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, TataMcGraw-Hill, New Delhi, 2003.

Throughout the manual, the numbers associated with referenced Equations,Figures, Tables, Examples, Review Examples and Sections are pointers to thematerial in the textbook.

The detailed solutions of the problems have been avoided; only assistance hasbeen provided.

CHAPTER 2 SIGNAL PROCESSING IN DIGITAL CONTROL

2.1 (a) x1, x2 : outputs of unit delayers, starting at the right and proceeding tothe 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)

F =

0 1

0 368 1 368−

�

�

��

�

�

��. .

; g =

0

1

�

�

��

�

�

��

; c = [0.264 0.368]

(b)Y zR z

( )

( ) =

Pi iD

D

∑

−1 =

0 264 0 368

1 1 368 0 368

2 1

1 2

. .

. .

z z

z z

� ��

� ��

2.2 (a) x1, x2 : outputs of unit delayers, starting at the right and proceeding tothe 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 = 0 1

3 5- -

��

��; g =

1

3-

��; c = [1 0]

(c)Y z

R z

� �

� � =

z

z z�

� �

25 32

2.3 (a) x1 : output of the unit delayer

x1(k + 1) = – 12

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

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

Y z

R z

� �

� � =

� �

�

z

z

212

SOLUTION MANUAL 5

(b) R (z) = 1; Y(z) = �

��

�

z

z z12

212

y (k) = –� ��

12

k

� (k) + 2 � ��

�12

1k

� (k – 1)

(c) R (z) = zz

Y z

z�

� �

1; =

�

��

�

53

12

23

1z z

y(k) = ��

53

12

k

� (k) � 23

� (k)

2.4 (a) y (k + 1) = �y (k) + �r (k);Y z

R z

� �

� � =

�

�z�

R(z) = A; Y(z) = A

z�

��; y(k) = A� (�)k–1 � (k – 1)

(b) R (z) = Az

z

Y z

z�

� �

1; =

Az z

�

�� 1;

Y (z) =

Az

z

Az

z

�

�

�

�

��

��

�

�

1 11

; y(k) = A�

�1� [1 – (�) k] � (k)

(c) R (z) = A z

zY z

z��

� �

1 2 ; = A

z z

�

�� 1 2

Y (z) =A z

z

z

zz

z

�

� �

�

1

1

1 12 2� ��

�

��

�

��

��

� �

� �

y (k) =A

kkb

aa a

11 12-( )

+ -( ) -( )[ ]; k � 0

(d) r(k) = A cos (�k) = Re{Aej�k}; ��{ej�k} = zz e j� �

The output,

y(k) = Re � -

-( ) -

��

��

��

��1 A z

z z e j

b

a W� � = Re

A

e

A

eej

kj

j k�

��

�

��

�

��

��

�

= Re A

eej

k j k�

��

��

��

��

��

6 DIGITAL CONTROL AND STATE VARIABLE METHODS

2.5 (a) Given difference equation in shifted form:

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

Y (z) = zz z2 3 2� �

= zz

zz�

��1 2

y (k) = (– 1)k – (– 2) k; k � 0

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

1 1� �z

Y (z) = zz z z

3

21 2 2 1� � ��

= zz

z z

z z��

� �

� �1 2 2 1

2

2 = zz

z z

z z

z

z z��

�

� ��

� �112

0 5

0 512

0 5

0 5

2

2 2

.

.

.

.

� ��

Therefore,

y (k) = (l)k – 12

12

��

k

cos k k

p

412

12

�� +

�� sin

k �

4

�� ; k � 0

2.6 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)

Y (z) =0 3679 0 2642

1 3679 0 36792. .

. .

z

z z

�

� �R (z)

R (z) = 1 + 0.2142z –1 – 0.2142z –2

Hence,

Y (z) =0 3679 0 2642

1 3679 0 36791 0 2142 0 21422

1 2. .. .

. .z

z zz z

�

� ��

� � � �� = 0.3679z – 1 + 0.8463z – 2 + z – 3 + z – 4 + z – 5 + �

y(0) = 0; y (1) = 0.3679; y (2) = 0.8463;y (k) = 1, k � 3.

2.7 Taking z-transform of the given equation:

Y zR z

� �

� � = z

z z

2

2 3 2� � = z

z z

2

1 2� ��

R(z) = 1 + z – 1 = z

z

� 1

Y zz( ) =

( )( )( )

zz z

+

- -

12 1

= 32

21z z-

--

y(k) = 3(2)k – 2(1) k; k � 0

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) = 2 3

0 5 0 3z

z z z

�

� �� . .

Y z

z

� � = 2 3

0 5 0 32

z

z z z

�

� �� . .

= � � ��

��

40 20 100 5

500 32z z z z. .

y(k) = – 40� (k) + 20� (k – 1) – 10(0.5) k + 50(– 0.3) k; k � 0

(b) R (z) = zz �1

; Y (z) = � �

� � �

��

��

6

1 12

14

12

14

2z z

z z j z j

Y z

z

� � =

�

��

�

� ��

�

� �

161

8 412

14

8 412

14

z

j

z j

j

z j

= –16

1

16 65

162z

z

z z-+

-

- +

y(k) = – 16(1)k + (0.56)k [7.94 sin (0.468k) + 16 cos (0.468k)]; k � 0

2.9 (a) R (z) = 1 + z – 2 + z – 4 + � = zz

2

2 1�

Y z

z

� � = z

z z z z� � � �� 1 1 0 0 3.5 .

=�

��

�

��

��

�

0 8330 5

0 410 3

0 4761

0 7691

..

..

. .z z z z

y(k) = – 0.833(0.5)k – 0.41(– 0.3)k + 0.476(– 1)k + 0.769(1)k; k � 0

(b) R(z) = z

z �1;

Y z

z

� � = 1

0 0 1 12( .5) ( . )( )z z z- - -

=�

��

��

�

��

��

5

0 5

2 50 5

6 940 1

4 4412z z z z.

..

..

.

y(k) = – 10k(0.5)k + 2.5(0.5)k – 6.94(0.1)k + 4.44(1)k; k � 0

2.10 y(�) = limz1

(z – 1) Y (z) = K a a

a a

1 1

1 1

2

2

� �

� �

� �

� �

� �� = K

For y (�) = 1, K = 1


2.11 Refer Eqn (2.45)

Y (z) = zz m��

; y (�) = limz m

z z

z

�

�

� �

� �1

1

� = 0, if |�| < 1.

(i) R(z) = zz �1

;Y z

z

� � = 1

1 12z z� �� =

12

1

12

1212z

z

z��

�

�

y(k) = 12

(1)k – 12

cos k �

2

�� – 1

2 sin

k �

2

�� ; k � 0

(ii) r(k) = {1, 0, – 1, 0, 1, 0, – 1, �} = cos k �

2

�� ; R(z) = z

z

2

2 1�;

Y (z) = z

z

2

2 21��

The output y (k) is bounded because of matching of system poles withexcitation poles.

2.13 (a) (z) = z3 – 1.3z2 – 0.08 z + 0.24 = 0

(i) (1) = – 0.14 < 0

The first condition of Jury’s criterion is violated. The system is un-stable; we may stop the test here.

(b) (z) = z4 – 1.368 z3 + 0.4z2 + 0.08z + 0.002 = 0

(i) (1) = 0.114 > 0; satisfied

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

Form Jury’s table; from which you will find that (conditions (2.50b)):

|a4| < |a0|; 0.002 < 1; satisfied

|b3| > |b0|; | – 1| > | – 0.083|; satisfied

|c2| > |c0|; 0.993 > 0.512; satisfied.

The system is stable.

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

(r) =1

1

4�

�

�

��

r

r + 0.5

1

1

3�

�

�

��

r

r – 0.2

1

1

2�

�

�

��

r

r +

11�

�

r

r + 0.4

=� � � � �

��

0 3 3 4 8 8 1 4 2 7

1

4 3 2

4. . . . .r r r r

r

Form the Routh table, from which you will find that one root of (r)lies in the right-half plane. Therefore, three poles of G (z) lie insidethe 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 + 2�wns + w2n

wn = 22

; ;w ps

T= > 2 T < �

2

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

� {cos 50kT} = � cos50 2

50

k� �� = � {cos 2k�}

=1

1 1� �z = 1 + z –1 + z –2 + �

It can be seen that the noise signal cos 50t sampled at 2

50

� sec be-

comes unity at every sampling instant. Thus the noise signal can gen-erate an undesirable d.c. component in the output.

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 ejbT

(ii) s = a � j�0; z = eaT e� j�0T

2.20 Ga(s) = 1

1s �; Gh0(s) =

1� �e

s

sT

G(z) = �{Gh0G(s)} = (1 – z –1) � 11s s �

��

= 1�

�

�

�

e

z e

T

T

R(z) = zz

Y zz�

� �

1; =

1

1

�

� �

�

� � �

e

z e z

T

T� � = 11

1z z e T�

��

� �

y(k) = 1 – e–kT; k � 0

2.21 Ga(s) = 102s s �

��

; G (z) = (1 – z –1) � 1022s s �

��

G(z) = 5(1 – z –1)T z

zz

zz

z e T��

��

�

��

��

� � � � �1 2 1 22 2� �

For T = 0.4 sec, G(z) = 10 0 76

16 1 0 46z

z z�

� �

� �

� � � �

..

2.22 (i) Refer Eqn (2.90b)

(ii) Refer Eqn. (2.107)


2.23 Refer Section 2.12; Fig. 2.41a.

2.24 Refer Section 2.12; Figs 2.44 and 2.37.

2.25u k u k

T

( )- -( )1

= Ke k e k

T Te k

T

Te k e k e kc

I

D� � � ��

� ��

��

��

1 1 2 1 22

U(z) = Kc 1 11

111�

�

�

��

��

��

�TT z

T

Tz

I

D � � E(z)

2.26 Required

� = 0.45, �M = 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 = 2 11T

zz�

�

D(z) = 0.4047 zz�

�

0 93910 9752

.

.

2.27 U (s) = Kc 1 1� ��

��T s

T sI

D E(s)

U(z) = Kc 1 1

2 11

1

+ +�

��

�

��=

-

+=

-T s

T sI

sT

z

z

D

szzT

( )

E (z)

= Kc 12

11

1�

�

��

��

��

TT

z

z

T

T

z

zI

D E (z)

u(k) = uP(k) + u1(k) + uD(k)where

uP(k) = Kce(k)

u1(k) – u1(k – 1) =K T

Tc

I2 [e(k) + e(k – 1)]

u1(k) =K T

T

e i e ic

I

- + ( )Â

1

2� �

uD(k) =K T

Te k e kc D ( ) - - 1� �

Therefore,

u(k) = Kc e k TT

e i e i T

Te k e k

I i

kD( ) +

- + ( )+ ( ) - -

��

��=

Â1

21

1

� � � ��

SOLUTION MANUAL 11

2.28 (a)dy t

dt

� � + ay(t) = r(t); y(t) = y(0) – a y

t

0� () d + r

t

0� () d

y(k) = y(k – 1) – a yk T

kT

��

�1

() d + rk T

kT

��

�1

() d

= y(k – 1) – aTy(k) + Tr(k)

y(k) =1

1� aTy(k – 1) +

TaT1�

r(k)

(b) y(k + 1) = y(k) – a ykT

k T��

�1

() d + rkT

k T��

�1

() d

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

2.29 (a)d y

dta

dy

dtby

2

2 � � = 0

12T

[y(k) – 2y(k – 1) + y(k – 2)] + aT

[y(k) – y(k – 1)] + by(k) = 0

b aT T

� �

��

12

y(k) – aT T

�

��

22

y(k – 1) + 12T

y(k – 2) = 0;

y(0) = �; y(– 1) = � – T�

(b) x1 = y; x2 = �y; �x1 = x2; �x2 = – bx1 – ax2

A =0 1

- -

��

��b a

Using Eqn (2.113) we obtain,

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

b

��

CHAPTER 3 MODELS OF DIGITAL CONTROL DEVICES ANDSYSTEMS

3.1 D(z) = {Gh0(s)G1(s)} = Gh0G1(z)

Y zR z

� �

� � =

D z G G z

D z G G H zh

h

� � � �

� � � ��0 2

0 21 =

G G z G G z

G G z G G H zh h

h h

0 1 0 2

0 1 0 21

� � � �

� � � ��

3.2Y z

R z

� �

� � =

G G z

G G z H zh

h

0

01

� �

� � � ��3.3 Y(z) = Gh0G2(z) U(z)

U(z) = G1R(z) – Gh0G2HG1(z) U(z)

Y(z) =G G z G R z

G G HG zh

h

0 2 1

0 2 11

� � � �

� ��

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

Y(z) = GpH2R(z) + D z G G z

D z G G zh p

h p

( ) ( )

( ) ( )+

0

01 [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

D z G G z D z G G G zh h

� � � �

� � � � � � � �� 1 0 1 0 1 2

; Y(z) = Gh0G1G2(z) U(z)

Y zR z

� �

� � =

G G G z D z

D z G G z G G G zh

h h

0 1 2

0 1 0 1 21

( ) ( )

( ) ( ) ( )+ +[ ]

SOLUTION MANUAL 13

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

X z

R z

� �

� � =

G G z

D z G G z G G G zh

h h

0 1

0 1 0 1 21

( )

( ) ( ) ( )+ +[ ]

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

D z G G zh

� �

� � � ��1 0

3.7 G(s) = 1

1s s �� ;�

�L

R

z

z

� �

� � =

G G z

G G zh

h

0

01

� �

� ��;

Gh0G(z) = (1 – z –1) � 112s s �

��

��

= z T e e Te

z z e

T T T

T

� � � � �

� �

� � �

��

1 1

1

� � � ��

For T = 0.25 sec,

Gh0G(z) =0 0288 0 92

1 0 7788. .

.z

z z�

� �

� �

� � � �

3.8 Plant transfer function is,

G(s) =185

0 025 1. s ��

Gh0G(z) = (1 – z –1) � 1850 025 1s s. �

��

��

= 185 1 40

40

�

�

�

�

e

z e

T

T

� �

Let x(k) be the input to D/A block.

x(k) = KF�r (k) + KP[�r(k) – � (k)]

X(z) = KF�r(z) + KP[�r(z) – �(z)]; �(z) = Gh0G(z) X(z)

�

�

zzr

� �

� � =

K K G G z

K G G zF P h

P h

��

�

� �

� �

0

01


3.9 The filter is described by the following difference equation,

u(k) = u(k – 1) + 0.5e(k);U zE z

� �

� � = 0.5 z

z �1

Gh0G(z) = (1 – z –1) � 112s s �

��

��

= T e z e Te

z z e

T T T

T

� � � � �

� �

� � �

��

1 1

1

� � � ��

fs = 5 Hz = 1T

; T = 0.2 sec

Gh0G(z) =0 019 0 0175

1 0 819. .

.z

z z�

� �� ;

Y zR z

� �

� � =

G G z D z

G G z D zh

h

0

01

� � � �

� � � �� =

0 0095 0 922 81 2 65 0 8193 2

. .. . .

z z

z z z

�

� � �

� �

3.10 (i) Gh0G(z) = (1 – z–1) � es s

s�

��

��

0 4

1

.

Using transform pairs of Table 3.1, we obtain,

Gh0G(z) =1

1

0 6 0 6 1

1

� � �

� �

�

�

�

� � �

��

e z e e

z z e

. .� ��

(1 – z –1) = 0 45 0 181

0 368. .

.z

z z�

��

Y zR z

� �

� � =

G G z

G G zh

h

0

01

� �

� �� =

0 45 0 1810 081 0 1812

. .. .z

z z

�

� �

(ii) Gh0G(z) = 0 45 0 181

0 3682. .

.

z

z z

�

�� ;

Y zR z

� �

� � =

0 45 0 181

0 368 0 45 0 1813 2. .

. . .

z

z z z

�

� � �

3.11 Refer Section 3.6.

3.12 Refer Gopal M., Digital Control Engineering, New Delhi, Wiley East-ern, 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) ��51 22s s s� �

��

��

=2 5

1.

z � – 3.75 +

5 10 3679z

z�

�

� �

. – 1.25

zz

�

�

� �10 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 15

3.14 z3 – 0.1z2 + 0.2Kz – 0.1K = �(z)

�(1) = 0.1K + 0.9 > 0�(– 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) � 13s s �

��

��

= K e

z e

T

T31 3

3�

�

��

��

�

�

�(z) = 1 + Gh0G(z) = z – e –3T + K3

(1 – e –3T) = 0

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

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

3.16 Gh0G(z) = (1 – z–1) � K

s s2 1+

��

��

=K z

z

z T e z e Te

z z e

T T T

T

� � � � � �

� �

�

�

�

� �

� �

� � �

�

1 1 1

1 2

� � � ��

�(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)

Put z = 11�

�

rr

, then r2 (1 + a + b) + 2r(1 – b) + 1 – a + b = 0

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 < 2 1

2 2

�

� � �

�

� �

e

T e Te

T

T T

� �

K < 1

1

�

� �

�

� �

e

e Te

T

T T

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

3.17 For T = 1 sec,

Gh0G(z) = K0 368 0 264

1 368 0 3682. .

. .

z

z z z

�

� �

� �

� �


�(z) = z3 – 1.368z2 + 0.368(1 + K)z + 0.264K = 0�(1) = – 0.368 + 0.368(1 + K) + 0.264K > 0

�(– 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) = 4500361 2

Ks s �� .

; K = 14.5; � = 0.707; �n = 255.44

y(t) = 1 – e nt�

�

� �

�1 2 sin � �

�

�n t1

12 12

� ��

��

�� tan

(b) T = 0.01 sec,

Gh0G(z) = 1 3198 0 4379

1 027 0 0272

. .. .

;z

z zY zR z

�

� �

� �

� � =

1 3198 0 43790 0 46492

. ..2929 .

z

z z

�

� �

By dividing the numerator polynomial by the denominator polyno-mial, 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) = 0 029 0 0257

1 697 0 6972

. .. .

;z

z zY zR z

�

� �

� �

� � =

0 029 0 02571 668 0 72262

. .. .

z

z z

�

� �

Dividing the numerator polynomial by the denominator polynomial,we can obtain the response.

3.19 Y zR z

� �

� � =

G G z

G GH zh

h

0

01

� �

� ��

Gh0G(z) = (1 – z –1) � 11s s +

��

��

= 1 1

1�

�

�

�

e

z e;

Gh0GH(z) = (1 – z –1) � 1

12s s +

��

��

= e z e

z z e

� �

�

� �

� ��

1 1

1

2 1

1 � �

Y zR z

� �

� � =

1 1

1

1

2 1

� �

� � �

�

�

� �e z

z z e

� �; R(z) = z

z�1; Y(z) = 0 632

0 6322.

.z

z z� �

SOLUTION MANUAL 17

a k sin (�k) ��

� �

� �

a z

z a z a

sin

cos

�

�2 22; a2 = 0.632;

cos� = 12a

= 0.629; � = 0.89 rad; y(k) = 1.02(0.795)k sin (0.89k)

3.20 Gh0G(z) = (1 – z–1) � 11s s +

��

��

= 0 6320 368.

.;

zY zR z�

� �

� � = 0 632

0.

.264z +;

R(z) = zz �1

Y(z) = 0 6321 0

..264

zz z� ��

; y(k) = 0.5 [1 – (– 0.264)k] �(k)

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

�Y (z) = ��{Gh0(s) G(s)e –�Ts)} E(z); E(z) = R(z) – Y(z);

Y(z) = ��{Gh0(s) G(s)} E(z)

E(z) =R zG s G sh

� �

� � � �� 1 0�

�Y (z) =�

�

[ ]

[ ]

G s G s e

G s G sh

Ts

h

0

01

� � � �

� � � �

�

�

D

R(z); ��{Gh0(s) G(s)} = 0 632

0 368.

.z �

��[Gh0(s) G(s) e –�Ts] = 0 393 0 239

0 368. .

.;�z

z zY zR z

�

��

� �

� � =

0 393 00 264

. .239.

zz z

�

�� ;

R(z) = zz �1

�y(k) = [0.5 – 0.107 (– 0.264)k–1] �(k – 1)

�y(1) = 0.393 = y(0.5T); �y(2) = 0.528 = y(1.5T)


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

3.21 (i) D(z) = 4z

z

z z

z z

�

�

��

��

� �

� �

��

��

10 1

1 1

0 3 0 8

2

2..2

. .

= 41

1 0 1

1 1

1 0 3 0 8

1

1

1 2

1 2�

�

�

� �

� �

� �

�

� �

�

�

� �

� �

z

z

z z

z z.

.2

. .

Refer Figs 3.19 – 3.20

(ii)D z

z

� � = � �

��

�

� �

50 46 620 1

7 38 4 05

0 3 0 82z z

z

z z

..

. .

. .

D(z) = –50 ��

��

� ��

�

� �

46 621 0 1

7 38 4 051 0 3 0 81

1

1 2

..

. .. .z

z

z z

Refer Figs 3.21 – 3.22

3.22 D(z) = 10 1

0 0 8

2

2

z z

z z z

� �

� ��

� �.5 .

= 233 330

127 080 8

25 106.252

..5

..z z z z�

��

� �

= ��

��

�

�

�

�233 331 0

127 081 0 8

1

1

1

1..5

..

zz

zz

+ (106.25 + 25z –1)z –1

SOLUTION MANUAL 19

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

mQ = 30

1 = 30�C/(kg/min)

�(t) = 0.283�ss at t1 = 25 min

�(t) = 0.632�ss at t2 = 65 min

Therefore (refer Eqns (3.53b))

�D + 13� = t1 = 25

�D + � = t2 = 65

This gives

�D = 5 min, � = 60 min

(b) �CD = �D + 12

T = 5.5

Kc = 1.5�/K�CD = 0.545

TI = 2.5�CD = 13.75 min

TD = 0.4�CD = 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


3.25

TIM000

TIM002

TIM001

TIM003

TIM001

10001

TIM000

TIM002

TIM003

10002

10002

10003

TIM002

TIM000

10000

10000

10001

# 0030

# 0015

# 0060

# 0030

I/O Assignment:

10000–N(Green); 10001–N (Red)

10002–S(Green); 10003–S (Red)

TIM000–Timer for 30 sec delay




SOLUTION MANUAL 21

3.26

10001

00001

00000

00000 00002

10003

10000

CNT000

00002

10000

1000100003

10002

10003

10000

10001

CNT000

# 0005

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


CNT000–Counter with a Set Value of 5.

3.27

1000000002

10001

10001

00000

CNT000

TIM001CNT000

00001

10000

10001

10000

CNT000

TIM001

# 0005

# 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


SOLUTION MANUAL 23

3.28

00002

00000

00001

TIM000

TIM001

TIM001

00001

00000

10001

01000

10000

10001

01000

10000

10000

01000

TIM000

TIM001

# 0020

# 0020

I/O Assignment:

00000–S1; 00001–S2

00002–S3; 10000–M1

10001–M2; 01000–Work-bit




3.29

00000

00001

00001

00003

00003

00002

10001

10000

00002

10000

10000

TIM000

TIM000

10001

10001

# 0007

I/O Assignment:

00000–PB1; 00001–PB2

00002–LS1; 00003–LS2

10000–Forward motor

10001–Reverse motor


CHAPTER 4 DESIGN OF DIGITAL CONTROL ALGORITHMS

4.1 Steady-state error can be calculated from the corresponding continuous-time system as sampling does not affect steady-state performance of acontinuous-time system.

+ +R s( ) E s( ) q( )s

K1

K2

Js s

1 1

G(s) =q( )

( )

s

E s =

K

s Js K1

2( )+

Kp = limsÆ0

G(s) = •

Kv = limsÆ0

sG(s) = K

K1

2

Ka = limsÆ0

s2G(s) = 0

4.2 Plant model:

G(s) = 157

1

.

( )s s +; Gh0( jw) G( jw) =

1 57

1

2.

( )

e

j j

jT

-

+

w

w w; T = 1.57 sec

Bode plot analysis:Phase margin (without ZOH) = 45 degPhase margin (with ZOH) = 0 degSpecification: fM = 45 degLet us use a lag compensator.From the Bode plot of the uncompensated system, we see that the phasemargin of 45 deg may be realized if the gain cross over frequency is movedfrom the present value (1 rad/sec) to a frequency of 0.5 rad/sec. To accom-modate phase lag, we take wc2 = 0.4 rad/sec20 log b = 8; b = 2.5

1

t =

w c2

10 = 0.04;

1

bt = 0.016; D1(s) =

1

1

+

+

s

s

t

bt =

1 25

1 62 5

+

+

s

s.

Phase margin of the compensated system is about 45 deg.


Bilinear transformation: s = 2 1

1T

z

z

-

+ = 1.274

z

z

-

+

1

1;

D1(z) = 0 4( 0 939

0 975

. . )

( . )

z

z

-

-

Kv of the original analog system is given by, Kv = limsÆ0

s 11

.57( )s s +

= 1.57

Kv of the equivalent digital system is:

Kv =1

1T zlim�

(z – 1) D1(z) Gh0G(z)

where Gh0G(z) = 1.57 1 57

1

0 792

0 208

. .

.z z--

-

��

��

, Kv = 1.57

4.3 Kp = limzÆ1

Gh0G(z) = •; Kv = 1

1T zlimÆ

(z – 1) Gh0G(z) = 3.041

Ka =1

2TlimzÆ1

(z – 1)2 Gh0G(z) = 0

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 nT-zw = 0.965 = r ; zwnT = – ln r; wnT 1 2- z = q

From the above equations, we get

z

z1 2- =

-In r

q; z = -

+

lnr

rln2 2q

wn =1 2 2

TrIn + q ; z = 0.199; wn = 8.93

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

( ) ( )

D z D z D z G G z R z

D z G G zh

h

2 1 3 0

1 01

+

+

(ii) R(s) = 0

Y2(z) = GW(z) – Gh0G(z) D1(z) Y2(z)

Y2(z) = GW z

D z G G zh

( )

( ) ( )1 1 0+; Y(z) = Y1(z) + Y2(z)

(b) D3(z) = D2(z) G�0G(z)

SOLUTION MANUAL 27

Y1(z) = [ ( ) ( )] ( ) ( )

( ) ( )

1

11 0 3

1 0

+

+

D z G G z D z R z

D z G G zh

h

= D3(z) R(z)

Y2(z) = GW z

D z G G zh

( )

( ) ( )1 1 0+; Y(z) = Y1(z) + Y2(z)

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

(i) D(s) = K1 + K

s2 ; D(s) G(s) =

sK K

s s1 2

1

+

+( )

Kv = K2;1

Kv

= 0.01 = 1

2K

(ii)

Y sW s

( )( )

= ss s K s K( )+ + +1 1 2

; Y(s) = 11 1 2s s K s K( )+ + +

; yss = 0

Thus a PI compensator meets the requirements.

4.7 G(s) = Ks� �1

; Gh0G(z) = K e

z e

T

T

( )1-

-

-

-

t

t

;

For K = t = 1, Gh0G(z) = 0 3930 607.

.z -; S(z) = 1

1 0+ G G zh ( ) = z

z-

-

0 6070

.

.214

S(e jwT) = cos( ) . sin( )cos( ) .214 sin( )

w w

w w

T j TT j T

- +

- +

0 6070

; |S(e jwT)|2

=1 3685 11 0458 0 428. .214cos( ). . cos( )

-

-

w

w

TT

w s

2 =

22

p

T = 6.28 rad/sec


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 – l2) � (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 thenew polynomial is,

D(lk + Dlk, K + DK) = D(lk, K) + ∂

∂ =

D

z z kl

�lk

+∂

∂ =

D

K z kl

DK + � = 0

Neglecting higher order terms, we obtain,

Dlk = – ∂ ∂

∂ ∂

��

�� =

D

D

/

/

K

z z kl

DK

∂

∂ =

D

Kkl l

= B(lk) = – 1K

Azk

z k

( );l

l

∂

∂ =

D = P

i kk i

π-( )l l

SKkl =

D

D

l k

K K/ =

A k

i kk i

( )

( )

l

l lPπ

-

(b) G(s) = Ks + 1

; Gh0G(z) = 0 3930 607

..

Kz -

; D(z, K) = 1 + Gh0G(z) = 0

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; d z

dz z

D ( )

= l

= 1; SKl = – 0.394

SOLUTION MANUAL 29

For a 10% change in nominal gain, the change in the root location is

given by, D(l k) = SKl ∂ K

K = – 0.0393

(c) G(s) = 1

1t s +; Gh0G(z) = 1

12

12

-

-

-

-

e

z e

t

t

; D(z, t ) = 1 + Gh0G(z) = 0;

z + 1 – 21

2e-

t = 0

Let p = e-

12t ; ∂

∂

p

t = 1

2 2

12

t

te-

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

Nominal value of p = e-

12 = 0.6065

Closed-loop pole is at z = 0.213 = l

A(l) = 1.213; d z

dz z p

D( )

=

= 1; Spl = 1.213 =

∂

∂

( )/l

p p

Now∂pp

= 1

2 2

12

t

tte

p

- ∂ = 1

2t

t

t

∂ = 0.5 ∂t

t; S

t

l = 0.6065

For 10% change in the nominal value of t, the change in root locationis given by,

D(lk) = St

l t

t

∂ = 0.06065

4.9 Gh0G(z) = 0 368 01 368 0 3682

. .264. .

zz z

+

- +; T = 1 sec; z = 1 0

1 0+

-

.5

.5ww

Gh0G(w) = - - +

+

0 0381 2 12 140

. ( )( . )( .924)w w

w w

Gh0G( jg) = 1

21

12 14

10

-�� +��

+��

j j

j j

g g

gg

.

.924

From the Bode plot and Nichols chart, we obtain,fM = 28º; GM = 8 dB; wb = 1.35 rad/sec; gb = 1.6 rad/sec

4.10 G(s) = 12s s( )+

; T = 0.1 sec

Gh0G(z) = 0.004683 ( .9355)

( )( . )z

z z+

- -

01 0 8187

; z = 1 0 051 0 05

+

-

.

.ww


Gh0G(w) = 0.5 ( . )( . )

( .5016 )1 0 001666 1 0 05

1 0+ -

+

w ww w

(a) limwÆ0

wGh0G(w) = Kv = 0.5K; K = 10

Gh0G(w) = 5 1 0 001666 1 0 05

1 0( . )( . )

( .5016 )+ -

+

w ww w

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

(b) Phase lead design: D(w) = 1

1

112

+

+

w

w.994

.5(obtained by standard design procedure),

fM = 55º, Kv = 5, GM = 12.4 dB

(c) D(w) = K w

w

( )( )

11

+

+

t

bt; Kv = 5

Kv = limwÆ0

wD(w) Gh0G(w), gives K = 10.

The Bode plot of G(w) = 5 1 0 001666 1 0 05

1 0( . )( . )

( .5016 )+ -

+

w ww w

(uncompensated system; K = 10 has been used in the plant model)gives fM = 30º

Required phase margin = 55º + 15º (error compensation).

Crossover frequency wc2 = 0.7

D(w) = 1

1+

+

t

bt

w

w; 1t

= w c2

22( ) = 0.18; t = 5.71

Corresponding gain = 20 log10 b = 17; b = 7.0;

D(w) = 0 14 0 18

0 02. ( . )( . )

ww

+

+

w = 2 11T

zz

-

+ gives, D(z) =

0 141 00

. ( .98)( .998)

zz

-

-

(d) From Nichols chart we find that bandwidth values gb for three designscorresponding 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

T = 62.83.

SOLUTION MANUAL 31

4.11 G(s) = 12s

, T = 0.1, Gh0G(z) = 0 005 1

1 2. ( )( )

z

z

+

-

z = 1 0 051 0 05

+

-

.

.ww

; Gh0G(w) = 1 0 052

- . ww

; Gh0G( jg) = 1 0 05

2

- j

j

.

( )

g

g

Bode plot analysis:fM = – 2 deg.In the low frequency range, –Gh0G( jg) is about –180º. Therefore, a lagcompensator cannot fulfil the requirement of 50º phase margin.

The lead compensator D(w) = 64 ( )

( )w

w+

+

116

satisfies the requirements.

The gain crossover frequency = 4fM = 50.62º, GM = 13 dB, Kv = •, Ka = 4.

4.12 (a) K = 50

Gh0G(z) = 0 0043 0 85

1 0 61. ( . )( )( . )

K zz z

+

- -;

Gh0G(w) = 10 1

201

246.67

14 84

-�� +��

+��

w w

w w.

Bode plot/Nichols chart analysis:

Gain crossover frequency = 6.6 rad/sec.

fM = 20º, gb = 10 rad/sec, wb = 9.27 rad/sec.

(b) Lead compensator D(w) = 0 10 06 1.219.

ww

+

+ results in gb = 18 rad/sec,

wb = 14.65 rad/sec.

Lag compensator D(w) = 1 6 16.4 1. w

w+

+ results in gb = 5 rad/sec,

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 bebrought down by 9 dB.20 log b = 9; b @ 3

1t

= 322.2 = 0.8

D1(w) = 1 13 75 1.25.

ww

+

+ results in f M = 54º, gb = 4.3.

We constrain the lead section design by taking a = 1b

= 0.333. The

lag compensated system has a dB of


– 10 log 1a

= – 4.77 at 4 rad/sec.

1t

= 4 a = 2.3; 1a t

= 6.92

D2(w) = 0 4347 10 144 1..

ww

+

+Lag-lead compensator results in fM = 60º, g b = 8, wb = 7.61 rad/sec.

(d) D1(z) = 0.342 zz

-

-��

00

.923

.973; D2(z) = 2.49 z

z-

-��

0 7930 484

.

.Refer Figs 3.19–3.22 for realization schemes.

4.13 (a)

Kp = 0.8K; 11 0 8+ . K

= 0.02; K = 61.25; G(s) = 493 1s +

(b) Gh0G(z) = 70 8465.5215

.z -Closed-loop pole: z + 6.675 = 0The system is unstable.

(c) Gh0G(w) = 49 1 4

1 3( / )( )

-

+

ww

Lag compensator D(w) = 11 10

+

+

ww

satisfies the requirements.

w = 4 zz

-

+��

11

; D(z) = 0.122 zz

-

-��

0 60 951

..

(d) Lead compensation will increase the gain. Since gain is to be reducedto stabilize the system, lead compensation cannot be employed.

4.14 (a) Refer Example 4.5.

(b) Gh0G(z) = K z

z

( .9048)( )-

-

01 2

Sketch a root locus plot; complex roots lie on a circle. Using magni-tude condition, we obtain the value of K at the closed-loop polez = – 1. It is 2.1. Therefore, the system is stable for 0 < K < 2.1. Thereis a double pole at z = 0.81. The value of K at this point is obtained as0.38.

4.15 Gh0G(z) = K T e z e Te z

z e z

T T T

T[( ) ( ) ]

( )( )- + + - -

- -

- - - - -

- - -

1 11 1

1 2

1 1

SOLUTION MANUAL 33

(i) T = 1 sec,

Gh0G(z) = 0 3679 0 7181

1 0 3679. ( . )( )( . )

K zz z

+

- -

The root locus is a circle; the breakaway points are at z = 0.6479 andz = – 2.0841.

At the point of intersection of the root locus with unit circle, we findby magnitude condition, K = 2.3925. This gain results in marginalstability.

(ii) T = 2 sec, Gh0G(z) = 11353 0

1 0 1353). ( .5232)( )( .

K zz z

+

- -

Breakaway points are at z = 0.4783, – 1.5247; Critical gainK = 1.4557.

(iii) T = 4 sec, Gh0G(z) = 3 0183 0 3010

1 0 0183). ( . )( )( .

K zz z

+

- -

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

4.16 Gh0G(z) = K z

z z( . )

( )( . )+

- -

0 7171 0 368

The root locus is a circle with breakaway point at z = 0.648, – 2.08.

(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 getK = 0.18. The line through intersection point makes an angle of 32ºwith real axis. Therefore

wnT 1 2- z = 32º = 0.558 rad; wn = 0.644 rad/sec

4.17 z2 + 0.2Az – 0.1A = 0; 1 + 0 02

.2 ( .5)A zz

- = 0; Gh0G(z) = K zz

( .5)- 02 ;

= K = 0.2 A

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


4.18 G(s) = 11s +

; T = 0.1

H(s) = PT2p

= 0.9554; Gh0GH(z) = 0 0910

..9048

Kz -

The root locus is a circle.

(a) 1 + Gh0GH(z) = 0; z – 0.9048 + 0.091K = 0

For K = 1, z – 0.8138 = 0

e–T/t = 0.8138; Tt

= 0.206; t = 0.4854 sec

(b) Required time-constant = 0 48544

. = 0.12136

e–T/0.12136 = 0.43867; 0.091K = 0.46613; K = 5.1223

4.19 Gh0G(z) = 0 01873 0

1 0 8187. ( .9356)( )( . )

K zz z

+

- -

(a) z = 0.5; wn = 4.5; z = e nT-zw –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. Wechoose 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 con-dition at P. This gives

D(z) = zz

-

-

0 81870 1595..

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

D1(z) = 13 0 8187

0 1595.934( . )( . )

zz

-

-

(b) Kv = 11T z

limÆ

(z – 1) D(z) Gh0G(z) = 3.

(c) D2(z) = z

z

-

-

b

b

1

2

;1

11

2

-

-

b

b = 3

Let, b2 = 0.98, then b1 = 0.94; D2(z) = zz

-

-

00.94.98

From the root locus plots of 1 + D1(z) Gh0G(z) = 0 and1 + D2(z) Gh0G(z) = 0, it is seen that lag compensator decreases themargin of stability.

(d) Refer Figs 3.19 – 3.22 for realization of D2(z) D1(z).

SOLUTION MANUAL 35

4.20 Gh0G(z) = 0 0 368

0 368 0 315.2 ( . )

( . )( . )K z

z z+

- -

The root locus of the uncompensated system is a circle. At point P corre-sponding to the intersection of root locus with z = 0.5 locus, we get0.2 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

70

.5.957

= 7.837; D(z) = z

z

-

-

b

b

1

2

;1

11

2

-

-

b

b = 7.837

Let, b2 = 0.98, then b1 = 0.84

wn is slightly decreased with lag compensator (as seen from root-locusplot of the lag-compensated system), which is acceptable.

4.21 Gh0G(z) = KT z

z

2

221

1+

-( ); T = 1 sec

z = 0.7, wn = 0.3, z = e nT-zw–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 crite-rion gives K/2 = 0.18.

Gh0G(z) D(z) = 0 362

11

0 82

.( )

.zz

zz

�

�

� ;

Ka = 12 1T z

lim�

(z – 1)2 Gh0G(z) D(z) = 0.072

Corresponding to K/2 = 0.18, we find from the root locus that third pole islocated at z = 0.2. It slows down the response.

4.22 G(s) = 4040

120es s

s�

�

/

( ); T = 1

120 sec; Gh0G(z) =

0 00133 0 751 0 72

. ( . )( )( . )

zz z z

�

� �

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) isfound at z = – 0.4. The magnitude criterion at the point of intersection ofthe root locus with the circle of radius 0.56 gives K = 0.2.

Therefore, D(z) = 0 0 72

0 00133 0 4.2( . )

. ( . )z

z�

� = 150

( . )( . )zz�

�

0 720 4


The third pole corresponding to K = 0.2 lies inside the circle of radius0.56.

Gh0G(z) D(z) = 0 0 75

1 0 4.2( . )

( )( . )z

z z z�

� �

Kp = limz�1

Gh0G(z) D(z) = •; ess* = 0

4.23 (a) Refer Examples 4.9–4.10.

(b) G(s) = 1

1s s( )�; T = 0.1 sec; Gh0G(z) =

0 0048 01 0

. ( .9833)( )( .9048)

zz z

�

� �

We select an M(z) that has pole excess of at least equal to that ofGh0G(z), and unstable (or critically stable) poles of Gh0G(z) are in-cluded in 1–M(z) as zeros. M(z) = z–1 satisfies these requirements.

D(z) = 110G G z

M zM zh ( )( )

( )�

��

��

= 208.33 zz�

��

��

00.9048.9833

4.24 G(s) = 11s s( )�

; T = 0.1 sec

(a) z = 0.8, wn = 2

10

p

T; z = e nT-zw

– wnT 1 2- z

This gives z1,2 = 0.55 ± j0.22

Gh0G(z) = ( )( )

( )( )

T e z a

z z e

T

T

- + +

- -

-

-

1

1; a =

1

1

- -

- +

- -

-

e Te

T e

T T

T

= 4.8 × 10–3 ( . )

( )( . )

z

z z

+

- -

0 9833

1 0 9048

D(z) = z2 – 1.1z + 0.3509

We select an M(z) that has pole excess of at least equal to that ofGh0G(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

z z

-

- +

a

2 1 1 0 3509. .

E(z) = R(z) [1 – M(z)]

e*ss(unit-step) = limz�1

(z – 1) E(z) = 0

e*ss(unit-ramp) = limz�1

(z – 1) Tz

z -( )1 2 (z – 1) F(z) = TF(1) = 1Kv

SOLUTION MANUAL 37

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

F(1) =1

1 1 1 0 3509�

� �

�

. .

� = 0.7491 satisfies steady-state accuracy requirements.

M(z) =0 6491 0 3982

1 1 0 35092

. .. .

z

z z

�

� �

� �

D(z) = 110G G z

M zM zh ( )( )

( )�

��

��

= 135.227 ( . ) ( . )

( . ) ( . )

z z

z z

- -

+ -

0 9048 0 6135

0 9833 0 7491

(b) y(k) = 0, 0.5, 1, 1, �

Y(z) = 0.5z –1 + z –2 + z –3 + �

Y zR z

� �

� � = (0.5z –1 + z –2 + z –3 + �) (1 – z –1)

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

zz z

�

� �

0 9833)1 0 9048

D(z) = 104.17 ( . )

( . ( . )z z

z z

- +( )

+ +

0 9048 10 9833) 0 5

E(z) = R(z) [1 – M(z)]

ess = limz�1

(z – 1) E(z) = 0.15.

4.25 G(s) = 110 1s �

; T = 2 sec; Gh0G(z) = 0 18

0 82.

.z�

y(t) = 1 – e– t; Y(s) = 1 11s s

��

Y(z) = zz

zz�

��1 0 14.

= 0 86

1 0 14.

( ) ( . )z

z z� �;

M(z) = Y zR z z

( )( )

..

��

0 860 14

D(z) = 110G G z

M zM zh ( )( )

( )�

��

�� =

4 8 3 9

1

1

1. .�

�

�

�

z

z

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)


U zE z

( )( )

= D(z) = 10 6

1 0 75

1

1�

�

�

�

z

z.

4.27 G(s) = 21 2s s� ��

; T = 1 sec

Gh0G (z) = 0.4 z

z z�

� �

� �

� � � �

0 3680 368 0 135

.. .

; R (z) = 11 1� �z

M(z) may be chosen as z–1; but, as can be examined, the response willexhibit intersample ripples. We therefore take,

M(z) = �1z–1 + � 2z–2

E(z) = R(z) [1 – M(z)]

limz�1

(z – 1) R(z) [1 – M(z)] = 0 gives �1 + �2 = 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 + �]

Dividing U(z) by R(z),

U zR z

( )( )

= u(0) + [u(1) – u(0)]z–1 + [u(2) – u(1)]z–2 = �0 + �1z–1 + �2z–2

Gh0G(z) = Y zU z

( )( )

= Y z R zU z R z

( ) / ( )( ) / ( )

=

�

�

�

�

�

�

�

�

1

0

1 2

0

2

1

0

1 2

0

21

z z

z z

� �

� �

�

� �

=0 4 0 1472

1 0 503 0 04968

1 2

1 2

. .

. .

z z

z z

- -

- -

+

- +

Comparing the coefficients, we get

a1 = 0.731, a2 = 0.269,

�0 = 1.8275, �1 = – 0.919, �2 = 0.09

D(z) = 110G G z

M zM zh ( )

� �

� ��

��

= U z R zY z R z

� � � �

� � � ��

//1

=1 8275 0 0 09

1 0 731 0

1 2

1 2. .919 .

. .269

� �

� �

� �

� �

z z

z z

4.28 G(s) = 1

1s s �� ; T = 1 sec

SOLUTION MANUAL 39

Gh0G(z) = 0 3679 0

1 1 3679 0 3679

1 2

1 2. .2642

. .z z

z z

- -

- -

+

- +; M(z) = a1z–1 + a2z

–2 = Y zR z

( )

( )

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

R(z) = 1

1 1- -z; U z

R z

( )

( ) = b0 + b1z–1 + b2z–2

Gh0G(z) = Y z R zU z R z

( ) ( )

( ) ( )

//

=

�

�

�

�

�

�

�

�

1

0

1 2

0

2

1

0

1 2

0

21

z z

z z

� �

� �

�

� �

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) = 110G G z

M zM zh ( )� �

� ��

��

��

= U z R zY z R z

� � � �

� � � ��

//1

= 1 582 0 582

1 0 418

1

1

. .

.

-

+

-

-

z

z

Output sequence,

Y zR z

( )

( ) = 0.582z–1 + 0.418z–2; R(z) = 1

1 1- -z

y(k) = 0, 0.582, 1, 1, º

4.29 G(s) = es

s-

+

5

10 1; T = 5 sec; Gh0G(z) =

0 39351 0 6065

2

1.

.z

z

-

--

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 + � = 0.6225z–2 + z–3 11 1- -z

Y zR z

( )

( ) = M(z) = 0.6225z–2 + 0.3775z–3

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

D(z) = 110G G z

M zM zh ( )

� �

� ��

��

= 1.582 1 0 3678

1 0 6225 0 3775

2

2 3-

- -

-

- -

.. .

z

z z

U(z) = Y z R z

G G z R zh

( ) ( )

( ) ( )

//0

= 1.582 1 0 3678

1

2

1

-

-

-

-

. z

z

� ��


= 1.582 + 1.582z–1 + z–2 + z–3 + �

Since u(k) is constant for k ≥ 2, there are no intersample ripples in theoutput of the system after the settling time is reached.

4.30 Gh0G(z) = 0 3935

1 0 6065

2

1.

.z

z

-

--

Pole excess = 2; second-order model.

Y zR z

( )

( ) = a2z

–2

U zR z

( )

( ) = u(0) + [u(1) – u(0)]z–1 + [u(2) – u(1)]z–2 = b0 + b1z–1 + b2z–2

Y z R zU z R z

( ) ( )

( ) ( )

//

=

a

b

b

b

b

b

2

0

2

1

0

1 2

0

21

z

z z

-

- -+ + = Gh0G(z)

From the steady-state error requirement, a2 = 1. This gives

b0 = 2.541, b1 = – 1.541, b2 = 0

D(z) = 110G G z

M zM zh ( )( )

( )-

��

��

= U z R z

Y z R z

� � � �

� � � ��

//1

= 2.541 1 0 6065

1 1

1

1 1

-

- +

-

- -

. z

z z

� ��

It is physically realizable.

Y(z) = z–2 11 1- -z

= z–2 + z–3 + z–4 + �

y(k) = 0, 0, 1, 1, 1, �

u(0) = b0 = 2.541

u(1) = b1 + u(0) = 1, u(2) = b2 + u(1) = 1

u(k) = 2.541, 1, 1, 1 �

CHAPTER 5 CONTROL SYSTEM ANALYSIS USING STATEVARIABLE METHODS

5.1

Je = n2J = 0.4 ; Be = n2B = 0.01

�x1 = x2 ; 0.4 �x2 + 0.01x2 = 1.2 x3

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

or �x = Ax + bu

A =

0 1 0 0

0 0 025 3 0

0 12 190 1000

0 0 0 4 2

-

- -

-

�

�

��

�

�

��

.

.

; b =

0

0

0

0 2.

�

�

��

�

�

��

y = qL = nx1 = 0.5 x1

5.2

x3 x2 x1

ifu qq

KA KTs f fL + R + 1

1 1

Js + B

1

s

�x1 = x2 ; 0.5 �x2 + 0.5x2 = 10x3

20 �x3 + 100x3 = 50u ; y = x1

A =

0 1 0

0 1 20

0 0 5

-

-

�

�

��

�

�

��

; b =

0

0

2 5.

�

�

��

�

�

��

; c = [1 0 0]


5.3 x1 = qM ; x2 = �q M , x3 = ia ; y = qL

�x1 = x2

2 �x2 + x2 = 38 x3

2 �x3 + 21x3 = ea – 0.5 x2

ea = k1(qR – qL) – k2�q M = k1qR –

k1

20x1 – k2x2

A =

0 1 0

0 0 5 19

40

0 5

2212

1 2

-

- -+

-

�

�

��

�

�

��

.( . )k k

; b =

0

0

21k

�

�

��

�

�

��

; c = 1

200 0�

��

��

5.4 x1 = w ; x2 = ia

J �w + Bw = KT ia

Raia + Ladi

dta = ea – Kbw

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

A =-

- + - +

�

�

��

�

�

��

B

J

K

Jk K K K

L

R k K

L

T

t c b

a

a c

a

( ) ( )1 2 ;

b =0

1k K

Lc

a

�

�

��

�

�

��

; c = [1 0]

5.5 A = P–1 AP = -

-

�

��

�

��

11 6

15 8

b = P–1 b = 1

2�

��

�� ; c = cP = [2 –1]

Y s

U s

( )

( )=

P1 1D

D =

s

s s

-

- -- - -

2

1 21 3 2( ) =

1

3 22s s+ +

SOLUTION MANUAL 43

X2X = Y1

s–1

–2

–3

11U

s–1

Y s

U s

( )

( )=

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

D

+ + +

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

1 11 8 15 6 11 8

1 1 2 1 1 1 1

1 1 1 1 1 1

s s s s s s s

s s s s s s

- - - - - - -

- - - - - -

- + + - + +

- - + + - + -

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

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

=1

3 22s s+ +

X2

X1

Y

s–1

–1

6

2

–15–11

2

1

8

U

s–1

5.6 A = 0 1

0 0�

��

�

�� ; b =

0

1�

��

��

A= P–1 AP = 1 1

–1 –1�

��

�

�� ; b = P–1 b =

0

1�

��

��

|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

s s s

s s s

s s

( )

( )

+ +

- +

- -

�

�

��

�

�

��

3 3 1

1 3

1 2

; H(s) = 1

1

2Ds

s

�

�

��

�

�

��

D = s3 + 3s2 + 1


5.8

5.9 Taking outputs of integrators as state variables, we get (x1 being theoutput of rightmost integrator),

�x1 = x2

�x2 = – 2x2 + x3

�x3 = – x3 – x2 – y + u

y = 2x1 – 2x2 + x3

A =

0 1 0

0 2 1

2 1 2

-

- -

�

�

��

�

�

��

; b =

0

0

1

�

�

��

�

�

��

; c = [2 –2 1]

5.10 Taking outputs of integrators as state variables (x1 and x2 are outputs oftop two integrators from left to right; x3 and x4 are corresponding vari-ables for other integrators):�x1 = – 4x4 + 3u1; �x2 = x1 – 3x2 + u1 + 2u2;

�x3 = – x2 + 3u2; �x4 = – 4x4 + x3;

A =

0 0 0 4

1 3 0 0

0 1 0 0

0 0 1 4

-

-

-

-

�

�

��

�

�

��

; B =

3 0

1 2

0 3

0 0

�

�

��

�

�

��

; C = 0 1 0 0

0 0 0 1�

��

�

��

5.11 (a) G(s) = c (sI – A)–1 b

=s

s s

+

+ +

3

1 2( )( )

SOLUTION MANUAL 45

(b) G(s) =1

1 2( )( )s s+ +

5.12 a1 = – tr (A) = – 4

Q2 = A + a1I =

- -

-

- -

�

�

��

�

�

��

2 1 0

1 3 2

1 0 3

; a2 = –1

2tr(AQ2) = 6

Q3 = AQ2 + a2I =

1 1 2

3 2 4

1 1 3

-

- -

�

�

��

�

�

��

;

a3 = – 1

3tr(AQ3) = – 5

As a numerical check, we see that the condition: 0 = AQ3 + a3I is satis-fied. Therefore, D(s) = s3 – 4s2 + 6s – 5.

(sI – A)+ = Q1s2 + Q2s + Q3 = Q(s)

G(s) =CQ B( )

( )

s

sD =

1

D( )s

- + -

- + - +

�

��

�

��

3 5 4( 3

2 2 3 12 2

s s

s s s s

)

( )

5.13 �x1 = – 3x1 + 2x2 + [– 2x1 – 1.5x2 – 3.5x3]

�x2 = 4x1 – 5x2

�x3 = x2 – r

A =

- -

-

�

�

��

�

�

��

5 0 5 3 5

4 5 0

0 1 0

. .

; b =

0

0

1-

�

�

��

�

�

��

; c = [0 1 0]

G(s) = c(sI – A)–1 b = 14

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

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

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

�x1 + 2x1 = x2 ; �x2 + x2 = – x1 + u

y = x2 + (– x1 + u)

A =-

- -

�

��

�

��

2 1

1 1 ; b =

0

1�

��

�� ; c = [–1 1]; d = 1


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

x2 = output of lag 1/s

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

�x1 + 2x1 = y ; �x2 = – x1 + u

�x3 + x3 = – x1 + u ; y = x2 + x3

A =

-

-

- -

�

�

��

�

�

��

2 1 1

1 0 0

1 0 1

; b =

0

1

1

�

�

��

�

�

��

; c = [0 1 1]

5.15 Taking outputs of pseudo-integrators as state variables (x1, x2, x3, x4: fromtop to bottom):

�x1 + x1 = u1; �x2 + 5x2 = 5u2; �x3 + 0.5x3 = 0.4u1; �x4 + 2x4 = 4u2

u1 = K1r1 – K1y1; u2 = K2r2 – K2y2

y1 = x1 + x2; y2 = x3 + x4

Writing the state equations we get,

�x = Ax + Bu

A =

- - -

- - -

- - -

- - -

�

�

��

�

�

��

1 0 0

0 5 5 5

0 4 0 4 0 0

0 0 4 2 4

1 1

2 2

1 1

2 2

K K

K K

K K

K K

. . .5;

B =

K

K

K

K

1

2

1

2

0

0 5

0 4 0

0 4

.

�

�

��

�

�

��

; C = 1 1 0 0

0 0 1 1�

��

�

��

5.16 (i)Y s

U s

( )

( )=

s

s s

+

+ +

3

1 2( )( ) = 2

1

1

2s s++

+

–

LLLLL =-

-

�

��

�

��

1 0

0 2 ; b =

1

1�

��

�� ; c = [2 –1]

SOLUTION MANUAL 47

–1

2

–2

u +

+

+

+

+

+

x2

x1

y

–1

(ii) Y s

U s

( )

( ) =

b

a a a

3

s s s31

22 3+ + +

= 5

4 5 23 2s s s+ + +

From Eqns (5.56), the second companion form of the state model is givenbelow.

A =

0 0 2

1 0 5

0 1 4

-

-

-

�

�

��

�

�

��

; b =

5

0

0

�

�

��

�

�

��

; c = [0 0 1]

2u + + +

– – –

x2x1 x3 = y

2 5 4

5

(iii) Y s

U s

( )

( )=

b b b b

a a a

03

12

2 33

12

2 3

s s s

s s s

+ + +

+ + + =

s s s

s s s

3 2

3 2

8 17 8

6 11 6

+ + +

+ + +

From Eqns (5.54), the state model in second companion form is givenbelow.


A =

0 1 0

0 0 1

6 11 6- - -

�

�

��

�

�

��

; b =

0

0

1

�

�

��

�

�

��

; c = [2 6 2]

6

8

y

81

6

17

11

u +

+ + +

+ + +

+ +

++

–

x2 x1x3

z z z

5.17 (i)Y s

U s

( )

( )=

s

s s s

+

+ +

1

3 23 2 = b b

a a a

2 33

12

2 3

s

s s s

+

+ + +

From Eqns (5.56):

A =

0 0 0

1 0 2

0 1 3

-

-

�

�

��

�

�

��

; b =

1

1

0

�

�

��

�

�

��

; c = [0 0 1]

(ii)Y s

U s

( )

( )=

1

6 11 63 2s s s+ + + =

b

a a a

33

12

2 3s s s+ + +

From Eqns (5.54):

A =

0 1 0

0 0 1

6 11 6- - -

�

�

��

�

�

��

; b =

0

0

1

�

�

��

�

�

��

; c = [1 0 0]

(iii)Y s

U s

( )

( ) =

s s s

s s s

3 2

3 2

8 17 8

6 11 6

+ + +

+ + + = 1

1

1

2

2

1

3+

-

++

++

+s s s

SOLUTION MANUAL 49

LLLLL =

-

-

-

�

�

��

�

�

��

1 0 0

0 2 0

0 0 3

; b =

1

1

1

�

�

��

�

�

��

; c = [–1 2 1] ; d = 1

5.18 (a)Y s

U s

( )

( )=

1000 5000

52 1003 2

s

s s s

+

+ + =

b b

a a a

2 33

12

2 3

s

s s s

+

+ + +

From Eqns (5.54):

A =

0 1 0

0 0 1

0 100 52- -

�

�

��

�

�

��

; b =

0

0

1

�

�

��

�

�

��

; c = [5000 1000 0]

(b)Y s

U s

( )

( ) =

1000 5000

52 1003 2

s

s s s

+

+ + =

50 3125

2

18 75

50s s s+

-

++

-

+

. .

LLLLL =

0 0 0

0 2 0

0 0 50

-

-

�

�

��

�

�

��

; b =

50

31 25

18 75

-

-

�

�

��

�

�

��

.

.

; c = [1 1 1]

5.19Y s

U s

( )

( ) =

2 6 5

1 2

2

2

s s

s s

+ +

+ +( ) ( ) =

1

1

1

1

1

22( )s s s++

++

+

From Eqns (5.65):

LLLLL =

-

-

-

�

�

��

�

�

��

1 1 0

0 1 0

0 0 2

; b =

0

1

1

�

�

��

�

�

��

; c = [1 1 1]

x(t) =0

t

� eLLLLL(t – t) bu(t)dt

e–LLLLLt = ��–1 [(sI – LLLLL)–1] =

e te

e

e

t t

t

t

- -

-

-

�

�

��

�

�

��

0

0 0

0 0 2


e u dtt

L ( ) ( )-�

tt tb

0

=

( ) ( )

( )

( )

t e d

e d

e d

tt

tt

tt

-�

�

��

�

�

��

- -

- -

- -

�

�

�

t t

t

t

t

t

t

0

0

2

0

=

1

11

21 2

- -

-

-

�

�

��

�

�

��

- -

-

-

e te

e

e

t t

t

t( )

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

u

y

++

+

+

+

+

+

–

x3

1

1

1

–1

–2

–1

1

x2x1

5.20 (i) A = 1 1

0 2�

��

�

�� ; |lI – A| = (l – 1) (l – 2) = 0

l1 = 1; l2 = 2

adj(lI – A) = l

l

-

-

�

��

�

��

2 1

0 1

For l1 = 1: adj(l1I – A) = -�

��

�

��

1 1

0 0; v1 =

1

0�

��

��

For l2 = 2; adj(l1I – A) = 0 1

0 1�

��

�

��; v2 =

1

1�

��

��

SOLUTION MANUAL 51

(ii) A =-

-

�

��

�

��

3 2

1 0; |lI – A| = l2 + 3l + 2

l1 = – 1, l2 = – 2; v1 = 1

1�

��

�� , v2 =

2

1�

��

��

(iii) A =

0 1 0

3 0 2

12 7 6- - -

�

�

��

�

�

��

; |lI – A| = (l + 1) (l + 2) (l + 3) = 0

l1 = – 1, l2 = – 2, l3 = – 3

(l1I – A)v1 = 0 gives

- -

- - -

�

�

��

�

�

��

1 1 0

3 1 2

12 7 5

n

n

n

11

12

13

�

�

��

�

�

��

=

0

0

0

�

�

��

�

�

��

v1 = [1 – 1 – 1]T


v2 = [1 – 2 1/2]T


v3 = [1 – 3 3]T

5.21 (a)

A =

0 1 0 0

0 0 1 0

0 0 0 1

1 2 1

. . .

. . .

. . . . . . .

. . . . . . .

. . .

. . .- - - -

�

�

��

�

�

��-a a a an n

(liI – A) =

l

l

l

a a a l a

i

i

i

n n i i

-

-

-

+

�

�

��

�

�

��-

1 0 0

0 1 0 0

0 0 1

1 2

. . .

. .

.

.

. . .

. . .


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 bydeleting last row and first column] = (– 1)n+1 [(– 1)n–1] = 1

cn2 = (– 1)n+2 [determinant obtained by deleting last row and secondcolumn] = (– 1)n+2 [li(– 1)n–2] = li

cn3 = (– 1)n+3 [l2i (– 1)n–3] = l2

i

.

.

.

cnn = l in-1

P =

1 1 1

1 2

12

22 2

11

21 1

. . .

. . .

. . .

. . . . . .

. . .

l l l

l l l

l l l

n

n

n nnn- - -

�

�

��

�

�

��

(b)

A =

0 1 0

0 0 1

24 26 9- - -

�

�

��

�

�

��

; |lI – A| = 0 yields: l1 = –2, l2 = –3, l3 = – 4

P =

1 1 1

2 3 4

4 9 16

- - -

�

�

��

�

�

��

= [v1 v2 v3]

5.22 (a) A =

0 1 0

0 0 1

2 4 3- - -

�

�

��

�

�

��

The characteristic equation is,

l3 + 3l

2 + 4l + 2 = 0

l1 = – 1 + j1, l2 = – 1 – j1, l3 = – 1.

SOLUTION MANUAL 53

P =

1 1 1

1 1 1 1 1

2 2 1

- + - - -

-

�

�

��

�

�

��

j j

j j

;

P–1 AP = LLLLL =

- +

- -

-

�

�

��

�

�

��

1 1 0 0

0 1 1 0

0 0 1

j

j

(b) Q =

1 2 1 2 0

1 2 1 2 0

0 0 1

/ /

/ /

-�

�

��

�

�

��

j

j ; Q–1 LLLLLQ =

-

- -

-

�

�

��

�

�

��

1 1 0

1 1 0

0 0 1

5.23 A = -

-

�

��

�

��

4 3

6 5; |lI – A| = (l + 1) (l – 2) = 0

adj(lI – A) = l

l

-

- +

�

��

�

��

5 3

6 4;

l1 = – 1, adj(l1I – A) = -

-

�

��

�

��

6 3

6 3; v1 =

1

1�

��

��

l2 = 2, adj(l2I – A) = -

-

�

��

�

��

3 3

6 6; v2 =

1

2�

��

��

P = 1 1

1 2�

��

�

�� ; J = P–1 AP =

-�

��

�

��

1 0

0 2

eAt = PeJt P–1 = Pe

e

t

t

-

-

�

��

�

��

0

0 2P–1 =

2

2 2 2

2 2

2 2

e e e e

e e e e

t t t t

t t t t

- -

- -

- - +

- - +

�

��

�

��

5.24 (a) A = 0 3

1 4

-

-

�

��

�

�� ; (sI – A)–1 =

s

s s s s

s s

s

s s

+

+ +

-

+ +

+ + + +

�

�

��

�

�

��

4

1 3

3

1 31

1 3 1 3

( )( ) ( )( )

( )( ) ( )( )

eAt =

32

12

32

32

12

12

12

32

3 3

3 3

e e e e

e e e e

t t t t

t t t t

- - - -

- - - -

--

+

--

+

�

�

��

�

�

��


(b) eAt =

32

12

12

12

32

32

12

32

3 3

3 3

e e e e

e e e e

t t t t

t t t t

- - - -

- - - -

- -

-+

-+

�

�

��

�

�

��

5.25 (a) A = 0 1

6 5- -

�

��

�

�� ; |lI – A| = l2 + 5l + 6 = 0; l1 = – 2, l2 = – 3

g(l) = b0 + b1l; f(A) = eAt; f(l) = elt

f(li) = g(li) gives

e–2t = b0 – 2b1; e–3t = b0 – 3b1

Solving we get,

b1 = e–2t – e–3t, b0 = 3e–2t – 2e–3t

eAt = b0I + b1A = 3 2

6 6 2 3

2 3 2 3

2 3 2 3

e e e e

e e e e

t t t t

t t t t

- - - -

- - - -

- -

- + - +

�

��

�

��

(b) A = 0 2

2 4- -

�

��

�

�� ; f(A) = eAt, l1 = l2 = – 2; g(l) = b0 + b1l

f(l1) = g(l1) gives e–2t = b0 – 2b1

d

dlf ( )l

l = -2

= d

dlg( )l

l= -2

gives

b1 = te–2t, b0 = (1 + 2t)e–2t

eAt = ( )

( )

1 2 2

2 1 2

2 2

2 2

+

- -

�

��

�

��

- -

- -

t e te

te t e

t t

t t

5.26

SOLUTION MANUAL 55

G(s) =G s G s

G s G s11 12

21 22

( ) ( )

( ) ( )�

��

�

�� ; H(s) =

H s

H s1

2

( )

( )�

��

�

��

X s

x1

10

( )= G11(s) =

s s

s s s s s

- -

- - - - -

+

- - - + + - -

1 1

1 1 2 1 1

1 2

1 2 2 2 2

( )

( ) ( )( )

=s s- -+1 11 2( )

D =

1 2

1

1 2

3

/ /

s s++

+

X s

x1

20

( )= G12(s) =

s-2 1( )

D =

1 2

1

1 2

3

/ /

s s++

-

+

X s

x2

10

( )= G21(s) =

s-2 1( )

D =

1 2

1

1 2

3

/ /

s s++

-

+

X s

x2

20

( )= G22(s) =

s s- -+1 11 2( )

D =

1 2

1

1 2

3

/ /

s s++

+

X s

U s1( )

( )= H1(s) =

s s s- - -+ +2 1 11 1 2( ) ( )

D =

1

1s +

X s

U s2( )

( )= H2(s) =

1

1s +

Zero-input response:

x(t) = eAt x0 = � –1 [G(s)x0]

=1

2

3 3

3 310

20

e e e e

e e e e

x

x

t t t t

t t t t

- - - -

- - - -

+ -

- +

�

��

�

��

��

��

Zero-state response:

x(t) = e dtt

A b( ) ( )-�

t u t t

0

= ��–1 [H(s)U(s)] =

1

1

-

-

�

��

�

��

-

-

e

e

t

t

5.27 A =

0 1 0

0 0 1

6 11 6- - -

�

�

��

�

�

��

; b =

0

0

2

�

�

��

�

�

��

|lI – A| = (l + 1) (l + 2) (l + 3) = 0


P =

1 1 1

1 2 3

12

22

32

l l l

l l l

�

�

��

�

�

��

=

1 1 1

1 2 3

1 4 9

- - -

�

�

��

�

�

��

; LLLLL = P–1 AP =

-

-

-

�

�

��

�

�

��

1 0 0

0 2 0

0 0 3

;

b = P–1b =

1

2

1

-

�

�

��

�

�

��

; c = cP = [1 1 1]

The state model in Jordan canonical form:

�x = LLLLLx + bu; y = c x

The transformed initial vector is:

x0 = P–1 x0 =

1

2

1

-

�

�

��

�

�

��

The solution is,

x t1( ) = e– t x10 + e dt

t- -� ( )t

t

0

= 1

x2 (t) = e–2t x20 + e dt

t- -� 2

0

( )tt = – 1 – e–2t

x3 (t) = e–3t x30 + e dt

t- -� 3(

0

tt

) = 1

3

2

3+ e–3t

x(t) = Px (t)

x1(t) = 1

3 – e–2t +

2

3e–3t; x2(t) = 2(e–2t – e–3t)

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

5.28 A =0 1

2 3- -

�

��

�

�� ; eigenvalues are l1 = – 1, l2 = – 2

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

=s

s s s s s

+

+ ++

+ +

4

1 2

1

1 2( )( ) ( )( )

SOLUTION MANUAL 57

=3

1

2

2

1 2 1

1

1 2

2s s s s s+-

++ -

++

+

/ /

y(t) =1

22

3

22+ -- -e et t

5.29 Y(s) = C(sI – A)–1 BU(s) = 1 0

1 1�

��

�

��

s

s

+

-

�

��

�

��

3 1

2

2 1

0 1�

��

�

��

1

1 3

/

/( )

s

s +

�

��

�

��

=

3 16 18

1 2 34 6

2 3

2s s

s s s ss

s s s

+ +

+ + ++

+ +

�

�

��

�

�

��

( )( )( )

( )( )

y(t) = y t

y t1

2

( )

( )�

��

�

�� =

35

2

1

21 2

2 3

2 3

- - +

+ -

�

�

��

�

�

��

- - -

- -

e e e

e e

t t t

t t

5.30 �x1 = – 3x1 + 2x2 + [7r – 3x1 – 1.5 x2]

�x2 = 4x1 – 5x2

A =-

-

�

��

�

��

6 0 5

4 5

. ; b =

7

0�

��

�� ; c = [0 1]

eAt = ��–1 [(sI – A)–1]

=

1

3

2

3

0 5

3

0 5

34

3

4

3

2

3

1

3

4 7 4 7

4 7 4 7

e e e e

e e e e

t t t t

t t t t

- - - -

- - - -

+ -

- +

�

�

��

�

�

��

. .

y(t) = x2(t) = 74

3

4

34 7

0

e e dt tt

- - - --�

��

��( ) ( )t t

t

=28

3

1

41

1

714 7( ) ( )- - -�

��

��- -e et t

5.31 eAt = e– t t t

t t t

te te

te e te

+

- -

�

��

�

��

- -

- - -

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

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


Manipulation of the equation gives

x1(2) = 2e–1 x1(1) + e–1 x2(1) = 2

If x2(1) = 2k, then x1(1) = e1 – k

Thus

e k

k

1

2

-�

��

�

�� is a possible set of states

x

x1

2

1

1

( )

( )�

��

�

�� for any k π 0

5.32 (a) A =

0 1 0

3 0 2

12 7 6- - -

�

�

��

�

�

��

l1 = – 1, l2 = – 2, l3 = – 3

Modes: e–t, e–2t, e–3t

(b) Eigenvectors:

v1 =

1

1

1

-

-

�

�

��

�

�

��

; v2 =

1

2

1 2

-

�

�

��

�

�

��/

; v3 =

1

3

3

-

�

�

��

�

�

��

; P =

1 1 1

1 2 3

1 1 2 3

- - -

-

�

�

��

�

�

��/

x = Px ; �x = P–1 APx =

-

-

-

�

�

��

�

�

��

1 0 0

0 2 0

0 0 3

x

x (0) = P–1 x(0)

x (t) =

e

e

e

t

t

t

-

-

-

�

�

��

�

�

��

0 0

0 0

0 0

2

3

x

x

x

1

2

3

0

0

0

( )

( )

( )

�

�

��

�

�

��

x1(0) = x2 (0) = 0. x3 (0) = k π 0

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

x(0) = Px (0) = P

0

0

k

�

�

��

�

�

��

=

k

k

k

-

�

�

��

�

�

��

3

3

; k π 0

SOLUTION MANUAL 59

5.33 (a) �x = 0 1

2 1�

��

�

�� x; y = [1 2]x

l1 = 2, l2 = –1.

Modes are e2t and e–t

(b) adj(lI – A) = l

l

-�

��

�

��

1 1

2

For l = 2, adj(lI – A) = 1 1

2 2�

��

�

�� ; v1 =

1

2�

��

��

For l = –1, adj(lI – A) = -

-

�

��

�

��

2 1

2 1; v2 =

1

1-

�

��

��

P = 1 1

2 1-

�

��

�

��

x = Px gives

�x = P–1 APx = 2 0

0 1-

�

��

�

��

x

x1

2

�

��

�� ; y(t) = cPx = [5 –1]x

x t

x t1

2

( )

( )�

��

�

�� =

e

e

t t

t t

2 0

0

0

0

( )

( )

-

- -

�

��

�

��

x t

x t1 0

2 0

( )

( )�

��

�

��

If x t1 0( ) = 0, the mode e2t will be suppressed in y(t) = 5x1 – x2 .

x(t0) = Px (t0) = P0

K�

��

�� =

K

K-

�

��

�� ; K π 0

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

5.34e

e

t

t

-

--

�

��

�

��

2

22 =

a b

c d�

��

�

�� -

�

��

��

1

2;

e

e

t

t

-

--

�

��

�

�� =

a b

c d�

��

�

�� -

�

��

��

1

1

This gives

eAt =a b

c d�

��

�

�� =

2

2 2 2

2 2

2 2

e e e e

e e e e

t t t t

t t t t

- - - -

- - - -

- -

- -

�

��

�

��


= ��–1 [(sI – A)–1]

(sI – A)–1 =

2

1

1

2

1

1

1

22

2

2

1

2

2

1

1

s s s s

s s s s

+-

+ +-

+

+-

+ +-

+

�

�

��

�

�

��

(sI – A) =s

s

-

+

�

��

�

��

1

2 3

A =0 1

2 3- -

�

��

�

��

5.35 V =

c

cA

cA

�

n-

�

�

��

�

�

��1

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

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

5.37 None of the rows of the B matrix corresponding to the Jordan blocks forl1 = – 1, l2 = – 2, l3 = – 3 is zero row. The system is completely control-lable.

First column of C matrix which corresponds to the eigenvalue –1, is azero column. Other columns are nonzero. The system is not completelyobservable.

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

0 1

-�

��

�

�� ; r(U) = 2

Completely controllable

V =c

cA�

��

�� =

1 1

3 3

-

-

�

��

�

�� ; r(V) = 1

Not completely observable.

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

(iii) The system is in Jordan canonical form; it is both controllable andobservable.

SOLUTION MANUAL 61

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

r (V) = r

c

cA

cA2

�

�

��

�

�

��

= 3

The system is completely observable.

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

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

The system is completely controllable.

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

2s +

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

(ii) G(s) =s

s s

+

+ +

4

2 3( )( )

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

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

The system is unstable.

(b) G(s) = c(sI – A)–1 b

=1

1 2( )( )s s+ +

The G(s) is stable

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


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

= b

a a

22

1 2s s+ +

A =0 1

0 1-�

��

�� ; b =

0

1�

��

�� ; c = [10 0]

(b) Let us obtain controllable companion form realization of

G(s) = 10 2

1 2

( )

( )( )

s

s s s

+

+ + =

10 20

3 23 2

s

s s s

+

+ + =

b b

a a a

2 33

12

2 3

s

s s s

+

+ + +

A =

0 1 0

0 0 1

0 2 3- -

�

�

��

�

�

��

; b =

0

0

1

�

�

��

�

�

��

; c = 20 10 0

(c) Let us obtain observable companion form realization of

G(s) =10 2

1 2

( )

( )( )

s

s s s

+

+ +

A =

0 0 0

1 0 2

0 1 3

-

-

�

�

��

�

�

��

; b =

20

10

0

�

�

��

�

�

��

; c = 0 0 1

5.42 Refer Section 5.4.

CHAPTER 6 STATE VARIABLE ANALYSIS OF DIGITALCONTROL SYSTEMS

6.1 F =

�

�

�

�

�

��

�

�

��

3 1 0

4 0 1

1 0 0

; g =

�

�

�

�

��

�

�

��

3

7

0

(zI – F) =

z

z

z

� �

�

�

�

��

�

�

��

3 1 0

4 1

1 0

; |zI – F| = z3 + 3z2 + 4z + 1 = �

(zI – F)–1 z =1

�

z z

z z z z

z z z

2 1

4 1 3 3

1 3 4

� � � �

� � � �

�

�

��

�

�

��

( ) ( )

( )

z = G(z)

H(z) = (zI – F)–1 g = 1

�

� �

� � �

�

�

�

��

�

�

��

3 7

7 9 3

3 7

2

2

z z

z z

z

6.2


6.3 F = 2 51

21

�

�

�

��

�

��; g =

1

0�

��

��; c = [2 0]; |zI – F| = z2 – z +

1

2 = �

G(z) = c(zI – F)–1 g = 2 2

12

2

z

z z

�

� �

6.4 C = 2 0

1 1�

�

��

��; G =

1 2 0

0 1 3

��

��

��; F =

2 51

21

�

�

�

��

�

��; D =

0 4 0

0 0 2�

�

��

��

� = |zI – F| = z2 – z + 1

2

G(z) = C(zI – F)–1 G + D

= 1

�

2 2 4 4 14 301

23 4 2 3 9

z z

z z z

� � � �

� � � � � �

�

��

�

��

�

�

6.5 x1, x2 and x3 : outputs of unit delayers starting at the top and proceedingtowards the bottom.

x1(k + 1) = 1

2x1(k) +

1

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

x2(k + 1) = – 1

2x2(k) – x3(k) – u(k)

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

3x3(k) + 2u(k)

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

F =

1

2

1

42

01

21

0 31

3

��

�

�

��

�

�

��

; g =

1

1

2

-

�

�

��

�

�

��; c = [5 6 –7]; d = 8

6.6 x1, x2 and x3 : outputs of unit delayers starting at the top and proceedingtowards 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 65

F =

0 1 0

3 0 2

12 7 6� � �

�

�

��

�

�

��

; G =

1 0

0 0

0 1

�

�

��

�

�

��

; C = 0 2 0

0 0 1�

��

��; D =

2 0

0 1�

��

��

6.7 (i)Y z

R z

( )( )

= 3 3

13

23

2

2

z z

z z

� �

� �

= b b b

a a

02

1 22

1 2

z z

z z

+ +

+ +

F = 0 12

3

1

3

��

��

�

��

; g = 0

1�

��

��; c = [– 1 – 2]; d = 3

(ii)Y z

R z

( )( )

= � � � �

� � �

2 2 234

3 2

3 2

z z z

z z z =

� � � �

� � �0

31

22 3

31

22 3

z z z

z z z

� � �

� � �

F =

0 034

1 0 1

0 1 1�

�

�

��

�

�

��

; g =

0 5

3

4

.

�

�

�

��

�

�

��

; c = [0 0 1]; d = – 2

6.8 (i)Y z

R z

( )( )

= 1 – 1

1z� +

2

1z� +

1

3z�

F =

�

�

�

�

�

��

�

�

��

1 0 0

0 2 0

0 0 3

; g =

1

1

1

�

�

��

�

�

��

; c = [– 1 2 1]; d = 1

(ii)Y z

R z

( )( )

= 5

13

3

z��

+ �

��

2

13

2

z

+ 3

13

z��

F =

1

31 0

01

31

0 01

3

�

�

��

�

�

��

; g =

0

0

1

�

�

��

�

�

��

; c = [5 – 2 3]; d = 0


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)

F =

0 0 3

1 0 7

0 1 5

�

�

�

�

�

��

�

�

��

; g =

0

0

0

�

�

��

�

�

��

; c = [0 0 1]; d = 0

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

Y z

R z

( )

( ) =

5 3

3 22

z

z z

�

� � =

�

�

2

1z +

7

2z�; F =

-

-��

��

1 0

0 2; g =

1

1�

��

��;

c = [–2 7]; d = 0

(iii) y(k + 3) + 5y(k + 2) + 7y(k + 1) + 3y(k) = r(k + 1) + 2r(k)

Y z

R z

( )

( ) =

� �

� � �2 3

31

22 3

z

z z z

�

� � � =

z

z z z

�

� � �

2

5 7 33 2

F =

0 1 0

0 0 1

3 7 5- - -

�

�

��

�

�

��; g =

0

0

1

�

�

��

�

�

��

; c = [2 1 0]; d = 0

6.10 F = 0 1

3 4-��

��; g(�) = �0 + �1�

|�I – F| = �2 – 4� + 3 = 0; �1 = 1, �2 = 3

f (�) = �k; f (1) = 1 = g(1) = �0 + �1; f (3) = 3k = g(3) = �0 + 3�1

�0 = 1.5 – 0.5(3)k; �1 = 0.5[(3)k – 1]

Fk = �0I + �1F = 1 5 0 5 3 0 5 3 1

1 5 3 1 0 5 15 3

. . ( ) . [( ) ]

. [( ) ] . . ( )

� �

� � �

�

��

�

��

k k

k k

6.11 Refer Example 6.3

(zI – F)–1 = 1

0 2 0 8( . )( . )z z� �

z

z

�

�

�

��

��1 1

0 16.

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

z�1

SOLUTION MANUAL 67

X(z) =

-

++

++

-

++

-

++

-

�

�

��

�

�

��

176

0 2

229

0 8

2518

13 46

0 2

17 690 8

718

1

z

z

z

z

z

z

z

z

z

z

z

z

. ..

.

.

.

x(k) =

��

� � � �

�

�

��

�

�

��

17

60 2

22

90 8

25

183 4

60 2

17 6

90 8

7

18

( . ) ( . )

.( . )

.( . )

k k

k k; y(k) = cx(k) = x1(k)

6.12 x1(0) = – 5; x2(0) = 1

zX1(z) + 5z = 3

2X1(z) – X2(z) + 3U(z)

zX2(z) – z = X1(z) – X2(z) + 2U(z)

U(z) = z

z�1 2/

X1(z) = �

�

51

2

z

z + 2

12

z

z + +

�

�

2

1

z

z; X2(z) =

�

�

21

2

z

z +

41

2

z

z� +

�

�

z

z 1

x1(k) = – 51

2��

k

+ 2��

1

2

k

– 2; x2(k) = –21

2��

k

+ 4��

1

2

k

– 1

y1(k) = – 3x1(k) + 4x2(k) – 2u(k) = 51

2��

k

+ 10��

1

2

k

+ 2

y2(k) = – x1(k) + x2(k) = 31

2��

k

+ 2��

1

2

k

+ 1

6.13 F =

-

-

-

�

�

��

�

�

��

1 1 0

0 1 0

0 0 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

(b) x(k) = Fkx(0) =

( ) ( )

( )

( )

� �

�

�

�

�

��

�

�

��

�1 1 0

0 1 0

0 0 2

1k k

k

k

k 0

1

1

�

�

��

�

�

��

=

k k

k

k

( )

( )

( )

�

�

�

�

�

��

�

�

��

�1

1

1

1


6.14 (a) Ga(s) = 1

2s s( )�; G(z) = (1 – z–1) �

1

22s s( )�

��

��

=0 2838 0 1485

11353 0 13532

. .

. .

z

z z

�

� �

F = 0 1

01353 11353�

�

��

��. .; g =

0

1�

��

��; c = [0.1485 0.2838]; d = 0

(b) A = 0 1

0 2-��

��

; b = 0

1�

��

��; c = [1 0]

F = eAT = 1

1

21

0

2

2

( )��

�

��

�

�

��

�

�

e

e

T

T; g = e dtt

TA

0��

��

�

��

b =

1

2

1

21

21

2

2

( )

( )

Te

e

T

T

��

�

�

�

��

�

�

��

�

�

Thus for T = 1 sec,

F = 1 0 4323

0 0 1353

.

.�

��

��; g =

0 2838

0 4323

.

.�

��

��; c = [1 0]

6.15 F = eAt = 0 696 0 246

0 123 0 572

. .

. .�

��

��; g = (eAT – I)A–1b =

��

��

��0 021

0 747

.

.;

c = [2 – 4]; d = 6

6.16 Let x1 be the output of the block 1

s and x2 be the output of the block

1

10 1s�

�x1 = x2; �x2 = – 0.1x2 + u + 0.1w

A = 0 1

0 0 1�

�

��

��.; b =

0

1�

��

��; b1 =

0

0 1.�

��

��

(sI – A)–1 =

1 1

0 1

01

0 1

s s s

s

( . )

.

�

�

�

�

��

�

�

��; eAt =

1 10 1

0

0 1

0 1

( ).

.

��

��

�

��

�

�

e

e

t

t

F = 1 10 1

0

0 1

0 1

( ).

.

-�

��

�

��

-

-

e

e

T

T; g = e d

TA b� �

0� =

10 100 100

10 10

0 1

0 1

T e

e

T

T

� �

�

�

��

�

��

�

�

.

.

SOLUTION MANUAL 69

g1 = e dT

A b� �1

0� =

T e

e

T

T

� �

�

�

��

�

��

�

�

10 10

1

0 1

0 1

.

.

For T = 0.1 sec,

F = 1 0 1

0 0 99

.

.�

��

��; g =

0 005

01

.

.�

��

��; g1 =

0

0 01.�

��

��;

x k

x k1

2

1

1

( )

( )

�

�

�

��

�� =

1 0 1

0 0 99

.

.�

��

��x(k) +

0 005

0 1

.

.�

��

��u(k) +

0

0 01.�

��

��w(k)

6.17 (sI – A)–1 =

1

0 10

0 1

01

1

0 12

s

s s

�

� �

�

�

��

�

�

��

..

( . ) .

; eAt = e

te e

t

t t

�

� �

�

��

�

��

0 1

0 1 0 1

0

0 1

.

. ..; T = 3 sec

F = eAt = 0 741 0

0 222 0 741

.

. .�

��

��; G = e d

TA B� �

0� =

259182 0

36 936 259182

.

. .�

��

��;

C = 1 0

0 1�

��

��

6.18 (a) G(z) = (1 – z–1) �1

12s s( )�

�

��

�� =

0 3679 0 7181

1 0 3679

. ( . )

( )( . )

z

z z

�

� �

(b)Y z

R z

( )( )

= G z

G z

( )( )1�

= 0 3679 0 2642

0 63212

. .

.

z

z z

�

� �

F = 0 1

0 6321 1�

�

��

��.; g =

0

1�

��

��; c = [0.2642 0.3679]

6.19 Ga(s) = e

s

s�

�

0 4

1

.

(a) G(z) = (1 – z–1) �e

s s

s�

�

��

��

0 4

1

.

( ) = (1 – z–1) �

e

s

e

s

s s� �

��

��

��

0 4 0 4

1

. .

T = 1; �D = 0.4 = �T; m = 1 – � = 0.6

From Table 3.1, we obtain

G(z) = (1 – z–1) �1

1

0 6

1z

e

z e��

�

��

��

�

�

.

= 0 4512 0 1809

0 36792

. .

.

z

z z

�

�


F = 0 1

0 0 3679.�

��

��; g =

0

1�

��

��; c = [0.1809 0.4512]

(b)Y s

U s

( )( )

= e

s

s�

�

0 4

1

.

State model,

�x1(t) = – x1(t) + u(t – 0.4); �D = 0.4, N = 0, � = 0.4, m = 0.6

F = e–1 = 0.3679

g2 = e d��

0

0 6.

= 0.4512; g1 = e–0.6 e d��

0

0 4.

= 0.1809

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 = 0 3679 01809

0 0

. .�

��

��; g =

0 4512

1

.�

��

��; c = [1 0]

6.20 �x (t) = – x(t) + u(t – 2.5); �D = 2.5 = NT + �T; T = 1N = 2, � = 0.5, m = 0.5

F = e–1 = 0.3679; g2 = e d��

0

0 5.

= 0.3935;

g1 = e–0.5 e d��

0

0 5.

= 0.2387

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)

F =

0 3679 0 2387 0 3935 0

0 0 1 0

0 0 0 1

0 0 0 0

. . .�

�

��

�

�

��

; g =

0

0

0

1

�

�

��

�

�

��

6.21Y s

U s

( )

( ) = Ga(s) =

e

s

s D� �

2 ; 0 £ tD £ T

x1 = y, x2 = �y ; �x1 = x2, �x2 = u(t – tD)

SOLUTION MANUAL 71

A = 0 1

0 0�

��

��; b =

0

1�

��

��; c = [1 0]

F = eAT = I + AT + A2 2

2

T

! + � =

1

0 1

T�

��

��

tD = �T = (1 – m)T

g2 = e dmT

A b��

0� =

( ) /T

TD

D

-

-

�

��

�

��

t

t

2 2; g1 = eAmT e d

TA b�

�

0

�

� = tt

t

DD

D

T( )-�

��

�

��

2

N = 0 (Refer Eqn. (6.35))

x(k + 1) = 1

0 1

T�

��

��x(k) + g1u(k – 1) + g2u(k)

Introduce x3 = u(k – 1). Then,

x(k + 1) = Fx(k) + gu(k); y = cx(k)

F =

12

0 1

0 0 0

T TDD

D

tt

t

-��

�

�

��

�

�

��

; g =

( )T

T

D

D

-

-

�

�

��

�

�

��

t

t

2

2

1

; c = [1 0 0]

6.22 For this problem, the solution comes from Review Example 6.3 withK = 2.

6.23 x1, x2 and x3: outputs of the blocks 1

s,

1

1s� and

1

2s� respectively.

(a)

�x1 = x2; �x2 = – x2 + x3; �x3 = – 2x3 + u

A =

0 1 0

0 1 1

0 0 2

-

-

�

�

��

�

�

��

; b =

0

0

1

�

�

��

�

�

��

Eigenvalues of A are �1 = 0, �2 = –1, �3 = –2

g(�) = �0 + �1� + �2�2; f (�) = e�T

For � = 0, e�T = 1 = �0

For � = –1, e–T = �0 – �1 + �2

For � = – 2, e–2T = �0 – 2�1 + 4�2


�0 = 1; �1 = 1

2 (3 + e–2T – 4e–T); �2 =

1

2 (1 + e–2T – 2e–T)

eAT = �0I + �1A + �2A2

=

1 11

21 2

0

0 0

2

2

2

� � �

�

�

�

��

�

�

��

� � �

� � �

�

e e e

e e e

e

T T T

T T T

T

( )

;

g = e dT

A b� �

0� =

��

� �

�

�

�

��

�

�

��

� �

� �

�

3

4

1

2

1

41

2

1

21

2

1

2

2

2

2

T e e

e e

e

T T

T T

T

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

F =

74

12

14

1 12

1 2

12

12

12

12

0

2 2

2 2

2 2

- - + - + -

- + - -

- +

�

�

��

�

�

��

- - - - -

- - - - -

- -

T e e e e e

e e e e e

e e

T T T T T

T T T T T

T T

( )

(b) x(t) = eA(t – kT) x(kT) + e u dt

kT

tA( ) ( )-� t

t tb ; kT £ t < (k + 1)T

6.24 Eigenvalues of A are �1 = –1, �2 = –2.

e�T = �0 + �1�

e–T = �0 – �1; e–2T = �0 – 2�1

�1 = e–T – e–2T; �0 = 2e–T – e–2T;

eAt = �0I + �1A = 2

2 2 2

2 2

2 2

e e e e

e e e e

T T T T

T T T T

� � � �

� � � �

� �

� � � �

�

��

�

�� ;

g = e dT

A b� �

0� =

0 5 0 5 2

2

. .� �

�

�

��

�

��

� �

� �

e e

e e

T T

T T

SOLUTION MANUAL 73

For T = 1 sec, eAT = 0 6 0 233

0 465 0 0972

. .

. .� �

�

��

��; g =

0 2

0 233

.

.�

��

��

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 = 0 4 0 233

0 698 0 0972

. .

. .� �

�

��

��; g =

0 2

0 233

.

.�

��

��; c = [1 0]

6.25 Plant model,

x(k + 1) = 0 6 0 233

0 465 0 0972

. .

. .� �

�

��

��x(k) +

0 2

0 233

.

.�

��

��e2(k);

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)

F =

0 6 0 233 0 2

0 465 0 0972 0 233

1 0 2

. . .

. . .� �

� �

�

�

��

�

�

��

; g =

0

0

1

�

�

��

�

�

��

; c = [1 0 0]

6.26 D(s) = 9 + 4.1s = U s

E s

( )

( ); 4.1

de t

dt

( ) + 9e(t) = u(t)

By backward-difference approximation,

4.1e k e k

T

( ) ( )� ��

��

1 + 9e(k) = u(k); u(k) = – 41e(k – 1) + 50e(k)

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 con-troller. The plant difference equations are given by (solution to Problem6.16)

x(k + 1) = 1 0 1

0 0 99

.

.�

��

��x(k) +

0 005

0 1

.

.�

��

��u(k)


The augmented state model becomes,

x k

x k

x kc

1

2

1

1

1

( )

( )

( )

�

�

�

�

�

��

�

�

��

=

1 0 1 0

0 0 99 0

0 0 0

.

.�

�

��

�

�

��

x k

x k

x kc

1

2

( )

( )

( )

�

�

��

�

�

��

+

0

0

1

�

�

��

�

�

��

e(k)

+

0 005

0 1

0

.

.�

�

��

�

�

��

[– 41xc(k) + 50(r(k) – x1(k))]

=

0 75 0 1 0 205

5 0 99 4 1

1 0 0

. . .

. .

�

� �

�

�

�

��

�

�

��

x(k) +

0 25

5

1

.�

�

��

�

�

��

r(k)

6.27 (a) U = [g Fg] = 1 1 1 1

0 0 1 1

- -

-��

��; � (U) = 2; V =

c

cF�

��

��; � (V) = 2

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

observable.

6.28 A = 0 1

1 0-��

��

; b = 0

1�

��

��; c = [1 0]

U = [b Ab] = 0 1

1 0�

��

��; � (U) = 2; V =

c

cA�

��

�� =

1 0

0 1�

��

��; � (V) = 2

The continuous-time system is both controllable and observable.

Comparing the given state model with Eqn (6.59), we observe that thegiven system is a harmonic oscillator with frequency � = 1 or periodT = 2.

Therefore, T = n; n = 1, 2, � will lead to hidden oscillations.

For T = , for example, the discrete-time model is given by (refer Eqns(6.60))

x(k + 1) = -

-��

��

1 0

0 1x(k) +

2

0�

��

��u(k); y(k) = [1 0]x(k)

The system is obviously both uncontrollable and unobservable.For T = 2, we have the following discrete-time model;

x(k + 1) = 1 0

0 1�

��

��x(k) +

0

0�

��

��u(k); y(k) = [1 0]x(k)

The system is both uncontrollable and unobservable. In general, forT = n, n = 1, 2, � the system is uncontrollable and unobservable.

SOLUTION MANUAL 75

6.29 The poles are: s1 = –1, s2,3 = –0.02 ± j

For Re(�i) = Re(�j)

Im(�i – �j) = 2

T �2

2

n

� n; T � 1, 2, 3, �

6.30 (a) |�I – F| = � ��

��

1

4

1

2

Eigenvalues of F are: �1 = 1

4, �2 =

1

2

(b) G(z) = c(zI – F)–1 g = 1

14

z��

(c) Since there is a pole-zero cancellation, the system is either uncontrol-lable or unobservable or both.

Given state model is in controllable canonical form. Therefore, themodel is controllable but not observable.

CHAPTER 7 POLE-PLACEMENT DESIGN AND STATEOBSERVERS

7.1Y s

U s

( )( )

= b b b

a a

11

22

11

s s

s s

n nn

n nn

- -

-

+ + +

+ + +

�

�

�x =

0 1 0 0

0 0 1 0 0

0 0 1

1 1

. .

.

. . . . . .

. . .

. . .- - -

�

�

��

�

�

��-a a an n

x +

0

0

1

.

.

�

�

��

�

�

��

u

y = [�n �n–1 � �1] x; u = – kx + r; k = [k1 k2 � kn]

�x = [A – bk] x + br; y = cx

A – bk =

0 1 0 0

0 0 1 0 0

0 0 1

1 1 2 1

. .

.

. . . . . .

. . .

. . .- - - - - -

�

�

��

�

�

��-a a an n nk k k

G(s) = c[sI – (A – bk)]–1 b = c

*1

*2

.

.

*n

�

�

��

�

�

��

0

0

1

.

.

�

�

��

�

�

��

× 1

�

*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]

A – bk =

0 1 0

0 0 1

6 11 61 2 3� � � � � �

�

�

��

�

�

��k k k

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]

(b) ~x = x – �x

~�x = (A – mc)~x ; mT = [m1 m2 m3]

A – mc =

�

�

� � � �

�

�

��

�

�

��

m

m

m

1

2

3

1 0

0 1

6 11 6

|sI – (A – mc)| = s3 + (6 + m1)s2 + (11 + 6m1 + m2)s + 11m1 + 6m2+ 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 inFig. 7.6.

(c) �x =

0 1 0

0 0 1

6 11 6- - -

�

�

��

�

�

��

x +

0

0

1

�

�

��

�

�

��

u

y = [1 0 0]x

Aee = 0 1

11 6- -��

��

; ale = [1 0]

~�x e = (Aee – male)~xe ; mT = [m1 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 k3]

A – bk =

� � � �

�

�

�

�

��

�

�

��

k k k1 2 36

1 0 11

0 1 6


|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

(c)

�

�

�

x

x

x

3

1

2

�

�

��

�

�

��

=

�

�

�

�

�

��

�

�

��

6 0 1

6 0 0

11 1 0

x

x

x

3

1

2

�

�

��

�

�

��

+

0

1

0

�

�

��

�

�

��

u; y = [1 0 0]

x

x

x

3

1

2

�

�

��

�

�

��

Aee = 0 0

1 0�

��

��; ale = [0 1]; mT = [m1 m2]

|sI – (Aee – male)| = (s + 2 + j3.464) (s + 2 – j3.464)

m1 = 16, m2 = 47.4 Estimation error should decay in less than 4 sec.

Settling time = 4

zw n

£ 4; zwn ≥ 1

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 = -��

1

3

��x = (A – mc) �x + (B – md)u + my

7.5 The estimation error should decay as fast as e–10t at least. Hence, the poleshould be at s = – 10. Desired characteristic polynomial: s + 10

�x = Ax + bu; y = cx

A = 1 0

0 0�

��

�

�� ; b =

1

1�

��

�� ; c = [2 –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 79

Q = 1 1

1 2�

��

�

�� ; Q–1 =

2 1

1 1

-

-��

��

;

�q = 2 2

1 1- -��

��

q + 1

0�

��

��u; y = [1 0]q

~�q2 = (Aee – male)~q2 = (– 1 – 2m) ~q2

Characteristic polynomial: s + 1 + 2m = s + 10; m = 4.5

Refer Eqn (7.49b)

�q2¢ = – 10 �q2 – 10y – 4.5u; �q2 = q¢2 + my; �q = �

�

q

q1

2

�

��

�� =

q

q1

2�

�

��

��

�x = Q �q = y q

y q

+

+

�

��

�

��

�

�

2

22

7.6 (a) A = 0 9

1 0�

��

�

�� ; b =

9

0�

��

�� ; c = [0 1]

(b) k = [k1 k2]

(A – bk) = - -�

��

�

��

9 9 9

1 01 2k k

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

Desired polynomial: (s + 3 + j3) (s + 3 – j3) = s2 + 6s + 18.

k = [2/3 3]

(c) m = [m1 m2]T

A – mc = 0 9

11

2

-

-

�

��

�

��

m

m; |sI – (A – mc)| = s2 + m2s + m1 – 9

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

m = 81

12�

��

��

(d) A = 0 9

1 0�

��

�

�� ; b =

9

9�

��

�� ; c = [0 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 = 1

9

2

9��

��

;

A – bk = -

-��

��

1 7

0 2; c(A – bk) = [0 –2]

V = c

c A bk( )-

�

��

�

�� =

0 1

0 2-��

��

; r (V) = 1. System is unobservable

7.7 ��y + w 02 y = u

(a) �x = 0 1

002-

�

��

�

��

wx +

0

1�

��

��u

(b) wn = 2w0; z = 1; s = – 2w0

Characteristic polynomial: (s + 2w0)2 = s2 + 4w0s + 4w0

2

A – bk = 0 1

02

1 2- - -

�

��

�

��

w k k; |sI – (A – bk)| = s2 + k2s + w0

2 + k1

k = [3w02 4w0]

(c) m = [m1 m2]T

|sI – (A – mc| = (s + 10w0)2 gives m = 20

990

02

w

w

�

��

�

�� .

Figure 7.6 gives a block diagram of the observer-based state-feed-back control system.

(d) A = 0 1

002-

�

��

�

��

w; b =

0

1�

��

�� ; c = [1 0]; ~�x2 = (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.

� ¢x2 = – 10w0 ¢x2 – 101w 02 y + u; �x2 = ¢x2 + 10w0y

7.8 (a) k = [k1 k2]

(A – bk) = 0 1

20 6 1 2. - -

�

��

�

��k k

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 = 16

84 6.�

��

�

��

(c)U s

Y s

( )

( )- = k [sI – (A – bk – mc)]–1 m =

77816 3690 72

19 6 151 22

. .

. .

s

s s

+

+ +

(d) �x = Ax – bk �x; ��x = (A – mc – bk) �x + my

= -

- -

�

��

�

��

16 1

93 6 3 6. .�x +

16

84 6.�

��

�

��x1

7.9 (a) A = 0 1

0 0�

��

�

�� ; b =

0

1�

��

�� ; c = [1 0]

(b) wn = 1; z = 1

2;

|sI – (A – bk)| = s2 + 2zwns + w n2 gives k = [k1 k2] = [1 2 ]

(c) wn = 5; z = 0.5; s2 + 2zwns + w n2 = s2 + 5s + 25

|sI – (A – mc)| = s2 + 5s + 25 gives m = m

m1

2

�

��

�� =

5

25�

��

��

(d)U s

Y s

( )

( )- = k [sI – (A – mc – bk)]–1 m =

40 4( 0 619

6 414 33 072

. . )

. .

s

s s

+

+ +

(e) A = 0 1

0 0�

��

�

�� ; b =

0

1�

��

�� ; c = [1 0]

~�x2 = (Aee – male)~x2 = –m~x2 ; s + m = s + 5; m = 5

Refer Eqns (7.38) – (7.49).

� ¢x2 = – 5 �x2 + u; �x2 = ¢x2 + 5y

(f) � ¢x2 = – 5 ¢x2 – 25y + u; ¢X2 (s) = 1

5s + [– 25Y (s) + U (s)];

u = – k1x1 – k2 [ ¢x2 + 5y]


U s

Y s

( )

( )- =

8 07 0 62

6 41

. ( . )

( . )

s

s

+

+

7.10 (a) Desired polynomial: (s + 2) (s + 1 + j1) (s + 1 – j1)= s3 + 4s2 + 6s + 4

�x1 = x2; �x2 = – x2 + x3; �x3 = – 2x3 + u; y = x1

~x1 = x1 – r; ~x2 = x2;~x3 = x3

~�x = A~x + bu

A =

0 1 0

0 1 1

0 0 2

-

-

�

�

��

�

�

��

; b =

0

0

1

�

�

��

�

�

��

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

U = [b Ab A2b] =

0 0 1

0 1 3

1 2 4

-

-

�

�

��

�

�

��

p1 = [1 0 0]

P =

p

p A

p A

1

1

12

�

�

��

�

�

��

=

1 0 0

0 1 0

0 1 1-

�

�

��

�

�

��

; |sI – A| = s3 + 3s2 + 2s

k = [4 6–2 4–3]; P = [4 3 1]

(b) Desired polynomial: s3 + 8s2 + 24s + 32

�z = AT

z + cTh; h = mT

z

U = [cT ATcT (AT)2cT] =

1 0 0

0 1 1

0 0 1

-

�

�

��

�

�

��

p1 = [0 0 1]; P =

p

p A

p A

1

1

12

T

T( )

�

�

��

�

�

��

=

0 0 1

0 1 2

1 3 4

-

-

�

�

��

�

�

��

|sI – A| = s3 + 3s2 + 2s

mT = [32 24 – 2 8 – 3] P = [5 7 8]

SOLUTION MANUAL 83

7.11

�

�

�

x

x

z

1

2

�

�

��

�

�

��

=

-

-

�

�

��

�

�

��

1 0 0

0 2 0

1 3 0

x

x

z

1

2

�

�

��

�

�

��

+

1

1

0

�

�

��

�

�

��

u +

0

0

1-

�

�

��

�

�

��r

U = [b Ab A2b] =

1 1 1

1 2 4

0 4 7

-

-

-

�

�

��

�

�

��; |U| = – 5

Conditions for existence of integral control solution are satisfied.

p1 = [– 0.8 0.8 0.2]

P =

p

p A

p A

1

1

12

�

�

��

�

�

��

=

-

-

-

�

�

��

�

�

��

0 8 0 8 0 2

1 1 0

1 2 0

. . .

Desired polynomial: s3 + 4s2 + 8s + 8

|sI – A| = s3 + 3s2 + 2s; k = [8 8 – 2 4 – 3] P = [– 1.4 2.4 1.6]

u(t) = 1.4x1 – 2.4x2 – 1.6 � (y – r)dt

x1 = q; x2 = �q ; x3 = ia

�x = Ax + bu

A =

0 1 0

0 1 1

0 1 10

-

- -

�

�

��

�

�

��

; b =

0

0

10

�

�

��

�

�

��

;

|sI – (A – bk)| = s3 + (11 + 10k3) s2 + (11 + 10k2 + 10k3)s + 10k1

Characteristic polynomial:

(s2 + 2zwns + w n2 ) (s + 10) = (s2 + 2s + 4) (s + 10) = s3 + 12s2 + 24s + 40

k = [4 1.2 0.1]

7.13 x1 = w; x2 = ia

�x1 = – x1 + x2; �x2 = – x1 – 10x2 + 10u

A = -

- -

�

��

�

��

1 1

1 10; b =

0

10�

��

��

z = ( ) ;y r dtt

-�0

�z = y – r = x1 – r


A =

-

- -

�

�

��

�

�

��

1 1 0

1 10 0

1 0 0

; b =

0

10

0

�

�

��

�

�

��

|sI – (A – bk)| = 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 0.1 4]

7.14 c = [1 0 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 m2 m3]T = [5 6 5]T

7.15 x1 = q, x2 = �q , x3 = 0.1ia

KT ia = J��q ; ��q = 50ia

�x1 = x2; �x2 = 5x3;

u – Kb�q = La

di

dta + (Ra + 0.1)ia

This gives, �x3 = – 2x3 – 20x2 + 20u

A =

0 1 0

0 0 5

0 20 2- -

�

�

��

�

�

��

; b =

0

0

20

�

�

��

�

�

��

|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 = 0.5��q + 0.5 �q

u = 100if + 20di

dtf

SOLUTION MANUAL 85

�x1 = x2; �x2 = –x2 + 20x3; �x3 = – 5x3 + 1

20u

A =

0 1 0

0 1 20

0 0 5

-

-

�

�

��

�

�

��; b =

0

01

20

�

�

��

�

�

��

¢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

7.17Y s

U s

( )

( ) =

b

as s( )+

(a) �x1 = x2; �x2 = – a x2 + b u

A = 0 1

0 -

�

��

�

��

a; b =

0

b

�

��

��

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

b k1 = a2; k1 = a2

b; a + bk2 = a1; k2 =

a1 - a

b; N = k1

7.18 x1 = w; x2 = ia

�x = 0 50

200 200- -

�

��

�

��x +

0

200�

��

��u +

-�

��

��

50

0TL

�z = y – r = x1 – r,

�

�

�

x

x

z

1

2

�

�

��

�

�

��

=

0 50 0

200 200 0

1 0 0

- -

�

�

��

�

�

��

x

x

z

1

2

�

�

��

�

�

��

+

0

200

0

�

�

��

�

�

��u +

0

0

1-

�

�

��

�

�

��

r +

-�

�

��

�

�

��

50

0

0

TL

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


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 = -

-��

��

3 2

4 5; b =

1

0�

��

�� ; c = [0 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 = -

-

�

��

�

��

6 0 5

4 5

.; �x =

-

-

�

��

�

��

6 0 5

4 5

.x +

1

0�

��

��w

At steady state, �x = 0; xls = 5

4x2s. Hence, x2s =

1

7w

(c) �z = y = x2; A =

-

-

�

�

��

�

�

��

3 2 0

4 5 0

0 1 0

; b =

1

0

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 1.5 3.5]; �x = [A – bk] x + bw

A – bk =

- -

-

�

�

��

�

�

��

5 0 5 3 5

4 5 0

0 1 0

. .

St steady state, �x = 0; ys = x2s = 0

A = -

-��

��

3 2

4 5; b =

1

0�

��

�� ; c = [0 1]

(a) |sI – (A – bk)| = (s + 4) (s + 7), gives k = [3 1.5]

(b)1

N = – c [A – bk]–1b; N = 7

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

SOLUTION MANUAL 87

�x1 = – 6x1 + 0.5x2 + 7r; �x2 = 4x1 – 5x2

As t Æ • , �x1 = �x2 = 0

4x1s – 5x2s = 0; x1s = 5

4x2s;

– 6 × 5

4x1s + 0.5x2s = – 7r; x2s = r = y (•)

(c) �x1 = – 3x1 + 2x2 – 3x1 – 1.5x2 + w; �x2 = 4x1 – 5x2

x2s (•) = y (•) = 1

7w

(d) �z = y – r = x2 – r

�

�

�

x

x

z

1

2

�

�

��

�

�

��

=

-

-

�

�

��

�

�

��

3 2 0

4 5 0

0 1 0

x

x

z

1

2

�

�

��

�

�

��

+

1

0

0

�

�

��

�

�

��

u +

0

0

1-

�

�

��

�

�

��r +

1

0

0

�

�

��

�

�

��

w

k = [2 1.5 3.5]

�

�

�

x

x

z

1

2

�

�

��

�

�

��

=

- -

-

�

�

��

�

�

��

5 0 3

4 5 0

0 1 0

.5 .5 x

x

z

1

2

�

�

��

�

�

��

+

1

0

0

�

�

��

�

�

��

w +

0

0

1-

�

�

��

�

�

��r

Y s

W s

( )

( ) =

4

1 2 7

s

s s s( )( )( )+ + +; y(•) = 0;

Y s

R s

( )

( ) =

14

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

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

10 1.r

(c) �z = w – r

�

�

w

z

�

��

�� =

-�

��

�

��

0 5 0

1 0

. w

z�

��

�� +

100

0�

��

�� u +

0

1-��r


A – bK = - - -��

��

0 100 100

1 01 2.5 K K

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

K1 = 0.105, K2 = 0.5

u(t) = – 0.105w(t) – 0.5 ( ( ) )w t r dtt

-�0

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

7.22 F =

0 1 0

0 0 1

4 2 1- - -

�

�

��

�

�

��

; g =

0

0

1

�

�

��

�

�

��

Desired characteristic equation: z z j+ +�

�

12

12

z j+ -�

�

12

12

= z3 + z2 + 1

2z

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

k = - -��

��

43

20

7.23 F =

12

1 0

1 0 1

0 0 0

-

�

�

��

�

�

��

; c = [1 0 0]

|zI – (F – mc)| = z3 + m112

-�

� z

2 + (1 + m2)z + m3

Desired characteristic polynomial: z z j+ +�

�

12

14

z j+ -�

�

12

14

= z3 + z2 + 5

16z.

m = 3

2

11

160-�

��

T

7.24 F =

0 1 0

0 0 1

0 5 0 2 11- -

�

�

��

�

�

��. . .

; g =

0

0

1

�

�

��

�

�

��

SOLUTION MANUAL 89

|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 = 0 1

0 16 1- -

�

��

�

��.

; c = [1 1]

The current observer equations are:

x (k + 1) = F �x(k) + gu(k)

�x(k + 1) = x (k + 1) + m[y(k + 1) – cx (k + 1)]

State error equation is, ~x (k + 1) = (F – mcF) ~x (k)

|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 x(k + 1) =

0 1 0

0 0 1

0 5 0 2 11- -

�

�

��

�

�

��. . .

x(k) +

0

0

1

�

�

��

�

�

��

u(k)

x k

x k

x k

2

1

3

1

1

1

( )

( )

( )

+

+

+

�

�

��

�

�

�� =

0 0 1

1 0 0

0 0 11- -

�

�

��

�

�

��.2 .5 .

x k

x k

x k

2

1

3

( )

( )

( )

�

�

��

�

�

��

+

0

0

1

�

�

��

�

�

��

u(k)

y(k) = [1 0 0]

x k

x k

x k

2

1

3

( )

( )

( )

�

�

��

�

�

��

~xe(k + 1) = (Fee – mf1e)~xe (k)

Fee = 0 0

0 5 11-

�

��

�

��. .; f1e = [0 1]

|zI – (Fee – mf1e)| = z2 + (m2 – 1.1) z – 0.5m1

Desired characteristic polynomial: z2

m = [0 1.1]T; �x2 (k) = x2(k) = y(k); �x e(k) = � ( )

� ( )

x k

x k1

3

�

��

�

��

Refer Eqns (7.91) – (7.97).

�x e(k + 1) = 0 0

0 5 0-

�

��

�

��.�x e(k) +

0

1�

��

��u(k) +

1

0 2-

�

��

�

��.y(k) +

0

11.�

��

��y(k + 1)


7.27 (a) F = 2 1

1 1

-

-��

��

; g = 4

3�

��

��

U = [g Fg] = 4 5

3 1-

�

��

�

��

p1 = 1

19 [3 – 4]; P =

p

p F1

1

��

= 1

19

3 4

10 7

-

-

�

��

�

��

|zI – F| = z2 – 3z + 1; |zl – (F – gk)| = z j+��

12

z j-��

12

= z2 + 1

4

k = -��

��

3

43 P =

111

76

18

19

-��

��

(b) �x(k + 1) = (F – mc) �x(k) + (g – md)u(k) + my(k); c = [1 1]z (k + 1) = FT

z(k) + cTh(k); h(k) = mT

z(k)

U = [cT FTcT] = 1 1

1 0�

��

�

��; p1 = [1 – 1]

P = p

p F1

1T

�

��

�

�� =

1 1

3 2

-

-

�

��

�

��

|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 �x1(k) + 3.79 �x2 (k)

x2(k + 1) = – x1(k) + x2(k) – 4.38 �x1(k) + 2.84 �x2 (k)

and

�x1(k + 1) = – 11.84 �x1(k) – 5.21 �x2 (k) + 8[x1(k) + x2(k)]

�x2 (k + 1) = – 0.38 �x1(k) + 8.84 �x2 (k) – 5[x1(k) + x2(k)]

7.28 (a)Y z

U z

( )

( ) =

1

0 162z z+ + .

x(k + 1) = Fx(k) + gu(k); y(k) = cx(k)

SOLUTION MANUAL 91

F = 0 1

0 16 1- -��

��.

; g = 0

1�

��

�� ; c = [1 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 –2.2]

(c) F = 0 1

0 16 1- -��

��.

; g = 0

1�

��

�� ; c = [1 0]

�x2 (k + 1) = Fee �x2 (k) + fe1y(k) + geu(k) + m(y(k + 1) – f11 y(k)

– g1u(k) – f1e �x2 (k))

= (– 1 – m) �x2 (k) – 0.16y(k) + my(k + 1) + u(k)

m = – 1 gives dead-beat response.

�x2 (k + 1) = – 0.16y(k) – y(k + 1) + u(k)

(d) u(k) = – 0.36x1(k) + 2.2 �x2 (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

Y z

( )

( )- =

2 56 1 0 1375

1 2 2

1

1

. ( . )

( . )

+

-

-

-

z

z

7.29 Let us transform c from [1 1] to [1 0].

x(k) = Qq(k); Q = 1 1

0 1

-��

��

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

F =0 1

0 16 1- -��

��.

; g = 0

1�

��

��

y(k) = cQq(k) = cq(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 = 0 1

0 16 1- -��

��.

; g = 0

1�

��

�� ; c = [1 0]

�q2 (k + 1) = Fee �q2 (k) + fe1y(k) + geu(k) + m(y(k + 1) – f11y(k) – g1u(k)

– f1e �q2 (k))

= – �q2 (k) – 0.16y(k) + my(k + 1) – mu(k) – m �q2 (k)

~q2 (k + 1) = (Fee – mf1e)~q2 (k) = (– 1 – m) ~q2 (k)

m = – 1 gives dead-beat response.

�q2 (k + 1) = – 0.16y(k) – y(k + 1) + u(k)

u(k) = – 0.36q1(k) + 2.2 �q2 (k) = – 2.56y(k) – 0.352y(k – 1)+ 2.2u (k – 1)

U z

Y z

( )

( )- =

2 56 1 0 1375

1 2 2

1

1

. ( . )

.

+

-

-

-

z

z

7.30 Double integrator plant:

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

y(k) = cx(k)

F = 1

0 1

T�

��

�

��; g =

T

T

2 2/�

��

�

�� ; c = [1 0]

Regulator: z = 0.5; wn = 4, z1,2 = e en nT j T- ± -zw w z1 2

= 0.77 ± j0.278

Characteristic polynomial: z2 – 1.54z + 0.67

|zI – (F – gk)| = z2 + Tk T k2

2

122+ -�

��

z + T

k2

12 – Tk2 + 1

k = [13 3.95]

Prediction observer:

Result follows from Example 7.14.

m = 2

10�

��

�� ;

U z

Y z

( )

( )- = k [zI – (F – gk – mc)]–1 m =

65 5 0 802

0 46 0 262

. ( . )

. .

z

z z

-

+ +

7.31 x(k + 1) = 1 0 0952

0 0 905

.

.�

��

�

�� x(k) +

0 00484

0 0952

.

.�

��

�

�� u(k); y(k) = [1 0] x(k)

SOLUTION MANUAL 93

f(F) = 1 0 0952

0 0 905

2.

.�

��

�

�� ;

U = [g Fg] = 0 00484 0 0139

0 0952 0 0862

. .

. .�

��

�

��

Using Ackerman’s formula:

k = [0 1]U–1f(F) = [105.01 14.648]

Observer design:

z (k + 1) = FTz(k) + cT

h(k); h(k) = mTz(k)

f(FT) = 1 0

0 0952 0

2

. .905��

��

U = [cT FTcT] = 1 1

0 0 0952.��

��

mT = [0 1] U–1f(FT) = [19 8.6]

Reduced order dead-beat observer:

F = 1 0 0952

0 0 905

.

.�

��

�

��; g =

0 00484

0 0952

.

.�

��

�

��; c = [1 0]

~x2 (k + 1) = (Fee – mf1e)~x2 (k); Fee = 0.905, f1e = 0.0952

Desired characteristic equation: z = 0; m = 9.51

Refer Equations (7.91)– (7.97)

�x2 (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;

eAT = 0.368; g = e dT

-� tt

0

= 0.632

y(k + 1) = 0.368y(k) + 0.632u(k) + 0.632w(k)

(b) Time constant = 0.5 sec; z = eT-

t = 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

N = – c (F – gk – 1)–1 g; N = 1.37

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

0 865

.

.r @ r

(c) At steady state, with r = 0;

ys = 0.135ys + 0.632w; ys = 0 6320 865..

w = 0.73w

(d) Integral control:

z = 0.5, wn = 4

re±jq = e en nT j T- ± -zw w z1 2

= – 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

y k

v k

( )

( )

+

+

�

��

�

��

1

1 =

0 368 0

0 368 1

.

.�

��

�

��

y k

v k

( )

( )�

��

�

�� +

0 632

0 632

.

.�

��

�

��u(k) +

0 632

0 632

.

.�

��

�

��w(k) +

0

1-��r

~x = y y

v vs

s

-

-

�

��

�

��;

~x (k + 1) = F x~ (k) + g~u (k)

F = 0 368 0

0 368 1

.

.�

��

�

�� ; g =

0 632

0 632

.

.�

��

�

�� ;

|zI – F | = z2 + 0.256z + 0.018

U = [g F g ] = 0 632 0 2326

0 632 0 8646

. .

. .�

��

�

��

p1 = [– 1.58 1.58]

P = p

p F1

1

�

��

�� =

-�

��

�

��

1 58 158

0 158

. .

.

k = [– 0.35 1.624]P = [0.553 2.013]

SOLUTION MANUAL 95

(e) F – g k = 0 018 1 27

0 018 0 27

. .

. .

-

-

�

��

�

��

With w = 0,

y

vs

s

�

��

�� =

0 018 1 27

0 018 0 27

. .

. .

-

-

�

��

�

��

y

vs

s

�

��

�� +

0

1-��r

Solving the above set, we get ys = r

With r = 0:

y

vs

s

�

��

�� =

0 018 1 27

0 018 0 27

. .

. .

-

-

�

��

�

��

y

vs

s

�

��

�� +

0 632

0 632

.

.�

��

�

��w

Solving the above set, we get ys = 0.

CHAPTER 8 LYAPUNOV STABILITY ANALYSIS

8.1 A = 0 1

1 2- -�

��

��. Let us take Q =

4 0

0 0�

��

�

�� =

2

0�

��

�� [2 0] = HHT

rH

H A

T

T

�

��

�

�� = r

2 0

0 2�

��

�

�� = 2

ATP + PA = – Q gives P = 5 2

2 1�

��

�

��

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.

8.2 A = 0 1

1 1-�

��

��; ATP + PA = – I

Solving for P, we get: P =

-

-

�

�

��

�

�

��

3

2

1

21

21

P is not a positive definite matrix; the system is unstable.

8.3 �x1 = x2, �x2 = – x1 – x2 + 2

Equilibrium state xe = x

x

e

e1

2

�

��

��

0 = x e2 ; 0 = – x e

1 – x e2 + 2; xe =

2

0�

��

��

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

A = 0 1

1 1- -�

��

��; ATP + PA = – I gives P =

3

2

1

21

21

�

�

��

�

�

��

P is positive definite matrix. Therefore equilibrium state [2 0]T isasymptotically stable.

SOLUTION MANUAL 97

8.4 A = -

-

�

��

�

��

4 4

4 6

K K

K K; ATP + PA = – I; P =

7

40

1

101

10

3

20

K K

K K

�

�

��

�

�

��

P is positive definite for K > 0; the system is asymptotically stable forK > 0.

8.5 From Fig. P8.5, we have

�x1 = – x1– Kx3; �x2 = x1 – x2; �x3 = x2 – x3

�x = Ax; A =

- -

-

-

�

�

��

�

�

��

1 0

1 1 0

0 1 1

K

; ATP + PA = – Q;

Take Q =

0 0 0

0 0 0

0 0 2

�

�

��

�

�

��

= HHT; HT = [0 0 2 ]

From Lyapunov equation, we obtain,

P = 1

1 8( )( )K K+ -

- - -

- - + - +

- - + - +

�

�

��

�

�

��

3 3 2

3 4 4

2 4 5 8

K

K K

K K K

( ) ( )

( ) ( )

For P to be positive definite, -

+ -

3

1 8( )( )K K > 0;

3 4 9

1 82 2

( )

( ) ( )

K

K K

+ -

+ - > 0,

and

1

1 8( )( )K K+ -

- - -

- - + - +

- - + - +

�

�

��

�

�

��

3 3 2

3 4 4

2 4 5 8

K

K K

K K K

( ) ( )

( ) ( )

> 0

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 beexamined that Routh criterion gives the same result.


8.6 F = 0 1

1 1

.5

- -�

��

��; FTPF – P = – I gives P =

3 5 2 25

2 25 3 875

. .

. .�

��

�

��

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

8.7 F = 0 0 5

0 5 1

.

.- -

�

��

�

�� ; FTPF – P = – I, gives,

P =

5227

4027

4027

10027

�

�

��

�

�

��

P is a positive definite matrix. Therefore, the system is asymptoticallystable.

8.8 �x1 = x2; �x2 = – (1 – |x1|) x2 – x1

Take V(x) = x12 + x2

2

Then, �V (x) = 2x1 �x1 + 2x2 �x2 = – 2x22 (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; �x2 = – x1 – x12 x2

V(x) = x12 + x2

2 ; V(x) Æ • as ||x|| Æ •

�V (x) = – 2x12 x2

2 < 0

Therefore origin is the equilibrium state and is asymptotically stable in-the-large.

8.10 �x1 = – 3x1 + x2; �x2 = x1 – x2 – x23; J =

-

- -

�

��

��3 1

1 1 3 22x

Take P = I. Then Q = – (JT + J) = 6 2

2 2 6 22

-

- +

�

��

��x

|Q| = 36x22 + 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 = 0 1

1 2- -��

��z

x = Ax; x(0) = 1

0�

��

��

J = 0

•

� e2(t)dt = 1

20

•

� 2 x12 (t)dt; Q =

2 0

0 0�

��

�

�� ; R = 0; K = 0

The Lyapunov equation becomes,

ATP + PA + Q = 0

Solving for P, we get: P = 2

1

21

11

2

zz

z

+�

�

��

�

�

��

J = xT(0) Px(0) = z + 1

4z

d J

dz = 1 –

1

4 2z

= 0; z = 0.5; d Jd

2

2z

> 0; Jmin = 1

9.2 The closed-loop transfer function Y s

R s

( )

( ) =

K

s s K2 + +a

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

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

�x = 0 1

- -

�

��

�

��K a

x; x(0) = 1

0�

��

��

J = 0

•

� e2(t)dt = 0

•

� x12 (t)dt; Q =

2 0

0 0�

��

�

�� ; R = 0; K = 0

Lyapunov equation: ATP + PA + Q = 0

Solving for P, we get


P =

a

a

a

K K

K K

+�

�

��

�

�

��

1 1

1 1 ; J = 1

2xT (0) Px(0) =

a

a2

1

2K+

(i) K = constant: ∂

∂

J

a =

1

2

1

2 2K-

a; a = K ,

∂

∂

2

2J

a > 0

(ii) a = constant: ∂

∂

J

K =

a

2-�

�

12K

= 0; K Æ •

9.3 A = 0 1

0 0�

��

�

�� ; B =

0

1�

��

�� ; K = [1 k]; Q =

1 0

0 1�

��

�

�� ;

R = 0; (A – BK) = 0 1

1- -��

��k

; x(0) = 1

1�

��

��

(A – BK)TP + P(A – BK) + Q = 0

Solving for P we get,

P =

k

k

k

2 2

2

1

21

2

1

+�

�

��

�

�

��

; J = 1

2xT (0) Px (0) =

k k

k

2 2 44

+ +

∂

∂

J

k = 0 gives k = 2; Jmin =

3

2

|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 =

D

D

J J

k k

/

/ =

0 04 1 5

0 5 2

. / .

. / = 0.107

9.4Y s

U s

( )

( ) = G(s) =

1002s

; ��y = 100u; y(0) = �y(0) = 0

x1 = y, x2 = �y; �x = 0 1

0 0�

��

�

��x +

0

100�

��

�� u

u = r – y(t) – K �y(t)

Let ~x1 = y – r = x1 – r, ~x2 = �y = x2

SOLUTION MANUAL 101

Then u = – ~x1 – Kx~2 ; K = [1 K]

~�x = Ax~ + Bu; ~x (0) = -��

1

0; Q =

2 0

0 0�

��

�

�� ; R = 0

(A – BK)TP + P(A – BK) + Q = 0

Solving for P, we get,

P =

100 1

100

1

1001

100

1

10

2

4

K

K

K

+�

�

��

�

�

��

J = 1

2~xT (0) P~x (0) =

100 1200

2K

K

+;

∂

∂

J

K = 0 gives K = 0.1;

Jmin = 0.1, ∂

∂

2

2

J

K > 0

For K = 0.1, J = 0.1.

For K = 0.09, J = 0.100556

SKopt =

D

D

J J

K K

/

/ =

0 000556 01

0 01 0 1

. / .

. / . = 0.0556

9.5 A = 0 1

0 0�

��

�

�� ; B =

0

1�

��

�� ; K = [k1 k2]; (A – BK) =

0 1

1 2- -

�

��

�

��k k

Closed-loop system: �x = 0 1

1 2- -

�

��

�

��k k

x or �x1 = x2; �x2 = – k1x1 – k2x2

Eliminating x2 from this equation yields ��x1 + k2 �x1 + k1x1 = 0

Since the undamped natural frequency is specified as 2 rad/sec, we obtaink1 = 4.

Therefore, (A – BK) = 0 1

4 2- -

��

��k

; R = 0; Q = 1 0

0 1�

��

�

�� ; x(0) =

1

0�

��

��

(A – BK)TP + P(A – BK) + Q = 0


P =

52 8

18

18

58

2

2

2

k

k

k

+�

�

��

�

�

��

; J = 1

2xT(0) Px(0) =

1

2

5

2 82

2

k

k+

�

��

�

�� ;

∂

∂

J

k2

= 0


gives k2 = 20 ; ∂

∂

2

22J

k > 0

For this value of k2 : Jmin = 5

4

9.6 A = 0 1

0 0�

��

�

�� ; B =

0

1�

��

�� ; K = [k k]; (A – BK) =

0 1

- -

�

��

�

��k k

Q = 2 0

0 2�

��

�

�� ; R = 0; x(0) =

1

0�

��

��

(A – BK)TP + P (A – BK) + Q = 0

Solving for P we get

P =

1 2 1

1 2 1

2 2

+

+

�

�

��

�

�

��

k

k k

k

k

k

( ) ; J = 1

2xT (0) Px (0) = 1 +

1

2k;

∂

∂

J

k = 0

gives k Æ •

9.7 (A – BK)TP + P(A – BK) + KTRK + Q = 0; Q = 2 0

0 2�

��

�

�� ; R = 2

Solving Lyapunov equation, we get

P =

( )

( )( )

k

k

k

kk

k

k k

k

+ +

+ + +

�

�

��

�

�

��

1 1

1 1 1

2 2

2 2

2

; J = ( )1

2

2+ kk

;∂

∂

J

k = 0 gives k = 1; Jmin = 2

For k = 1, J = 2.

For k = 0.9, J = 2.0055.

Skopt =

D

D

J J

k k

/

/ =

0 0055 2

01 1

. /

. / = 0.0275

9.8 A = 0 1

0 0�

��

�

�� ; B =

0

1�

��

�� ; u = – Kx; Q =

2 0

0 2�

��

�

�� ; R = 2

{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 = 2 3 2

2 2 3

�

��

�

�� ; K = R–1BP = [1 3 ]

SOLUTION MANUAL 103

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

Q = 2 0

0 0�

��

�

�� = HTH =

2

0

�

��

�� [ 2 0]; r

H

HA�

��

�� = 2

{A, H} pair is observable. Sufficient conditions satisfied.

Solving the Ricatti equation, we get

P = 2 2 2

2 2 2

�

��

�

�� ; K = R–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. Theoptimal closed-loop system is therefore stable.

9.10 A = 0 0

0 1�

��

�

�� ; B =

1

1�

��

�� ; Q =

2 0

0 0�

��

�

�� ; R = 2

{A, B} pair is controllable.

Q = HTH = 2

0

�

��

�� [ 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 = 6 8

8 16

-

-��

��

; K = R–1BTP = [– 1 4]

|lI – (A – BK)| = l2 + 2l + 1

Optimal closed-loop system is asymptotically stable:u = – Kx = – [– 1 4]x

9.11 A = -��

��

1 0

1 0; B =

1

0��; Q =

2 0

0 0�

��

�

�� ; R = 2

{A, B} pair is controllable.

Q = HTH = 2

0

�

��

�� [ 2 0]

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

2(– p11 + p12) – 1

2p11

2 + 2 = 0; – p12 + p22 – 1

2p11p12 = 0;

-1

2p12

2 = 0

This gives p11 = p22 = 0. Therefore a positive definite solution of theRicatti equation is not possible. Hence an optimal solution to the controlproblem does not exist.


9.12 �x = 0 1

0 2-��

��

x + 0

20�

��

��u; y = x1; J =

1

20

•

� [y(t) – 1]2 + u2dt; yd = r = 1

Let ~x1 = x1 – r = x1 – 1; ~x2 = x2. Then

~�x = 0 1

0 2-��

��

~x + 0

20�

��

��u = A~x + Bu;

J = 12 1

2

0

(~x•

� + u2)dt; Q = 1 0

0 0�

��

�

�� ; R = 1

{A, B} pair is controllable

Q = 1

0�

��

�� [1 0] = HTH. {A, H} pair is observable.

ATP + PA – PBR–1BTP + Q = 0


P =

6 63

200 05

0 054 63

400

..

..

�

�

��

�

�

��

; K = R–1BTP = [1 0.23]

u = –K~x = – x1 – 0.23x2 + r

9.13 A = 0 1

0 0�

��

�

�� ; B =

0

1�

��

�� ; C =

1 0

0 2�

��

�

��

J = 1

22

0

•

� ( y12 + y2

2 + u2)dt = 1

22

0

•

� (yTy + u2)dt = 1

20

•

� (xTCTCx + u2)dt

Q = 2 0

0 8�

��

�

�� ; R = 2

{A, B} pair is controllable. Q is positive definite.

Solving Ricatti equation, ATP + PA – PBR–1BTP + Q = 0, we get

P = 24 2

2 24

�

��

�

�� ; K = R–1BTP = [1 6 ]; u = – x1 – 6 2x

9.14 A = 0 1

0 1-��

��

; B = 0

1�

��

�� ; Q =

2 0

0 2�

��

�

�� ; R = 2


SOLUTION MANUAL 105

P = 4 2

2 2��

��

; K = R–1BTP = [1 1]

Equation of the state observer:

�

�x = (A – MC) �x + Bu + My: M = m

m1

2

�

��

�� ; C = [1 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 characteristicequation is l2 + 6l + 9 = 0. Comparing the coefficients, we get

m1 = 5, m2 = 4; u = – �x1 – �x2

�

�x = A �x + Bu + M[y – C �x]; M = 5

4�

��

��

9.15 State variable description of the system, obtained from Fig. P9.15, is givenby

�x = 0 1

0 5-��

��x +

0

1�

��

��u; x(0) =

0

0�

��

�� ; y = [1 0]x

In terms of the shifted state variables, ~x1 = x1 – qr;~x2 = x2, the model

becomes

~�x = 0 1

0 5-��

��

~x + 0

1�

��

��u = A~x + Bu

The problem is to determine optimal control law, u = – K~x , that mini-mizes

J = ~x u12 2

0

1

2+�

�

•

� dt

From this J, we obtain the following matrices:

Q = 2 0

0 0�

��

�

�� ; R = 1

The pair {A, B} is controllable.

Q = HTH = 2

0

�

��

�� [ 2 0]. The pair {A, H} is observable.

Solving the matrix Ricatti equation, ATP + PA – PBR–1BTP + Q = 0,we get


P = 7 495 2

2 0 275

.

.

�

��

�

��

The optimal gain matrix is K = R–1BTP = [ 2 0.275]

The matrix (A – BK) is stable.

The minimum value of J for an initial condition ~x (0) = [– 5 0]T is,

J0 = 1

20 0~ ( ) ~( )x PxT = 93.2375

9.16 A = 0 1

0 5-

��

��; B =

0

1�

��

�� ; Q =

2 0

0 0�

��

�

�� ; R = 2r.


P =

2 252

2

2 2 5 252

1 2 1 2

1 2

rr

r

r rr

/ /

/

+�

�

�

- + +�

�

�

�

�

��

�

�

��

Case (i): r = 0.1

P = 3 5397 0 6325

0 6325 01194

. .

. .�

��

�

�� ; K = R–1BTP = [3.1623 0.5968]

Poles: – 0.6377, – 4.9592

Case (ii): r = 0.01

P = 1 3416 0 2

0 2 0 0342

. .

. .�

��

�

�� ; K = [10 1.7082]

Poles: – 2.2361, – 4.4721

Case (iii): r = 0.001

P = 0 5941 0 0632

0 0632 0 0088

. .

. .�

��

�

�� ; K = [31.6228 4.3939]

Poles: – 4.6970 ± j3.0921

9.17 ~�x = A~x + Bu; u = –K~x = – [k1 k2]~xThe constraint of partial state feedback can be met by setting k2 = 0 in theoptimization 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,

P =

( )( )25 2

10

2

22

2

2

10

1 12

1

12

1

12

1

12

1

+ + +

+ +

�

�

��

�

�

��

k k

k

k

kk

k

k

k

For an initial condition ~x (0) = [– 5 0]T, the performance index has thevalue:

J = 1

2~x T (0)P~x (0) =

25 25 2

201 1

2

1

( )( )+ +k k

k;

∂

∂

J

k1

= 0 gives k1 = 1.345;

∂

∂

2

12

J

k > 0

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 = (~ ~ )y u dt2 2

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 = 1

2; N = 2

(c) From Fig. P9.18, Y s

W s

( )

( ) =

1

1s K+ + =

1

2s +; W(s) =

1

s;

Y(s) = 1

2s s( )+

Steady-state error due to disturbance: lim ( )s

sY sÆ0

= 1

2.


(d) �y = – y + u; �z = y – r

�

�

y

z�

��

�� =

-��

��

1 0

1 0

y

z +

1

0��

u + 0

1-��r

At steady state �y = �z = 0, and therefore

0

1-��

r = – -��

��

1 0

1 0

y

zs

s +

1

0�

��

��us

Substituting in state equation, we get

~�

~�y

z

�

��

�� =

-��

��

1 0

1 0

~

~y

z +

1

0�

��

��~u

~y = y – ys;~z = z – zs;

~u = u – us

J = (~ ~ ~ )y z u dt2 2 2

0

+ +

•

�

A = -��

��

1 0

1 0; B =

1

0�

��

�� ; Q =

2 0

0 2�

��

�

�� ; R = 2

This shifted regulator problem has already been solved in Example9.8. The matrix, K = [1 1]

Therefore, u(t) = – y(t) – 0

t

� [y(t) – r]dt

(e) Block diagram of the control scheme is shown in the figure thatfollows:

Block diagram manipulation gives (for r = 0)

Y s

W s

( )

( ) =

s

s( )+ 1 2 ; W(s) = 1

s; Y(s) =

1

1 2( )s +

Steady-state error: lim ( )s

sY sÆ0

= 0

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:

~�w = – 0.5 ~

w + 100~u

where, ~w = w – ws;

~u = u – us

J = (~ ~ )w2 2

0

100+

•

� u dt

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.

Now, ~u = – 0.095 ~w ; u = – 0.095w + 0.095ws + us; us + 0.095ws = Nr

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

10 1.r

(c) �z = w – r

�

�

w

z

�

��

�� =

-�

��

�

��

��

��

0 5 0

1 0

. w

z +

100

0�

��

�� u +

0

1-��r

At steady-state: �w = 0, �z = 0. Therefore,

0

1-��r = –

-�

��

�

��

��

��

0 5 0

1 0

. w s

sz –

100

0�

��

�� us

where ws, us, zs are steady-state values. Substituting in the stateequations,

~�

~�w

z

�

��

�

�� =

-�

��

�

��

��

��

0 5 0

1 0

. ~

~w

z +

100

0�

��

��

~u

~w = w – ws,

~z = z – zs,~u = u – us


J = (~ ~ ~ )w2 2 2

0

100+ +

•

� z u dt; Q = 2 0

0 2�

��

�

�� ; R = 200; A =

-�

��

�

��

0 5 0

1 0

.;

B = 100

0�

��

��

ATP + PA – PBR–1BTP + Q = 0, gives

P = 0 21 0 2

0 2 2 2

. .

. .�

��

�

�� ; K = R–1BTP = [0.105 0.1]

u(t) = – 0.105w(t) – 0.10

t

� [w (t) – r]dt

The control scheme is shown in figure that follows.

For G(s) = 100

s a+, the closed-loop transfer function is:

w( )

( )

s

R s =

10 5

10 10 5

.

( ) .s s a+ + +

For R(s) = 1

s, we have w (•) = lim ( )

ss s

Æ0w = 1

Therefore, for A + dA = 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

2

2

2

+

-

. K

K K

J = 1

2xT(0) Px (0) =

1 0 75

2

2

2

+

-

. K

K K{x(0)}2

∂

∂

J

K = 0 gives K =

2

3.

SOLUTION MANUAL 111

For K = 0.67, J = 1.499.

For K = 0.57, J = 1.526.

SKopt =

D

D

J J

K K

/

/ =

0 027 1 499

0 1 0 67

. / .

. / . = 0.1207

9.21 x(k + 1) = 0.368x(k) + 0.632u(k)

J = k =

•

Â0

[x2(k) + u2(k)]; F = 0.368, G = 0.632, Q = 2, R = 2

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) �1

1s s( )+

��

��

= 1-

-

-

-

e

z e

T

T

For T = 0.5, Gh0G(z) = 0 4

0 6

.

.z - =

Y z

U z

( )

( ); y(k + 1) = 0.6y(k) + 0.4u(k)

(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) + 0.4~u (k)

where ~y = y – ys,~u = u – us

J = 1

22

0

~ ( )y kk =

•

Â + ~u 2 (k); F = 0.6, G = 0.4, Q = 1, R = 1

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

0 5108

.

.r = 0.217r

(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


y(•) = 0 5108

0 5108

.

.r = r

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)

where ~x (k) = x(k) – xs,~u (k) = u(k) – us

J = 1

22 2

0

~ ( ) ~ ( )x k u kk

+=

•

Â ; F = 0.5, G = 2, Q = 1, R = 1

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

(b) ~u (k) = – K ~x (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

11

.

.r

(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. Acontrol law of the form, u(k) = – K1x(k) – K2v(k) will provide therequired 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 inSection 10.4 and Section 10.11.

10.2 Revisiting Example 10.1 (Fig. 10.19) will be helpful.

G(jw) =4

11

2j j N Ew w+( )-

( ); =

- p E

M4

– 90∞ – 2 tan–1w1 = – 180∞

This gives w1 = 1 rad/sec

-1

1N E� � = |G(jw1)| = 2

This gives E1 = 8M/p

y(t) = – e(t) = – 8Mp

sin t


G(jw) = 51 0 1 2j jw w+( ).

; N(E) = 4 10 1 2

p E E- ��

��

.

– 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

1N E� � = |G(jw1)| = 0.25

This gives E1 = 0.3

Amplitude of limit cycle is 0.3 and frequency is 10 rad/sec.


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

N E( ) for any value of E

The intersection of the two curves does not occur; therefore no limit cycleexists.Intersection of the two curves occurs for


p D

2 < 0.206 or D < 0.131

10.5 G(jw) = 10

1 0 4 1 2+( ) +( )j j. w w

N (E) = 4 12

p EHE

j HE

- �� -

�

�

��

; H = 0.2

The following figure shows G(jw)-plot and -

( )

1

N E locus.

E H=

Im

Re

G j( )w

E increasing -1

pH

N E( )

4

At the point of intersection of G(jw)-plot with -

( )

1N E

locus, measure the

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

1N E� � =

p E1

4 = 0.33

This gives E1 = 0.42

10.6 G( jw) = Kj jw w1 2+( )

N(E) = 22

1 1 1 112

p

p- - - ��

��

�

�

��

-sinE E E

SOLUTION MANUAL 115

= 1 2 1 1 1 112- + -

��

��

-

psin

E E E = 1 – Nc(E)

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 -

( )

1N E

locus

(ii) G(jw)-plot intersects the -

( )

1N E

locus; but the point of intersection

corresponds to an unstable limit cycle.

10.7 Revisiting Review Example 10.3 (Fig. 10.40) will be helpful.

G(jw) = 101 1 0 5j j jw w w+ +� � � �.

N(E) = 2 1 1 1 112

psin- + -��

��E E E

= Nc(E)

The function Nc(E) is listed in Table 10.3.

–G(jw1) = – 180° gives w1 = 2

|G(jw1)| =- 1

1N E� � gives E1 = 4.25

The limit cycle with amplitude 4.25 and frequency 2 rad/sec is a stablelimit cycle.


G(jw) =K

j j jw w w1 1 0 5+( ) +( ).

For K = 1, G(jw)-plot does not intersect -

( )

1N E

locus. For K = 2, two

intersection points are found. One corresponds to unstable limit cycle andthe other corresponds to a stable limit cycle of amplitude 3.75 andfrequency 1 rad/sec.

10.9 (a) Characteristic equation has two real, distinct roots in the left half ofs-plane.

The origin in the (y, �y ) plane is a stable node.

(b)d y

dt

d y

dt

2

2

15

1-( )+

-( ) + 6(y – 1) = 0


The singular point is located at (1, 0) in the (y, �y) plane. The charac-teristic equation is

s2 + 5s + 6 = 0 = (s + 3) (s + 2)

The singular point is a stable node.

(c)d y

dt

d y

dt

2

2

28

2--

-� � � � + 17(y – 2) = 0

The singular point is located at (2, 0) in the (y, �y ) plane. The charac-teristic equation is

s2 – 8s + 17 = 0 with complex roots in right half s-plane. The singularpoint is an unstable focus.

10.10 e = f(qR – q) = sin (qR – q) = sin (– q)

��q + a �q + K sin q = 0

With x1 = q and x2 = �q , we have

�x1 = x2

�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 singularpoints.

Linearized equation around singular point (0, 0):

��q + a �q + Kq = 0

The characteristic equation is

s2 + as + K = 0

The singular point is stable and is either a node or a focus depending uponthe magnitudes of a and K.

Linearization about the singular point (p, 0) gives

��q + a �q – Kq = 0

The characteristic equation is

s2 + as – K = 0

The singular point is a saddle point.

10.11d y

dt

d y

dt

2

2

12

1-( )+

-( )z + y – 1 = 0

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.


J��q + Tc sgn �q � = KA K1 e; e = qR – q

Therefore

�� sgn �eKJ

eT

Jec+ + ( ) = 0; K = KAK1

�� sgn �e eT

Ken

c+ +��

��

( )w2 = 0; wn = K J/

With x1 = e and x2 = �e/wn, we have

�x1 = wn x2

�x2 = – wn xT

Kxc

n1 2+��

��

sgn w� �

For Tc = 0, we have

�x1 = wn x2; �x2 = – wn x1

d x

d x2

1

= – x

x1

2

; this gives x21 + x 2

2 = x21(0)

The effect of the Coulomb friction is to shift the centre of the circles inphase plane to + Tc/K for x2 < 0 and to – Tc/K for x2 > 0. The steady-stateerror 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

�x2 =

- - <

- - - >

- - + < -

�

��

��

x x

x x x

x x x

2

2 1 1

2 1

1

2 1 1

2 1 1

;

;

;

Region I 1 <

Region II

Region III

1

1

� ��

From the phase portrait shown in the figure below, it is seen that thesystem is stable in the equilibrium zone.


10.14 Revisiting Review Example 10.6 (Fig. 10.42) will be helpful. The systemequation in the linear range is

��e + �e + e = 0The characteristic equation

s2 + s + 1 = 0

has complex roots in left half of s-plane. Therefore, on the (e, �e) plane, theorigin 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 onwhether the error is +ve or – ve.

From the phase trajectories plotted from the given initial conditions (withand without saturation) we find that velocity is always greater for trajecto-ries without saturation; thus the error saturation has a slowing down effecton the transient.

10.15 With x1 = e, x2 = �e and u = f (e), we have

�x1 = x2

�x2 = – 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 inthe figure below.

SOLUTION MANUAL 119

10.16 (a) ��e = – u = – f(e); x1 = e, x2 = �e

The trajectories corresponding to x1 > 0 are given by (refer Eqns(10.24))

x1(t) = – 1

2 22x (t) + x1(0) +

1

2 22x (0)

and trajectories for x1 < 0 are given by

x1(t) = 1

2 22x (t) + x1(0) –

1

2 22x (0)

The system response is a limit cycle as shown in the figure.

(b) ��e = – u = – f (e + KD �e); x1 = e, x2 = �e

The trajectories corresponding to (x1 + KD x2) > 0 are given by

x1(t) = – 1

2 22x (t) + x1(0) +

1

2 22x (0)

and the trajectories corresponding to (x1 + KDx2) < 0 are given by

x1(t) = 1

2 22x (t) + x1(0) –

1

2 22x (0)

With derivative feedback, the limit cycle gets eliminated as shown inthe figure.


(c) Constructing a trajectory for the case of large KD proves the point. Anillustrative trajectory is shown in the figure.

Slope = -1

KDTrajectory

Trajectory

Trajectory

Switching

line

Switching

line

x2

x2

x2

x1

x1

x1

10.17 ��q + 0.5 �q = 2 sgn(e + 0.5 �q )With x1 = e and x2 = �e, we have

�x1 = x2

�x2 = – 0.5x2 – 2 sgn(x1 + 0.5 x2)As seen from the phase trajectory in the figure, the system has good damp-ing, no oscillations but exhibits chattering behaviour. Steady-state error iszero.

SOLUTION MANUAL 121

+e

+

- -

qR = const

x1

x2

q q1 1

s + 0.5

0.5

s

2

-2

1

2

x x1 + 0.5 = 02

10.18 (a) With x1 = e, x2 = �e and u = f(e), we have

�x1 = x2

�x2 = – x2 – f(x1)

The system oscillates with everdecreasing amplitude and everin-creasing frequency (refer Fig. 10.32).


(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

�x2 = – x2 – f x x1 21

3+�

��

A rough sketch of the phase trajectory is shown in the figure. Byderivative control action (i) settling time is reduced, but (ii) chatteringeffect appears.

10.19 Section 10.10 provides solution to this problem. Optimum switching curveis given by Eqns (10.48). Figure 10.37 shows a few trajectories.

10.20 ��e + �e = – u; x1 = e, x2 = �eFor u = +1 (refer Eqns (10.26))

x1 – x1(0) = – (x2 – x2(0)) + ln 1

1 02

2

+

+

��

��

x

x ( )

The trajectories are asymptotic to the ordinate – 1.

For u = – 1

x1 – x1(0) = – (x2 – x2(0)) – ln 1

1 02

2

-

+

��

��

x

x ( )

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 + x

x2

2| | ln 1 2

2

2

+��

��

x

x| |

The figure given below shows the switching curve and a few typical mini-mum-time trajectories.

CHAPTER 11 NEURAL NETWORKS FOR CONTROL

11.4 E = 12

0 4 3 0 8 202

02

. ( )) ( . ( )- - + + - +s sw w w w� �

∂

∂

Ew

= 0

= – [(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]

∂

∂

Ew0

= 0 = – [x + y]

This gives x = y = 0

0 4 3 1 3 3 0

0

1

0 0 0. ( ) ( ) ( )

( )

( )

- - + - - + - + =

= Æ -•

= Æ +•

s s s

s

s

w w w w w w

a a

a a

� �� only at

only at Therefore

s (– 3w + w0) = 0.4which gives

–3w + w0 = – 0.41The equation y = 0 gives

s (2w + w0) = 0.8or, 2w + w0 = 1.386Solving for w and w0, we get

w = 0.36, w0 = 0.666

11.5 zl = s w x wlih

i

n

i lh1

101

=

Â +��

�

tr = s w z wrlh

l

m

l rh1

101

=

Â +��

�

�yj = v t vjr r jr

p

+=

Â 01

� �

vjr (k + 1) = vjr (k) + h tr ej (k)vj0 (k + 1) = vj0 (k) + h ej (k)

ej (k) = yj - �y j (k)

w krlh2 1( )+ = w k t t z v e krl

hr r l jr j

j

q2

1

1( ) ( ) ( )+ -=Âh

SOLUTION MANUAL 125

w krh02 1( )+ = w k t t v e kr

hr r jr j

j

q

02

1

1( ) ( ) ( )+ -=

Âh

w klih1 1( )+ = w k x z z e v w t tli

hi l l j jr rl

hr r

r

p

j

q1 2

11

1 1( ) ( ) ( )+ - -�

�

�

��==

ÂÂh

w klh01 1( )+ = w k z z e v w t tl

hl l j jr rl

hr r

r

p

j

q

01 2

11

1 1( ) ( ) ( )+ - -�

�

�

��==

ÂÂh

11.6 (a) s (a) = 21

1 11+

- =-

+-

-

-eeea

a

a

s ¢(a) =( ( ))( ( )) ( )1 1

2- +

=s s sa a d a

daNetwork Output:

�( )y k = s w x xi ii

n

=

Â��

�

=0

0 1;

Weight update rule:

wi (k + 1) = w k y y k y k xi i( ) �( ) � ( )+ -( ) -h

21 2� �

(b) a(1) = w xi ii

( )1

1

4

=

Â = 1 + 2 – 0.5 = 2.5

�( )y 1 = s [a(1)] = 0.848

w (1) = w (0) + h

21 0 848 1 0 848 2 1- - -( ). ( . ) ( )� �x

= [0.974, – 0.948, 0, 0.526]�

(c) �( )y 1 = s w x w x w x w x1 11

2 21

3 31

4 41 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

w1(1) = w1(0) + h x eyp p

p

p1

2

1

3 1

2( ) ( )

( )( � )-�

�

�

��

=Â

� �

= 1.0518�

11.7 (a) zl = s w x wli i li

+��

�

==

Â 01

2

; l 1, 2


�y = s v z vl ll

+��

�=

Â 01

2

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

2

2- ( )=

s s( ) ( )a d ada

zl = s (wl x + wl0); l = 1, 2, …, m

�y = s v z vl ll

m

+��

�=

Â 01

e(k) = y – �y (k)

vl (k + 1) = v k e k y k z kl l( ) ( ) � ( ) ( )+ -h

21 2� �

v0 (k + 1) = v k e k y k02

21( ) ( ) � ( )+ -

h� �

wl (k + 1) = w k e k y k v k z k x kl l l( ) ( ) � ( ) ( ) ( ) ( )+ - -h

41 12 2� � � �

wl0 (k + 1) = w k e k y k v k z kl l l02 2

41 1( ) ( ) � ( ) ( ) ( )+ - -

h� � � �

CHAPTER 12 FUZZY CONTROL

12.7

x

y

mR

12.11 z* =3 1

44 1

45 1

46 1

27 1

27 3

48 3

49 3

814

14

14

12

12

34

34

38

�� + �

�� + ��

�� + �

�� + �

�� + �

�� + ��

�� + �

��

�� + ��

�� + ��

�� + ��

�� + ��

�� + ��

�� + ��

�� + ��

��

.5

= 6.7612.12 Crisp value for speed =

1600 0 1700 0 1800 0 1900 0 8 2000 0 72100 0 2200 0 3) 2300 0 15 2400 0

0 0 0 0 8 0 7 0 0 3 0 15

( .9) ( .9) ( .9) ( . ) ( . )( .5) ( . ( . ) ( )

.9 .9 .9 . . .5 . .

+ + + ++ + + +

+ + + + + + += 1857.28

12.13 x = 4; y = 8

a1 = min ( ), ( ) min ,~ ~

m mA B1 24 8 2

31 2

3��

��

= ��

�� =

a2 = min ( ), ( ) min ,~ ~

m mA B2 24 8 1

323

13

��

��

= ��

�� =

min , ( )~

a m1 1C z��

��

= min , ( )~

23 1

mC z��

��

min ( , ( ))~

a m2 2C z = min , ( )~

13 2

mC z��

��

magg (z) = m m mC C Cz z z1 1 2

23

13~ ~ ~

( ) max min , ( ) , min , ( )=��

��

��

��

��

��


z* = 2 1

33 2

34 2

35 2

36 1

37 1

38 1

313

23

23

23

13

13

13

�� + ��

�� + �

�� + �

�� + �

�� + �

�� + ��

��

�� + ��

�� + ��

�� + ��

�� + ��

�� + ��

�� + ��

��

= 4.712.14 x = distance D = 27

y = vehicle speed V = 55

m PS~

( )55 = 0 55 0 55 0 75 ; ; m mPM PL~ ~

( ) .25 ( ) .= =

m PS~

( )27 = 0 38 27 0 62 27 0. ( ) . ( )~ ~

; ; m mPM PL= =

a1 = min ( ), ( ) .25~ ~

m mPS PM27 55 0��

�� =

a2 = min ( ), ( ) .~ ~

m mPM PL27 55 0 62��

�� =

z = breaking force B

min , ( )~

a m1 PL z��

�� = min .25, ( )

~0 m PL z��

��

min , ( )~

a m2 PM z��

�� = min . , ( )

~0 62 m PM z��

��

magg (z) = max min .25, ( ) , min . , ( )~ ~

0 0 62m mPL PMz z��

��

��

��

��

z* =

10 0 20 0 4 30 0 6 40 0 62 50 0 6260 0 62 70 0 6 80 0 4 90 0 100 0

0 0 4 0 6 0 62 0 62 0 62 0 6 0 4 0 0

( .2) ( . ) ( . ) ( . ) ( . )( . ) ( . ) ( . ) ( .25) ( .25)

.2 . . . . . . . .25 .25

+ + + ++ + + + +

+ + + + + + + + + = 53.18

12.15 m m mSD MD LD~ ~ ~

( ) ( ) ( )60 0 60 45

60 15

= = = ; ;

m m mNG MG LG~ ~ ~

( ) ( ) ( )70 0 70 35

70 25

= = = ; ;

a1 = min ( ), ( )~ ~

m mMD MG60 70 35

��

�� =

a2 = min ( ), ( )~ ~

m mMD LG60 70 25

��

�� =

a3 = min ( ), ( )~ ~

m mLD MG60 70 15

��

�� =

a4 = min ( ), ( )~ ~

m mLD LG60 70 15

��

�� =

~

SOLUTION MANUAL 129

min , ( ) min , ( )

min , ( ) min , ( )

min , ( ) min , ( )

min , ( ) min , ( )

~ ~

~ ~

~ ~

~ ~

a m m

a m m

a m m

a m m

1

2

3

4

35

25

15

15

M M

L L

L L

VL VL

z z

z z

z z

z z

��

�� = �

��

��

�� = �

��

��

�� = �

��

��

�� = �

��

magg (z) = max min , ( ) , min , ( ) , min , ( )~ ~ ~

35

25

15

m m mM L VLz z z��

��

��

��

��

��

��

��

z* =

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 60 00 0 325 0 6 0 6 0 6 0 4 0 4 0 4 0 4 0 0

( ) ( . ) ( . ) ( . ) ( . ) ( . )

( . ) ( . ) ( . ) ( .25) ( .2). . . . . . . . .25 .2

+ + + + +

+ + + + +

+ + + + + + + + + + = 34.40

12.16 With 10 bits we can get solution accuracy of 6 0

2 110-

-

( )

� or 0.006 in the

interval (0,6).

Let the first string of initial population be :

11001000001110010000

The first substring decodes to value equal to (29 + 28 + 25) = 800 . Thus

corresponding parameter value is 06 0 800

10234 692+

- ¥=

( ). . Similarly for

second substring, the parameter value is equal to 5.349. Thus the firststring 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

fitness function F [x(1)] = 11 959 680

0 001+

=.

. .

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Solution Manual Digital Control and State Variable Methods

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