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10.1  Assume we want to decimate an x(n) timedomain sequence by four.
 
10.2  Assume we have a 72sample sinusoidal x(n) timedomain sequence, the first 36 samples of which are shown in Figure P10–2(a). Next we decimate x(n) by two to generate 36 samples of the y(m) sequence shown in Figure P10–2(b). Sequence y(m) is also sinusoidal, as we should expect, but its frequency appears to be double the frequency of x(n). Explain that apparent frequency difference.
Figure P10–2.
 
10.3  Assume we collected 2048 samples of a sinewave whose frequency is 128 Hz using an f_{s} sample rate of 1024 Hz, and we call those samples w(n). The first 20 samples of w(n) are shown in Figure P10–3. Next we perform a 2048point FFT on w(n) to produce a W(m) sequence.
Figure P10–3.
 
10.4  In this chapter we’ve portrayed decimation by an integer factor M with the block diagram shown in Figure P10–4, that is, a lowpass decimation filter followed by a downsampler (the “↓M” symbol) that discards all but every Mth filter output sample. In this problem we explore the changes in signal timedomain amplitude and frequencydomain magnitude caused by decimation.
Figure P10–4.
For this problem, our assumptions are:
 
10.5  Given the x_{r}(n) input signal in Figure P10–5(a), whose X_{r}(f) magnitude spectrum is shown in Figure P10–5(b), draw a rough sketch of the X_{c}(f) spectrum of the system’s complex x_{c}(m) = x_{I}(m) + jx_{Q}(m) output sequence. The frequency magnitude responses of the complex bandpass h_{BP}(k) filter and the realvalued highpass h_{HP}(k) filters are provided in Figures P10–5(c) and P10–5(d).
Figure P10–5.
 
10.6  Assume we want to design the decimation by M = 30 system shown in Figure P10–6(a). The desired LPF_{0} lowpass filter’s frequency magnitude response is the solid lines shown in Figure P10–6(b). The filter’s stopband attenuation is 50 dB. (The dashed lines are the spectral replication of the lowpass filter’s frequency response.) The onesided passband width of the lowpass filter is B’ = 1.7 kHz.
Figure P10–6.
 
10.7  Here is a interesting problem. In Chapter 5 we discussed the transient response of tappeddelay line FIR filters and stated that an FIR filter’s output samples are not valid until the filter’s delay line is filled with input data samples. Assuming that the 23rd output sample of LPF_{1} is the first sample applied to LPF_{2}, how many x_{old}(n) input samples must be applied to the twostage decimation filter shown in Figure P10–7 to fill the LPF_{1} and LPF_{2} lowpass filters with input data?
Figure P10–7.
 
10.8  Assume we want to interpolate an x(n) timedomain sequence by three.
 
10.9  Let’s make sure we fully understand the spectral effects of interpolation by considering the 8sample, singlecycle, x(n) sinewave sequence in Figure P10–9(a). That sequence’s X(m) DFT spectral magnitude samples are shown in Figure P10–9(b). If we upsample x(n) by a factor of three, by inserting two zerovalued samples between each x(n) sample, we produce the 24sample y(p) time sequence shown in Figure P10–9(c).
Figure P10–9.
 
10.10  Assume we have a timedomain sequence of realvalued samples, x_{old}(n), whose spectral magnitude is shown in Figure P10–10. (We represent spectral replications by the dashed lines.) There we see that the frequency points of spectral symmetry of X_{old}(f), represented by the bold down arrows, can be described by
Figure P10–10.
where k is an integer. If we upsample x_{old}(n) by two, by inserting a zerovalued sample between each x_{old}(n) sample, to generate a new time sequence x_{new}(m), what is the expression for the frequency points of spectral symmetry of X_{new}(f)?  
10.11  Texas Instruments Inc. produces a digital filter chip, Part #GC2011A, used in cell phones for frequency upconversion. The process, described in their AN9804 application note document, is depicted in Figure P10–11(a). The lowpass filter’s 1 MHzwide passband covers the frequency range shown in Figure P10–11(b). (The lowpass filter block comprises two separate realvalued 1 MHzwide filters, filtering the real and imaginary parts of the complex signal at node B.) If the spectral magnitude of the x(n) input is that shown by the solid curves in Figure P10–11(c), where we represent spectral replications by the dashed curves, draw the spectral magnitudes of the complex sequences at nodes A, B, C, and the real part of the y(m) output sequence.
Figure P10–11.
Hint: In Chapter 8 we learned that multiplying a time sequence by e^{–j2n/4} = 1, –j, –1, j, ..., translates the signal’s spectrum down in frequency.  
10.12  Here is a fun interpolation problem. Figure P10–12(a) shows a simple digital filtering system. Assume that the analog x(t) signal applied to the analogdigital (A/D) converter contains a 9 kHz sinusoid and an 11 kHz sinusoid. The spectral magnitude of the sampled x(n) sequence is given in Figure P10–12(b). The system’s function is to filter out the 11 kHz tone and provide a y(m) output sequence that is a 9 kHz sinusoid at a sample rate of f_{s} = 32 kHz. The dashed curve in Figure P10–12(b) indicates the unitygain bandpass filter’s frequency magnitude response, while the spectrum of our desired filter output, whose magnitude is K, is given in Figure P10–12(c).
Figure P10–12.
Now, assume that the system is constrained to use an A/D converter whose clock rate is 8 kHz (instead of 32 kHz), as shown in Figure P10–12(d).
 
10.13  In this chapter we discussed various forms of interpolation. There is a wellknown interpolation process called linear interpolation. It’s an interpolationbytwo method for estimating sample values of a continuous function between some given x(n) sample values of that function. For the x(n) time samples in Figure P10–13(a), linear interpolation is the process of computing the intermediate y(n) samples shown as the black squares in Figure P10–13(b). That is, the interpolated sample y(1) is the value lying at the center of the straight line connecting x(0) and x(1), the interpolated sample y(2) is the value lying at the center of the straight line connecting x(1) and x(2), and so on. Given this process of linear interpolation:
Figure P10–13.
 
10.14  Assume we must convert a compact disc (CD) audio signal, whose sample rate is f_{s}_{,CD} = 44.1 kHz, to a digital audio tape (DAT) signal whose sample rate is f_{s}_{,DAT} = 48 kHz. If we interpolate that CD signal by a factor of L = 160, by what factor M must we decimate the interpolated signal to obtain a final sample rate of 48 kHz?  
10.15  Consider the x_{o}(n) time sequence in Figure P10–15(a), whose sample rate is f_{s} = 1 kHz. If we decimate x_{o}(n) by two, we obtain the x_{D}(m_{D}) sequence shown in Figure P10–15(b), where the oddn samples of x_{o}(n) have been discarded. Next, if we interpolate x_{o}(n) by two, we obtain the x_{I}(m_{I}) sequence shown in Figure P10–15(c), where the interpolated samples are shown as white dots. Comment on how decimation and interpolation affect the time duration of the decimated and interpolated sequences relative to the time duration of the original x_{o}(n) sequence.
Figure P10–15.
 
10.16  Fill in the following table. When complete and correct, the table shows the timedomain and frequencydomain gain of the two processes: decimation by M, and interpolation by L.
Here, decimation means lowpass filtering (by a unitygain filter) NM time samples followed by the discarding of every Mth filter output sample to obtain N time samples. By “interpolation” we mean upsampling by inserting L–1 zerovalued samples between adjacent samples of an Nlength timedomain sequence followed by lowpass filtering using a unitygain lowpass filter to obtain NL time samples. Assume the sample rate change factors M and L are integers.
 
10.17  Here is an interesting, and educational, problem because it shows the spectral effects of upsampling a downsampled sequence. Think about the sample rate change process in Figure P10–17(a). The upsampling operation “↑4” means insert three zerovalued samples between each q(m) sample. Assume the spectral magnitude of the x(n) sequence is the X(f) shown in Figure P10–17(b).
Figure P10–17.
 
10.18  One way to implement a secure telephone communications channel is shown in Figure P10–18(a). Anyone monitoring the telephone line will not be able to understand the audio speech signal on that line. The scrambling network is shown in Figure P10–18(b), where the two identical L(f) digital lowpass filters have passbands that extend from –2 kHz to +2 kHz. The two identical H(f) digital highpass filters have passbands that extend from –6 kHz to –2 kHz, and 2 kHz to 6 kHz.
Figure P10–18.
 
10.19  In Section 10.7 we depicted a polyphase filter, used in an interpolationbyfour process, with the structure shown in Figure P10–19–I. The H_{k}(z) blocks represent tappeddelay line FIR polyphase subfilters containing unitdelay elements, multipliers, and adders.
Figure P10–19–I.
 
10.20  Occasionally in the literature of DSP you’ll encounter documentation that uses a drawing like that in Figure P10–20 to illustrate some concept, or principle, regarding multirate systems. Notice that the cascaded elements are not our standard “z^{–1}” delaybyonesample elements but, instead, are advancebyonesample elements indicated by a “z” (z^{+1}).
Figure P10–20.
Show how you would implement the system in Figure P10–20, in our universe where we cannot look forward in time, to provide the appropriate four timedomain sequences to the “Some useful processing” subsystem’s input ports?  
10.21  In the text we discussed decimation by M = 3 and showed two equivalent realizations of such a decimation process as those in Figures P10–21(a) and P10–21(b). Assume that all six subfilters in Figure P10–21 are tappeddelay lines containing four multipliers, and that f_{s} = 30 samples/second.
Figure P10–21.
 
10.22  The decimationbyfour (lowpass filtering followed by downsampling) process shown in Figure P10–22(a) is inefficient because three out of every four computational results are discarded. A more efficient decimation process is shown in Figure P10–22(b), where the switches driving the multipliers close once, for one sample time period only, upon the arrival of every fourth x_{old}(n) sample. This way, no unnecessary computations are performed. Likewise, in polyphase decimation filtering no unnecessary computations are performed. In realtime hardware implementations, explain the fundamental difference between the computations performed, from a timedomain standpoint, in the Figure P10–22(b) decimation filter and a polyphase decimationbyfour filter having 12 multipliers?
Figure P10–22.
 
10.23  In Section 10.7 we depicted a polyphase filter, used in a decimationbyfour process, with the structure shown in Figure P10–23–I. The H_{k}(z) blocks represent tappeddelay line FIR polyphase subfilters containing unitdelay elements, multipliers, and adders.
Figure P10–23–I.
Figure P10–23–II.
The correct solution to this problem will show a polyphase structure with which you should become familiar. That structure is often used in the DSP literature of multirate systems to depict polyphase decimation filters. Hint: Given some x(n) sequence, write the x(n), x(n–1), x(n–2), etc., sample sequences on the four lines driving the H_{k}(z) subfilters in Figure P10–23–I. Then determine how to obtain those same sample sequences for routing to the subfilters in Figure P10–23–II(b).  
10.24  This problem is related to the material in the text’s Section 10.10. Assume we are resampling a time sequence by the rational factor 5/4 using a fiveposition commutating filter output switch as shown in Figure P10–24.
Figure P10–24.
 
10.25  Think about the multirate decimation system, employing lowpass halfband filters, in Figure P10–25(a). If the spectrum of the wideband x(n) noise sequence is that shown in Figure P10–25(b), the spectrum of the a(n) noise sequence is as shown in Figure P10–25(c). Draw the spectra, with appropriate frequencyaxis labeling in Hz, of the b(n), c(m), and y(p) sequences.
Figure P10–25.
 
10.26  The zdomain transfer function of a CIC filter’s comb subfilter having a delay line length of N = 8, shown in Figure P10–26(a), is
H_{comb}(z) = 1 –z^{–8}, and its frequency magnitude response is shown on a linear scale in Figure P10–26(b).
Figure P10–26.
 
10.27  In the text we stated that the interpolation CIC filter in Figure P10–27(a) has an impulse response, when its differential delay D = 5, equal to that shown in Figure P10–27(c). We also stated that swapping Figure P10–27(a)’s comb and integrator resulted in a decimation CIC filter as shown in Figure P10–27(b). Prove that the decimation CIC filter in Figure P10–27(b) also has an impulse response equal to that shown in Figure P10–27(c).
Figure P10–27.
 
10.28  Here is an important problem with regard to implementing two theoretically equivalent digital filters. We illustrate our point using the CIC filters shown in Figures P10–28(a) and P10–28(b). Because they are linear, we can swap the comb and integrator stages of the CIC filter used for interpolation to obtain a CIC filter used for decimation. The two CIC filters have identical timedomain impulse responses.
 
10.29  Here is a typical problem faced by engineers who use CIC filters. As of this writing, Intersil Corp. makes a decimating digital filter chip (Part #HSP43220) that contains a 5thorder CIC filter. When used for decimation by a factor of R = 6, and the internal comb filters have a differential delay of D = 6, the CIC filter’s frequency magnitude response is shown in Figure P10–29(a).
Figure P10–29.
 
10.30  There are digital filtering schemes that use the process conceptually shown in Figure P10–30(a). In that network the input is lowpass filtered to generate the sequence w(n). The network’s y(n) output is the x(n) input sequence minus the lowpassfiltered w(n) sequence. The actual implementation of such a process is shown in Figure P10–30(b) where the multielement delay line in the upper path of Figure P10–30(b) is needed for time alignment to compensate for the time (group) delay of the CIC filter. If we had to implement this parallelpath filter with a CIC filter whose differential delay is D = 9, how many unitdelay elements would we use in the upper path of Figure P10–30(b)? Show how you obtained your solution.
Figure P10–30.
