Can we reconstructs a signal after it is being multiplied (not added) by a random noise (not with the normal distribution)?

Lets say I have S=10*sin(2*pi*5*t) for t=[0:.001:1] and I have a random noise V for the same time t. How Can I reconstruct S from X=S.*V (element wise product)?

Is is theoretically possible? I read this post about the Multiplicative noise but still I need help.

If the noise is truly random, doesn't mean that it can ruin the signal completely?


Background of my question: there is a machine learning method called "Non-negative matrix factorization" that can be used to find the components of a signal (assume signal is a combination of those components). I have seen this method as able to find the right components even after the combined signal was multiplied by a white noise!!

  • 1
    $\begingroup$ It is impossible to obtain $S$ from $X$. Even if you know $V$ perfectly, it could equal zero, and then $S$ would be indeterminate. It may still be possible to tell something about the signal (such as its "components", whatever that is), but not its precise value. $\endgroup$ – MBaz Jun 21 '18 at 18:10
  • $\begingroup$ @MBaz the same is true for additive noise $\endgroup$ – Marcus Müller Jun 21 '18 at 19:04
  • 1
    $\begingroup$ Can we assume the signal is Sine Wave? $\endgroup$ – Royi Jun 21 '18 at 20:06
  • 1
    $\begingroup$ @MarcusMüller That is true, but that wasn't the question :-P $\endgroup$ – MBaz Jun 21 '18 at 22:44
  • 1
    $\begingroup$ Non negative matrix factorisation is not a machine learning method. Such factorisations pop up every where. It is used in machine learning for sure but it is not a machine learning method. Rant over. If you leave the question open as it is then no, you cannot reconstruct, but, in some cases you can (still this might be not what you are after though :/ ). $\endgroup$ – A_A Jun 22 '18 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.