∀n,n∈ℤ:φt=∑nhn2φ2t-n
n
n
φ
t
n
n
h
n
2
φ
2
t
n
(1)
Φω=∏k=1∞12Hω2kΦ0
Φ
ω
k
1
1
2
H
ω
2
k
Φ
0
(2)
From the basic recursion equations, it is apparent that
hn
h
n
completely characterizes the scaling function
φt
φ
t
. In order for a solution for
φt
φ
t
to exist,
hn
h
n
must satisfy a set of sufficient conditions.
If the following conditions on
hn
h
n
hold, then
φt
φ
t
exists.
-
∑nhn=2
n
n
h
n
2
-
hn
h
n
has finite support or decays fast enough so
that
∑n|hn|1+|n|ε<∞
n
n
h
n
1
n
ε
for some
ε>∞
ε
.
This is the weakest possible condition, and therefore does
not necessarily provide a useful expansion system. The
resulting scaling function can be very poorly behaved and
not support a multiresolution analysis.
Let
hn=0.08001.07330.03820.1169-0.01940.1252
h
n
0.0800
1.0733
0.0382
0.1169
-0.0194
0.1252
, which satifies
∑nhn=2
n
n
h
n
2
and none of the other sufficient conditions.
The magnitude of the frequency response of filter
hn
h
n
is shown in Figure 1.
The filter is a poorly behaved filter in that it is not
even a low-pass filter and therefore
φt
φ
t
can not be calculated and does not support a
multiresolution analysis.
If the following conditions on
hn
h
n
hold, then
φt
φ
t
exists.
-
∑nh2n=∑nh2n+1=12
n
n
h
2
n
n
n
h
2
n
1
1
2
-
hn
h
n
has finite support or decays fast enough so
that
∑n|hn|1+|n|ε<∞
n
n
h
n
1
n
ε
for some
ε>∞
ε
.
In addition to having a solution for
φt
φ
t
, the following sum holds
∑kφt-k=1
k
k
φ
t
k
1
(3)
By forcing the sum of the odd coefficients be equal to the
sum of the even coefficients, it guarantees that
Hπ=0
H
0
, which means
hn
h
n
is a low-pass filter. This condition is much
stronger than sufficient condition 1 and is called the
Fundamental Condition. From the product
formula, Equation 2, requiring
Hπ=0
H
0
gives a better behaved
Φω
Φ
ω
.
Let
hn=0.35260.13280.30180.37020.05270.2041
h
n
0.3526
0.1328
0.3018
0.3702
0.0527
0.2041
, which satifies
∑nh2n=∑nh2n+1=12
n
n
h
2
n
n
n
h
2
n
1
1
2
and therefore also
∑nhn=2
n
n
h
n
2
. The magnitude of the frequency response of
filter
hn
h
n
is shown in Figure 2.
It yields a better conditioned filter than in Figure 1. The scaling function
φt
φ
t
can also be calculated and is shown. The
Fundamental Condition guarantees that a scaling filter can
be calculated and that an expansion system exists. It says
nothing about whether the expansion is orthogonal or even
a tight frame, which will be important in transforming to
and from the wavelet domain.
If the following conditions on
hn
h
n
hold, then
φt
φ
t
exists and generates a wavelet system that is a
tight frame in
L2
L
2
.
-
∑nhn=2
n
n
h
n
2
-
∑nhnhn-2k=δk
n
n
h
n
h
n
2
k
δ
k
-
hn
h
n
has finite support or decays fast enough so
that
∑n|hn|1+|n|ε<∞
n
n
h
n
1
n
ε
for some
ε>∞
ε
.
If the following conditions on
hn
h
n
hold, then
φt
φ
t
exists and generates an orthonormal basis in
L2
L
2
.
-
hn
h
n
has compact support.
-
∑nhn=2
n
n
h
n
2
-
∑nhnhn-2k=δk
n
n
h
n
h
n
2
k
δ
k
-
Hω≠0
H
ω
0
for
-π3≤ω≤π3
3
ω
3
If the following conditions on
hn
h
n
hold, then
φt
φ
t
has finite support.
-
hn
h
n
has finite support.
-
φt∈L1
φ
t
L
1
Let
hn=0.33270.80690.4599-0.1350-0.08540.0352
h
n
0.3327
0.8069
0.4599
-0.1350
-0.0854
0.0352
, which satifies Sufficient Conditions 1-5.
Figure 3 shows a plot of the
frequency response of the scaling filter
|Hω|
H
ω
as well as the scaling function
φt
φ
t
.
Hω
H
ω
is a well behaved low-pass filter.
φt
φ
t
is a well defined function. This wavelet system
is known as Daubechies 6 and is used in many
singal processing applications. The basis formed by this
system is orthonormal.
These conditions give a workable basis set with some useful
properties. A
hn
h
n
that satisfies Sufficient Conditions 1-5 will have
N2-1
N
2
1
degrees of freedom, where
NN is the length of filter
hn
h
n
. The remaining degrees of freedom will permit
more conditions to be placed on the scaling functions. Click
here for a Labview program that will allow you to use the remaining degrees of freedom to generate an orthogonal wavelet system.
