Modern Control System

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15 Νοε 2013 (πριν από 3 χρόνια και 9 μήνες)

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Modern Control System


05105


Information Technology


SAMPLE QUESTIOS


For



Chapter3

-

state variable Models



Chapter4

-

Feedback Control system characteristic



Chapter5

-

The Performance of Feedback Control Systems



Chapter
6

-

The stability of Linear Feedback Systems



Chapter7

-

The Root Lows Method



Chapter8

-

Frequency Response Method



Chapter9

-

Stability in the frequency Domain



Chapter13
-

Digital Control Systems


















SAMPLE QUESTIO
N
S


Chapter

(3)


1. A s
ingle


loop control system is shown in figure.


(a)Determine closed
-
loop transfer function T(s) =
).
(
)
(
s
R
s
Y
.


(b)Sketch the state model flow graph for the system and determine the matrix differential
equation for the phase variable
format.





<16 marks >










2. A single
-
input, single
-
output system has the matrix equations.


<16 marks >


x
.

=








4
1
3
0
x +



1
0



u


and
, Y=

10


0
x


Determine the transfer function

G(s) =
).
(
)
(
s
U
s
Y




3. An RLC circuits is shown in figure.

(a)

Identify a suitable set of state

variables.

(b)

Obtain the of first
-
order differential equations in teams of the state variables.

(c)

Write the state differential equation.





<16 marks >

V
Voltage
source
R
c
c
v
+
-
i
DC
L









G(s)=3)
)
4
)(
2
(
)
3
)(
1
(
2




s
s
s
s
s


Y(s)

R(s)

+

-



4. A closed
-
loop control system is shown in figure.


(a) Dete
rmine the closed
-
loop transfer function T(s)=
)
(
)
(
s
R
s
Y
.


(b) Sketch the state model flow graph for the system and determine the matrix
differential equation for phase variable format
.

















<16 mark>




Chapter4



5. A robot uses feedback to control the orientation of each joint axis. The load effect
varies due

to varying load objects and extended position of the asm. The system will be
deflected by the load carried
in the gripper. The system may be represented in figure,
where the load Torque is
S
D
. Assume R(s) =0 (a) what is the effect of T
L
(s) on Y(s)
? (b)
Determine the sensitivity of the closed
-
loop to k
2
.


















<16 mark>












6.An automatic speed control system will be necessary for passenger cars traveling on the automatic
highways of the future. A model of a feedback speed control system for a standard vehicle is shown in
figure.

(a) Determine the sensitivity of the syste
m to change in the engine gain ke .

(b) What is the steady

state error when R(s)

=
.
1
s
.




<
16 marks>

controller

)
6
(
)
1
(


s
s

)
1
(
1

s

s
1


+

-

+

R(s)

controlle
r


K1

)
1
(
2

rs
s
k

k
3
+k
4
s

Desired


joint

angle

R(s)
+

-

Load Disturbance


T
l
(s)

Y(s)


Actual


joint angle

Y(s)

position












7.
One small submersible

vehicle has a depth
-
control system as illustrated in figure.

<16 marks >


(a) Determine the cl
osed
-
loop transfer function T(s)=
)
(
)
(
s
R
s
Y
.


(b) Determine the sensitivity
t
k
s
1

and
t
k
s
.

















Chapter 5



8. A unity negative feedback control system has the plant G(s) =
)
2
(
k
s
s
K


.















(a)

Determine the percent overshoot and settling time (using a 2% criterion) due to a
unit step input.

(b)

For what range of k is the settling times less than 1 second.




<16 marks
>



-

V(s)




Spee
d



1
)
(


s
k
s
G
ehicle
Engineandv
e
e


1
hom

t
k
eter
tac

G
1
(s) =
1
1

s
k



Throttlecontro
ller

speed


setting

R(s) +

kg

Load Torque

)
(
s
D


Desired


depth

R(s)

K
2

Y(s)



Actual


depth

k
1


K

-

+

s
1

k

Disturbance

D(s)

-

+

+

-

9. For the syst
em with unity feedback shown in figure. Determine the steady
-
state error
for a step and ramp input when G(s) =
50
14
10
2


s
s
.


















<16 marks>









10. A feedback system is shown in figure



(a) Determine the steady
-
state error for a unit step when K=0.4 and Gp(s)=1
.


(b) Select an appropriate value for Gp(s) so that the steady
-
state error is equal to zero
for the unit step input.











<16 marks>













Chapter 6


11. A system has a characteristic equation s

3
+
9s
2

+263+24=0.


Using the Routh
-
Hurwitz criterion, show that the system is stable.


<16 marks>


12. A control system has the structure shown i
n figure.


Determine the gain at which the system will become stable
.



<16 marks>















R(s)

G(s
)

+


-

+


-

R(s)

)
2
(

s
s
k

)
1
.
0
(
)
3
(


s
s

Gp(s)

Y(s)



Y(s)





1
2

s

)
2
(

s
s
k


-

+

R(s)


+

-

13. Designers have developed small,

fast,

vertical
-
take off fighter aircraft that are invisible
to radar.This aircraft concept uses quickly

turning jet norrles to steer the airplane .The
control system for the heading or direction control is shown in figure.

Determine the
maximum gain of the system for stable operation.












<16 marks>











14.Large welding robots are used in toda
y’s autoplants.The welding head is moved to
different positions on the auto body and rapid accurate response is required .A block
diagram of a welding head positioning system is shown in figure.

Determine the range of
K and

for which the system is stable
.




















<16 marks>











Chapter
-
7




15. Consider a feedback system with a loop transfer function

GH(s) =
)
3
)(
2
(
)
1
(



s
s
s
s
k

(a)Find the asymptotes and draw then in s
-
plane.

(b)Find the angle of depa
rture from the complex poles.

(c)Determine the gain when two roots lie on the imaginary axis

(d)Sketch the root locus.



16. A unity feedback system has a plant G(s) =
)
3
)(
2
(
)
1
(



s
s
s
s
k

(a)Find the asymptotes and draw them in the s
-
plane

(b)Find the a
ngle of departure from the complex poles

(c)Determine the gain when two roots lie on the imaginary axis

(d)Sketch the root locus.





+

R (s)

Y(s)


Heading








2
)
10
(
)
20
(


s
s
s


k
controller

-

Desired

position

Y(s)


Data head
position








)
3
)(
2
(
1


s
s
s


)
1
(
)
(


s
a
s
k

-

+

R (s)

17.

Consider a unity feedback system with G(s) =
5
4
)
1
(
2



s
s
s
k

(a)

Find the angle of departure of the root focus
from the complex poles
.

(b)

Find the entry point for the root locus as it enters the real axis
.



18. A closed
-
loop negative unity feedback system is us
ed to control the yaw of the


A
-
6
in
tru
de attack jet.





G(s)

=
)
2
2
)(
3
(
2



s
s
s
s
k

.

Determine (a)

the root locus breakaway point



(b)

the value of the roots on the jw
-
axis and the gain required for those
roots.

Sketch the root locus.


Chapter8




19. Draw the polar plot for transfer function of an RC Filter
.





)
(
1
s
V
+
-
+
-
C
R


20. Dr
aw the polar plot for the transfer function

G(s) =
2
)
1
(

s
k
, when k=4

Calculate the phase and magnitude at w = 0.5, 1, 2




21. Draw the Bode diagram for the transfer function

G(jw)

=
)
)
50
(
50
6
.
0
1
)(
5
.
0
1
(
)
1
.
0
1
(
5
2
jw
w
j
w
j
jw
w
j





G(jw)

=
)
)
50
(
50
6
.
0
1
)(
2
1
1
(
)
10
1
1
(
5
2
jw
w
j
w
j
jw
w
j







Corner frequency 2,10,50

)
(
jw
G

W<2=
)
(
jw
G
=
dec
db
20


2<w<10
=
)
(
jw
G
=
-
20+(
-
20)+20
=
dec
db
40


10<w<50
=
)
(
jw
G
=
-
20+(
-
20)+20
=
dec
db
20


w>50
=
)
(
jw
G
=
-
20+(
-
20)+20+(
-
40)
=
dec
db
60



Starting point w=0.1
=
)
(
jw
G
=
1
.
0
5
=
20log50=34db


)
(

j
G













W<0.2=
-
90
.

0.2<w<1
=
dec
45


1<w<5
=
dec
.
0

5<w<20
=
dec
.
90


20<w<100
=
dec
.
45



100<q<500
=
dec
.
90



w>500=
-
270
.




22.

Draw the Bode diagram for the transfer function


G(s) =
15434
386
2572
2


s
s



=
)
341
)(
3
.
45
(
2572


s
s

Show that

the magnitude of a (jw) is
-
15.6dB at w=10 and
-
30dB at w=200

Also show that the phase is
-
150
.
at w =
700.







<16 marks>





.
90


dec
.
45


dec
90


1

5

10

50

100

500

0.2

dec
.
45


2

20

Chapter
-
9


23.

A transfer function is

GH(s)

=
2
1

s
.













Using the contour in t
he s
-
plane .Determine the corresponding contou
r

in the F(s)
-
plane.











<16 marks>


24. A single loop control system









Where GH(s)

=
)
1
(

Ys
s
k

(a)Draw the Nyquist contour and mapping for GH(s)

(b)Determine whether

the system is stable by utilizing the Nyquist criterion.




Chapter13




25.

Determine the z transform of f(t)=sin

wt for t

0





Sin

wt

=
j
e
e
jwt
jwt
2









<16 marks>




26. Determine the
z
-
transform of F(t)

=
at
e


and t

0.









<8 marks>











A

B

C

D

E

F

G

H

0


j



-
1

1

0



Chapter
-
3


No.1 A single
-
loop control system is shown in figure.

(a)

Determine closed
-
loop transfer function

)
(
s
T
)
(
)
(
s
R
s
Y
.

(b)


Sketch the stable model flo
w graph for the system and determine the matrix
differential equation for the phase variable format.









Solution


(a) Transfer function
, T(s)

=

GH
G

1








=

1
*
)
4
)(
2
(
)
3
)(
1
(
2
1
)
4
)(
2
(
)
3
)(
1
(
2









s
s
s
s
s
s
s
s
s
s








=
)
4
)(
2
(
)
3
)(
1
(
2
)
4
)(
2
(
)
4
)(
3
(
)
3
)(
1
(
2











s
s
S
s
s
s
s
s
s
s
s
s
s









=
)
3
)(
2
2
(
)
4
)(
25
(
)
3
)(
1
(
2
2







s
s
s
s
s
s






=
6
2
6
2
8
2
4
6
2
6
2
2
2
2
3
2










s
s
s
s
s
s
s
s
s
s





T(s
)

=

6
16
8
6
8
2
2
3
2





s
s
s
s
s

(b)

T(s)

=
6
16
8
6
8
2
2
3
2





s
s
s
s
s


=
)
6
16
8
(
1
6
8
2
3
2
1
3
2
1












s
s
s
s
s
s



=
)
(
1
3
2
1
3
3
2
2
1
1
L
L
L
P
P
P












G(s)=
)
4
)(
2
(
)
3
)(
1
(
2




s
s
s
s
s

+

-

R(s)

Y(s)















x

=










3
2
1
x
x
x



,x=





1
1
1
x
x
x




2
2
2
x
x
x









3
3
3
x
x
x





A

=







6
0
0

18
0
1








8
1
0
,

B

=











1
0
0


Y

=

cx


=c










3
2
1
x
x
x
=

6

8

2










3
2
1
x
x
x


Bu
Ax
x






=





6
0
0

18
0
1








8
1
0
x+










1
0
0
u


No.2 A single
-
input, single
-
output system has the matrix equation







3
0
x






4
1
x+






1
0
u


and y

=

10


0
x


Determine the transfer function G(s)

=Y(s)/U(s).



Solution





x


Ax+Bu


A=




3
0





4
1

,

B

=






1
0


Y=

cx


c=

10


0


sx(s)

=



A
sI


=

BU(s)




Y(s)

=

cx(s)

R(s
)

Y(s
)

1

1

1

1

6

s
1

2
x


3
x


3
x

2
x

1
x


1
x

s
1

s
1


SI
-
A=



0
s




s
0
-




3
0





4
1
=



3
s






4
1
s




)
(
1
s
A
SI





s
(

)=






3
4
)
(
1
s
s




s
1
=










)
(
3
)
(
4
s
s
s









)
(
)
(
1
s
s
s

)
(
s

=
3
4
2


s
s


Y(s)=c

(s)BU(s)



10
)
(
)
(

s
U
s
Y


0










)
(
3
)
(
4
s
s
s









)
(
)
(
1
s
s
s






1
0



=
0
)
(
)
4
(
10






s
s






0
)
(
10
s






1
0
=
)
(
10
s

=
3
4
10
2


s
s



No.3 An RLC circuit is shown in figure.

(a)

Identify a suitable set of state variables.

(b)

Obtain the set of first
-
order differential equations in terms of the state
variables.

(c)

Write the state differential equation.



DC



Solution


(a)

i
=
)
(
1
t
x

,
)
(
2
t
x
v
c


(b)

v(t)

=
)
(
)
(
2
1
t
x
dt
t
dx
L
R
i



1
)
(
1
2
1
)
(
1
)
(
L
R
-
v(t)
1
x
dt
dx
t
x
L
t
x
L
t





1



dt
t
dx
c
t
x
dt
dv
c
i
c
c
)
(
)
(
2
1




Voltage


source


V

i



c
v

c

R

L




)
(
1
1
t
x
c
dt
t
dx
)
(
2
=x
2
x



2


(c)

st
ate differential equation












2
1
x
x
x
Bu
Ax
x







x=



1
1
x
x




2
2
x
x


A=







c
L
R
1





0
1
L

,

B

=








0
1
L





No.4 A closed
-
loop control

system is shown in figure
.

(a)

Determine the closed
-
loop transfer function


)
(
s
T
)
(
)
(
s
R
s
Y
.

(b)

Sketch the state model flow graph for the system and determine the matrix
differential equation for phase variable format.














Solution


(a)

)
(
s
T
)
(
)
(
s
R
s
V
=
GH
G

1





=
1
*
)
1
)(
6
(
1
1
)
1
)(
6
(







s
s
s
s
s
s
s
a
s





=
)
1
(
)
1
)(
6
(
1





s
s
s
s







1
6
5
1
2
3






s
s
s
s
s


)
6
(
)
1
(


s
s
controller

)
1
(
1

s

s
1

Voltage

)
(
1
s
v

Velocity
)
(
s
v

+

-

Y(s)

position





=
1
5
5
1
2
3




s
s
S
S



(b) T(s) =
1
5
5
1
2
3




s
s
S
S




=
)
5
5
(
1
3
2
1
3
2









s
s
s
s
s




=
)
(
1
3
2
1
2
2
1
1
L
L
L
P
P






















matrix differential equation



Bu
Ax
x

















3
2
1
x
x
x
x





,x=





1
1
1
x
x
x

2
2
2
x
x
x






3
3
3
x
x
x

,

u

=


)
(
s
R



A=






1
0
0

5
0
1








5
1
0

,

B

=










1
0
0






Bu
Ax
x







x







1
0
0

5
0
1








5
1
0
x+










1
0
0
u



y=cx




=c





1
1
1
x
x
x

2
2
2
x
x
x






3
3
3
x
x
x
=

1

1



0










3
2
1
x
x
x




Y(s
)

R(s
)

1

1

1

1

1

s
1

2
x


3
x


3
x

2
x

1
x


1
x

s
1

s
1

Chapter 4


5. A robot
uses feedback to control the orientation of each joint axis. The load effect varies
due to varying load objects and extended position of the arm. The system will be deflected
by the load carried in the gripper. The system may be represented in figure, wher
e the load
Torque is
S
D
. Assume R(s)

=

0. (a) What is the effect of
)
(
s
T
l
on Y(s)?

(b) Determine the sensitivity of the closed loop to
2
K
.















(a)

Load Disturbance
s
D
s
T
l

)
(












Y(s)

=Transfer*Input

Y(s)

=T(s)*Input

T(s)

=
)
(
*
)
1
(
1
)
1
(
4
1
3
1
2
2
s
K
K
K
K
s
s
K
s
s
K








=

s
K
K
K
K
K
K
s
s
K
4
2
1
3
2
1
2
)
1
(





Y(s)

=

s
K
K
K
K
K
K
s
s
K
4
2
1
3
2
1
2
)
1
(




*(
-
s
D
)


(b)

Closed
-
loop T(s)

0
)
(

s
T
l

1
k
controller

)
1
(

s
s
k


Load
Disturbanc
e

)
(
s
T
L

+

-

Y(s
)

s
k
k
4
3


R(s)

Desired

joint angle

Angle

Joint angle

R(s)

=0

)
1
(

s
s
k


1
k

s
k
k
4
3


Y(s)

-

-

)
(
s
T
L









T
K
K
T
K
K
T
T
s
T
K
2
2
2
2
*
2







T(s)=
)
(
*
)
1
(
1
)
1
(
4
3
2
1
2
1
s
K
K
s
s
K
K
s
s
K
K








=
)
)
1
(
(
4
2
1
3
2
1
2
1
s
K
K
K
K
K
K
s
s
K
K





2
K
T


=


2
4
2
1
3
2
1
4
1
3
1
2
1
1
4
2
1
3
2
1
)
)
1
(
(
)
(
*
)
1
(
s
K
K
K
K
K
K
s
s
s
K
K
K
K
K
K
K
s
K
K
K
K
K
K
s
s












=
2
4
2
1
3
2
1
1
)
)
1
(
(
)
1
(
s
K
K
K
K
K
K
s
s
s
s
K







T
K
s
2
=
2
1
2
4
2
1
3
2
1
2
2
4
2
1
3
2
1
1
)
)
1
(
(
*
)
)
1
(
(
)
1
(
K
K
s
K
K
K
K
K
K
s
s
K
s
K
K
K
K
K
K
s
s
s
s
K












=
s
K
K
K
K
K
K
s
s
s
s
4
2
1
3
2
1
)
1
(
)
1
(








6. A

automatic speed control system will be necessary for passenger cars traveling on the
automatic highways of the future. A model of a feedback speed control system for a
standard ve
hicle is shown in figure. (a) Determine the sensitivity of the system to cha
nge
in the engine gain
e
K
. (b) What is the steady
-
state error when R(s)

=
s
1

and
.
0
)
(


s
D
















)
1
(

s
s
k


k

s
k
k
4
3


Y(s)

-

+

)
(
s
R

speed

+

-

V(s)

1
hom

t
k
eter
Tac

R(s)

Speed

setting

Load Torque

)
(
s
D



Throt t le cont roller


1
)
(
1
1
1


s
k
s
G


Engine and vehic le

1
)
(


s
e
k
s
G


kg

(a)











T(s

)=
1
*
)
1
)(
1
(
1
)
1
)(
1
(
1
1
1
1





s
s
K
K
s
s
K
K
e
e
e
e






=

e
e
e
K
K
s
s
K
K
1
1
1
)
1
)(
1
(






e
K
T


=




2
1
1
1
1
1
1
1
)
1
)(
1
(
*
)
1
)(
1
(
e
e
e
e
e
K
K
s
s
K
K
K
K
K
K
s
s













=




2
1
1
1
1
)
1
)(
1
(
)
1
)(
1
(
e
e
e
K
K
s
s
K
s
s











T
K
e
s
=
e
K
T


*
T
K
e


=




2
1
1
1
1
)
1
)(
1
(
)
1
)(
1
(
e
e
e
K
K
s
s
K
s
s









*


e
e
e
e
K
K
K
K
K
s
s
1
1
1
)
1
)(
1
(







=
e
e
e
K
K
s
s
s
s
1
1
1
)
1
)(
1
(
)
1
)(
1
(













(b)

E(s)=Input
-
Outpu
t


=R(s)
-
V(s)



=R(s)
-
T(s)R(s)



=R(s)
-


)
(
1
s
T



)
(
0
s
sE
Lim
e
s
ss




=
s
Lim
s
0

*


)
(
1
1
s
T
s



=


)
0
(
1
T



=1
-
e
e
K
K
K
K
1
1
1



=
e
e
e
K
K
K
K
K
K
1
1
1
1
1




1
)
(
1
1


s
k
s
G


1

t
k

1
)
(


s
k
s
G
e
e


V(s)

-

+

)
(
s
R


e
ss
K
K
e
1
1
1





7.

One small submersible vehicle has a depth
-
control system as illustrated in figure.


(a) Determine the closed
-
loop transfer function T(s)=
)
(
)
(
s
R
s
Y
.


(b) Determine the sensitivity
T
K
s
1
and
T
K
s
.














(a)

s
K
K
s
K
s
T
2
1
1
1
1
)
(





=
2
1
1
K
K
s
K











K
K
s
T
T
s
T
*
*
)
(
1
)
(
1
1




=
2
1
1
1
1
1
K
K
s
KK
KK
s
KK





=
1
2
1
1
KK
K
K
s
KK




Y(s)

Actual
depth

+

-

k

Disturbance
D(s)

)
(
s
D


1
k

2
k

s
1

k

-

E(s
)

+

+

-

R(s)

Desire
d

depth

)
(
s
R

Y(s)

-

+

)
(
1
s
T

k

k

(b)
T
K
s
1
=
1
K
T


*
T
K
1


1
K
T



=
2
1
2
1
2
1
1
2
1
)
(
)
(
)
(
KK
K
K
s
K
K
KK
K
KK
K
K
s








=
2
1
2
1
1
2
2
1
1
2
2
1
)
(
KK
K
K
s
K
K
K
KK
K
K
K
KK
Ks








=
2
1
2
1
)
(
KK
K
K
s
Ks




T
K
s
1
=
2
1
2
1
)
(
KK
K
K
s
Ks



*
1
1
2
1
1
)
(
KK
KK
K
K
s
K




=
1
2
1
KK
K
K
s
s






T
K
s
=
K
T


*
T
K


K
T



=
2
1
2
1
2
1
1
1
2
1
)
(
)
(
KK
K
K
s
KK
K
KK
K
K
s









=
2
1
2
1
2
1
1
)
(
)
(
KK
K
K
s
K
K
s
K





T
K
s

=
2
1
2
1
2
1
1
)
(
)
(
KK
K
K
s
K
K
s
K




*
1
1
2
1
)
(
KK
KK
K
K
s
K





=
)
(
1
2
1
2
1
KK
K
K
s
K
K
s







8. A unity negative feedback contro
l system has the plant G(s) =
.
2
(
k
s
s
k



(a) Determine the percent overshoot and setting time (using a 2% setting criterion) due
to a unit step input.


(b) For what range of k is the setting time less than 1 second?








(a)
T(s) =
GH
G

1



+

Output

-

)
(
s
G

Input

Unity feedback

1


=
1
*
)
2
(
1
)
2
(
k
s
s
k
k
s
s
k






=
k
k
s
s
k


)
2
(


=
k
s
k
s
k


2
2


S
tandard notation =
2
2
2
2
n
n
n
w
s
zw
s
w





2
n
w
=k



k
w
n




2z
k
w
n
2



2z
k
k
2




z =
k
k
2
2

=
2
2



P.O = 100

2
1
z
z
e







= 100
4
2
1
2
2



e




= 100
2
2
2
2


e





= 100


e




= 4.3 %



n
s
zw
T
4





=
k
*
2
2
4





=
k
2
2
*
4





=
k
2
8

(b)
1

s
T




1
2
8

k




8

k
2




64

2k





k

32




Range of k

32


9. For the system with unity feedback shown in fig, determine the steady
-
state error for a
step and a ramp input when
G(s)

=
50
14
10
2


s
s
.











T(s) =
GH
G

1


=
50
14
10
1
50
14
10
2
2





S
S
S
S


=
10
50
14
10
2



S
S


=
60
14
10
2


S
S




Chapter


5



For step input
s
A

,

E(s) = Input


Output





= Input


T(s)* Input





= Input
-



)
(
1
s
T





)
(
0
s
sE
Lim
e
s
ss







=
0

s
Lim
s*
s
A


)
(
1
s
T





= A


)
0
(
1
T






=
A
6
5
= 0.8333A


For ramp input
2
s
A



E(s) =Input (1
-
T(s))



E(s) =

2
s
A
(1
-
T(s))



)
(
0
s
sE
Lim
e
s
ss



Output

-

50
14
10
2


s
s

Input

1

+





)
(
1
*
2
0
s
T
s
A
s
Lim
s








=
6
5
*
0
A




=




10. A feedback system is shown in Fig.


(a) Determine the steady
-
state error for a unit step when k=0.4 and Gp(s)=1.


(b) Select an appropriate value for Gp(s) so that the steady
-
state error is equal to zero



for the unit step input.










(a)

k=0.4 , Gp(s)=1







T(s) =
GH
G

1

=
1
.
0
3
*
)
2
(
1
)
2
(





s
s
s
s
k
s
s
k







=
k
ks
s
s
s
s
k
3
)
1
.
0
)(
2
(
)
1
.
0
(









=
k
ks
s
s
s
k
ks
3
2
.
0
1
.
2
1
.
0
2
2








=
2
.
1
6
.
0
1
.
2
04
.
0
4
.
0
2
2



s
s
s
s


E(s) = R(s
)


Y(s)




= R(s)
-

R(s) Gp(s) T(s)




= R(s)


)
(
)
(
1
s
T
s
Gp




= R(s)


)
(
1
s
T


)
(
0
s
sE
Lim
e
s
ss






=
0

s
Lim

s*


)
(
1
1
s
T
s







= 1
-
T(0)





= 1
-

0.
033





= 0.96

Y(s)

-

+

)
2
(

s
s
k

)
1
.
0
(
)
3
(


s
s

Gp(s)

Gp(s)

T(s)

Y(s)

R(s)



(b)
0

ss
e


E(s)

= R(s)


)
(
)
(
1
s
T
s
Gp





=


)
(
)
(
1
1
s
T
s
Gp
s




)
(
0
s
sE
Lim
e
s
ss





0 =
0

s
Lim
s*


)
(
)
(
1
1
s
T
s
Gp
s





0 =



)
(
)
(
1
s
T
s
Gp




0 = 1
-
Gp(s) 0.033



Gp(s)0.033


= 1



Gp(s)

=
033
.
0
1



Gp(s)

= 30



Chapter
-
6



11. A system has a characteristic equation
.
0
24
26
9
2
3




s
s
s

Using the Routh
-
Hurwitz criteri
on, show that the system is stable.


Solution


q(s)


=
24
26
9
2
3



s
s
s


Using the Routh
-
Hurwitz
criterion,






0
1
2
3
s
s
s
s
24
26
9
1

0
0
24
26


No sign change in
st
1

column 30 the sys
tem is stable.


12. A control system has the structure shown in figure. Determine the gain at which the

system will become stable.













+

Y(s)

-

1
2

s

)
2
(

s
s
k

R(s)

+

Solution










)
1
(
1
2



s



=
1
1
1
1
2






s
s
s
s









G(s) =
s
s
s
Ks
K
s
s
s
Ks
K
2
3
)
2
)(
1
(
2
3








Characteristic equation,

1+GH = 0

1+
0
2
3
2
3




s
s
s
Ks
K

0
)
2
(
3
2
3





k
s
k
s
s




0
1
2
3
s
s
s
s
b
a
3
1

0
0
2
k
k



a =
3
4
6
3
3
6
3
)
2
(
3
k
k
k
k
k








b =
k
a
k
a


0
*


For stable system, no sign cha
nge in
st
1

column,



0

a
and
0

b



0
3
4
6


k

and
0

k



6
-
4k
0


and
0

k



6
k
4


and
0

k



k
5
.
1


and
0

k



Range of k, 0
5
.
1


k


+

-

1
2

s

1

+

-

1
1


s
s

-

)
2
(

s
s
k

Y(s)

R(s)

13. Designers have developed small, fast, vertical
-
take off fighter aircraft that are invisible
to radar. This

aircraft concept uses quickly turning jet nozzles to steer the airplane. The
control system for the heading or direction control is shown in figure.

Determine the maximum gain of the system for stable operation.










Solution


G(s) =
2
)
10
(
)
20
(


s
s
s
k



=
)
100
20
(
20
2



s
s
s
k
ks



=
s
s
s
k
ks
100
20
20
2
3





Characteristic equation,



1+GH = 0



1+
0
1
*
100
20
20
2
3




s
s
s
k
ks



s
s
s
100
20
2
3


+ks+20k

=0



0
20
)
100
(
20
2
3




k
s
k
s
s

Routh
-
Hurwitz,






0
1
2
3
s
s
s
s
b
a
20
1

0
0
20
100
k
k






a =
20
20
)
100
(
20
k
k


=
20
20
20
100
*
20
k
k








= 100



b =
a
k
a
0
20
*





=20k


The system is stable, no sign change in
st
1

column,



b
0




20k
0




k
0


Range of k, k
0

.



+

-

k
controller

-

2
)
10
(
)
20
(


s
s
s

Y(s)

R(s)

Headin
g

14. Large welding robots are used in today’s autoplants. The welding heo is moved to
different positions on the auto body and rapid accum r
esponse is required. A block
diagram of a welding head positioning system is shown in figure. Determine the range of k
and a for which the system is stable.





Solution;



G(s) =
)
3
)(
2
)(
1
(
)
(




s
s
s
s
a
s
k



=
)
6
2
3
)(
(
2
2





s
s
s
s
s
ka
ks




=
s
s
s
s
ka
ks
6
11
6
2
3
4






Characteristic equation,



1+GH(s) = 0



1+
0
1
*
6
11
6
2
3
4





s
s
s
s
ka
ks


0
6
11
6
2
3
4






ka
ks
s
s
s
s



Using Routh Hurwitz,






0
1
2
3
4
s
s
s
s
s
a
k
c
b
6
1

0
0
6
11
a
k
k


0
0
0
0
a
k





b=
6
60
6
6
66
6
)
6
(
66
k
k
k










c =
b
k
k
b
a
6
)
6
(



The system is stable, no sign change in
st
1

column.


0

b

and
0

c


6
60
k

0


and

b
ka
k
b
6
)
6
(


0



60
-
k
0


and b(k+6)
-
6ka
0



60
k


and b(k+6)
kia
6



k
60


and b
6
6


k
ka


k
60


and
6
6
6
60



k
ka
k

Data head

position

+

-

1
)
(


s
a
s
k

-

)
3
)(
2
(
1


s
s
s

Y(s)

R(s)

Desired

position


k
60


and (60
-
k)(k+6)
ka
36



k
60


and
a
k
k
k



36
)
6
)(
60
(

if k=40, a
639
.
0





Chapter
-
7



15.

Consider a feedback system with a loo
p transfer function GH(s) =
)
3
)(
2
(
)
1
(



s
s
s
s
k


(a) Find the asymptotes and draw them in s
-
plane.


(b) Find the root lows breakaway point and the gain k for this point.


GH(s) =
)
3
)(
2
(
)
1
(



s
s
s
s
k

1+kp(s) = 0

(a)

1+
)
3
)(
2
(
)
1
(



s
s
s
s
k
= 0


















Asymptote center,
z
p
zero
pole
A
n
n












=
1
3
)
1
(
)
3
(
)
2
(
0












=
2
4
2
1
5




=
-
2

Asymptote angle
,....
2
,
1
,
0
,
180
*
1
2




q
n
n
q
z
p
A






=
180
*
1
3
1
0
*
2






j



k



k

0

k

2


1


3


0

k



k

2


A


Breakaway=
-
2.5

pt





=
180
*
2
1
=
0
9



(b)

Breakaway point,

0
)
3
)(
2
(
)
1
(
1





s
s
s
s
k

1
6
5
)
1
(
2
3





s
s
s
s
k



k =
)
1
(
)
6
5
(
2
3




s
s
s
s



0
)
1
(
1
*
)
6
5
(
)
6
10
3
)(
1
(
2
2
3
2








s
s
s
s
s
s
s
ds
dk






0
6
5
6
10
3
6
10
3
2
3
2
2
3







s
s
s
s
s
s
s
s







0
6
10
8
2
2
3




s
s
s











s =
-
2.5,
-
0.76
0





breakaway point

s =
-
2.5


breakaway point
maz
k















Routh
-
Hurwitz



0
6
5
)
1
(
1
2
3





s
s
s
s
k



s
s
s
6
5
2
3


+ks+1=0



0
1
2
3
s
s
s
s
k
k
5
5
30
5
1


0
0
1
6
k



0
5
30
0
5
5
30




k
k



5k =
-
30



k =
-
6


U(s) = 5s
2
+k


= 5s
6
2



=
1
.
1



s

-
2.1

-
2.2


k

0.12

0.29

0.37


0.4
1

-
2.3

0.48

0.39

-
2.4

-
2.5

-
2.6


16. A uni
ty feedback system has a plant
)
5
2
(
2



s
s
s
k
G
.


(a) Find the asymptotes and draw them in the s
-
plane.


(b) Find the angle of departure from the complex poles.


(c) Determine the gain when two roots lie on the imaginary axis.


(d) S
ketch the root focus.


Solution


0
)
2
1
)(
2
1
(
1
0
)
5
2
(
1
0
)
(
1
)
5
2
(
)
(
1
)
(
)
5
2
(
)
(
2
2
2



















j
s
j
s
s
k
s
s
s
k
s
kp
s
s
s
k
s
GH
s
H
s
s
s
k
s
G




















(a)

Asymptote center
0
3
)
2
1
(
)
2
1
(
0













j
j
n
n
z
p
zero
pole
A



=
67
.
0
3
2
3
2
1
2
1







j
j

1


A



j



k

0

k

2


1


37
.
0

A


-
1+2j

-
2j

+
2j

-
j

+
j

j
5


j
5




Asymptote angle
,...
2
,
1
,
0
9
,
0
18
*
1
2





z
p
A
n
n
q








=
0
6
0
18
*
3
1
0
18
*
0
3
1
0
*
2























(b)

Departure angle


5
4
5
25
0
18
0
18
)
0
9
5
13
(
0
36
0
18
)
(
0
36
0
18
2
1


























d
d
d
d
pole
zero








(c)

Routh
-
Hurwrtz

0
5
2
0
5
2
1
2
3
2
3







s
s
s
s
s
s
k

0
1
2
3
s
s
s
s
k
k
2
10
2
1


0
0
5
k


k
s
s
U
k
k
k







2
2
)
(
10
0
10
0
2
10

1



j



k


-
1+2j

-
2j

+
2j



d


0
9
2




5
13
1




-
1
-
2j

A




=
18
.
32
23
.
2
)
5
2
(
0
5
2
1
0
10
5
2
10
)
5
)(
5
(
2
)
5
(
2
10
2
2
3
2
3
2
3
2
2





















k
s
s
s
s
s
k
s
s
s
k
s
s
s
k
s
s
s
s
s
s


17. Consider the unity feed
back system with
5
4
)
1
(
)
(
2




s
s
s
k
s
G


(a) Find the angle of departure of the root locus from the complex poles


(b) Find the entry point for the root locus as it enters the real axis.



Solution


5
4
)
1
(
)
(
2




s
s
s
k
s
G

,H(s)

=1


5
4
)
1
(
)
(
2




s
s
s
k
s
G


1+kp(s)

=0


0
)
2
)(
2
(
)
1
(
1
0
5
4
)
1
(
1
2












j
s
j
s
s
k
s
s
s
k


















0

k

A


-
1+2j

-
1
-
2j

-
1

-
2

-
3

-
4



k

0

k

0



j

Asymptote center
3
1
2
)
1
(
2
2















j
j
n
n
z
p
zeros
poles
A


Asympto
te angle
0
18
0
18
*
1
2
1
0
*
2
,...
2
,
1
,
0
,
0
18
*
1
2











A
z
p
A
q
n
n
q



Departure angle


0
18
)
0
9
(
5
13
0
36
0
18
)
(
0
36
0
18
2
1



















d
d
pole
zero








5
22
5
22






d
d




Entry point,





0
2
0
5
4
4
6
2
)
1
(
)
1
)(
5
4
(
)
4
2
)(
1
(
0
)
1
(
)
5
4
(
0
5
4
)
1
(
1
0
2
2
2
2
2
2
2





























I
s
s
s
s
s
s
s
s
s
s
s
ds
dk
ds
dk
s
s
s
k
s
s
s
k
ds
dk


s=
4
.
2
,
4
.
0






Entry
min
k














Entry point s=
-
2.4




82
.
0
min

k


k

-
2.1

0.92

s

0.87

-
2.2

-
2.3

0.84

-
2.4

0.82

-
2.5

0.83

18. A closed
-
loop negative unity feedback system is used to control the
yaw
of the A
-
6
Intruder attack jet.
)
2
2
)(
3
(
)
(
2




s
s
s
s
k
s
G


Determine (a) the root locus breakaway point

(b)

the value of the roots on the j
-
axis and the gain required for those roots.

Sketch the root locus.


Solution


H(s)

=1


0
)
1
)(
1
)(
3
(
1
0
)
(
1
)
2
2
)(
3
(
)
(
2













j
s
j
s
s
s
k
s
kp
s
s
s
s
k
s
GH


















Asymptote center
z
p
zeros
poles
A
n
n






=
4
)
1
(
)
1
(
)
3
(
0
j
j















=
25
.
1
4
5




Asymptote angle
,...
2
,
1
,
0
,
0
18
*
1
2




q
n
n
q
z
p
A






=
5
4
0
18
4
1
0
18
*
4
1
0
*
2






Breakaway point s =
-
229




j

1


A




k

0

k

2


1


25
.
1


A


-
1+j

-
2j

+
2j

-
j

j

3


-
1
-
j

0

k


























Departure angle


0
18
0
9
0
9
5
13
0
36
0
18
0
36
0
18
3
2
1






















d
d
pole
zero









d


=
5
7




(c)

Routh
-
Hurwitz

0
6
8
5
0
6
8
5
1
2
3
4
2
3
4








k
s
s
s
s
s
s
s
s
k




s
4

1

8

K

s
3

5

6

0

s
2

6.8

K

0

s
1

8
.
6
5
8
.
40
K


0

0

s
0

K

0

0



k

-
2.1

4.18

s

4.29

-
2.2

-
2.3

4.33

-
2.4

4.26

-
2.5

4.1

1



j



k






d


0
9
2




5
13
1




A


-
3

30
3







8
.
6
5
8
.
40
K


= 0




40.8


5K = 0






40.8 = 5K





K = 8.16






U(s)

=

6.8s
2

+ K





=

6.8s
2
+ 8.16





=

6.8 ( s
2

+ 1.2 )





=

6.8 ( s + 1.1j ) ( s


1.1j )




K

=

)
6
8
5
(
2
3
4
s
s
s
s








s

=

j
1
.
1





K

=

8.22


19
-
Draw the polar plot for transfer function of an RC filter.












Solu:

Transfer function G(s)




G(s) =
input
output

=
)
(
)
(
1
2
s
V
s
V





)
(
2
s
V


=
cs
1

* I(s)




)
(
1
s
V

= R I(s) +
cs
1

I(s)





= I(s)







cs
R
1






= I(s)







cs
Rcs
1




G(s)

=
)
(
)
(
1
2
s
V
s
V

=








cs
Rcs
s
I
s
I
cs
1
*
)
(
)
(
*
1

=
1
1

Rcs




G(jw)

=

)
1
(
1

jw
R

R
+
+
_
_
C
)
(
2
s
V
)
(
1
s
V




=

1
1
1

Rc
jw




Rc
1



constant





w


frequency




Rc
1

=
1
w




G(jw)

=
1
1
1

w
jw





=
)
1
(
1
1

w
jw

*
)
1
(
)
1
(
1
1


w
jw
w
jw





=
1
)
(
1
2
1
2
1


w
w
j
w
jw





=
1
)
)(
1
(
1
2
1
1



w
w
w
jw

=











1
)
(
)
1
(
)
1
)(
1
(
2
1
1
w
w
w
jw






G(
jw)

= R(w) + j X(w)





=







1
)
(
1
2
1
w
w

+ j








1
)
(
)
(
2
1
1
w
w
w
w





w = 0


R(w) = 1 , X(w) = 0



w =




R(w) =

1

= 0 , X(w) =
-
2



=
-

1

= 0



w =
1
w


R(w) =
2
1

, X(w) =
-
2
1




)
(
w
G

=




2
2
)
(
)
(
w
X
w
R





w = 0


)
(
w
G

=
2
2
0
1


= 1



w =




)
(
w
G

=
2
2
0
0


= 0



w =
1
w


)
(
w
G


=
2
2
)
2
1
(
)
2
1
(








=
)
4
1
(
)
4
1
(









=
4
2






=
2
1




G(jw) =
1
1
1

w
w
j







)
(
w




=

1
1
1



w
w
j










=
1
1
1



w
w
j








=
1
1
tan
0
w
w











=
0


-

1
tan
1
1



w
w








=
1
1
tan



w
w




)
(
w



=
1
1
tan



w
w



w = 0


)
(
w


=
1
tan


0 = 0



w =




)
(
w


=
1
tan




=
-
90
˙




w =
1
w


)
(
w


=
1
tan


1


=
-
45
˙



























R(w)

X(w)



Real
1
w
w

1

Im

























19.


Draw the polar plot for the transfer function

G(s)

=

2
)
1
(

s
K

, when K = 4




Calculate the phase and magnitude at w = 0.5, 1, 2



Solu:


G(s)

=

2
)
1
(

s
K




G(jw)

=

2
)
1
(

jw
K





=

1
2
2
2


jw
w
j
K




=

1
2
2



jw
w
K





=

jw
w
K
2
)
1
(
2







=

jw
w
K
2
)
1
(
2



*
jw
w
jw
w
2
)
1
(
2
)
1
(
2
2









=

2
2
2
2
2
2
)
1
(
2
)
(
w
j
w
Kw
j
Kw
K









=

2
2
2
2
4
)
1
(
2
)
(
w
w
Kw
j
Kw
K





w=


(0,0)

-
45
˙

2
1

-

˙

2
1


w㴰

(ㄬ〩

)
2
1
,
2
1
(





G(jw)

=

R(w) + jX(w)





G(jw)

=

2
2
2
2
4
)
1
(
2
)
(
w
w
Kw
j
Kw
K





+ j
2
2
2
4
)
1
(
)
2
(
w
w
Kw







w = 0 , R(w) = 4


X(w) =
1
0

= 0



w = 0.5, R(w) = 1
.92


X(w) =
-
2.56



w = 1 , R(w) = 0


X(w) =
-
2



w = 2 , R(w) =
-
0.48

X(w) =
-
0.64



w =


, R(w) = 0


X(w) = 0



w = 0 ,
)
(
jw
G

=




2
2
)
(
)
(
w
X
w
R


=
2
2
0
4


= 4



w = 0.5 ,
)
(
jw
G

=
2
2
)
56
.
2
(
)
92
.
1
(



= 3.2



w = 1 ,
)
(
jw
G

=
2
2
)
2
(
0



= 2



w = 2 ,
)
(
jw
G

=
2
2
)
64
.
0
(
)
48
.
0
(




= 0.
8



w =


,
)
(
jw
G

=
2
2
0
0


= 0







G(jw)

=

2
)
1
(
4

jw




)
(
w


=
)
1
(
)
1
(
4





jw
jw





G(jw)

=
w
w
1
1
tan
tan
0











=
w
w
1
1
tan
tan
0












=

-
2
w
1
tan




w = 0 ,
)
(
w


=
-
2
0
tan
1


= 0



w = 0.5,
)
(
w


=
-
2
5
.
0
tan
1


=
-
53.13



w = 1 ,
)
(
w


=

-
2
1
tan
1


=
-
90
˙



w = 2 ,
)
(
w


=
-
2
2
tan
1


=
-
127
˙



w =


,
)
(
w


=
-
2


1
tan

=

-
2 * 90
˙

=
-
180
˙
































20.


Draw the Bode diagram for the transfer function

G(s)

=

15434
3865
2572
2


s


=

)
341
)(
3
.
45
(
2572


s
s




Show that the magnitude of a (jw) is
-
15.6dB at w=10 and
-
30dB at w=200.


Also show that the phase is
-
150
˙

at w=70
0.


Solu:



G(s)

=

)
341
)(
3
.
45
(
2572


s
s




G(jw)

=

)
341
)(
3
.
45
(
2572


jw
jw





=

)
1
341
)(
1
3
.
45
(
341
*
3
.
45
2572


jw
jw





=

)
1
341
)(
1
3
.
45
(
17
.
0


jw
jw




Corner frequency
c
w

= 45.3, 341


w < 45 =
dec
dB
0



w < w <

340 =
dec
dB
20


-
127
˙

X(w)

-
180
˙




(〬〩

(
-
0.48,
-
0.64)


w=2

(0,2)

w=2

127
˙

53.13
˙

-
90
˙

(1.92,
-
2.56)


w=0.5

w=0

(4,0)

R(w)



w > 340 =
dec
dB
40




w = 1


)
(
jw
GH

=
2
2
2
2
)
341
1
(
1
*
)
3
.
45
1
(
1
17
.
0









= 0.17







=


20 log 0.17







=

-
15.4dB




Corner frequency
c
w

= 45.3, 341
















w < 4.5

= 0



4.5 < w < 34


=
dec
5
4





34 < w < 450 =
dec
0
9






450 < w < 3410 =
dec
5
4






w < 3410

=
-
180
˙





Chapter
-
9


21.

A transfer function is



GH(S) =
__1__


S+2




1+L(S) =F(S)




1 + _
__1


=F(S)


S+2




S
+2+1

= F(S)


S+2



4.5

34

45

341

450

3400

dec
5
4



dec
5
4




F(S)
= S+3 __


S+2





Point A




Point B



Point
C____



S=б+jω

-
1+j0

-
1+j1

0+j1




GH(S) = υ+j v


2

+1.5
-
0.5j

+1.4
-
0.2j

__________
________________________________
______
_________






Point D Point E

Point F Point G


Point



S
=б+jω


+1+j1


1+j0

+1
-
j1

0
-
j1

-
1
-
j




GH(S) = υ+j v


1.3
-
0.1j

1.3+j0

1.3+0.1j

1.4+0.2j


1.5+1

______________________________________________________




















23.

A single
-
loop control system

where H(S) =K S
(
S+1

)


(a)Draw
the Nyquist contour and mapping for GH(s).


(b) Determine whether the system is stable by utilixing the Nyquist Cripesion.



GH(s) =
)
1
(

s
s
k



jv

u

1
j


1
j


0

1


(a)

The origin of the S
-
plane



s
=E e



0
lim

E
GH(s) =
0
lim

E


i
Ee
k



=
0
lim

E


i
e
E
k

)
(


-
90<Ф<+90



Ф=90
ׁ

at
ω=0
+
GH(s)

=
-
90



Ф=
-
90
ׁ

at
ω=0
-

GH
(s
)

=
-
+90


Magni
tude of GH(s) is
infinite.


(b) The portion from
ω=0
+

to

ω=+∞




GH(s
)

j
s

=

GH(jω)





lim
GH(jω)

=




lim

)
1
(



j
j
k






=




lim



1
2
tan
2




k



ω = +∞


mag
nitude

=

0




phase

=
-
180
ׁ





ω =

1
,

magnitude=
K




phase =
-
135
ׁ


(c) The portion from

ω = +∞
to
ω =
-






r
Lim

GH(s
)


=


r
Lim
2
r
K


e
-
2





S=
re
j

Ф



at

ω=+∞ ,

Ф
(
ω)
=

-
180

at

ω=
-
∞ ,
Ф
(
ω)
= +
180







































Nyquist contour








Z

=N+P


=0+0


= 0


This system is stable.








Radius E

s
-
plane

Radius

r







j

*


1

Radius=






*













1




0




0


u

Chapter
-
13


24
-

Determine the z
-
transform of f(t)=sin

ω
t for t ≥ 0






sin

ω
t =
e
j
ω
t
-

e

j
ω
t



2j




sin
ω
t =
{
e
j
ω
t


-

e

j
ω
t

}


2j 2j

f(z) =
1

(


__z__


-

____
z
__
)


2j z
-
e
j
ω
t

z
-
e

j
ω
t



=
1

(
z ( z
-
e

j
ω
t

)
-

z (z
-
e
j
ω
t

)

)



2j z
2
-

ze

j
ω
t

-

ze
j
ω
t

+1





=
1


(
z ( z
-
e

j
ω
t

)
-

z
(z
-
e
j
ω
t

)
)



2j z
2
-

ze

j
ω
t

-

ze
j
ω
t

+1



Z sin
ω
t




= __________________________________


z
2
-
z
(
cos
ω
t

jsin
ω
t +cos
ω
t +jsin
ω
t
)



Z sin
ω
t

F(z) = ___________________________



z
2
-
2z cos
ω
t+1




25
-
Determine the Z
-
transform of f(t)=e

at

t ≥ f(t)=e

at




Z
{

e

at
}
=

F(z) = ∑
e

ak
T

Z
-
k


K=0






F(z) = ∑ (
Z e
+
aT
)

k



K=0




F(z)

=

1+
(
Z e
+
aT
)

1
+
(
Z e
+
aT
)

2
+
---------------
+




F(z
)
=
_____1_____

1
-

(
Z e
aT
)

1




F(z
)
=
_____1_____

1
-
(
Z

1

e
-
aT
)





F(z
)
=



Z
_____

Z
-
e

aT