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
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