Due to the minimum-phase spectral factorization, orthogonal PR-FIR
filterbanks will not have linear-phase analysis and
synthesis filters. Non-linear phase may be undesirable for
certain applications. "Bi-orthogonal" designs are closely
related to orthogonal designs, yet give linear-phase filters.
The analysis-filter design rules for the bi-orthogonal case
are
-
Fz
F
z
: zero-phase real-coefficient halfband such
that
Fz=∑n=-N-1N-1fnz-n
F
z
n
N
1
N
1
f
n
z
n
, where NN is even.
-
z-N-1Fz=
H
0
z
H
1
-z
z
N
1
F
z
H
0
z
H
1
z
It is straightforward to verify that these design choices
satisfy the FIR perfect reconstruction condition
detHz=cz-l
H
z
c
z
l
with
c=1
c
1
and
l=N-1
l
N
1
:
detHz=
H
0
z
H
1
-z-
H
0
-z
H
1
z=z-N-1Fz--1-N-1z-N-1F-z=z-N-1Fz+F-z=z-N-1
H
z
H
0
z
H
1
z
H
0
z
H
1
z
z
N
1
F
z
1
N
1
z
N
1
F
z
z
N
1
F
z
F
z
z
N
1
(1)
Furthermore, note that
z-N-1Fz
z
N
1
F
z
is causal with real coefficients, so that both
H
0
z
H
0
z
and
H
1
z
H
1
z
can be made causal with real coefficients. (This
was another PR-FIR requirement.) The choice
c=1
c
1
implies that the synthesis filters should obey
G
0
z=2
H
1
-z
G
0
z
2
H
1
z
G
1
z=-2
H
0
-z
G
1
z
-2
H
0
z
From the design choices above, we can see that bi-orthogonal
analysis filter design reduces to the factorization of a
causal halfband filter
z-N-1Fz
z
N
1
F
z
into
H
0
z
H
0
z
and
H
1
z
H
1
z
that have
both real
coefficients and linear-phase. Earlier we saw that
linear-phase corresponds to root symmetry across the unit
circle in the complex plane, and that real-coefficients
correspond to complex-conjugate root symmetry. Simultaneous
satisfaction of these two properties can be accomplished by
quadruples of roots. However, there are
special cases in which a root pair, or even a single root, can
simultaneously satisfy these properties. Examples are
illustrated in
Figure 1:
The design procedure for the analysis filters of a
bi-orthogonal perfect-reconstruction FIR filterbank is
summarized below:
-
Design a zero-phase real-coefficient filter
Fz=∑n=-N-1N-1fnz-n
F
z
n
N
1
N
1
f
n
z
n
where N is a positive even integer (via,
e.g., window designs, LS, or
equiripple).
-
Compute the roots of
Fz
F
z
and partition into a set of root groups
G
0
G
1
G
2
…
G
0
G
1
G
2
…
that have both complex-conjugate and
unit-circle symmetries. Thus a root group may have one of
the following forms:
G
i
=
a
i
a
i
¯1
a
i
1
a
i
¯
G
i
a
i
a
i
1
a
i
1
a
i
∀
a
i
,|
a
i
|=1:
G
i
=
a
i
a
i
¯
a
i
a
i
1
G
i
a
i
a
i
∀
a
i
,
a
i
∈ℝ:
G
i
=
a
i
1
a
i
a
i
a
i
G
i
a
i
1
a
i
∀
a
i
,
a
i
=±1:
G
i
=
a
i
a
i
a
i
±
1
G
i
a
i
Choose
a subset of root groups and construct
H
^
0
z
H
^
0
z
from those roots. Then construct
H
^
1
-z
H
^
1
z
from the roots in the remaining root groups. Finally,
construct
H
^
1
z
H
^
1
z
from
H
^
1
-z
H
^
1
z
by reversing the signs of odd-indexed coefficients.
-
H
^
0
z
H
^
0
z
and
H
^
1
z
H
^
1
z
are the desired analysis filters up to a
scaling. To take care of the scaling, first create
H
~
0
z=a
H
^
0
z
H
~
0
z
a
H
^
0
z
and
H
~
1
z=b
H
^
1
z
H
~
1
z
b
H
^
1
z
where aa and
bb are selected so that
∑n
h
~
0
n=1=∑n
h
~
1
n
n
h
~
0
n
1
n
h
~
1
n
. Then create
H
0
z=c
H
~
0
z
H
0
z
c
H
~
0
z
and
H
1
z=c
H
~
1
z
H
1
z
c
H
~
1
z
where cc is selected so that
the property
z-N-1Fz=
H
0
z
H
1
-z
z
N
1
F
z
H
0
z
H
1
z
is satisfied at DC (i.e.,
z=ⅇⅈ0=1
z
0
1
). In other words, find
cc so that
∑n
h
0
n∑m
h
1
n-1m=1
n
h
0
n
m
h
1
n
1
m
1
.
