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I got curious based on this question here, but basically, is there ever a real-life signal that exists where its Fourier transform does not exist? If a signal is not finite energy, then its Fourier Transform does not exist, so what might be an example, (if any), of such a signal in real-life?

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    $\begingroup$ All real-life signals are finite-energy signals since they began when you turned on the equipment when you walked into the lab this morning or since the last time Windows crashed or since the Big Bang occurred. Now, you might use your imagination and consider the possibility that a finite power signal such as a pure sinusoid will continue on for ever, Armageddon and similar events notwithstanding, but this is a leap of faith since you are unlikely to be around to verify that this does actually happen. So, No, there is no real-life signal that is not a finite-energy signal. $\endgroup$ Jul 31, 2013 at 14:08
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    $\begingroup$ Your app has turned off exception handling. The summer intern just checked in some bogus code that divides by zero, and so you're trying to feed your FFT a vector full on NaNs. $\endgroup$
    – hotpaw2
    Jul 31, 2013 at 14:32
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    $\begingroup$ @DilipSarwate You should make that an answer. $\endgroup$
    – Jim Clay
    Jul 31, 2013 at 14:39
  • $\begingroup$ @JimClay and TheGrapeBeyond Thanks for the suggestion but I will leave it the way it is as a comment and not make it into an answer. Five people thus far have gone on record as finding the comment interesting; I doubt it would have garnered as many upvotes had it been posted as an answer. $\endgroup$ Jul 31, 2013 at 15:58
  • $\begingroup$ @JimClay What is the main reason that our professors tells us about the cases where fourier transform does not exist then? I mean, for what purpose does it serve in real life in this case? $\endgroup$ Aug 1, 2013 at 14:28

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All real-life signals are finite energy. The universe contains a fixed (and finite) quantity of energy, which has been unchanged since it came into being.

A signal's energy is given by

$E =\int_{-\infty}^{\infty}|x(t)|^2dt$

Thus, the only way to make a signal's energy go to infinity is to allow it to continue for infinite time or reach an infinite peak level. While useful in mathematical and/or physics theory, neither of these is possible in reality.

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