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Generally we apply a filter to reduce noise from signal. For example if the received signal is $S$ we would get a signal $Y$ with reduced noise applying a filter $W$ with the operation Y= convolution(S,W). In other words filter is nothing but convolution. Is there a known method or example of getting a useful $Y$ such that $Y=S.*W$ where $.*$ means element wise multiplication. An example of element wise multiplication is demodulation.

Can anyone give an insight if element wise multiplication can be used to reduce noise under some circumstances?

Edit: Yes, product in time domain means convolution in frequency domain. Still the question under what situation convolution in frequency domain or product in time domain can reduce noise? Is there any such application? Did anyone used such a method?

Edit: There may be a situation where it is much easier to find a vector $W$ such that the element wise multiplication does reduce noise. It may difficult to find the corresponding filter.

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    $\begingroup$ it's what happens in the frequency domain. convolution in the time domain can be accomplished with multiplication in the frequency domain. $\endgroup$
    – robert bristow-johnson
    Commented Jul 25, 2019 at 1:46
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    $\begingroup$ The best way to reduce noise is the element-wise product with the zero vector - all noise gone. What this observation tells you is that you need to specify more than that. You'll have some constraints to your noise reduction regarding your desired signal. You'll have to formulate these rigorously, otherwise we have no chance to understand what exactly you are trying to do. $\endgroup$
    – Florian
    Commented Jul 25, 2019 at 9:16
  • $\begingroup$ @Florian, Thank you for your interest. Can you tell what is not clear(?), as your example shows you have understood the question very clearly. You may say you have not seen anyone doing it and you may say that you do not know where it may be applicable. Where and what I am tying to do is not at all related to the question. $\endgroup$
    – Creator
    Commented Jul 25, 2019 at 10:00
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    $\begingroup$ What's unclear is what are your constraints with respect to the desired signal. Must it be completely unharmed in the process (this would imply you can only touch the samples where the desired signal is zero)? You want a "useful $Y$", what makes a $Y$ useful or useless to you? The way you formulate it now, the optimal solution (in terms of how strongly noise is reduced) is still $W=0$. $\endgroup$
    – Florian
    Commented Jul 25, 2019 at 10:48
  • $\begingroup$ @Florian W is never 0 It is number between 0 and 1. $\endgroup$
    – Creator
    Commented Jul 25, 2019 at 18:51

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Classic filteration is indeed done using convolution. Though I have seen broader definition of filtering as shaping the signal in its frequency domain which can be done in many other methods as well.

Of course you can create meaningful operations using element wise operations and even specifically multiplication.

Think of the case you have a noise with the following properties:

  1. It is non negative.
  2. It is signal dependent.

Then operation like thresholding would make sense and it is applied using element wise multiplication. This case is very similar to $ {L}_{1} $ Denoising.

Pay attention that in this case the filter depends on the data.

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  • $\begingroup$ Not sure wha you mean by noise is non negative. The weight should be non negative. Any idea where it has been used, I undertstand it is not in the question. $\endgroup$
    – Creator
    Commented Aug 4, 2020 at 18:09
  • $\begingroup$ Imagine the added noise is random [0, a] where a depends on the signal per sample. Then denoising can be done using Threshold which is a multiplication. Just like you asked in the question. $\endgroup$
    – Royi
    Commented Aug 5, 2020 at 5:01
  • $\begingroup$ I mean the first condition. Even if the noise is negative, one can reduce the noise. Consider signal value is 1 and the noise added is -.1 so the value is .9 so for this sample the corresponding weight is 1. For some other sample the noise added is .2 then the correspondin weigth is .75. The output of the whole signal is just differ by a constant. The point is: the first condition you mentioned " it is non negative" is redundant. $\endgroup$
    – Creator
    Commented Aug 5, 2020 at 23:23
  • $\begingroup$ If the noise is negative the case I showed above (Soft Thresholding) won't work. Of course you can build other "Denoisers" for other cases. You asked for example, I gave example of a real world case. I think for the case the noise is negative as well as positive multiplication based denoisers won't work well at all unless you know the sign of the noise per sample. $\endgroup$
    – Royi
    Commented Aug 6, 2020 at 6:11
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As far my knowledge goes, the following are a few points over your question:

  • Noise filtering process involves identifying the spectral behavior of noise (because that is how noise is characterized) and then suppressing it using appropriate filters, which selectively mitigate its energy across desired frequency bands.
  • You are asking whether or not element wise multiplication (which is dot product) of a vector with the signal vector would help in noise reduction. Multiplication of a function in time domain would convolve its spectrum with the signal spectrum.
  • Convolution operation always would yield a wideband signal and cannot be selective in removing noise by design.

So, in my opinion, a dot product won't be able to yield the results you are expecting.

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  • $\begingroup$ As you said convolution is always smoothing, further filtering referes to reduce noise, so by convolution in frequency you can smooth (or wide) the spiked frequency which is reduced noise (spiked noise in frequency), so your answer fails. $\endgroup$
    – Creator
    Commented Jul 29, 2019 at 22:02

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