Biomedical Engineering Reference
In-Depth Information
B
1
e
45:84
t
þ
B
2
e
0:94
t
þ
B
3
e
0:18
t
þ
e
1:1
t
þ
e
0:91
t
gives
which with
q
3
¼
2101
0
:
88
0
:
031
460
:
23
1548
:
80
q
3
ð
0
Þ¼
24
:
75
¼
2101
B
1
þ
0
:
88
B
2
þ
0
:
031
B
3
þ
460
:
23
þ
1548
:
80
To solve for the unknown constants, we evaluate
2
4
3
5
2
4
3
5
¼
2
4
3
5
1
1
1
2250
:
7
B
1
B
2
B
3
45
:
84
0
:
94
0
:
18
2120
:
3
2101
0
:
88
0
:
031
1984
:
3
which gives
2
4
3
5
¼
2
4
3
5
:
004
B
1
B
2
B
3
2259
:
51
8
:
844
Thus,
u
ð
t
Þ
e
45:84
t
e
0:94
t
þ
e
0:18
t
þ
e
1:1
t
þ
e
0:91
t
q
3
¼
0
:
004
2259
:
51
8
:
84
380
:
39
1870
:
41
ð
7
:
122
Þ
which is plotted in Figure 7.34. Note that if the oral dose involved T4 instead of T3, the model
would need to be changed by adding three more compartments for T4 (lower part of
Figure 7.31). We still need the three T3 compartments, since T4 transforms into T3.
Another way to solve for the response is to directly analyze the system using the D-Operator
matrix approach on Eqs. (7.113)-(7.117), which appears easier, since the input is 0 and there is
no forced response. However, considerable additional work is required to calculate the two extra
initial conditions (
) needed to solve for the extra two terms in the natural response,
which is not trivial. Thus, we have
__
q
1
ð
0
Þ
and
____
q
1
ð
0
Þ
6
5
4
3
2
1
0
0
5
10
15
20
25
30
Time
FIGURE 7.34
Illustration of the quantity in compartment 3 in Example Problem 7.18.
