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
Any other combinations may be used, as desired.
. Thus with P(z) - P(-z) = 2z-1, results in one sample delay.
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.
The factor for the other set will be k = 3/256 and m = 5 samples delay.
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.
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.
Use G0(z) = H1(-z) and G1(z) = -H0(-z) to derive the synthesis filters
See pages 84 and 88
33 bits
29 bits