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# Scaling Filter Sufficient Conditions

Module by: Jeremy Pearce. E-mail the author

Summary: In order to find a solution to the basic reqursion equations, the scaling filter must satisfy a set of sufficient conditions. Certain conditions on the scaling filter will directly result in a wavelet expansion system with certain properties.

## Basic Recursion Equations

n,nZ:φt=nnhn2φ2tn n n φ t n n h n 2 φ 2 t n
(1)
Φω=k=1(12Hω2k)Φ0 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.

### Sufficient Condition 1

If the following conditions on hn h n hold, then φt φ t exists.

1. nnhn=2 n n h n 2
2. hn h n has finite support or decays fast enough so that nn|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.

#### Example 1

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 nnhn=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.

### Sufficient Condition 2

If the following conditions on hn h n hold, then φt φ t exists.

1. nnh2n=nnh2n+1=12 n n h 2 n n n h 2 n 1 1 2
2. hn h n has finite support or decays fast enough so that nn|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
kkφtk=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 .

#### Example 2

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 nnh2n=nnh2n+1=12 n n h 2 n n n h 2 n 1 1 2 and therefore also nnhn=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.

### Sufficient Condition 3

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 .

1. nnhn=2 n n h n 2
2. nnhnhn2k=δk n n h n h n 2 k δ k
3. hn h n has finite support or decays fast enough so that nn|hn|1+|n|ε< n n h n 1 n ε for some ε> ε .

### Sufficient Condition 4

If the following conditions on hn h n hold, then φt φ t exists and generates an orthonormal basis in L2 L 2 .

1. hn h n has compact support.
2. nnhn=2 n n h n 2
3. nnhnhn2k=δk n n h n h n 2 k δ k
4. Hω0 H ω 0 for π3ωπ3 3 ω 3

### Sufficient Condition 5

If the following conditions on hn h n hold, then φt φ t has finite support.

1. hn h n has finite support.
2. φtL1 φ t L 1

#### Example 3

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 N21 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.

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