Biomedical Engineering Reference
In-Depth Information
EXAMPLE PROBLEM 7.15
Consider the mammillary three-compartment model with the source compartment shown in
Figure 7.24. The input is
f
1
(
t
)
¼d(
t
). Assume that the initial conditions are zero. Solve for the quan-
tity in each compartment.
Solution
Once again, the input is transformed into a change in initial condition for compartment 1,
q
1
(0)
¼
1. The equations describing this model are
q
1
¼
K
12
q
1
ð
7
:
97
Þ
q
2
¼
K
12
q
1
K
20
þ
K
23
ð
Þ
q
2
þ
K
32
q
3
ð
7
:
98
Þ
q
3
¼
K
23
q
2
K
32
q
3
ð
7
:
99
Þ
Þ
e
K
12
t
u
ð
t
Þ¼
e
K
12
t
u
ð
t
Þ:
Since Eq. (7.97) involves only
q
1
, it is easily solved as
q
1
¼
q
1
ð
0
Substitut-
ing the solution for
q
1
into Eqs. (7.98) and (7.99), we now have two equations as
q
2
¼
K
12
e
K
12
t
u
ð
t
Þ
K
20
þ
K
23
ð
Þ
q
2
þ
K
32
q
3
ð
7
:
100
Þ
q
3
¼
K
23
q
2
K
32
q
3
ð
7
:
101
Þ
The D-Operator gives the reconstructed differential equations for
q
2
and
q
3
as
>>
syms D K20 K23 K32
>>
¼
þ
A
[-(K20
K23) K32;K23 -K32];
>>
det(D*eye(2)-A)
ans
¼
D
^
2
þ
D*K32
þ
K20*D
þ
K20*K32
þ
K23*D
>>
adj
¼
det(D*eye(2)-A)*inv(D*eye(2)-A)
adj
¼
[D
þ
K32, K32
]
þ
þ
[ K23,
D
K20
K23]
and
Þ
e
K
12
t
q
2
þð
K
32
þ
K
20
þ
K
23
Þ
q
2
þ
K
20
K
32
q
2
¼
K
12
K
32
K
12
ð
ð
7
:
102
Þ
q
3
þð
K
32
þ
K
20
þ
K
23
Þ
q
3
þ
K
20
K
32
q
3
¼
K
23
K
12
e
K
12
t
ð
7
:
103
Þ
The roots are
>>
eig(A)
ans
¼
-1/2*K20-1/2*K23-1/2*K32
þ
1/2*(K20
^
2
þ
2*K20*K23-
2*K20*K32
þ
K23
^
2
þ
2*K32*K23
þ
K32
^
2)
^
(1/2)
-1/2*K20-1/2*K23-1/2*K32-1/2*(K20
^
2
þ
2*K20*K23-
2*K20*K32
þ
K23
^
2
þ
2*K32*K23
þ
K32
^
2)
^
(1/2)
Continued
