1
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.