Biomedical Engineering Reference
In-Depth Information
. Substituting
e
0:2
t
Solving Eq. (7.67) gives
q
2
u
¼
400 1
q
2
u
into Eq. (7.66) gives
q
1
u
¼
e
0:2
t
80 1
0
:
2
q
1
u
ð
7
:
68
Þ
q
1
n
¼
B
1
e
0:2
t
. The input in Eq. (7.66)
has the same term as in the natural solution, and so the forced response is
The root for Eq. (7.68) is
s
¼
0.2, and has a natural solution
q
1
f
¼
B
3
þ
B
2
te
0:2
t
.
Substituting
q
1
f
into Eq. (7.68) gives
B
2
¼
80 and
B
3
¼
400. The complete response is
q
1
u
¼
q
1
n
þ
q
1
f
¼
B
1
e
0:2
t
þ
te
0:2
t
400
80
ð
7
:
69
Þ
B
1
is found using the initial condition
q
1
(0)
¼
0 and
t
¼0
¼
B
1
þ
¼
B
1
e
0:2
t
þ
te
0:2
t
q
1
u
ð
0
Þ¼
0
400
80
400
B
1
¼
400. Thus,
and
u
ð
t
Þ
e
0:2
t
te
0:2
t
q
1
u
¼
400
400
80
ð
7
:
70
Þ
80u(t30) Input
Next, consider the
80
u(t
30)
input. By the property of a linear system, then
e
0:2
t
30
ð
Þ
Þ
e
0:2
t
30
ð
Þ
q
1
u
30
¼
400
400
80
ð
t
30
ut
ð
30
Þ
e
0:2
t
30
ð
Þ
q
2
u
30
¼
400 1
u
ð
t
30
Þ
Complete Solution
The complete response is
u
ð
t
Þ
q
1
¼
q
1
d
þ
q
1
u
þ
q
1
u
30
¼
4
te
0:2
t
u
ð
t
Þþ
400
400
e
0:2
t
80
te
0:2
t
ut
400
400
e
0:2
t
30
ð
Þ
80
ð
t
30
Þ
e
0:2
t
30
ð
Þ
ð
30
Þ
u
ð
t
Þ
e
0:2
t
u
ð
t
Þþ
e
0:2
t
e
0:2ð
t
30Þ
q
2
¼
q
2
d
þ
q
2
u
þ
q
2
u
30
¼
20
400 1
400 1
u
ð
t
30
Þ
which is plotted in Figure 7.18.
400
q
2
q
1
200
0
0
10
20
30
40
50
60
Time
FIGURE 7.18
Plot of the solute quantities for Example Problem 7.10.
