1. | Derive the analysis and synthesis subband filters of the product filter P(z) defined by its z transform as:
|
|
2. | The z transform of a pair of lowpass and highpass analysis filters is given by:
|
|
3. | The product filter for wavelets with n zeros at z = -1 is given by:
Use eqn. 4.7 to calculate the weighting factor of the product filter P(z) and the corresponding end-to-end coding delay, for:
|
|
4. | Show that the low and highpass analysis filters of a in problem 3 are in fact the (5, 3) and (4, 4) subband filters of LeGall and Tabatabai. |
|
5. | Show that the filters for b in problem 3 are the 9/3 Daubechies filter pairs. |
|
6. | Derive the corresponding pairs of the synthesis filters of problems 4 and 5. |
|
7. | List major similarities and differences between EZW and SPIHT. |
|
8. | Consider a simple wavelet transform of 8 x 8 image shown in Figure 4.21. Assume it has been generated by a three-level wavelet transform:
|
|
9. |
|
|
Answers
1. | P(z) = H0 (z)H1 (-z) should be factorised into two terms. The given P(z) is zero at z-1 = -1, hence it is divisible by 1 + z-1. Divide as many times as possible, that gives: Thus in (i) the lowpass analysis filter will be:
and the high pass analysis filter is:
In (ii) and the highpass:
which are the (5,3) subband filter pairs. In (iii) and the highpass
which gives the second set of (4,4) subband filters. In (iv) and
Any other combinations may be used, as desired. |
2. |
|
3. |
|
4. | In problem 3, P(z) is in fact type (iv) of problem 1. Hence it leads not only to the two sets of (5,3) and (4,4) filter pairs, but also to two new types of filter, given in (i) and (iv) of problem 1. |
5. | With
retaining H1(-z) = (1 + z-1)2 = 1 + 2z-1 + z-2, that gives the three-tap highpass filter of H1(z) = 1 - 2z-1 + z-2 and the remaining parts give the nine-tap lowpass filter
or
Had we divided the highpass filter coefficients, H1(z), by , and hence multiplying those of lowpass H0(z) by this amount, we get the 9/3 tap filter coefficients of Table 4.2. |
6. | Use G0(z) = H1(-z) and G1(z) = -H0(-z) to derive the synthesis filters |
7. | See pages 84 and 88 |
8. | 33 bits |
9. | 29 bits |