A very important class of wavelet systems are those with
compact support. These give rise to simple finite impulse
response (FIR) filters with conventient time-localization
properties.
Minimal requirements for these compact FIR filters are:
-
The length of the scaling filter
hn
h
n
must be even.
-
∑nhn=2
n
n
h
n
2
-
∑nhnhn-2k=δk
n
n
h
n
h
n
2
k
δ
k
After the
NN linear constraint of
(ii) and the
N2
N
2
bilinear constraints of (iii), there are
N2-1
N
2
1
remaining degrees of freedom that we can adjust to
give a valid
hn
h
n
.
Requirements:
-
N=2
N
2
-
h0+h1=2
h
0
h
1
2
-
h20+h21=1
h
0
2
h
1
2
1
Degrees of freedom:
N2-1=0!
N
2
1
0
After the minimal requirements for the scaling filter are
satisfied, there are no degrees of freedom left to give us
flexibility in the design of
hn
h
n
. There is only one set of possible coefficients:
h
D2
=h0h1=1212
h
D2
h
0
h
1
1
2
1
2
(1)
The length-2 scaling coefficient vector is also known as the
Haar or Daubechies-2 coefficients and will be discussed in further
length
later.
Requirements:
-
N=4
N
4
-
h0+h1+h2+h3=2
h
0
h
1
h
2
h
3
2
-
h20+h21+h23=1
h
0
2
h
1
2
h
3
2
1
and
h0h2+h1h3=0
h
0
h
2
h
1
h
3
0
Degrees of freedom:
N2-1=1
N
2
1
1
With a length-4 scaling vector, there is still one degree of
freedom remaining after the minimal requirements have been
satisfied. Letting
αα
represent this degree of freedom parameter, we can formulate
scaling filter coefficient equations such that:
h0=1-cosα+sinα22
h
0
1
α
α
2
2
(2)
h1=1+cosα+sinα22
h
1
1
α
α
2
2
(3)
h2=1+cosα-sinα22
h
2
1
α
α
2
2
(4)
h3=1-cosα-sinα22
h
3
1
α
α
2
2
(5)
We can adjust
αα to give
us a wavelet system with the desired properties. However,
most values of
αα do not
lead to a useful wavelet. The Daubechies wavelet with
filter length 4 arises from
α=π3
α
3
.
h
D4
=1+3423+3423-3421-342
h
D4
1
3
4
2
3
3
4
2
3
3
4
2
1
3
4
2
(6)
Note that for
α∈0π23π2π
α
0
2
3
2
, we get the length-2 Haar coefficients.
Requirements:
-
N=6
N
6
-
h0+h1+h2+h3+h4+h5=2
h
0
h
1
h
2
h
3
h
4
h
5
2
-
h20+h21+h23+h24+h25=1
h
0
2
h
1
2
h
3
2
h
4
2
h
5
2
1
and
h0h2+h1h3+h2h4+h3h5=0
h
0
h
2
h
1
h
3
h
2
h
4
h
3
h
5
0
Degrees of freedom:
N2-1=2
N
2
1
2
We now have two degrees of freedom. Defining our freedom
parameters as
αα and
ββ, our resulting
coefficient vector becomes:
h0=1+cosα+sinα1-cosβ-sinβ+2sinβcosα42
h
0
1
α
α
1
β
β
2
β
α
4
2
(7)
h1=1+cosα+sinα1+cosβ-sinβ+2sinβcosα42
h
1
1
α
α
1
β
β
2
β
α
4
2
(8)
h2=1+cosα-β+sinα-β22
h
2
1
α
β
α
β
2
2
(9)
h3=1+cosα-β-sinα-β22
h
3
1
α
β
α
β
2
2
(10)
h4=12-h0-h2
h
4
1
2
h
0
h
2
(11)
h5=12-h1-h3
h
5
1
2
h
1
h
3
(12)
The length-6 Daubechies wavelet is generated with
α=1.35980373244182
α
1.35980373244182
, and
β=-0.78210638474440
β
-0.78210638474440
. Note that for
α=β
α
β
, we get the length-2 Haar coefficients. Length-4
Daubechies coefficients are found if
α=π3
α
3
, and
β=0
β
0
. Realizing that the values of these parameters is
unusual, the formula used to calculate each parameter is
given below:
α=arctan2h20+h21-1+h2+h322
α
2
h
0
2
h
1
2
1
h
2
h
3
2
2
(13)
β=α-arctanh2-h3h2+h3-12
β
α
h
2
h
3
h
2
h
3
1
2
(14)
Given these equations, we can work both forwards and
backwards to determine the filter coefficients of a
particular system. Because
α∧β∈-ππ
α
β
, all possible
hn
h
n
can be generated.
More about Daubechies filters will be discussed later.
