Image Processing Reference
In-Depth Information
10.17 Appendix
Proof of lemma 10.1 Noting that
=1, we can choose
u
so that it is orthogonal
to
k
. By juxtaposing these two column vectors side by side, we can construct the
orthogonal matrix
U
=[
k
,
u
], i.e.,
UU
T
k
=
1
. Recalling that
r
=(
x, y
)
T
, and
=(
ω
x
,ω
y
)
T
, the 2D Fourier transform of a linearly symmetric image then yields:
F
(
ω
x
,ω
y
)=
f
(
x, y
) exp(
−i
ω
ω
T
r
)
dxdy
=
g
(
k
T
r
) exp(
T
r
)
dxdy
−
i
ω
(10.102)
where we have substituted the 2D linearly symmetric function
f
(
x, y
) with its ver-
sion constructed from the 1D function via
g
(
k
T
r
). We can now perform a variable
substitution to the effect that
r
=
U
T
r
, where
r
=(
x
,y
)
T
. The differential term
dxdy
can be replaced by
dx
dy
without further consideration because
U
is orthog-
onal and causes the Jacobian determinant to become 1. Because of the orthogonality
of
U
, we also have
r
=
Ur
. Consequently, the Fourier transform reduces to
F
(
ω
x
,ω
y
)=
g
(
x
) exp(
T
Ur
)
dx
dy
−
i
ω
=
g
(
x
) exp(
T
[
k
,
u
]
r
)
dx
dy
−
i
ω
=
g
(
x
) exp(
ω
T
k
,
ω
T
u
]
r
)
dx
dy
−
i
[
=
T
k
x
)
dx
g
(
x
) exp(
−i
ω
T
u
y
)
dy
exp(
−i
ω
=
G
(
k
T
)
δ
(
u
T
ω
ω
)
(10.103)
Proof of theorem 10.2 We will first express the error or the inertia given by Eq.
(10.17) by means of complex numbers instead of vectors. To be precise, the distance
defined in Eq. (10.16) can be computed by replacing the scalar product in the 2D
plane with complex multiplications via
x
2
)=
1
x
T
1
x
1
x
1
x
1
x
2
)
∗
]
x
2
=
(
2
[
x
2
+(
(10.104)
where
·
is the operator that constructs a complex number from a real 2D vector:
x
=(
x, y
)
T
,
x
=
x
+
iy,
when
(10.105)
and
=1and
S
is positive semidefinite ,
k
maximizing
k
T
Sk
is the eigenvector belonging to the
largest eigenvalue of
S
. By using Eqs. (10.23) and Eq. (10.29) we obtain:
(
x
) equals to
x
, the real part of the complex number
x
. Since
k
