EE 464 Band-Pass Sampling Example Fall 2018

Transcription

1 EE 464 Band-Pass Sampling Example Fall 2018 Summary This example demonstrates the use of band-pass sampling. First, a band-pass signal is onstruted as a osine modulated speeh signal. This is a double sideband modulation, (double sideband, suppressed arrier amplitude modulation) so the bandpass signal spetrum has twie the bandwidth of the baseband signal. Then, the band-pass waveform is sampled and the translated spetrum shown to lie in the base-band region. Next, a band-pass sampling rate is seleted and the bandpass sampled signal generated. The spetrum is ompared to the original baseband signal, and double sideband harateristi is evident. Finally, the original speeh signal is reovered from the band-pass sampled waveform. The proess is desribed as general equations and speifi Matlab ommands for a given speeh file. Step 1 Generate the Bandpass Signal Let s(t) be a base-band signal, bandlimited to 0.5/T Hz and sampled at rate 1/T samples/se. Denote the samples as s (n), with the sampling rate omitted. A band-pass signal is generated, for this example, by multiplying by a osine, forming the band-pass signal x( t) = s( t)os(2π f t). For the example, 1/T = 8,000 Hz, and f = 106, 000 Hz The signal x (t) then lies in the frequeny band [ f, f + ] = [102,000, 110,000] Hz. To represent the signals using Matlab, we begin with a speeh.wav file, represented as the vetor, s, and then use the resample funtion to inrease the signal sampling rate to a rate muh larger than the osine arrier frequeny (speifially, at least twie as large as the highest frequeny in the bandpass signal). For the example, the higher sampling rate is seleted as 1 MHz. The resample funtion generates the samples of the analog signal that would have been generated if the original analog signal were sampled at the higher sampling rate. After the up-sampled signal is formed, its spetrum is omputed. This is ompared to the spetrum of the original disrete-time signal. Sine the up-sampled signal is a bandpass signal, to ompare the spetra it is neessary to zoom in on the plot. It is then evident that the bandpass signal is a double sideband modulated version of the original, with a arrier frequeny of 106 khz. Using Matlab, the band-pass signal is formed by multiplying the up-sampled speeh waveform by a osine (sampled at the 1 MHz rate). The Matlab ode is as follows, with plots shown in Figure 1. Note that the plots are labeled using the signal frequeny domain notation XX(ff), rather than XX(ee jj2ππππππ ). >> s1=wavread('sent1.dat')'; >> s1=s1/0.1; % sale amplitude T T

2 >> m1=resample(s1,125); % inrease sampling rate from 8 khz to 1 MHz >> t=[0:length(m1)-1]/ ; % time samples >> x1=m1.*os(2*pi*106000*t); % form band-pass signal >> M1=fft(m1); >> S1=fft(s1); >> X1=fft(x1); % take fft of eah signal >> f1=[0:length(s1)-1]*8000/length(s1); >> fm=[0:length(m1)-1]* /length(m1); >> figure(1) >> subplot(3,1,1) >> plot(f1,abs(s1)) % plot spetrum of original speeh sentene >> xlabel('frequeny, f, Hz') >> subplot(3,1,2) >> plot(fm,abs(m1)) % plot spetrum of up-sampled signal >> xlabel('frequeny, f, Hz') >> subplot(3,1,3) >> plot(fm,abs(x1)) % plot spetrum of band-pass signal >> xlabel('frequeny, f, Hz') Note in the figure below that the bottom two plots have a muh larger frequeny range than the top plot, and that the seond plot shows the same spetrum as the top plot, but the spetrum is ondensed into a very narrow range near the origin for the enlarged frequeny range. The bottom shows the band-pass nature of the modulated signal. Figure 1. Spetra of original, up-sampled original, and band-pass signals.

3 Figure 2 shows the spetra of the original speeh signal (top) and the spetrum of the band-pass signal with zoomed in frequeny range. The bottom figure learly shows the double sideband nature of the band-pass signal and that the spetrum ranges from about 102 khz to 110 khz. Figure 2. Spetra of original and band-pass signals. Step 2 Determine Bandpass Sampling Rate Next, the band-pass sampling rate is determined, as follows. Let f1 = 100,000, f 2 = 112,000 Hz, and define W = f f 12, 000 Hz. Selet integer 2 1 = f1 100,000 K = = 4 2 = 2 12,000. The band-pass sampling rate is then W 1 f1 100,000 = = = 25,000 Hz. The 4 th translate of X ( f ) will lie in the base-band T bp K 4 frequeny region [0, 12.5 khz]. Sine x (t) is already band-limited to the interval [100, 112] khz, there is no need to first apply a band-pass filter. We an sub-sample diretly to form the band-pass sampled signal. The ratio of bandpass sample rate, 1 MHz, to bandpass sample rate, 25 khz, is an integer, 40, so we an diretly subsample the bandpass signal by just taking every 40 th sample. Alternatively, using Matlab, we ould use the resample funtion (and would need

4 to if the ratio of sampling rates were not an integer). The Matlab ommands are as follows. >> xsub=x1(1:40:end); % sub-sampling diretly >> tsub=[0:length(xsub)-1]/25000; % time index of sub-sampled signal >> Xsub=fft(xsub); % Compute the DFT >> fsub=[0:length(xsub)-1]*25000/length(xsub); % Compute frequeny range The spetrum of the band-pass signal is shown below in Figure 3. Note that the spetrum of the band-pass signal shown in the bottom of Figure 2 is now learly visible in the lower-frequeny range (0 to 10 khz) for Figure 3. The band-pass signal is still a double sideband modulated version of the original speeh waveform, but the effetive arrier frequeny has now been moved to 6 khz. (The original arrier frequeny was 106 khz, and 106 khz 4 25 khz = 6 khz. K = 4 is the number of translations of the bandpass spetrum down to the baseband frequeny range of [0, 12.5] khz.) If the goal of the signal proessing is to use bandpass sampling and determine the spetrum of the bandpass signal, then the goal has been aomplished. Figure 3. Band-pass signal (top), band-pass sampled signal (middle) with sampling rate 25 khz, band-pass signal spetrum (bottom).

5 Step 3 Reover the Original Signal In this example, the bandpass sampling an atually be used to help demodulate the bandpass AM signal. To reover the original speeh signal we need to remove the 6 khz modulating osine, and then apply a low-pass filter. Multiplying by a osine of 6 khz shifts the spetrum down, but also introdues a spetral lobe entered at twie the modulating frequeny, that is at 12 khz. Applying a subsequent lowpass filter retrieves only the portion of the signal below 4 khz (orresponding to the original signal being bandlimited to 4 khz). Impliitly, this is oherent demodulation, sine the phase of the multiplying osine is exatly mathed to the (translated) arrier osine at 6 khz. The Matlab ommands are as follows. >> ysub=xsub.*os(2*pi*6000*tsub); >> Ysub=fft(ysub); >> fsub=[0:length(ysub)-1]*25000/length(ysub); >> s1_25=resample(s1,25,8); >> subplot(4,1,1) >> t1_25=[0:length(s1_25)-1]/25000; >> plot(t1_25,s1_25) >> subplot(4,1,2) >> plot(tsub,ysub) >> S1_25=fft(s1_25); >> f1_25=[0:length(s1_25)-1]*25000/length(s1_25); >> plot(f1_25,abs(s1_25)) >> subplot(4,1,3) >> plot(tsub,ysub) >> subplot(4,1,4) >> plot(fsub,abs(ysub)) Figure 4 shows the original speeh signal (re-sampled at the 25 khz rate) and its spetrum, as well as the reovered speeh signal, but before any final low-pass filtering. Note that although the time waveform (third plot in Figure 4) looks similar to the original (top), the spetrum (bottom) has signifiant power at frequenies between 8 and 12.5 khz. The high-frequeny power is easily removed by filtering using a linear-phase FIR filter with utoff frequeny at about 6 khz, resulting in the signal shown in Figure 5. Note that this signal spetrum (bottom plot in Figure 5) shows that the high-frequeny power (shown in seond plot in Figure 5, and entered around 12.5 khz) has been removed, and the original speeh signal reovered.

6 Figure 4. Original speeh and spetrum, and reovered signal and spetrum before final low-pass filtering. Figure 5. Reovered signal and spetrum, and low-pass filtered version.