10
$\begingroup$

I am fond of Fourier series & Fourier transform.

But every approach has some outcomes and some shortcomings. It's limitations lead to innovation of new approach. So, can anybody explain about

  1. The limitations/ shortcomings of the Fourier Series?

  2. The limitations/ shortcomings of the Fourier Transform?

$\endgroup$
13
  • $\begingroup$ What is the "Fourier Theorem?" $\endgroup$ Commented May 5, 2015 at 17:07
  • $\begingroup$ Fourier series have the benefit of being discrete which makes it easy to do computationally. However it requires that your signal be on a finite domain. In practice this isn't a problem so much. However the functional analytic properties of Fourier series are not that nice. Fourier transforms deal with signals that don't have compact support and can be thought of as a translation between functions of the same type: it's a unitary map on an inner product space. Fourier series don't have this property which makes them so much harder to study in full detail. $\endgroup$ Commented May 6, 2015 at 6:09
  • $\begingroup$ In some sense they're the same thing and you can view the Fourier transform as being a continuum certain of Fourier series under certain assumptions. The beauty of the Fourier transform is its analytic properties. $\endgroup$ Commented May 6, 2015 at 6:13
  • 1
    $\begingroup$ A limitation of the Fourier transform is that it's not truly realizable in practice - we can never sample a function for every $x\in\Bbb R$! This can be mitigated by the way we do integrals numerically: we do Riemann sums, which naturally require sampling your function anyway! So we might not get the true Fourier transform of a recorded sound for instance but we can get pretty close! $\endgroup$ Commented May 6, 2015 at 6:24
  • 1
    $\begingroup$ @pandu this question is too broad. Some of the transforms you mention liked the Z are traditionally used for analyzing processes (IIR filters, stability of my E&M code), wavelet transforms are used for band-extraction, filtering. Integral style Fourier transforms are like spectral transforms are encountered in nature with a clear meaning ascribed to the spectral domain, for example position and momentum being Fourier transform pairs. Unlike these transforms, the sided Laplace transform has different integration limits, and is often used to interpret system that are not BIBO stable. $\endgroup$
    – Mikhail
    Commented May 7, 2015 at 23:58

2 Answers 2

8
+50
$\begingroup$

Here is my biased and probably incomplete take on the advantages and limitations of both Fourier series and the Fourier transform, as a tool for math and signal processing.

Advantages

  • Fourier series and the Fourier transform hold a unique place in the analysis of many linear operators, essentially because the complex exponentials are the eigenvectors/eigenfunctions of linear, shift-invariant operators. In signal processing this is illustrated via the convolution theorem, though the theory goes much deeper (see: Pseduo-differential operators). Related to this is the role of the Fourier transform in the mathematical foundations of quantum mechanics - Fourier analysis is directly related to "momentum", since the eigenfunctions of the momentum operator $-i\partial_x$ are the complex exponentials.
  • In this same vein, Fourier analysis leads to an extremely powerful theory of smoothness, because of the correspondence between differentiability and decay of the Fourier coefficients. See Sobolev spaces.
  • Fourier analysis is very powerful in the study of generalized functions.
  • From a numerical analysis and signal processing point of view, the accuracy of Fourier based methods have the advantage of being limited only by the smoothness of the underlying function. This means several things: Fourier methods are very good at approximating very smooth things, but perhaps not so good at approximating less smooth things. See "disadvantages".
  • The general techniques we learn from Fourier, like expanding functions in an orthonormal basis, are extremely powerful. See spectral theory.

Disadvantages

  • First off, from a numerical standpoint, issues of convergence play a massive role. See Gibbs Phenomenon. This leads to a secondary issue that Fourier series are not "efficient" at resolving discontinuous or multi-scale functions. This is illustrated, for example, by the vast difference between original JPEG image compression, which is based on Fourier series, and modern image compression techniques like JPEG2000, which are based on more multi-scale techniques like Wavelets.
  • Related to the above fact is that Fourier series give no information on the spatial/temporal localization of features. A Fourier series or transform can tell you that there is a discontinuity, but it can't tell you where it is. Think of a musical score: having just the Fourier transform is like knowing which notes you need to play, but not when to play them. Not very useful if you want to hear music! This is partially what inspired the study of phase-space/time-frequency/wavelet representations (which incidentally are playing an increasing role in quantum theory).
  • Classical Fourier analysis is less generally applicable for nonlinear and nonstationary/transient phenomenon (although it is still hugely powerful in some cases!)
$\endgroup$
5
  • $\begingroup$ >>but not when to play them .. This isn't true when you have the FT you have real and imaginary components (phase), the imaginary part can give you the when and the real gives you the how loud. $\endgroup$
    – Mikhail
    Commented May 12, 2015 at 22:08
  • $\begingroup$ @Mikhail: The power of each frequency is the square module of the coefficient of that frequency. What do you exactly mean by "how loud" and how is it related to the real part of the coefficient (or transformed function)? The same question relates to the imaginary part and the "when". $\endgroup$
    – Hans
    Commented May 14, 2015 at 16:41
  • $\begingroup$ @Hans certainly. There are two frequently used conventions, one is to use two terms A_n and B_n another is to combine that information into C_n where C_n can be imaginary. When using C_n the location is encoded in the phase, if you IFT you get back exactly what you started with. When we take the conjugate squared (conj(x).x or <x|x>) you loose this information, and only get something related to the absolute value. From a cursory Google search: mathworks.com/matlabcentral/fileexchange/… $\endgroup$
    – Mikhail
    Commented May 15, 2015 at 4:38
  • $\begingroup$ @Mikhail: I am still not very clear of what you are saying. Certainly, if the original function is real, the coefficients of the conjugate terms of the same frequency is conjugate to each other. One can certainly separate each coefficient into real and imaginary parts. Is that what you mean by $A_n$, $B_n$ and $C_n$. Otherwise, it is not clear how you define these coefficients. Taking inverse Fourier transform certainly gets back the original function. There does not seem to be anything special about it. Could you please provide a reference which provide all the specifics and theorems? Thanks. $\endgroup$
    – Hans
    Commented May 15, 2015 at 16:01
  • $\begingroup$ @Mikhail: I think icurays1 is saying that one can not pick a finite number of frequency space components of a Fourier expansion and reconstruct a part of the original time domain function localized near a given time point. Is your comment saying otherwise? $\endgroup$
    – Hans
    Commented May 15, 2015 at 20:14
2
$\begingroup$

The integral of a canonical Fourier transform must converge, meaning the bandwidth of the signal is somewhat limited. Now consider, the difficulty in interpreting the Fourier transform for even the most common functions, such as cosine, or more interestingly functions like rand(x).

$\endgroup$
4
  • $\begingroup$ This is not quite true. Because the Fourier transform maps the Schwarz class to itself, we can exploit duality to define it on on any tempered distribution. (The idea is that we pick the unique extension which is unitary with respect to the pairing between a distribution and a function.) This includes all $L^p$ functions ($1 \leq p \leq \infty$) as well as all compactly supported distributions, such as the Dirac delta. But not all of the nice properties are retained when we extend to tempered distributions. $\endgroup$
    – Ian
    Commented May 7, 2015 at 1:14
  • $\begingroup$ >such as cosine fourier transform does not have any difficulty at all with cosine $\endgroup$ Commented May 16, 2015 at 15:46
  • $\begingroup$ What's $\mathtt{rand}(x)$? $\endgroup$
    – wlad
    Commented May 16, 2015 at 16:09
  • $\begingroup$ @user3491648 A function which picks a random number from a uniform distribution, critically the integral for such a thing strictly diverges, nevertheless it is useful to form a generalized Fourier transform. $\endgroup$
    – Mikhail
    Commented May 17, 2015 at 7:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .