How can we separate two frequency component when they get mixed means when signal get mixed with noise how receiver can judge which part is signal and which is noise ?

  • $\begingroup$ In case of a single sinusoid, which doesn't exist in reality, simply by its gain being above the noise-average. Also, depending on the receiver there could be a physical notch-filter applied on its input at the corresponding frequencies, which would reduce inter-modulation products resulting from, e.g. white gaussian noise on other channels. As a side note, with things being relative, for the noise itself the signal is noise. With a Fourier Transform, it being orthogonal, therefore preserving information, noise in one domain becomes noise in another domain. There are different FTs. $\endgroup$ – Starhowl Aug 21 '18 at 17:49
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    $\begingroup$ elaborate please. Would you mind putting some mathematical framework to your problem ? $\endgroup$ – Fat32 Aug 21 '18 at 18:52
  • $\begingroup$ Hi! Thanks for posting this as a separate question. But when I recommended that, I also asked you to mathematically define mixed, as, as I said, this word has multiple meanings. Now we have to ask you again for the meaning of your words. $\endgroup$ – Marcus Müller Aug 22 '18 at 7:04

You asked "how" without first asking "if", and, if so, under what circumstances.

If you add two completely unknown numbers A + B and get 100, can you un-mix the sum to the original two numbers A and B?

But if you know B = 7, you might have a solution.

In the receiver case, if you know the 2 spectrums are disjoint (or other properties, such as redundancy in modulation/coding, etc.), judging might be possible. Or the answer might be statistical.

  • $\begingroup$ I think you have solution of my question but please try to describe me sir $\endgroup$ – user48391 Aug 22 '18 at 17:24
  • $\begingroup$ he described all he could based on the little info you're giving in the question. Please read the commends you've got under your question – they ask for clarification. It's up to you to ask a question with enough information to allow us to help you! We really can't do that for you. $\endgroup$ – Marcus Müller Aug 22 '18 at 20:36

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