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Example Problems and Solutions 235
It follows that
From the block diagram we have
from which
Therefore, the values of TandKare determined as
A–5–4. Determine the values of Kandkof the closed-loop system shown in Figure 5–53 so that the maximum
overshoot in unit-step response is 25%and the peak time is 2 sec. Assume that J=1kg-m^2.
Solution.The closed-loop transfer function is
By substituting J=1kg-m^2 into this last equation, we have
Note that in this problem
The maximum overshoot Mpis
which is specified as 25%. Hence
from which
zp
21 - z^2
=1.386
e-zp^21 - z
2
=0.25
Mp=e-zp^21 - z
2
vn= 1 K, 2 zvn=Kk
C(s)
R(s)
=
K
s^2 +Kks+K
C(s)
R(s)
=
K
Js^2 +Kks+K
K =v^2 n T=1.14^2 *1.09=1.42
T =
1
2 zvn
=
1
2 0.41.14
=1.09
vn=
A
K
T
, 2 zvn=
1
T
C(s)
R(s)
=
K
Ts^2 +s+K
vn=1.14
+
- +–
R(s) C(s)
k
1
s
K
Js
Figure 5–53
Closed-loop system.