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In-Depth Information
[
,
]=
[
,
]
Substituting g
i
k
f
i
a
k
b
in ( 6.5 ) yields:
M
1
i = 0
N
1
k = 0
1
MN
e j (
M + n N
i
)
m
[
,
]=
[
,
]
G
m
n
f
i
a
k
b
(6.6)
M
a
u = a
1
N
b
v = b
1
1
MN
e j (
m u + M +
n v + b
N
)
=
f
[
u
,
v
]
(6.7)
e j (
m M +
n N
) F
=
[
m
,
n
]
(6.8)
The relation between the first line and the second line is obtained by substituting
index i and k with i
=
u
+
a and k
=
v
+
b , and F
[
m
,
n
]
is the Fourier transform of
. Taking the complex conjugate F [
image
{
f
[
i
,
k
] }
m
,
n
]
to multiply the relation in
Eq. ( 6.8 ) produces:
e j (
m M +
n N
) F
F [
F [
G
[
m
,
n
]
m
,
n
]=
[
m
,
n
]
m
,
n
]
(6.9)
e j (
m M +
n N
) |
2
=
F
[
m
,
n
] |
(6.10)
e j (
a
M +
n N
) |
m
=
G
[
m
,
n
] ||
F
[
m
,
n
] |
(6.11)
The relation from the second line to the third line is obtained by substitution of
|
F
[
m
,
n
] |
with
|
F
[
m
,
n
] | = |
G
[
m
,
n
] |
. Equation ( 6.11 ) can be written in the form of a
cross-power spectrum as:
e j (
)
m M +
n N
F [
G
[
m
,
n
]
m
,
n
]
] | =
(6.12)
|
G
[
m
,
n
] ||
F
[
m
,
n
According to Gonzalez and Woods [ 173 ], we can obtain the inverse Fourier
transform of the cross-power spectrum using:
MN
M 1
m = 0
N 1
n = 0 e j ( m i M + n k b
,
i
=
a
,
k
=
b
) =
(6.13)
N
0
,
i
=
a
,
k
=
b
=
MN
· ʴ [
i
a
,
k
b
]
(6.14)
where
ʴ [
i
a
,
k
b
]
is equal to zero at index i
=
a and k
=
b , whereas
ʴ [
i
a
,
k
b
]
is equal to one at index i
b . At this position, we can find the parameters
for image translation between the two images in the row direction by a pixels and
in the column direction by b pixels.
=
a and k
=
6.2.2
Parameter Estimation for Rotation
Let
0 and translates by a
pixels in the row direction and b pixels in the column direction. The relation of the
gray scale corresponding image can be written as:
{
g
[
i
,
k
] }
be the reference image
{
f
[
i
,
k
] }
that rotates by
ʸ
 
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