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I am new in the field of compressive sensing, I've read many papers explaining that compressive sensing is used widely in sparse signal reconstruction. I've tried to understand how compressive sensing is used in signal processing, but I couldn't understand the concept.

According to the paper On some common compressive sensing recovery algorithms and applications - Review paper in page 2, THE MATHEMATICAL BACKGROUND OF THE COMPRESSIVE SENSING CONCEPT, I got the best description for compressing sensing mathematical concept. Assuming we are dealing with signal $x$ with length $N$, that is sparse in time domain. So according to equation 5, we can write the system equation as $Y = AX$, where $A$ is the CS matrix. My questions,

1- What is the signal $x$ with length $N$ which we want to reconstruct ? Is it $X$ or $Y$ ? And what does $A$ represent?

2- Is there any tutorial or file which explain that point with details and in easy way?

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  • $\begingroup$ I strongly suggest this video to clear you doubts youtube.com/watch?v=RvMgVv-xZhQ $\endgroup$ – MimSaad Aug 29 '18 at 18:50
  • $\begingroup$ Thank you so much .. I've watched that video .. It's A great one $\endgroup$ – Fatima_Ali Aug 30 '18 at 8:18
  • $\begingroup$ Any more details needed? $\endgroup$ – Laurent Duval Apr 26 at 23:14
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If the signal $x$ is already sparse in the original/time domain, then it does not need to be transformed, in other words the transform $\mathcal{J}$ can be taken to be the identity. So $x$ and $X=\mathcal{J}x$ are equal. What we want is to recover, from observed $y$, the unknown $x$, via the recovery of $X$, since $x=\mathcal{J}^{-1}X$. In that case, $A$ is just the "random" sensing matrix.

The above paper does not seem too complicated, and I am not sure there are many "much simpler" explanation. You can try:

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  • $\begingroup$ OK .. thank you so much .. So, you mean y is gotten signal, and we need to estimate x based on y and the random sensing matrix A. BUT, what's about J ? why have you inserted it ? what does it represent ? $\endgroup$ – Fatima_Ali Aug 30 '18 at 8:21
  • $\begingroup$ Because sometimes $x$ itself is not sparse, and is transformed into a sparser transform version before acquisition $\endgroup$ – Laurent Duval Nov 28 '18 at 19:37

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