Chapter 4

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:

   1. 

   2. 

   3. 

   4. 

   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

   

   Any other combinations may be used, as desired.

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

   2. 

      

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

   

   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