In the above analysis, we used the factorisation of Pt ⁢Z Pt Z to give us H0 H0 and G0 G0 . This always gives unbalanced factors if terms of Pt Pt in even powers of ZZ are zero.

However each of these factors in ZZ may itself be factorised into a pair of factors in zz, since:

For each factor of Pt ⁢Z Pt Z , we may allocate one of its zz subfactors to H0 ⁢z H0 z and the other to G0 ⁢z G0 z . Where roots of Pt ⁢Z Pt Z are complex, the subfactors must be allocated in conjugate pairs so that H0 H0 and G0 G0 remain purely real.

Since the subfactors occur in reciprocal pairs (roots at z=α z α and α-1 α ), we find that

Therefore the frequency responses are related by G0 ⁢ei⁢ω⁢ Ts = H0 ⁢e−(i⁢ω⁢ Ts ) G0 ω Ts H0 ω Ts .

Hence the magnitudes of the frequency responses are the same, and their phases are opposite. It may be shown that this is sufficient to obtain orthogonal wavelets, but unfortunately the separate filters are no longer zero (or linear) phase. (Linear phase is zero phase with an arbitrary delay z−k z k .)

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

If Pt ⁢Z Pt Z is of order 2⁢K−1 2 K 1 , then it may have KK zeros at Z=-1 Z -1 such that

Equation 10 is the K=2 K 2 solution to Equation 17. Therefore, Rt ⁢Z=1−12⁢Z Rt Z 1 1 2 Z so β=−12 β 1 2 and, from Equation 15, the factors of R⁢z R z are R⁢z=(α⁢z+1)⁢(1+α⁢z-1)1+α2 R z α z 1 1 α z 1 α 2 where α=3−2 α 3 2 . Also 1+Z2=12⁢Z+12⁢12⁢1+z-12 1 Z 2 1 2 Z 1 2 1 2 1 z 2 Hence H0 ⁢z=12⁢1+α2⁢1+z-12⁢(1+α⁢z-1)=0.4830+0.8365⁢z-1+0.2241⁢z-2−0.1294⁢z-3 H0 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 H1 ⁢z=z-3⁢ G0 ⁢−z=z-3⁢ H0 ⁢−z-1=0.1294+0.2241⁢z-1−0.8365⁢z-2+0.4830⁢z-3 H1 z z -3 G0 z z -3 H0 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 Figure 3. It is clear that the wavelets and scaling function are no longer linear phase and are less smooth than those for the LeGall 3,5-tap filters. The frequency responses also show worse sidelobes. The G0 G0 , G1 G1 filters give the time reverse of these wavelets and identical frequency responses.

Figure 3: 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 result in more visible coding artefacts than linear phase filters, which distribute any artefacts equally on either side of sharp edges 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 convolution with the filters. Symmetric extension assumes that the image is reflected by mirrors at each edge, so that an infinitely tessellated plane of reflected images is generated. Reflections avoid unwanted edge discontinuities. If the filters are linear phase, then the DWT coefficients also form reflections and no increase in size of the transformed image is necessary to accomodate convolution effects.

To ensure that the filters H0 H0 , H1 H1 and G0 G0 , G1 G1 are linear phase, the factors in ZZ must be allocated to H0 H0 or G0 G0 as a whole and not be split, as was done for the Daubechies filters. In this way the symmetry between zz and z-1 z is preserved in all filters.

Perfect balance of frequency responses between H0 H0 and G0 G0 is then not possible, if PR is preserved, but we have found a factorisation of Pt ⁢Z Pt Z which achieves near balance of the responses.

This is:

It turns out that c=−27 c 2 7 gives good similarity and when substituted into Equation 19, Equation 20 and Equation 18 gives:

Figure 4: 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 Figure 5 which shows the alternative 7,5-tap versions (i.e. with HH and GG swapped). It is quite difficult to spot the minor differences between these figures.

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

In all of the above designs we have used the substitution Z=12⁢(z+z-1) Z 1 2 z z . However other substitutions may be used to create improved wavelets. To preserve PR, the substitution should contain only odd powers of zz (so that odd powers of ZZ should produce only odd powers of zz), and to produce zero phase, the coefficients of the substitution should be symmetric about z0 z 0 .

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

The 2nd 2nd order factor in Pt ⁢Z Pt Z now produces terms from z6 z6 to z-6 z-6 and the 3rd 3rd order factor produces terms from z9 z9 to z-9 z-9 . Hence the filters become 13 and 19 tap filters, although 2 taps of each are zero and the outer two taps of the 19-tap filter are very small ( ≈⁢10-4 ≈ 10 -4 ).

Figure 6 shows the wavelets and frequency responses of the 13,19-tap filters, obtained by substituting Equation 22 into Equation 21. Note the smoother wavelets and scaling function and the much lower sidelobes in the frequency responses from these higher order filters.

Figure 6: Impulse responses and frequency responses of the 4-level tree of near-balanced 13,19-tap filters.

Figure 7 demonstrates that the near balanced properties of Equation 21 are preserved in the high order filters.

Figure 7: 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 be near optimum for image compression.