Image Processing Reference
In-Depth Information
1
4
π
2
1
4
π
2
k
T
Sk
=
k
T
(
T
f
)
k
d
x
=
(
k
T
f
)(
k
T
∇
f
j
)(
∇
∇
∇
f
)
d
x
(
1
4
π
2
(
k
∗
∇
f
))
2
d
x
=
Now, with ordinary algebraic manipulations applied to complex numbers, we
obtain
1
4
(
k
2
)
∗
(
∇
|
2
d
x
1
4
π
2
f
)
2
+[(
k
2
)
∗
(
∇
| ∇
k
T
Sk
=
f
)
2
]
∗
+2
f
1
4
π
2
1
|
2
d
x
+
1
2
| ∇
[(
k
2
)
∗
(
∇
f
)
2
]
d
x
=
f
2
(10.106)
where
∂x
+
i
∂
∂
∇
=
D
x
+
iD
y
=
(10.107)
∂y
By construction,
∇
f
and
k
are the complex interpretations of the real 2D vectors
f
and
k
. Since the first term of the sum in Eq. (10.106) is free of
k
, the complex
number
k
2
that maximizes Eq. (10.106) is the same as the one that maximizes the
second term:
∇
(
k
2
)
∗
f
)
2
d
x
1
2
(
∇
1
4
π
2
1
2
1
4
π
2
[(
k
2
)
∗
(
∇
f
)
2
]
d
x
=
(10.108)
1
2
[(
k
2
)
∗
I
20
]
=
(10.109)
) and
·
where we have used the fact that
are linear operators that commute with
each other, and
I
20
represents the complex scalar:
(
·
(
∇
f
)
2
d
x
=
(
ω
x
+
iω
y
)
2
|
1
4
π
2
|
2
d
I
20
=
F
ω
.
(10.110)
The choice of the subscript in
I
20
is justified because the second integral in Eq.
(10.110) is a complex moment.
Identifying the scalar product in Eq. (10.109) via Eq. (10.104) and remembering
that
|
k
2
|
=1, we interpret the expression geometrically. The vector that corresponds
to the complex scalar
I
20
is projected onto a line with the direction vector that cor-
responds to
k
2
. The projection, a real scalar, is given by Eq. (10.109), which is to be
maximized. Evidently, the direction
k
2
max
that maximizes this projection is the same
as the direction of
I
20
:
k
2
max
=
I
20
/
|
I
20
|
.
(10.111)
and the projection result is the (positive real) scalar
. Similarly, the complex
number
k
2
min
that minimizes Eq. (10.106) is given by the (negative real) scalar
−
k
2
max
yielding the negative scalar
|
I
20
|
. Calling the (positive real) scalar expres-
sion in the first term of Eq. (10.106) as
I
11
:
−|
I
20
|
