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I went through a paper in random demodulator. It states that If the sampling rate is $R$ ($R$ is less then $W$, $W=$band limit of a signal in Hz), and assume that $W$ is divided by $R$. Then each sample is the sum of $W/R$ consecutive entries of the demodulated signal. Does it mean that the signal is compressed by taking $W/R$ samples at a time?

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  • $\begingroup$ Welcome to DSP.SE! This question is very broad. You might be better off explaining what you currently understand about how the random demodulator in matrix form works in compressive sensing and asking questions about where your understanding is vague or fails. Otherwise, you are better off Googling or looking up tutorial articles. $\endgroup$ – Peter K. Oct 1 '15 at 16:46
  • $\begingroup$ @PeterK. I have edited the question please have a look. $\endgroup$ – J Cian Oct 1 '15 at 23:49
  • $\begingroup$ That seems more reasonable. Let's see if that gets any answers. Thanks for the update. $\endgroup$ – Peter K. Oct 2 '15 at 0:09
  • $\begingroup$ Can you add a link to the paper you went through? That might also help. $\endgroup$ – Peter K. Oct 2 '15 at 2:50
  • $\begingroup$ Here is the link for the paper. arxiv.org/pdf/0902.0026.pdf $\endgroup$ – J Cian Oct 2 '15 at 3:11
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The random demodulator (in this definition) compresses the signal by taking samples that are each linear combinations of W/R Nyquist rate samples. So it does not take W/R samples at a time; it takes one sample at a time, each of which contain information corresponding to several (W/R) Nyquist rate samples. It is then up to a reconstruction algorithm to subsequently "disentangle" these into the original signal (approximately) at Nyquist rate.

The random demodulator works by "smearing" the content of the input signal all over the spectrum (by multiplication with the spreading sequence). Provided that the input signal is sufficiently sparse, we can now reconstruct it from just a small portion of the resulting spectrum (selected by the low-pass filter).

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  • $\begingroup$ Then can we say that when the signal as shown in Fig. 3 is passed through the low-pass filter, the portion that passes through the low pass filter (as shown in Fig. 2) will be the combination of blue and red signal? $\endgroup$ – J Cian Oct 2 '15 at 16:41
  • $\begingroup$ As I read it, the signal formed in figure 2 is passed through the low-pass filter (lower right). Two different examples of the outcome of this (zoomed in on the passband) are shown in figure 3. $\endgroup$ – Thomas Arildsen Oct 2 '15 at 20:25
  • $\begingroup$ But I understood that it is for a signal which is the sum of two signals with different tones (frequency). If that is not the case, then what will happen if a signal which is sum of two signals is passed through the random demodulator? $\endgroup$ – J Cian Oct 2 '15 at 23:35
  • $\begingroup$ If the signal is the sum of two single tones this signal is very sparse in frequency and it will work. It will also work for signals that are sums of more than two tones, i.e. less sparse. How many tones can be reconstructed depends on your W/R. $\endgroup$ – Thomas Arildsen Oct 3 '15 at 5:21
  • $\begingroup$ Thank you for your answer. I also have the following questions from paper: 1. From equation (2) we have $${ a_\omega: \omega \in \Omega}$$ where $$\Omega \subset {0, \pm 1, \pm 2, \ldots ( W/2-1), \pm W/2 }$$ Does it mean that $a_\omega$ will be zero if $\omega \notin \Omega$. 2. Also, it says that all the frequency components of signal f lies below W/2 of band limit W Hz. Does it mean that no frequency component lies above W/2 Hz of bandwidth W Hz. 3. I Couldn't understand why there is R in $$y_m=R\int_\frac{m}{R}^\frac{(m+1)}{R}y(t)dt \hspace{5mm},m=0,1,\ldots,R-1$$ $\endgroup$ – J Cian Oct 4 '15 at 4:45

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