Biomedical Engineering Reference
In-Depth Information
2
1
2
z
@
K
@
2
ð
K
K
þ
K
K
þ
K
K
Þ
K
ð
þ
K
Þ
K
ð
þ
K
þ
K
Þ
K
ð
K
þ
K
K
þ
K
K
Þ
12
23
23
31
31
12
23
31
12
23
31
12
23
23
31
31
12
12
¼
4
ð
K
K
23
þ
K
23
K
31
þ
K
31
K
12
Þ
12
2
ð
K
K
þ
K
K
þ
K
K
Þ
K
ð
þ
K
Þ
K
ð
þ
K
þ
K
Þ
12
23
23
31
31
12
23
31
12
23
31
¼
¼
0
3
2
4
ð
K
12
K
23
þ
K
23
K
31
þ
K
31
K
12
Þ
The minimum occurs when the numerator is zero—that is,
2
ð
K
þ
K
K
þ
K
31
K
12
Þ ¼
K
ð
þ
K
Þ
K
ð
þ
K
þ
K
Þ
K
12
23
23
31
23
31
12
23
31
2
23
2
31
¼
K
K
þ
K
þ
K
K
þ
K
K
þ
K
K
þ
K
23
12
23
31
31
12
31
23
Simplifying, we have
2
23
2
31
K
12
K
23
þ
K
31
ð
Þ ¼
K
þ
K
or
23
31
K
12
¼
K
þ
K
ð
K
þ
K
Þ
23
31
2
31
K
12
þ
K
31
2
12
2
23
K
12
þ
K
23
2
12
z
@
K
23
¼
@
¼
K
þ
K
z
@
K
31
¼
@
¼
K
þ
K
Repeating for
0, we get
, and for
0, we get
.
K
K
23
31
ð
Þ
ð
Þ
The only way these relationships are valid is if
K
12
¼
K
23
¼
K
31
¼
K
, and from Eq. (7.95),we
3
2
p ¼
find z
¼
0
:
866, which does not have a very noticeable oscillatory behavior. We will
see in the next section that a more noticeable oscillatory response is possible with more than
three compartments.
EXAMPLE PROBLEM 7.14
Consider the unilateral three-compartment model shown in Figure 7.22 with no loss of solute
to the environment from any compartments and an input for compartment 3 only. Additionally,
K
12
¼
K
23
¼
K
31
¼
2, and
f
3
ð
t
Þ¼
5d
ð
t
Þ:
Assume that the initial conditions are zero. Solve for the
quantity in each compartment.
Solution
As before, we transform the input,
f
3
(
t
)
¼
5d(
t
), into a change in initial condition for compart-
ment 3 to
q
3
(0)
¼
5 and no input. The conservation of mass for each compartment yields
q
1
¼
K
31
q
3
ð
K
10
þ
K
12
Þ
q
1
¼
2
q
1
þ
2
q
3
q
2
¼
K
12
q
1
ð
K
20
þ
K
23
Þ
q
2
¼
2
q
1
2
q
2
ð
7
:
96
Þ
q
3
¼
K
23
q
2
ð
K
30
þ
K
31
Þ
q
3
¼
2
q
2
2
q
3
