4.4 Good filters / wavelets  (Page 3/3)

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Therefore the frequency responses are related by ${G}_{0}(e^{i\omega {T}_{s}})={H}_{0}(e^{-(i\omega {T}_{s})})$ .

Hence the magnitudes of the frequency responses are the same, and their phases are opposite. It may be shown that this issufficient to obtain orthogonal wavelets, but unfortunately the separate filters are no longer zero (orlinear) phase. (Linear phase is zero phase with an arbitrary delay $z^{-k}$ .)

Daubechies wavelets may be generated in this way, with the added constraint that the maximum number of zeros of ${P}_{t}(Z)$ are placed at $Z=-1$ (producing pairs of zeros of $P(z)$ at $z=-1$ ), consistent with terms in even powers of $Z$ being zero.

If ${P}_{t}(Z)$ is of order $2K-1$ , then it may have $K$ zeros at $Z=-1$ such that

${P}_{t}(Z)=(1+Z)^{K}{R}_{t}(Z)$
where ${R}_{t}(Z)$ is of order $K-1$ and its $K-1$ roots may be chosen such that terms of ${P}_{t}(Z)$ in the $K-1$ even powers of $Z$ are zero.

is the $K=2$ solution to . Therefore, ${R}_{t}(Z)=1-\frac{1}{2}Z$ so $\beta =-\left(\frac{1}{2}\right)$ and, from , the factors of $R(z)$ are $R(z)=\frac{(\alpha z+1)(1+\alpha z^{(-1)})}{1+\alpha ^{2}}$ where $\alpha =\sqrt{3}-2$ . Also $(1+Z)^{2}=\frac{1}{2}(Z+1)^{2}\frac{1}{2}(1+z^{(-1)})^{2}$ Hence ${H}_{0}(z)=\frac{1}{2\sqrt{1+\alpha ^{2}}}(1+z^{(-1)})^{2}(1+\alpha z^{(-1)})=0.4830+0.8365z^{(-1)}+0.2241z^{-2}-0.1294z^{-3}$ and ${H}_{1}(z)=z^{-3}{G}_{0}(-z)=z^{-3}{H}_{0}(-z^{(-1)})=0.1294+0.2241z^{(-1)}-0.8365z^{-2}+0.4830z^{-3}$ The wavelets and frequency responses for these filters are shown in . It is clear that the wavelets and scaling function are no longer linearphase and are less smooth than those for the LeGall 3,5-tap filters. The frequency responses also show worsesidelobes. The ${G}_{0}$ , ${G}_{1}$ filters give the time reverse of these wavelets and identical frequency responses.

Higher order Daubechies filters achieve smooth wavelets but they still suffer from non-linear phase. This tends to resultin more visible coding artefacts than linear phase filters, which distribute any artefacts equally on either side of sharpedges in the image.

Linear phase filters also allow an elegant technique, known as symmetric extension, to be used at the outer edges of images,where wavelet filters would otherwise require the size of the transformed image to be increased to allow for convolutionwith the filters. Symmetric extension assumes that the image is reflected by mirrors at each edge, so that an infinitelytessellated plane of reflected images is generated. Reflections avoid unwanted edge discontinuities. If the filters are linearphase, then the DWT coefficients also form reflections and no increase in size of the transformed image is necessary toaccomodate convolution effects.

Filters with linear phase and nearly balanced frequency responses:

To ensure that the filters ${H}_{0}$ , ${H}_{1}$ and ${G}_{0}$ , ${G}_{1}$ are linear phase, the factors in $Z$ must be allocated to ${H}_{0}$ or ${G}_{0}$ as a whole and not be split, as was done for the Daubechies filters. In this way the symmetry between $z$ and $z^{(-1)}$ is preserved in all filters.

Perfect balance of frequency responses between ${H}_{0}$ and ${G}_{0}$ is then not possible, if PR is preserved, but we have found a factorisation of ${P}_{t}(Z)$ which achieves near balance of the responses.

This is:

${P}_{t}(Z)=(1+Z)(1+aZ+bZ^{2})(1+Z)(1+cZ)$
This is a ${5}^{\mathrm{th}}$ order polynomial, and if the terms in $Z^{2}$ and $Z^{4}$ are to be zero, there are two constraints on the 3 unknowns 
a b c
so that one of them (say $c$ ) may be regarded as a free parameter. These constraints require that:
$a=-\left(\frac{(1+2c)^{2}}{2(1+c)^{2}}\right)$
and
$b=\frac{c(1+2c)}{2(1+c)^{2}}$
$c$ may then be adjusted to give maximum similarity between the left and right pairs of factorsin as $Z$ varies from 1 to -1 ( $\omega {T}_{s}$ varies from 0 to $\pi$ ).

It turns out that $c=-\left(\frac{2}{7}\right)$ gives good similarity and when substituted into , and gives:

${P}_{t}(Z)=\frac{1}{50}(50+41Z-15Z^{2}-6Z^{3})\frac{1}{7}(7+5Z-2Z^{2})$
We get ${G}_{0}(z)$ and ${H}_{0}(z)$ by substituting $Z=\frac{1}{2}(z+z^{(-1)})$ into these two polynomial factors. This results in 5,7-tap filters whose wavelets and frequency responses areshown in .

The near balance of the responses may be seen from which shows the alternative 7,5-tap versions (i.e. with $H$ and $G$ swapped). It is quite difficult to spot the minor differences between these figures.

Smoother wavelets

In all of the above designs we have used the substitution $Z=\frac{1}{2}(z+z^{(-1)})$ . However other substitutions may be used to create improved wavelets. To preserve PR, the substitution shouldcontain only odd powers of $z$ (so that odd powers of $Z$ should produce only odd powers of $z$ ), and to produce zero phase, the coefficients of thesubstitution should be symmetric about $z^{0}$ .

A substitution, which can give much greater flatness near $z=±(1)$ while still satisfying $Z=±(1)$ when $z=±(1)$ , is:

$Z=pz^{3}-\frac{1}{2}(z+z^{(-1)})+pz^{-3}$
$Z$ then becomes the following function of frequency when $z=e^{i\omega {T}_{s}}$ :
$Z=1\cos (\omega {T}_{s})+2p\cos (3\omega {T}_{s})$
A high degree of flatness (with some ripple) is achieved near $\omega {T}_{s}=0$ and $\pi$ , when $p=-\left(\frac{3}{32}\right)$ . This is equivalent to more zeros near $z=-1$ for each ( $Z+1$ ) factor than when $Z=\frac{1}{2}(z+z^{(-1)})$ is used.

The ${2}^{\mathrm{nd}}$ order factor in ${P}_{t}(Z)$ now produces terms from ${z}^{6}$ to ${z}^{-6}$ and the ${3}^{\mathrm{rd}}$ order factor produces terms from ${z}^{9}$ to ${z}^{-9}$ . Hence the filters become 13 and 19 tap filters, although 2 taps of each are zero and the outer two taps of the19-tap filter are very small ( $\approx (10^{-4})$ ).

shows the wavelets and frequency responses of the 13,19-tap filters, obtained bysubstituting into . Note the smoother wavelets and scaling function and the much lower sidelobes in the frequencyresponses from these higher order filters.

demonstrates that the near balanced properties of are preserved in the high order filters.

There are many other types of wavelets with varying features and complexities, but we have found the examples given to benear optimum for image compression.

do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
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it is a goid question and i want to know the answer as well
Maciej
Abigail
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
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Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
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Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
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Cied
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what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
I'm interested in nanotube
Uday
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preparation of nanomaterial
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Stotaw
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Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
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Azam
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Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
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Azam
Hello
Uday
I'm interested in Nanotube
Uday
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Prasenjit
how did you get the value of 2000N.What calculations are needed to arrive at it
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