Image Processing Reference
In-Depth Information
x
2
+
y
2
2
σ
2
1
2
πσ
2
Q
(
D
x
+
iD
y
)
exp(
−
)=
(11.113)
N−
1
x
2
+
y
2
2
σ
2
1
2
πσ
2
a
n
(
−
x
σ
2
+
i
−
y
σ
2
)
n
exp(
=
−
)
n
=0
x
2
+
y
2
2
σ
2
=
Q
(
−
x
σ
2
+
i
−
y
σ
2
)
1
2
πσ
2
exp(
−
)
(11.114)
in which the linearity of derivative operators and Eq. (11.98) have been used.
Proof of theorem 11.4
We can ignore nearly all steps below to conclude that the theorem holds, provided
that the following conditions (to be proven) are granted: (i) convolution and sym-
metry derivatives are distributive, i.e., (
D
x
+
iD
y
)[
f
∗
g
]=[(
D
x
+
iD
y
)
f
]
∗
g
=
f
[(
D
x
+
iD
y
)
g
], and (ii) the Gaussian convolutions are variance-additive, i.e.,
Γ
{
0
,σ
1
}
∗
∗
Γ
{
0
,σ
2
}
=
Γ
{
0
,σ
1
+
σ
2
}
.
Convolution in the Fourier transform domain reduces to an ordinary multiplica-
tion:
Γ
{p
1
,σ
1
}
∗
Γ
{p
2
,σ
2
}
↔F
[
Γ
{p
1
,σ
1
}
]
[
Γ
{p
2
,σ
2
}
]
F
(11.115)
By using Eq. (11.101) we note that
[
Γ
{p
1
,σ
1
}
](
ω
x
,ω
y
)=2
π
−
p
1
σ
1
Γ
{p
1
,
σ
1
}
(
ω
x
,ω
y
)
i
σ
1
F
(11.116)
Due to Eq. (11.98), we can write:
[
Γ
{p
1
,σ
1
}
]
[
Γ
{p
2
,σ
2
}
]
F
·F
(11.117)
=2
π
−
p
1
σ
1
Γ
{p
1
,
σ
1
}
(
ω
x
,ω
y
)
2
π
−
p
2
σ
2
Γ
{p
2
,
σ
2
}
(
ω
x
,ω
y
)
i
σ
1
i
σ
2
·
exp
−
p
1
(
ω
x
+
ω
y
2
i
σ
1
1
2
πσ
1
=2
πσ
1
σ
1
)
p
1
(
ω
x
+
iω
y
)
p
1
σ
1
−
−
···
−
p
2
(
ω
x
+
ω
y
2
i
σ
2
1
2
πσ
2
2
πσ
2
σ
2
)
p
2
(
ω
x
+
iω
y
)
p
2
σ
2
×
−
exp
−
σ
2
)
p
2
(
ω
x
+
iω
y
)
p
1
+
p
2
exp
=
−
p
1
−
p
2
(
ω
x
+
ω
y
2
i
σ
1
i
σ
2
σ
1
)
p
1
(
−
−
−
1
σ
1
+
σ
2
Here, we have used the definition of
Γ
{p,
σ
}
to obtain the Gaussian terms. Cancelling
the redundant terms allows us to write the product as:
