0.5 The scaling function and scaling coefficients, wavelet and  (Page 12/13)

Because of the iterative form of this algorithm, applying the same process over and over, it is sometimes called the cascade algorithm [link] , [link] .

Iterating the filter bank

An interesting method for calculating the scaling function also uses an iterative procedure which consists of the stages of the filterstructure of Chapter: Filter Banks and the Discrete Wavelet Transform which calculates wavelet expansions coefficients (DWT values) at one scale from those at another. A scalingfunction, wavelet expansion of a scaling function itself would be a single nonzero coefficient at the scale of $j=1$ . Passing this single coefficient through the synthesis filter structure of Figure: Two-Stage Two-Band Synthesis Tree and [link] would result in a fine scale output that for large $j$ would essentially be samples of the scaling function.

Successive approximation in the frequency domain

The Fourier transform of the scaling function defined in [link] is an important tool for studying and developing wavelet theory. It could beapproximately calculated by taking the DFT of the samples of $\phi \left(t\right)$ but a more direct approach is available using the infinite product in [link] . From this formulation we can see how the zeros of $H\left(\omega \right)$ determine the zeros of $\text{Φ}\left(\omega \right)$ . The existence conditions in Theorem 5 require $H\left(\pi \right)=0$ or, more generally, $H\left(\omega \right)=0$ for $\omega =(2k+1)\pi$ . Equation [link] gives the relation of these zeros of $H\left(\omega \right)$ to the zeros of $\text{Φ}\left(\omega \right)$ . For the index $k=1$ , $H(\omega /2)=0$ at $\omega =2(2k+1)\pi$ . For $k=2$ , $H(\omega /4)=0$ at $\omega =4(2k+1)\pi$ , $H(\omega /8)=0$

at $\omega =8(2k+1)\pi$ , etc. Because [link] is a product of stretched versions of $H\left(\omega \right)$ , these zeros of $H(\omega /2^j)$ are the zeros of the Fourier transform of $\phi \left(t\right)$ . Recall from  Theorem 15 that $H\left(\omega \right)$ has no zeros in $-\pi /3<\omega <\pi /3$ . All of this gives a picture of the shape of $\text{Φ}\left(\omega \right)$ and the location of its zeros. From an asymptotic analysis of $\text{Φ}\left(\omega \right)$ as $\omega \to \infty$ , one can study the smoothness of  $\phi \left(t\right)$ .

A Matlab program that calculates $\text{Φ}\left(\omega \right)$ using this frequency domain successive approximations approach suggested by [link] is given in Appendix C . Studying this program gives further insight into the structure of $\text{Φ}\left(\omega \right)$ . Rather than starting the calculations given in [link] for the index $j=1$ , they are started for the largest $j=J$ and worked backwards. If we calculate a length-N DFT consistent with $j=J$ using the FFT, then the samples of $H(\omega /2^j)$ for $j=J-1$ are simply every other sample of the case for $j=J$ . The next stage for $j=J-2$ is done likewise and if the original $N$ is chosen a power of two, the process in continued down to $j=1$ without calculating any more FFTs. This results in a very efficient algorithm. The details are in theprogram itself.

This algorithm is so efficient, using it plus an inverse FFT might be a good way to calculate $\phi \left(t\right)$ itself. Examples of the algorithm are illustrated in [link] where the transform is plotted for each step of the iteration.

The dyadic expansion of the scaling function

The next method for evaluating the scaling function uses a completely different approach. It starts by calculating the values of the scalingfunction at integer values of $t$ , which can be done exactly (within our ability to solve simultaneous linear equations). Consider the basicrecursion equation [link] for integer values of $t=k$