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
8 DIGITAL CONTROL AND STATE VARIABLE METHODS
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)
10 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
14 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
�
� �
� �
� �
16 DIGITAL CONTROL AND STATE VARIABLE METHODS
�(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)
18 DIGITAL CONTROL AND STATE VARIABLE METHODS
�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
20 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
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
22 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
TIM001–Timer for 2 sec delay
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
TIM000–Timer for 20 sec delay
TIM001–Timer for 20 sec delay
24 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
TIM000–Timer for 7 sec delay
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.
26 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
28 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
30 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
32 DIGITAL CONTROL AND STATE VARIABLE METHODS
– 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.
34 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
36 DIGITAL CONTROL AND STATE VARIABLE METHODS
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)
38 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
� �� �
40 DIGITAL CONTROL AND STATE VARIABLE METHODS
= 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]
42 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
44 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
46 DIGITAL CONTROL AND STATE VARIABLE METHODS
(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.
48 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
50 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
(l2I – A)v2 = 0 gives
v2 = [1 – 2 1/2]T
(l3I – A)v3 = 0 gives
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
. . .
. .
.
.
. . .
. . .
52 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
- - - -
- - - -
--
+
--
+
�
�
���
�
�
���
54 DIGITAL CONTROL AND STATE VARIABLE METHODS
(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
56 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
58 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
- - - -
- - - -
- -
- -
�
��
�
��
60 DIGITAL CONTROL AND STATE VARIABLE METHODS
= ��–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
62 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
64 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
66 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
68 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
�
�
70 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
72 DIGITAL CONTROL AND STATE VARIABLE METHODS
�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)
74 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
78 DIGITAL CONTROL AND STATE VARIABLE METHODS
|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.
80 DIGITAL CONTROL AND STATE VARIABLE METHODS
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]
82 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
84 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
86 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
88 DIGITAL CONTROL AND STATE VARIABLE METHODS
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)
90 DIGITAL CONTROL AND STATE VARIABLE METHODS
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]
92 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
94 DIGITAL CONTROL AND STATE VARIABLE METHODS
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.
98 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
100 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
Solving for P, we get
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
102 DIGITAL CONTROL AND STATE VARIABLE METHODS
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.
104 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
Solving for P, we get
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
Solving Ricatti equation, ATP + PA – PBR–1BTP + Q = 0, we get
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
106 DIGITAL CONTROL AND STATE VARIABLE METHODS
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.
Solving Ricatti equation, ATP + PA – PBR–1BTP + Q = 0, we get
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.
108 DIGITAL CONTROL AND STATE VARIABLE METHODS
(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
110 DIGITAL CONTROL AND STATE VARIABLE METHODS
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 =
•
 + ~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
112 DIGITAL CONTROL AND STATE VARIABLE METHODS
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
+=
•
 ; 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
10.3 Revisiting Example 10.1 (Fig. 10.22) will be helpful.
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.
10.4 Revisiting Example 10.1 (Fig. 10.22) will be helpful.
For a limit cycle to exist, 2 D < E < • ; D = 0.2
–G(jw1) = – 180∞ for w1 = 10.95. |G(jw1)| = 0.206
|G(jw1)| π -1
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
114 DIGITAL CONTROL AND STATE VARIABLE METHODS
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.
10.8 Revisiting Example 10.2 (Fig. 10.24) will be helpful.
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
116 DIGITAL CONTROL AND STATE VARIABLE METHODS
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.
10.12 Revisiting Example 10.4 (Fig. 10.34) will be helpful.
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.
118 DIGITAL CONTROL AND STATE VARIABLE METHODS
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.
120 DIGITAL CONTROL AND STATE VARIABLE METHODS
(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).
122 DIGITAL CONTROL AND STATE VARIABLE METHODS
(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
=
 +���
�
tr = s w z wrlh
l
m
l rh1
101
=
 +���
�
�yj = v t vjr r jr
p
+=
 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
=
���
�
=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
=
 = 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
+���
�
==
 01
2
; l 1, 2
126 DIGITAL CONTROL AND STATE VARIABLE METHODS
�y = s v z vl ll
+���
�=
 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
+���
�=
 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 , ( )=���
���
���
���
��
��
128 DIGITAL CONTROL AND STATE VARIABLE METHODS
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+
=.
. .