Biomedical Engineering Reference
In-Depth Information
EXAMPLE PROBLEM 7.13
Consider the mammillary three-compartment model shown in Figure 7.21, with a loss of solute
to the environment only from compartment 1 and input only from compartment 2. Additionally,
K
12
¼
2,
K
21
¼
1.5,
K
10
¼
0.5,
K
23
¼
1.3,
K
32
¼
0.4, and
f
2
(
t
)
¼
10d(
t
). Assume that the initial conditions are
zero. Solve for the quantity in each compartment.
Solution
With the input
f
2
(
t
)
¼
10d(
t
) transformed into a change in initial condition for compartment 2 to
q
2
(0)
¼
10 and no input, conservation of mass for each compartment yields
q
1
¼
K
21
q
2
ð
K
10
þ
K
12
Þ
q
1
¼
2
:
5
q
1
þ
1
:
5
q
2
q
2
¼
K
12
q
1
ð
K
21
þ
K
23
Þ
q
2
þ
K
32
q
3
¼
2
q
1
2
:
8
q
2
þ
0
:
4
q
3
ð
7
:
91
Þ
q
3
¼
K
23
q
2
ð
K
30
þ
K
32
Þ
q
3
¼
1
:
3
q
2
0
:
4
q
3
Using the D-Operator method with MATLAB, we get
>>
syms D
>>
¼
A
[-2.5 1.5 0;2 -2.8 0.4;0 1.3 -0.4];
>>
det(D*eye(3)-A)
ans
¼
^
þ
^
þ
þ
D
3
57/10*D
2
28/5*D
3/10
and
28
5
q
1
þ
q
1
þ
__
q
1
þ
5
:
7
0
:
3
q
1
¼
0
28
5
q
2
þ
__
q
2
þ
5
:
7
q
2
þ
0
:
3
q
2
¼
0
28
5
q
3
þ
q
3
þ
__
q
3
þ
5
:
7
0
:
3
q
3
¼
0
Using the
“
eig(A)
”
command gives the roots as
4.46,
1.18, and
0.06. Thus, we have
q
1
¼
B
1
e
4:46
t
þ
B
2
e
1:18
t
þ
B
3
e
0:06
t
q
2
¼
B
4
e
4:46
t
þ
B
5
e
1:18
t
þ
B
6
e
0:06
t
q
3
¼
B
7
e
4:46
t
þ
B
8
e
1:18
t
þ
B
9
e
0:06
t
(since the forced response is zero). Note that since there is no input, all we needed to do was
define the matrix
A
and then use the
“
eig(A)
”
command (i.e., no need to use the
“
det
”
com-
mand). However, we shall use the
“
det
”
command because it gives the intermediate result.
The initial conditions are
q
1
(0)
¼
0,
q
2
(0)
¼
10, and
q
3
(0)
¼
0. To determine the initial conditions for
the derivative terms, we use Eq. (7.91) and get
q
1
ð
0
Þ¼
2
:
5
q
1
ð
0
Þþ
1
:
5
q
2
ð
0
Þ¼
15
q
2
ð
0
Þ¼
2
q
1
ð
0
Þ
2
:
8
q
2
ð
0
Þþ
0
:
4
q
3
ð
0
Þ¼
28
q
3
ð
0
Þ¼
1
:
3
q
2
ð
0
Þ
0
:
4
q
3
ð
0
Þ¼
13
Continued
