4.8 Problems

4.8 Problems

1. 

Derive the analysis and synthesis subband filters of the product filter P(z) defined by its z transform as:

 p ( z ) = h 0 ( z ) h 1 ( -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: a. b. c. d. thus in (i) the lowpass analysis filter will be: and the high pass analysis filter is: * h 1 ( z ) = 1 - z -1 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. 

The z transform of a pair of lowpass and highpass analysis filters is given by:

  1. Calculate the product filter, and deduce the amount of end to end encoding/decoding delay

  2. Derive the corresponding pairs of the synthesis filters.

 a. . thus with p ( z ) - p ( -z ) = 2 z -1 , results in one sample delay. b.

3. 

The product filter for wavelets with n zeros at z = -1 is given by:

  • P(z) = (1 + z-1)2nQ(z)

Use eqn. 4.7 to calculate the weighting factor of the product filter P(z) and the corresponding end-to-end coding delay, for:

  1. n = 2 and Q(z) = -1 + 4z-1 - z-2

  2. n = 3 and

 a. with a weighting factor of k , p ( z ) = k (1 + z -1 ) -4 (-l + 4 z -1 - z -2 ), and using p ( z ) - p ( -z ) = 2 z -m , gives k = 1/16 and m = 3. b. the factor for the other set will be k = 3/256 and m = 5 samples delay.

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.

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. 

Show that the filters for b in problem 3 are the 9/3 Daubechies filter pairs.

with retaining h 1 ( -z ) = (1 + z -1 ) 2 = 1 + 2 z -1 + z -2 , that gives the three-tap highpass filter of h 1 ( z ) = 1 - 2 z -1 + z -2 and the remaining parts give the nine-tap lowpass filter or had we divided the highpass filter coefficients, h 1 ( z ), by , and hence multiplying those of lowpass h 0 ( z ) by this amount, we get the 9/3 tap filter coefficients of table 4.2 .

6. 

Derive the corresponding pairs of the synthesis filters of problems 4 and 5.

use g 0 ( z ) = h 1 ( -z ) and g 1 ( z ) = -h 0 ( -z ) to derive the synthesis filters

7. 

List major similarities and differences between EZW and SPIHT.

see pages 84 and 88

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:

  1. Show different steps in the first dominant pass (most significant bit plane) of the EZW algorithm

  2. Assuming that EZW uses two bits to code symbols in the alphabet {POS, NEG, ZT, Z} and one bit to code the sign bit, calculate total number of bits outputted in this pass.

click to expand
Figure 4.21

33 bits

9. 

  1. In Figure 4.21, calculate the number of bits outputted in the first significant pass of the SPIHT algorithm

  2. Comment on the results of 8b and 9a.

29 bits

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:

  • H1(z) = 1 - z-1

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

click to expand

Any other combinations may be used, as desired.

2. 

  1. click to expand. Thus with P(z) - P(-z) = 2z-1, results in one sample delay.

3. 

  1. With a weighting factor of k, P(z) = k(1 + z-1)-4(-l + 4z-1-z-2), and using P(z) - P(-z) = 2z-m, gives k = 1/16 and m = 3.

  2. The factor for the other set will be k = 3/256 and m = 5 samples delay.

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

click to expand

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

click to expand

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



Standard Codecs(c) Image Compression to Advanced Video Coding
Standard Codecs: Image Compression to Advanced Video Coding (IET Telecommunications Series)
ISBN: 0852967101
EAN: 2147483647
Year: 2005
Pages: 148
Authors: M. Ghanbari

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