A two-channel analysis/synthesis filter bank has the following
form (Figure 1 and Figure 2).
Usually,
h
0
n
h
0
n
and
g
0
n
g
0
n
are lowpass filters, and
h
1
n
h
1
n
and
g
1
n
g
1
n
are highpass filters. In the discussion of Two-Channel filter banks, we looked at
orthonormal filter banks. An orthonormal filter bank satisfies
two properties:
-
The perfect reconstruction condition
yn=xn
y
n
x
n
(or the more general form
yn=xn-L
y
n
x
n
L
).
-
The synthesis filters are time-reversed versions of the
analysis filters (
g
i
n=
h
i
-n
g
i
n
h
i
n
or
g
i
n=
h
i
L-n
g
i
n
h
i
L
n
).
However, a filterbank does not have to be an orthonormal filter
bank to satisfy the perfect reconstruction property. Many of the
most useful filter banks satisfy the perfect reconstruction (PR)
condition (
yn=xn
y
n
x
n
or
yn=xn-L
y
n
x
n
L
) but are not orthonormal. For these PR filter banks,
the synthesis and analysis filters are not time-reversed versions
of one another.
We will derive the conditions that the four filters
H
0
z
H
0
z
,
H
1
z
H
1
z
,
G
0
z
G
0
z
, and
G
1
z
G
1
z
must satisfy, so that the filter bank has the perfect
reconstruction property
yn=xn-L
y
n
x
n
L
Equivalently,
Yz=z-LXz
Y
z
z
L
X
z
We will allow a delay of LL
samples. Using multirate identities, the
ZZ-transform of
ynyn
is given by the following expression.
Yz=12
G
0
z
H
0
zXz+
H
0
-zX-z+12
G
1
z
H
1
zXz+
H
1
-zX-z
Y
z
1
2
G
0
z
H
0
z
X
z
H
0
z
X
z
1
2
G
1
z
H
1
z
X
z
H
1
z
X
z
or
Yz=12
G
0
z
H
0
z+
G
1
z
H
1
zXz+12
G
0
z
H
0
-z+
G
1
z
H
1
-zX-z
Y
z
1
2
G
0
z
H
0
z
G
1
z
H
1
z
X
z
1
2
G
0
z
H
0
z
G
1
z
H
1
z
X
z
So for
xn=yn-L
x
n
y
n
L
we need
G
0
z
H
0
z+
G
1
z
H
1
z=2z-L
G
0
z
H
0
z
G
1
z
H
1
z
2
z
L
(1)
G
0
z
H
0
-z+
G
1
z
H
1
-z=0
G
0
z
H
0
z
G
1
z
H
1
z
0
(2)
Suppose we chose the highpass filters to be related to the
lowpass filters as:
H
1
z=
G
0
-z
H
1
z
G
0
z
(3)
G
1
z=-
H
0
-z
G
1
z
H
0
z
(4)
or equivalently as
h
1
n=-1n
g
0
n
h
1
n
1
n
g
0
n
g
1
n=--1n
h
0
n
g
1
n
1
n
h
0
n
Then condition
Equation 2 becomes
G
0
z
H
0
-z-
H
0
-z
G
0
z=0
G
0
z
H
0
z
H
0
z
G
0
z
0
but the left hand side is always zero, regardless of what the
filters
H
0
z
H
0
z
,
G
0
z
G
0
z
are. With the lowpass filters chosen as in (
Equation 3,
Equation 4), condition
Equation 1
becomes
H
0
z
G
0
z-
H
0
-z
G
0
-z=2z-L
H
0
z
G
0
z
H
0
z
G
0
z
2
z
L
Defining the product filter
P
0
z≔
H
0
z
G
0
z
≔
P
0
z
H
0
z
G
0
z
(5)
or equivalently
p
0
z=
h
0
z*
g
0
z
p
0
z
h
0
z
g
0
z
,
we get
P
0
z-
P
0
-z=2z-L
P
0
z
P
0
z
2
z
L
(6)
Note that the left hand side has only
odd
powers of
zz. Therefore
LL must be odd (otherwise
P
0zP
0z
can not satisfy this condition). If we take the inverse
ZZ-transform of
Equation 7 we get an equivalent condition in the time
domain:
p
0
n--1n
p
0
-n=2δn-L
p
0
n
1
n
p
0
n
2
δ
n
L
(7)
which in turn is equivalent to
p
0
2n+L=δn
p
0
2
n
L
δ
n
This means that
p
0
n
p
0
n
is a halfband filter centered at
n=L
n
L
. To see this clearly, we can set
nn to various values in
Equation 7:
n=L
p
0
L+
p
0
L=2δ0=2
p
0
L=1
n
L
p
0
L
p
0
L
2
δ
0
2
p
0
L
1
(Remember
LL is odd)
n=L+1
p
0
L+1-
p
0
L+1=2δ1=00=0
n
L
1
p
0
L
1
p
0
L
1
2
δ
1
0
0
0
n=L+2
p
0
L+2+
p
0
L+2=2δ2=0
p
0
L+2=0
n
L
2
p
0
L
2
p
0
L
2
2
δ
2
0
p
0
L
2
0
When
nn is even, we get from
Equation 7 that
0=0
0
0
, which does not tell us much. When
nn is odd, we get
p
0
n=0
p
0
n
0
except when
n=L
n
L
in which case
p
0
n=1
p
0
n
1
. That means
p
0
n
p
0
n
is a halfband filter centered at
n=L
n
L
.
So for the filter bank to have the perfect reconstruction
property, the product of the two lowpass filters
should be a halfband filter. Therefore, one approach
to the design of PR filter banks is to design a halfband filter
p
0
n
p
0
n
, and then factorize it as in
Equation 5. It is not required that
|
H
0
ⅇⅈω|=|
G
0
ⅇⅈω|
H
0
ω
G
0
ω
. This kind of factorization is more general than
spectral factorization. Its advantage is that it is a more
flexible factorization so that (non-orthonormal) solutions with
additional properties can be obtained. For example, it turns
out that if it is desired that the impulse responses of the
filters be symmetric, then the only FIR solution is the Haar
solution which lacks smoothness. By giving up orthonormality,
one can obtain a perfect reconstruction filter bank consisting
of filters
h
i
n
h
i
n
and
g
i
n
g
i
n
with symmetric impulse responses. We will see
examples.
For
xn=yn-
n
0
x
n
y
n
n
0
(PR with a delay of n
0n
0 samples), we need
G
0
z
H
0
z+
G
1
z
H
1
z=2z-
n
0
G
0
z
H
0
z
G
1
z
H
1
z
2
z
n
0
(8)
G
0
z
H
0
-z+
G
1
z
H
1
-z=0
G
0
z
H
0
z
G
1
z
H
1
z
0
(9)
Suppose we choose the synthesis filters to be related to the
analysis filters as:
G
0
z=z-d
H
1
-z
G
0
z
z
d
H
1
z
(10)
G
1
z=-z-d
H
0
-z
G
1
z
z
d
H
0
z
(11)
or equivalently as
g
0
n=-1d-1n
h
1
n-d
g
0
n
1
d
1
n
h
1
n
d
g
1
n=--1d-1n
h
0
n-d
g
1
n
1
d
1
n
h
0
n
d
Notice that this is more general than in the last section. If
d=0
d
0
then we get the derivation given in the last section.
Then condition
Equation 9 becomes
G
0
z
H
0
-z-z-d
H
0
-zzd
G
0
z=0
G
0
z
H
0
z
z
d
H
0
z
z
d
G
0
z
0
or
G
0
z
H
0
-z-
H
0
-z
G
0
z=0
G
0
z
H
0
z
H
0
z
G
0
z
0
But the left hand side is always zero because the terms
cancel, regardless of what the filters
H
0
z
H
0
z
,
G
0
z
G
0
z
are. With the synthesis filters chosen as in (
Equation 10,
Equation 11), condition
Equation 8
becomes
G
0
z
H
0
z-z-d
H
0
-z-zd
G
0
-z=2z-
n
0
G
0
z
H
0
z
z
d
H
0
z
z
d
G
0
z
2
z
n
0
or
H
0
z
G
0
z--1d
H
0
-z
G
0
-z=2z-
n
0
H
0
z
G
0
z
1
d
H
0
z
G
0
z
2
z
n
0
(12)
Defining the product filter
Pz≔
H
0
z
G
0
z
≔
P
z
H
0
z
G
0
z
(13)
or equivalently
pn=
h
0
n*
g
0
n
p
n
h
0
n
g
0
n
, we get
Pz--1dP-z=2z-
n
0
P
z
1
d
P
z
2
z
n
0
or equivalently
pn--1d-1npn=2δn-
n
0
p
n
1
d
1
n
p
n
2
δ
n
n
0
(14)
which in turn is equivalent to
p2n+
n
0
=δn
p
2
n
n
0
δ
n
(15)
provided dd
and
n0n0
have opposite parity. This equation says that
pn
p
n
is a halfband filter centered at
n=
n
0
n
n
0
.
If dd and
n
0
n
0
have the same parity, then there is no solution to Equation 14.
