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Running sine100hz.m plots the first 10 periods of the sine wave. Each period lasts 0.01 seconds, and each period contains 100 points,as can be verified by looking at w(1:100). Changing the variables time or Ts displays different numbers of cycles of the same sine wave, while changing f plots sine waves with different underlying frequencies.

What must the sampling rate be so that each period of the wave is represented by 20 samples?Check your answer using the program above.

Let Ts=1/500 . How does the plot of the sine wave appear? Let Ts=1/100 , and answer the same question. How large can Ts be if the plot is to retain the appearance of a sine wave?Compare your answer to the theoretical limit. Why are they different?

When the sampling is rapid compared to the underlying frequency of the signal (for instance, the program sine100hz.m creates 100 samples in each period), then the plot appears and acts much like an analog signal, even though it isstill, in reality, a discrete time sequence. Such a sequence is called oversampled relative to the signal period. The following program simulates the process of sampling the100 Hz oversampled sine wave. This is downsampling, as shown in [link] .

f=100; time=0.05; Ts=1/10000; t=Ts:Ts:time;   % freq and time vectors w=sin(2*pi*f*t);                              % create sine wave w(t)ss=10;                                        % take 1 in ss samples wk=w(1:ss:end);                               % the "sampled" sequencews=zeros(size(w)); ws(1:ss:end)=wk;           % sampled waveform ws(t) plot(t,w)                                     % plot the waveformhold on, plot(t,ws,'r'), hold off             % plot "sampled" wave
sine100hzsamp.m simulated sampling of the 100 Hz sine wave (download file)

Running sine100hzsamp.m results in the plot shown in [link] , where the “continuous” sine wave w is downsampled by a factor of ss=10 ; that is, all but one of each ss samples is removed. Thus, the waveform w represents the analog signal that is to be sampled at the effective sampling interval ss*Ts . The spiky signal ws corresponds to the sampled signal w s ( t ) , while the sequence wk contains just the amplitude values at the tips of the spikes.

Removing all but one of each N points from an oversampled waveform simulates the sampling process.
Removing all but one of each N points from an oversampled waveform simulates the sampling process.

Modify sine100hzsamp.m to create an oversampled sinc wave, and then sample this with ss=10 . Repeat this exercise with ss=30 , ss=100 , and ss=200 . Comment on what is happening. Hint: In each case, what is theeffective sampling interval?

Plot the spectrum of the 100 Hz sine wave when it is created with different downsampling rates ss=10 , ss=11 , ss=30 , and ss=200 . Explain what you see.

Interpolation and reconstruction

The previous sections explored how to convert analog signals into digital signals. The central result is thatif the sampling is done faster than the Nyquist rate, then no information is lost.In other words, the complete analog signal w ( t ) can be recovered from its discrete samples w [ k ] . When the goal is to find the complete waveform,this is called reconstruction ; when the goal is to find values of the waveform at particular pointsbetween the sampling instants, it is called interpolation . This section explores bandlimited interpolation andreconstruction in theory and practice.

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Source:  OpenStax, Software receiver design. OpenStax CNX. Aug 13, 2013 Download for free at http://cnx.org/content/col11510/1.3
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