Image Processing Reference
In-Depth Information
The corresponding eigenvectors are
T
Ax
1
¼ l
1
x
1
,
therefore x
1
¼
½
1
j
T
Ax
2
¼ l
2
x
2
,
therefore x
2
¼
½
1
j
Hence,
"
#
e
(1j)t
"
#
11
"
#
1
11
0
(t)M
1
w
(t)
¼ ML
¼
j
j
e
(1þj)t
j
j
0
"
#
"
#
e
(1j)t
e
(1þj)t
0
:
5
0
:
5j
¼
je
(1j)t
je
(1þj)t
0
:
50
:
5j
"
#
0
:
5e
(1j)t
þ
0
:
5e
(1þj)t
0
:
5je
(1j)t
þ
0
:
5je
(1þj)t
¼
0
:
5je
(1j)t
0
:
5je
(1þj)t
0
:
5e
(1j)t
þ
0
:
5e
(1þj)t
"
#
e
t
cos t e
t
sin t
¼
e
t
sinte
t
cos t
(b)
Use of Laplace Transform
: Using Laplace transform, we have
1
1
s þ
1
1
s þ
11
(sI A)
1
F
(s)
¼
¼
¼
1
s þ
1
1)
2
1
s þ
1
(s þ
þ
1
or
2
4
3
5
s þ
1
1
1)
2
1)
2
(s þ
þ
1
(s þ
þ
1
F
(s)
¼
1
s þ
1
1)
2
1)
2
þ
1
þ
1
(s þ
(s þ
Taking inverse Laplace transform, we have
e
t
cos t e
t
sin t
e
t
sinte
w
(t)
¼
t
cos t
4.5.3 C
OMPLETE
S
OLUTION OF
S
TATE
E
QUATION
The complete solution of the state equation (Equation 4.62) is the sum of the
homogeneous and particular solutions and is given by
ð
t
0
w(
t
t)
Bu
(t)dt
x
(
t
) ¼ w(
t
)
x
(
0
) þ
(
4
:
79
)
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