Biomedical Engineering Reference
In-Depth Information
131
moves into the hepatic
duct, where it is absorbed within the bile. The feedback control by the pituitary gland and
the involvement of the kidneys are ignored in this example.
131
. The
Within the liver, some of the T4 is converted into T3 and
I
I
EXAMPLE PROBLEM 7.16
Consider a three-compartment thyroxine transport model as shown in Figure 7.25. The input is
f
1
ð
t
Þ¼
10
3
dð
t
Þ
0.3.
Assume that the initial conditions are zero. Solve for the quantity of radioactive iodine-T4 in the
plasma compartment.
0
:
1
g of radioactive iodine-T4. Additionally,
K
12
¼
0.6,
K
21
¼
0.5, and
K
23
¼
Solution
Since there is no output to another compartment for compartment 3, this compartment model
has a sink for compartment 3. As before, rather than solving the problem with a bolus input, the
initial condition is changed for compartment 1 to
10
3
, with zero input. The conser-
q
1
ð
0
Þ¼
0
:
1
vation of mass for each compartment yields
q
1
¼
K
21
q
2
K
12
q
1
¼
0
:
6
q
1
þ
0
:
5
q
2
q
2
¼
K
12
q
1
ð
K
21
þ
K
23
Þ
q
2
¼
0
:
6
q
1
0
:
8
q
2
ð
7
:
105
Þ
q
3
¼
K
23
q
2
¼
0
:
3
q
2
Using the D-Operator method with MATLAB, we get
>>
syms D
>>
A
¼
[-0.6 0.5 0;0.6 -0.8 0;0 0.3 0];
>>
det(D*eye(3)-A)
ans
¼
^
þ
^
þ
D
3
7/5*D
2
9/50*D
and
7
5
q
1
þ
9
50
q
1
¼
__
q
1
þ
0
The
“
eig(A)
”
command gives the roots as 0,
1.26, and
0.14. Thus, we have
q
1
¼
B
1
þ
B
2
e
1:26
t
þ
B
3
e
0:14
t
(since the forced response is zero). The initial conditions are
10
3
,
q
1
ð
0
Þ¼
0
:
1
q
2
ð
0
Þ¼
0, and
q
3
(0)
¼
0. To determine the initial conditions for the derivative terms, we use Eq. (7.105), which
gives
10
3
q
1
ð
0
Þ¼
0
:
6
q
1
ð
0
Þþ
0
:
5
q
2
ð
0
Þ¼
0
:
06
10
3
To determine the initial conditions for the second derivative, we take the derivative of
Eq. (7.105) and set t
q
2
ð
0
Þ¼
0
:
6
q
1
ð
0
Þ
0
:
8
q
2
ð
0
Þ¼
0
:
06
¼
0, giving
10
5
q
1
ð
0
Þ¼
0
:
6
q
1
ð
0
Þþ
0
:
5
q
2
ð
0
Þ¼
6
:
6
10
5
q
2
ð
q
1
ð
q
2
ð
0
Þ¼
0
:
6
0
Þ
0
:
8
0
Þ¼
8
:
4
Continued
