Image Processing Reference
In-Depth Information
is the only truly complex-valued coefficient, where its real and imaginary parts always
possess identical moduli. All other coefficients are either purely imaginary or real-valued.
Obviously, all frequency domain symmetry properties, including also those related to strict
complementarity, are retained in the respective frequency-shifted versions, cf. Subsection 2.3.1
and [Göckler & Damjanovic (2006a)].
3.2 Linear-phase directional separation filter
We start with the presentation of the FDMUX approach [Göckler & Groth (2004);
Göckler & Eyssele (1992)] followed by the investigation of a synergetic combination of two
COHBF [Göckler (1996a;c); Göckler & Damjanovic (2006a)].
3.2.1 FDMUX approach
Using time-domain convolution, the
I
4 potentially required complex output signals,
decimated by 2 and related to the channel indices
o
=
∈{
0, 1, 2, 3
}
, are obtained as follows:
N
1
κ
=
0
x
(
2
m
−
κ
)
h
o
(
κ
−
−
N
−
1
y
o
(
mT
d
)
:
=
y
o
(
m
)=
)
,
(43)
2
where the complex impulse responses of channels
o
are introduced in causal (realizable) form.
Replacing the complex impulse responses with the respective modulation forms (39), and
setting the constant to
a
=(
2
o
+
1
)(
N
−
1
)
/2, we get:
N
1
κ
=
0
x
(
2
m
−
κ
)
h
(
κ
−
−
N
−
1
e
j
4
κ
(
2
o
+
1
)
,
y
o
(
m
)=
)
(44)
2
[
−
(
−
)
]
where
h
represents the real HBF prototype (9) in causal form. Next, in order
to introduce an
I
-component polyphase decomposition for efficient decimation, we split the
convolution index
k
N
1
/2
κ
into two indices:
κ
=
+
=
+
rI
p
4
r
p
,
(45)
=
−
=
=
(
−
)
=
(
−
)
where
p
0, 1, 2,
I
1
3and
r
0,1,...,
N
1
/
I
N
1
/4
.Asaresul ,it
follows from (44):
p
=
0
N
−
1
3
4
e
j
4
(
4
r
+
p
)(
2
o
+
1
)
.
r
=
0
N
−
1
)
·
y
o
(
)=
(
−
−
)
(
+
−
(46)
m
x
2
m
4
r
p
h
4
r
p
2
Rearranging the exponent of the exponential term according to
4
(
4
r
+
p
)(
2
o
+
1
)=
2
π
ro
+
p
4
+
2
4
op
, (46) can compactly be rewritten as [Oppenheim & Schafer (1989)]:
+
π
r
3
p
=
0
v
p
(
m
)
·
e
j
2
4
op
y
o
(
m
)=
=
4
·
IDFT
4
{
v
p
(
m
)
}
,
(47)
where the quantity
N
−
1
4
N
−
1
r
e
j
p
4
r
=
0
v
p
(
m
)=
x
(
2
m
−
4
r
−
p
)
h
(
4
r
+
p
−
)(
−
1
)
(48)
2