Solution Manual Digital Control and State Variable Methods

SOLUTION MANUAL TO DIGITAL CONTROL AND STATE VARIABLE METHODS

CHAPTER 1

OUTLINE OF THE SOLUTION MANUAL

This Solution Manual has been designed as a supplement to the textbook: Gopal, M, Digital Control and State Variable Methods, 2nd Edition, Tata McGraw-Hill, New Delhi, 2003. Throughout the manual, the numbers associated with referenced Equations, Figures, Tables, Examples, Review Examples and Sections are pointers to the material in the textbook. The detailed solutions of the problems have been avoided; only assistance has been provided.

CHAPTER 2

SIGNAL PROCESSING IN DIGITAL CONTROL

2.1 (a) x1, x2 : outputs of unit delayers, starting at the right and proceeding to the left. x1(k + 1) = x2(k); x2(k + 1) = – 0.368x1(k) + 1.368x2(k) + r(k) y(k) = 0.264x1(k) + 0.368x2(k)

LM 0 F= M MN−0.368 (b)

OP LM0OP PP ; g = MM PP ; c = [0.264 1.368Q N1 Q 1

∑ Pi D i 0 . 264 z-2 + 0 . 368 z-1 Y (z ) = = 1 2 R (z) 1− D 1 - 1. 368 z - - 0 . 368 z -

e

0.368]

j

2.2 (a) x1, x2 : outputs of unit delayers, starting at the right and proceeding to the left. x1(k + 1) = x2(k) + r (k) x2(k + 1) = – 5x2(k) – 3x1(k) – 3r (k) y(k) = x1(k) x1(k + 2) = x2(k + 1) + r (k + 1) Substituting for x2(k + 1) from the above set, x1(k + 2) = – 5x1(k + 1) – 3x1(k) + r(k + 1) + 2r (k) y(k + 2) + 5y(k + 1) + 3y(k) = r (k + 1) + 2r (k) (b) F =

(c)

LM 0 1 OP ; g = LM 1 OP ; c = [1 N-3 -5Q N-3Q

0]

Y (z) z+2 = 2 R( z) z + 5z + 3

2.3 (a) x1 : output of the unit delayer

1 x (k) + r(k); y(k) = 2x1(k) – x1(k + 1) = 2.5x1(k) – r(k) 2 1 y(k + 1) = 2.5x1(k + 1) – r(k + 1) = – 0.5y(k) + 2r(k) – r(k + 1)

x1(k + 1) = –

Y (z) -z + 2 = 1 R( z) z+ 2

SOLUTION MANUAL 5

(b) R (z) = 1; Y(z) =

y (k) = –

FH -1 IK 2

-z 2 + 1 1 z+ z+ 2 2 k

m (k) + 2

FH -1 IK 2

k -1

m (k – 1)

2 -5 Y (z) z 3 + 3 ; (c) R (z) = = 1 z -1 z z -1 z+ 2 y(k) = -

FH IK

5 -1 3 2

k

m (k) +

2.4 (a) y (k + 1) = ay (k) + b r (k); R(z) = A; Y(z) = (b) R (z) =

2 m (k) 3

Y (z) b = R( z) z -a

Ab ; y(k) = Ab (a) k–1 m (k – 1) z -a

A z Y (z) Ab ; = ; ( z - a) ( z -1) z -1 z

Ab Ab z z 1 -a Ab a -1 Y (z) = ; y(k) = [1 – (a) k] m (k) + z -a z -1 1 -a (c) R (z) =

Az Ab Y(z) ; = 2 ( z - 1) ( z - a) ( z - 1) 2 z

LM N

OP Q

(d) r(k) = A cos (Wk) = Re {AejWk}; Z {ejWk} =

z z - e jW

(1 - a) z Ab z + - z 2 (1 - a) ( z - 1) 2 z -a z -1 Ab [ (a ) k + (1 - a ) k - 1] ; k ³ 0 y (k) = (1 - a ) 2

Y (z) =

The output,

LM N

RS A b z T (z - a)b z - e L A b ba = Re M N a -e

y(k) = Re Z -1

jW

UVOP = Re L A b g WQ MN a - e O - e gP Q

jW

k

jWk

jW

ak +

Ab e jWk e jW - a

OP Q

6

DIGITAL CONTROL AND STATE VARIABLE METHODS

2.5 (a) Given difference equation in shifted form: y (k) + 3y (k – 1) + 2y (k – 2) = 0 Y (z) + 3z–1 Y (z) + 3y (– 1) + 2z – 2 Y (z) + 2z – 1 y (– 1) + 2y (– 2) = 0

z z z = z z 1 2 + + 3 2 z + z+ k k y (k) = (– 1) – (– 2) ; k ³ 0

Y (z) =

2

(b) 2Y (z) – 2z – 1 Y (z) + z – 2 Y (z) = Y (z) =

=

z3 ( z - 1) 2 z 2 - 2 z + 1

a

f

az

-z 2 + z z 1 z = + 2 z -1 2 z -2 z +1 z -1 2

Therefore, y (k) = (l)k – 2.6

1 1 - z-1

1 2

FH 1 IK 2

k

cos

az

2

2

-0 . 5 z

1

-z + 0 . 5

FH k p IK + 1 FH 1 IK 4 2 2

k

sin

f+2

0 . 5z z 2 -z + 0 . 5

FH k p IK ; k ³ 0 4

From the given difference equation we get, y (k + 1) – 1.3679 y (k) + 0.3679 y (k – 1) = 0.3679r (k) + 0.2642 r (k – 1) 0 . 3679 z + 0 . 2642 R (z) Y (z) = 2 z - 1. 3679 z + 0 . 3679 R (z) = 1 + 0.2142z –1 – 0.2142z –2 Hence, ( 0 . 3679 z + 0 . 2642 ) Y (z) = 2 1 + 0 . 2142 z -1 - 0 . 2142 z-2 z - 1. 3679 z + 0 . 3679 = 0.3679z – 1 + 0.8463z – 2 + z – 3 + z – 4 + z – 5 + L y(0) = 0; y (1) = 0.3679; y (2) = 0.8463; y (k) = 1, k ³ 3.

a

2.7

f

Taking z-transform of the given equation:

Y ( z) z2 z2 = 2 = R ( z) (z - 1) (z - 2 ) z - 3z + 2 z +1 R(z) = 1 + z – 1 = z ( z + 1) Y (z) 3 2 = = z - 2 z -1 ( z - 2 )( z - 1) z y(k) = 3(2) k – 2(1) k; k ³ 0

f

SOLUTION MANUAL 7

Final value theorem will not give correct value since a pole of Y (z) lies outside the unit circle. 2.8 (a) R (z) = z – 1; Y (z) =

2z - 3 z ( z - 0 . 5) ( z + 0 . 3)

Y (z) 2 z -3 = 2 ( z z z - 0 . 5) ( z + 0 . 3) 40 20 10 50 =+ 2 + z z - 0 .5 z + 0 .3 z y(k) = – 40d (k) + 20d (k – 1) – 10(0.5) k + 50(– 0.3) k; k ³ 0

(b) R (z) =

- 6z2 + z z ; Y (z) = 1 1 1 1 z -1 ( z - 1) z z- - j +j 2 4 2 4

FH

IK FH

IK

8+ j4 8- j4 - 16 Y (z) + + = 1 1 1 1 z -1 z z- + j z- -j 2 4 2 4 16 z - 6 16 + =– z-1 z2 - z + 5 16 y(k) = – 16(1)k + (0.56)k [7.94 sin (0.468k) + 16 cos (0.468k)]; k ³ 0 2.9 (a) R (z) = 1 + z – 2 + z – 4 + ¼ =

z2 z -1 2

Y (z) z = (z - 1) ( z + 1) ( z - 0 .5) ( z + 0 . 3) z 0 . 476 0 . 769 - 0 . 833 - 0 . 41 = + + + z - 0. 5 z + 0.3 z +1 z -1 k k y(k) = – 0.833(0.5) – 0.41(– 0.3) + 0.476(– 1)k + 0.769(1)k; k ³ 0 (b) R(z) =

z ; z -1

Y (z) 1 = 2 z ( z - 0.5) ( z - 0.1)( z - 1) - 6 . 94 2. 5 4 . 44 -5 + + + = ( z - 0 . 5) 2 z - 0. 5 z - 0 .1 z -1 k k k y(k) = – 10k(0.5) + 2.5(0.5) – 6.94(0.1) + 4.44(1)k; k ³ 0

2.10

y(¥) = lim (z – 1) Y (z) = z ®1

For y (¥) = 1, K = 1

a

f

K 1 - a 2 (1 - a )

a1 - a f

2 ( 1 - a)

=K

8


2.11

Refer Eqn (2.45) Y (z) =

( z - 1) z z = 0, if |a| < 1. m ; y (¥) = lim ( m z ®1 z - a) ( z - a)

1 1 1 z+ Y (z) z 1 ; = 2 = 2 - 22 2 (i) R(z) = z -1 z -1 z z + 1 ( z - 1) z +1

a

FH k p IK – 1 sin FH k p IK ; k ³ 0 2 2 2 kp z r(k) = {1, 0, – 1, 0, 1, 0, – 1, ¼} = cos F H IK ; R(z) = y(k) =

(ii)

f

1 1 (1)k – cos 2 2

z2 +1

2

Y (z) =

az

z2 2

f

+1

2

;

2

The output y (k) is bounded because of matching of system poles with excitation poles. 2.13 (a) D(z) = z3 – 1.3z2 – 0.08 z + 0.24 = 0 (i) D(1) = – 0.14 < 0 The first condition of Jury’s criterion is violated. The system is unstable; we may stop the test here. (b) D(z) = z4 – 1.368 z 3 + 0.4z 2 + 0.08z + 0.002 = 0 (i)

D(1) = 0.114 > 0; satisfied

(ii)

D(– 1) = 2.69 > 0; (n even); satisfied

Form Jury’s table; from which you will find that (conditions (2.50b)): |a 4| < |a 0|; 0.002 < 1; satisfied |b 3| > |b 0|; | – 1| > | – 0.083|; satisfied |c 2| > |c 0|; 0.993 > 0.512; satisfied. The system is stable. 2.14

D(z) = z 4 + 0.5 z 3 – 0.2 z 2 + z + 0.4 = 0

F 1 + r IJ D(r) = G H 1-r K

4

F 1 + r IJ + 0.5 G H 1 -r K

3

F 1 + r IJ – 0.2 G H 1-r K

2

+

1+ r + 0.4 1-r

- 0 . 3 r 4 + 3. 4 r 3 + 8 . 8 r 2 + 1. 4 r + 2 . 7 4 (1 - r ) Form the Routh table, from which you will find that one root of D(r) lies in the right-half plane. Therefore, three poles of G (z) lie inside the unit circle. =

SOLUTION MANUAL 9

2.15

Refer Section 2.8; Eqn (2.56)

2.16 (a) Refer Section 2.9 (b) s2 + 2s + 2 = s2 + 2zw ns + w 2n wn =

2;

ws p p = > 2; T < 2 T 2

(c) Noise signal: n(t) = cos 50t

{

50 k ´ 2 p 50

Z {cos 50kT} = Z cos =

} = Z {cos 2kp}

1 = 1 + z –1 + z –2 + L -1 1- z

2p sec be50 comes unity at every sampling instant. Thus the noise signal can generate an undesirable d.c. component in the output.

It can be seen that the noise signal cos 50t sampled at

2.17

Refer Section 2.10; Fig. 2.36.

2.18

Refer Section 2.12; Eqns (2.72)

2.19

(i) s = – a + jb; z = e –aT e jbT (ii) s = a ± jw0; z = eaT e± jw0T

2.20

Ga(s) =

1 - e-sT 1 ; Gh0(s) = s s+1

G(z) = Z{Gh0G(s)} = (1 – z –1) Z R(z) =

-T

(

)

–kT

;k³0

a

f

RS 10 UV ; G (z) = (1 – z ) Z RS 10 UV T s (s + 2 ) W T s (s + 2 ) W L T z - z + z OP G(z) = 5(1 – z ) M N z -1 2 z -1 2 a z - e f Q –1

Ga (s) =

2

–1

(

For T = 0.4 sec, G(z) = 2.22

-T

Y (z) 1 - e-T -1 z 1 ; = = + -T ( ) z -1 z z -1 z - e-T z -e z -1

y(k) = 1 – e 2.21

RS 1 UV = 1 - e T s s + 1 W z -e

(i) Refer Eqn (2.90b) (ii) Refer Eqn. (2.107)

)2

(

)

10 ( z + 0 . 76) 16 ( z - 1) ( z - 0 . 46 )

-2 T

10


2.23

Refer Section 2.12; Fig. 2.41a.

2.24

Refer Section 2.12; Figs 2.44 and 2.37.

2.25

u( k ) - u( k - 1) T = Kc

LM e k -e k -1 + 1 e k + T e k -2e k -1 + e k -2 OP T T Q N T LM1 + T FG 1 IJ + T a1 - z fOP E(z) Q N T H 1- z K T ( )

(

)

D { ( ) 2

( )

(

)

(

)}

I

U(z) = K c

I

2.26

Required z = 0.45, fM = 45 deg, T = 1.57 sec. (i)

D(s) is a phase lag compensator that meets the requirements.

(ii)

D(z) = D(s) with s =

LM N

U (s) = Kc 1 +

2 z -1 T z +1

z - 0 . 9391 z - 0 . 9752

D(z) = 0.4047

2.27

-1

D

-1

OP Q

1 + TD s E(s) TI s

LM 1 MM T s N LM1+ T N 2T

U(z) = Kc 1 +

I

= Kc

I

+ TD s s=

2 ( z -1) T z +1

s=

z -1 zT

OP Q

OP E (z) PP Q

z +1 TD z -1 + E (z) z -1 T z

u(k) = uP(k) + u1(k) + uD(k) where uP(k) = Kce(k) K T u1(k) – u1(k – 1) = c [e(k) + e(k – 1)] 2 TI

Therefore,

LM MN

a f

e i - 1 + e (i ) 2

u1(k) =

Kc T TI

uD(k) =

Kc TD e( k ) - e k - 1 T

u(k) = Kc e( k ) +

T TI

Â

a

a f

f

e i - 1 + e(i ) T + D T 2 i =1 k

Â

ke(k ) - eak - 1fpOPP Q

SOLUTION MANUAL

z

11

z

t

t

dy (t ) 2.28 (a) + ay(t) = r(t); y(t) = y(0) – a y (t) dt + r (t) dt dt 0 0

z

z

kT

y(k) = y(k – 1) – a

kT

y (t) dt +

( k -1) T

r (t) dt

( k -1 ) T

= y(k – 1) – aTy(k) + Tr(k) y(k) =

1 T y(k – 1) + r(k) 1 + aT 1 + aT (k

(b) y(k + 1) = y(k) – a

z

+ 1) T

(k

z

+ 1) T

y (t) dt +

kT

r (t) dt

kT

= (1 – aT) y(k) + Tr (k) y(k) = (1 – aT) y(k – 1) + Tr (k – 1) 2.29 (a)

d2 y dy +a + by = 0 dt dt 2 1 a [y(k) – 2y(k – 1) + y(k – 2)] + [y(k) – y(k – 1)] + by(k) = 0 T T2 1 a + 12 y(k) – a + 22 y(k – 1) + 2 y(k – 2) = 0; b+ T T T T T y(0) = a; y(– 1) = a – Tb

FH

IK

FH

IK

(b) x1 = y; x2 = y& ; x&1 = x2; x& 2 = – bx1 – ax2

LM 0 N- b

OP Q

1 -a Using Eqn (2.113) we obtain, A=

x(k + 1) = x(k) + T)x(k) = (I + AT) x(k); x(0) =

LMa OP Nb Q

CHAPTER 3 MODELS OF DIGITAL CONTROL DEVICES AND SYSTEMS 3.1 D(z) = {Gh0(s)G1(s)} = Gh0G1(z)

D( z )Gh 0 G2 ( z ) Gh 0 G1 ( z )Gh 0 G2 ( z ) Y ( z) = = 1+ D( z )Gh 0 G2 H ( z ) 1+ Gh 0 G1 ( z )Gh 0 G2 H ( z ) R ( z) Gh 0 G( z ) Y ( z) 3.2 = ( ) 1 R z + Gh 0 G( z ) H ( z ) 3.3 Y(z) = Gh0G2(z) U(z) U(z) = G1R(z) – Gh0G2HG1(z) U(z) Y(z) =

Gh 0 G2 ( z )G1 R( z ) 1+ Gh 0 G2 HG1 ( z )

3.4 Reduced form of the block diagram in Fig. P3.4:

Y(z) = GpH2 R(z) +

D ( z ) Gh 0 G p ( z ) 1 + D ( z ) Gh 0 G p ( z )

[H1R(z) – GpH2R(z)]

This is the required answer. 3.5 Y(z) = Gh0G1G2(z)U(z); X(z) = Gh0G1(z) U(z) E(z) = R(z) – X(z) – Y(z); U(z) = D(z) E(z) = D(z) [R(z) – Gh0G1(z) U(z) – Gh0G1G2(z) U(z)] U(z) =

D ( z ) R( z ) ; Y(z) = Gh0G1G2(z) U(z) 1+ D( z )Gh 0 G1 ( z ) + D( z )Gh 0 G1G2 ( z )

Gh 0 G1G2 ( z ) D ( z ) Y ( z) = 1 + D( z )[ Gh 0 G1 ( z ) + Gh 0 G1G2 ( z ) ] R ( z)

SOLUTION MANUAL

13

X(z) = Gh0G1(z) U(z)

Gh 0 G1 ( z ) X (z) = 1 + D( z )[ Gh 0 G1 ( z ) + Gh 0 G1G2 ( z ) ] R(z) 3.6 For r = 0, the block diagram reduces to the following:

Y(z) = WG(z) – D(z)Gh0G(z) Y(z); Y(z) =

WG ( z ) 1 + D ( z )Gh 0 G ( z )

Gh 0 G( z ) q ( z) 1 ; L = ; 1 + Gh 0 G( z ) s( s + 1) q R ( z ) z T -1+ e-T + 1- e-T - Te-T 1 = Gh0G(z) = (1 – z –1) Z 2 ( z -1) z - e- T s (s + 1)

3.7 G(s) =

UV a W

RS T

f a a

For T = 0.25 sec, Gh0G(z) =

0 . 0288( z + 0 . 92 ) ( z - 1) ( z - 0 . 7788)

3.8 Plant transfer function is, G(s) =

185 (0 . 025s + 1)

RS 185 UV = 185a1-e f T s (0 . 025s + 1) W z - e - 40 T

Gh0G(z) = (1 – z –1) Z

- 40 T

Let x(k) be the input to D/A block. x(k) = KF wr (k) + KP [wr (k) – w (k)] X(z) = KF wr (z) + KP[wr(z) – w(z)]; w(z) = Gh0G(z) X(z)

(K F + K P )Gh 0 G( z ) w( z ) = 1 + K P Gh 0 G ( z ) wr ( z )

f

f

14


3.9 The filter is described by the following difference equation, u(k) = u(k – 1) + 0.5e(k);

Gh0G(z) = (1 – z –1) Z fs = 5 Hz = Gh0G(z) =

U (z) z = 0.5 z -1 E(z )

RS 1 UV = aT -1+ e fz +a1-e -Te f z -1 a z - e f T s (s + 1) W -T

2

(

-T

-T

-T

)

1 ; T = 0.2 sec T

0 . 019 z + 0 . 0175 ; ( z - 1) ( z - 0 . 819 )

Gh 0 G( z ) D( z ) 0 . 0095z ( z + 0 . 92 ) Y (z) = = 3 1 + Gh 0 G( z ) D( z ) R( z ) z - 2 . 81z 2 + 2 . 65z - 0 . 819 3.10

(i) Gh0G(z) = (1 – z–1) Z

RS e UV T s (s + 1) W -0 . 4 s

Using transform pairs of Table 3.1, we obtain,

LM a1-e fz + e -e OP (1 – z N z - 1 az - e f Q -0 . 6

Gh0G(z) =

(

-0 . 6

-1

-1

)

–1

)=

0 . 45z + 0 .181 z ( z - 0 . 368)

Gh 0 G( z ) 0 . 45 z + 0 .181 Y (z) = = 2 1 + Gh 0 G( z ) R( z ) z + 0 . 081z + 0 .181 (ii) Gh0G(z) =

0 . 45z + 0 .181 Y ( z ) 0.45 z + 0.181 ; = 3 2( z z - 0 . 368) R( z ) z - 0.368z 2 + 0.45 z + 0.181

3.11 Refer Section 3.6. 3.12 Refer Gopal M., Digital Control Engineering, New Delhi, Wiley Eastern, 1988. 3.13 (a) 1 + G(s) = 0; s3 + 3s2 + 2s + 5 = 0 From Routh table, we find that the closed-loop system is stable. (b) Gh0G(z) = (1 – z–1) Z

RS UV 5 T s (s + 1) (s + 2) W 2

( z - 1) 2.5 5( z - 1) – 3.75 + – 1.25 z -1 z - 0 . 3679 z - 0 .1353 The characteristic equation is: z2 + 2.12z + 0.234 = 0; z1, 2 = – 2, – 0.12

=

A pole lies outside the unit circle; the system is unstable.

SOLUTION MANUAL

3.14 z3 – 0.1z2 + 0.2Kz – 0.1K = D(z) D(1) = 0.1K + 0.9 > 0 D(– 1) = – 1.1 – 0.3 K < 0 From Jury’s table, we get the following conditions. (refer conditions (2.50b)) |– 0.1K| < 1; True for 0 < K < 10 |0.01K2 – 1| > |– 0.19K|; |0.01K2 – 1| > |0.19K| K2 + 19K – 100 = 0 The system is stable for 0 < K < 4.293. 3.15 Gh0G(z) = K(1 – z–1) Z

RS 1 UV = K FG 1 - e IJ T s (s + 3 ) W 3 H z - e K -3T

-3T

K (1 – e –3T) = 0 3

D(z) = 1 + Gh0G(z) = z – e –3T +

(i) For T = 0.5, system stable for 0 < K < 4.723 (ii) For T = 1, system stable for 0 < K < 3.315

RS K UV T s a s + 1f W K z - 1 L zkaT - 1 + e f z + a1 - e - Te fp O = MN PQ z z - 1 az - e f

3.16 Gh0G(z) = (1 – z–1) Z (

2

-T

)

-T

)2

(

-T

-T

D(z) = 1 + Gh0G(z) = 0; z2 – az + b = 0 where,

a = e –T + 1 – K(T – 1 + e –T); b = e –T + K(1 – e –T – Te –T) 1+ r Put z= , then r2 (1 + a + b) + 2r(1 – b) + 1 – a + b = 0 1- r System is stable for, 1 + a + b > 0; 1 – b > 0; 1 – a + b > 0

Substituting for a and b and solving for K yields: K<

a

2 1 + e- T T - 2 + 2e

-T

f

+ Te-T

1 - e- T 1 - e-T - Te-T T(1 – e –T) > 0 or e–T < 1 which implies T > 0. K<

3.17 For T = 1 sec, Gh0G(z) = K

a

(0 . 368 z + 0 . 264 )

f

z z 2 - 1. 368z + 0 . 368

15

16


D(z) = z3 – 1.368z2 + 0.368(1 + K)z + 0.264K = 0 D(1) = – 0.368 + 0.368(1 + K) + 0.264K > 0 D(– 1) = – 2.368 – 0.368(1 + K) + 0.264K < 0 From Jury’s table, we get the following conditions (refer conditions (2.50b)): |0.264K| < 1, gives K < 3.79 |0.07K2 – 1| < |0.361K + 0.368 + 0.368K| This gives, K < 0.785. The system is stable for 0 < K < 0.785. 3.18 (a) G(s) =

4500 K ; K = 14.5; z = 0.707; wn = 255.44 s (s + 361. 2 )

y(t) = 1 –

e-z wn t 1- z2

sin

F dw GH

n

i

1 - z 2 t + tan -1

1- z 2 z

I JK

(b) T = 0.01 sec, Gh0G(z) =

1. 3198z + 0 . 4379 Y ( z ) 1.3198z + 0.4379 ; = 2 2 z -1.027 z + 0.027 R( z ) z + 0.2929 z + 0.4649

By dividing the numerator polynomial by the denominator polynomial, we obtain, y(T) = 1.3198, y(2T) = 1.3712, y(3T) = 0.7426, y(4T) = 0.9028, y(5T) = 1.148 T = 0.001 sec, Gh0G(z) =

3.19

0.029 z + 0.0257 0.029 z + 0.0257 Y(z) ; = 2 z 2 - 1.697 z + 0.697 R( z ) z - 1.668 z + 0.7226

Dividing the numerator polynomial by the denominator polynomial, we can obtain the response. Gh 0 G( z ) Y (z) = 1 + Gh 0 GH ( z ) R( z )

RS 1 UV = 1 - e ; T sa s + 1f W z - e R 1 UV = e z - 2e + 1 )Z S T s a s + 1f W z - 1 az - e f -1

Gh0G(z) = (1 – z –1) Z Gh0GH(z) = (1 – z –1

a

f

-1

-1

2

(

-1

)

-1

1- e-1 ( z -1) Y (z) z 0.632 z = 2 ; R(z) = ; Y(z) = 2 z -1 R( z ) z - z + 0.632 z - z +1- e-1

SOLUTION MANUAL

a k sin (Wk) «

(a

z

2

sin W) z ; a2 = 0.632; 2 ) cos W z + a

1 = 0.629; W = 0.89 rad; y(k) = 1.02(0.795)k sin (0.89k) 2a

cos W =

3.20 Gh0G(z) = (1 – z–1) Z

R(z) =

- (2a

17

RS 1 UV = 0.632 ; Y z T sa s + 1f W z -0.368 R z

( ) ( )

=

0.632 ; z + 0.264

z z -1

Y(z) =

0.632 z ; y(k) = 0.5 [1 – (– 0.264)k] m(k) ( z - 1) ( z + 0.264 )

y(0) = 0; y(1) = 0.632; y(2) = 0.465; y(3) = 0.509 ¼

Y (z) = Z {Gh0(s) G(s)e –DTs)} E(z); E(z) = R(z) – Y(z); Y(z) = Z {Gh0(s) G(s)} E(z) E(z) =

R( z ) 1 + Z {Gh 0 ( s ) G( s )}

Z [ Gh 0 ( s ) G( s ) e- DTs ] 0 . 632 Y (z) = R(z); Z {Gh0(s) G(s)} = z - 0 . 368 1 + Z [ Gh 0 ( s ) G( s )] Z [Gh0(s) G(s) e –DTs] = R(z) =

0 . 393z + 0 . 239 Y ( z ) 0.393z + 0.239 ; = ; z ( z - 0 . 368) R( z ) z ( z + 0 . 264 )

z z -1

y (k) = [0.5 – 0.107 (– 0.264)k –1] m(k – 1) y (1) = 0.393 = y(0.5T); y (2) = 0.528 = y(1.5T)

18


y (3) = 0.493 = y(2.5T); y (4) = 0.5019 = y(3.5T)

FG z -1 IJ FG z +1.2 z +1 IJ H z + 0.1 K H z - 0.3z + 0.8 K L 1- z OP LM 1+1.2 z + z OP = 4M N 1+ 0.1z Q N 1- 0.3z + 0.8z Q 2

3.21

(i) D(z) = 4

2

-1

-1

-1

-1

-2

-2

Refer Figs 3.19 – 3.20 (ii)

46 . 62 7 . 38 z + 4 . 05 D( z ) 50 =+ + 2 z z + 0 .1 z z - 0 . 3z + 0 . 8 D(z) = –50 +

46 . 62 7 . 38 + 4 . 05 z-1 + 1 + 0 .1z-1 1 - 0 . 3z -1 + 0 . 8 z- 2

Refer Figs 3.21 – 3.22 10 z 2 + z +1 233.33 127.08 25 106.25 3.22 D(z) = 2 = + + 2 + z - 0.5 z - 0.8 z z z (z - 0.5) (z - 0.8) 233.33z -1 127.08z -1 =+ + (106.25 + 25z –1)z –1 1- 0.5z -1 1- 0.8 z-1

a

f

SOLUTION MANUAL

3.23 (a) Process steady-state gain, K =

qss Qm

q(t) = 0.283qss at t1 = 25 min q(t) = 0.632qss at t2 = 65 min Therefore (refer Eqns (3.53b)) tD +

1 t = t1 = 25 3

tD + t = t2 = 65 This gives tD = 5 min, t = 60 min

1 T = 5.5 2 Kc = 1.5t/KtCD = 0.545

(b) tCD = tD +

TI = 2.5tCD = 13.75 min TD = 0.4tCD = 2.2 min 3.24 Ultimate gain, Kcu = 5 Ultimate period, Tu = 34 sec Kc = 0.45 Kcu = 2.13 TI = Tu /1.2 = 666.66 sec

=

30 = 30°C/(kg/min) 1

19

20


3.25 TIM001

TIM000 # 0030

TIM000

TIM001 # 0060

TIM000

10000

10000

10001

10001

TIM002 # 0015

TIM002

TIM003 # 0030

TIM002 TIM003

10002

10002

10003

I/O Assignment: 10000–N(Green); 10001–N (Red) 10002–S(Green); 10003–S (Red) TIM000–Timer for 30 sec delay TIM001–Timer for 60 sec delay TIM002–Timer for 15 sec delay TIM003–Timer for 30 sec delay

SOLUTION MANUAL

3.26

00000 10003 10000

10000 00002

10000

00000

10002

00002 00003

10001

10001 10001

CNT000 # 0005

00001 CNT000

10003

I/O Assignment: 00000–Start Push button (PB1) 00001–Stop Push button (PB2) 00002–Upper level sensor: 1 if liquid level above LL1; otherwise 0 00003–Lower level sensor: 1 if liquid level above LL2; otherwise 0 10000–Liquid supply valve (V1) 10001–Drain valve (V2) 10002–Stirring motor (M) 10003–Buzzer

21

22


CNT000–Counter with a Set Value of 5. 3.27 00000 10000

CNT000 10000

00001

00002

10000

CNT000 # 0005

10001 CNT000

10001

TIM001

10001

TIM001 # 0002

I/O Assignment: 00000–Start push button 00001–Stop push button 00002–Product proximity sensor 10000–Conveyor motor 10001–Solenoid CNT000–Counter with a Set Value of 5 TIM001–Timer for 2 sec delay

SOLUTION MANUAL

3.28 00002 TIM000

10001

TIM001

10000

00001

TIM000

10001 00001

10000 10000

# 0020 00000 TIM001

01000

00000

TIM001

01000

01000

# 0020

I/O Assignment: 00000–S1; 00001–S2 00002–S3; 10000–M1 10001–M2; 01000–Work-bit TIM000–Timer for 20 sec delay TIM001–Timer for 20 sec delay

23

24


3.29 00000 00001 10000

10001 00003

10000 00002

00003

TIM000 # 0007

TIM000 00001

10001

10000 00002 10001

I/O Assignment: 00000–PB1; 00001–PB2 00002–LS1; 00003–LS2 10000–Forward motor 10001–Reverse motor TIM000–Timer for 7 sec delay

CHAPTER 4 DESIGN OF DIGITAL CONTROL ALGORITHMS 4.1 Steady-state error can be calculated from the corresponding continuoustime system as sampling does not affect steady-state performance of a continuous-time system. R(s) +

E(s)

K1

+

1 Js

1 s

q(s)

K2

G(s) =

K1 q( s ) = s( Js + K 2 ) E ( s)

Kp = lim G(s) = • sÆ0

Kv = lim sG(s) = sÆ0

K1 K2

Ka = lim s2G(s) = 0 sÆ0

4.2 Plant model: -j

G(s) =

wT 2

1.57 1.57e ; G ( jw) G( jw) = ; T = 1.57 sec s( s + 1) h0 jw ( jw + 1)

Bode plot analysis: Phase margin (without ZOH) = 45 deg Phase margin (with ZOH) = 0 deg Specification: f M = 45 deg Let us use a lag compensator. From the Bode plot of the uncompensated system, we see that the phase margin of 45 deg may be realized if the gain cross over frequency is moved from the present value (1 rad/sec) to a frequency of 0.5 rad/sec. To accommodate phase lag, we take wc2 = 0.4 rad/sec 20 log b = 8; b = 2.5

w c2 1 1 + st 1 1 + 25s = = 0.04; = 0.016; D1(s) = = 10 t bt 1 + sbt 1 + 62.5s Phase margin of the compensated system is about 45 deg.

26 DIGITAL CONTROL AND STATE VARIABLE METHODS

2 z -1 z -1 = 1.274 ; T z +1 z +1 0.4( z - 0.939) D1(z) = ( z - 0.975)

Bilinear transformation: s =

Kv of the original analog system is given by, Kv = lim s sÆ0

1.57 = 1.57 s ( s + 1)

Kv of the equivalent digital system is: 1 lim (z – 1) D1(z) Gh0G(z) Kv = T z ®1 where Gh0G(z) = 1.57

LM 1.57 - 0.792 OP , K = 1.57 N z - 1 z - 0.208 Q v

4.3 Kp = lim Gh0G(z) = •; Kv = zÆ1

Ka =

1 lim (z – 1) Gh0G(z) = 3.041 T z Æ1

1 lim (z – 1)2 Gh0G(z) = 0 T 2 zÆ1

ess (unit step) = 0; ess (unit ramp) = 0.33; ess (unit acceleration) = • 4.4 z2 – 1.9z + 0.9307 = 0; z = 0.95 ± j0.168 = 0.965e± j0.175 = re± jq

e -zw n T = 0.965 = r ; zwnT = – ln r; wnT 1 - z 2 = q From the above equations, we get

z 1-z wn =

2

=

-In r ;z= q

- ln r ln 2 r + q 2

1 In 2 r + q 2 ; z = 0.199; wn = 8.93 T

4.5 (a) (i) W(s) = 0; Y1(z) = Gh0G(z) [R(z)D2(z) + {R(z)D3(z) – Y1(z)}D1(z)] Y1(z) =

[ D2 ( z ) + D1 ( z ) D3 ( z )]Gh 0 G( z ) R( z ) 1 + D1 ( z )Gh 0 G( z )

(ii) R(s) = 0 Y2(z) = GW(z) – Gh0G(z) D1(z) Y2(z) Y2(z) =

GW ( z ) ; Y(z) = Y1(z) + Y2(z) 1 + D1 ( z )Gh 0 G( z)

(b) D3(z) = D2(z) GD0G(z)

SOLUTION MANUAL

Y1(z) =

[1 + D1 ( z )Gh 0 G( z )] D3 ( z ) R( z ) = D3(z) R(z) 1 + D1 ( z )Gh 0 G( z )

Y2(z) =

GW ( z ) ; Y(z) = Y1(z) + Y2(z) 1 + D1 ( z )Gh 0 G( z)

27

(c) D1(z) can be made large to reject the disturbances. 4.6 Consider the corresponding continuous-time system.

(i) D(s) = K1 + Kv = K2;

sK1 + K 2 K2 ; D(s) G(s) = s( s + 1) s

1 1 = 0.01 = K2 Kv

(ii)

Y (s) s 1 = ; Y(s) = ; yss = 0 s( s + 1) + K1s + K2 s( s + 1) + K1s + K2 W (s) Thus a PI compensator meets the requirements. -

T

K K (1 - e t ) 4.7 G(s) = ; Gh0G(z) = ; T J s +1 z-e t For K = t = 1, Gh0G(z) = S(e jwT) =

1 0.393 z - 0.607 ; S(z) = = z - 0.607 z - 0.214 1 + Gh 0 G( z )

cos(wT ) - 0.607 + j sin(wT ) ; |S(e jwT)|2 cos(wT ) - 0.214 + j sin(wT ) =

ws 2p = = 6.28 rad/sec 2 2T

1.3685 - 1.214 cos(wT ) 1.0458 - 0.428 cos(wT )


w

|S|2

|S|

0

0.25

0.5

1

0.45

0.67

2

0.8752

0.9355

2.5

1.0827

1.04

3.14

1.3086

1.144

4

1.53

1.23

6.28

1.753

1.324

| S | < 1 for 0 £ w £ 2. 4.8 (a) D(z, K) = A(z) + KB(z) = (z – l1) (z – l 2) L (z – lk) We consider the effect of parameter K on the root lk. By definition, D(lk, K) = 0. If K is changed to (K + DK), then lk also changes and the new polynomial is, D(lk + Dlk, K + DK) = D(lk, K) + +

∂D ∂K

∂D ∂z

z=lk

Dlk

DK + L = 0 z= lk

Neglecting higher order terms, we obtain, Dlk = –

∂D ∂K

LM ∂ D / ∂K OP N ∂D / ∂z Q

l =lk

SKl k =

DK z = lk

= B(lk) = –

∂D 1 A( l k ); ∂z K

z = lk

= P (l k - l i ) iπk

A( l k ) Dl k = P (l k - l i ) DK / K iπk

(b) G(s) =

K 0.393K ; Gh0G(z) = ; D(z, K) = 1 + Gh0G(z) = 0 s +1 z - 0.607

A(z) + KB(z) = (z – 0.607) + K(0.393) = 0 Nominal value of K = 1. Closed-loop pole is at z= 0.213 = l A(l) = 0.213 – 0.607 = – 0.394;

dD ( z ) dz

= 1; SKl = – 0.394 z=l

SOLUTION MANUAL

29

For a 10% change in nominal gain, the change in the root location is given by, D(l k) = SKl

∂K = – 0.0393 K -

1

1 1 - e 2t (c) G(s) = ; Gh0G(z) = ; D(z, t ) = 1 + Gh0G(z) = 0; 1 t s +1 t 2 z-e z + 1 – 2e 1 2t

-

1 2t

=0

1

∂p 1 - 2t = e ∂t 2t 2 Characteristic equation is, z + 1 + p(– 2) = 0; A(z) + p(B(z)) = 0

Let p = e

-

;

-

1

Nominal value of p = e 2 = 0.6065 Closed-loop pole is at z = 0.213 = l A(l) = 1.213;

dD( z ) dz

z= p

= 1; Spl = 1.213 =

∂( l ) ∂p / p

1

∂p 1 - 2 t ∂t 1 ∂t ∂t e Now = = = 0.5 ; Stl = 0.6065 2 2t t t p p 2t For 10% change in the nominal value of t, the change in root location is given by, D(l k ) = Stl 4.9

∂t = 0.06065 t

Gh0G(z) =

0.368 z + 0.264 1 + 0.5w ; T = 1 sec; z = 1 - 0.5w z 2 - 1.368z + 0.368

Gh0G(w) =

-0.0381( w - 2 )( w + 12.14 ) w( w + 0.924)

Gh0

FH1 - j g IK FH1 + j g IK 2 12.14 G( jg) = g I F jg 1 + j H 0.924 K

From the Bode plot and Nichols chart, we obtain, fM = 28º; GM = 8 dB; wb = 1.35 rad/sec; gb = 1.6 rad/sec 1 ; T = 0.1 sec 4.10 G(s) = s( s + 2 ) Gh0G(z) = 0.004683

( z + 0.9355) 1 + 0.05w ;z= 1 - 0.05w ( z - 1)( z - 0.8187)


Gh0G(w) = 0.5

(1 + 0.001666 w )(1 - 0.05w ) w (1 + 0.5016 w )

(a) lim wGh0G(w) = Kv = 0.5K; K = 10 wÆ0

Gh0G(w) =

5(1 + 0.001666 w )(1 - 0.05w ) w (1 + 0.5016 w )

Bode plot analysis: f M = 30º (crossover at 2.6 rad/sec)

w 1 .994 (b) Phase lead design: D(w) = w 1+ 12.5 (obtained by standard design procedure), 1+

fM = 55º, Kv = 5, GM = 12.4 dB (c) D(w) =

K (1 + t w ) ; Kv = 5 (1 + bt w )

Kv = lim wD(w) Gh0G(w), gives K = 10. wÆ0

5(1 + 0.001666 w )(1 - 0.05w ) w (1 + 0.5016 w ) (uncompensated system; K = 10 has been used in the plant model) gives f M = 30º The Bode plot of G(w) =

Required phase margin = 55º + 15º (error compensation). Crossover frequency wc2 = 0.7 D(w) =

w 1+ t w 1 ; = c22 = 0.18; t = 5.71 1 + bt w t ( 2)

Corresponding gain = 20 log10 b = 17; b = 7.0; D(w) = w=

0.14( w + 0.18) ( w + 0.02 )

0.141( z - 0.98) 2 z -1 gives, D(z) = T z +1 ( z - 0.998)

(d) From Nichols chart we find that bandwidth values gb for three designs corresponding to parts (a), (b) and (c) are respectively 4.8 rad/sec, 9.8 rad/sec and 1.04 rad/sec. (e) Reasonable sampling rates are 10 to 30 times the bandwidth. ws =

2p = 62.83. T

SOLUTION MANUAL

4.11 G(s) =

31

0.005( z + 1) 1 , T = 0.1, Gh0G(z) = s2 ( z - 1) 2

1 - j 0.05g 1 + 0.05w 1 - 0.05w ; Gh0G(w) = ; Gh0G( jg) = 2 1 - 0.05w w ( jg ) 2 Bode plot analysis: f M = – 2 deg. In the low frequency range, – Gh0G( jg) is about –180 º. Therefore, a lag compensator cannot fulfil the requirement of 50º phase margin. ( w + 1) satisfies the requirements. The lead compensator D(w) = 64 ( w + 16) The gain crossover frequency = 4 f M = 50.62º, GM = 13 dB, Kv = •, Ka = 4. z=

4.12 (a) K = 50 Gh0G(z) =

0.0043K ( z + 0.85) ; ( z - 1)( z - 0.61)

FH

IK FH1 + w IK 246.67 w I F w 1+ H 4.84 K

10 1 Gh0G(w) =

w 20

Bode plot/Nichols chart analysis: Gain crossover frequency = 6.6 rad/sec. f M = 20º, gb = 10 rad/sec, wb = 9.27 rad/sec. 0.219w + 1 results in gb = 18 rad/sec, (b) Lead compensator D(w) = 0.06 w + 1 wb = 14.65 rad/sec. 1.6 w + 1 results in gb = 5 rad/sec, Lag compensator D(w) = 6.4 w + 1 wb = 4.9 rad/sec. (c) For partial compensation, we design lag section by selecting a cross over frequency of 3.2 rad/sec. The uncompensated plot has to be brought down by 9 dB. 20 log b = 9; b @ 3 1 3.2 = 2 = 0.8 t 2 1.25w + 1 results in f M = 54º, gb = 4.3. D1(w) = 3.75w + 1 1 = 0.333. The We constrain the lead section design by taking a = b lag compensated system has a dB of


1 = – 4.77 at 4 rad/sec. a 1 1 = 4 a = 2.3; = 6.92 at t

– 10 log

0.4347 w + 1 0.144 w + 1 Lag-lead compensator results in f M = 60º, g b = 8, wb = 7.61 rad/sec. D2(w) =

(d) D1(z) = 0.342

FH z - 0.923 IK ; D (z) = 2.49 FH z - 0.793 IK z - 0 .973 z - 0.484 2

Refer Figs 3.19–3.22 for realization schemes. 4.13 (a)

1 49 = 0.02; K = 61.25; G(s) = 1 + 0.8 K 3s + 1 7.5215 (b) Gh0G(z) = z - 0.8465 Closed-loop pole: z + 6.675 = 0 The system is unstable. 49(1 - w / 4) (c) Gh0G(w) = (1 + 3w ) Kp = 0.8K;

Lag compensator D(w) = w=4

1+ w satisfies the requirements. 1 + 10 w

FH z - 1 IK ; D(z) = 0.122 FH z - 0.6 IK z +1 z - 0.951

(d) Lead compensation will increase the gain. Since gain is to be reduced to stabilize the system, lead compensation cannot be employed. 4.14 (a) Refer Example 4.5.

K ( z - 0.9048) ( z - 1) 2 Sketch a root locus plot; complex roots lie on a circle. Using magnitude condition, we obtain the value of K at the closed-loop pole z = – 1. It is 2.1. Therefore, the system is stable for 0 < K < 2.1. There is a double pole at z = 0.81. The value of K at this point is obtained as 0.38.

(b) Gh0G(z) =

4.15 Gh0G(z) =

K [( T - 1 + e - T ) z -1 + (1 - e - T - Te - T ) z -2 ] (1 - z -1 )(1 - e - T z -1 )

SOLUTION MANUAL

33

(i) T = 1 sec, Gh0G(z) =

0.3679 K ( z + 0.7181) ( z - 1)( z - 0.3679)

The root locus is a circle; the breakaway points are at z = 0.6479 and z = – 2.0841. At the point of intersection of the root locus with unit circle, we find by magnitude condition, K = 2.3925. This gain results in marginal stability. (ii) T = 2 sec, Gh0G(z) =

11353 . K ( z + 0.5232) ( z - 1)( z - 0.1353)

Breakaway points are at z = 0.4783, – 1.5247; Critical gain K = 1.4557. (iii) T = 4 sec, Gh0G(z) =

3.0183K ( z + 0.3010 ) ( z - 1)( z - 0.0183)

Breakaway points are at z = 0.3435, – 0.9455; Critical gain K = 0.9653. The smaller the sampling period, the larger the critical gain for stability.

K ( z + 0.717) ( z - 1)( z - 0.368) The root locus is a circle with breakaway point at z = 0.648, – 2.08.

4.16 Gh0G(z) =

(a) At the point of intersection with the unit circle, K = 0.88 and z1,2 = 0.244 ± j0.97 = 1 – 1.33 rad = e jwT; w = 1.33 rad/sec. (b) The value of K at this point is 0.072. e–T/t = 0.648, t = 2.3 sec. (c) At the point of intersection of the root locus with z = 0.5 locus, we get K = 0.18. The line through intersection point makes an angle of 32º with real axis. Therefore wnT 1 - z 2 = 32º = 0.558 rad; wn = 0.644 rad/sec 4.17 z2 + 0.2Az – 0.1A = 0; 1 +

0 .2 A( z - 0.5) K ( z - 0.5) = 0; Gh0G(z) = ; 2 z z2

= K = 0.2 A The root locus is a circle. At the point of intersection with unit circle, we find by magnitude criterion, K = 0.666. Therefore, A = 3.33.


1 ; T = 0.1 s +1 PT 0.091K = 0.9554; Gh0GH(z) = H(s) = 2p z - 0.9048 The root locus is a circle.

4.18 G(s) =

(a) 1 + Gh0GH(z) = 0; z – 0.9048 + 0.091K = 0 For K = 1, z – 0.8138 = 0 e–T/t = 0.8138;

T = 0.206; t = 0.4854 sec t

(b) Required time-constant =

0.4854 = 0.12136 4

e–T/0.12136 = 0.43867; 0.091K = 0.46613; K = 5.1223 4.19 Gh0G(z) =

0.01873 K ( z + 0.9356) ( z - 1)( z - 0.8187)

(a) z = 0.5; wn = 4.5; z = e -zw nT – wnT(1 – z 2) This gives z1,2 = 0.6354 – ± 45º = 0.4493 ± j0.4493. Angle deficiency at point P corresponding to z1 is – 72.25 deg. We choose the zero of the controller to cancel the pole at z = 0.8187. Then the pole of the controller is determined to satisfy the angle condition at P. This gives D(z) =

z - 0.8187 ; |D(z)Gh0G(z)| = 1, requires K = 13.934 z - 0.1595

D1(z) =

13.934( z - 0.8187) ( z - 0.1595)

(b) Kv =

1 lim (z – 1) D(z) Gh0G(z) = 3. T zÆ1

(c) D2(z) =

z - b1 1 - b1 ; =3 z - b2 1- b2

z - 0.94 z - 0.98 From the root locus plots of 1 + D1(z) Gh0G(z) = 0 and 1 + D2(z) Gh0G(z) = 0, it is seen that lag compensator decreases the margin of stability. (d) Refer Figs 3.19 – 3.22 for realization of D2(z) D1(z). Let, b2 = 0.98, then b1 = 0.94; D2(z) =

SOLUTION MANUAL

4.20 Gh0G(z) =

35

0.2 K ( z + 0.368) ( z - 0.368)( z - 0.315)

The root locus of the uncompensated system is a circle. At point P corresponding to the intersection of root locus with z = 0.5 locus, we get 0.2 K = 0.3823 or K = 1.91. At this point, wn = 1.65, Kp = 0.957. We will use lag compensator. Kp is to be increased by a factor of

z - b1 1 - b 1 7.5 = 7.837; D(z) = ; = 7.837 0.957 z - b2 1- b2 Let, b2 = 0.98, then b1 = 0.84 wn is slightly decreased with lag compensator (as seen from root-locus plot of the lag-compensated system), which is acceptable. 4.21 Gh0G(z) = K

T2 z +1 ; T = 1 sec 2 ( z - 1) 2

z = 0.7, wn = 0.3, z = e -zw nT – wnT(1 – z 2) This gives z1,2 = 0.78 ± j0.18 Place a compensator zero at 0.8. Angle criterion gives compensator pole location as z = 0. Magnitude criterion gives K/2 = 0.18. Gh0G(z) D(z) = Ka =

0.36 z + 1 z - 0.8 ; 2 ( z -1) 2 z

1 lim (z – 1)2 Gh0G(z) D(z) = 0.072 T 2 z®1

Corresponding to K/2 = 0.18, we find from the root locus that third pole is located at z = 0.2. It slows down the response. 4.22 G(s) =

0.00133( z + 0.75) 40e-s /120 1 ;T= sec; Gh0G(z) = + s ( s 40 ) 120 z ( z -1)( z - 0.72 )

The closed-loop poles are required to lie inside the circle of radius 0.56. Let us try a lead compensator. Cancel the pole at z = 0.72 by zero of D(z). By angle criterion, at a point on circle of radius 0.56, the pole of D(z) is found at z = – 0.4. The magnitude criterion at the point of intersection of the root locus with the circle of radius 0.56 gives K = 0.2. Therefore, D(z) =

0.2( z - 0.72 ) ( z - 0.72 ) = 150 0.00133( z + 0.4) ( z + 0.4)


The third pole corresponding to K = 0.2 lies inside the circle of radius 0.56.

0.2( z + 0.75) z( z -1)( z + 0.4)

Gh0G(z) D(z) =

Kp = lim Gh0G(z) D(z) = •; e*ss = 0 z®1

4.23 (a) Refer Examples 4.9–4.10. (b) G(s) =

1 0.0048( z + 0.9833) ; T = 0.1 sec; Gh0G(z) = s( s + 1) ( z -1)( z - 0.9048)

We select an M(z) that has pole excess of at least equal to that of Gh0G(z), and unstable (or critically stable) poles of Gh0G(z) are included in 1–M(z) as zeros. M(z) = z–1 satisfies these requirements. D(z) = 4.24 G(s) =

LM N

M( z) 1 Gh 0 G( z ) 1 - M ( z )

OP = 208.33 L z - 0.9048 O NM z + 0.9833 QP Q

1 ; T = 0.1 sec s( s + 1)

(a) z = 0.8, wn =

2p -zw T ; z = e n – wnT 1 - z 2 10 T

This gives z1,2 = 0.55 ± j0.22 Gh0G(z) =

( T - 1 + e - T )( z + a) 1 - e - T - Te - T ; a = ( z - 1)( z - e - T ) T - 1 + e-T

= 4.8 × 10–3

( z + 0.9833) ( z - 1)( z - 0.9048)

D(z) = z2 – 1.1z + 0.3509 We select an M(z) that has pole excess of at least equal to that of Gh0G(z), unstable poles of Gh0G(z) are included in 1 – M(z) as zeros, and satisfies transient accuracy requirements. 1 – M(z) = (z – 1) F(z); F(z) =

z -a z - 1.1z + 0 . 3509 2

E(z) = R(z) [1 – M(z)] e*ss(unit-step) = lim (z – 1) E(z) = 0 z ®1

e*ss(unit-ramp) = lim (z – 1) z ®1

Tz 1 2 (z – 1) F(z) = TF(1) = ( z - 1) Kv

SOLUTION MANUAL

37

Kv = 5, T = 0.1: F(1) = 2

1-= 1 - 1.1 + 0 . 3509 = = 0.7491 satisfies steady-state accuracy requirements.

F(1) =

M(z) = D(z) =

( 0 . 6491z - 0 . 3982 )

z 2 - 1.1z + 0 . 3509

LM N

M( z) 1 Gh 0 G( z ) 1 - M ( z )

= 135.227

OP Q

( z - 0 .9048) ( z - 0.6135) ( z + 0.9833) ( z - 0.7491)

(b) y(k) = 0, 0.5, 1, 1, K Y(z) = 0.5z –1 + z –2 + z –3 + L

Y ( z) = (0.5z –1 + z –2 + z –3 + L) (1 – z –1) R ( z) M(z) = (0.5z–1 + 0.5z–2); Gh0G(z) = 4.8 × 10–3

( z + 0.9833) ( z - 1) ( z - 0.9048)

( z - 0.9048) ( z + 1) ( z + 0.9833) ( z + 0 . 5) E(z) = R(z) [1 – M(z)]

D(z) = 104.17

ess = lim (z – 1) E(z) = 0.15. z ®1

4.25 G(s) =

0 .18 1 ; T = 2 sec; Gh0G(z) = 10 s + 1 z - 0 . 82 y(t) = 1 – e–t; Y(s) = Y(z) =

1 1 s s +1

0 . 86 z z z = ; z - 1 z - 0 .14 ( z - 1) ( z - 0 .14)

M(z) =

0 . 86 Y (z) = R(z) z - 0 .14

D(z) =

4 . 8 - 3 . 9 z -1 M( z) 1 = Gh 0 G( z ) 1 - M ( z ) 1 - z-1

LM N

OP Q

4.26 Parallel to Review Example 4.3. Result: u(k) = 10e(k) – 6e(k – 1) – 0.75u (k – 1) U(z) = 10E(z) – 6z–1 E(z) – 0.75z–1 U(z)


10 - 6 z-1 U(z) = D(z) = E( z ) 1 + 0 . 75 z-1 2

4.27 G(s) =

( s + 1) (s + 2 )

Gh0G (z) = 0.4

; T = 1 sec ( z + 0 . 368)

( z - 0 . 368) ( z - 0 .135)

; R (z) =

1 1 - z-1

M(z) may be chosen as z–1; but, as can be examined, the response will exhibit intersample ripples. We therefore take, M(z) = =1z–1 + = 2 z–2 E(z) = R(z) [1 – M(z)] lim (z – 1) R(z) [1 – M(z)] = 0 gives = 1 + = 2 = 1 z ®1

For no intersample ripples, we require (refer Eqn. (4.81)) U(z) = u(0) + u(1)z–1 + u(2) [z–2 + z–3 + L] Dividing U(z) by R(z),

U(z) = u(0) + [u(1) – u(0)]z–1 + [u(2) – u(1)]z–2 = >0 + >1z–1 + >2z–2 R( z )

=1 -1 =2 -2 z + z >0 >0 Y (z) Y ( z ) / R( z ) Gh0G(z) = = = > > U(z) U ( z ) / R( z ) 1 + 1 z-1 + 2 z-2 >0 >0 =

0 . 4 z - 1 + 0 .1472 z - 2 1 - 0 . 503z - 1 + 0 . 04968z - 2

Comparing the coefficients, we get a1 = 0.731, a2 = 0.269, >0 = 1.8275, >1 = – 0.919, >2 = 0.09 D(z) =

=

LM N

OP Q

M (z ) 1 U ( z ) / R( z ) = 1-Y ( z ) / R( z ) Gh 0 G( z ) 1- M ( z ) 1.8275- 0.919 z-1 + 0.09 z-2 1 - 0.731z-1 - 0.269 z -2

1 4.28 G(s) = s (s + 1) ; T = 1 sec

SOLUTION MANUAL

Gh0G(z) =

0.3679 z -1 + 0.2642 z -2 Y (z) –1 –2 -1 -2 ; M(z) = a1z + a2z = 1 - 1. 3679 z + 0 . 3679 z R( z )

U(z) = u(0) + u(1)z–1 + u(2) [z–2 + z–3 + L] R(z) =

1 U ( z) ; = b0 + b1z–1 + b2z–2 1 - z -1 R( z )

=1 -1 =2 -2 z + z >0 >0 Y ( z ) / R( z ) Gh0G(z) = = > > U ( z ) / R( z ) 1 + 1 z-1 + 2 z-2 >0 >0 a1 + a2 = 1 Comparing the coefficients of Gh0G(z), we get a1 = 0.582, a2 = 0.418; M(z) = 0.582z–1 + 0.418z–2 D(z) =

LM N

OP Q

1.582 - 0 .582 z - 1 M(z) U ( z ) / R( z ) 1 = = 1-Y ( z ) / R( z ) Gh 0 G( z ) 1- M ( z ) 1 + 0 . 418 z - 1

Output sequence,

1 Y ( z) = 0.582z–1 + 0.418z–2; R(z) = 1 - z -1 R( z ) y(k) = 0, 0.582, 1, 1, º -2

4.29 G(s) =

0 . 3935z e- 5s ; T = 5 sec; Gh0G(z) = 10 s + 1 1 - 0 . 6065z - 1

y(0) = 0, y(1) = 0, y(2) = 1.582(1 – e–0.5) = 0.6225 y(k) = 1; k ≥ 3 Y(z) = 0.6225z–2 + z–3 + z– 4 + L = 0.6225z–2 + z–3

1 1 - z -1

Y ( z) = M(z) = 0.6225z–2 + 0.3775z–3 ( ) Rz

Pole excess of M(z) = 2 = pole excess of Gh0G(z)

LM N

OP Q

D(z) =

1 - 0 . 3678z - 2 M (z ) 1 = 1.582 1 - 0 . 6225z - 2 - 0 . 3775z - 3 Gh 0 G( z ) 1- M ( z )

U(z) =

1 - 0 . 3678 z - 2 Y ( z ) / R( z ) = 1.582 1 - z -1 Gh 0 G( z ) / R( z )

c

c

h

h

39


= 1.582 + 1.582z–1 + z–2 + z–3 + L Since u(k) is constant for k ≥ 2, there are no intersample ripples in the output of the system after the settling time is reached. 4.30 Gh0G(z) =

0 . 3935z - 2 1 - 0 . 6065z - 1

Pole excess = 2; second-order model. Y ( z) = a2z–2 R( z ) U ( z) = u(0) + [u(1) – u(0)]z–1 + [u(2) – u(1)]z–2 = b0 + b1z–1 + b2z–2 R( z )

a 2 -2 z b0 Y ( z ) / R( z ) = = Gh0G(z) b b U ( z ) / R( z ) 1 + 1 z -1 + 2 z - 2 b0 b0 From the steady-state error requirement, a2 = 1. This gives b0 = 2.541, b1 = – 1.541, b2 = 0

LM N

OP = U z / R z Q 1-Y z / R z c1 - 0 . 6065z h = 2.541 c1- z hc1 + z h

D(z) =

M(z) 1 Gh 0 G( z ) 1 - M ( z )

( )

( )

( )

-1

-1

-1

It is physically realizable. Y(z) = z–2

1 = z–2 + z–3 + z–4 + L 1 - z -1

y(k) = 0, 0, 1, 1, 1, K u(0) = b0 = 2.541 u(1) = b1 + u(0) = 1, u(2) = b2 + u(1) = 1 u(k) = 2.541, 1, 1, 1 K

( )

CHAPTER 5

CONTROL SYSTEM ANALYSIS USING STATE VARIABLE METHODS

5.1

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

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

or

A =

LM0 MM00 MN0

LM MM MN

OP PP PQ

0 1 0 0 0 - 0.025 3 0 ;b= 0 -12 -190 1000 0 0 0.2 - 4.2

OP PP PQ

y = qL = nx1 = 0.5 x1 5.2 u

KA

1 sLf + Rf + 1

if x3

1 Js + B

KT

q x2

1 s

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 -5 2.5

0 0]

q x1


5.3 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 1

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 - 0.5 ( k + 0.5) - 2 2

0

1

5.4 x1 = w ; x2 = ia Jw& + 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 = [2 N2 Q c

1

a

5.5

A =

b =

–1

–1

–1]

Y (s ) PD s -2 1 = 1 1 = = 2 -1 -2 U ( s) D 1 - ( -3s - 2s ) s + 3s + 2

0 0

OP Q

SOLUTION MANUAL

s –1

1

U

X2

s –1

1

X1 = Y

–3 –2

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

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

=

1 s 2 + 3s + 2 s –1

X1

1

2 –15

–11

Y

U 6 –1

2 X2

s –1 8

5.6

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 5.7 X(s) = (sI – A)–1 x0 + (sI – A)–1 b U(s) = G(s)x0 + H(s) U(s)

G(s) =

1 D

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

D = s3 + 3s2 + 1

OP PP Q

LM MM N

1 s+3 1 1 s( s + 3) s ; H(s) = s D 2 2 -1 s s

OP PP Q

43


5.8

5.9

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 =

LM 0 MM-02 N

OP PP Q

LM OP MM PP NQ

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

–2

1]

5.10 Taking outputs of integrators as state variables (x1 and x2 are outputs of top two integrators from left to right; x3 and x4 are corresponding variables for other integrators): x&1 = – 4x4 + 3u1; x& 2 = x1 – 3x2 + u1 + 2u2; x& 3 = – x2 + 3u2; x& 4 = – 4x4 + x3;

LM0 1 A= M MM0 N0

5.11 (a)

0 -3 -1 0

OP PP PQ

0 -4 0 0 ;B= 0 0 1 -4

LM3 MM10 MN0

G(s) = c (sI – A)–1 b =

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

OP PP PQ

0 2 0 1 0 0 ;C= 3 0 0 0 1 0

LM N

OP Q

SOLUTION MANUAL

(b) 5.12

45

1 ( s + 1)(s + 2) a1 = – tr (A) = – 4 G(s) =

LM-2 -1 0 OP Q = A + a I = 1 -3 2 ; a MM -1 0 -3PP N Q LM 1 1 - 2OP Q = AQ + a I = -3 2 - 4 ; MM 1 1 3 PP N Q 2

1

3

2

a3 = –

2

=–

1 tr(AQ2) = 6 2

2

1 tr(AQ3) = – 5 3

As a numerical check, we see that the condition: 0 = AQ3 + a3I is satisfied. Therefore, D(s) = s3 – 4s2 + 6s – 5. (sI – A)+ = Q1s2 + Q2s + Q3 = Q(s) G(s) = 5.13

LM N

4( s - 3) -3s + 5 1 CQ( s )B = 2 D( s) - s + 2s 2( s 2 - 3s + 1) D( s)

OP Q

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

A =

LM-5 MM 40 N

G(s) = c(sI – A)–1 b = 5.14 (a)

OP PP Q

LM MM N

OP PP Q

0.5 -3.5 0 0 ; b = 0 ; c = [0 -5 1 0 -1

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 = [–1 N -1 -1Q N1Q

1]; d = 1

0]


(b)

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

LM-2 MM --11 N

A =

OP PP Q

LM OP MM PP NQ

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

1]

5.15 Taking outputs of pseudo-integrators as state variables (x1, x2, x3, x4: from top to bottom): x&1 + x1 = u1; x& 2 + 5x2 = 5u2; x& 3 + 0.5x3 = 0.4u1; x& 4 + 2x4 = 4u2

u1 = K1r1 – K1y1; u2 = K2r2 – K2y2 y1 = x1 + x2; y2 = x3 + x4 Writing the state equations we get, x& = Ax + Bu

LM-1 - K 0 A= M MM -0.4 K N 0

1

1

LM K 0 B= M MM0.4K N 0 1

5.16 (i)

Y (s ) U ( s)

1

=

L =

- K1 -5 -0.4 K1 0

OP PP PQ

0 -5 K 2 -0.5 -4 K 2

OP PP ; 0 -2 - 4 K PQ 0 -5 K 2

0 5K 2 1 1 0 0 ;C= 0 0 0 1 1 4K 2

LM N

2

OP Q

–1 s+3 2 + = s+1 s+ 2 ( s + 1)(s + 2)

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

–1]

SOLUTION MANUAL

x1

u + +

47

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 (5.56), 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 1 -4 0

2

5

0

1]

x3 = y

x2 + –

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 (5.54), the state model in second companion form is given below.


A =

LM 0 MM-06 N

OP PP Q

LM OP MM PP NQ

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

z

u + –

+

+

+

+

+ 17

z

x3

+

z

x2

x1

6

+

+

+

b 2s + b 3 Y (s ) s +1 = 3 = 3 2 U ( s) s + 3s + 2s s + a 1s 2 + a 2 s + a 3 From Eqns (5.56):

A

(ii)

LM0 = 1 MM0 N

OP PP Q

LM OP MM PP NQ

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

0

1]

b3 Y (s ) 1 = 3 = 3 2 2 U ( s) s + a 1s + a 2 s + a 3 s + 6 s + 11s + 6 From Eqns (5.54):

A =

(iii)

LM 0 MM-06 N

1 0 - 11

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

y

8

11

6

5.17 (i)

2]

+

8

1

6

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

0]

SOLUTION MANUAL

L =

5.18 (a)

LM-1 MM 00 N

OP PP Q

LM OP MM PP NQ

0 0 1 -2 0 ; b = 1 ; c = [–1 2 0 -3 1

1] ; d = 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 (5.54):

A =

(b)

OP PP Q

LM OP MM PP NQ

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

1000

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

L =

5.19

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 0 -50 -18.75

1

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 (5.65):

L =

LM-1 MM 00 N

z

OP PP Q

LM OP MM PP NQ

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

1

1]

t

x(t) =

eL (t – t) bu(t)dt

0

LMe ]= M 0 MN 0

-t

e

–L Lt

= L

–1

[(sI – L)

–1

te -t e -t 0

0 0 e -2 t

OP PP Q

1]

0]

49


z t

e

L ( t -t )

bu(t ) dt

=

0

LM MM MM MM MM N

z z z t

(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

-2t

0 t

e -2( t - t ) dt

0

OP PP PQ

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

1

u

1

1

+

x2 +

+

–

x1

–1

–1 +y +

x3

+

1

+ –2

5.20 (i)

A=

LM1 1OP ; |lI – A| = (l – 1) (l – 2) = 0 N0 2 Q

l1 = 1; l2 = 2

OP l - 1Q L-1 1OP ; v = LM1OP = 1: adj(l I – A) = M N 0 0 Q N0 Q L0 1OP ; v = LM1OP = 2; adj(l I – A) = M N0 1Q N1Q

adj(lI – A) =

For l1

For l2

LMl - 2 N 0

1

1

1

1

2

SOLUTION MANUAL

LM-3 2OP ; |lI – A| = l + 3l + 2 N-1 0Q L1O L2O l = – 1, l = – 2; v = M P , v = M P N1Q N1Q LM 0 1 0 OP A= 3 MM-12 -07 -26PP ; |lI – A| = (l + 1) (l + 2) (l + 3) = 0 N Q 2

(ii) A =

1

(iii)

2

1

2

l1 = – 1, l2 = – 2, l3 = – 3 (l1I – A)v1 = 0 gives

LM-1 MM12-3 N

-1 0 -1 -2 7 5

v1 = [1

OP LMn PP MMnn QN

11 12 13

OP L0O PP = MMM0PPP Q N0 Q

– 1]T

–1

(l2I – A)v2 = 0 gives v2 = [1

– 2 1/2]T

(l3I – A)v3 = 0 gives v3 = [1 5.21 (a)

LM 0 MM 0. A= M MM . MMN-a0

3]T

–3

n

1 0 . . 0 -a n -1

LM l MM 0. (l I – A) = M MM . MMa0 N

i

0 1 . . 0 . -1 li

. . . . . .

. . . . . . . . . . . -a 2 0 . -1 0

. .

0 0 . . 1 -a 1 . .

i

n

0 a n -1

. .

. .

. li . a2

OP PP PP PP PQ

OP PP PP -1 P P l + a PQ 0 0

i

i

51


vi = [ck1 ck2 .

ckn]T; the cofactors of the k th row.

.

Let k = n cn1 = (– 1)n+1 [determinant of (n – 1) × (n – 1) matrix obtained by deleting last row and first column] = (– 1)n+1 [(– 1)n–1] = 1 cn2 = (– 1)n+2 [determinant obtained by deleting last row and second column] = (– 1)n+2 [li(– 1)n–2] = li cn3 = (– 1)n+3 [l2i (– 1)n–3] = l2i . . . cnn = ln-1 i

LM 1 l M P= M l MM . MNl

1 2 1

1 l2 l22

n -1 1

ln2-1

.

. . . . . . . . .

1 ln l2n

. . . . n -1 . . . ln

OP PP PP PQ

(b)

LM 0 A= 0 MM-24 N LM MM N

1 0 - 26

OP P -9 PQ 0

1 ; |lI – A| = 0

OP PP Q

yields: l1 = –2, l2 = –3, l3 = – 4

1 1 1 P = -2 -3 -4 = [v1 v2 v3] 4 9 16

5.22 (a)

LM 0 A= 0 MM-2 N

1 0 0 1 -4 -3

OP PP Q

The characteristic equation is, l3 + 3l2 + 4l + 2 = 0 l1 = – 1 + j1, l2 = – 1 – j1, l3 = – 1.

SOLUTION MANUAL

LM 1 P = -1 + j1 MM - j2 N

OP PP Q

1 1 -1 - j1 -1 ; 1 j2

LM-1 + j1 0 0 OP P AP = L = MM 00 -1 0- j1 -01PP N Q LM 1/ 2 - j1/ 2 0OP LM-1 1 (b) Q = 1/ 2 MM 0 j10/ 2 10PP ; Q LQ = MM-01 -01 N Q N L-4 3OP ; |lI – A| = (l + 1) (l – 2) = 0 5.23 A = M N-6 5Q Ll - 5 3 OP ; adj(lI – A) = M N -6 l + 4 Q L-6 3OP ; v = LM1OP l = – 1, adj(l I – A) = M N-6 3Q N1Q L-3 3OP ; v = LM1OP l = 2, adj(l I – A) = M N-6 6Q N2Q L1 1OP ; J = P AP = LM-1 0OP P= M N1 2Q N 0 2Q Le 0 OP P = LM 2e - e e = Pe P = P M N 0 e Q N2 e - 2 e –1

–1

1

1

2

OP PP Q

0 0 -1

1

2

2

–1

At

Jt

-t

–1

–1

-2 t

5.24 (a)

L0 A= M N1 At

e

-t

2t

-t

OP Q

LM MM MN

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

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

2t

s+4 -3 ( s + 1)( s + 3) ; (sI – A)–1 = 1 -4 ( s + 1)( s + 3)

LM 3 e = M2 MN 12 e

-t -t

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

-3 ( s + 1)( s + 3) s ( s + 1)( s + 3)

OP PP Q

OP PP PQ

OP Q

53


1 1 LM 3 e - 1 e e - e O PP 2 2 2 e =M2 3 3 -3 -1 e + e P Q NM 2 e + 2 e 2 2 L 0 1 OP ; |lI – A| = l + 5l + 6 = 0; l A= M N-6 -5Q -t

(b)

5.25 (a)

-3 t

-t

-3 t

At

-t

-3 t

-t

-3 t

2

1

= – 2, l2 = – 3

g(l) = b0 + b1l; f(A) = eAt; f(l) = elt f(li) = g(li) gives e–2t = b0 – 2b 1; e–3t = b0 – 3b1 Solving we get, b1 = e–2t – e–3t, b0 = 3e–2t – 2e–3t eAt = b0I + b1A =

(b) A =

LM 3e N-6e

-2 t

- 2e -3t -2 t + 6 e -3 t

LM 0 2 OP ; f(A) = e N-2 -4Q

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

At

, l1 = l2 = – 2; g(l) = b0 + b1l

f(l1) = g(l1) gives e–2t = b0 – 2b1 d d g(l ) f (l ) = gives dl dl l = -2 l = -2

b1 = te–2t, b0 = (1 + 2t)e–2t eAt =

5.26

LM(1 + 2t )e N -2te

-2 t

-2 t

OP Q

2te -2 t (1 - 2t )e -2 t

OP Q

SOLUTION MANUAL

G(s) =

LMG NG

11 ( s )

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

G12 ( s )

21 ( s ) G22

1 2

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

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

X1 ( s) s -2 (1) 1 / 2 -1 / 2 = G (s) = = + 12 D s +1 s + 3 x 20 X 2 (s) x10

= G21(s) =

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

X 2 (s) x 20

= G22(s) =

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

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

= H2(s) =

1 s +1

Zero-input response: x(t) = eAt x0 = L =

LM N

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

–1

[G(s)x0]

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

OP LMx OP Q Nx Q 0 1 0 2

Zero-state response:

z t

x(t) =

e A (t -t ) bu(t ) dt = L –1 [H(s)U(s)] =

0

5.27

LM 0 A= 0 MM-6 N

OP PP Q

LM OP MM PP NQ

LM1 - e OP N1 - e Q -t -t

1 0 0 0 1 ; b = 0 |lI – A| = (l + 1) (l + 2) (l + 3) = 0 -11 -6 2

55


LM 1 P = Ml MNl

1

–1

l2

1 2 1

OP L 1 1 1 O LM-1 M P l P = -1 -2 -3 ; L = P AP = 0 M P MM 0 l PQ MN 1 4 9 PQ N LM 1 OP b = -2 ; c = cP = [1 1 1] MM 1 PP N Q 1

3 2 3

l22

b = P–1

The state model in Jordan canonical form:

x& = L x + b u; y = c x The transformed initial vector is: 0

–1

x =P x

LM 1 OP = -2 MM 1 PP N Q

0

The solution is,

z z z t

–t

x1 (t ) = e

x10

+

e - ( t -t ) dt = 1

0

t

x 2 (t) = e–2t x 20 +

e -2 ( t -t ) dt = – 1 – e–2t

0 t

x 3 (t) = e

–3t

x 30

+

0

e -3( t -t ) dt =

1 2 –3t + e 3 3

x(t) = P x (t) x1(t) =

1 2 – e–2t + e–3t; x2(t) = 2(e–2t – e–3t) 3 3

x3(t) = – 2(2e–2t – 3e–3t); y(t) = x1(t) 5.28

A =

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

1

= – 1, l2 = – 2

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

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

OP PP Q

0 0 -2 0 ; 0 -3

SOLUTION MANUAL

=

y(t) =

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

LM1 0OP LMs + 3 1OP LM2 1OP LM 1/ s OP N1 1Q N -2 sQ N0 1Q N 1/(s + 3)Q LM 3s + 16s + 18 OP = M s( s + 1)( s + 2)( s + 3) P MM 4s + 6 PP N s(s + 2)(s + 3) Q L y (t ) OP = LM3 - 25 e - e + 12 e OP y(t) = M Ny (t)Q MN 1 + e - 2e PQ

5.29 Y(s) = C(sI – A)–1 BU(s) = 2

-t

1

-2 t

-2 t

2

-3 t

-3 t

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

5.30

x& 2 = 4x1 – 5x2

A =

LM-6 N4

OP Q

LM OP NQ

0.5 7 ;b= ; c = [0 0 -5

1]

eAt = L –1 [(sI – A)–1]

=

LM 1 e MM 34 e N3

-4t -4 t

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

+

z LMN t

y(t) = x2(t) = 7

0

=

5.31

eAt =

LMe + te N -te –t

-t

-t

LM N

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

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

OP Q

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

OP Q

x(t) = e A ( t - t0 ) x ( t 0 ) Given: x1(2) = 2 ; t0 = 1, t = 2

OP Q

OP PP Q

57


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 π 0 MN x (1)PQ N 2k Q LM 0 1 0 OP 5.32 (a) A = 3 MM-12 -07 -26PP N Q 1

1 2

l1 = – 1, l2 = – 2, l3 = – 3 Modes: e–t, e–2t, e–3t (b) Eigenvectors:

v1

LM 1 OP = -1 ; v MM-1PP N Q

2

LM 1 OP = -2 ; v MM1/ 2 PP N Q

x = P x ; x& = P–1

LM 1 OP LM 1 = -3 ; P = -1 MM 3 PP MM-1 N Q N LM-1 0 0 OP AP x = 0 -2 0 x MM 0 0 -3PP N Q 3

1 -2 1/ 2

1 -3 3

OP PP Q

x (0) = P–1 x(0)

LMe x (t) = M 0 MN 0

-t

0 e

-2 t

0

0 0 e -3 t

OP L x (0)O PP MMx (0)PP Q MN x (0)PQ 1 2

3

x1 (0) = x 2 (0) = 0. x 3 (0) = k π 0

With these initial conditions, only the mode corresponding to e–3t will be excited.

LM0OP LM k OP x(0) = P x (0) = P 0 = -3k ; k π 0 MMkPP MM 3k PP NQ N Q

SOLUTION MANUAL

5.33 (a) x& =

LM0 1OP x; y = [1 N2 1Q

2]x

l1 = 2, l2 = –1. Modes are e2t and e–t (b) adj(lI – A) =

LMl - 1 1 OP N 2 lQ

LM1 1OP ; v = LM1OP N2 2Q N2Q L-2 1 OP ; v = LM 1 OP For l = –1, adj(lI – A) = M N 2 -1Q N-1Q L1 1 OP P= M N2 -1Q For l = 2, adj(lI – A) =

1

2

x = P x gives

LM N

OP LM x OP ; y(t) = cP x = [5 Q Nx Q 0 O L x (t ) O PQ MN x (t )PQ

2 0 x& = P–1 AP x = 0 -1

LM x (t)OP = LMe Nx (t )Q N 1

2 ( t - t0 )

2

0

1

–1] x

2

e - ( t - t0 )

1

0

2

0

If x1 (t 0 ) = 0, the mode e2t will be suppressed in y(t) = 5 x1 – x 2 . x(t0) = P x (t0) = P

LM 0 OP = LM K OP ; K π 0 N K Q N- K Q

With this initial condition, the mode e2t will be hidden from y(t).

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 -2 t

5.34

-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

59


= L

(sI – A)

–1

=

(sI – A) =

A =

5.35

LM c cA V= M MM M NcA

–1

[(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

5.36

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.

5.37

None of the rows of the B matrix corresponding to the Jordan blocks for l1 = – 1, l2 = – 2, l3 = – 3 is zero row. The system is completely controllable. First column of C matrix which corresponds to the eigenvalue –1, is a zero column. Other columns are nonzero. The system is not completely observable.

5.38 (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 Jordan canonical form; it is both controllable and observable.

SOLUTION MANUAL

61

(iv) 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. (v) 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. 5.39 (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 5.40

(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. 5.41 (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


5.42 (a)

G(s) =

A =

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

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

0]

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

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

A =

LM0 MM00 N

LM OP MM PP NQ

OP PP Q

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 =

5.42

Refer Section 5.4.

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

CHAPTER 6

6.1

STATE VARIABLE ANALYSIS OF DIGITAL CONTROL SYSTEMS

LM-3 F = -4 MM-1 N

OP PP Q

LM MM N

OP PP Q 0O -1P ; |zI – F| = z P z QP

1 0 -3 0 1 ; g = -7 0 0 0

LMz + 3 -1 (zI – F) = MM 41 0z N L z 1 M -(4 z + 1) (zI – F) z = DM MN -z 2

–1

6.2

+ 3z2 + 4z + 1 = D

OP z( z + 3) z + 3 P z = G(z) z( z + 3) + 4 PQ -1 L -3z - 7z OP 1 M -7z - 9z + 3P g= DM MN 3z + 7 PQ 1

z

2

H(z) = (zI – F)–1

3

2


LM MN

OP PQ

LM OP NQ

2 -5 1 6.3 F = 1 -1 ; g = ; c = [2 0 2 G(z) = c(zI – F)–1 g =

6.4

C=

LM2 N1

OP Q

0]; |zI – F| = z2 – z +

1 =D 2

2z + 2 z2 - z +

LM N

1 2

OP Q

LM MN

OP PQ

LM N

2 -5 0 1 -2 0 0 4 0 ;G= ; F = 1 -1 ; D = 0 1 3 0 0 -2 -1 2

D = |zI – F| = z2 – z +

OP Q

1 2

G(z) = C(zI – F)–1 G + D

LM MN

OP PQ

-30 1 2 z + 2 4D - 4 z - 14 1 -3z - 4 -2D - 3z - 9 D z+ 2 6.5 x1, x2 and x3 : outputs of unit delayers starting at the top and proceeding towards the bottom. =

x1(k + 1) =

1 1 x1(k) + x2(k) + 2x3(k) + u(k) 2 4

x2(k + 1) = –

1 x2(k) – x3(k) – u(k) 2

x3(k + 1) = 3x2(k) +

1 x3(k) + 2u(k) 3

y(k) = 5x1(k) + 6x2(k) – 7x3(k) + 8u(k)

LM 1 2 M F = M0 MM N0

OP PP PP Q

1 2 1 4 -1 -1 ; g = -1 ; c = [5 2 2 1 3 3

LM MM N

OP PP Q

6 –7]; d = 8

6.6 x1, x2 and x3 : outputs of unit delayers starting at the top and proceeding towards the bottom. x1(k + 1) = x2(k) + u1(k); x2(k + 1) = 3x1(k) + 2x3(k) x3(k + 1) = – 12x1(k) – 7x2(k) – 6x3(k) + u2(k)

SOLUTION MANUAL

LM 0 F= 3 MM-12 N 6.7

(i)

LM MN

(i)

LM N

OP Q

OP 3 PQ 1

-1 ; g =

LM0OP ; c = [– 1 N1 Q

LM OP N Q

– 2]; d = 3

OP PP PQ

LM MM N

OP PP Q

3 0.5 4 0 1 ; g = -3 ; c = [0 0 1 -1 4

0

1]; d = – 2

1 2 1 Y (z) =1– + + R( z ) z +1 z +1 z + 3

LM-1 F= 0 MM 0 N (ii)

OP PP Q

b z 3 + b1 z 2 + b2 z + b3 Y ( z ) -2 z 3 + 2 z 2 - z + 2 = = 03 3 R( z ) z + a1 z 2 + a 2 z + a 3 z3 + z2 - z 4

LM0 F = M1 MM0 N 6.8

LM MM N

b z 2 + b 1z + b 2 Y (z) 3z 2 - z - 3 = = 02 R( z ) z 2 + 1 z - 2 z + a 1z + a 2 3 3 0 F= 2 3

(ii)

OP PP Q

1 0 1 0 0 2 0 2 0 ;D= 0 2 ;G= 0 0 ;C= 0 0 1 0 1 0 1 -7 -6

Y (z) = R( z )

LM 1 3 M F = M0 MM N0

OP PP Q

LM OP MM PP NQ

0 0 1 -2 0 ; g = 1 ; c = [– 1 0 -3 1 5

-2 1 z3

3

2

1]; d = 1

F z - 1 I F I FH z - 1 IK 3 H 3K H K 1 0O PP L0O 1 1 P ; g = M0P ; c = [5 – 2 3]; d = 0 MM1PP 3 1P NQ P 0 3Q 3

+

2

+

65


6.9

(i) y(k + 3) + 5y(k + 2) + 7y(k + 1) + 3y(k) = 0 Refer Fig. 6.3. x1(k + 1) = – 3x3(k) x2(k + 1) = – 7x3(k) + x1(k) x3(k + 1) = – 5x3(k) + x2(k)

LM0 F= 1 MM0 N

OP PP Q

LM OP MM PP NQ

0 -3 0 0 -7 ; g = 0 ; c = [0 0 1 -5 0

1]; d = 0

(ii) y(k + 2) + 3y(k + 1) + 2y(k) = 5r(k + 1) + 3r(k)

LM N

OP Q

LM OP NQ

-1 0 1 5z + 3 7 -2 Y (z) = 2 = + ;F= ;g= ; 0 -2 1 R( z ) z + 3z + 2 z +1 z + 2

c = [–2 7]; d = 0 (iii) y(k + 3) + 5y(k + 2) + 7y(k + 1) + 3y(k) = r(k + 1) + 2r(k)

z+2 b 2 z + b3 Y (z) = 3 = 3 2 R( z ) z + a1 z + a 2 z + a 3 z + 5z 2 + 7z + 3

6.10

LM 0 1 0 OP LM0OP F= 0 MM-3 -70 -51PP ; g = MM01PP ; c = [2 N Q NQ L 0 1OP ; g(l) = b + b l F= M N-3 4Q 0

1 0]; d = 0

1

|lI – F| = l2 – 4l + 3 = 0; l1 = 1, l2 = 3 f (l) = lk; f (1) = 1 = g(1) = b 0 + b 1; f (3) = 3k = g(3) = b 0 + 3b 1 b 0 = 1.5 – 0.5(3)k; b 1 = 0.5[(3)k – 1]

LM1.5 - 0.5(3) N1.5[(3) -1] k

Fk = b 0I + b 1F =

k

0.5[(3) k - 1] -0.5 + 1.5(3)k

OP Q

6.11 Refer Example 6.3 (zI – F)–1 =

LM N

z +1 1 1 ( z + 0.2)( z + 0.8) -0.16 z

OP Q

X(z) = (zI – F)–1 zx(0) + (zI – F)–1 gU(z); U(z) =

z z -1

SOLUTION MANUAL

67

LM -17 z 22 z 25 z OP 6 + 9 + 18 P M z 0 . 2 z 0 . 8 z -1 P + + X(z) = M 3 . 4 17 . 6 MM 6 z - 9 z 187 z PP MN z + 0.2 + z + 0.8 + z - 1 PQ LM-17 (-0.2) + 22 (-0.8) + 25 OP 9 18 ; y(k) = cx(k) = x (k) x(k) = M 6 3.4 17.6 MN 6 (-0.2) - 9 (-0.8) + 187 PPQ k

k

k

1

k

6.12 x1(0) = – 5; x2(0) = 1 zX1(z) + 5z =

3 X1(z) – X2(z) + 3U(z) 2

zX2(z) – z = X1(z) – X2(z) + 2U(z) U(z) =

X1(z) =

z z - 1/ 2

-5 z 1 z2

2z

+

1 x (k) = – 5 F I H 2K

z+

1 2

-2 z -z 4z -2 z ; X2(z) = + + 1 1 z -1 z -1 zz+

+

2

1 -1 -1 + 2 F I – 2; x (k) = –2 F I + 4 F I – 1 H2K H 2K H 2 K 1 -1 y (k) = – 3x (k) + 4x (k) – 2u(k) = 5 F I + 10 F I + 2 H 2K H 2 K 1 -1 y (k) = – x (k) + x (k) = 3 F I + 2 F I + 1 H 2K H 2 K LM-1 1 0 OP F = 0 -1 0 MM 0 0 -2 PP N Q k

k

k

1

1

1

1

k

2

k

2

k

2

k

6.13

2

k

2

Eigenvalues of matrix F are – 1, – 1 and – 2. (a) Therefore the modes of the free response are (– 1)k, k(– 1)k and (– 2)k

LM(-1) (b) x(k) = F x(0) = M 0 MN 0 k

k

k (-1) k-1 (-1) 0

k

OP L0O LMk(-1) 0 P M1 P = M (-1) MP (-2) PQ MN1 PQ MN (-1)

k -1

0

k

k k

OP PP Q


6.14 (a) Ga(s) =

RS T

1 1 ; G(z) = (1 – z–1) Z 2 s( s + 2) s (s + 2) 0.2838z + 0.1485 z - 11353 . z + 0.1353

=

2

1 O LM 0 L0 O ; g = M P ; c = [0.1485 P . N-0.1353 11353 Q N1 Q L0 1OP ; b = LM0OP ; c = [1 0] (b) A = M N0 -2Q N1Q F=

F = eAT

UV W

L1 =M MN0

OP PQ

1 (1 - e-2 T ) ;g= 2 e- 2 T

LM MN

z

0.2838]; d = 0

OP LM 12 (T + e 2 -1)OP dt b = M PQ M 1 (1- e ) PP N 2 Q -2T

T

e At

0

-2 T

Thus for T = 1 sec,

LM1 0.4323OP ; g = LM0.2838OP ; c = [1 0] N0 0.1353Q N0.4323Q L0.696 0.246OP ; g = (e – I)A b = LM-0.021OP ; =M N 0.123 0.572Q N 0.747 Q

F=

6.15

F = eAt

AT

c = [2 6.16

–1

– 4]; d = 6

Let x1 be the output of the block

1 1 and x2 be the output of the block 10s + 1 s

x&1 = x2; x& 2 = – 0.1x2 + u + 0.1w A=

LM0 N0

OP Q

(sI – A)–1

F=

LM1 N0

LM OP NQ

LM OP N Q

1 0 0 ;b= ; b1 = 1 0.1 -0.1

LM1 = Ms MN 0

OP PP Q

LM N

OP Q

e A q b dq =

LM10T -100 + 100e OP N 10 -10e Q

1 -0.1t ) s( s + 0.1) ; eAt = 1 10(1 - e - 0.1t 1 0 e s + 0.1

10(1 - e -0.1T ) e -0.1T

OP ; g = Q

z T

0

-0 .1T

- 0.1T

SOLUTION MANUAL

z T

g1 =

e Aq b1 dq =

0

69

LMT -10 + 10e OP N 1- e Q - 0.1T

- 0.1T

For T = 0.1 sec,

LM OP N Q LM x (k + 1) OP = LM1 0.1 OP x(k) + LM0.005OP u(k) + LM 0 OP w(k) N 0.1 Q N0.01Q N x (k + 1)Q N0 0.99Q LM 1 0 O P Le 0 O 0.1 (sI – A) = M s + PQ ; T = 3 sec 0.1 1 P ; e = M0.1te e N MN (s + 0.1) s + 0.1PQ L0.741 0 OP ; G = e B dq = LM259.182 0 OP ; F=e = M N0.222 0.741Q N 36.936 259.182Q L1 0OP C= M N0 1 Q L 1 OP = 0.3679(z + 0.7181) (a) G(z) = (1 – z ) Z M N s (s + 1) Q (z - 1)(z - 0.3679) F=

LM1 N0

OP Q

LM N

OP Q

0.1 0.005 0 ;g= ; g1 = ; 0.99 0.1 0.01

1 2

6.17

–1

- 0.1t

At

- 0.1t

- 0.1t

2

z T

At

Aq

0

6.18

–1

2

(b)

0.3679 z + 0.2642 Y (z) G( z ) = = 2 R( z ) 1 + G( z ) z - z + 0.6321 F=

6.19

Ga(s) =

LM 0 1OP ; g = LM0OP ; c = [0.2642 N-0.6321 1Q N1Q

0.3679]

e -0 . 4 s s +1

(a) G(z) = (1 – z–1) Z

RS e UV = (1 – z T s(s + 1) W -0 . 4 s

–1

)Z

RS e T s

- 0. 4 s

-

e- 0. 4 s s +1

T = 1; tD = 0.4 = DT; m = 1 – D = 0.6 From Table 3.1, we obtain G(z) = (1 – z–1) Z

RS 1 - e UV = 0.4512 z + 0.1809 T z -1 z - e W z - 0.3679z - 0.6

-1

2

UV W


F=

(b)

LM0 N0

OP Q

LM OP NQ

1 0 ;g= ; c = [0.1809 0.3679 1

0.4512]

Y ( s ) e- 0.4 s = U ( s) s +1 State model, x&1 (t) = – x1(t) + u(t – 0.4); tD = 0.4, N = 0, D = 0.4, m = 0.6

F = e–1 = 0.3679

z

0. 6

g2 =

z

0. 4 -s

e ds = 0.4512; g1 = e

–0.6

0

e-s ds = 0.1809

0

x1(k + 1) = 0.3679x1(k) + 0.1809u(k – 1) + 0.4512u(k) Let us introduce a new state, x2(k) = u(k – 1); x(k + 1) = Fx(k) + gu(k); y(k) = cx(k) F= 6.20

LM0.3679 N 0

OP Q

LM N

OP Q

0.1809 0.4512 ;g= ; c = [1 0 1

0]

x& (t) = – x(t) + u(t – 2.5); tD = 2.5 = NT + DT; T = 1 N = 2, D = 0.5, m = 0.5

z

0. 5

F = e–1 = 0.3679; g2 =

e-s ds = 0.3935;

0

z

0. 5

g1 = e

–0.5

e-s ds = 0.2387

0

x(k + 1) = 0.3679x(k) + 0.2387u(k – 3) + 0.3935u(k – 2) x2(k) = u(k – 3), x3(k) = u(k – 2), x4(k) = u(k – 1) x(k + 1) = Fx(k) + gu(k)

LM0.3679 0 F= M MM 0 N 0

6.21

0.2387 0.3935 0 1 0 0 0 0

OP PP PQ

0 0 ;g= 1 0

s Y (s ) e- t D = Ga(s) = ; 0 £ tD £ T U ( s) s2 x1 = y, x2 = y& ; x&1 = x2, x& 2 = u(t – tD)

LM0OP MM00PP MN1PQ

A=

LM0 1OP ; b = LM0OP ; c = [1 N0 0 Q N1 Q

F = eAT = I + AT +

g2 =

z

e

As

LM N

1 T A2T 2 +L= 0 1 2!

LMt MN

OP PQ

OP Q

L (T - t ) / 2 OP ; g = e b ds = M N T -t Q

z

DT

2

D

AmT

1

D

0

71

0]

tD = D T = (1 – m)T mT

SOLUTION MANUAL

e A s b ds =

0

D

(T tD

tD ) 2

N = 0 (Refer Eqn. (6.35)) x(k + 1) =

LM1 T OP x(k) + g u(k – 1) + g u(k) N0 1 Q 1

2

Introduce x3 = u(k – 1). Then, x(k + 1) = Fx(k) + gu(k); y = cx(k)

LM1 F = M0 MM0 MN

F H

T tD T 1 0

tD 2

tD 0

I OP LM (T - t KP M 2 PP ; g = MM T -1t QP NM

D D

)2

OP PP ; c = [1 PP Q

0

0]

6.22 For this problem, the solution comes from Review Example 6.3 with K = 2. 6.23 x1, x2 and x3: outputs of the blocks

1 1 1 , and respectively. s+2 s s +1

(a) x&1 = x2; x& 2 = – x2 + x3; x& 3 = – 2x3 + u

LM 0 A= 0 MM 0 N

OP PP Q

LM OP MM PP NQ

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

Eigenvalues of A are l1 = 0, l2 = –1, l3 = –2 g(l) = b 0 + b 1l + b 2l2; f (l) = elT For l = 0, elT = 1 = b 0 For l = –1, e–T = b 0 – b 1 + b 2 For l = – 2, e–2T = b 0 – 2b 1 + 4b 2


b 0 = 1; b 1 =

1 1 (3 + e–2T – 4e–T ); b 2 = (1 + e–2T – 2e–T) 2 2

eAT = b 0I + b 1A + b 2A2

LM1 = M0 MM0 MN

z

e -T

e-2T

0

LM-3 + 1 T + e - 1 e 4 2 4 M 1 1 bdq = M -e + e MM 2 1 1 2 - e 2 2 N -T

T

g=

OP PP PP Q

1 (1 + e -2 T - 2e - T ) 2 ; e -T - e -2 T

1 - e-T

e Aq

0

-T

-2T

-2T

-2 T

OP PP PP Q

x(k + 1) = eAT x(k) + g[r(k) – x1(k)] = Fx(k) + gr(k)

LM 7 - 1 T - e + 1 e 4 M 4 12 1 F= M - +e - e MM 2- 1 + 1 e2 N 2 2 -T

-T

-2 T

e -T

1 (1 + e -2 T - 2 e - T ) 2 e - T - e -2 T

0

e -2 T

1 - e -T

-2 T

-2 T

z t

(b) x(t) = eA(t – kT) x(kT) +

e A( t -t ) bu(t )dt ; kT £ t < (k + 1)T

kT

6.24

Eigenvalues of A are l1 = –1, l2 = –2. elT = b 0 + b 1l e–T = b 0 – b 1; e–2T = b 0 – 2b 1 b 1 = e–T – e–2T; b 0 = 2e–T – e–2T; eAt = b 0I + b 1A =

z T

g=

0

e

Aq

LM 2e N-2e

-T

OP Q

T e -T - e - 2 T - e-2 ; T T T -T + 2e-2 -e- + 2e-2

L0.5 - e b dq = M N e

-T

+ 0.5e-2 T -T - e- 2 T

OP Q

OP PP PP Q

SOLUTION MANUAL

For T = 1 sec, eAT =

LM 0.6 N-0.465

OP Q

LM N

73

OP Q

0.233 0.2 ;g= 0.233 -0.0972

Discrete-time model of the plant is, x(k + 1) = eAT x(k) + gu(k); y(k) = x1(k) For the feedback system, e(k) = r(k) – x1(k) = u(k) x(k + 1) = eAT x(k) + g[r(k) – x1(k)] Discrete-time model of the closed-loop system is, x(k + 1) = Fx(k) + gr(k); y = cx(k) F=

LM 0.4 N-0.698

OP Q

LM N

OP Q

0.233 0.2 ;g= ; c = [1 0.233 -0.0972

0]

6.25 Plant model, x(k + 1) =

LM 0.6 N-0.465

OP Q

LM N

OP Q

0.233 0.2 x(k) + e (k); 0.233 2 -0.0972

e2(k + 1) = – 2e2(k) + e1(k) Let e2(k) = x3(k), then x3(k + 1) = – 2x3(k) + [r(k) – x1(k)] Discrete-time model of the closed-loop system is, x(k + 1) = Fx(k) + gr(k); y = cx(k)

LM 0.6 F = -0.465 MM -1 N

OP PP Q

LM OP MM PP NQ

0.233 0.2 0 -0.0972 0.233 ; g = 0 ; c = [1 0 1 -2

6.26 D(s) = 9 + 4.1s =

0

0]

U ( s) de(t ) ; 4.1 + 9e(t) = u(t) E ( s) dt

By backward-difference approximation, 4.1

LM e(k) - e(k -1) OP + 9e(k) = u(k); u(k) = – 41e(k – 1) + 50e(k) N T Q

Choosing e(k – 1) as controller state variable, xc(k), we obtain, xc(k + 1) = e(k), u(k) = – 41xc(k) + 50e(k) as the state model of the controller. The plant difference equations are given by (solution to Problem 6.16) x(k + 1) =

LM1 N0

OP Q

LM N

OP Q

0.1 0.005 x(k) + u(k) 0.99 0.1


The augmented state model becomes,

LM x (k + 1) OP L1 MM xx ((kk ++ 11))PP = MMM00 N Q N

OP LM x (k) OP LM0OP 0.99 0 x ( k ) + 0 e(k) PM P MP 0 0 QP MN x ( k ) PQ MN1 PQ LM0.005OP + 0.1 [– 41x (k) + 50(r(k) – x (k))] MM 0 PP N Q LM0.75 0.1 -0.205OP LM0.25OP = -5 0.99 -4.1 x(k) + 5 r(k) MM -1 0 0 PP MM 1 PP N Q N Q L1 -1 1 -1OP ; r (U) = 2; V = LM c OP ; r (V) = 2 (a) U = [g Fg] = M N0 0 1 -1Q NcFQ 1

0.1

0

1

2

2

c

c

c

6.27

1

Therefore system is both controllable and observable. (b) Given system is in Jordan canonical form. It is both controllable and observable. 6.28

LM 0 1OP ; b = LM0OP ; c = [1 0] N - 1 0 Q N1 Q L0 1OP ; r (U) = 2; V = LM c OP = LM1 0OP ; r (V) = 2 U = [b Ab] = M N1 0 Q NcAQ N0 1Q

A=

The continuous-time system is both controllable and observable. Comparing the given state model with Eqn (6.59), we observe that the given system is a harmonic oscillator with frequency w = 1 or period T = 2p. Therefore, T = np; n = 1, 2, K will lead to hidden oscillations. For T = p, for example, the discrete-time model is given by (refer Eqns (6.60)) x(k + 1) =

LM -1 0 OP x(k) + LM2OP u(k); y(k) = [1 N 0 -1Q N0 Q

0]x(k)

The system is obviously both uncontrollable and unobservable. For T = 2p, we have the following discrete-time model; x(k + 1) =

LM1 0OP x(k) + LM0OP u(k); y(k) = [1 N0 1 Q N0 Q

0]x(k)

The system is both uncontrollable and unobservable. In general, for T = np, n = 1, 2, K the system is uncontrollable and unobservable.

SOLUTION MANUAL

75

6.29 The poles are: s1 = –1, s2,3 = – 0.02 ± jp For Re(li) = Re(lj) Im(li – lj) = 2p T¹

2 np ¹ n; T ¹ 1, 2, 3, K 2p

F H

6.30 (a) |lI – F| = l -

1 4

I Fl - 1 I K H 2K

Eigenvalues of F are: l1 = (b) G(z) = c(zI – F)–1 g =

1 1 , l2 = 4 2

1

Fz - 1I H 4K

(c) Since there is a pole-zero cancellation, the system is either uncontrollable or unobservable or both. Given state model is in controllable canonical form. Therefore, the model is controllable but not observable.

CHAPTER 7

7.1

POLE-PLACEMENT DESIGN AND STATE OBSERVERS

b s n -1 + b 2 s n - 2 + L + b n Y (s ) = 1n U ( s) s + a 1s n -1 + L + a n

LM 0 M0 &x = M . MM 0 MN- a y = [>n

1

n

0

0 . 0 - a n -1

>n–1

.

.

1 0 . . . . . . . .

L

.

.

OP LM0OP 0 P M0 P . Px + M . P u 1 P PP MM . PP -a Q MN1PQ 0

1

>1] x; u = – kx + r; k = [k1 k2 L

x& = [A – bk] x + br; y = cx

LM 0 0 M . A – bk = M MM 0 MN- a - k n

1 0 . 0 - a n -1 - k2

1

G(s) = c[sI – (A – bk)]–1

0 . . 0 1 0 . 0 . . . . . . . 1 . . . - a 1 - kn

LM M b=c M MM NM

OP LM0OP 0 M P P 1 . P M.P × D . P M.P M P P *n QP NM1 QP *1 *2

*1 = Co-factor of – =n – k1, is independent of k. *2 = Co-factor of – =n–1 – k2, is independent of k. . . Zeros are not affected by k. 7.2 (a) k = [k1

k2

k3]

LM 0 A – bk = MM-6 0- k N

1

1 0 0 1 -11- k2 -6 - k3

OP PP PP PQ

OP PP Q

kn]

SOLUTION MANUAL

77

Charteristic equation is, s3 + (6 + k3)s2 + (11 + k2)s + 6 + k1 = 0 Desired characteristic equation is (s + 5) (s + 2 + j3.464) (s + 2 – j3.464) = s3 + 9s2 + 36s + 80 = 0 k = [74

25

3]

x = x – x$ (b) ~ ~ x ; mT = [m1 x& = (A – mc) ~

LM -m A – mc = -m MM-6 - m N

m2 m3 ]

1

1

0

0 1 -11 -6

2

3

OP PP Q

|sI – (A – mc)| = s3 + (6 + m1)s2 + (11 + 6m1 + m2)s + 11m1 + 6m 2 + m3 + 6 (s + 5) (s + 2 + j3.464) (s + 2 – j3.464) = s3 + 9s2 + 36s + 80 m1 = 3, m2 = 7, m3 = – 1 Observation equation is Eqn (7.31) and the structure is given in Fig. 7.6.

(c)

LM 0 x& = 0 MM-6 N y = [1 Aee =

OP PP Q

LM OP MM PP NQ

1 0 0 0 1 x+ 0 u 1 -11 - 6 0

LM 0 N -11

0]x 1 -6

OP ; a Q

le

= [1

0]

~ x e ; mT = [m1 x& e = (Aee – male) ~

m2]

|sI – (Aee – male)| = s2 + (6 + m1)s + 6m1 + m2 + 11 (s + 2 + j3.464) (s + 2 – j3.464) = s2 + 4s + 16 m1 = – 2, m2 = 17 Observer equations may be obtained from Eqns (7.38) – (7.49). 7.3 (a) k = [k1

k2

LM-k A – bk = 1 MM 0 N

k3]

1

-k2 -6 - k3 0 -11 1 -6

OP PP Q


|sI – (A – bk)| = (s + 5) (s + 2 + j3.464) (s + 2 – j3.464) k1 = 3, k2 = 7, k3 = – 1 (b) mT = [m1 m2 m3] |sI – (A – mc)| = (s + 5) (s + 2 + j3.464) (s + 2 – j3.464) m1 = 74, m2 = 25, m3 = 3

LM x& OP L -6 0 (c) x& = M -6 0 M x& P MM-11 1 NM PQ N L0 0OP ; a A = M N1 0 Q 3

1 2

ee

le

1 0 0

OP LM x OP LM0OP PP MM xx PP + MM10PP u; y = [1 QN Q NQ

= [0

3

1

3

0

2

1]; mT = [m1

LM x OP 0] x Mx P NM PQ 1 2

m2]

|sI – (Aee – male)| = (s + 2 + j3.464) (s + 2 – j3.464) m1 = 16, m2 = 4 7.4 Estimation error should decay in less than 4 sec. Settling time =

4 £ 4; zwn ≥ 1 zw n

The observer poles may be fixed at s = – 2, – 3. Characteristic polynomial: s2 + 5s + 6. |sI – (A – mc)| = s2 + (2 + m2)s + 2m2 + 1 + m1 Comparing and solving the resulting equations: m =

LM -1OP N 3Q

x&$ = (A – mc) x$ + (B – md)u + my

7.5 The estimation error should decay as fast as e–10t at least. Hence, the pole should be at s = – 10. Desired characteristic polynomial: s + 10 x& = Ax + bu; y = cx

A=

LM1 0OP ; b = LM1OP ; c = [2 N0 0Q N1Q

–1]

We obtain a transformation: x(t) = Qq(t), that transforms c from [2 –1] to [1 0] q& = Q–1 AQq + Q–1bu; y = cQq

[2

–1] Q = [1

0]

SOLUTION MANUAL

LM1 1OP ; Q = L 2 -1O ; MN -1 1PQ N1 2Q L 2 2 OP q + LM1OP u; y = [1 q& = M N-1 -1Q N0Q

79

–1

Q=

0]q

q~& 2 = (Aee – male) q~2 = (– 1 – 2m) q~2

Characteristic polynomial: s + 1 + 2m = s + 10; m = 4.5 Refer Eqn (7.49b)

LM OP LM OP N Q N Q

$ q q& 2¢ = – 10 q$ 2 – 10y – 4.5u; q$ 2 = q¢2 + my; q$ = q1 = 1 $q 2 q$ 2

LM y + q$ OP N y + 2q$ Q L0 9OP ; b = LM9OP ; c = [0 (a) A = M N1 0 Q N0 Q x$ = Q q$ =

2

2

7.6

1]

(b) k = [k1 k2] (A – bk) =

LM-9k N1

1

OP Q

9 - 9 k2 ; |sI – (A – bk)| = s2 + 9k1s + 9k2 – 9 0

Desired polynomial: (s + 3 + j3) (s + 3 – j3) = s2 + 6s + 18. k = [2/3

3]

(c) m = [m1 m2]T A – mc =

LM0 N1

9 - m1 - m2

OP ; |sI – (A – mc)| = s Q

2

+ m2s + m1 – 9

Desired polynomial: (s + 6 + j6) (s + 6 – j6) = s2 + 12s + 72.

LM81OP N12Q L0 9OP ; b = LM9OP ; c = [0 (d) A = M N1 0Q N9Q m=

1]

Select gain k which places one of the closed-loop poles at s = – 1. Pole-zero cancellation makes the feedback system unobservable.


Desired characteristic polynomial: (s + 1) (s + 2) = s2 + 3s + 2 |sI – (A – bk| = s2 + 3s + 2, gives k =

LM 1 2 OP ; N9 9 Q

LM-1 7 OP ; c(A – bk) = [0 –2] N 0 -2Q L c OP = L0 1O ; r (V) = 1. System is unobservable V=M Nc(A - bk)Q MN0 -2 PQ A – bk =

7.7 && y + w 02 y = u (a) x& =

LM 0 N-w

OP x + LM0OP u 0Q N1 Q 1

2 0

(b) wn = 2w0; z = 1; s = – 2w0 Characteristic polynomial: (s + 2w0)2 = s2 + 4w0s + 4w 02

LM 0 N-w - k

A – bk = k=

2 0

[3w 02

OP ; |sI – (A – bk)| = s -k Q 1

1

2

+ k2s + w 02 + k1

2

4w0]

(c) m = [m1 m2]T |sI – (A – mc| = (s + 10w0)2 gives m =

LM20w OP . N99w Q 0 2 0

Figure 7.6 gives a block diagram of the observer-based state-feedback control system. (d) A =

LM 0 N-w

OP ; b = LM0OP ; c = [1 0Q N1 Q 1

2 0

0]; ~ x& 2 = (Aee – male) ~ x2 = – m ~ x2

Characteristic polynomial: s + m. Desired polynomial: s + 10w0 m = 10w0 Refer Eqns (7.38) – (7.49) and Fig. 7.7. x& 2¢ = – 10w0 x 2¢ – 101w 20 y + u; x$ 2 = x 2¢ + 10w0y

7.8 (a) k = [k1 k2] (A – bk) =

LM 0 N20.6 - k

1

OP -k Q 1

2

SOLUTION MANUAL

81

Characteristic polynomial: |sI – (A – bk)| = s2 + k2s – 20.6 + k1 Desired polynomial: (s + 1.8 + j2.4) (s + 1.8 – j2.4) = s2 + 3.6s + 9. k = [29.6

3.6]

(b) m = [m1 m2]T |sI – (A – mc)| = (s + 8)2 gives m =

(c)

LM 16 OP N84.6Q

U ( s) 778.16 s + 3690.72 = k [sI – (A – bk – mc)]–1 m = 2 -Y ( s) s + 19.6 s + 151.2

(d) x& = Ax – bk x$ ; x$& = (A – mc – bk) x$ + my 1 16 -16 x$ + x = 84.6 1 -93.6 -3.6 7.9

LM OP LM OP N Q N Q L0 1OP ; b = LM0OP ; c = [1 0] (a) A = M N0 0 Q N1 Q

(b) wn = 1; z =

1 ; 2

|sI – (A – bk)| = s2 + 2zwns + w 2n gives k = [k1 k2] = [1 (c) wn = 5; z = 0.5; s2 + 2zwns + w 2n = s2 + 5s + 25 |sI – (A – mc)| = s2 + 5s + 25 gives m =

LM m OP = LM 5 OP Nm Q N25Q 1 2

(d)

40.4(s + 0.619) U ( s) = k [sI – (A – mc – bk)]–1 m = 2 s + 6.414 s + 33.07 -Y ( s)

(e) A =

LM0 1OP ; b = LM0OP ; c = [1 N0 0 Q N1 Q

0]

~ x& 2 = (Aee – male) ~ x 2 = –m ~ x 2 ; s + m = s + 5; m = 5

Refer Eqns (7.38) – (7.49). x& 2¢ = – 5 x$ 2 + u; x$ 2 = x 2¢ + 5y

(f) x& 2¢ = – 5 x 2¢ – 25y + u; X2¢ (s) = u = – k1x1 – k2 [ x 2¢ + 5y]

1 [– 25Y (s) + U (s)]; s+5

2]


U ( s) 8.07(s + 0.62) = -Y ( s) ( s + 6.41) 7.10 (a) Desired polynomial: (s + 2) (s + 1 + j1) (s + 1 – j1) = s3 + 4s2 + 6s + 4 x&1 = x2; x& 2 = – x2 + x3; x& 3 = – 2x3 + u; y = x1 ~ x1 = x1 – r; ~ x 2 = x2; ~ x 3 = x3 ~ x + bu x& = A ~

LM0 A= 0 MM0 N

LM OP MM PP NQ

OP PP Q

1

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

x = – k1x1 – k2x2 – k3x3 + k1r; N = k1; u = – k~ U = [b

Ab

p1 = [1

0

LM0 A b] = 0 MM1 N

OP PP Q

0]

LM p OP L1 0 MM PP = MM0 1 Np A Q MN0 -1

OP PP Q

0 0 ; |sI – A| = s3 + 3s2 + 2s 1

1

P = p1 A

2

1

k = [4

0 1 1 -3 -2 4

2

6–2

4–3]; P = [4

3

3

1]

2

(b) Desired polynomial: s + 8s + 24s + 32 z& = ATz + cT h; h = mTz

U = [c

T

LM1 0 0 OP A c (A ) c ] = 0 1 -1 MM0 0 1 PP Q N LM p OP L0 0 0 1]; P = M p A P = M0 1 MNp (A ) PQ MMN1 -3 T T

T 2 T

1

p1 = [0

T

1

T 2

1

|sI – A| = s3 + 3s2 + 2s mT = [32

24 – 2

8 – 3] P = [5

7 8]

1 -2 4

OP PP Q

SOLUTION MANUAL

LM x& OP L-1 MM x&z& PP = MMM 10 N Q N

OP LM x OP LM1OP LM 0OP -2 0 x + 1 u + 0 r P M P M P MM-1PP 3 0 PQ MN z PQ MN0PQ N Q LM1 -1 1OP Ab A b] = 1 -2 4 ; |U| = – 5 MM0 4 -7PP Q N

1

7.11

83

2

0

0

1 2

2

U = [b

Conditions for existence of integral control solution are satisfied. p1 = [– 0.8 0.8

0.2]

LM p OP L-0.8 P= MpAP = M 1 MNp A PQ MMN -1

OP 0 P 0 PQ

0.8 0.2

1

-1

1

2

2

1

Desired polynomial: s3 + 4s2 + 8s + 8 |sI – A| = s3 + 3s2 + 2s; k = [8

z

u(t) = 1.4x1 – 2.4x2 – 1.6 x1 = q; x2 = q& ; x3 = ia x& = Ax + bu

LM0 A= 0 MM0 N

LM MM N

OP PP Q

8 – 2 4 – 3] P = [– 1.4

2.4

1.6]

(y – r)dt

OP PP Q

1 0 0 -1 1 ;b= 0 ; 10 -1 -10

|sI – (A – bk)| = s3 + (11 + 10k3) s2 + (11 + 10k2 + 10k3)s + 10k1 Characteristic polynomial: (s2 + 2zwns + w 2n ) (s + 10) = (s2 + 2s + 4) (s + 10) = s3 + 12s2 + 24s + 40 k = [4

1.2

0.1]

7.13 x1 = w; x2 = ia x&1 = – x1 + x2; x& 2 = – x1 – 10x2 + 10u

A=

LM-1 N-1

z

OP Q

LM OP N Q

1 0 ;b= 10 -10

t

z=

0

( y - r )dt; z& = y – r = x1 – r


LM-1 A = -1 MM 1 N

OP LM 0 OP 0 ; b = 10 P MM 0 PP 0 PQ N Q

1 0 -10 0

|sI – ( A – b k)| = s3 + (11 + 10k2)s2 + (11 + 10k1 + 10k2)s + 10k3 = (s + 10) (s + 1 + j 3 ) (s + 1 – j 3 ) = s3 + 12s2 + 24s + 40 k = [1.2 7.14 c = [1

0.1 0

4]

0]

|sI – (A – mc)| = s3 + (11 + m1)s2 + (11 + 11m1 + 11m2)s + 11m1 + 10m2 + m3 Desired polynomial: (s + 10) (s + 3 + j 3 ) (s + 3 – j 3 ) = s3 + 16s2 + 72s + 120. m = [m1

m3]T = [5

m2

6

5]T

7.15 x1 = q, x2 = q& , x3 = 0.1ia KT ia = Jq&& ; q&& = 50ia x&1 = x2; x& 2 = 5x3;

u – Kbq& = La

dia + (Ra + 0.1)ia dt

This gives, x& 3 = – 2x3 – 20x2 + 20u

LM0 A= 0 MM0 N

OP PP Q

1 0

LM MM N

0 0 5 ;b= 0 20 -20 -2

OP PP Q

|sI – (A – bk¢)| = s3 + (20k¢3 + 2)s2 + (100 + 100k¢2)s + 100k¢1 Desired polynomial: (s + 3 + j3) (s + 3 – j3) (s + 20) = s3 + 26s2 + 138s + 360. k¢ = [3.6

0.38

1.2]

k1¢ = k1KA = KA k2¢ = k2KA; k2 = 0.11

k3¢ = k3KA; k3 = 0.33 7.16 10if = 0.5q&& + 0.5q& u = 100if + 20

di f dt

SOLUTION MANUAL

x&1 = x2; x& 2 = –x2 + 20x3; x& 3 = – 5x3 +

LM0 A= 0 MM0 N

1 -1 0

1 u 20

OP LM 0 OP 20 ; b = M 0 P P M1P -5PQ MN 20 PQ 0

k1¢ = k1KA = KA; k2¢ = k2KA; k3¢ = k3KA

|sI – (A – bk¢)| = s3 + (6 + 0.05 k3¢ )s2 + (5 + 0.05 k3¢ + k2¢ )s + k1¢ Desired polynomial: (s2 + 2s + 4) (s + 10) = s3 + 12s2 + 24s + 40. k¢ = [40

13

120]

KA = 40 k2 = 0.325 k3 = 3 7.17

Y (s ) b = U ( s) s( s + a ) (a) x&1 = x2; x& 2 = – a x2 + b u A=

LM0 N0

LM OP NQ

OP Q

1 0 ;b= -a b

|sI – (A – bk)| = s2 + (a + b k2)s + b k1 = s2 + a1s + a2 b k1 = a2; k1 = 7.18

a2 a -a ; a + b k2 = a1; k2 = 1 ; N = k1 b b

x1 = w; x2 = ia x& =

LM 0 N-200

OP LM OP LM OP Q N Q N Q

50 0 -50 x+ u+ TL -200 200 0

z& = y – r = x1 – r,

LM x& OP LM 0 MM x&z& PP = MM-200 N Q N1 1 2

50 0 -200 0 0 0

OP LM x OP LM 0 OP LM 0 OP LM-50OP u+ 0 r+ 0 T PP MM xz PP + MM200 P M P M P Q N Q N 0 PQ MN-1PQ MN 0 PQ 1 2

L

|sI – (A – bk)| = s3 + 200 (1 + k2)s2 + 10000(1 + k1)s + 10000k3

85


Desired polynomial: (s + 10 + j10) (s + 10 – j10) (s + 300) = s3 + 320s2 + 6200s + 60,000. k = [– 0.38

0.6

6]

Resulting system is of type-1. Steady-state error to constant disturbance is zero, w(t) Æ r. 7.19 (a) A =

LM-3 2 OP ; b = LM1OP ; c = [0 N 4 -5Q N0Q

1]

|sI – (A – bk)| = s2 + (8 + k1)s + 7 + 5k1 + 4k2 Desired polynomial: (s + 4) (s + 7) = s2 + 11s + 28. k = [3

1.5]

(b) A – bk =

(c)

LM-6 N4

OP Q

LM N

OP LM OP Q NQ

0.5 1 -6 0.5 ; x& = x+ w 4 -5 0 -5

At steady state, x& = 0; xls =

5 1 x2s. Hence, x2s = w 4 7

LM-3 z& = y = x ; A = 4 MM 0 N

OP PP Q

2

LM OP MM PP NQ

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

|sI – (A – bk)| = s3 + (8 + k1)s2 + (7 + 5k1 + 4k2)s + 4k3 Desired polynomial: (s + 1) (s + 2) (s + 7) = s3 + 10s2 + 23s + 14. k = [2

3.5]; x& = [A – bk] x + bw

1.5

LM-5 A – bk = 4 MM 0 N

0.5 -3.5 -5

0

1

0

OP PP Q

St steady state, x& = 0; ys = x2s = 0 A=

LM -3 N4

OP ; b = LM1OP ; c = [0 -5 Q N0 Q 2

1]

(a) |sI – (A – bk)| = (s + 4) (s + 7), gives k = [3 (b)

1.5]

1 = – c [A – bk]–1b; N = 7 N

With u = – kx + Nr, the feedback system becomes,

SOLUTION MANUAL

87

x&1 = – 6x1 + 0.5x2 + 7r; x& 2 = 4x1 – 5x2 As t Æ • , x&1 = x& 2 = 0 4x1s – 5x2s = 0; x1s = –6×

5 x2s; 4

5 x1s + 0.5x2s = – 7r; x2s = r = y (•) 4

(c) x&1 = – 3x1 + 2x2 – 3x1 – 1.5x2 + w; x& 2 = 4x1 – 5x2 x2s (•) = y (•) =

1 w 7

(d) z& = y – r = x2 – r

LM x& OP L-3 MM x&z& PP = MMM 40 N Q N

2 0 -5 0 1 0

k = [2

3.5]

1 2

1.5

LM x& OP L-5 MM x&z& PP = MMM 40 N Q N 1 2

OP LM x OP LM1OP LM 0OP LM1OP PP MM xz PP + MM00PP u + MM-01PP r + MM00PP w QN Q NQ N Q NQ

0.5 -3.5 0 -5 1 0

1 2

OP LM x OP LM1OP LM 0OP PP MM xz PP + MM00PP w + MM-01PP r QN Q NQ N Q 1 2

Y ( s) 4s Y ( s) 14 = ; y(•) = 0; = ; W ( s) ( s + 1)( s + 2)( s + 7) R( s) ( s + 1)( s + 2)( s + 7) y(•) = r 7.21

w& = – 0.5w + 100u; u = – Kw + Nr; w& = (– 0.5 – 100K)w + 100 Nr (a) Time constant = 0.1 sec. s + 0.5 + 100K = s + 10; K = 0.095 N–1 = – c (A – bK)–1b = 10; N = 0.1; At steady-state w& Æ 0 . This gives w(•) = r (b) A = – 0.5; A + dA = – 0.6

w& = – 0.6w + 100u = – 0.6w + 100 (– 0.095w + 0.1r) = – 10.1w + 10r w(•) =

10 r 10.1

(c) z& = w – r

LMw& OP = LM-0.5 0OP LMw OP + LM100OP u + L 0O r N z& Q N 1 0Q N z Q N 0 Q MN -1PQ


A – bK =

LM- 0.5 - 100 K N 1

-100 K 2 0

1

OP Q

|sI – (A – bK)| = s2 + (0.5 + 100K1)s + 100K2 = s2 + 11s + 50 K1 = 0.105, K2 = 0.5

z t

u(t) = – 0.105w(t) – 0.5

(w ( t ) - r )dt

0

At steady-state, z& Æ 0; this gives w(•) = r.

7.22

LM 0 F= 0 MM-4 N

1 0 -2

OP LM0OP 1 ;g= 0 P MM1PP -1PQ NQ 0

FH

Desired characteristic equation: z z +

1 1 + j 2 2

IK FH z + 1 - j 1 IK 2 2 = z3 + z2 +

1 z 2

|zI – (F – gk)| = z3 + (1 + k3) z2 + (2 + k2) z + 4 + k1

LM N LM 1 2 F = M -1 MM 0 N

k = -4 -

7.23

3 2

0

OP Q

OP 0 1 P ; c = [1 0 0] P 0 0 PQ 1 |zI – (F – mc)| = z + FH m - IK z 1 0

3

2

1

2

+ (1 + m2)z + m3

FH

Desired characteristic polynomial: z z +

1 1 + j 2 4

IK FH z + 1 - j 1 IK 2 4 = z3 + z2 +

LM 3 - 11 0OP N 2 16 Q LM 0 1 0 OP LM0OP F= 0 MM- 0.5 - 00.2 111. PP ; g = MM01PP N Q NQ T

m=

7.24

5 z. 16

SOLUTION MANUAL

|zI – (F – gk)| = z3 + (k3 – 1.1) z2 + (k2 + 0.2) z + k1 + 0.5 Desired polynomial: z3. k = [– 0.5 – 0.2 1.1] 7.25 F =

LM 0 1 OP ; c = [1 N- 0.16 -1Q

1]

The current observer equations are: x (k + 1) = F x$ (k) + gu(k)

x$ (k + 1) = x (k + 1) + m[y(k + 1) – c x (k + 1)] x (k + 1) = (F – mcF) ~ x (k) State error equation is, ~ |zI – (F – mcF)| = z2 + (1 – 0.16m1) z + 0.16 (1 – m1 – m2) Desired characteristic polynomial: z2 m = [6.25 – 5.25]T

7.26

LM 0 1 0 OP LM0OP x(k + 1) = 0 MM-0.5 -00.2 111. PP x(k) + MM01PP u(k) N Q NQ LM x (k + 1)OP L 0 0 1 O LM x (k)OP L0O MM x (k + 1)PP = MMM 1 0 0 PPP MM x (k)PP + MMM0PPP u(k) N x (k + 1)Q N-0.2 -0.5 11. Q N x (k)Q N1Q LM x (k)OP y(k) = [1 0 0] M x ( k ) P MN x (k)PQ 2

2

1

1

3

3

2 1 3

~ x e (k + 1) = (Fee – mf1e) ~ x e (k) Fee =

LM 0 0 OP ; f N-0.5 11. Q

1e

= [0

1]

|zI – (Fee – mf1e)| = z2 + (m2 – 1.1) z – 0.5m1 Desired characteristic polynomial: z2 m = [0

1.1]T; x$ 2 (k) = x2(k) = y(k); x$ e (k) =

LM x$ (k ) OP N x$ (k )Q 1

3

Refer Eqns (7.91) – (7.97).

x$ e (k + 1) =

LM 0 0OP x$ (k) + LM0OP u(k) + LM 1 OP y(k) + LM 0 OP y(k + 1) N-0.5 0Q N1Q N- 0.2Q N11.Q e

89


LM 2 -1OP ; g = LM4OP N - 1 1 Q N 3Q L4 5 OP U = [g Fg] = M N3 -1Q 1 L p OP = 1 LM 3 - 4OP p = [3 – 4]; P = M 19 Np FQ 19 N10 -7Q 1 1 |zI – F| = z – 3z + 1; |zl – (F – gk)| = F z + j I F z - j I = z H KH K

7.27 (a) F =

1

1

1

2

k=

2

LM -3 3OP P = LM111 N 4 Q N 76

-18 19

2

OP Q

2

(b) x$ (k + 1) = (F – mc) x$ (k) + (g – md)u(k) + my(k); c = [1 1] z (k + 1) = FTz(k) + cTh(k); h(k) = mTz(k) U = [cT

P=

LM1 1OP ; p N1 0Q OP = LM1 -1OP Q N3 -2Q

FTcT] =

LM p Np F 1

T

1

1

= [1

– 1]

|zI – FT| = z2 – 3z + 1 |zI – (FT – mTcT)| = z2 mT = [– 1

3] P = [8

– 5]

(c) x(k + 1) = Fx(k) + gu(k); u(k) = – k x$ (k)

x$ (k + 1) = (F – mc) x$ (k) + gu(k) + my(k); y(k) = cx(k) + du(k) x1(k + 1) = 2x1(k) – x2(k) – 5.84 x$1 (k) + 3.79 x$ 2 (k) x2(k + 1) = – x1(k) + x2(k) – 4.38 x$1 (k) + 2.84 x$ 2 (k) and x$1 (k + 1) = – 11.84 x$1 (k) – 5.21 x$ 2 (k) + 8[x1(k) + x2(k)] x$ 2 (k + 1) = – 0.38 x$1 (k) + 8.84 x$ 2 (k) – 5[x1(k) + x2(k)]

7.28 (a)

Y (z) 1 = 2 U ( z ) z + z + 0.16 x(k + 1) = Fx(k) + gu(k); y(k) = cx(k)

+

1 4

SOLUTION MANUAL

F=

LM 0 N- 0.16

OP ; g = LM0OP ; c = [1 -1Q N1 Q 1

91

0]

(b) |zI – (F – gk)| = z2 + (1 + k2)z + 0.16 + k1 Desired characteristic equation: (z – 0.6 + j0.4) (z – 0.6 – j0.4) = z2 – 1.2z + 0.52 k = [0.36 (c) F =

–2.2]

LM 0 1OP ; g = LM0OP ; c = [1 N- 0.16 -1Q N1Q

0]

x$ 2 (k + 1) = Fee x$ 2 (k) + fe1 y(k) + geu(k) + m(y(k + 1) – f11 y(k)

– g1u(k) – f1e x$ 2 (k)) = (– 1 – m) x$ 2 (k) – 0.16y(k) + my(k + 1) + u(k) m = – 1 gives dead-beat response. x$ 2 (k + 1) = – 0.16y(k) – y(k + 1) + u(k)

(d) u(k) = – 0.36x1(k) + 2.2 x$ 2 (k) = – 0.36y(k) + 2.2 [– y(k) – 0.16y(k – 1) + u(k – 1)] = – 2.56y(k) – 0.352y(k – 1) + 2.2u(k – 1)

U(z ) 2.56(1 + 0.1375z -1 ) = (1 - 2.2 z -1 ) -Y ( z) 7.29 Let us transform c from [1 1] to [1 x(k) = Qq(k); Q =

LM1 N0

0].

OP Q

-1

1

q(k + 1) = Q–1 FQq(k) + Q–1 gu(k) = F q(k) + g u(k)

F =

LM 0 1OP ; g = LM0OP N-0.16 -1Q N1Q

y(k) = cQq(k) = c q(k) = [1

0] q(k)

Controller design: |zI – ( F – g k)| = z2 + (1 + k2)z + 0.16 + k1 Desired characteristic polynomial: z2 – 1.2z + 0.52; k = [0.36

– 2.2]


Observer design:

F =

LM 0 1OP ; g = LM0OP ; c = [1 N- 0.16 -1Q N1Q

0]

q$ 2 (k + 1) = Fee q$ 2 (k) + fe1 y(k) + geu(k) + m(y(k + 1) – f11y(k) – g1u(k) – f1e q$ 2 (k))

= – q$ 2 (k) – 0.16y(k) + my(k + 1) – mu(k) – m q$ 2 (k) q~2 (k + 1) = (Fee – mf1e) q~2 (k) = (– 1 – m) q~2 (k) m = – 1 gives dead-beat response. q$ 2 (k + 1) = – 0.16y(k) – y(k + 1) + u(k)

u(k) = – 0.36q1(k) + 2.2 q$ 2 (k) = – 2.56y(k) – 0.352y(k – 1) + 2.2u (k – 1)

U(z ) 2.56(1 + 0.1375z -1 ) = -Y ( z) 1 - 2.2 z -1 7.30 Double integrator plant: x(k + 1) = Fx(k) + gu(k); y(k) = cx(k)

LM1 T OP ; g = LM T / 2 OP ; c = [1 N0 1 Q N T Q 2

F=

Regulator: z = 0.5; wn = 4, z1,2 = e

0] -zw n T ± jw n T 1-z 2

e

= 0.77 ± j0.278

2

Characteristic polynomial: z – 1.54z + 0.67

LM N

|zI – (F – gk)| = z2 + Tk2 + k = [13

OP Q

T2 T2 k1 – Tk2 + 1 k1 - 2 z + 2 2

3.95]

Prediction observer: Result follows from Example 7.14.

LM 2 OP ; U(z) = k [zI – (F – gk – mc)] m = 65.5( z - 0.802) z + 0.46 z + 0.26 N10Q -Y(z) L1 0.0952OP x(k) + LM0.00484OP u(k); y(k) = [1 0] x(k) x(k + 1) = M N0 0.905 Q N 0.0952 Q

m=

7.31

–1

2

SOLUTION MANUAL

L1 f(F) = M N0 U = [g

OP ; 0.905 Q L0.00484 Fg] = M N 0.0952 0.0952

93

2

OP 0.0862Q 0.0139

Using Ackerman’s formula: k = [0

1]U–1 f(F) = [105.01

14.648]

Observer design: z (k + 1) = FTz(k) + cTh(k); h(k) = mTz(k) f(FT) =

LM 1 N0.0952

OP Q

0 0.905

LM1 N0

2

1 0.0952

U = [cT

FTcT] =

mT = [0

1] U–1 f(FT) = [19

OP Q 8.6]

Reduced order dead-beat observer: F=

LM1 N0

OP Q

LM N

OP Q

0.0952 0.00484 ;g= ; c = [1 0.905 0.0952

0]

~ x 2 (k + 1) = (Fee – mf1e) ~ x 2 (k); Fee = 0.905, f1e = 0.0952

Desired characteristic equation: z = 0; m = 9.51 Refer Equations (7.91)– (7.97)

x$ 2 (k + 1) = 9.51y(k + 1) – 9.51y(k) + 0.05u(k) 7.32 (a) y& + y = u + w; T = 1 sec; A = – 1, b = 1;

z T

AT

e

= 0.368; g =

e -t dt = 0.632

0

y(k + 1) = 0.368y(k) + 0.632u(k) + 0.632w(k) (b) Time constant = 0.5 sec; z =

-T et

= 0.135

Desired characteristic equation: z – 0.135 = 0 F – gk = 0.368 – 0.632K |z – (F – gk)| = z – (0.368 – 0.632K) = 0 Comparison with desired characteristic equation gives, K = 0.3687


1 = – c (F – gk – 1)–1 g; N = 1.37 N With u(k) = – 0.3687y(k) + 1.37r we have y(k + 1) = 0.135y(k) + 0.866r + 0.632w(k) At steady state, with w = 0; ys = 0.135ys + 0.866r; ys =

0.866 r @r 0.865

(c) At steady state, with r = 0; ys = 0.135ys + 0.632w; ys =

0.632 w = 0.73w 0.865

(d) Integral control: z = 0.5, wn = 4 re±jq = e

-zw n T ± jw n T 1-z 2

e

= – 0.128 ± j0.043

(z + 0.128 + j0.043) (z + 0.128 – j0.043) = z2 + 0.256z + 0.018 Desired characteristic equation: z2 + 0.256z + 0.018 v(k) = v(k – 1) + y(k) – r

LM y(k + 1)OP = LM0.368 0OP LMy(k)OP + LM0.632OP u(k) + LM0.632OP w(k) + L 0O r Nv(k + 1)Q N0.368 1Q Nv(k)Q N0.632Q N0.632Q MN -1PQ L y - y OP ; ~x (k + 1) = F ~x (k) + gu~ (k) ~ x =M Nv - v Q L0.368 0OP ; g = LM0.632OP ; F =M N0.368 1Q N0.632Q s

s

|zI – F | = z2 + 0.256z + 0.018

Fg]=

U = [g

p1 = [– 1.58 P=

LM0.632 N0.632

1

k = [– 0.35

OP Q

1.58]

LM p OP = LM-1.58 Np F Q N 0 1

0.2326 0.8646

OP Q

1.58 1.58

1.624]P = [0.553

2.013]

SOLUTION MANUAL

(e) F – g k =

LM0.018 N0.018

- 1.27

OP Q

- 0.27

With w = 0,

LM y OP = LM0.018 Nv Q N0.018 s

s

OP LM y OP + L 0O r - 0.27Q N v Q MN -1PQ - 1.27

s

s

Solving the above set, we get ys = r With r = 0:

LM y OP = LM0.018 Nv Q N0.018 s

s

OP LM y OP + LM0.632OP w - 0.27Q N v Q N0.632Q - 1.27

s

s

Solving the above set, we get ys = 0.

95

CHAPTER 8

LYAPUNOV STABILITY ANALYSIS

LM 0 1OP . Let us take Q = LM4 0OP = LM2OP [2 N -1 -2 Q N0 0 Q N 0 Q L H OP = r LM2 0OP = 2 rM N H A Q N0 2 Q

0] = HHT

8.1 A =

T

T

ATP + PA = – Q gives P =

LM5 2OP N2 1 Q

Since P is positive definite, the system is asymptotically stable. Also, V(x) = xTPx Æ • as ||x|| Æ • Therefore the system is asymptotically stable in-the-large.

LM 0 1OP ; A P + PA = – I N-1 1Q LM -3 Solving for P, we get: P = M 2 MN 12

8.2 A =

T

OP PP Q

1 2 -1

P is not a positive definite matrix; the system is unstable. 8.3 x1 = x2, x 2 = – x1 – x2 + 2 Equilibrium state xe =

LM x OP Nx Q e 1 e 2

0 = x 2e ; 0 = – x1e – x 2e + 2; xe =

LM2OP N0 Q

x Let ~ x = A ~ x1 = x1 – 2, ~ x 2 = x2; then ~

LM 3 0 1O L 2 A= M N-1 -1PQ ; A P + PA = – I gives P = MM 1 N2 T

1 2 1

OP PP Q

P is positive definite matrix. Therefore equilibrium state [2 asymptotically stable.

0] T is

SOLUTION MANUAL

8.4

L- 4 K A= M N 4K

L7 OP ; A P + PA = – I; P = MM 40 K - 6KQ MN 101K 4K

T

1 10 K 3 20 K

97

OP PP Q

P is positive definite for K > 0; the system is asymptotically stable for K > 0. 8.5 From Fig. P8.5, we have x1 = – x1– Kx3; x 2 = x1 – x2; x 3 = x2 – x3

LM-1 x = Ax; A = 1 MM 0 N L0 0 Take Q = M0 0 MM0 0 N

OP PP Q

0 -K 0 ; ATP + PA = – Q; -1 1 -1

OP P 2 PQ 0

0 = HHT; HT = [0

0

2]

From Lyapunov equation, we obtain,

1 P= ( K + 1)( K - 8)

LM - 3 MMK--32 N

K -2 -3 - ( K + 4) - ( K + 4) - ( K + 4) - (5K + 8)

For P to be positive definite,

OP PP Q

3( K + 4) - 9 -3 > 0; > 0, ( K + 1) 2 ( K - 8) 2 ( K + 1)( K - 8)

and

1 ( K + 1)( K - 8)

LM - 3 MMK--32 N

-3 - ( K + 4) - ( K + 4)

OP >0 P - (5 K + 8)PQ K -2

- ( K + 4)

These conditions reduce to the following: (K + 1) (K – 8) < 0

or

K > – 1; K < 8

K>–1 (K + 1)2 (K – 8) < 0; K < 8 Therefore, – 1 < K < 8 will guarantee system stability. It can easily be examined that Routh criterion gives the same result.

98


8.6 F =

LM 0.5 1OP ; F PF – P = – I gives P = LM 3.5 N -1 -1Q N2.25

2.25

T

OP Q

3.875

P is a positive definite matrix. Therefore system is asymptotically stable.

LM 0 0.5OP ; F PF – P = – I, gives, N- 0.5 -1Q 40 O LM 52 27 27 P P= M 40 100 P MN 27 27 PQ

8.7 F =

T

P is a positive definite matrix. Therefore, the system is asymptotically stable. 8.8 x1 = x2; x 2 = – (1 – |x1|) x2 – x1 Take V(x) = x12 + x 22 Then, V (x) = 2x1 x1 + 2x2 x 2 = – 2 x 22 (1 – |x1|) For V (x) to be negative definite, (1 – |x1|) > 0 for x2 π 0 or |x1| < 1 V (x) is not identically zero anywhere other than the origin. Therefore, for |x1| < 1, origin is the equilibrium state and is asymptotically stable.

8.9 x1 = x2; x 2 = – x1 – x12 x2 V(x) = x12 + x 22 ; V(x) Æ • as ||x|| Æ • V (x) = – 2 x12 x 22 < 0

Therefore origin is the equilibrium state and is asymptotically stable inthe-large. 8.10 x1 = – 3x1 + x2; x 2 = x1 – x2 – x 23 ; J = Take P = I. Then Q = – (JT + J) =

LM 6 N-2

LM-3 N1

1 -1 - 3 x 22

-2 2 + 6 x 22

OP Q

OP Q

|Q| = 36 x 22 + 8 > 0 Q is positive definite; therefore origin is asymptotically stable.

CHAPTER 9

LINEAR QUADRATIC OPTIMAL CONTROL

9.1 The system of Fig. P9.1 is governed by the differential equation: y + 2z y + y = 0 Defining state variables as x1 = y, x2 = y , we obtain the following state equations for the system, x =

LM 0 N-1

z

OP Q

•

J=

LM OP NQ L2 0OP ; R = 0; K = 0 2 x (t)dt; Q = M N0 0 Q

1 1 x = Ax; x(0) = -2z 0 1 2

e2(t)dt =

0

z

•

0

2 1

The Lyapunov equation becomes, ATP + PA + Q = 0

LM2z + 1 2z Solving for P, we get: P = M MM 1 N J = xT(0) Px(0) = z +

OP 1 P P 2z PQ 1

1 4z

dJ 1 d2J =1– = 0; z = 0.5; > 0; Jmin = 1 dz 4z 2 dz 2 9.2 The closed-loop transfer function

Y ( s) K = 2 R ( s) s + a s + K

y + a y + Ky = Kr; y(0) = y (0) = 0

e + a e + Ke = 0; x1 = e, x2 = e e = r – y; x =

LM 0 N- K

z

•

J=

0

OP Q

LM OP NQ

1 1 x; x(0) = -a 0

z

•

e2(t)dt =

x12 (t)dt; Q =

0

LM2 0OP ; R = 0; K = 0 N0 0 Q

Lyapunov equation: ATP + PA + Q = 0 Solving for P, we get


LM a + 1 P = MK a MN K1

OP PP Q

1 K ; J = 1 xT (0) Px(0) = a + 1 1 2 2 K 2a Ka

(i) K = constant:

K,

∂2 J >0 ∂a 2

I = 0; K Æ • K L0 1OP ; B = LM0OP ; K = [1 k]; Q = LM1 0OP ; A=M N0 0 Q N1 Q N0 1 Q L 0 1 OP ; x(0) = LM1OP R = 0; (A – BK) = M N-1 - k Q N1Q

(ii) a = constant:

9.3

∂J 1 1 = ;a= ∂a 2 K 2a 2

F H

1 ∂J a = - 2 K ∂K 2

(A – BK)TP + P(A – BK) + Q = 0 Solving for P we get,

LM k + 2 P = M 2k MN 12 2

OP PP Q

1 2 2 ; J = 1 xT (0) Px (0) = k + 2k + 4 1 4k 2 k

∂J 3 = 0 gives k = 2; Jmin = 2 ∂k |lI – (A – BK)| = l2 + 2l + 1; l1 = – 1, l2 = – 1. The system is stable. For k = 1.5, J = 1.54. For k = 2, J = 1.5 Skopt =

9.4

0.04 / 1.5 DJ / J = = 0.107 0.5 / 2 Dk / k

Y (s ) 100 = G(s) = 2 ; y = 100u; y(0) = y (0) = 0 U ( s) s x1 = y, x2 = y ; x =

LM0 1OP x + LM 0 OP u N0 0Q N100Q

u = r – y(t) – K y (t) Let ~ x1 = y – r = x1 – r, ~ x 2 = y = x2

SOLUTION MANUAL 101

x1 – Kx~2 ; K = [1 Then u = – ~

K]

LM OP N Q

LM OP N Q

2 0 -1 ~ + Bu; ~ ~ x (0) = ;Q= ;R=0 x = Ax 0 0 0 (A – BK)TP + P(A – BK) + Q = 0 Solving for P, we get,

LM100K + 1 P = M 100 K MN 1001 2

J=

1 100 1 10 4 K

OP PP Q

100 K 2 + 1 ∂ J 1 ~T x (0) = ; = 0 gives K = 0.1; x (0) P ~ 200 K ∂K 2 Jmin = 0.1,

∂2 J >0 ∂K2

For K = 0.1, J = 0.1. For K = 0.09, J = 0.100556 SKopt =

0.000556 / 0.1 DJ / J = = 0.0556 0.01 / 0.1 DK / K

LM0 1OP ; B = LM0OP ; K = [k N0 0 Q N1 Q L0 Closed-loop system: x = M N- k

9.5 A =

k2]; (A – BK) =

1

1

LM 0 N-k

1

OP x or x -k Q 1

1

OP -k Q 1

2

= x2; x 2 = – k1x1 – k2x2

2

Eliminating x2 from this equation yields x1 + k2 x1 + k1x1 = 0 Since the undamped natural frequency is specified as 2 rad/sec, we obtain k1 = 4. Therefore, (A – BK) =

LM 0 N- 4

T

OP ; R = 0; Q = LM1 0OP ; x(0) = LM1OP -k Q N0 1 Q N0 Q 1

2

(A – BK) P + P(A – BK) + Q = 0 Solving for P, we get

LM 5 2k P= M MM N

+ 2

1 8

k2 8

OP PP PQ

LM N

OP Q

1 k 1 T 1 5 ∂J 8 + 2 ; ;J= x (0) Px(0) = =0 5 2 2 2 k2 ∂ k2 8 8 k2


20 ;

gives k2 =

∂2 J >0 ∂ k 22

For this value of k2 : Jmin =

LM0 N0 L2 Q= M N0

9.6 A =

5 4

OP ; B = LM0OP ; K = [k k]; (A – BK) = LM 0 0Q N1 Q N- k 0O 1 P2Q ; R = 0; x(0) = LMN0OPQ 1

1 -k

OP Q

(A – BK)TP + P (A – BK) + Q = 0 Solving for P we get

LM1 + 2k P= M k MN 1k

OP PP Q

1 1 1 ∂J k ; J = xT (0) Px (0) = 1 + ; =0 2( k + 1) 2 2k ∂ k 2k 2

gives k Æ •

9.7 (A – BK)TP + P(A – BK) + KTRK + Q = 0; Q =

LM2 0OP ; R = 2 N0 2 Q

Solving Lyapunov equation, we get

LM (k + 1) P= M k MM 1 + k N k

OP PP PQ

1+ k2 (1 + k ) 2 ∂ J k ;J= ; = 0 gives k = 1; Jmin = 2 2 (1 + k )(1 + k ) ∂k 2k k2

2

2

For k = 1, J = 2. For k = 0.9, J = 2.0055. Skopt =

9.8 A =

0.0055 / 2 DJ / J = = 0.0275 0.1 / 1 Dk / k

LM0 1OP ; B = LM0OP ; u = – Kx; Q = LM2 0OP ; R = 2 N0 0 Q N1 Q N0 2 Q

{A, B} pair is controllable, and Q is positive definite. Matrix Ricatti equation is: ATP + PA – PBR–1BTP + Q = 0 Solving for P, we get, P=

LM2 3 N2

2 2

OP ; K = R 3Q

–1

BP = [1

3]

SOLUTION MANUAL 103

9.9 The matrix Ricatti equation is: ATP + PA – PBR–1BTP + Q = 0 Q=

LM2 0OP = H H = LM 2 OP [ N0 0 Q N0Q T

2

0]; r

LM H OP = 2 NHAQ

{A, H} pair is observable. Sufficient conditions satisfied. Solving the Ricatti equation, we get P=

LM2 2 N2

2 2

OP ; K = R 2Q

–1

BP = [1

2 ];

|lI – (A – BK)| = l2 + 2 l + 1 The roots of this equation are in the left half of the complex plane. The optimal closed-loop system is therefore stable. 9.10 A =

LM0 0OP ; B = LM1OP ; Q = LM2 0OP ; R = 2 N0 1Q N1Q N0 0Q

{A, B} pair is controllable. Q = HTH =

LM 2 OP [ N0Q

2

0]

{A, H} pair is not observable. Sufficient conditions not satisfied. Solving the Ricatti equation, ATP + PA – PBR–1BTP + Q = 0, we get P=

LM 6 N-8

OP Q

-8 ; K = R–1BTP = [– 1 4] 16

|lI – (A – BK)| = l2 + 2l + 1 Optimal closed-loop system is asymptotically stable: u = – Kx = – [– 1 4]x 9.11 A =

LM-1 0OP ; B = LM1OP ; Q = LM2 0OP ; R = 2 N 1 0 Q N0 Q N0 0 Q

{A, B} pair is controllable. Q = HTH =

LM 2 OP [ N0Q

2

0]

{A, H} pair is not observable. Sufficient conditions not satisfied. Solving Ricatti equation, A TP + PA – PBR–1BTP + Q = 0, we get

1 2 1 -1 2 p11 + 2 = 0; – p12 + p22 – p11p12 = 0; p12 = 0 2 2 2 This gives p11 = p22 = 0. Therefore a positive definite solution of the Ricatti equation is not possible. Hence an optimal solution to the control problem does not exist.

2(– p11 + p12) –


9.12 x =

LM0 N0

OP LM OP Q N Q

0 1 1 x+ u; y = x1; J = -2 20 2

z

•

[y(t) – 1]2 + u2dt; yd = r = 1

0

Let ~ x1 = x1 – r = x1 – 1; ~ x 2 = x2. Then

LM N

OP Q

LM OP N Q

0 0 1 ~ ~ x + x + Bu; u = A~ x = 0 -2 20 J=

z

LM OP N Q

• 1 0 1 ~2 ( x1 + u2)dt; Q = ;R=1 20 0 0

{A, B} pair is controllable Q=

LM1OP [1 N0 Q

0] = HTH. {A, H} pair is observable.

ATP + PA – PBR–1BTP + Q = 0 Solving for P, we get

LM 6.63 P = M 20 MN 0.05

OP 4.63 P ; K = R P 400 Q 0.05

–1 T

B P = [1

0.23]

x = – x1 – 0.23x2 + r u = –K ~ 9.13 A =

LM0 1OP ; B = LM0OP ; C = LM1 0OP N0 0 Q N1 Q N0 2 Q

z

z

•

•

1 1 1 2 ( y12 + y 22 + u2)dt = 2 (yTy + u2)dt = J= 2 2 2 0

Q=

0

LM2 0OP ; R = 2 N0 8 Q

z

•

(xTCTCx + u2)dt

0

{A, B} pair is controllable. Q is positive definite. Solving Ricatti equation, A TP + PA – PBR–1BTP + Q = 0, we get

LM 24 2 OP ; K = R B P = [1 6 ]; u = – x N 2 24 Q L0 1OP ; B = LM0OP ; Q = LM2 0OP ; R = 2 A= M N0 -1Q N1Q N0 2Q

P=

9.14

–1 T

1

–

6 x2

Solving Ricatti equation, A TP + PA – PBR–1BTP + Q = 0, we get

SOLUTION MANUAL 105

P=

LM 4 2OP ; K = R N2 2Q

–1 T

B P = [1

Equation of the state observer:

1]

LM OP N Q

m1 ; C = [1 x = (A – MC) x + Bu + My: M = m2

0]

|lI – (A – MC)| = l2 + (1 + m1)l + (m1 + m2) = 0 If the poles of the observer are to lie at s = – 3, the desired characteristic equation is l2 + 6l + 9 = 0. Comparing the coefficients, we get m1 = 5, m2 = 4; u = – x1 – x 2

LM OP NQ

5 x = A x + Bu + M[y – C x ]; M = 4

9.15 State variable description of the system, obtained from Fig. P9.15, is given by x =

LM0 N0

OP x + LM0OP u; x(0) = LM0OP ; y = [1 -5 Q N1 Q N0 Q 1

0]x

In terms of the shifted state variables, ~ x1 = x1 – qr; ~ x 2 = x2, the model becomes

LM N

LM OP NQ

OP Q

0 0 1 ~ ~ x + x + Bu u = A~ x = 0 -5 1 x , that miniThe problem is to determine optimal control law, u = – K ~ mizes

z FH •

J=

0

I K

1 ~ x12 + u 2 dt 2

From this J, we obtain the following matrices: Q=

LM2 0OP ; R = 1 N0 0 Q

The pair {A, B} is controllable. Q = HTH =

LM 2 OP [ N0Q

2

0]. The pair {A, H} is observable.

Solving the matrix Ricatti equation, ATP + PA – PBR–1BTP + Q = 0, we get


P=

LM7.495 N 2

OP 0.275Q 2

The optimal gain matrix is K = R–1BTP = [ 2 The matrix (A – BK) is stable.

0.275]

x (0) = [– 5 The minimum value of J for an initial condition ~ J0 = 9.16 A =

0]T is,

1 ~T ~( 0 ) = 93.2375 x ( 0)Px 2

LM0 1 OP ; B = LM0OP ; Q = LM2 0OP ; R = 2r. N0 -5Q N1Q N0 0Q

Solving Ricatti equation, A TP + PA – PBR–1BTP + Q = 0, we get

LM F 2 I 2r GH 25 + r JK M P= M MM 2r NM 1/ 2

1/ 2

Case (i): r = 0.1 P=

LM3.5397 N0.6325

2 r 1/ 2

F GH

2 r - 5 + 25 +

OP PP 2 IP JP r KP Q

OP Q

0.6325 ; K = R–1BTP = [3.1623 0.5968] 0.1194

Poles: – 0.6377, – 4.9592 Case (ii): r = 0.01 P=

LM1.3416 N 0.2

OP Q

0.2 ; K = [10 0.0342

1.7082]

Poles: – 2.2361, – 4.4721 Case (iii): r = 0.001 P=

LM0.5941 N0.0632

OP Q

0.0632 ; K = [31.6228 0.0088

4.3939]

Poles: – 4.6970 ± j3.0921

x + Bu; u = –K ~ x = – [k1 k2] ~ x 9.17 ~ x = A ~ The constraint of partial state feedback can be met by setting k2 = 0 in the optimization problem. For the problem under consideration, the Lyapunov equation is, (A – BK)TP + P(A – BK) + KTRK + Q = 0

SOLUTION MANUAL 107

Solving for P we get,

LM (25 + k )(2 + k ) 10k P= M 2 MM + k N 2k 2 1

1

1 2 1

1

2 + k12 2 k1 2 + k12 10 k1

OP PP PQ

x (0) = [– 5 For an initial condition ~ value: J=

0]T, the performance index has the

25( 25 + k1 )( 2 + k12 ) ∂ J 1 ~T x (0) = ; = 0 gives k1 = 1.345; x (0)P ~ 20k1 2 ∂ k1

∂2 J >0 ∂ k12 Jmin = 93.26 The matrix (A – BK) is stable. 9.18 From Fig. P9.18, we obtain, y = – y + u + w; y(0) = 0

Let ys and us be steady-state values (assuming w = 0 for the time being). At steady state: 0 = – ys + us The steady-state equation can now be expressed as, ~ y = – ~ y + u~ , where ~ y = y – ys, u~ = u – us; J =

z

•

(~ y 2 + u~ 2 )dt

0

(a) A = – 1, B = 1, Q = 2, R = 2 ATP + PA – PBR–1BTP + Q = 0, gives P2 + 4P – 4 = 0 P = 2 2 – 2; K = R–1BTP = 2 – 1 (b) u – us = – K(y – ys) u = – Ky + Nr; Kys + us = Nr N –1 = – C (A – BK)–1 B = (c) From Fig. P9.18, Y(s) =

1 ;N= 2

2

Y ( s) 1 1 1 = = ; W(s) = ; s W ( s) s + 1 + K s + 2

1 s( s + 2 )

Steady-state error due to disturbance: lim sY (s) = s Æ0

1 . 2


(d) y = – y + u; z = y – r

LMyOP = L-1 0O L yO + L1 O u + L 0O r NzQ MN 1 0PQ MN z PQ MN0PQ MN-1PQ

At steady state y = z = 0, and therefore

LM 0OP r = – LM-1 0OP LM y OP + LM1OPu N-1Q N 1 0Q N z Q N0Q s

s

s

Substituting in state equation, we get

LM~y OP = L-1 0O L~y O + L1Ou~ N~z Q MN 1 0PQ MN~z PQ MN0PQ

~ z = z – zs; u~ = u – us y = y – ys; ~

z

•

J=

(~ y2 + ~ z 2 + u~ 2 ) dt

0

A=

LM-1 0OP ; B = LM1OP ; Q = LM2 0OP ; R = 2 N 1 0 Q N0 Q N0 2 Q

This shifted regulator problem has already been solved in Example 9.8. The matrix, K = [1 1]

z t

Therefore, u(t) = – y(t) –

[y(t) – r]dt

0

(e) Block diagram of the control scheme is shown in the figure that follows:

Block diagram manipulation gives (for r = 0)

Y ( s) s 1 1 = ; W(s) = ; Y(s) = 2 s ( s + 1) 2 W ( s) ( s + 1) Steady-state error: lim sY (s) = 0 s Æ0

SOLUTION MANUAL 109

9.19 (a) w = – 0.5w + 100u Let ws and us be the steady-state values. At steady state: 0 = – 0.5ws + 100us The state equation can now be expressed as:

~ + 100 u~ ~ = – 0.5w w ~ = w – w ; u~ = u – u where, w s s

z

•

J=

~ 2 + 100u~ 2 ) dt (w

0

A = – 0.5, B = 100, Q = 2, R = 200 Solving Ricatti equation for P: ATP + PA – PBR–1BTP + Q = 0, we get – P – 50P2 + 2 = 0. This gives P = 0.19; K = R–1BTP = 0.095. ~ ; u = – 0.095w + 0.095w + u ; u + 0.095w = Nr Now, u~ = – 0.095w s

s

s

s

N –1 = – C(A – BK)–1 B = 10, N = 0.1 The control law, therefore, is u = – 0.095w + 0.1r The closed-loop system becomes: w = – 0.5w + 100u = – 10w + 10r At steady-state: w = 0, w (•) = r (b) A = – 0.5, A + dA = – 0.6

w = – 0.6w + 100u = – 0.6w + 100(– 0.095w + 0.1r) = – 10.1w + 10r At steady-state: w (•) =

10 r 10.1

(c) z = w – r

LMw OP = LM- 0.5 0OP LMw OP + LM100OP u + L 0O r N z Q N 1 0Q N z Q N 0 Q MN -1PQ

At steady-state: w = 0, z = 0. Therefore,

LM 0OP r = – LM- 0.5 0OP LMw OP – LM100OP u N-1Q N 1 0Q N z Q N 0 Q s

s

s

where ws, us, zs are steady-state values. Substituting in the state equations,

LMw~ OP = L- 0.5 0O Lw~ O + L100O u~ MN ~z PQ MN 1 0PQ MN ~z PQ MN 0 PQ

~ = w – w , ~z = z – z , u~ = u – u w s s s


z

•

J=

0

B=

LM OP N Q

LM N

OP Q

2 0 - 0.5 0 ~2 + ~ (w z 2 + 100u~ 2 ) dt; Q = ; R = 200; A = ; 0 2 1 0

LM100OP N0Q

ATP + PA – PBR–1BTP + Q = 0, gives P=

LM0.21 N 0.2

0.2 2.2

OP ; K = R Q

–1 T

B P = [0.105 0.1]

z t

u(t) = – 0.105w(t) – 0.1 [w (t) – r]dt 0

The control scheme is shown in figure that follows.

For G(s) =

100 , the closed-loop transfer function is: s+a

10.5 w( s ) = s( s + 10 + a) + 10.5 R( s) For R(s) =

1 , we have w (•) = lim sw ( s) = 1 s Æ0 s

Therefore, for A + d A = a, the steady-state error is zero. 9.20 F = 1, G = 1, Q = 2, R = 1.5 From Lyapunov equation, (F – GK)TP(F – GK) – P + KTRK + Q = 0, we get P=

2 + 1.5 K 2 2K - K 2

J=

1 + 0.75K 2 1 T x (0) Px (0) = {x(0)}2 2 2K - K 2

∂J 2 = 0 gives K = . 3 ∂K

SOLUTION MANUAL 111

For K = 0.67, J = 1.499. For K = 0.57, J = 1.526. SKopt =

0.027 / 1.499 DJ / J = = 0.1207 0.1 / 0.67 DK/ K

9.21 x(k + 1) = 0.368x(k) + 0.632u(k) •

J=

Â [x2(k) + u2(k)]; F = 0.368, G = 0.632, Q = 2, R = 2

k= 0

From matrix Ricatti equation, P = Q + FTPF – FTPG(R + GTPG)–1GTPF, we get 0.3994P2 + 0.9304P – 4 = 0 P = 2.207; K = (R + GTPG)–1GTPF = 0.178; u(k) = – 0.178x(k) 9.22 Gh0G(z) = (1 – z–1) Z For T = 0.5, Gh0G(z) =

RS 1 UV = 1 - e T s(s + 1) W z - e

-T -T

Y (z) 0.4 = ; y(k + 1) = 0.6y(k) + 0.4u(k) z - 0.6 U ( z )

(a) Let ys and us be the steady-state-values. At steady-state, ys = 0.6ys + 0.4us Substituting in the state equation, ~ y (k + 1) = 0.6 ~ y (k) + 0.4u~ (k) where ~ y = y – ys, u~ = u – us •

J=

Â

1 ~ y 2 ( k ) + u~ 2 (k); F = 0.6, G = 0.4, Q = 1, R = 1 2 k=0

The Ricatti equation P = Q + FTPF – FTPG(R + GTPG)–1GTPF gives, 0.16P2 + 0.48P – 1 = 0; P = 1.415 K = (R + GTPG)–1GTPF = 0.277 (b) The closed-loop system is described by the equation, y(k + 1) = 0.6y(k) + 0.4[0.277 (r – y(k)] = 0.4892y(k) + 0.1108r At steady-state, y(•) = ys =

0.1108 r = 0.217r 0.5108

(c) N–1 = – C(F – GK – I )–1G = – (0.6 – 0.4 × 0.277 – 1)–1 × 0.4; N = 1.277 The closed-loop system is, y(k + 1) = 0.6y(k) + 0.4[– 0.277y(k) + 1.277r] = 0.4892y(k) + 0.5108r


0.5108 r=r 0.5108 9.23 (a) x(k + 1) = 0.5x(k) + 2u(k) = Fx(k) + Gu(k) Let xs and us be the steady-state values. At steady state, xs = 0.5xs + 2us Substituting in the state equation, we get ~ x (k + 1) = 0.5 ~ x (k) + 2 u~ (k) y(•) =

x (k) = x(k) – xs , u~ (k) = u(k) – us where ~ •

Â

1 ~ x 2 ( k ) + u~ 2 ( k ) ; F = 0.5, G = 2, Q = 1, R = 1 J= 2 k=0 The matrix Ricatti equation, P = Q + FTPF – FTPG(R + GTPG)–1GTPF yields, 4P2 – 3.25P – 1 = 0; P = 1.0505; K = (R + GTPG)–1 GTPF = 0.2 x (k) (b) u~ (k) = – K ~ u(k) = – Kx(k) + Kxs + us = – Kx(k) + Nr where, Kxs + us = Nr N–1 = – C(F – GK – I)–1G; N = 0.45 u(k) = – 0.2x(k) + 0.45r The closed-loop system becomes: x(k + 1) = 0.1x(k) + 0.9r. At steady-state, x(•) = r (c) Let F + dF = 0.3 x(k + 1) = 0.3x(k) + 2u(k) then, x(k + 1) = 0.3x(k) – 0.4x(k) + 0.9r = – 0.1x(k) + 0.9r At steady state, x(•) =

0.9 r 11 .

(d) Add to the plant equation, an integrator equation, v(k) = v(k – 1) + x(k) – r where v(k) is the state of the integrator. A control law of the form, u(k) = – K1x(k) – K2v(k) will provide the required robustness. Refer Fig. 7.17 for the control structure.

CHAPTER 10

NONLINEAR CONTROL SYSTEMS

10.1 Describing functions of some entries of Table 10.2 have been derived in Section 10.4 and Section 10.11. 10.2 Revisiting Example 10.1 (Fig. 10.19) will be helpful. G(jw) =

-pE 4 1 ;= 2 4M N ( E) jw (1 + jw )

– 90∞ – 2 tan–1 w1 = – 180∞ This gives w1 = 1 rad/sec

-

1 N E1

a f

= |G(jw1)| = 2

This gives E1 = 8M/p

8M sin t p 10.3 Revisiting Example 10.1 (Fig. 10.22) will be helpful. y(t) = – e(t) = –

G(jw) =

5 4 ; N(E) = 2 pE jw (1 + j 0 .1w )

1-

FH 0 .1 IK E

2

– 90∞ – 2 tan–1 0.1w1 = – 180∞ This gives w1 = 10 rad/sec. For a stable limit cycle, 2 D < E < • ; D = 0.1

–

1 N E1

a f

= |G(jw1)| = 0.25

This gives E1 = 0.3 Amplitude of limit cycle is 0.3 and frequency is 10 rad/sec. 10.4 Revisiting Example 10.1 (Fig. 10.22) will be helpful. For a limit cycle to exist,

2 D < E < • ; D = 0.2

–G(jw1) = – 180∞ for w1 = 10.95. |G(jw1)| = 0.206 |G(jw1)| π -

1 for any value of E N ( E)

The intersection of the two curves does not occur; therefore no limit cycle exists. Intersection of the two curves occurs for

114


pD < 0.206 or D < 0.131 2 10 10.5 G(jw) = (1 + j 0 . 4w ) (1 + j 2w ) 4 pE

N (E) =

LM MN

1-

FH H IK E

2

- j

OP PQ

H ; H = 0.2 E

The following figure shows G(jw)-plot and

-1 locus. N ( E)

Im

Re pH 4

E increasing

-1 N(E)

E=H G( jw)

At the point of intersection of G(jw)-plot with

-1 locus, measure the N(E)

angle of G(jw). This comes out to be –115.7∞. –G(jw1) = – tan–1 0.1w1 – tan–1 2w1 = – 115.7∞ This gives w1 = 5.9 rad/sec |G(jw1)| = 0.33

1 N E1

a f

-

=

p E1 = 0.33 4

This gives E1 = 0.42 10.6 G( jw) =

N(E) =

2 p

K jw (1 + jw ) 2

LM p - sin MN 2

-1

1 1 E E

1-

FH 1 IK E

2

OP PQ

SOLUTION MANUAL 115

=1-

LM N

2 1 1 + sin -1 p E E

1-

OP = 1 – N (E) Q

1 E2

c

The function Nc(E) is listed in Table 10.3. For any K > 0, one of the following can happen. (i) G(jw)-plot does not intersect

-1 locus N(E)

-1 locus; but the point of intersection N(E) corresponds to an unstable limit cycle.

(ii) G(jw)-plot intersects the

10.7 Revisiting Review Example 10.3 (Fig. 10.40) will be helpful. G(jw) =

N(E) =

10 jw 1 + jw 1 + j 0 .5w

a 2 L sin p MN

fa

-1

1 1 + E E

f

1-

1 E2

OP = N (E) Q c

The function Nc(E) is listed in Table 10.3. –G(jw1) = – 180° gives w1 = |G(jw1)| =

-1 N E1

a f

2

gives E1 = 4.25

The limit cycle with amplitude 4.25 and frequency 2 rad/sec is a stable limit cycle. 10.8 Revisiting Example 10.2 (Fig. 10.24) will be helpful. G(jw) =

K jw (1 + jw ) (1 + j 0 . 5w )

-1 locus. For K = 2, two N(E) intersection points are found. One corresponds to unstable limit cycle and the other corresponds to a stable limit cycle of amplitude 3.75 and frequency 1 rad/sec.

For K = 1, G(jw)-plot does not intersect

10.9 (a) Characteristic equation has two real, distinct roots in the left half of s-plane. The origin in the (y, y ) plane is a stable node. (b)

d 2 ( y - 1) d ( y - 1) + 6(y – 1) = 0 +5 2 dt dt

116


The singular point is located at (1, 0) in the (y, y ) plane. The characteristic equation is s2 + 5s + 6 = 0 = (s + 3) (s + 2) The singular point is a stable node. (c)

a

f

a

f

d2 y - 2 d y-2 + 17(y – 2) = 0 -8 2 dt dt The singular point is located at (2, 0) in the (y, y ) plane. The characteristic equation is s2 – 8s + 17 = 0 with complex roots in right half s-plane. The singular point is an unstable focus.

10.10 e = f(qR – q) = sin (qR – q) = sin (– q)

q + aq + K sin q = 0 With

x1 = q and x2 = q , we have

x 1 = x 2 x 2 = – K sin x1 – ax2 Singular points are given by the solution of the equations x2 = 0 – K sin x1 – a x2 = 0 This gives x1 = kp; k in an integer. We therefore have multiple singular points. Linearized equation around singular point (0, 0): q + aq + Kq = 0

The characteristic equation is s2 + as + K = 0 The singular point is stable and is either a node or a focus depending upon the magnitudes of a and K. Linearization about the singular point (p, 0) gives

q + aq – Kq = 0 The characteristic equation is s2 + as – K = 0 The singular point is a saddle point. 10.11

d 2 ( y - 1) d ( y - 1) +y–1=0 + 2z 2 dt dt

SOLUTION MANUAL 117

(i) z = 0; singularity (1, 0) on (y, y ) plane is a centre (ii) z = 0.15; singularity (1, 0) on (y, y ) plane is a stable focus. 10.12 Revisiting Example 10.4 (Fig. 10.34) will be helpful.

ch

Jq + Tc sgn q = KA K1 e; e = qR – q Therefore

T K e + e + c sgn ( e ) = 0; K = KAK1 J J

LM N

e + w 2n e +

OP Q

Tc sgn ( e ) = 0; wn = K

K/J

With x1 = e and x2 = e /wn, we have x1 = wn x2

LM N

x 2 = – wn x1 + For

a

Tc sgn w n x 2 K

fOPQ

Tc = 0, we have x1 = wn x2; x 2 = – wn x1

d x2 x = – 1 ; this gives x12 + x 22 = x 21 (0) d x1 x2 The effect of the Coulomb friction is to shift the centre of the circles in phase plane to + Tc/K for x2 < 0 and to – Tc/K for x2 > 0. The steady-state error found from phase trajectory is – 0.2 rad. From the phase trajectory, we see that (a) the system is obviously stable for all initial conditions and inputs; and (b) maximum steady-state error = ± 0.3 rad. 10.13 With x1 = y and x2 = y , we get x1 = x2

x 2

R|- x = S- x |T- x

2 2 2

a - 1 < x < 1f - 2a x - 1f ; Region II a x > 1f - 2a x + 1f ; Region III a x < - 1f ; Region I

1

1

1

1

1

From the phase portrait shown in the figure below, it is seen that the system is stable in the equilibrium zone.

118


10.14 Revisiting Review Example 10.6 (Fig. 10.42) will be helpful. The system equation in the linear range is

e + e + e = 0 The characteristic equation s2 + s + 1 = 0 has complex roots in left half of s-plane. Therefore, on the (e, e ) plane, the origin is a stable focus. The system equation in saturation region is e + e = sgn (e) The trajectories are asymptotic to the ordinates – 1 and + 1 depending on whether the error is +ve or – ve. From the phase trajectories plotted from the given initial conditions (with and without saturation) we find that velocity is always greater for trajectories without saturation; thus the error saturation has a slowing down effect on the transient. 10.15 With x1 = e, x2 = e and u = f (e), we have x1 = x2 x 2 = – x2 – f (x1)

Without hysteresis, the system behaviour is oscillatory with decreasing (tending towards zero) period and amplitude of oscillations (refer Fig. 10.32). With hysteresis, the system enters into a limit cycle, as is shown in the figure below.

SOLUTION MANUAL 119

e = – u = – f(e); x1 = e, x2 = e 10.16 (a) The trajectories corresponding to x1 > 0 are given by (refer Eqns (10.24)) x1(t) = –

1 2 1 x 2 (t) + x1(0) + x 22 (0) 2 2

and trajectories for x1 < 0 are given by x1(t) =

1 2 1 x 2 (t) + x1(0) – x 22 (0) 2 2

The system response is a limit cycle as shown in the figure.

e = – u = – f (e + KD e ); x1 = e, x2 = e (b) The trajectories corresponding to (x1 + KD x2) > 0 are given by x1(t) = –

1 2 1 x 2 (t) + x1(0) + x 22 (0) 2 2

and the trajectories corresponding to (x1 + KD x2) < 0 are given by x1(t) =

1 2 1 x 2 (t) + x1(0) – x 22 (0) 2 2

With derivative feedback, the limit cycle gets eliminated as shown in the figure.

120


(c) Constructing a trajectory for the case of large KD proves the point. An illustrative trajectory is shown in the figure. x2

x2 Switching line

x1

Trajectory

x1

Trajectory

Slope = -

1 KD

x2

Switching line x1 Trajectory

10.17 q + 0.5q = 2 sgn(e + 0.5q ) With x1 = e and x2 = e , we have x1 = x2 x 2 = – 0.5x2 – 2 sgn(x1 + 0.5 x2) As seen from the phase trajectory in the figure, the system has good damping, no oscillations but exhibits chattering behaviour. Steady-state error is zero.

SOLUTION MANUAL 121

qR = const +

e +

-

2

1 s + 0.5

-2

-

q

q

1 s

0.5

x2

2 1

x1

x1 + 0.5 x2 = 0

10.18 (a) With x1 = e, x2 = e and u = f(e), we have x1 = x2 x 2 = – x2 – f(x1)

The system oscillates with everdecreasing amplitude and everincreasing frequency (refer Fig. 10.32).

122


(b) A rough sketch of the phase trajectory is shown in the figure.

(i) Deadzone provides damping; oscillations get reduced. (ii) Deadzone introduces steady-state error; maximum error = ± 0.2. (c) x1 = x2

SOLUTION MANUAL 123

F H

x 2 = – x2 – f x1 +

1 x2 3

I K

A rough sketch of the phase trajectory is shown in the figure. By derivative control action (i) settling time is reduced, but (ii) chattering effect appears. 10.19 Section 10.10 provides solution to this problem. Optimum switching curve is given by Eqns (10.48). Figure 10.37 shows a few trajectories. e + e = – u; x1 = e, x2 = e 10.20 For u = +1 (refer Eqns (10.26)) x1 – x1(0) = – (x2 – x2(0)) + ln

FG 1 + x IJ H 1 + x ( 0) K 2

2

The trajectories are asymptotic to the ordinate – 1. For u = – 1 x1 – x1(0) = – (x2 – x2(0)) – ln

FG 1 - x IJ H 1 + x ( 0) K 2

2

The trajectories are asymptotic to the ordinate +1. For x1(0) = x2(0) = 0, x1 = – x2 + ln (1 + x2); u = + 1 x1 = – x2 – ln (1 – x2); u = – 1 Switch curve is given by x1 = – x2 +

FG H

x2 x2 ln 1 + 2 | x2 | | x2 |

IJ K

The figure given below shows the switching curve and a few typical minimum-time trajectories.

CHAPTER 11

11.4 E =

NEURAL NETWORKS FOR CONTROL

b

g

2 1 0.4 - s ( -3w + w0 )) 2 + (0.8 - s (2 w + w0 ) 2 ∂E = 0 ∂w = – [(0.4 – s (–3w + w0)) s (–3w + w0) (1 – s (–3w + w0)) (–3) + (0.8 – s (2w + w0)) s (2w + w0) (1 – s (2w + w0))(2)] = – [–3x + 2y]

∂E = 0 = – [x + y] ∂w0 This gives x = y = 0

a0.4 - s (-3w + w )fa1 - s (-3w + w )fs ( -3w + w ) = 0 0

0

0

s ( a ) = 0 only at a Æ -• s ( a ) = 1 only at a Æ +• Therefore s (– 3w + w0) = 0.4 which gives –3w + w0 = – 0.41 The equation y = 0 gives s (2w + w0) = 0.8 or, 2w + w0 = 1.386 Solving for w and w0, we get w = 0.36, w0 = 0.666

FG Â w H n

11.5 zl = s

h1 li x i

+ wlh01

i =1

IJ K

FG Â w z + w IJ H K Â bv t + v g m

tr = s y j =

h1 rl l

h1 r0

l =1

p

jr r

j0

r =1

vjr (k + 1) = vjr (k) + h tr ej (k) vj0 (k + 1) = vj0 (k) + h ej (k) ej (k) = yj - y j (k) q

wrlh2 ( k + 1) = wrlh2 ( k ) + h t r (1 - tr ) zl Â v jr e j ( k ) j =1

SOLUTION MANUAL 125 q

wrh02 ( k + 1) = wrh02 ( k ) + h tr (1 - tr ) Â v jr e j ( k ) j =1

wlih1 ( k

h1 + 1) = wli ( k ) + h xi zl (1 - zl )

wlh01 ( k + 1) = wlh01 ( k ) + h zl (1 - zl )

LMÂ e Â v MN q

p

j

j =1

r =1

LMÂ e Â v MN q

p

j

j =1

h2 jr wrl tr (1 - tr )

h2 jr wrl tr (1 - t r )

r =1

1 - e-a 2 1 -a - = 1 + e-a 1+ e

11.6 (a) s (a) =

s ¢(a) =

(1 - s ( a ))(1 + s ( a )) ds ( a ) = 2 da

Network Output: y ( k ) = s

FG Â w x IJ ; x H K n

i i

=1

0

i= 0

Weight update rule: wi (k + 1) = wi ( k ) + (b) a(1) =

b

g

h ( y - y ( k ) ) 1 - y 2 ( k ) x i 2

4

Â w i xi (1 )

= 1 + 2 – 0.5 = 2.5

i =1

y (1) = s [a(1)] = 0.848

c

h

h ( -1 - 0.848 ) 1 - ( 0.848) 2 x (1) 2 = [0.974, – 0.948, 0, 0.526] M

w (1) = w (0) +

b

g

(c) y (1) = s w1 x1( 1) + w2 x 2 (1) + w3 x 3 (1) + w4 x 4 (1) = 0.8483 e(1) = y(1) – y (1) = –1.8483

y ( 2 ) = – 0.7616; e(2) = – 0.2384 y ( 3 ) = – 0.8483; e(3) = 1.8483

LM Â x N 3

w1(1) = w1(0) + h = 1.0518 M

FG Â w x + w IJ ; l = 1, 2 H K 2

11.7 (a) zl = s

li i

i =1

l0

p =1

( p) ( p) e 1

b1 - ( y

( p) 2

2

)

g OP Q

OP PQ

OP PQ

126


FG Â v z + v IJ H K 2

y = s

l l

0

l =1

e = y – y vl (k + 1) = vl (k) + 0.1e (k) zl (k) y (k) [1 – y (k)] v0 (k + 1) = v0 (k) + 0.1e (k) y (k) [1 – y (k)] wli (k + 1) = wli (k) + 0.1xi (k) e (k) y (k) [1 – y (k)] zl (k) [1 – zl (k)] wlo (k + 1) = wlo(k) + 0.1e (k) y (k) [1 – y (k)] zl (k) [1 – zl (k)] 11.8 s ¢(a) =

1 - (s ( a )) 2 ds ( a ) = 2 da zl = s (wl x + wl0); l = 1, 2, …, m y = s

FG Â v z + v IJ H K m

l l

0

l =1

e(k) = y – y (k)

b g h v (k + 1) = v ( k ) + e( k )b1 - y ( k ) g 2 h w (k + 1) = w ( k ) + e( k ) b1 - y ( k )g v ( k )b1 - z ( k ) g x ( k ) 4 h w (k + 1) = w ( k ) + e( k )b 1 - y ( k ) g v ( k )b 1 - z ( k ) g 4 vl (k + 1) = vl ( k ) +

h e( k ) 1 - y 2 ( k ) zl ( k ) 2 2

0

0

l

l

2

2

l

l

2

l0

l0

2

l

l

CHAPTER 12

FUZZY CONTROL

12.7 mR

y

x

3 *

12.11 z =

FH 1 IK + 4 FH 1 IK + 5 FH 1 IK + 6 FH 1 IK + 7 FH 1 IK + 7.5 FH 3 IK + 8 FH 3 IK + 9 FH 3IK 4 4 4 2 2 4 4 8 1 1 1 1 1 3 3 3 FH IK + FH IK + FH IK + FH IK + FH IK + FH IK + FH IK + FH IK 4 4 4 2 2 4 4 8

= 6.76 12.12 Crisp value for speed =

1600 ( 0.9) + 1700 ( 0.9) + 1800 ( 0.9) + 1900 (0.8) + 2000 (0.7) + 2100 ( 0.5) + 2200 ( 0.3) + 2300 ( 0.15) + 2400 ( 0 ) 0.9 + 0.9 + 0.9 + 0.8 + 0.7 + 0.5 + 0.3 + 0.15 = 1857.28 12.13 x = 4; y = 8

FG IJ F I H K H K F I a = min G m ( 4 ), m (8)J = min F 1 , 2 I = 1 H ~ ~ K H 3 3K 3 F I F2 I min G a , m ( z )J = min G , m ( z )J H ~ K H3 ~ K F1 I min (a , m ( z )) = min G , m ( z )J H3 K ~ R F2 I F1 IU m (z) = m ( z ) = max S min G , m ( z )J , min G , m ( z )J V ~ T H 3 ~ K H 3 ~ KW a1 = min m A1 ( 4 ), m B2 ( 8) = min 2 , 1 = 2 3 3 ~ ~ 2

agg

A2

B2

1

C1

C1

2

C2

C2 ~

C1

C1

C2

128


2 z* =

FH 1 IK + 3 FH 2 IK + 4 FH 2 IK + 5 FH 2 IK + 6 FH 1 IK + 7 FH 1 IK + 8 FH 1 IK 3 3 3 3 3 3 3 FH 1 IK + FH 2 IK + FH 2 IK + FH 2 IK + FH 1 IK + FH 1 IK + FH 1 IK 3 3 3 3 3 3 3

= 4.7 12.14 x = distance D = 27 y = vehicle speed V = 55

m PS ( 55) = 0 ; m PM ( 55) = 0.25 ; m PL ( 55) = 0.75 ~

~

~

m PS ( 27) = 0.38 ; m PM ( 27) = 0.62 ; m PL ( 27 ) = 0 ~

~

~

F I H ~ K ~ min F m (27), m ( 55)I = 0.62 H ~ K ~

a1 = min m PS (27), m PM (55) = 0.25 a2 =

FH min F a H

PM

PL

z = breaking force B

F I IK H K ~ , m ( z )I = min F 0.62, m ( z )I K H K ~ ~ ~ m (z) = max RSmin F 0.25, m ( z )I , min F 0.62, m ( z )I UV K H KW ~ ~ T H

min a 1 , m PL ( z ) = min 0.25, m PL ( z ) ~

2

PM

PM

agg

PL

PM

10 ( 0.2) + 20 ( 0.4) + 30 ( 0.6 ) + 40 ( 0.62 ) + 50 ( 0.62 ) + 60 ( 0.62 ) + 70 ( 0.6 ) + 80 ( 0.4 ) + 90 ( 0.25) + 100 ( 0.25) z* = 0.2 + 0.4 + 0.6 + 0.62 + 0.62 + 0.62 + 0.6 + 0.4 + 0.25 + 0.25 = 53.18 4 1 ; m LD ( 60) = 12.15 m SD ( 60 ) = 0 ; m MD ( 60 ) = ~ ~ ~ 5 5 3 2 ; m LG ( 70) = m NG ( 70 ) = 0 ; m MG (70 ) = ~ ~ ~ 5 5

F I 3 H ~ K 5 ~ 2 min F m ( 60), m ( 70 )I = H ~ K 5 ~ 1 min F m (60 ), m ( 70 )I = H ~ K ~ 5 1 min F m ( 60 ), m ( 70 )I = H ~ K 5 ~

a1 = min m MD ( 60), m MG ( 70 ) = a2 = a3 = a4 =

MD

LG

LD

MG

LD

LG

SOLUTION MANUAL 129

F I F3 I H ~ K H 5 , m ~ ( z )K 2 min F a , m ( z )I = min F , m ( z )I H ~ K H5 ~ K 1 min F a , m ( z )I = min F , m ( z )I H ~ K H5 ~ K 1 min F a , m ( z )I = min F , m ( z )I H ~ K H5 ~ K 3 2 1 m (z) = max RS min F , m ( z )I , min F , m ( z )I , min F , m ( z )I UV H K H K H 5 ~ 5 ~ KW T 5 ~ min a 1 , m M ( z ) = min

M

2

L

L

3

L

L

4

VL

agg

VL

M

L

VL

10 ( 0 ) + 15 ( 0.325) + 20 ( 0.6 ) + 25 ( 0.6 ) + 30 ( 0.6) + 35 ( 0.4 ) + 40 ( 0.4 ) + 45 ( 0.4) + 50 ( 0.4 ) + 55 ( 0.25) + 60 ( 0.2) z* = 0 + 0.325 + 0.6 + 0.6 + 0.6 + 0.4 + 0.4 + 0.4 + 0.4 + 0.25 + 0.2 = 34.40 12.16 With 10 bits we can get solution accuracy of interval (0,6).

(6 - 0)

b2

10

g or 0.006 in the

-1

Let the first string of initial population be : 11001000001110010000 The first substring decodes to value equal to (29 + 28 + 25) = 800 . Thus ( 6 - 0 ) ¥ 800 = 4.692 . Similarly for corresponding parameter value is 0 + 1023 second substring, the parameter value is equal to 5.349. Thus the first string in initial population corresponds to the point x(1) = [4.692 5.349]T. The function value at this point is equal to f [x(1)] = 959.680, and the 1 = 0.001. fitness function F [x(1)] = 1 + 959.680

Solution Manual Digital Control and State Variable Methods

Recommend Documents