Biomedical Engineering Reference
In-Depth Information
rate and the fraction released immediately experimentally, the pill is dissolved in a solution
similar to the stomach, and the concentration is measured. From this data, parameter values
can be determined.
EXAMPLE PROBLEM 7.10
Consider the two-compartment model shown in Figure 7.15 with
K
12
¼
K
20
¼
0,
K
21
¼
K
10
¼
0.2,
f
1
(
t
)
¼
0, and
f
2
(
t
)
¼
20d(
t
)
þ
80(
u
(
t
)
u
(
t
30)). Assume that the initial conditions are zero
(not including that provided by 20d(
t
)). Solve for the quantity in each compartment.
Solution
Since
0, this model has a source compartment. The solution is carried out using superpo-
sition, separating the input
K
12
¼
f
2
(
t
) into 20d(
t
), 80
u
(
t
), and 80
u
(
t
30), and then summing the individ-
ual responses to get the complete response.
The conservation of mass for each compartment is
q
1
¼
0
:
2
q
2
0
:
2
q
1
ð
7
:
61
Þ
q
2
¼
f
2
0
:
2
q
2
ð
7
:
62
Þ
20
d
(t) Input
First, consider the 20d(
) input. As in Example Problem 7.8, we treat the impulse input as
a change in an initial condition, yielding
t
q
2
d
¼
20
e
0:2
t
u
ð
t
Þ:
Substituting this result into Eq. (7.61)
gives
q
1
d
¼
e
0:2
t
4
0
:
2
q
1
d
ð
7
:
63
Þ
q
1
n
¼
B
1
e
0:2
t
. The input in
Eq. (7.63) has the same form as the natural solution (expected since (
The root for Eq. (7.63) is
s
¼
0.2, and has a natural solution
K
20
þ
K
21
) equals (
K
10
þ
q
1
f
¼
B
2
te
0:2
t
:
K
12
) and
K
12
¼
0), and so the forced response is
Substituting
q
1
f
into Eq. (7.63) gives
B
2
¼
4. The complete response is
q
1
d
¼
q
1
n
þ
q
1
f
¼
B
1
e
0:2
t
þ
te
0:2
t
4
ð
7
:
64
Þ
B
1
is found using the initial condition
q
1
(0)
¼
0 and
t
¼0
¼
B
1
¼
B
1
e
0:2
t
þ
te
0:2
t
q
1
d
ð
0
Þ¼
0
4
Thus,
te
0:2
t
u
ð
t
Þ
q
1
d
¼
4
ð
7
:
65
Þ
80u(t) Input
Next, consider the 80
u(t)
input. The conservation of mass equations are
q
1
u
¼
0
:
2
q
2
u
0
:
2
q
1
u
ð
7
:
66
Þ
q
2
u
¼
80
0
:
2
q
2
u
ð
7
:
67
Þ
Continued