Basic Recursion Equations n n φ t n n h n 2 φ 2 t n Φ ω k 1 1 2 H ω 2 k Φ 0 From the basic recursion equations, it is apparent that h n completely characterizes the scaling function φ t . In order for a solution for φ t to exist, h n must satisfy a set of sufficient conditions.

Sufficient Condition 1 If the following conditions on h n hold, then φ t exists. n n h n 2 h n has finite support or decays fast enough so that n n h n 1 n ε for some 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 h n 0.0800 1.0733 0.0382 0.1169 -0.0194 0.1252 , which satifies n n h n 2 and none of the other sufficient conditions. The magnitude of the frequency response of filter h n is shown in . The filter is a poorly behaved filter in that it is not even a low-pass filter and therefore φ t can not be calculated and does not support a multiresolution analysis.

Frequency response of the scaling filter that satisfies Sufficient Condition 1.

Sufficient Condition 2 If the following conditions on h n hold, then φ t exists. n n h 2 n n n h 2 n 1 1 2 h n has finite support or decays fast enough so that n n h n 1 n ε for some ε . In addition to having a solution for φ t , the following sum holds k k φ t k 1 By forcing the sum of the odd coefficients be equal to the sum of the even coefficients, it guarantees that H 0 , which means 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, , requiring H 0 gives a better behaved Φ ω . Let h n 0.3526 0.1328 0.3018 0.3702 0.0527 0.2041 , which satifies n n h 2 n n n h 2 n 1 1 2 and therefore also n n h n 2 . The magnitude of the frequency response of filter h n is shown in . It yields a better conditioned filter than in . The scaling function 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.

Frequency response of the scaling filter that satisfies Sufficient Condition 1 and 2.

Sufficient Condition 3 If the following conditions on h n hold, then φ t exists and generates a wavelet system that is a tight frame in L 2 . n n h n 2 n n h n h n 2 k δ k h n has finite support or decays fast enough so that n n h n 1 n ε for some ε .

Sufficient Condition 4 If the following conditions on h n hold, then φ t exists and generates an orthonormal basis in L 2 . h n has compact support. n n h n 2 n n h n h n 2 k δ k H ω 0 for 3 ω 3

Sufficient Condition 5 If the following conditions on h n hold, then φ t has finite support. h n has finite support. φ t L 1 Let h n 0.3327 0.8069 0.4599 -0.1350 -0.0854 0.0352 , which satifies Sufficient Conditions 1-5. shows a plot of the frequency response of the scaling filter H ω as well as the scaling function φ t . H ω is a well behaved low-pass filter. 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.

Frequency response of the scaling filter that satisfies Sufficient Conditions 1-5. These conditions give a workable basis set with some useful properties. A h n that satisfies Sufficient Conditions 1-5 will have N 2 1 degrees of freedom, where N is the length of filter 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.