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486 Chapter 9 SAMPLING RATE CONVERSION
lowpass filter has attenuated x(n) prior to downsampling. Recall that the cutoff
frequency of the lowpass filter is set to 0.8π/D =0.1π which eliminates x(n).
If we had used the downsampling operation on x(n) instead of decimation, the
resulting sequence would be y(m) =1, which is an aliased signal. Thus, the
lowpass filtering is necessary.
□
9.3 INTERPOLATION BY A FACTOR I
An increase in the sampling rate by an integer factor of I—i.e., F y =
IF x —can be accomplished by interpolating I − 1 new samples between
successive values of the signal. The interpolation process can be accomplished
in a variety of ways. We shall describe a process that preserves
the spectral shape of the signal sequence x(n). This process can be accomplished
in two steps. The first step creates an intermediate signal at
the high rate F y by interlacing zeros in between nonzero samples in an
operation called upsampling. Inthe second step, the intermediate signal
is filtered to “fill in” zero-interlaced samples to create the interpolated
high-rate signal. As before, we will first study the time- and frequencydomain
characteristics of the upsampled signal and then introduce the
interpolation system.
9.3.1 THE UPSAMPLER
Let v(m) denote the intermediate sequence with a rate F y = IF x , which
is obtained from x(n) byadding I − 1 zeros between successive values of
x(n). Thus,
{
x(m/I), m =0, ±I,±2I,...
v(m) =
(9.26)
0, otherwise
and its sampling rate is identical to the rate of v(m). The block diagram
of the upsampler is shown in Figure 9.9. Again, any system containing
the upsampler is a time-varying system (Problem P9.1).
x(n)
Rate F x
↑I
v(m)
Rate IF x = F v
FIGURE 9.9
An upsampling element
□ EXAMPLE 9.3 Let I =2and x(n) ={1, 2, 3, 4}. Verify that the upsampler is time varying.
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