Daubechies wavelet filters utilize all the remaining
N2-1
N
2
1
degrees of freedom to get the maximum possible number of
vanishing wavelet moments. In this case,
K=N2
K
N
2
, where NN is the length
of the filter.
Hz=1+z-12KQz
H
z
1
z
2
K
Q
z
Now since we know
K=N2
K
N
2
, we only need to get
Qz
Q
z
under the unitariness constraint to decide
Hz
H
z
, and equivalently
hn
h
n
.
Hω=1+ⅇⅈ2KQω
H
ω
1
ω
2
K
Q
ω
satisfies
|Hω|2+|Hω+π|2=2
H
ω
2
H
ω
2
2
If and only if
|Qω|2
Q
ω
2
can be written as
|Qω|2=Psin2ω2
Q
ω
2
P
ω
2
2
where
Py=∑k=0K-1K-1+kkyk+yKR12-y
P
y
k
0
K
1
K
1
k
k
y
k
y
K
R
1
2
y
and
Ry
R
y
is an odd polynomial chosen so that
Py≥0
P
y
0
for
0≤y≤1
0
y
1
.
Since the length of
hn
h
n
is NN,
Hz
H
z
is an
N-1
N
1
degree polynomial of
zz. For
K=N2
K
N
2
,
Qz
Q
z
is an
N2-1
N
2
1
degree of polynomial in
zz, and
|Qz|2=Psin2ω2
Q
z
2
P
ω
2
2
is an
N-2
N
2
degree polynomial in
zz. Also we have
sin2ω2=121-cosω121-12z+z-1
ω
2
2
1
2
1
ω
1
2
1
1
2
z
z
So
Psin2ω2=∑k=0K-1K-1+kk12-14z+z-1k+12-14z+z-1KR14z+z-1
P
ω
2
2
k
0
K
1
K
1
k
k
1
2
1
4
z
z
k
1
2
1
4
z
z
K
R
1
4
z
z
The first item, when
K=N2
K
N
2
, is a polynomial of degree
N-2
N
2
in zz, so
R=0
R
0
. If
K<N2
K
N
2
,
PY
P
Y
must have higher degree and
R≠0
R
0
in that case.
So for Daubechies filters we have
|Qz|2=∑k=0N2-1N2-1+kk12-14z+z-1k
Q
z
2
k
0
N
2
1
N
2
1
k
k
1
2
1
4
z
z
k
We can get
Qz
Q
z
using spectral factorization, where there is a
degree of freedom to choose which roots to use for
factorization. Note that
|Qz|2=QzQz-1
Q
z
2
Q
z
Q
z
, so the roots are in pairs of reciprocals. We can
use a minimal phase factorization (choose roots within the
unit circle), or a maximal phase factorization (choose roots
outside of the unit circle), or a mixed solution.
For
N=8
N
8
,
QzQz-1=∑k=03K+3k12-14z+z-1k=-516z3+52z2-1318z+13-1318z-1+52z-2-516z-3
Q
z
Q
z
k
0
3
K
3
k
1
2
1
4
z
z
k
5
16
z
3
5
2
z
2
131
8
z
13
131
8
z
5
2
z
2
5
16
z
3
(1)
Solve
QzQz-1=0
Q
z
Q
z
0
, we get the roots
z=0.32890.2841+0.2432ⅈ2.0311+1.7390ⅈ3.0407
z
0.3289
0.28410.2432
2.03111.7390
3.0407
. We will need
33
roots to factorize
Qz
Q
z
. Note that these roots are in pairs of
reciprocals, and since the coefficients of
Qz
Q
z
are real numbers, the roots are either real
numbers or in conjugate pairs (see
Figure 1).
The parts of Figure 2 are
described below.
-
Minimal phase factorization: Take
Qz=z-0.3289z-0.2841+0.2432ⅈz-0.2841+0.2432ⅈ
Q
z
z
0.3289
z
0.28410.2432
z
0.28410.2432
, after normalizing to
2
2
, we get
h=0.23040.71480.6309-0.0280-0.18700.03080.0329-0.0106
h
0.2304
0.7148
0.6309
-0.0280
-0.1870
0.0308
0.0329
-0.0106
-
Maximal phase factorization: Take
Qz=z-3.0407z-2.0311+1.739ⅈz-0.2841+1.739ⅈ
Q
z
z
3.0407
z
2.03111.739
z
0.28411.739
,
after normalizing to
2
2
, we get
h=-0.01060.03290.0308-0.1870-0.02800.63090.71480.2304
h
-0.0106
0.0329
0.0308
-0.1870
-0.0280
0.6309
0.7148
0.2304
-
Mixed phase factorization 1: Take
Qz=z-0.3289z-2.0311+1.739ⅈz-2.0311+1.739ⅈ
Q
z
z
0.3289
z
2.03111.739
z
2.03111.739
,
after normalizing to
2
2
, we get
h=0.0322-0.0126-0.09920.29790.80370.4976-0.0296-0.0758
h
0.0322
-0.0126
-0.0992
0.2979
0.8037
0.4976
-0.0296
-0.0758
-
Mixed phase factorization 2: Take
Qz=z-3.0407z-0.2841+0.2432ⅈz-0.2841+0.2432ⅈ
Q
z
z
3.0407
z
0.28410.2432
z
0.28410.2432
, after normalizing to
2
2
, we get
h=-0.0758-0.02960.49760.80370.2979-0.0992-0.01260.0322
h
-0.0758
-0.0296
0.4976
0.8037
0.2979
-0.0992
-0.0126
0.0322
