I have tried a code to compress a signal using Compressed Sensing(CS). The input signal is $x$ and the compressed signal $y$ is given by : $y=Φ*x$ where $Φ$ is the sensing matrix.
I have used the following matlab function to compute the energy of $x$ and the energy of $y$ .

energy_x= sum(x(1:384,:).^2);
energy_y = sum(y(1:192,:).^2);

The length of $x$ is 384 , and the length of $y$ is 192.
I found that the energy of $x$ is 446910 and the energy of $y$ is 77651282 .
Is it is reasonable to obtain higher energy for the output than the input?

  • $\begingroup$ i think that these is related to scale. is there a difference between ones(1,100) and 50*ones(1,100)? information wise $\endgroup$
    – user28715
    Oct 25, 2018 at 15:23

1 Answer 1


Without information on $Φ$, you can obtain almost anything, since $\lambda Φ$ could be a valid CS matrix as well. Generally, one imposes structure contraints, such as unit energy for their rows or columns.

This being said, compressive sensing does not compress data in a strict sense, and energy is a poor measure of compressibility. Entropy and norm ratios could be more interesting measures.

  • $\begingroup$ Thanks a lot for your reply. I have fould the following function to calculate the entropy, but still the entropy of the output higher than the entropy of the input. Can you please indicate the reason, and help me with the suitable matlab function? Entropy link $\endgroup$ Oct 25, 2018 at 17:59
  • $\begingroup$ It really depends on the signals and the matrix. Can you share them, or the program that generate them? $\endgroup$ Oct 25, 2018 at 18:18
  • $\begingroup$ In genral, please, how can I compare two compression methods in terms of transferring most of the energy content of the original signal, and not neglecting significant signal information @ Laurent Duval $\endgroup$ Oct 25, 2018 at 22:10
  • $\begingroup$ Compare two approximations of functions remains an open question to me. Typically, I am still trying to find metrics for sparse functions. Without a knowledge on the operator, the question is difficult. Entropies could be useful $\endgroup$ Oct 30, 2018 at 16:33

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