# Motivation of time-frequency analysis

Can anyone give me an example of two signals with different temporal waveforms having the same Fourier transform (FT)?

Would the inverse Fourier transform still be able to recover correctly each signal?

Actually, I tried to check the question above, in matlab, using two chirp signals (same duration), the first one having a frequency increasing from 10 to 50 Hz as time increases and the other one having a frequency decreasing from 50 to 10 Hz as time increases. The result was that the moduli of the FT were identical but not the phases (that were symmetric wrt frequency = 0) even with the addition of noise.

What surprised me is that the inverse FT was able to recover correctly both signals when I was expecting it to not be able to do it since we usually say that the FT makes it so that we lose the time localization.

I am aware that the time localization information is still contained in the phase part of the FT but if this is the case then what do we really need the time-frequency representations for? Is it just to easily extract this information since it's so hard doing it using the phase of the FT?

Any thoughts on that would be appreciated.

• TIme-frequency representations are a load of codswallop (AKA BS). They are an interesting thought experiment, but anything in that area is more rigorously done using wavelets. :-) The motivation? Publishing papers! – Peter K. May 24 '17 at 2:03
• @PeterK. Really! a load of codswallop! I think I'll read the motivation paper you gave me before I say anything :) Just one question though, do you agree with me that the FT does not discard the time localization information but keeps it in its phase? which to say is the inverse of the argument usually put up front to motivate time-freq representations against the FT!!! – Learn_and_Share May 24 '17 at 2:39
• yes, I agree. The FT is a unitary invertible transform: the inverse perfectly reconstructs the time signal. – Peter K. May 24 '17 at 2:50
• @PeterK. Okay. Then, I really need to check my understanding for the motivation of the time-frequency representations. – Learn_and_Share May 24 '17 at 3:13
• All answers and comments were great, Peter K., Fat32, Laurent Duval and hotpaw2, and each one added something new to the discussion. I wish I could choose all the answers as accepted but unfortunately I can't, I had to choose and I went with the first answer (having the most votes too). – Learn_and_Share May 26 '17 at 10:21

What a tricky question to overlook. Indeed I'm one of those who would immedieately press that Fourier transforms do lose time localization of the events as the comments stated. Yet it's certainly (mathematically and practically) true that any (transformable) signal waveform is exactly preserved under this reversible transform including all of its time distribution information as well. Your example of the reversed chirp case clearly demonstrates this. This needs an answer. At least to clarify why then there is the well accepted idea that FT do not preserve time localization of events?

Now, the idea that the Fourier transforms are losing time localization of the signals come from the observation that its bases are of infinite extent sinusoidals. And a sine wave of infinite extent and constant amplitude will not have a time localization. On the other hand it has a perfect frequency localization, being a pair of impulses at an exact frequency. The bases of wavelets, for example, are partly local in both domains.

This is the essence of time-frequency analysis. A base that's exactly local in frequency will loose all time localization and a base that's exactly local in time will loose any frequency localization. And those in between are the transforms that provide a compromise between exact localization and no localization at all.

A consequence of the fact that Fourier bases are of infininte extent is the following: Assume part of a signal contains a short duration high frequency spike, and the remaining parts are of quite still low frequency variations. When a Fourier decomposition or synthesis is used to analyse or construct such a signal, the high frequency spike (which is higly local in time) is to be created with high frequency sine bases that not only exist at the position of the spike but also exist along all the rest of the signal. Because those sines will extend from the beginning to the end of the whole signal duration. This creates the problem that local processing is quite inefficient with Fourier bases for transient signals. And transient signals have the prime importance in many mathematical, scientific and engineering fields.

Wavelets are kind of the optimal (probably in some sense?) transforms that provide a maximum amount of simultaneous time-frequency localizations (resolutions). And this is clearly apparent from the wave-packet (or the gaussian, or the maxican hat, or a shapeless daubechies...) shape of its bases. Wavelets therefore can provide an improved solution the problem associated with the previous paragraph.

A spectogram is partly providing similar information to a wavelet time-frequency analysis, and gives you frequency contents of the signal over its time duration.

Note that you can still go with Fourier analysis for transient cases as well but that requires what Peter K. referred to as a codswallop work...

• So to sum up, and correct me if I'm wrong, it's not that the FT loses the time information contained in the signal but instead it's just not fully adapted to analyzing highly localized (in time) signals? – Learn_and_Share May 25 '17 at 15:49
• Also, what do you mean by "local processing is quite inefficient with Fourier bases for transient signals". Is it for the example you gave the fact that when we do the Fourier decomposition we will need to somehow correct for the part of the high freq sinusoids that exists despite it is not needed outside the spike duration? – Learn_and_Share May 25 '17 at 15:53
• Your both comments are practically right. Position information (chronological order of events) in a given signal is encoded in the phase of the associated Fourier transform. Magnitude carries no such information. Therefore the chirp and the reversed-chirp will have the same Fourier magnitude spectrum but different phase spectrums. On the other hand I cannot easily quantify the inefficiency of local processing due to FT, as this will highly dependent on the application under concern. – Fat32 May 25 '17 at 20:09

Fourier transforms generally yield complex spectral data. Under some technical conditions, they are bijections. From the Fourier transform, you can uniquely recover one single signal. However, when looking at spectra, the situation is different: signals can have the same amplitude spectrum and very different phases, as in the example below:

Being bijections, Fourier transforms don't lose time localization; however, amplitude spectra somehow do. One issue is that the phase encodes time or space localization in a way that it is not easy to read or decipher in general. One interpretation is that, for less stationary data, the phase can change very fast, and being known in $]-\pi,\pi]$, it looks like a truncated signal, difficult to unwrap. This can be seen especially for images. Below you have a row for Lenna, and one for the Cornouaille boat.

The second column is the amplitude spectrum. Shaky, but readable. The third column is the unwrapped phase. It seems to have not structure at all. However, it encodes most of the changes in the original image. You can test this by swapping moduli and phases: this is the fourth column. If you take the modulus of Lenna spectrum and the phase of the boat, and do an inverse Fourier transform, you get a ghostly boat, almost no trace of Lenna. You can see the converse on the bottom row. A phase component carries a lot of information, but it is not easy to make sense of it when it is not localized.

One way to localize spectra is to use a window $h$, and you easily get short-time Fourier transforms:

$$S_s(\tau,f;h) = \int s(t)h^*(t-\tau) e^{-\imath 2 \pi f t} dt$$

but if you consider $h$ as an infinite constant window, then $h^*(t-\tau)$ is constant, and you recover the standard Fourier transform:

$$S_s(\cdot,f;1) = \int s(t) e^{-\imath 2 \pi f t} dt$$

But people are still trying to understand complex phase patterns, even with short-time Fourier. Windowing does not unveil everything.

[EDIT] Finding a best among time-frequency or time-scale representations is a difficult matter for several reasons (not exhaustive):

• "best" often relate to objective metrics. The traditional trade-off (Weyl-Pauli-Heisenberg) is stated as lower bounds in $L_2$ norms, which is well-suited to dual time/frequency vector spaces. However, it does not tell you much about how the representation is simple, sparse, and many other information metrics (entropies, divergences) are much much informative to that respect, but way much difficult to compute.
• "best" for all data is not so meaningful, one ought to restrict to more specific data models and spaces, and to derive "best" within such classes: piecewise polynomials, sum of peaks...
• digital signals require a discretization of the above continuous schemes, and the way one discretizes has a lot of impact.
• actual measurements do not obey standard assumptions: all is not linear, realizations don't precisely follow theoretical distributions, short data don't get asymptotic properties.
• ultimately, a theoretical best is not the one you can easily parameterize to get reproducible results.

If the morphology of your data is diverse (like spikes + oscillations), you may having to try mastering different tools: frames, unions of bases, data transformations, etc.

• Insightful answer, thanks. What is your take, then, on wavelet transform compared to windowing and time-frequency representations? Can we say that there is one "best" way to encode time/space information (localization) without disregarding the frequency localization? I guess people agree that a trade-off is inevitable, still I want to confirm whether this is like an inescapable faith for the time/frequency couple :-) as long as there is a need to be together or is it something we can bend somehow... – Learn_and_Share May 26 '17 at 5:09
• That's a fair share of information you've added, thanks. You've already highlighted what to consider if you face such a question and that's enough for me :-) – Learn_and_Share May 26 '17 at 10:14
• Sure: I have been working a lot with wavelets (they help me understand the powerful Fourier, still a mystery to me), yet I don't think many discrete signals are really multiscale. The whole framework of oversampled complex filter banks is convenient to me, yet designing them for tailored purposes remains a challenge – Laurent Duval May 26 '17 at 10:25

An FT (being invertible) does preserve all transient event time information, however this time locality information is usually preserved by distributed it as varying changes to the entire phase spectrum, which obscures (or is almost like encrypting) all the time locality information. Humans have to decrypt (inverse FT) the phase to make any sense out of transient events in terms of a time (early, late, when, etc.)

Spectrogram or STFT-like Time-Frequency representations pre-window the data to pre-preserve some time locality information (in the window parameters) before all the frequency transforms converts (obscures/encrypts) time locality details into a hidden or much less intuitive form. In the case of wavelets, a form of windowing to pre-preserve time locality can be done within each transform basis vector, instead of to the signal or data before an FT.

• Does the fact that the FT preserves all event time information mean that we can't find two FTs that are identical (modulus and phase) unless they are the FTs of the same signal? – Learn_and_Share May 25 '17 at 15:55
• Note that IFT(FT(x)) = x – hotpaw2 May 25 '17 at 15:57
• I'll take that as a yes :-) – Learn_and_Share May 25 '17 at 16:03
• Which is better (if a comparison is possible)? time freq. representations that pre-window the "data" or wavelets that uses localized basis? – Learn_and_Share May 25 '17 at 16:13
• "Better" depends on your exact requirements. Perhaps both. Perhaps neither. – hotpaw2 May 25 '17 at 16:28