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Therefore the frequency responses are related by G 0 ω T s H 0 ω 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 2 K 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 1 2 Z so β 1 2 and, from , the factors of R z are R z α z 1 1 α z 1 α 2 where α 3 2 . Also 1 Z 2 1 2 Z 1 2 1 2 1 z 2 Hence H 0 z 1 2 1 α 2 1 z 2 1 α z 0.4830 0.8365 z 0.2241 z -2 0.1294 z -3 and H 1 z z -3 G 0 z z -3 H 0 z 0.1294 0.2241 z 0.8365 z -2 0.4830 z -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.

Impulse responses and frequency responses of the 4-level tree of Daubechies 4-tap filters.

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 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 a Z b Z 2 1 Z 1 c Z
This is a 5 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 1 2 c 2 2 1 c 2
and
b c 1 2 c 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 ( ω T s varies from 0 to ).

It turns out that c 2 7 gives good similarity and when substituted into , and gives:

P t Z 1 50 50 41 Z 15 Z 2 6 Z 3 1 7 7 5 Z 2 Z 2
We get G 0 z and H 0 z by substituting Z 1 2 z z into these two polynomial factors. This results in 5,7-tap filters whose wavelets and frequency responses areshown in .

Impulse responses and frequency responses of the 4-level tree of near-balanced 5,7-tap filters.

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.

Impulse responses and frequency responses of the 4-level tree of near-balanced 7,5-tap filters.

Smoother wavelets

In all of the above designs we have used the substitution Z 1 2 z z . 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 p z 3 1 2 p z z p z -3
Z then becomes the following function of frequency when z ω T s :
Z 1 2 p ω T s 2 p 3 ω T s
A high degree of flatness (with some ripple) is achieved near ω T s 0 and , when p 3 32 . This is equivalent to more zeros near z -1 for each ( Z 1 ) factor than when Z 1 2 z z is used.

The 2 nd order factor in P t Z now produces terms from z 6 to z -6 and the 3 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 ( 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.

Impulse responses and frequency responses of the 4-level tree of near-balanced 13,19-tap filters.

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

Impulse responses and frequency responses of the 4-level tree of near-balanced 19,13-tap 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.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Image coding. OpenStax CNX. Jan 22, 2004 Download for free at http://cnx.org/content/col10206/1.3
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