Image Processing Reference
In-Depth Information
Therefore,
e
t
11
11
0
e
At
(t)M
1
¼ ML
¼
e
2t
1
2
0
1
2
2e
t
e
2t
e
t
e
2t
¼
2e
t
2e
2t
e
t
2e
2t
þ
þ
Since matrix A is nonsingular, we have
1
ð
t
01
2e
t
e
2t
e
t
e
2t
10
01
e
At
d
t¼A
1
(e
At
I)
¼
2e
t
2e
2t
e
t
2e
2t
2
3
þ
þ
0
2e
t
1
5
10
:
5
0
:
e
2t
1
e
t
e
2t
¼
2e
t
2e
2t
e
t
2e
2t
þ
þ
1
2e
t
5e
2t
e
t
5e
2t
þ
0
:
þ
1
:
5
0
:
¼
2e
t
e
2t
e
t
e
2t
1
(b) In this case, matrix A is singular, and
2e
t
2e
t
1
þ
2
e
At
¼
þ e
t
e
t
1
2
Therefore, we use direct integration:
2
4
3
5
¼
Ð
t
t
Ð
t
2e
t
)d
2e
t
)d
ð
t
(
1
þ
(2
t
2e
t
2t þ
2e
t
t
2
þ
2
e
At
d
0
0
t ¼
Ð
Ð
t
t
t
1
þe
t
2t þ
1
e
t
þe
t
)d
e
t
)d
(
1
t
(2
t
0
0
0
3.11.6 C
OMPUTING
e
At
U
SING
L
APLACE
T
RANSFORM
Laplace transform can be used to compute e
At
. In this method, e
At
is computed by
solving the DE describing e
At
using the Laplace transform. Let
w(
t
) ¼
e
At
, then
dw(
t
)
d
t
¼
Ae
At
¼
A
w(
t
)
(
3
:
195
)
) ¼
e
A0
with
¼
I
Taking the Laplace transform from both sides of Equation 3.195, we have
w(
0
s
F(
s
) w(
) ¼
A
F(
s
)
(
:
)
0
3
196
Since
w(
0
) ¼
I, then
(
sI
A
)F(
s
) ¼
I
(
3
:
197
)
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