(c)
For stability, K
D
> 0,
0 .5
+ K P > 0.
Thu s,
K
P
> 0 .5
S tability Region:
6 -16 S tate diagram:
= 1 + s + Ks
2
C haracteristic equation:
s
S tability requirement:
2
+s+K =0
K>0
83
Chapter 7
TIME-DOMAIN ANALYSIS OF
CONTROL SYSTEMS
7 -1 (a) 0 . 707
(e) A ngle of arrival
K < 0:
1 + 6 2 3 4 5 = 360
1 + 90 135
o
1 = 108
8 -3 (a)
o
o
.435
135
o
o
o
45 26 . 565
o
= 360
(b)
(c)
(d)
122
o
8 -4 (a) B reakaway-point Equation:
2s
5
+ 20
s
4
+ 74
0 . 7275
B reakaway Points:
(b) B reakaway-point Equation:
(c)
N=3 C losed-Loop Transfer Function:
M
l
e
= 0 . 3333
1
= 2 d 2 d1 =
2
2
Thus, d
H
=
2
2
f
4
(s)
l
f
=
L
(s)
s
+ 11 s + 10 s + 30
l
2
+ 0 . 333
s
+ 0 . 3667
d
1
2
1
= 2 d2
+ 0 .2186
s
f
= 0 .2186
+ 0 . 335
2
s
2
+ 0 . 0333
s
3
s
2
= d2 =
2
4
f
= 2 l1
(b)
= 0 .456
For maximum overshoot = 0.2,
Settling time t
s
=
3 .2
n
.
3 .2
=
0 .456
K
t
=
5
+ 500
= 0 . 01
n
0 . 912
K
t
= 2 n = 0 . 912 n
n =
sec.
n 5
3 .2
0 . 01
0 .456
= 701 . 75
rad / sec
= 1.27
500
S ystem Transfer Function:
Y (s)
R( s)
=
492453
1 =
3
1 2 1
3
1
3 s 1 = 0
B reakaway-point Equation:
s
B reakaway Points;
8 -6 (a)
= 2
0.3473, 1.532,
Q(s ) = s + 5
(
1.879
)
P (s ) = s s + 3s + 2 = s (s +1)( s + 2)
2
o
A symptotes:
K > 0:
90 ,
I ntersect of Asymptotes:
270
o
K < 0:
1 =
B reakaway-point
8 -7 (c) A symptotes:
K > 0:
180
o
B reakaway-point Equation:
B reakaway Points:
1.727
When
s
4
+ 4 s + 10
3
s
+ 300
2
K = 9.65
= 0 . 707
,
(on the RL)
146
s
+ 500 = 0
8 -7 (d)
K > 0:
90
o
,
270
o
I ntersect of Asymptotes:
1 =
When
= 0 . 707
,
2 2 5 6
4
M
r
r
1
=
2
= 2 . 06
1
1 2
= 9 . 33 rad
2
+ 0 . 0589 = 0
4
n
/ sec
2
GL ( s) =
9 -24 (a)
(b)
M
M
r
r
s ( s + 2 n )
= 2 . 96 , r = 666
1
=
2
r
=
1 2
s( s + 4.987)
. 67 rad / sec,
=
9 . 33
=
s(1 + 0.2005 s)
=
is
= 0.25.
rad / sec
B W = 15.21 rad/sec
BW = 1
For stability, Z = 0.
11 = ( 0.5 P + P ) 180 = 0
o
0
<
K
< 2500
11 = 0
K
o
S table
11 = 360
< 625
625 < K < 0
o
11 = 180
> 2500
K
o
o
U nstable
U nstable
o
11 = 0
S table
The system is stable for 625 < K < 2 500.
9 -12
(
)
(
)
s s + 2s + s +1 + K s + s
8 -22
Let the angle of the vector drawn from the zero at s
=
j12
to a point s on the root locuss near the zero
1
.
1 =
Let
angle
of th e vect or dra wn fro m the
pole a t j 10 to s .
2 =
angle
of th e vect or dra wn fro
pole a t 0 to s .
3 =
angle
of th e
8 -14 (c)
P (s)
= s + 12 . 5 s + 1666 . 67 = ( s + 17 . 78
Q ( s ) = 0 . 02 s ( s + 12 . 5)
3
2
A symptotes:
K
o
> 0:
180
)( s
2 . 64 +
j 9 . 3)( s
2 . 64
j 9 . 3)
o
B reakaway-point Equation: 0 . 02 s + 0 . 5 s + 3 .125 s
B reakaway Point: (RL)
5.797