# Random demodulator in matrix form

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?

• 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. – Peter K. Oct 1 '15 at 16:46
• @PeterK. I have edited the question please have a look. – J Cian Oct 1 '15 at 23:49
• That seems more reasonable. Let's see if that gets any answers. Thanks for the update. – Peter K. Oct 2 '15 at 0:09
• Can you add a link to the paper you went through? That might also help. – Peter K. Oct 2 '15 at 2:50
• Here is the link for the paper. arxiv.org/pdf/0902.0026.pdf – J Cian Oct 2 '15 at 3:11

• 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 $$y_m=R\int_\frac{m}{R}^\frac{(m+1)}{R}y(t)dt \hspace{5mm},m=0,1,\ldots,R-1$$ – J Cian Oct 4 '15 at 4:45