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