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I have a question regarding the inverse Fourier transform and its relevance to non-stationary signals. And by non-stationary signal, I'm talking about a signals whose frequency content varies with time. I'll provide the example that made me originally ask this question.

Say for example I have a sinusoidal signal with a single frequency that goes on for ever. I could take the Fourier transform of this signal to obtain its frequency content. And because this signal is a single frequency that goes on forever, this signal is considered stationary.

However, now let's say I have a signal that is zero except for in some window in which the signal is sinusoidal. This nonzero window exists from t = 0 to 1 second. If I were to take the Fourier Transform of this signal over a finite duration (such that it acts on both the nonzero and zero part of the signal), I would get a frequency content that shows there exists a peak frequency content due to the nonzero window.

Now let's say I were to do the same thing, except this time the nonzero window exists from t = 6 to 7 seconds. If I were to take the Fourier transform, over the same finite duration, I believe I would yield the exact same frequency content (both phase and magnitude) from the earlier experiment.

Assuming what I said was correct, it should thus be IMPOSSIBLE to recover the exact signal from the Fourier transform since two different time domain signals have mapped to the same frequency domain signal. Am I correct with this statement? Furthermore, I believe the inverse Fourier transform will end up producing a stationary signal. As in, the frequency that was once contained just within the nonzero window will now be shown to be infinite from the inverse Fourier transform.

TLDR:

  • Does the inverse Fourier transform only produce stationary time signals?
  • Thus if I have non-stationary signal, take the Fourier transform and then take the inverse Fourier transform, I will now have a signal that is stationary?
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The FT of a finite length sine wave burst consists of more than just the one frequency of that sine wave. A finite width time domain signal has an infinite width FT. Same for an FFT, a rectangular window produces a full width periodic Sinc in the FFT result. So even if the phase of the 1 sine wave result bin stays constant, the phase of all the other frequencies will rotate as the position of the window changes. Two different rectangular windows will NOT map to the same FFT, but convolve against any single sine wave differently.

Therefore an inverse FT or IFFT of the full spectrum can produce a signal whose statistics can vary significantly over different portions of the result, by reproducing a window which can be zero in parts of the result, and non zero in others.

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To be honest, "stationary" as a term is already occupied by stochastic processes and means "the stochastic moments don't change with time"; it's a bit unlucky you're re-using the term; however

Now let's say I were to do the same thing, except this time the nonzero window exists from t = 6 to 7 seconds. If I were to take the Fourier transform, over the same finite duration, I believe I would yield the exact same frequency content (both phase and magnitude) from the earlier experiment.

Is plain wrong: The Fourier transform has the time shift property, which applies here; so, your signal will be the same in absolute, but not the same in phase in frequency domain.

Assuming what I said was correct, it should thus be IMPOSSIBLE to recover the exact signal from the Fourier transform since two different time domain signals have mapped to the same frequency domain signal.

Luckily, what you said was wrong!

Am I correct with this statement?

Ex falso sequitur quodlibet; you can infer anything from a wrong premise.

So, if you had a Transform that acted like the Fourier transform but had no phase information, then your statement would've been right. The FT does, however, maintain that information, and hence has an inverse.

I'm a bit surprised you didn't actually try to just feed your example into the Fourier Transform and then back through the Inverse Fourier Transform; you should've noticed that you get a different result each time, just by looking at the formula.

Furthermore, I believe the inverse fourier transform will end up producing a stationary signal.

Really, your definition of "stationary" isn't compatible with what I'd usually use it for. In fact, there's other terms that would describe what you mean, for example "the Short-Time Fourier Transform is time-invariant".

However, that would imply the signal is constant, because even for the STFT you'd get a phase rotation depending at which phase of your observed periodic signal you start.

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  • $\begingroup$ I used to reserve the term stationary for random signals only, but I've keep hearing reference to the term when talking about time varying frequency content within non-random signals. This question came to be after I read this article - wseas.us/e-library/conferences/2008/cairo/CD-CIMMACS/… Which states that the fourier transform doesn't encode information regarding when a frequency occurs within a time varying signal. This is why I assumed that having a frequency exist at two different times within a finite duration would yield the same frequency content. $\endgroup$ – Izzo Sep 3 '16 at 7:03
  • $\begingroup$ I really don't have the time to read that article in whole, but I skimmed it and it does say the exact opposite of what you claim , for example "The STFT is a Fourier-related transform that is used to determine the sinusoidal frequency and the phase content of the local sections of a signal.." (emphasis by me) $\endgroup$ – Marcus Müller Sep 3 '16 at 7:06
  • $\begingroup$ From the introduction "But, Fourier analysis has also some other serious drawbacks. One of them may be that time information is lost in transforming to the frequency domain. When looking at a Fourier transform of a signal, it is impossible to tell when a particular event has taken place" That's the part that I'm primarily referencing. $\endgroup$ – Izzo Sep 3 '16 at 7:08
  • $\begingroup$ and also, this sentence from the paper: "One of them may be that time information is lost in transforming to the frequency domain." is just plain wrong. If you know the researchers, let them know. $\endgroup$ – Marcus Müller Sep 3 '16 at 7:08
  • $\begingroup$ come on, you've worked with filters, and other more complex stuff, we know that from your other questions. You know that a shift in time domain yields a multiplication with $e^{j2\pi f \Delta t}$ in frequency domain; that claim from the paper must be wrong! $\endgroup$ – Marcus Müller Sep 3 '16 at 7:09

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