set

1 at

= 8.4

Therefore the compensator D(s) =

0.2 s + 1 0.0668s + 1

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



SOLUTION MANUAL

101

10.7 From the Bode plot of G(jw) =

20 jw ( j 0.5w + 1)

we find that the phase margin of the uncompensated system is 18º. To realize a phase margin of 45º, the gain crossover frequency should be moved to w g¢ where the phase angle of the uncompensated system is: – 180º + 45º + 5º = – 130º. From the Bode plot of uncompensated system, we find that w g¢ = 1.5. The attenuation necessary to cause w g¢ 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 a decade below w g¢ , we have 1/t = 0.15. Therefore, the lag compensator D(s) =

6.66 s + 1 66.6 s + 1

As a final check, we numerically evaluate the phase margin and the bandwidth of the compensated system. It is found that that FM = 45º and wb = 2.5 rad/sec. 10.8 From the steady-state requirement, we set K = 100. From the Bode plot of G(jw) =

100 jw ( j 0.1w )( j 0.2w + 1)

we find that the phase margin is – 40º, which means that the system is unstable. The rapid decrease of phase of G(jw) at the gain crossover frequency wg = 17 rad/sec, implies that single-stage lead compensation may be ineffective for this system. (The reader should, in fact, try a single-stage lead 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 portion of the desired phase margin improvement by each compensator stage. We first add a single-stage lead compensator that will provide a phase lead of about 42.5º, i.e., the compensator will improve the phase margin to about 2.5º.



102


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

1 - sin 42.5º 1 + sin 42.5º

a =

will not yield a phase margin of 2.5º. In fact, when we set a = 0.08 and t = 0.1, the single-stage compensator improves the phase margin to 3º. Therefore D1(s) =

0.1s + 1 0.008s + 1

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

100( 0.1s + 1) s( 0.2 s + 1)( 0.008s + 1) 2

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) =

2500 K s ( s + 25)

K = 1 satisfies the steady-state performance requirement. The phase margin of the uncompensated system, read at the gain crossover frequency wg = 47 rad/sec, is 28º. The phase lead required at the gain crossover frequency of the compensated magnitude curve = 45º – 28º + 8º = 25º a =

1 - sin 25º = 0.405 1 + sin 25º

The frequency at which the uncompensated system has a magnitude of – 20 log (1/ a ) = – 3.93 dB is 60 rad/sec. Selecting this frequency as gain crossover frequency of the compensated magnitude curve, we set

1 at

= 60

The transfer function of the lead compensator is D(s) =

0.026 s + 1 0.01s + 1



SOLUTION MANUAL

103

The phase margin of the compensated system is found to be 47.5º. By Nichols chart analysis, it is found that the lead compensator reduces the resonance peak from 2 to 1.25 and increases the bandwidth from 74 to 98 rad/sec. 10.10 The phase margin of the uncompensated system 2500 K ;K=1 s ( s + 25)

G(s) =

read at the gain crossover frequency wg = 47 rad/sec is 28º. To realize a phase 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 w g¢ = 25. Since the lag compensator contributes a small negative phase when the upper corner frequency of the compensator is placed at 1/10 of the value of w g¢ , it is a safe measure to choose w g¢ 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 w g¢ , we have 1/t = 2. Therefore the lag compensator D(s) =

0.5s + 1 2.5s + 1

From the Bode plot–Nichols chart analysis we find that the compensated system has FM = 51º; Mr = 1.2 and wb = 27.5 rad/sec. 10.11 Kv = lim sG(s) = 5 s®0

From the Bode plot we find that the uncompensated system has a phase margin of – 20º; the system is therefore unstable. We attempt a lag compensator. This choice is based on the observation made from Bode plot of uncompensated system that there is a rapid decrease in phase of G(jw) near wg. Lead compensator will not be effective for this system.



104


To realize a phase margin of 40º, the gain crossover frequency should be moved 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 w g¢ = 0.5 rad/sec. The attenuation necessary to cause w g¢ 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 Since we have taken a large safety margin of 12°, we can place the upper corner frequency of the compensator at 0.1 rad/sec, i.e., 1/t = 0.1. Thus the transfer function of the lag compensator becomes D(s) =

10 s + 1 100 s + 1

From the Bode plot of compensated system we find that the phase margin is about 40° and the gain margin is about 11 dB. 10.12 It easily follows that K = 30 satisfies the specification on Kv. From the Bode plot of G(jw) =

30 jw ( j 0.1w + 1)( j 0.2w + 1)

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 still further, resulting in an undesirable system which will be sensitive to noise. If lag compensation is attempted, the bandwidth will decrease sufficiently so as to fall short of the specified value of 12 rad/sec, resulting in a sluggish system. These facts can be verified by designing lead and lag compensators. We thus find that there is need to go in for lag-lead compensation. Since the full lag compensator will reduce the system bandwidth excessively, the lag section of the lag-lead compensator must be designed so as to provide partial compensation only. The lag section, therefore, should move the gain crossover frequency to a value higher than the gain crossover frequency of the fully lag-compensated system. We make a choice of new gain crossover frequency as w g¢ = 3.5 rad/sec. The attenuation necessary to cause w g¢ 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 the transfer function of the lag section as D1(s) =

t 1s + 1 s +1 = bt 1s + 1 10 s + 1

It is found that the lag-section compensated system has a phase margin of 24º. We now proceed to design the lead section. The implementation of the laglead compensator is simpler if a and b parameters of the lead and lag compensators, respectively, are related as a = 1/b. Let us first make this choice. 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 this frequency as gain crossover frequency of the lag-lead compensated magnitude curve, we set

1 at2

= 7.5

The transfer function of the lead-section becomes

t 2s + 1 0 .422 s + 1 = 0.0422 s + 1 at 2 s + 1

D2(s) =

The analysis of the lag-lead compensated system gives FM = 48º and wb = 13 rad/sec. 10.13 z 2 =

(ln M p ) 2 (ln M p ) 2 + p 2

M p = 0.2 ® z = 0.456 p tp = wn 1-z2 tp = 0.1 ® wn = 35.3 wb = wn [1 - 2z 2 + 2 - 4z 2 + 4z 4 ]1/ 2 = 46.576 FM = tan–1

2z 4z + 1 - 2z 2 4

~ - 48º



106


In order to meet the specification of Kv = 40, K must be set at 1440. From the Bode plot of G(jw) =

144000 jw ( jw + 36)( jw + 100)

we find that the gain crossover frequency is 29.7 rad/sec and phase margin is 34º. The phase lead required at the gain crossover frequency of the compensated magnitude curve = 48º – 34º + 10º = 24º a =

1 - sin 24 º = 0.42 1 + sin 24 º

The frequency at which the uncompensated system has a magnitude of – 20 log (1 / a ) = – 3.77 dB is 39 rad/sec. Selecting this frequency as gain crossover frequency of the compensated magnitude curve, we set 1 = 39 at The transfer function of the lead compensator becomes 1440(0.04 s + 1) D(s) = 0.0168s + 1 It can easily be verified by Nichols chart analysis that the bandwidth of the compensated system exceeds the requirement. We assume the peak time specification is met. This conclusion about the peak time is based on a second-order approximation that should be checked via simulation. 10.14 K = 2000 meets the requirement on Kv. We estimate a phase margin of 65º to 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 we observe that there is a rapid decrease of phase at the gain crossover frequency. Since the requirement on phase lead is quite large, it is not advisable to compensate this system by a single-stage lead compensator (refer Fig. 10.15). (b) Allowing 10º for the lag compensator, we locate the frequency at which the phase angle of the uncompensated system is: – 180º + 65º + 10º = – 105º. This frequency is equal to 1.5 rad/sec. The gain crossover frequency w g¢ should be moved to this value. The necessary attenuation is 23 dB. The b parameter of the lag compensator 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 decade below w g¢ , we have 1/t = 0.15. Therefore the lag compensator D(s) =

6.66 s + 1 94.66 s + 1

The phase margin of the compensated system is found to be 67º. (c) The bandwidth of the compensated system is found to be wb = 2.08 rad/sec. 10.15 Approximate 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 be moved to w g¢ where the phase angle of the uncompensated system is: – 180º + 40º + 10º = – 130º. From the Bode plot of the uncompensated system we find that, w g¢ = 6. The attenuation necessary to cause w g¢ 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 w g¢ , we have 1/t = 1.5. The lag compensator D(s) =

0.67s + 1 2s + 1

Nichols chart analysis of the compensated system gives wb = 11 rad/sec. Using second-order approximation wb = wn 1 - 2z 2 + 2 - 4z 2 + 4z 4

1/ 2

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-order system 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



108


(b) Phase margin of the system with the second-order compensator is 9.47º. There is no effect of the compensator on steady-state performance of the system. (c) The lead compensator D(s) =

1 + 0.0378s 1 + 0.0012 s

meets the requirements on relative stability. 10.17 Kv = 4.8



CHAPTER 11

11.1 (a)

(b)

HARDWARE AND SOFTWARE IMPLEMENTATION OF COMMON COMPENSATORS

E( s) RD TD s = – =– ; a = R/RD ; TD = RD CD Y ( s) R + 1 / CD s aTD s + 1 E( s) R K (T s + 1) =– =– c D R1 + R2 / (1 + sR2 C) aTD s + 1 Y ( s) Kc =

R R1 ;a= ; TD = R2C R1 + R2 R1 + R2

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

E2 ( s ) Z ( s) = – 2 [ -1] E1 (s) Z1 ( s) =

t s +1 R2 at s + 1 ; t = R1C, a = R + R < 1 1 2

Refer Section 11.2 (Fig. 11.4) for the Bode plot and filtering properties of the lead compensator. 11.3 Z1(s) = R1 + R2 ; Z2(s) = R2 + R1/(1 + R1Cs)

E2 ( s) Z (s ) R1 R2 R + R2 t s +1 C, b = 1 =– 2 =– ;t= >1 E1 (s) Z1 (s) R1 + R2 R2 bt s + 1 Add an inverting amplifier of gain unity. Refer Section 11.2 (Fig. 11.7) for the 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]

FG H

Kc s +

LM OP N Q FG s + 1 H at

E2 ( s) Z2 (s ) R6 E1 (s) = – Z ( s) - R = 1 5

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

IJ FG s + 1 IJ KH t K IJ FG s + 1 IJ K H bt K

1 t1

2

1

2

R1 R + R4 ;b= 2 ; R1 + R3 R2



110


Kc =

FG H

R2 R4 R6 R1 + R3 R1 R3 R5 R2 + R4

IJ K

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

LM OP N Q

LM N

Z (s ) R4 E2 ( s) 1 + TD s =– 2 = Kc 1 + T1s E1 (s) Z1 ( s) R3

OP Q

Kc =

R4 ( R1C1 + R2 C2 ) = 39.42; TI = R1C1 + R2C2 = 3.077 R3 R1C2

TD =

R1 R2 C1C2 = 0.7692 R1C1 + R2 C2

These equations give R1 = R2 = 153.85 kW; R4 = 197.1 kW 11.6 Z1(s) = R1/(1 + R1C1s); Z2(s) = R2/(1 + R2C2s)

LM N

Z ( s) R4 E2 ( s ) = - 2 E1 (s) Z1 (s) R3 =

OP Q

FG H

IJ K

R4 R2 R1C1s + 1 0.345s + 1 = 2.51 0.185s + 1 R3 R1 R2 C2 s + 1

F H

I K

This equation gives R1 = 34.5 kW; R2 = 18.5 kW; R4 = 45.8 kW 11.7 By Trapezoidal rule for integration: kT

z

e(t )dt = T

0

LM e(0) + e(T ) + e(T ) + e(2T ) +...+ e(k - 1)T + e(kT ) OP 2 2 N 2 Q

LM e(i - 1)T + e(iT ) OP T MNå PQ k

=T

i =1

By backward-difference approximation for derivatives:

de(t ) e( kT ) - e(( k - 1)T ) = T dt t = kT A difference equation model of the PID controller is, therefore, given by



SOLUTION MANUAL

RS T

u(k) = Kc e( k ) +

T TI

111

UV W

e(i - 1) + e(i ) TD + [e( k ) - e( k - 1)] 2 T i =1 k

å

11.8 Taking sampling frequency twenty times the closed-loop natural frequency (which is a measure of bandwidth), we have ws = 20wn =

2p ; this gives T = 0.5 sec T

Settling time ts = 4/zwn = 13 1 th of ts is 1.3 sec; T = 0.5 sec is therefore a safe choice. 10

U ( s) 2.2( s + 0.1) = E( s) s + 0.01

u& (t) + 0.01 u(t) = 2.2 e& (t) + 0.22 e(t) 1 2.2 [u(k) – u(k – 1)] + 0.01 u(k) = [e(k) – e(k – 1)] + 0.22 e(k) T T

This gives u(k) =

2 4.62 4.4 u(k – 1) + e(k) – e(k – 1) 2.01 2.01 2.01

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 - 1) K [e(k) – e(k – 1)] + Kbe(k) + au( k ) = T T

By backward-rectangular rule for integration: t

t

z

z

0

0

u(t) = u(0) – a u(t )dt + Ke(t ) - Ke(0) + Kb e(t )dt 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 following computer algorithm. u(k) =

1 K (1 + bT ) K e( k ) e(k - 1) u( k - 1) + 1 + aT 1 + aT 1 + aT



112


(b) u(k) = u(k – 1) – aT

LM u(k) + u(k - 1) OP + Ke(k) – Ke(k – 1) N 2 Q

LM e(k) + e(k - 1) OP N 2 Q bT I bT I K F1 + K F1 H K H 2 2 K u(k – 1) + e(k) – e(k – 1)

+ KbT

aT 2 = aT 1+ 2 1-

1+

aT 2

1+

aT 2

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

U ( s) E ( s)

112 . s s2 +1 2 + 36 (36) 2 s +1 740

=

F H

I K

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

LM u(k) - 2u(k - 1) + u(k - 2) OP N Q T L u(k) - u(k - 1) OP + u(k) + 2.7 × 10 M N T Q e ( k ) - 2e ( k - 1) + e( k - 2) O 7.7 × 10 L MN PQ T e( k ) - e( k - 1) O + 0.03 LM N T PQ + e(k)

1.826 × 10–6

2

–3

=

–4

2

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-loop performance using longer value of sampling interval.



SOLUTION MANUAL

11.11

113

B(s) Kt s = Y ( s) t s + 1 t b& (t) + b(t) = Kt y& (t) t

b(t) = b(0) –

z

K 1 b(q )dq + t [ y(t ) - y(0)] t t 0

b(k) = b(k – 1) –

=

LM N

OP Q

K T b( k ) + b( k - 1) + t [ y( k ) - y( k - 1)] t t 2

1 - T / 2t Kt / t Kt / t y( k ) y( k - 1) b(k – 1) + 1 + T / 2t 1 + T / 2t 1 + T / 2t

11.12 The set-point control, proportional action and derivative action are all included in the op amp circuit of Fig. 11.43. Only the integral action is to be added. Connect the integral-action unit of Fig. 11.41 in cascade with the proportional action unit of Fig. 11.43; the output e4 of the proportionalaction unit of Fig. 11.43 becomes input to the integral-action unit (output e3). The resulting op amp circuit is governed by the following relation (refer Examples 11.6 and 11.7);

FG H

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

Kc =

1 TI s

IJ [Y(s) – R(s)] K

R2 ; TD = RDCD ; TI = RI CI R2¢

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

R2 R2¢

E4(s) = Kc [E3(s) – E2(s)] ; Kc = = 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 × 100 = 185% PB 0.54

TI = Tu/1.2 = 3.75 min or

1 = 0.266 repeats/min 3.75



114


(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. Process A is more controllable than processes B and C, which are equally controllable. 11.18 (a) From the tuning rules given in Table 11.2: K c¢ = 3 ; TI¢ = 4 min ; TD¢ = 1 min

(b) tCD = tD +

1 1 T = 2 min + × 8 sec = 2.067 min. 2 2

Replacing tD by tCD in the tuning formulas, we obtain K c¢ = 2.9 ; TI¢ = 4.13 min ; TD¢ = 1.03 min 1 T. Increase 2 in sampling time demands reduced Kc and increased TI and TD.

11.19 Replace tD in tuning formulas of Table 11.2 by tCD = 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 has directly been conducted on the digital loop.



CHAPTER 12 CONTROL SYSTEM ANALYSIS USING STATE VARIABLE METHODS 12.1 if

u

Kg + –

1 sLf + Rf

qM

ia 1 s (La + Lg ) + Ra +Rg

KT

1 sJe + Be

x3

x4

qM 1 s

x2

n x1

Kb

Je = n2J = 0.4 ; Be = n2B = 0.01 x&1 = x2 ; 0.4 x& 2 + 0.01x2 = 1.2x3

0.1 x& 3 + 19x3 = 100 x4 – 1.2x2 ; 5 x& 4 + 21x4 = 4 x& = Ax + bu

or

A =

LM0 MM00 MN0

OP PP PQ

LM MM MN

0 1 0 0 0 - 0.025 3 0 ;b= 0 -12 -190 1000 0 0 4 2 0 .2 . -

OP PP PQ

y = qL = nx1 = 0.5x1 12.2 u

KA

1 sLf + Rf + 1

if x3

1 Js + B

KT

q x2

x&1 = x2 ; 0.5 x& 2 + 0.5x2 = 10x3

20 x& 3 + 100x3 = 50u ; y = x1

A =

LM0 MM00 N

OP PP Q

LM MM N

OP PP Q

1 0 0 -1 20 ; b = 0 ; c = 1 0 0 0 -5 2.5


1 s

qL

q x1


116


12.3 x1 = qM ; x2 = q& M , x3 = ia ; y = qL x&1 = x2

2 x& 2 + x2 = 38 x3 2 x& 3 + 21x3 = ea – 0.5 x2

k e a = k1(qR – qL) – k2 q& M = k1qR – 1 x1 – k2x2 20

LM 0 A= M 0 MM k MN- 40

OP L 0 O M P 1 19 P ; b = M 0 P ; c = LM P 20 MM k PP N 21 - P N2Q 2 PQ

1

0

- 0.5 ( k + 0.5) - 2 2

1

1

12.4 x 1 = w ; x2 = ia J w& + Bw = KT ia Raia + La

dia dt

= ea – Kbw

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

A =

LM - B MM -(k K KJ + K ) MN L 1

t

c

a

b =

b

OP PP PQ

KT J -( Ra + k2 Kc ) ; La

LM k 0K OP MN L PQ ; c = 1 0 L-11 6OP P AP = M N-15 8Q L1 O P b = M P ; c = cP = N2 Q c

1

a

12.5

A =

b = Y (s ) U ( s)

=

–1

–1

2 -1

P1 D1 s -2 1 = = 2 -1 -2 D 1 - ( -3s - 2s ) s + 3s + 2


0 0

OP Q


SOLUTION MANUAL

s –1

1

U

X2

s –1

1

117

X1 = Y

–3 –2

P D + P2 D 2 + P3 D 3 + P4 D 4 Y (s ) = 1 1 U ( s) D =

2s -1 (1 - 8s -1 ) + 15s -2 (1) + ( -2s -1 )[1 + 11s -1 ] + 2s -1 (6s -1 )2[1] 1 - [ -11s -1 + 8s -1 + ( -15s -1 )(6 s -1 )] + [( -11s -1 )(8s -1 )]

=

1 s + 3s + 2 2

s –1

X1

1

2 –15

–11

Y

U 6 –1

2 X2

s –1 8

12.6

LM0 1OP ; b = LM0OP N0 0 Q N1 Q L 1 1 OP ; b = P A = P AP = M N–1 –1Q

A=

–1

–1

b=

LM0OP N1 Q

|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) 1 G(s) = D

LMs(s + 3) MM --1s N

s+3 s( s + 3) -1

OP L1O 1M P s s ; H(s) = P DM P MNs PQ s PQ 1

2

2



118


D = s3 + 3s2 + 1 12.8

u +

+

0 x3 + x3

+

0 x2 + x2

+

0 x1 + x1

+ +

–3

+ –1

12.9 Taking outputs of integrators as state variables, we get (x1 being the output of rightmost integrator), x&1 = x 2

x& 2 = – 2x2 + x3 x& 3 = – x3 – x2 – y + u y = 2x1 – 2x2 + x3

A =

12.10 (a)

(b) 12.11

LM 0 MM-02 N

OP PP Q

LM OP MM PP NQ

1 0 0 -2 1 ; b = 0 ; c = 2 -2 1 1 -2 1

G(s) = c (sI – A)–1 b =

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

G(s) =

1 ( s + 1)(s + 2)

x&1 = – 3x1 + 2x2 + [– 2x1 – 1.5x2 – 3.5x3] x& 2 = 4x1 – 5x2 x& 3 = x2 – r



SOLUTION MANUAL

A =

LM-5 MM 40 N

OP PP Q

G(s) = c(sI – A)–1 b = 12.12 (a)

LM MM N

OP PP Q

0.5 -3.5 0 0 ;b= 0 ;c= 0 1 0 -5 1 0 -1 14 ( s + 1)( s + 2)( s + 7)

x 1 = output of lag 1/(s + 2) x 2 = output of lag 1/(s + 1)

x&1 + 2x1 = x2 ; x& 2 + x2 = – x1 + u y = x2 + (– x1 + u) A =

LM-2 1 OP ; b = LM0OP ; c = N -1 -1Q N1Q

-1 1 ; d = 1

x 1 = output of lag 1/(s + 2)

(b)

x 2 = output of lag 1/s x 3 = output of lag 1/(s + 1) x&1 + 2x1 = y ; x& 2 = – x1 + u x& 3 + x3 = – x1 + u ; y = x2 + x3

A =

12.13 (i)

Y (s ) U ( s)

=

A =

LM-2 MM --11 N

OP PP Q

LM OP MM PP NQ

1 1 0 0 0 ;b= 1 ;c= 0 1 1 0 -1 1

–1 s+3 2 + = s+1 s+ 2 ( s + 1)(s + 2)

LM-1 0 OP ; b = LM1OP ; c = N 0 -2Q N1Q

2 -1


119


120


x1

u + +

2

+

–1

y

+ x2

+ +

–1

–2

(ii)

b3 Y (s ) 5 = 3 = 3 2 2 U ( s ) s + a 1s + a 2 s + a 3 s + 4 s + 5s + 2

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

A

u

5

LM0 MM10 N

=

LM OP MM PP NQ

+

x1 +

–

–

2

(iii)

OP PP Q

0 -2 5 0 -5 ; b = 0 ; c = 0 0 1 1 -4 0

2

x3 = y

x2 +

5

–

4

Y (s ) b s 3 + b 1s 2 + b 2 s + b 3 s 3 + 8s 2 + 17s + 8 = 03 = U ( s) s + a 1s 2 + a 2 s + a 3 s 3 + 6s 2 + 11s + 6

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



SOLUTION MANUAL

A =

LM 0 MM-06 N

OP PP Q

LM OP MM PP NQ

1 0 0 0 1 ;b= 0 ;c= 2 6 2 -11 -6 1

+

+

+

+

+

+ 17

8

1

u +

x3

z

–

z +

Y (s ) U ( s)

=

x2

+

+

z +

b 2s + b 3 s +1 = 3 2 s + 3s + 2s s + a 1s 2 + a 2 s + a 3

From Eqns (12.46):

A

(ii)

Y (s ) U ( s)

=

OP PP Q

LM OP MM PP NQ

0 0 1 0 -2 ; b = 1 ; c = 0 0 1 1 -3 0

b3 1 = 3 2 s + a 1s + a 2 s + a 3 s + 6 s + 11s + 6 3

2

From Eqns (12.44):

A =

(iii)

LM 0 MM-06 N

1 0 -11

OP LM0OP ;b= 0 ;c= 1 P MM1PP -6 PQ NQ 0 1

0 0

Y (s ) s 3 + 8s 2 + 17s + 8 2 1 -1 = 3 = 1+ + + U ( s) s + 6s 2 + 11s + 6 s +1 s + 2 s + 3


x1

6

3

LM0 = 1 MM0 N

y

8

11

6

12.14 (i)

121


122


A =

12.15 (a)

LM-1 MM 00 N

OP PP Q

LM OP MM PP NQ

0 0 1 -2 0 b = 1 c = -1 2 1 ; d = 1 0 -3 1

b 2s + b 3 Y (s ) 1000 s + 5000 = 3 = 3 2 U ( s) s + a 1s 2 + a 2 s + a 3 s + 52 s + 100 s From Eqns (12.44):

A =

(b)

OP PP Q

LM OP MM PP NQ

1 0 0 0 1 ; b = 0 ; c = 5000 1000 0 -100 -52 1

Y ( s) 1000s + 5000 50 -31.25 -18.75 + + = 3 = 2 s s+2 s + 50 U ( s) s + 52 s + 100 s

A =

12.16

LM0 MM00 N LM0 MM00 N

OP PP Q

LM MM N

OP PP Q

0 0 50 -2 0 ; b = -31.25 ; c = 1 1 1 0 -50 -18.75

Y (s ) 2s 2 + 6s + 5 1 1 1 = = + + U ( s) ( s + 1) 2 ( s + 2) ( s + 1) 2 s + 1 s + 2 From Eqns (12.55):

A =

LM-1 MM 00 N

OP PP Q

LM OP MM PP NQ

1 0 0 -1 0 ; b = 1 ; c = 1 1 1 0 -2 1

t

x(t) =

z

eA(t – t) bu(t)dt

0

LMe ]= M 0 MN 0

-t

e

–At

= L

–1

[(sI – A)

–1

te -t e -t 0


0 0 e -2 t

OP PP Q


SOLUTION MANUAL

LM MM MM MM MM N

t

z

e A( t -t ) bu(t )dt

=

0

t

z z z

(t - t )e -(t - t ) dt

OP PP L1 - e - te PP = MM1 - e PP MMN 12 (1 - e ) PP Q -t

0 t

e

- (t - t )

-t

-t

dt

-2 t

0 t

e -2( t - t ) dt

0

123

OP PP PQ

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

1

u

+

1

x2 +

z

+

1

–

–1

z

x1

–1 +y

+

1

+

+

x3

z –2

12.17

LM0.696 N 0.123

0.246 0.572

F

= eAT =

g

= (e–AT – I) A–1 b

=

LM– 0.304 N 0.123

OP Q

L -3 OP MM 4 Q M -41 N

0.246 -0.428

-1 2 -1 2

OP L-1O L-0.021O PP MN 5 PQ = MN 0.747 PQ Q

c = 2 -4 ; d = 6



124


Y (s ) e -0 .4 s = U ( s) s +1

12.18

x&1 (t ) = – x1(t) + u(t – 0.4)

tD = 0.4 ; therefore, N = 0, D = 0.4, m = 0.6 F = e–1 = 0.3679 0.6

g2 =

z

0. 4

e -s ds = 0.4512 ; g1 = e– 0.6

0

z

e -q dq = 0.1809

0

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 =

LM0.3679 N 0

LM N

OP Q

0.1809 0.4512 ;g= 0 1

OP Q

c = 1 0 12.19 1 0

0

x2

x2 s –1 u

s –1

1

s –1 s –1

1 x2

x1

–2

–2

1

G(s) =

LMG NG

OP ; H(s) = LM H ( s) OP ( s )Q N H (s)Q

11 ( s )

G12 ( s )

1

21 ( s )

G22

2

X1 ( s) s -1 (1 + 2s -1 ) = G (s) = 11 1 - ( -2s -1 - 2 s -1 + s -2 ) + ( -2s -1 )(-2 s -1 ) x10



SOLUTION MANUAL

=

s -1 (1 + 2 s -1 ) 1 / 2 1 / 2 = + s +1 s + 3 D

X1 ( s) s -2 (1) 1 / 2 -1 / 2 = G (s) = = + 12 0 s +1 s + 3 D x2 X2 ( s) x10

= G21(s) =

s -2 (1) 1 / 2 -1 / 2 = + s +1 s + 3 D

X2 ( s) x 20

= G22(s) =

s -1 (1 + 2 s -1 ) 1 / 2 1 / 2 = + s +1 s + 3 D

X1 ( s) s -2 (1) + s -1 (1 + 2 s -1 ) 1 = H1(s) = = ( ) U s s +1 D X2 ( s) U ( s)

= H2(s) =

1 s +1

Zero-input response: x(t) = eAt x0 = L-1 [G(s)x0] =

LM N

1 e -t + e -3t 2 e - t - e -3t

e - t - e -3t e - t + e -3t

OP LMx OP Q Nx Q 0 1 0 2

Zero-state response: t

x(t) =

z

e A(t -t ) bu(t ) dt = L

–1

0

12.20

A =

LM 0 1 OP ; eigenvalues are l N-2 -3Q

1

=

s+4 1 + ( s + 1)( s + 2) s( s + 1)( s + 2)

=

3 2 1/ 2 1 1/ 2 + + s +1 s + 2 s s +1 s + 2 1 3 + 2 e - t - e -2 t 2 2


-t

= – 1, l2 = – 2

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

y(t) =

LM1 - e OP N1 - e Q -t

[H(s)U(s)] =

125


126


x&1 = – 3x1 + 2x2 + [7r – 3x1 – 1.5 x2]

12.21

x& 2 = 4x1 – 5x2 A =

LM-6 N4

eAt = L

=

OP ; b = LM7OP ; c = -5 Q N0 Q

0.5

–1

LM 1 e MM 34 e N3

[(sI – A)–1]

-4 t -4 t

2 -7 t e 3 4 - e -7 t 3 t

z LMN 0

OP Q

LM N

LMe + te N -te -t

OP PP Q

4 -4 ( t -t ) 4 -7 ( t -t ) e dt - e 3 3

28 1 1 (1 - e -4 t ) - (1 - e -7t ) 3 4 7 –t

eAt =

12.22

0.5 -4t 0.5 -7t e e 3 3 2 -4 t 1 -7 t e + e 3 3

+

y(t) = x2(t) = 7

=

0 1

-t

te - t e

-t

- te

-t

OP Q

OP Q

x(t) = e A ( t - t0 ) x ( t 0 ) 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

LMe - k OP is a possible set of states L x (1) O for any k ¹ 0 MN x (1)PQ N 2k Q LM e OP = La bO L 1 O ; LM e OP = La b O L 1 O N-2e Q MNc d PQ MN-2PQ N-e Q MNc d PQ MN-1PQ 1

1 2

-2 t

12.23

-t

-2 t

-t

This gives eAt =

LMa bOP = LM 2e N c d Q N2 e

-t

- e -2 t -2 t - 2e - t

e - t - e -2 t 2 e -2 t - e - t


OP Q


SOLUTION MANUAL

= L

(sI – A)

–1

=

(sI – A) =

A =

12.24

LM c cA V= M MM M NcA

–1

127

[(sI – A)–1]

LM 2 - 1 MM s +2 1 - s +2 2 Ns + 2 s +1

1 1 s +1 s + 2 2 1 s + 2 s +1

OP PP Q

LM s -1 OP N2 s + 3Q LM 0 1 OP N-2 -3Q

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

n-1

i

12.25 U = [b Ab ... An–1 b] is a triangular matrix with diagonal elements equal to unity; |U| = (– 1)n for all ai’s. This proves the result. 12.26 (i)

U = [b Ab] =

LM1 N0

OP Q

-2 ; r(U) = 2 1

Completely controllable V =

LM c OP = L 1 -1O ; r(V) = 1 NcAQ MN-3 3 PQ Not completely observable.

(ii) The system is in Jordan canonical form; it is controllable but not observable. (iii) The system is in controllable companion form. The given system is therefore controllable. We have to test for observability property only.

r(V) =

LM c OP r M cA P = 3 MNcA PQ 2

The system is completely observable.



128


(iv) Given system is in observable companion form. We have to test for controllability 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 s+2

Given state model is in observable companion form. Since there is a pole-zero cancellation, the state model is uncontrollable. (ii)

G(s) =

s+4 ( s + 2)( s + 3)

The given state model is in controllable companion form. Since there is 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 ( s + 1)(s + 2)

The G(s) is stable (c) The unstable mode et of the free response is hidden from the transfer function representation 12.30 (a)

G(s) =

A =

b2 10 = 2 s + a 1s + a 2 s2 + s

LM0 1 OP ; b = LM0OP ; c = [10 N0 -1Q N1Q

0]

(b) Let us obtain controllable companion form realization of G(s)=

10( s + 2) 10 s + 20 = 3 = s( s + 1)(s + 2) s + 3s 2 + 2 s

b 2s + b 3 s + a 1s 2 + a 2 s + a 3


3


SOLUTION MANUAL

A =

LM0 MM00 N

OP PP Q

LM OP MM PP NQ

1 0 0 0 1 ; b = 0 ; c = 20 10 0 -2 -3 1

(c) Let us obtain observable companion form realization of G(s) =

A =

10( s + 2) s( s + 1)( s + 2)

LM0 MM10 N

OP PP Q

LM MM N

OP PP Q

0 0 20 0 -2 ; b = 10 ; c = 0 0 1 1 -3 0


129

Solution Manual - Control Systems by Gopal

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