Biomedical Engineering Reference
In-Depth Information
which simplifies to
q
K
32
þ
K
20
þ
K
23
s
1
,
2
¼
K
20
þ
K
23
þ
K
32
ð
Þ
1
2
2
ð
Þ
4
K
32
K
23
ð
7
:
104
Þ
2
Only real roots are possible with
K
ij
0
:
The natural response for
q
2
is
q
2
n
¼
B
1
e
s
1
t
þ
B
2
e
s
2
t
q
2
f
¼
B
3
e
K
12
t
, which after substituting into Eq. (7.103) gives
The forced response is
K
23
K
12
B
3
¼
12
K
K
12
K
32
þ
K
20
þ
K
23
ð
Þ þ
K
20
K
32
The complete response is then
q
2
¼
B
1
e
s
1
t
þ
B
2
e
s
2
t
þ
B
3
e
K
12
t
q
2
ð
We use the initial conditions,
q
2
ð
0
Þ¼
0 and
0
Þ¼
K
12
q
1
ð
0
Þ
, to solve for
B
1
and
B
2
as follows:
q
2
ð
0
Þ¼
B
1
þ
B
2
þ
B
3
¼
0
and with
q
2
¼
s
1
B
1
e
s
1
t
þ
s
2
B
2
e
s
2
t
K
12
B
3
e
K
12
t
,
we have
q
2
ð
0
Þ¼
K
12
¼
s
1
B
1
þ
s
2
B
2
K
12
B
3
To solve for
B
1
and
B
2
, we evaluate
¼
11
s
1
B
1
B
2
B
3
K
12
1
s
2
ð
þ
B
3
Þ
which gives
2
4
3
5
K
12
B
3
þ
K
12
þ
B
3
s
2
s
1
s
ð Þ
K
12
B
3
þ
K
12
þ
B
3
s
1
s
1
s
2
¼
B
1
B
2
ð
Þ
The final solution is
0
@
1
A
K
B
þ
K
þ
B
3
s
e
s
1
t
K
B
þ
K
þ
B
3
s
12
3
12
2
12
3
12
1
e
s
2
t
ð
s
1
s
2
Þ
ð
s
1
s
2
Þ
q
2
ð
t
Þ¼
u
ð
t
Þ
K
23
K
12
Þ þ
K
20
K
32
e
K
12
t
þ
2
12
K
K
12
K
32
þ
K
20
þ
K
23
ð
Repeating for
q
3
, we have
q
3
ð
t
Þ¼
B
3
K
12
þ
s
2
ð
Þ
e
s
1
t
B
3
K
12
þ
s
1
ð
Þ
K
23
K
12
e
s
2
t
þ
Þ þ
K
20
K
32
e
K
12
t
u
ð
t
Þ
2
12
ð
s
1
s
2
Þ
ð
s
1
s
2
Þ
K
K
12
K
32
þ
K
20
þ
K
23
ð
