Solution Manual Digital Control and State Variable Methods

SOLUTION MANUAL

TO

DIGITAL CONTROL

AND

CHAPTER 1 OUTLINE OF THE SOLUTION MANUAL

This Solution Manual has been designed as a supplement to the textbook: Gopal, M, Digital Control and State Variable Methods, 2nd Edition, Tata McGraw-Hill, New Delhi, 2003.

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

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

CHAPTER 2 SIGNAL PROCESSING IN DIGITAL CONTROL 2.1 (a) x₁, x₂: outputs of unit delayers, starting at the right and proceeding to

the left. x₁(k + 1) = x₂(k); x₂(k + 1) = – 0.368x₁(k) + 1.368x₂(k) + r(k) y(k) = 0.264x₁(k) + 0.368x₂(k) F = 0 1 0 368 1 368 −

L

N

MM

M

O

Q

PP

P

. . ; g = 0 1

L

N

MM

M

O

Q

PP

P

; c = [0.264 0.368] (b) Y z R z ( ) ( ) = P_i D_i D ∑ − 1 = 0 264 0 368 1 1 368 0 368 2 1 1 2 . . . . z z z z - ₊ --

e

- - -

j

2.2 (a) x₁, x₂ : outputs of unit delayers, starting at the right and proceeding to the left.

x₁(k + 1) = x₂(k) + r (k)

x₂(k + 1) = – 5x₂(k) – 3x₁(k) – 3r (k) y(k) = x₁(k)

x₁(k + 2) = x₂(k + 1) + r (k + 1) Substituting for x₂(k + 1) from the above set,

x₁(k + 2) = – 5x₁(k + 1) – 3x₁(k) + r(k + 1) + 2r (k) y(k + 2) + 5y(k + 1) + 3y(k) = r (k + 1) + 2r (k)

(b) F = 0 1 3 5 -

-L

NM

O

QP

; g = 1 3

-L

NM

O

QP

; c = [1 0] (c) Y z R z ( ) ( ) = z z z + + + 2 5 3 2

2.3 (a) x₁ : output of the unit delayer x₁(k + 1) = – 1 2x1(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 2 1 2

(b) R (z) = 1; Y(z) = -+ + + z z 1 z 2 2 1 2 y (k) = –

FH IK

-1 2 k m (k) + 2

FH IK

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

FH IK

-3 1 2 k m (k)+ 2 3 m (k) 2.4 (a) y (k + 1) = ay(k) + br(k); Y z R z ( ) ( ) = b a z -R(z) = A; Y(z) = A z b a - ; y(k) = Ab(a) k–1_{m (k – 1)} (b) R (z) = A z z Y z z -( ) 1; = A z z b a - -( ) ( 1); Y (z) = A z z A z z b a a b a -- + - -1 1 1 ; y(k) = Ab a 1- [1 – (a) k ]m (k) (c) R (z) = A z z Y z z -( ) ( ) 12 ; = A z z b a - -( ) ( 1)2 Y (z) = A z z z z z z b a a a 1 1 1 1 2 2 -

L

NM

- + -- - -

O

QP

( ) ( ) ( ) y (k) = Ab k k a a a 1- 2 1 1 ( ) [( ) + -( ) - ]; k³ 0 (d) r(k) = A cos (Wk) = Re{AejWk};Z {ejWk} = z z-ejW The output, y(k) = Re Z -( )

-RS

T

UV

W

L

NM

1 A z

O

QP

z z ej b a

b

W

g

= Re A e A e e j k j j k b a a b a - +

-L

NM

W W W

O

QP

= Re A ej e k j k b a- a

-L

NM

W

b

W

g

O

QP

2.5 (a) Given difference equation in shifted form: y (k) + 3y (k – 1) + 2y (k – 2) = 0 Y (z) + 3z–1Y (z) + 3y (– 1) + 2z– 2Y (z) + 2z– 1 y (– 1) + 2y (– 2) = 0 Y (z) = z z2+3z+2 = z z z z +1- +2 y (k) = (– 1)k – (– 2)k; k³ 0 (b) 2Y (z) – 2z– 1Y (z) + z– 2Y (z) = 1 1-z-1 Y (z) = z z z z 3 2 1 2 2 1 - - + ( )

a

f

= z z z z z z -1+ - +2 - +2 1 2 2 = z z z z z z z z z - -- + + - + 1 1 2 0 5 0 5 1 2 0 5 0 5 2 2 2 . . . .

a

f

a

f

Therefore, y (k) = (l)k – 1 2 1 2

FH IK

k cos k k p 4 1 2 1 2

FH IK

+

FH IK

sin kp 4

FH IK

; k³ 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 3679 2 . . . . z z z + - + R (z) R (z) = 1 + 0.2142 z–1 – 0.2142 z–2 Hence, Y (z) = 0 3679 0 2642 1 3679 0 3679 1 0 2142 0 2142 2 1 2 . . . . . . z z z z z + - + + -( )

a

_{- -}

f

= 0.3679z– 1 + 0.8463z– 2 + z– 3 + z– 4 + z– 5 + L y(0) = 0; y (1) = 0.3679; y (2) = 0.8463; y (k) = 1, k³ 3.

2.7 Taking z-transform of the given equation: Y z R z ( ) ( ) = z z z 2 2 3 2 - + = z z z 2 1 2 - -( ) ( ) R(z) = 1 + z– 1 = z z +1 Y z z ( ) = ( ) ( )( ) z z z + - -1 2 1 = 3 2 2 1 z- - z -y(k) = 3(2)k – 2(1)k; k³ 0

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 3 z z z z -- + ( . ) ( . ) Y z z ( ) = 2 3 0 5 0 3 2 z z z z -- + ( . ) ( . ) =- + -- + + 40 20 10 0 5 50 0 3 2 z z z . z . y(k) = – 40d (k) + 20d(k – 1) – 10(0.5)k + 50(– 0.3)k; k³ 0 (b) R (z) = z z-1; Y (z) = - + -

FH

- +

IK

FH

- -

IK

( ) 6 1 1 2 1 4 1 2 1 4 2 z z z z j z j Y z z ( ) = -- + _{- +}+ + _{- -} -16 1 8 4 1 2 1 4 8 4 1 2 1 4 z j z j j z j = – 16 1 16 6 5 16 2 z z z z - + -- +

y(k) = – 16(1)k + (0.56)k [7.94 sin (0.468k) + 16 cos (0.468k)]; k³ 0

2.9 (a) R (z) = 1 + z– 2 + z– 4 + ¼ = z z 2 2 1 -Y z z ( ) = z z- z+ z- z+ ( 1) ( 1) ( 0.5) ( 0 3. ) = -- + -+ + + + -0 833 0 5 0 41 0 3 0 476 1 0 769 1 . . . . . . z z z z y(k) = – 0.833(0.5)k – 0.41(– 0.3)k + 0.476(– 1)k + 0.769(1)k; k³ 0 (b) R(z) = z z-1; Y z z ( ) = 1 0 2 0 1 1 (z- .5) (z- . )(z- ) = -- + - + -- + -( ) 5 0 5 2 5 0 5 6 94 0 1 4 44 1 2 z . z z z . . . . . y(k) = – 10k(0.5)k + 2.5(0.5)k – 6.94(0.1)k + 4.44(1)k; k³ 0 2.10 y(¥) = lim z®1 (z – 1) Y (z) = K a a a a 1 1 1 1 2 2 - -- -( ) ( )

a

f

a

f

= K For y (¥) = 1, K = 1

2.11 Refer Eqn (2.45) Y (z) = z z- m ( a) ; y (¥) = limz m z z z ® -( ) ( ) 1 1 a = 0, if |a| < 1. (i) R(z) = z z-1; Y z z ( ) = 1 1 1 2 z + (z- )

a

f

= 1 2 1 1 2 1 2 1 2 z z z - -+ + y(k) = 1 2 (1) k – 1 2 cos kp 2

FH IK

– 1 2 sin kp 2

FH IK

; k³ 0 (ii) r(k) = {1, 0, – 1, 0, 1, 0, – 1, ¼} = cos kp 2

FH IK

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

a

f

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

2.13 (a) D(z) = z3 – 1.3z2 – 0.08 z + 0.24 = 0 (i) D(1) = – 0.14 < 0

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

(b) D(z) = z4 – 1.368 z3 + 0.4 z2+ 0.08z + 0.002 = 0 (i) D(1) = 0.114 > 0; satisfied

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

Form Jury’s table; from which you will find that (conditions (2.50b)): |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 D(z) = z4 + 0.5 z3 – 0.2 z2 + z + 0.4 = 0 D(r) = 1 1 4 +

-FHG IKJ

r r + 0.5 1 1 3 +

-FHG IKJ

r r – 0.2 1 1 2 +

-FHG IKJ

r r + 1 1 + -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 D(r) lies in the right-half plane. Therefore, three poles of G (z) lie inside the unit circle.

2.15 Refer Section 2.8; Eqn (2.56) 2.16 (a) Refer Section 2.9

(b) s2 + 2s + 2 = s2 + 2zw_ns + w2_n w_n = 2 2 ;ws p ; T = > 2 T < p 2 (c) Noise signal: n (t) = cos 50t

Z {cos 50kT} = Z cos 50 2 50 k´ p

{

}

= Z {cos 2kp} = 1 1-z-1 = 1 + z –1 + z–2 + L

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

p _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–aTej bT (ii) s = a± jw₀; z = eaTe±jw0T 2.20 G_a(s) = 1 1 s+ ; Gh0(s) = 1-e -s sT G(z) =Z{G_h0G(s)} = (1 – z–1)Z 1 1 s s+

RST UVW

( ) = 1 -e z e T T R(z) = z z Y z z -( ) 1; = 1 1 -- -- _{( )} e z e z T T

a

f

= 1 1 1 z- + -z-e-T y(k) = 1 – e–kT; k³ 0 2.21 G_a(s) = 10 2 s s+

RST UVW

( ) ; G (z) = (1 – z –1 )Z 10 2 2 s s+

RST

( )

UVW

G(z) = 5(1 – z–1) T z z z z z z e T - - - +

-L

NM

( 1)2 2( 1) 2

a

-2

f

O

QP

For T = 0.4 sec, G(z) = 10 0 76 16 1 0 46 z z z + - -( ) ( ) ( ) . . 2.22 (i) Refer Eqn (2.90b)

2.23 Refer Section 2.12; Fig. 2.41a.

2.24 Refer Section 2.12; Figs 2.44 and 2.37. 2.25 u k u k T ( )- ( -1) = K e k e k T T e k T T e k e k e k c I D ( ) ( ) ( ) { ( ) ( ) ( )} - - + + - - +

-L

NM

1 1 ₂ 2 1 2

O

QP

U(z) = K_c 1 1 1 1 1 1 +

-FHG IKJ

+

-L

NM

_TT _z- -

O

QP

T T z I D

a

f

_E(z) 2.26 Required z = 0.45, fM = 45 deg, T = 1.57 sec.

(i) D(s) is a phase lag compensator that meets the requirements. (ii) D(z) = D(s) with s = 2 1 1 T z z -+ D(z) = 0.4047 z z -0 9391 0 9752 . . 2.27 U (s) = K_c

L

1+ 1 +

NM

T s_I T sD

O

QP

E(s) U(z) = K_c 1 1 2 1 1 1 + +

L

N

MM

M

O

Q

PP

P

= -+ = -T sI T s s T z z D s z zT ( ) E (z) = K_c 1 2 1 1 1 + + - +

-L

NM

TT

O

QP

z z T T z z I D _{E (z)} u(k) = u_P(k) + u₁(k) + u_D(k) where u_P(k) = K_ce(k) u₁(k) – u₁(k – 1) = K T T c I 2 [e(k) + e(k – 1)] u₁(k) = K T T e i e i c I - + ( )

Â

a f

1₂ u_D(k) = K T T e k e k c D _{( ) -}

a f

_{- 1} Therefore, u(k) = K_c e k T T e i e i T T e k e k I i k D ( ) + - + ( ) + ( ) -

-L

NMM

Â

=

O

QPP

1 2 1 1

a f

_k

_{a f}

_p

2.28 (a) dy t dt

( )

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

z

(t) dt + r t 0

z

(t) dt y(k) = y(k – 1) – a y k T kT -(

z

1) (t) dt + r k T kT -(

z

1) (t) dt = y(k – 1) – aTy(k) + Tr(k) y(k) = 1 1+aT y(k – 1) + T aT 1+ r(k) (b) y(k + 1) = y(k) – a y kT k+ T ( )

z

1 (t) dt + r kT k+ T ( )

z

1 (t) dt = (1 – aT) y(k) + Tr (k)

y(k) = (1 – aT) y(k – 1) + Tr (k – 1) 2.29 (a) d y dt a dy dt by 2 2 + + = 0 1 2

T [y(k) – 2y(k – 1) + y(k – 2)] + a

T [y(k) – y(k – 1)] + by(k) = 0

b a T T + +

FH

1

IK

2 y(k) – a T + T

FH

2

IK

2 y(k – 1) + 1 2 T y(k – 2) = 0; y(0) = a; y(– 1) = a – Tb

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

A = 0 1

-

-L

NM

b a

O

QP

Using Eqn (2.113) we obtain,

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

b

L

NM

O

QP

CHAPTER 3 MODELS OF DIGITAL CONTROL DEVICES AND SYSTEMS 3.1 D(z) = {G_h0(s)G₁(s)} = G_h0G₁(z) Y z R z ( ) ( ) = D z G G z D z G G H z h h ( ) ( ) ( ) ( ) + 0 2 0 2 1 = G G z G G z G G z G G H z h h h h 0 1 0 2 0 1 0 2 1 ( ) ( ) ( ) ( ) + 3.2 Y z R z ( ) ( ) = G G z G G z H z h h 0 0 1 ( ) ( ) ( ) + 3.3 Y(z) = G_h0G₂(z) U(z) U(z) = G₁R(z) – G_h0G₂HG₁(z) U(z) Y(z) = G G z G R z G G HG z h h 0 2 1 0 2 1 1 ( ) ( ) ( ) +

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

Y(z) = G_pH₂R(z) + D z G G z D z G G z h p h p ( ) ( ) ( ) ( ) + 0 0 1 [H1R(z) – GpH2R(z)] This is the required answer.

3.5 Y(z) = G_h0G₁G₂(z)U(z); X(z) = G_h0G₁(z) U(z) E(z) = R(z) – X(z) – Y(z); U(z) = D(z) E(z) = D(z) [R(z) – G_h0G₁(z) U(z) – G_h0G₁G₂(z) U(z)] U(z) = D z R z D z G G z_h D z G G G z_h ( ) ( ) ( ) ( ) ( ) ( ) + + 1 _{0 1 0 1 2} ; Y(z) = Gh0G1G2(z) U(z) Y z R z ( ) ( ) = G G G z D z D z G G z G G G z h h h 0 1 2 0 1 0 1 2 1 ( ) ( ) ( ) ( ) ( ) + [ + ]

X(z) = G_h0G₁(z) U(z) X z R z ( ) ( ) = G G z D z G G z G G G z h h h 0 1 0 1 0 1 2 1 ( ) ( ) ( ) ( ) + [ + ]

3.6 For r = 0, the block diagram reduces to the following:

Y(z) = WG(z) – D(z)G_h0G(z) Y(z); Y(z) = WG z D z G G z_h ( ) ( ) ( ) + 1 ₀ 3.7 G(s) = 1 1 s s( + ); q qLR z z ( ) ( ) = G G z G G z h h 0 0 1 ( ) ( ) + ; G_h0G(z) = (1 – z–1)Z 1 1 2 s s+

RST UVW

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

a

f a

f

a

f

For T = 0.25 sec, G_h0G(z) = 0 0288 0 92 1 0 7788 . . . z z z + - -( ) ( ) ( )

3.8 Plant transfer function is, G(s) = 185 0 025. s+1 ( ) G_h0G(z) = (1 – z–1)Z 185 0 025 1 s . s+

RST

( )

UVW

= 185 1 40 40 -e z e T T

a

f

Let x(k) be the input to D/A block. x(k) = K_Fw_r(k) + K_P[w_r(k) – w (k)] X(z) = K_Fw_r(z) + K_P[w_r(z) – w(z)]; w(z) = G_h0G(z) X(z) w w z z r ( ) ( ) = K K G G z K G G z F P h P h + ( ) + ( ) ( ) 0 0 1

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

E z ( ) ( ) = 0.5 z z-1 G_h0G(z) = (1 – z–1)Z 1 1 2 s s+

RST UVW

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

a

f a

f

a

f

f_s = 5 Hz = 1 T; T = 0.2 sec G_h0G(z) = 0 019 0 0175 1 0 819 . . . z z z + - -( ) ( ); Y z R z ( ) ( ) = G G z D z G G z D z h h 0 0 1 ( ) ( ) ( ) ( ) + = 0 0095 0 92 2 81 2 65 0 819 3 2 . . . . . z z z z z + - + -( ) 3.10 (i) G_h0G(z) = (1 – z–1)Z e s s s -+

RST UVW

( ) 0 4 1 .

Using transform pairs of Table 3.1, we obtain,

G_h0G(z) = 1 1 0 6 0 6 1 1 - + --

-L

NM

O

QP

- - -( ) e z e e z z e . .

a

f

a

f

(1 – z–1) = 0 45 0 181 0 368 . . . z z z + -( ) Y z R z ( ) ( ) = G G z G G z h h 0 0 1 ( ) ( ) + = 0 45 0 181 0 081 0 181 2 . . . . z z z + + + (ii) G_h0G(z) = 0 45 0 181 0 368 2 . . . z z z + -( ) ; Y z R z ( ) ( ) = 0 45 0 181 0 368 0 45 0 181 3 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) G_h0G(z) = (1 – z–1)Z 5 1 2 2 s s+ s+

RST

( ) ( )

UVW

= 2 5 1 . z- – 3.75 + 5 1 0 3679 z z -( ) . – 1.25 z z -( 1) 0 1353.

The characteristic equation is: z2 + 2.12z + 0.234 = 0; z_{1, 2} = – 2, – 0.12 A pole lies outside the unit circle; the system is unstable.

3.14 z3 – 0.1z2 + 0.2Kz – 0.1K = D(z)

D(1) = 0.1K + 0.9 > 0 D(– 1) = – 1.1 – 0.3 K < 0

From Jury’s table, we get the following conditions. (refer conditions (2.50b))

|– 0.1K| < 1; True for 0 < K < 10

|0.01K2 – 1| > |– 0.19K|; |0.01K2 – 1| > |0.19K| K2 + 19K – 100 = 0

The system is stable for 0 < K < 4.293. 3.15 G_h0G(z) = K(1 – z–1)Z 1 3 s s+

RST UVW

( ) = K e z e T T 3 1 3 3

-F

HG

--

I

KJ

D(z) = 1 + Gh0G(z) = z – e –3T + K 3 (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 G_h0G(z) = (1 – z–1)Z K s2 s+1

RST UVW

a f

= K z z z T e z e Te z z e T T T T - - + + - --

-L

NM

O

QP

( ) ( ) - - -1 1 1 12

a

f a

f

k

p

a

f

D(z) = 1 + Gh0G(z) = 0; z 2 – az + b = 0 where, a = e–T + 1 – K(T – 1 + e–T); b = e–T + K(1 – e–T – Te–T) Put z = 1 1 + -r r , then r 2 _{(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

a

f

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, G_h0G(z) = K 0 368 0 264 1 368 0 368 2 . . . . z z z z + - + ( )

a

f

D(z) = z3

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

D(1) = – 0.368 + 0.368(1 + K) + 0.264K > 0 D(– 1) = – 2.368 – 0.368(1 + K) + 0.264K < 0

From Jury’s table, we get the following conditions (refer conditions (2.50b)):

|0.264K| < 1, gives K < 3.79

|0.07K2 – 1| < |0.361K + 0.368 + 0.368K| This gives, K < 0.785.

The system is stable for 0 < K < 0.785.

3.18 (a) G(s) = 4500 361 2 K s s( + . ); K = 14.5; z = 0.707; wn = 255.44 y(t) = 1 – e nt -z w z 1 2 sin w z z z n 1 t 1 2 1 2 - +

-F

HG

d

i

_tan-

I

KJ

(b) T = 0.01 sec, G_h0G(z) = 1 3198 0 4379 1 027 0 027 2 . . . . ; z z z Y z R z + - + ( ) ( ) = 1 3198 0 4379 0 0 4649 2 . . .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, G_h0G(z) = 0 029 0 0257 1 697 0 697 2 . . . . ; z z z Y z R z + - + ( ) ( ) = 0 029 0 0257 1 668 0 7226 2 . . . . z z z + - +

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

3.19 Y z R z ( ) ( ) = G G z G GH z h h 0 0 1 ( ) ( ) + G_h0G(z) = (1 – z–1)Z 1 1 s s+

RST UVW

a f

= 1 1 1 -e z e ; G_h0GH(z) = (1 – z–1)Z 1 1 2 s s+

RST UVW

a f

= e z e z z e - -- + - -( ) 1 1 1 2 1 1

a

f

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

a

f

_{; R(z) =} z z-1; Y(z) = 0 632 0 632 2 . . z z - +z

ak sin (Wk)« - + ( ) ( ) a z z a z a sin cos W W 2 ₂ 2 ; a 2 = 0.632; cosW = 1 2a = 0.629; W = 0.89 rad; y(k) = 1.02(0.795) k sin (0.89k) 3.20 G_h0G(z) = (1 – z–1)Z 1 1 s s+

RST UVW

a f

= 0 632 0 368 . . ; z Y z R z -( ) ( ) = 0 632 0 . .264 z+ ; R(z) = z z-1 Y(z) = 0 632 1 0 . .264 z z- z+ ( ) ( ); y(k) = 0.5 [1 – (– 0.264) k ]m(k) y(0) = 0; y(1) = 0.632; y(2) = 0.465; y(3) = 0.509 ¼

Y(z) =Z {G_h0(s) G(s)e–DTs)} E(z); E(z) = R(z) – Y(z); Y(z) =Z {G_h0(s) G(s)} E(z) E(z) = R z Gh s G s ( ) ( ) ( ) + { } 1 Z ₀ Y(z) = Z Z [ ] [ ] G s G s e G s G s h Ts h 0 0 1 ( ) ( ) ( ) ( ) -+ D R(z);Z {G_h0(s) G(s)} = 0 632 0 368 . . z -Z [Gh0(s) G(s) e –DTs_{] =} 0 393 0 239 0 368 . . . ;  z z z Y z R z + -( ) ( ) ( ) = 0 393 0 0 264 . .239 . z z z + + ( ) ; R(z) = z z-1 y(k) = [0.5 – 0.107 (– 0.264)k –1]m(k – 1) y(1) = 0.393 = y(0.5T); y(2) = 0.528 = y(1.5T)

y(3) = 0.493 = y(2.5T); y(4) = 0.5019 = y(3.5T) 3.21 (i) D(z) = 4 z z z z z z -+

FHG IKJ

+ + - +

F

HG

I

KJ

1 0 1 1 1 0 3 0 8 2 2 . .2 . . = 4 1 1 0 1 1 1 1 0 3 0 8 1 1 1 2 1 2 -+

L

NM

O

QP

L

NM

-+ ++

O

QP

-- -- -z z z z z z . .2 . . Refer Figs 3.19 – 3.20 (ii) D z z ( ) = - + + + - ++ 50 46 62 0 1 7 38 4 05 0 3 0 8 2 z z z z z . . . . . . D(z) = –50 + + - + - + + -- -46 62 1 0 1 7 38 4 05 1 0 3 0 8 1 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 + + - -( ) ( )

a

f

.5 . = 233 33 0 127 08 0 8 25 106.25 2 . .5 . . z- + z- + z + z = -- + -233 33 1 0 127 08 1 0 8 1 1 1 1 . .5 . . z z z z + (106.25 + 25 z –1 )z–1

3.23 (a) Process steady-state gain, K = qss m Q = 30 1 = 30°C/(kg/min) q(t) = 0.283qss at t1 = 25 min q(t) = 0.632qss at t2 = 65 min

Therefore (refer Eqns (3.53b))

tD + 1 3 t = t1 = 25 tD + t = t2 = 65 This gives tD = 5 min, t = 60 min (b) t_CD = t_D + 1 2 T = 5.5 K_c = 1.5t/Kt_CD = 0.545 T_I = 2.5t_CD = 13.75 min T_D = 0.4t_CD = 2.2 min 3.24 Ultimate gain, K_cu = 5

Ultimate period, T_u = 34 sec K_c = 0.45 K_cu = 2.13

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 TIM001–Timer for 60 sec delay TIM002–Timer for 15 sec delay TIM003–Timer for 30 sec delay

3.26 10001 00001 00000 00000 00002 10003 10000 CNT000 00002 10000 10001 00003 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 10000 00002 10001 10001 00000 CNT000 TIM001 CNT000 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 TIM001–Timer for 2 sec delay

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 TIM000–Timer for 20 sec delay TIM001–Timer for 20 sec delay

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 a continuous-time system. + + R s( ) E s( ) q( )s K1 K2 Js s 1 1 G(s) =q( ) ( ) s E s = K s Js K 1 2 ( + ) K_p = lim sÆ0 G(s) = • K_v = lim sÆ0 sG(s) = K K 1 2 K_a = lim sÆ0 s 2 G(s) = 0 4.2 Plant model: G(s) = 1 57 1 . ( ) s s+ ; Gh0( jw) G( jw) = 1 57 1 2 . ( ) e j j j T -+ w w w ; T = 1.57 sec

Bode plot analysis:

Phase margin (without ZOH) = 45 deg Phase margin (with ZOH) = 0 deg Specification:f M = 45 deg Let us use a lag compensator.

From the Bode plot of the uncompensated system, we see that the phase margin of 45 deg may be realized if the gain cross over frequency is moved from the present value (1 rad/sec) to a frequency of 0.5 rad/sec. To accom-modate phase lag, we take w_c2 = 0.4 rad/sec

20 log b = 8; b = 2.5 1 t = wc2 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 1 T z z -+ = 1.274 z z -+ 1 1; D₁(z) = 0 4( 0 939 0 975 . . ) ( . ) z z

-K_v of the original analog system is given by, K_v = lim sÆ0s 1 1 .57 ( ) s s+ = 1.57 K_v of the equivalent digital system is:

K_v = 1 1 Tlimz_® (z – 1) D1(z) Gh0G(z) where G_h0G(z) = 1.57 1 57 1 0 792 0 208 . . . z- -z

-L

NM

O

QP

, K_v = 1.57 4.3 K_p = lim zÆ1 Gh0G(z) = •; Kv = 1 1 Tlimz_Æ (z – 1) Gh0G(z) = 3.041 K_a = 1₂ T limz_Æ1 (z – 1) 2 G_h0G(z) = 0

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

e-zwnT _{= 0.965 = r ;zw}

nT = – ln r;wnT 1

2

- z = q From the above equations, we get

z z 1- 2 = -In r q ;z = -+ ln r r ln2 q2 w_n = 1 2 2 T In r+ q ;z = 0.199; wn = 8.93 4.5 (a) (i) W(s) = 0; Y₁(z) = G_h0G(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 z h h 2 1 3 0 1 0 1 + + (ii) R(s) = 0 Y₂(z) = GW(z) – G_h0G(z) D₁(z) Y₂(z) Y₂(z) = GW z D z G G z_h ( ) ( ) ( ) 1+ _{1 0} ; Y(z) = Y1(z) + Y2(z) (b) D₃(z) = D₂(z) G_D0G(z)

Y₁(z) = [ ( ) ( )] ( ) ( ) ( ) ( ) 1 1 1 0 3 1 0 + + D z G G z D z R z D z G G z h 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 s 2 _{; D(s) G(s) =} sK K s s 1 2 1 + + ( ) K_v = K₂; 1 K_v = 0.01 = 1 2 K (ii) Y s W s ( ) ( ) = s s s( + +1) K s₁ +K₂ ; Y(s) = 1 1 _{1 2} s s( + +) K s+K ; yss = 0 Thus a PI compensator meets the requirements.

4.7 G(s) = K s J +1; Gh0G(z) = K e z e T T (1- ) -t t ; For K = t = 1, G_h0G(z) = 0 393 0 607 . . z- ; S(z) = 1 1+ G G z_h0 ( ) = z z -0 6-07 0 . .214 S(ejwT) = cos( ) . sin( )

cos( ) .214 sin( ) w w w w T j T T j T - + - + 0 607 0 ; |S(e jwT_)|2 = 1 3685 1 1 0458 0 428 . .214 cos( ) . . cos( ) -w w T T w_s 2 = 2 2 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)L (z – l_k)

We consider the effect of parameter K on the root l_k. By definition,

D(lk, K) = 0. If K is changed to (K + DK), then lk also changes and the

new polynomial is,

D(lk + Dlk, K + DK) = D(lk, K) + ∂ ∂ ₌ D z _z k l Dlk + ∂ ∂ ₌ D K _z k l DK + L = 0

Neglecting higher order terms, we obtain,

Dlk = – ∂ ∂ ∂ ∂

L

NM

O

QP

= D D / / K z _z k l DK ∂ ∂ ₌ D K k l l = B(l_k) = – 1 K A k z _z k (l ); l ∂ ∂ ₌ D _{= P} iπk(lk -li) S_Klk ₌ D D l_k K K/ = A _k i k k i ( ) ( ) l l l P π -(b) G(s) = K s+ 1; Gh0G(z) = 0 393 0 607 . . K z- ;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; S_Kl = – 0.394

For a 10% change in nominal gain, the change in the root location is given by, D(l_k) = S_Kl ∂ K K = – 0.0393 (c) G(s) = 1 1 t s+ ; Gh0G(z) = 1 1 2 1 2 -e z e t t ;D(z, t ) = 1 + G_h0G(z) = 0; z + 1 – 2 1 2 e- t _{= 0} Let p = e -1 2t_; ∂ ∂ p t = 1 2 2 1 2 t t e -Characteristic equation is,

z + 1 + p(– 2) = 0; A(z) + p(B(z)) = 0 Nominal value of p = e -1 2 _{= 0.6065} Closed-loop pole is at z = 0.213 = l A(l) = 1.213; d z dz z p D( ) = = 1; Sp l = 1.213 = ∂ ∂ ( ) / l p p Now ∂p p = 1 2 2 1 2 t t t e p - _∂ = 1 2t t t ∂ _{= 0.5}∂t t ; St l = 0.6065 For 10% change in the nominal value of t, the change in root location is given by, D(lk) = Stl t t ∂ _{= 0.06065} 4.9 G_h0G(z) = 0 368 0 1 368 0 368 2 . .264 . . z z z + - + ; T = 1 sec; z = 1 0 1 0 + -.5 .5 w w G_h0G(w) = - - + + 0 0381 2 12 14 0 . ( )( . ) ( .924) w w w w G_h0G( jg) = 1 2 1 12 14 1 0

-FH IK

FH

+

IK

+

FH

IK

j j j j g g g g . .924

From the Bode plot and Nichols chart, we obtain,

fM = 28º; GM = 8 dB; w_b = 1.35 rad/sec; g_b = 1.6 rad/sec 4.10 G(s) = 1 2 s s( + ); T = 0.1 sec G_h0G(z) = 0.004683 ( .9355) ( )( . ) z z z + - -0 1 0 8187 ; z = 1 0 05 1 0 05 + -. . w w

G_h0G(w) = 0.5 ( . )( . ) ( .5016 ) 1 0 001666 1 0 05 1 0 + -+ w w w w (a) lim wÆ0 wGh0G(w) = Kv = 0.5K; K = 10 G_h0G(w) = 5 1 0 001666 1 0 05 1 0 ( . )( . ) ( .5016 ) + -+ w w w w

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

(b) Phase lead design: D(w) = 1 1 1 12 + + w w .994 .5 (obtained by standard design procedure), fM = 55º, K_v = 5, GM = 12.4 dB (c) D(w) = K w w ( ) ( ) 1 1 + + t bt ; Kv = 5 K_v = lim wÆ0 wD(w) Gh0G(w), gives K = 10.

The Bode plot of G(w) = 5 1 0 001666 1 0 05 1 0 ( . )( . ) ( .5016 ) + -+ w w w w

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

Required phase margin = 55º + 15º (error compensation). Crossover frequency w_c2 = 0.7 D(w) = 1 1 + + t bt w w; 1 t = w_c2 2 2 ( ) = 0.18; t = 5.71 Corresponding gain = 20 log₁₀b = 17; b = 7.0;

D(w) = 0 14 0 18 0 02 . ( . ) ( . ) w w + + w = 2 1 1 T z z -+ gives, D(z) = 0 141 0 0 . ( .98) ( .998) z z

-(d) From Nichols chart we find that bandwidth values g_b for three designs corresponding to parts (a), (b) and (c) are respectively 4.8 rad/sec, 9.8 rad/sec and 1.04 rad/sec.

(e) Reasonable sampling rates are 10 to 30 times the bandwidth. w_s = 2p

4.11 G(s) = 1₂ s , T = 0.1, Gh0G(z) = 0 005 1 12 . ( ) ( ) z z + -z = 1 0 05 1 0 05 + -. . w w ; Gh0G(w) = 1 0 05 2 - . w w ; Gh0G( jg) = 1 0 05 2 - j j . ( ) g g

Bode plot analysis: f M = – 2 deg.

In the low frequency range, –G_h0G( jg) is about –180 º. Therefore, a lag compensator cannot fulfil the requirement of 50º phase margin.

The lead compensator D(w) = 64 ( )

( ) w w + + 1

16 satisfies the requirements. The gain crossover frequency = 4

f M = 50.62º, GM = 13 dB, Kv = •, Ka = 4. 4.12 (a) K = 50 G_h0G(z) = 0 0043 0 85 1 0 61 . ( . ) ( )( . ) K z z z + - - ; G_h0G(w) = 10 1 20 1 246.67 1 4 84

-FH IK

FH

+

IK

+

FH

IK

w w w w . Bode plot/Nichols chart analysis: Gain crossover frequency = 6.6 rad/sec. f M = 20º, g_b = 10 rad/sec, w_b = 9.27 rad/sec. (b) Lead compensator D(w) = 0 1 0 06 1 .219 . w w + + results in gb = 18 rad/sec, w_b = 14.65 rad/sec. Lag compensator D(w) = 1 6 1 6.4 1 . w w + + results in gb = 5 rad/sec, w_b = 4.9 rad/sec.

(c) For partial compensation, we design lag section by selecting a cross over frequency of 3.2 rad/sec. The uncompensated plot has to be brought down by 9 dB. 20 log b = 9; b@ 3 1 t = 3 22 .2 = 0.8 D₁(w) = 1 1 3 75 1 .25 . w w + + results in f M = 54º, gb = 4.3.

We constrain the lead section design by taking a = 1

b = 0.333. The

– 10 log 1 a = – 4.77 at 4 rad/sec. 1 t = 4 a = 2.3; 1 a t = 6.92 D₂(w) = 0 4347 1 0 144 1 . . w w + +

Lag-lead compensator results in f M = 60º, gb = 8, wb = 7.61 rad/sec.

(d) D₁(z) = 0.342 z z

-FH

0

IK

0 .923 .973 ; D2(z) = 2.49 z z

-FH

0 793

IK

0 484 . . Refer Figs 3.19–3.22 for realization schemes. 4.13 (a) K_p = 0.8K; 1 1 0 8+ . K = 0.02; K = 61.25; G(s) = 49 3s+1 (b) G_h0G(z) = 7 0 8465 .5215 . z -Closed-loop pole: z + 6.675 = 0 The system is unstable.

(c) G_h0G(w) = 49 1 4 1 3 ( / ) ( ) -+ w w Lag compensator D(w) = 1 1 10 + + w

w satisfies the requirements. w = 4 z z -+

FH IK

1 1 ; D(z) = 0.122 z z

-FH

0 6

IK

0 951 . .

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

(b) G_h0G(z) = K z z ( .9048) ( ) -0 12

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 pole z = – 1. It is 2.1. Therefore, the system is stable for 0 < K < 2.1. There is a double pole at z = 0.81. The value of K at this point is obtained as 0.38. 4.15 G_h0G(z) = K T e z e Te z z e z T T T T [( ) ( ) ] ( )( ) - + + - -- -- - - - -- - -1 1 1 1 1 2 1 1

(i) T = 1 sec, G_h0G(z) = 0 3679 0 7181 1 0 3679 . ( . ) ( )( . ) K z z z + -

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

At the point of intersection of the root locus with unit circle, we find by magnitude condition, K = 2.3925. This gain results in marginal stability. (ii) T = 2 sec, G_h0G(z) = 1 1353 0 1 0 1353) . ( .5232) ( )( . K z z z + -

-Breakaway points are at z = 0.4783, – 1.5247; Critical gain K = 1.4557. (iii) T = 4 sec, G_h0G(z) = 3 0183 0 3010 1 0 0183) . ( . ) ( )( . K z z z + -

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

4.16 G_h0G(z) = K z z z ( . ) ( )( . ) + - -0 717 1 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

z_1,2 = 0.244 ± j0.97 = 1–1.33 rad = ejwT;w = 1.33 rad/sec.

(b) The value of K at this point is 0.072. e–T/t = 0.648, t = 2.3 sec.

(c) At the point of intersection of the root locus with z = 0.5 locus, we get K = 0.18. The line through intersection point makes an angle of 32º with real axis. Therefore

w_nT 1- z2 = 32º = 0.558 rad; w_n = 0.644 rad/sec 4.17 z2 + 0.2Az – 0.1A = 0; 1 + 0.2 (A z₂ 0.5) z - _{= 0; G} h0G(z) = K z z ( - 0.5) 2 ; = K = 0.2 A

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

4.18 G(s) = 1 1 s+ ; T = 0.1 H(s) = PT 2p = 0.9554; Gh0GH(z) = 0 091 0 . .9048 K z -The root locus is a circle.

(a) 1 + G_h0GH(z) = 0; z – 0.9048 + 0.091K = 0 For K = 1, z – 0.8138 = 0 e–T/t = 0.8138; T t = 0.206; t = 0.4854 sec (b) Required time-constant = 0 4854 4 . = 0.12136 e–T/0.12136 = 0.43867; 0.091K = 0.46613; K = 5.1223 4.19 G_h0G(z) = 0 01873 0 1 0 8187 . ( .9356) ( )( . ) K z z z + - -(a) z = 0.5; w_n = 4.5; z = e-zwnT –_w nT(1 – z 2 ) This gives z_1,2 = 0.6354 –± 45º = 0.4493 ± j0.4493.

Angle deficiency at point P corresponding to z₁ is – 72.25 deg. We choose the zero of the controller to cancel the pole at z = 0.8187. Then the pole of the controller is determined to satisfy the angle con-dition at P. This gives

D(z) = z z -0 8187 0 1595 . . ; |D(z)Gh0G(z)| = 1, requires K = 13.934 D₁(z) = 13 0 8187 0 1595 .934( . ) ( . ) z z -(b) K_v = 1 1 T limz_Æ (z – 1) D(z) Gh0G(z) = 3. (c) D₂(z) = z z -b b 1 2 ; 1 1 1 2 -b b = 3 Let,b₂ = 0.98, then b₁ = 0.94; D₂(z) = z z -0 0 .94 .98

From the root locus plots of 1 + D₁(z) G_h0G(z) = 0 and 1 + D₂(z) G_h0G(z) = 0, it is seen that lag compensator decreases the margin of stability.

4.20 G_h0G(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 get 0.2 K = 0.3823 or K = 1.91.

At this point,

w_n = 1.65, K_p = 0.957.

We will use lag compensator. K_p is to be increased by a factor of 7 0 .5 .957 = 7.837; D(z) = z z -b b 1 2 ; 1 1 1 2 -b b = 7.837 Let,b₂ = 0.98, then b₁ = 0.84

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

4.21 G_h0G(z) = KT z z 2 2 2 1 1 + -( ) ; T = 1 sec z = 0.7, w_n = 0.3, z = e-zwnT –_w nT(1 – z 2 ) This gives z_1,2= 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. G_h0G(z) D(z) = 0 36 2 1 1 0 8 2 . ( ) . z z z z + - - ; K_a = 1₂ 1 T limz® (z – 1) 2 G_h0G(z) D(z) = 0.072

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

4.22 G(s) = 40 40 120 e s s s -+ / ( ); T = 1 120 sec; Gh0G(z) = 0 00133 0 75 1 0 72 . ( . ) ( )( . ) z z 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) is found at z = – 0.4. The magnitude criterion at the point of intersection of the root locus with the circle of radius 0.56 gives K = 0.2.

Therefore, D(z) = 0 0 72 0 00133 0 4 .2( . ) . ( . ) z z -+ = 150 ( . ) ( . ) z z -+ 0 72 0 4

The third pole corresponding to K = 0.2 lies inside the circle of radius 0.56. G_h0G(z) D(z) = 0 0 75 1 0 4 .2( . ) ( )( . ) z z z z + - + K_p = lim z®1 Gh0G(z) D(z) = •;ess * = 0 4.23 (a) Refer Examples 4.9–4.10.

(b) G(s) = 1 1 s s( + ); T = 0.1 sec; Gh0G(z) = 0 0048 0 1 0 . ( .9833) ( )( .9048) z z z + -

-We select an M(z) that has pole excess of at least equal to that of G_h0G(z), and unstable (or critically stable) poles of G_h0G(z) are in-cluded in 1–M(z) as zeros. M(z) = z–1 satisfies these requirements. D(z) = 1 1 0 G G z M z M z h ( ) ( ) ( )

-L

NM

O

QP

= 208.33 z z -+

LNM

0

OQP

0 .9048 .9833 4.24 G(s) = 1 1 s s( + ); T = 0.1 sec (a) z = 0.8, w_n = 2 10 p T; z = e nT -zw _–w nT 1 2 - z This gives z_1,2 = 0.55 ± j0.22 G_h0G(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 of G_h0G(z), unstable poles of G_h0G(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) = lim z®1 (z – 1) E(z) = 0 e*_ss(unit-ramp) = lim z®1 (z – 1) Tz z -( 1)2 (z – 1) F(z) = TF(1) = 1 K_v

K_v = 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 3509 2 . . . . z z z -- + ( ) D(z) = 1 1 0 G G z M z M z h ( ) ( ) ( )

-L

NM

O

QP

= 135.227 ( . ) ( . ) ( . ) ( . ) z z z z - -+ -0 9-048 0 6135 0 9833 0 7491 (b) y(k) = 0, 0.5, 1, 1, K Y(z) = 0.5z–1 + z–2 + z–3 + L Y z R z ( ) ( ) = (0.5z –1 _{+ z}–2 _{+ z}–3 _{+ L) (1 – z}–1₎ M(z) = (0.5z–1 + 0.5z–2); G_h0G(z) = 4.8 × 10–3 ( . ( ) ( . ) z z z + - -0 9833) 1 0 9048 D(z) = 104.17 ( . ) ( . ( . ) z z z z - ( + ) + + 0 9048 1 0 9833) 0 5 E(z) = R(z) [1 – M(z)] e_ss =lim z®1 (z – 1) E(z) = 0.15. 4.25 G(s) = 1 10s+1; T = 2 sec; Gh0G(z) = 0 18 0 82 . . z -y(t) = 1 – e–t; Y(s) = 1 1 1 s- s+ Y(z) = z z z z -1- -0 14. = 0 86 1 0 14 . ( ) ( . ) z z- z- ; M(z) = Y z R z z ( ) ( ) . . = -0 86 0 14 D(z) = 1 1 0 G G z M z M z h ( ) ( ) ( )

-L

NM

O

QP

= 4 8 3 9 1 1 1 . - . -z z 4.26 Parallel to Review Example 4.3.

Result:

u(k) = 10 e(k) – 6e(k – 1) – 0.75u (k – 1) U(z) = 10E (z) – 6z–1E(z) – 0.75z–1U(z)

U z E z ( ) ( ) = D(z) = 10 6 1 0 75 1 1 -+ -z z . 4.27 G(s) = 2 1 2 s+ s+ ( ) ( ); T = 1 sec G_h0G (z) = 0.4 z z z + - -( ) ( ) ( ) 0 368 0 368 0 135 . . . ; R (z) = 1 1-z-1

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

M(z) = =₁z–1 + =₂z–2 E(z) = R(z) [1 – M(z)] lim

z®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 + L]

Dividing U(z) by R(z), U z

R z ( )

( ) = u(0) + [u(1) – u(0)]z

–1 + [u(2) – u(1)]z–2 = >₀ + >₁z–1 + >₂z–2 G_h0G(z) = Y z U z ( ) ( ) = Y z R z U z R z ( ) / ( ) ( ) / ( ) = = > => > > >> 1 0 1 2 0 2 1 0 1 2 0 2 1 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) = 1 1 0 G G z M z M z h ( ) ( ) ( )

-L

NM

O

QP

= U z R z Y 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) = _{s s(} 1₊₁₎; T = 1 sec

G_h0G(z) = 0 3679 0 1 1 3679 0 3679 1 2 1 2 . .2642 . . z z z z - --+ -- + ; M(z) = a1z –1 _{+ a} 2z –2 ₌ Y z R z ( ) ( ) U(z) = u(0) + u(1)z–1 + u(2) [z–2 + z–3 + L]

R(z) = 1 1-z-1 ; U z R z ( ) ( ) = b0 + b1z –1 + b2z –2 G_h0G(z) = Y z R z U z R z ( ) ( ) ( ) ( ) / / = = > => > > >> 1 0 1 2 0 2 1 0 1 2 0 2 1 z z z z - -- -+ + + a1 + a2 = 1

Comparing the coefficients of G_h0G(z), we get a₁ = 0.582, a₂ = 0.418; M(z) = 0.582z–1 + 0.418z–2 D(z) = 1 1 0 G G z M z M z h ( ) ( ) ( )

-L

NM

O

QP

= U z R z Y z R z ( ) ( ) ( ) ( ) -/ / 1 = 1 582 0 582 1 0 418 1 1 . . . -+ -z z Output sequence, Y z R z ( ) ( ) = 0.582z –1 + 0.418z–2; R(z) = 1 1-z-1 y(k) = 0, 0.582, 1, 1, º 4.29 G(s) = e s s -+ 5 10 1; T = 5 sec; Gh0G(z) = 0 3935 1 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 + L = 0.6225z–2 + z–3 1 1-z-1 Y z R z ( ) ( ) = M(z) = 0.6225z –2 + 0.3775z–3

Pole excess of M(z) = 2 = pole excess of G_h0G(z) D(z) = 1 1 0 G G z M z M z h ( ) ( ) ( )

-L

NM

O

QP

= 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_h R z ( ) ( ) ( ) ( ) / / 0 = 1.582 1 0 3678 1 2 1 -. z z

c

h

c

h

= 1.582 + 1.582z–1 + z–2 + z–3 + L

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

4.30 G_h0G(z) = 0 3935 1 0 6065 2 1 . . z z

-Pole excess = 2; second-order model. Y z R z ( ) ( ) = a2z –2 U z R z ( )

( ) = u(0) + [u(1) – u(0)]z

–1 + [u(2) – u(1)]z–2 = b₀ + b₁z–1 + b₂z–2 Y z R z U z R z ( ) ( ) ( ) ( ) / / = a b b b b b 2 0 2 1 0 1 2 0 2 1 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) = 1 1 0 G G z M z M z h ( ) ( ) ( )

-L

NM

O

QP

= U z R z Y z R z ( ) ( ) ( ) ( ) -/ / 1 = 2.541 1 0 6065 1 1 1 1 1 -- + -- -. z z z

c

h

c hc h

It is physically realizable. Y(z) = z–2 1 1-z-1 = z –2 + z–3 + z–4 + L y(k) = 0, 0, 1, 1, 1, K u(0) = b₀ = 2.541

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

CHAPTER 5 CONTROL SYSTEM ANALYSIS USING STATE VARIABLE METHODS 5.1 J_e = n2J = 0.4 ; B_e = 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 --

-L

N

MM

MM

O

Q

PP

PP

. . ; b = 0 0 0 0 2.

L

N

MM

MM

O

Q

PP

PP

y = q_L = nx₁ = 0.5 x₁ 5.2 x3 _{x2 x1} if u q _q K_{A s} K_T f f L + R + 11 Js + B1 1s &x1 = x2 ; 0.5&x2 + 0.5x2 = 10x3 20&x₃ + 100x₃ = 50u ; y = x₁ A = 0 1 0 0 1 20 0 0 5

-L

N

MM

M

O

Q

PP

P

; b = 0 0 2 5.

L

N

MM

M

O

Q

PP

P

; c = [1 0 0]

5.3 x₁ = q_M ; x₂ = &q_M, x₃ = i_a ; y = q_L &x1 = x2 2&x₂ + x₂ = 38 x₃ 2&x₃ + 21x₃ = e_a – 0.5 x₂ e_a = k₁(q_R – q_L) – k₂ &q_M = k₁q_R – k1 20 x1 – k2x2 A = 0 1 0 0 0 5 19 40 0 5 2 21 2 1 2 -- - +

-L

N

MM

MMM

O

Q

PP

PPP

. ( . ) k k ; b = 0 0 2 1 k

L

N

MM

MM

O

Q

PP

PP

; c = 1 20 0 0

L

NM

O

QP

5.4 x₁ = w ; x₂ = i_a J&w + Bw = K_Ti_a R_ai_a + L_a di dt a = e_a – K_bw e_a = K_ce_c = K_c [k₁ (e_r – K_tw) – k2ia} A = -- + - +

L

N

MM

MM

O

Q

PP

PP

B J K J k K K K L R k K L T t c b a a c a ( ₁ ) ( ₂ ) ; b = 0 1 k K L c a

L

N

MM

O

_Q

PP

; c = [1 0] 5.5 A = P–1AP =

-L

NM

1511 68

O

QP

b = P–1b = 1 2

L

NM

O

QP

; c = cP = [2 –1] Y s U s ( ) ( ) = P₁D₁ D = s s s -- -- -- -2 1 2 1 ( 3 2 ) = 1 3 2 2 s + s+

X2 X = Y1 s–1 –2 –3 1 1 U s–1 Y s U s ( ) ( ) = P₁D₁ P₂D₂ P₃D₃ P₄D₄ 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 2 2 s + s+ X2 X1 Y s–1 –1 6 2 –15 –11 2 1 8 U s–1 5.6 A = 0 1 0 0

L

NM

O

QP

; b = 0 1

L

NM

O

QP

A= P–1AP = 1 1 –1 –1

L

NM

O

QP

; b= P–1b = 0 1

L

NM

O

QP

|lI – A| = |lI – A| = l2

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

-L

N

MM

M

O

Q

PP

P

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

L

N

MM

M

O

Q

PP

P

D = s3 + 3s2 + 1

5.8

5.9 Taking outputs of integrators as state variables, we get (x₁ being the output 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 --

-L

N

MM

M

O

Q

PP

P

; b = 0 0 1

L

N

MM

M

O

Q

PP

P

; c = [2 –2 1]

5.10 Taking outputs of integrators as state variables (x₁ and x₂ are outputs of top two integrators from left to right; x₃ and x₄ are corresponding 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

-L

N

MM

MM

O

Q

PP

PP

; B = 3 0 1 2 0 3 0 0

L

N

MM

MM

O

Q

PP

PP

; C = 0 1 0 0 0 0 0 1

L

NM

O

QP

5.11 (a) G(s) = c (sI – A)–1b = s s s + + + 3 1 2 ( )( )

(b) G(s) = 1 1 2 (s+ )(s+ ) 5.12 a₁ = – tr (A) = – 4 Q₂ = A + a₁I = - --

-L

N

MM

M

O

Q

PP

P

2 1 0 1 3 2 1 0 3 ;a₂ = –1 2 tr(AQ2) = 6 Q₃= AQ₂ + a₂I = 1 1 2 3 2 4 1 1 3 --

-L

N

MM

M

O

Q

PP

P

; a3 = – 1 3 tr(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 s D = 1 D( )s - + -- + - +

L

NM

32 25 2 4(2 33 1

O

QP

s s s s s s ) ( ) 5.13 &x₁ = – 3x₁ + 2x₂ + [– 2x₁ – 1.5x₂ – 3.5x₃] &x2 = 4x1 – 5x2 &x3 = x2 – r A = -

-L

N

MM

M

O

Q

PP

P

5 0 5 3 5 4 5 0 0 1 0 . . ; b = 0 0 1

-L

N

MM

M

O

Q

PP

P

; c = [0 1 0] G(s) = c(sI – A)–1b = 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 = --

-L

NM

21 11

O

QP

; b = 0 1

L

NM

O

QP

; c = [–1 1]; d = 1

(b) x₁ = output of lag 1/(s + 2) x₂ = output of lag 1/s x₃ = output of lag 1/(s + 1) &x1 + 2x1 = y ; &x2 = – x1 + u &x3 + x3 = – x1 + u ; y = x2 + x3 A = --

-L

N

MM

M

O

Q

PP

P

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

L

N

MM

M

O

Q

PP

P

; c = [0 1 1]

5.15 Taking outputs of pseudo-integrators as state variables (x₁, x₂, x₃, x₄: from top 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 = - - -- - -- - --

-L

N

MM

MM

O

Q

PP

PP

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 .

L

N

MM

MM

O

Q

PP

PP

; C = 1 1 0 0 0 0 1 1

L

NM

O

QP

5.16 (i) Y s U s ( ) ( ) = s s s + + + 3 1 2 ( )( ) = 2 1 1 2 s+ + s+ – L LL L L =

-L

NM

01 02

O

QP

; b = 1 1

L

NM

O

QP

; c = [2 –1]