# 0.6 Regularity, moments, and wavelet system design  (Page 3/13)

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This is a very powerful result [link] , [link] . It not only ties the number of zero moments to the regularity but also to the degree ofpolynomials that can be exactly represented by a sum of weighted and shifted scaling functions.

Theorem 21 If $\psi \left(t\right)$ is $K$ -times differentiable and decays fast enough, then the first $K-1$ wavelet moments vanish [link] ; i.e.,

$\left|\frac{{d}^{k}}{d{t}^{k}},\psi ,\left(t\right)\right|<\infty ,\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}0\le k\le K$

implies

${m}_{1}\left(k\right)\phantom{\rule{0.277778em}{0ex}}=\phantom{\rule{0.277778em}{0ex}}0.\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}0\le k\le K$

Unfortunately, the converse of this theorem is not true. However, we can relate the differentiability of $\psi \left(t\right)$ to vanishing moments by

Theorem 22 There exists a finite positive integer $L$ such that if ${m}_{1}\left(k\right)=0$ for $0\le k\le K-1$ then

$\left|\frac{{d}^{P}}{d{t}^{P}},\psi ,\left(t\right)\right|<\infty$

for $L\phantom{\rule{0.166667em}{0ex}}P>K$ .

For example, a three-times differentiable $\psi \left(t\right)$ must have three vanishing moments, but three vanishing moments results in only one-timedifferentiability.

These theorems show the close relationship among the moments of ${h}_{1}\left(n\right)$ , $\psi \left(t\right)$ , the smoothness of $H\left(\omega \right)$ at $\omega =0$ and $\pi$ and to polynomial representation. It also states a loose relationship with thesmoothness of $\phi \left(t\right)$ and $\psi \left(t\right)$ themselves.

## Daubechies' method for zero wavelet moment design

Daubechies used the above relationships to show the following important result which constructs orthonormal wavelets with compact support with themaximum number of vanishing moments.

Theorem 23 The discrete-time Fourier transform of $h\left(n\right)$ having $K$ zeros at $\omega =\pi$ of the form

$H\left(\omega \right)={\left(\frac{1+{e}^{i\omega }}{2}\right)}^{K}\phantom{\rule{0.277778em}{0ex}}L\left(\omega \right)$

satisfies

${|H\left(\omega \right)|}^{2}+{|H\left(\omega +\pi \right)|}^{2}=2$

if and only if $L\left(\omega \right)={|L\left(\omega \right)|}^{2}$ can be written

$L\left(\omega \right)=P\left({sin}^{2}\left(\omega /2\right)\right)$

with $K\le N/2$ where

$P\left(y\right)=\sum _{k=0}^{K-1}\phantom{\rule{0.166667em}{0ex}}\left(\begin{array}{c}K-1+k\\ k\end{array}\right)\phantom{\rule{0.277778em}{0ex}}{y}^{k}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}+\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{y}^{K}\phantom{\rule{0.166667em}{0ex}}R\left(\frac{1}{2}-y\right)$

and $R\left(y\right)$ is an odd polynomial chosen so that $P\left(y\right)\ge 0$ for $0\le y\le 1$ .

If $R=0$ , the length $N$ is minimum for a given regularity $K=N/2$ . If $N>2\phantom{\rule{0.166667em}{0ex}}K$ , the second term containing $R$ has terms with higher powers of $y$ whose coefficients can be used for purposes other than regularity.

The proof and a discussion are found in Daubechies [link] , [link] . Recall from [link] that $H\left(\omega \right)$ always has at least one zero at $\omega =\pi$ as a result of $h\left(n\right)$ satisfying the necessary conditions for $\phi \left(t\right)$ to exist and have orthogonal integer translates. We are now placing restrictions on $h\left(n\right)$ to have as high an order zero at $\omega =\pi$ as possible. That accounts for the form of [link] . Requiring orthogonality in [link] gives [link] .

Because the frequency domain requirements in [link] are in terms of the square of the magnitudes of the frequency response, spectralfactorization is used to determine $H\left(\omega \right)$ and therefore $h\left(n\right)$ from ${|H\left(\omega \right)|}^{2}$ . [link] becomes

${|H\left(\omega \right)|}^{2}={\left|\frac{1+{e}^{i\omega }}{2}\right|}^{2K}\phantom{\rule{0.277778em}{0ex}}{|L\left(\omega \right)|}^{2}.$

If we use the functional notation:

$M\left(\omega \right)={|H\left(\omega \right)|}^{2}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\text{and}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}L\left(\omega \right)={|L\left(\omega \right)|}^{2}$

$M\left(\omega \right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}|{cos}^{2}{\left(\omega /2\right)|}^{K}\phantom{\rule{0.166667em}{0ex}}L\left(\omega \right).$

Since $M\left(\omega \right)$ and $L\left(\omega \right)$ are even functions of $\omega$ they can be written as polynomials in $cos\left(\omega \right)$ and, using $cos\left(\omega \right)=1-2\phantom{\rule{0.166667em}{0ex}}{sin}^{2}\left(\omega /2\right)$ , [link] becomes

$\stackrel{˜}{M}\left({sin}^{2}\left(\omega /2\right)\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}{|{cos}^{2}\left(\omega /2\right)|}^{K}\phantom{\rule{0.166667em}{0ex}}P\left({sin}^{2}\left(\omega /2\right)\right)$

which, after a change of variables of $y={sin}^{2}\left(\omega /2\right)=1-{cos}^{2}\left(\omega /2\right)$ , becomes

$\stackrel{˜}{M}\left(y\right)={\left(1-y\right)}^{K}\phantom{\rule{0.166667em}{0ex}}P\left(y\right)$

where $P\left(y\right)$ is an $\left(N-K\right)$ order polynomial which must be positive since it will have to be factored to find $H\left(\omega \right)$ from [link] . This now gives [link] in terms of new variables which are easier to use.

In order that this description supports an orthonormal wavelet basis, we now require that [link] satisfies [link]

${|H\left(\omega \right)|}^{2}+{|H\left(\omega +\pi \right)|}^{2}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}2$

$M\left(\omega \right)+M\left(\omega +\pi \right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}{\left(1-y\right)}^{K}\phantom{\rule{0.166667em}{0ex}}P\left(y\right)+{y}^{K}\phantom{\rule{0.166667em}{0ex}}P\left(1-y\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}2.$

Equations of this form have an explicit solution found by using Bezout's theorem. The details are developed by Daubechies [link] . If all the $\left(N/2-1\right)$ degrees of freedom are used to set wavelet moments to zero, we set $K=N/2$ and the solution to [link] is given by

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