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
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