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530 Chapter 9 SAMPLING RATE CONVERSION
the integer part of (i m + D)/I. The integer i m+1 should be saved to be
used in determining the subfilter from which the next sample is taken.
□ EXAMPLE 9.19 For the sampling rate converter designed in Example 9.15, specify the set of
time-varying coefficients {g(n, m)} used in the realization of the converter based
on the structure given in Figure 9.19. Also, specify the corresponding implementation
based in polyphase filters.
Solution The coefficients of the filter are given by (9.43)
( ⌊ ⌋ ) D
g(n, m) =h(nI +(mD) I)=h nI + mD −
I m I
By substituting I =5and D =2,weobtain
(
⌊ 2m
g(n, m) =h 5n +2m − 5
5
⌋)
By evaluating g(n, m) for n =0, 1,...,5 and m =0, 1,....,4weobtain the
following coefficients for the time-variant filter:
g(0,m)={h(0) h(2) h(4) h(1) h(3)}
g(1,m)={h(5) h(7) h(9) h(6) h(8)}
g(2,m)={h(10) h(12) h(14) h(11) h(13)}
g(3,m)={h(15) h(17) h(19) h(16) h(18)}
g(4,m)={h(20) h(22) h(24) h(21) h(23)}
g(5,m)={h(25) h(27) h(29) h(26) h(28)}
Apolyphase filter implementation would employ five subfilters, each of length
six. To decimate the output of the polyphase filters by a factor of D =2simply
means that we take every other output from the polyphase filters. Thus, the first
output y(0) is taken from p 0(n), the second output y(1) is taken from p 2(n),
the third output is taken from p 4(n), the fourth output is taken from p 1(n), the
fifth output is taken from p 3(n), and so on.
□
9.7 PROBLEMS
P9.1 Consider the upsampler with input x(n) and output v(m) given in (9.26). Show that the
upsampler is a linear but time-varying system.
P9.2 Let x(n) =0.9 n u(n). The signal is applied to a downsampler that reduces the rate by a
factor of 2 to obtain the signal y(m).
1. Determine and plot the spectrum X(ω).
2. Determine and plot the spectrum Y (ω).
3. Show that the spectrum in part (2) is simply the DTFT of x(2n).
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