Biomedical Engineering Reference
In-Depth Information
Using the D-Operator method with MATLAB, we get
>>
syms D
>>
¼
A
[-2 0 2;2 -2 0;0 2 -2];
>>
det(D*eye(3)-A)
ans
¼
D
^
3
þ
6*D
^
2
þ
12*D
and
__
q
1
þ
6
q
1
þ
12
q
1
¼
0
6
q
2
þ
12
q
2
¼
__
q
2
þ
0
__
q
3
þ
6
q
3
þ
12
q
3
¼
0
The roots from the characteristic equation are 0,
3
j
1
:
7321
:
The complete solution is the
natural solution, since the forced response is zero, and is given by
q
1
¼
B
1
þ
e
3
t
B
2
cos 1
ð
:
7321
t
þ
B
3
sin 1
:
7321
t
Þ
q
2
¼
B
4
þ
e
3
t
B
5
cos 1
ð
:
7321
t
þ
B
6
sin 1
:
7321
t
Þ
q
3
¼
B
7
þ
e
3
t
B
8
cos 1
ð
:
7321
t
þ
B
9
sin 1
:
7321
Þ
t
The initial conditions are
q
1
(0)
¼
0,
q
2
(0)
¼
0, and
q
3
(0)
¼
5. To determine the initial conditions for
the derivative terms, we use Eq. (7.96) and get
q
1
ð
0
Þ¼
2
q
1
ð
0
Þþ
2
q
3
ð
0
Þ¼
10
q
2
ð
0
Þ¼
2
q
1
ð
0
Þ
2
q
2
ð
0
Þ¼
0
q
3
ð
0
Þ¼
2
q
2
ð
0
Þ
2
q
3
ð
0
Þ¼
10
To determine the initial conditions for the second derivative, we take the derivative of
Eq. (7.96) with
t
¼
0, giving
q
1
ð
0
Þ¼
2
q
1
ð
0
Þþ
2
q
3
ð
0
Þ¼
40
q
2
ð
q
1
ð
q
2
ð
0
Þ¼
2
0
Þ
2
0
Þ¼
20
q
3
ð
0
Þ¼
2
q
2
2
q
3
¼
20
For
q
1
, we have
q
1
ð
0
Þ¼
0
¼
B
1
þ
B
2
q
1
ð
0
Þ¼
10
¼
3
B
2
þ
1
:
7321
B
3
q
1
ð
0
Þ¼
40
¼
6
B
2
10
:
4
B
3
5
3
,
5
3
, and
which gives
B
1
¼
B
2
¼
B
3
¼
2
:
9, and
5
3
e
3
t
5
3
q
1
¼
cos 1
:
7321
t
2
:
9 sin 1
:
7321
t
u
ð
t
Þ
Repeating for
q
2
and
q
3
, we have
0
@
0
@
1
A
1
A
u
ð
t
Þ
5
3
e
3
t
5
3
q
¼
cos 1
:
7321
t
þ
2
:
9 sin 1
:
7321
t
2
u
ð
t
Þ
5
3
e
3
t
q
3
¼
1
þ
2
cos 1
:
7321
t
Continued
