Image Processing Reference
In-Depth Information
Substituting Equation 2.33 into Equation 2.32 yields
2
4
3
5
e
j
(v
x
n
D
x
þv
y
m
D
y
)
1
1
1
1
1
F
(
u, v
)
e
j
(
un
D
x
þ
vm
D
y
)
d
u
d
v
F*
(v
x
,
v
y
) ¼
p
2
4
n
¼1
m
¼1
1
1
"
#
d
u
d
v
1
1
1
1
1
e
j
(
u
v
x
)
n
D
x
j
(
v
v
y
)
m
D
y
¼
F
(
u, v
)
4
p
2
n
¼1
m
¼1
1
1
1
1
1
e
j
(
v
v
y
)
m
D
y
1
n
¼1
1
e
j
(
u
v
x
)
n
D
x
¼
F
(
u, v
)
d
u
d
v
(
2
:
34
)
4
p
2
m
¼1
1
1
We now use the well-known identity
1
1
e
j2
p
xk
¼
d(
x
k
)
(
2
:
35
)
k
¼1
k
¼1
Then, Equation 2.34 can be reduced to
1
1
1
1
(
u
v
x
)D
x
2
1
F*
(v
x
,
v
y
) ¼
F
(
u, v
)
d
n
4
p
2
p
n
¼1
m
¼1
1
1
(
v
v
y
)D
y
2
d
m
d
u
d
v
(
2
:
36
)
p
Let u
0
¼
(
u
v
x
)D
x
2
and v
0
¼
(
v
v
y
)D
y
2
, then
p
p
1
1
1
1
1
4
p
2
p
D
x
u
0
,
v
y
þ
2
p
D
y
v
0
2
p
D
x
2
p
D
y
d
u
0
d
v
0
d[
u
0
n
]d[
v
0
m
]
F*
(v
x
,
v
y
)¼
F
v
x
þ
2
n
¼1
m
¼1
1
1
1
1
d[
u
0
n
]d[
v
0
m
]d
u
0
d
v
0
1
1
1
D
x
1
D
y
2
p
D
x
u
0
,
v
y
þ
2
p
D
y
v
0
¼
F
v
x
þ
n
¼1
m
¼1
1
1
¼
f
s
x
f
s
y
1
n
¼1
1
(
2
:
37
)
F
(v
x
þ
n2
p
f
s
x
,
v
y
þ
m2
p
f
s
y
)
m
¼1
Therefore, the Fourier transform of the sampled signal is the periodic extension of
the Fourier transform of the analog signal with period given by
(v
s
x
,
v
s
y
)
as shown in
Figure 2.9.
As a result of this, the original signal can be reconstructed by low-pass
filtering
the sampled signal, provided that there is no overlap between the spectrum compon-
ents in frequency domain or in other words there is no aliasing. This is possible only
if the sampling frequencies satisfy.
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