Automatic Control Systems, 9th Edition - Solutions Manual

947 Pages·2009·26.64 MB·English

by  Farid Golnaraghi, Benjamin C. Kuo

Ninth 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