Why is Fourier transform broken

Sami AldalahmehOctober 4, 20112 comments

Every engineer who took a basic signal processing course is familiar with the Gibbs phenomenon, however, not all know why it occurs, I mean really why!

The answer lies in the mathematical background that is almost always skipped in signal processing courses. Moreover, from my experience at least, many textbooks present the theory, e.g. the Fourier transform, as infallible and no discussion of the limitation of the topic is given.

The short answer is that the metric space of continuous function, ( C[a,b], dp ), is not complete. Here, C[a,b] is the space of continuous function on the inerval [a,b] and dp is the pth distance between two elements of the set, which are functions in this case, e.g. Euclidean distance for p=2 between the functions f(t) and g(t).

You probably are wondering what does that mean from an engineering prepective. Well, this means that an arbitrary continuous functions cannot be represented, constructed, or approximated by a sequence of other continuous functions. You might argue that many mathematical conditions are violated in practice and engineers get things done! Then I would say look at the Gibbs case, it's not working. 

Now we need another metric space that completes C[a,b] and contains it as well. Here the (Lp[a,b], dp) space comes to the rescue. The Lp[a,b] is defined as the space of functions, x(t), such that

 \int_a^b |x(t)|^p dt <\infty \;\text{where} \; 1\leq p<\infty


d_p(f(t),g(t))=\left [ \int_a^b |f(t)-g(t)|^pdt\right ]^{1/p}

In case of p=2 it is the space of finite energy signals and the distance being the root mean square distance, with slight abuse of terminology.

But this new space comes with few catches. The one of interest here is that there exist two different functions with zero distance between them, which might seems strange but is plausible when we recal that the distance is actually the mean distance, so we can think of it as the functions are equal in the mean sense. Formally speaking, the functions are equal almost everywhere, except at a countable set of points.

And this is what happens in the Gibbs phenomena; the sum of trignometric functions used in the Fourier transform are equal almost everywhere to the discontinuous functions except at the point of discontinuity, for which the Fourier approximation never converges.

Another drawback of the Fourier transform is not giving any kind of time localization, i.e., there is no information about the time location of a certain frequency component of the signal under consideration. 

Despite all that, the Fourier transform is widely popular and used in signal processing and other engineering and science areas. In the end, what I wanted to point out is engineers should be aware of the drawbacks and limitations of the tools that they use such as the celebrated Fourier transform.

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Comment by iVenkyNovember 4, 2011
Yes sir. You are right. I am doing undergraduation in Engineering and we learn Fourier formulas as it is and all of my friends accept it as a fact. I believe that Fourier representation do have some limitations.
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Comment by bbhattacNovember 9, 2011
Why do you say "finite energy signals" when the integration is carried out over a finite interval [a,b]?

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