Biomedical Engineering Reference
In-Depth Information
1
0.8
0.6
0.4
0.2
5
= 25
0
T p =2
0
5
10
15
20
25
30
t (s)
FIGURE 13.70 Graphical analysis for Example Problem 13.12.
From Figure 13.70, the time to peak overshoot occurs at approximately 2 s. Using Eq. (13.79),
we have
s 1 T p þ
ln
s ðÞ¼ s 2 T p þ
ln
s ðÞ
and MATLAB's “solve” command gives
s 1 .
>>
þ
¼
þ
¼
solve('s1*2
log(s1)
s2*2
log(s2)','s2
.2')
¼
s1: [2x1 sym]
s2: [2x1 sym]
>>
ans
s1
¼
s1
1.0093940626323415922798829136295
0.2
Notice that there are two roots for
s 1 , since both are solutions to the equation. Since
s 2 ¼
.2,
then
s 1 ¼
1. Finally, we have
2 and z ¼ s 1 þ s 2
2o n
o 2
n ¼ s 1 s 2 ¼
0
:
¼
3
:
13.11 EXERCISES
1. Consider the system shown in Figure 13.71 defined with
M 1 ¼
1 kg,
M 2 ¼
2 kg,
B 1 ¼
.5 N-s/m,
B 2 ¼
1 N-s/m,
K 1 ¼
3 N/m, and
K 2 ¼
2 N/m . Let
(
) be the applied force, and
x 1 and
x 2 be
f
t
the displacements from rest. (a) Find the transfer function. (b) Solve for
x 1 (
)if
(
)
¼
3
(
)N
t
f
t
u
t
and zero initial conditions. (c) Simulate the solution with SIMULINK if
(
)
¼
3
(
) N and
f
t
u
t
zero initial conditions. (c) Use MATLAB to draw the Bode diagram.
Continued
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