Table Of ContentNinth Edition
Farid Golnaraghi • Benjamin C. Kuo
Solutions Manual
Solutions Manual
, 9th Edition
Automatic Control Systems
Chapter 2 Solutions
Golnar
C
2
(
2
2
M
Chapter 2
2‐1 (a) Poles
Zeros
(c) Poles: s =
Zeros
2-2)
a)
b)
c)
2-3)
MATLAB code
s: s = 0, 0, −1, −
s: s = −2, ∞, ∞
0, −1 + j, −1 −
s: s = −2.
���� � �
���� � �
���� � �
e:
−10;
, ∞.
j;
�����
���� �������
�
��
����
�
�����
����
�������������
(b)
(d) Poles: s
2‐1
Poles: s = −2,
Zeros: s = 0.
The pole and
= 0, −1, −2, ∞
, −2;
zero at s = −1 c
.
aghi, Kuo
cancel each other.
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
clear all;
s = tf('s')
'Generated transfer function:'
Ga=10*(s+2)/(s^2*(s+1)*(s+10))
'Poles:'
pole(Ga)
'Zeros:'
zero(Ga)
'Generated transfer function:'
Gb=10*s*(s+1)/((s+2)*(s^2+3*s+2))
'Poles:';
pole(Gb)
'Zeros:'
zero(Gb)
'Generated transfer function:'
Gc=10*(s+2)/(s*(s^2+2*s+2))
'Poles:';
pole(Gc)
'Zeros:'
zero(Gc)
'Generated transfer function:'
Gd=pade(exp(-2*s),1)/(10*s*(s+1)*(s+2))
'Poles:';
pole(Gd)
'Zeros:'
zero(Gd)
2‐2
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
Poles and zeros of the above functions:
(a)
Poles: 0 0 ‐10 ‐1
Zeros: ‐2
(b)
Poles: ‐2.0000 ‐2.0000 ‐1.0000
Zeros: 0 ‐1
(c)
Poles:
0
‐1.0000 + 1.0000i
‐1.0000 ‐ 1.0000i
Zeros: ‐2
Generated transfer function:
(d) using first order Pade approximation for exponential term
Poles:
0
‐2.0000
‐1.0000 + 0.0000i
‐1.0000 ‐ 0.0000i
Zeros:
1
2‐3
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
2-4) Mathematical re
ation:
present
In all cases substitute � � �� and simplify. The use MATLAB to verify.
a)
3
1
2
2
2
2
2
2
2
2
2
2
2
10(
2)
(
1)(
10)
10(
2)
(
1)(
10)
(
1)(
10)
(
1)(
10)
10(
2)(
1)(
10)
(
1)(
100)
2
1
10
2
1
10
(
)
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
R
R e
e
e φ
φ
φ
ω
ω
ω
ω
ω
ω
ω +
ω
ω
ω
ω
ω
ω
ω
ω
ω ω
ω
ω
ω
ω
ω
ω
ω
+
−
+
+
+
−
+
−
=
×
−
+
+
−
+
−
+
+
−
+
−
+
=
−
+
+
+
−
+
−
+
=
+
+
+
=
2
2
2
2
2
2
2
2
2
2
1
1
2
2
2
1
2
2
2
2
1
3
2
2
1
2
3
10 2
1
10
;
(
1)(
100)
2
tan
2
2
1
tan
1
1
10
tan
10
10
R
ω
ω
ω
ω ω
ω
ω
ω
φ
ω
ω
ω
φ
ω
ω
ω
φ
ω
φ
φ
φ
φ
−
−
−
+
+
+
=
−
+
+
+
=
+
−
+
=
+
−
+
=
+
=
+
+
b)
3
1
2
2
2
2
2
2
2
2
10
(
1) (
3)
10
(
1)(
1)(
3)
(
1)(
1)(
3)
(
1)(
1)(
3)
10(
1)(
1)(
3)
(
1) (
9)
1
1
3
1
1
9
(
)
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
R
R e
e
e φ
φ
φ
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
+
+
−
+
−
+
−
+
=
×
+
+
+
−
+
−
+
−
+
−
+
−
+
−
+
=
+
+
−
+
−
+
−
+
=
+
+
+
=
2
2
2
2
2
2
1
1
2
2
1
2
2
2
1
3
2
1
2
3
10 1
9
;
(
1) (
9)
1
tan
1
1
1
tan
1
1
9
tan
3
9
R
ω
ω
ω
ω
ω
ω
φ
ω
ω
ω
φ
ω
ω
ω
φ
ω
φ
φ
φ
φ
−
−
−
+
+
=
+
+
−
+
=
+
−
+
=
+
−
+
=
+
=
+
+
2‐4
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
c)
2
2
2
2
2
2
2
2
2
2
2
2
10
( 2
2
)
10
(2
2 )
( 2
2
)
(2
2 )
10( 2
(2
) )
(4
(2
) )
2
(2
)
4
(2
)
(
)
j
j
j
j
j
j
j
j
j
R
R e φ
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
+
−
−
−
−
=
×
+
−
−
−
−
−
−
=
+
−
−
−
−
=
+
−
=
ω
2
2
2
2
2
2
2
2
2
2
2
2
1
2
2
10 4
(2
)
10
;
(4
(2
) )
4
(2
)
2
4
(2
)
tan
2
4
(2
)
R
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
φ
ω
ω
ω
−
+
−
=
=
+
−
+
−
− −
+
−
=
−
+
−
2
2
d)
3
1
2
2
2
2
2
2
/2
2
2
2
10
(
1)(
2)
(
1)(
2)
10 (
1)(
2)
2
1
2
1
(
)
j
j
j
j
j
j
j
e
j
j
j
j
j
j
e
j
j
R
e
R e
e
e
ω
ω
ω
π
φ
φ
φ
ω
ω
ω
ω
ω
ω ω
ω
ω
ω
ω
ω
−
−
−
−
+
+
−
−
+
−
+
=
+
+
−
+
−
+
=
+
+
=
2
2
2
2
2
1
1
2
2
2
1
2
2
1
2
3
1
;
10
2
1
2
tan
2
2
1
tan
1
1
R
ω
ω
ω
ω
ω
φ
ω
ω
ω
φ
ω
φ
φ
φ
φ
−
−
=
+
+
−
+
=
+
−
+
=
+
=
+
+
MATLAB code:
clear all;
s = tf('s')
'Generated transfer function:'
Ga=10*(s+2)/(s^2*(s+1)*(s+10))
figure(1)
2‐5
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
Nyquist(Ga)
'Generated transfer function:'
Gb=10*s*(s+1)/((s+2)*(s^2+3*s+2))
figure(2)
Nyquist(Gb)
'Generated transfer function:'
Gc=10*(s+2)/(s*(s^2+2*s+2))
figure(3)
Nyquist(Gc)
'Generated transfer function:'
Gd=pade(exp(-2*s),1)/(10*s*(s+1)*(s+2))
figure(4)
Nyquist(Gd)
Nyquist plots (polar plots):
Part(a)
2‐6
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
-300
-250
-200
-150
-100
-50
0
-15
-10
-5
0
5
10
15
Nyquist Diagram
Real Axis
Imaginary Axis
Part(b)
-1
-0.5
0
0.5
1
1.5
2
2.5
-1.5
-1
-0.5
0
0.5
1
1.5
Nyquist Diagram
Real Axis
Imaginary Axis
Part(c)
2‐7
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
-7
-6
-5
-4
-3
-2
-1
0
-80
-60
-40
-20
0
20
40
60
80
Nyquist Diagram
Real Axis
Imaginary Axis
Part(d)
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
Nyquist Diagram
Real Axis
Imaginary Axis
2‐8
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
2-5)
In all cases find the real and imaginary axis intersections.
a)
{
}
{
}
2
2
2
2
2
2
1
2
2
10
10(
2)
10
2
(
)
(
2)
(
4)
(
4)
(
4)
2
Re
(
)
cos
,
(
4)
Im
(
)
sin
,
(
4)
2
(
4)
tan
(
4)
10
(
4)
j
j
G j
j
G j
G j
R
;
ω
ω
ω
ω
ω
ω
ω
ω
φ
ω
ω
ω
φ
ω
ω
φ
ω
ω
ω
−
−
+
−
=
=
=
−
+
+
+
=
=
+
−
=
=
+
+
=
−
+
=
+
1
0
1
1
lim
(
)
5;
tan
90
0
0
lim
(
)
0;
tan
180
1
Real axis intersection @
0
Imaginary axis
er
b&c)
���
= 1
o
int
section does not exist.
G j
G j
j
ω
ω
ω
φ
ω
φ
ω
−
→
−
→∞
=
=
= −
−
=
=
= −
−
=
o
o
���
�����
� 0
�����∞ ����� = 0 � -180o
����� �
�
��� �
���
�
����� �
��� �
���� �
���
�
����� �
����
���� �
���
�
�
�
��� �
���
�
Therefore:
Re{ G(jω) } =
��� �
���
�
���� �
���
�
�
�
��� �
���
�
Im {G(jω)} = �
���� �
���
���� �
���
�
�
�
��� �
���
�
2‐9
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
If Re{G(jω )} = 0
�
�
� ��
If Im{ G(jω )} = 0
�
�
� � 0
� � 0
�
� ∞
If ω = ωn
�
�
� ����� �
� �� �� � � � 90�
�
If ω = ωn
and ξ = 1
�
������ �
If ω = ωn and ξ
�
� 0
������ �
If ω = ωn and ξ � ∞
�
������ � 0
d)
ω) =
����
����
G(j
�����
ω��
lim
G�jω� =
�
limω�∞ G�jω� =
- 90o
� -180o
e)
����| �
�����
|�
������ � �
�
√
�
�
��� �
� G(jω) = �
�
����� + � ����� = tan-1 (ω T) – ω L
2‐10
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
2‐6
MATLAB code:
clear all;
s = tf('s')
%Part(a)
Ga=10/(s-2)
figure(1)
nyquist(Ga)
%Part(b)
zeta=0.5; %asuuming a value for zeta <1
wn=2*pi*10 %asuuming a value for wn
Gb=1/(1+2*zeta*s/wn+s^2/wn^2)
figure(2)
nyquist(Gb)
%Part(c)
zeta=1.5; %asuuming a value for zeta >1
wn=2*pi*10
Gc=1/(1+2*zeta*s/wn+s^2/wn^2)
figure(3)
nyquist(Gc)
%Part(d)
T=3.5 %assuming value for parameter T
Gd=1/(s*(s*T+1))
figure(4)
nyquist(Gd)
2‐11
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
%Part(e)
T=3.5
L=0.5
Ge=pade(exp(-1*s*L),2)/(s*T+1)
figure(5)
hold on;
nyquist(Ge)
notes: In order to use Matlab Nyquist command, parameters needs to be assigned with values, and Pade
approximation needs to be used for exponential term in part (e).
Nyquist diagrams are as follows:
2‐12
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
Part(a)
-5
-4
-3
-2
-1
0
1
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
Nyquist Diagram
Real Axis
Imaginary Axis
Part(b)
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-1.5
-1
-0.5
0
0.5
1
1.5
Nyquist Diagram
Real Axis
Imaginary Axis
2‐13
Automatic Control Systems, 9th Edition
Chapter 2 Solutions
Golnaraghi, Kuo
Part(c)
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
Nyquist Diagram
Real Axis
Imaginary Axis
Part(d)
2‐14
