Biomedical Engineering Reference
In-Depth Information
Since the conservation of mass equation for
q
1
involves only
q
1
,
it
is easily solved as
e
1:1
t
u
ð
t
Þ:
q
1
¼
25
Substituting the solution for
q
1
into Eq. (7.114), gives
e
1:1
t
q
2
¼
27
:
5
0
:
91
q
2
ð
7
:
118
Þ
Equation (7.118) involves an input and only
q
2
, which is solved independently as
u
ð
t
Þ
Þ
K
12
K
20
þ
K
23
K
12
q
1
ð
0
e
K
20
þ
K
23
ð
Þ
t
e
K
12
t
e
0:91
t
e
1:1
t
q
2
¼
¼
144
:
74
ð
7
:
119
Þ
ð
Þ
The solution for
q
2
is substituted into Eq. (7.115), yielding
Þ
K
12
K
20
þ
K
23
K
12
q
1
ð
0
q
3
¼
K
23
e
K
20
þ
K
23
ð
Þ
t
e
K
12
t
15
:
5
q
3
þ
30
q
4
þ
0
:
4
q
5
ð
Þ
ð
7
:
120
Þ
e
0:91
t
e
1:1
t
¼
130
:
27
15
:
5
q
3
þ
30
q
4
þ
0
:
4
q
5
Equations (7.120), (7.116), and (7.117) are solved using MATLAB and the D-Operator, as
follows:
>>
syms D
>>
¼
A
[-15.5 30 0.4;15 -31 0;0.5 0 -0.45];
>>
det(D*eye(3)-A)
>>
¼
adj
det(D*eye(3)-A)*inv(D*eye(3)-A)
which gives the reconstructed differential equations for
q
3
as
e
1:1
t
e
0:91
t
__
q
3
þ
46
:
95
q
3
þ
51
:
225
q
3
þ
7
:
525
q
3
¼
2531
:
8
1803
:
12
ð
7
:
121
Þ
with roots
45.84,
0.94, and
0.18. The natural response is
q
3
n
¼
B
1
e
45:84
t
þ
B
2
e
0:94
t
þ
B
3
e
0:18
t
The forced response is
q
3
f
¼
B
4
e
1:1
t
þ
B
5
e
0:91
t
, which after substituting into Eq. (7.113) gives
B
4
¼
380
:
39 and
B
5
¼
1870
:
41
The complete response is then
q
3
¼
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
380
:
39
1870
:
41
We use the initial conditions,
q
3
ð
0
Þ¼
0,
q
3
ð
0
Þ¼
0, and
q
3
ð
0
Þ¼
24
:
75, to solve for
B
1
,
B
2
, and
B
3
as follows:
q
3
ð
0
Þ¼
0
¼
B
1
þ
B
2
þ
B
3
þ
380
:
39
þ
1870
:
41
and with
q
3
¼
45
:
84
B
1
e
45:84
t
0
:
94
B
2
e
0:94
t
0
:
18
B
3
e
0:18
t
418
:
39
e
1:1
t
1701
:
97
e
0:91
t
we
have
q
3
ð
0
Þ¼
0
¼
45
:
84
B
1
0
:
94
B
2
0
:
18
B
3
418
:
39
1701
:
97
Continued
