0
$\begingroup$

Why is it a problem to do frequency analysis on a non-stationary signal?

what makes the frequency interpretation incorrect?

$\endgroup$
1
  • $\begingroup$ terms like non-stationary, non-gaussian, non-linear, are non-specific. If we are truly honest, almost everything that is correct isn’t totally correct and interpretation can be a set of interpretations. Please elaborate $\endgroup$
    – user28715
    Oct 4, 2017 at 21:23

1 Answer 1

1
$\begingroup$

The frequency interpretation on a non-stationary signal is not "incorrect". For example, it still works for doing convolution/filtering. However, it becomes opaque with regard to reflecting features interesting for analysis.

As a simple example, a morse signal of fixed carrying frequency is basically signal times carrier in the time domain and since the carrier is represented by two pulses in the frequency domain, the total frequency representation is an "echoed" variant of the transform of the signal.

However, the signal is interesting as a sequence in time. Its Fourier transform is an obfuscation rather than a recovery of information.

For many non-stationary signals, the information you are looking for is contained in the sequence of parameter changes over time: an overall Fourier transformation hides those further under the surface than they were before.

There are "number theoretic transforms" also obeying the convolution theorem of the Fourier transform and based on powers of unit roots. They have FFT-like algorithms and data flow and can be used for convolving number sequences. They are utterly useless for any form of analysis since their "transforms" invariably look like pseudo-random noise.

In this case, we are talking about something which is "correct" and thus applicable in given situations but which is not useful as an analysis tool. To a lesser degree, that is the situation when using a Fourier transform encompassing all of a non-stationary signal.

$\endgroup$
1
  • $\begingroup$ how does the windowing help? $\endgroup$
    – Bob Burt
    Oct 4, 2017 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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