Image Processing Reference
In-Depth Information
ω
2
π
D
-π
ω
1
π
D
-π
FIGURE 3.12
Support of X
(
j
v
1
, j
v
2
)
.
S
OLUTION
Using inverse 2-D DSFT, we have
p
p
1
jv
2
)e
jn
1
v
1
e
jn
2
v
2
d
x(n
1
, n
2
)
¼
X( jv
1
,
v
1
d
v
2
4
p
2
p
p
The above integral is
ð
0
ð
0
p
p
1
1
4
e
jn
1
v
1
e
jn
2
v
2
d
e
jn
1
v
1
e
jn
2
v
2
d
x(n
1
, n
2
)
¼
v
1
d
v
2
þ
v
1
d
v
2
4
p
2
p
2
p
p
0
0
Upon integration, we have
e
jpn
1
jn
1
e
jpn
2
jn
2
e
jpn
1
e
jpn
2
1
1
1
1
1
1
x(n
1
, n
2
)
¼
þ
4
p
2
4
p
2
jn
1
jn
2
which can be simpli
ed to
1
1
e
jpn
1
jn
1
1
e
jpn
2
jn
2
1
e
jpn
1
1
e
jpn
2
1
x(n
1
, n
2
)
¼
þ
4
p
2
4
p
2
jn
1
jn
2
Therefore,
<
:
0
:
5
n
1
¼ n
2
¼
0
0
n
1
¼
0, n
2
6¼
0
x(n
1
, n
2
)
¼
0
n
1
6¼
0, n
2
¼
0
(1
cos n
1
p
)(1
cos n
2
p
)
otherwise
2
p
2
n
1
n
2
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