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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
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.
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:
It turns out that $c=-\left(\frac{2}{7}\right)$ gives good similarity and when substituted into , and gives:
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.
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 $z=\pm (1)$ while still satisfying $Z=\pm (1)$ when $z=\pm (1)$ , is:
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.
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