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# Parameterization of Scaling Coefficients

Module by: Kileen Cheng. E-mail the author

Summary: Given the conditions our wavelet system needs to satisfy, how do we construct a valid scaling filter?

The properties of the scaling filter hn h n can be used as criteria in the design of our wavelet system. Given a scaling filter that satisifies our desired properties, the scaling and wavelet functions can then be calculated. (see calculating the scaling function)

## Compact FIR Filters

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:

1. The length of the scaling filter hn h n must be even.
2. nnhn=2 n n h n 2
3. nnhnhn2k=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 N21 N 2 1 remaining degrees of freedom that we can adjust to give a valid hn h n .

### Length-2 Scaling Filter

Requirements:

1. N=2 N 2
2. h0+h1=2 h 0 h 1 2
3. h20+h21=1 h 0 2 h 1 2 1

#### Note:

Degrees of freedom: N21=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.

### Length-4 Scaling Filter

Requirements:

1. N=4 N 4
2. h0+h1+h2+h3=2 h 0 h 1 h 2 h 3 2
3. 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

#### Note:

Degrees of freedom: N21=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=1cosα+sinα2×2 h 0 1 α α 2 2
(2)
h1=1+cosα+sinα2×2 h 1 1 α α 2 2
(3)
h2=1+cosαsinα2×2 h 2 1 α α 2 2
(4)
h3=1cosαsinα2×2 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+34×23+34×2334×2134×2 h D4 1 3 4 2 3 3 4 2 3 3 4 2 1 3 4 2
(6)
Note that for α0π23π2π α 0 2 3 2 , we get the length-2 Haar coefficients.

### Length-6 Scaling Filter

Requirements:

1. N=6 N 6
2. h0+h1+h2+h3+h4+h5=2 h 0 h 1 h 2 h 3 h 4 h 5 2
3. 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

#### Note:

Degrees of freedom: N21=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+sinα)(1cosβsinβ)+2sinβcosα4×2 h 0 1 α α 1 β β 2 β α 4 2
(7)
h1=(1+cosα+sinα)(1+cosβsinβ)+2sinβcosα4×2 h 1 1 α α 1 β β 2 β α 4 2
(8)
h2=1+cosαβ+sinαβ2×2 h 2 1 α β α β 2 2
(9)
h3=1+cosα(βsinα)β2×2 h 3 1 α β α β 2 2
(10)
h4=12h0h2 h 4 1 2 h 0 h 2
(11)
h5=12h1h3 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=arctan2(h20+h21)1+h2+h322 α 2 h 0 2 h 1 2 1 h 2 h 3 2 2
(13)
β=αarctanh2h3h2+h312 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 αβ π π α β , all possible hn h n can be generated.

More about Daubechies filters will be discussed later.

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