# 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

where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
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Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
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write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
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