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Take for example one period of a sine:

$f(x) = \{\sin(\omega_1x) \; \mathrm{if} \; x \in [0, 2\pi) \;; \quad 0 \; \mathrm{elsewhere} \}$

If we now translate $f$, then according to the argument that the Fourier transform doesn't localize in time (answer stating this), we wouldn't be able to tell a difference.

However, the Fourier transform has an inverse, especially for a function like the above from $L^2$, so there must be way to tell $f$ and it's shifted variants apart.

I assume that the way the complex spectrum encodes position through phase difference with a lot of leakage in the case of a non-infinite sine is not useful to localize a signal in time and that's why the transform is said to not localize in time. In theory though all the information is encoded, just in an incomprehensible way. Is this correct?

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Sir Vivor
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  • The individual sines with amplitude and phase as calculated by the transform extend from 'minus infinity to infinity', hence no localization. Yet the interference of all sines makes that it is only non-zero between 0 and 2pi in your example. – Martijn Pot Jan 23 '15 at 20:36

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Translation of a signal is reflected by a (complex) rotation in the Fourier domain as the well-known property states ($\mathcal{F}(s(t-\tau))=e^{-j\omega\tau}\mathcal{F}(s(t))$). Therefore, the statement that the Fourier transform is not able to tell the difference due to time shifts is incorrect since, if the latter was the case, the translated signals would have the same Fourier representation.

The statement that the Fourier transform is not localized in time does not mean that it is unable to represent non-stationary signals and/or identify signal properties related to time shifts e.t.c. It means that the Fourier transform is not practical for easily identifying non-stationarities as the latter are masked in the phase of the transform. Ideally, we would like a transform that represents the time-frequency properties of a signal by examining only the magnitude of the transform.

Clearly, the Fourier transform fails in doing so (e.g., the magnitude of the Fourier transform of translated signals is exactly the same - it is only in this sense that the Fourier transform fails to discriminate between the two signals). For this reason, other transforms, notably, short time Fourier transform and wavelet transform are employed for representing and analyzing non-stationary signals.

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