-1
$\begingroup$

Is there any way to tell if a signal is decaying from its fourier transform?

$\endgroup$
  • $\begingroup$ Is it possible to post a plot with a view of your signal? $\endgroup$ – A_A Apr 8 at 4:36
  • $\begingroup$ I don't have a specific signal. I'm interested in any insight into what phenomena decay (monotonous) introduces into the fourier transform $\endgroup$ – Artemmm Apr 8 at 6:20
  • $\begingroup$ Be more specific give a mathematical formula for your problem then we can help you otherwise I have to flag it as unclear of what you are asking. $\endgroup$ – Ch.Siva Ram Kishore Apr 8 at 11:19
  • $\begingroup$ I edited to be more specific $\endgroup$ – Artemmm Apr 8 at 18:19
  • $\begingroup$ All signals that have finite energy and thus have Fourier transforms in the classical sense (no impulses a.k.a. Dirac deltas anywhere) must decay away to $0$ as $t \to \infty$ or $\to -\infty$. So a dead giveaway is "no impulses in Fourier transform? signal decays" $\endgroup$ – Dilip Sarwate Apr 10 at 4:12
-1
$\begingroup$

Is there any way to tell if a signal is decaying from its fourier transform?

From a single Fourier transform, no there is not, if we assume 'decay' to mean a decrease in amplitude or power over time. Amplitude and time are the variables which define a time domain signal, magnitude and phase are the variables which define a frequency domain signal (that of the Fourier transform).

If you have multiple Fourier transforms of a signal, each one from a different time period of the signal (for example with Short time fourier transforms) then you can determine the decay in the signal amplitude, by frequency, over time. Implicitly you could also derive the total loss of power in the signal and so its net 'decay'. It should however be noted that this is an inefficient method of determining the decay in signal power/amplitude compared to using the time domain representation.

$\endgroup$
  • $\begingroup$ This answer makes no sense. See my comment on the OP's question. $\endgroup$ – Dilip Sarwate Apr 10 at 4:14
  • $\begingroup$ Your understanding of the question is obviously different from mine, you should post your answer to expand your explanation rather than downvoting mine. I do not personally understand what "no impulses in Fourier transform" is supposed to mean. $\endgroup$ – PAK-9 Apr 10 at 21:04

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.