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

 Page 3 / 13

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

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years? Kala Reply lim x to infinity e^1-e^-1/log(1+x) given eccentricity and a point find the equiation Moses Reply 12, 17, 22.... 25th term Alexandra Reply 12, 17, 22.... 25th term Akash College algebra is really hard? Shirleen Reply Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table. Carole I'm 13 and I understand it great AJ I am 1 year old but I can do it! 1+1=2 proof very hard for me though. Atone hi Adu Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily. Vedant find the 15th term of the geometric sequince whose first is 18 and last term of 387 Jerwin Reply I know this work salma The given of f(x=x-2. then what is the value of this f(3) 5f(x+1) virgelyn Reply hmm well what is the answer Abhi If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10 Augustine how do they get the third part x = (32)5/4 kinnecy Reply make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be AJ how Sheref can someone help me with some logarithmic and exponential equations. Jeffrey Reply sure. what is your question? ninjadapaul 20/(×-6^2) Salomon okay, so you have 6 raised to the power of 2. what is that part of your answer ninjadapaul I don't understand what the A with approx sign and the boxed x mean ninjadapaul it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared Salomon I'm not sure why it wrote it the other way Salomon I got X =-6 Salomon ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6 ninjadapaul oops. ignore that. ninjadapaul so you not have an equal sign anywhere in the original equation? ninjadapaul hmm Abhi is it a question of log Abhi 🤔. Abhi I rally confuse this number And equations too I need exactly help salma But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends salma Commplementary angles Idrissa Reply hello Sherica im all ears I need to learn Sherica right! what he said ⤴⤴⤴ Tamia hii Uday hi salma hi Ayuba Hello opoku hi Ali greetings from Iran Ali salut. from Algeria Bach hi Nharnhar A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place. Kimberly Reply Jeannette has$5 and \$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Got questions? Join the online conversation and get instant answers!