Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$f(x) = \cos(x^2)$ and $g(k) = \sqrt\pi \cos((\pi k)^2 - \pi/4)$ are a Fourier pair.

I want to reproduce $g(k)$ by Fourier integrating $f(x)$ using FFT, i.e.

approximating Integrate[ f(x) * exp(2 pi * ikx), {x, -inf, inf} ]

with Sum[ fn * exp(2 pi * ik x_n), {n, 0, N-1} ] * Delta_x

However the result agrees with $g(k)$ only on very small $k$ ranges if it agrees at all (the same code works well for smooth Fourier pairs e.g. the Gaussian functions). I guess the problem is choosing appropriate values for N and Delta_x. Are there any established rules for how to choose them? Where can I find related topics in literature (I've read Numerical Recipe section 13.9 but it does not seem to solve my problem)?

share|cite|improve this question
What is a 'Fourier pair'? – copper.hat May 9 '12 at 16:32
@copper: one's a Fourier transform of the other, apparently... – J. M. May 9 '12 at 16:36
up vote 1 down vote accepted

I'd be surprised if you can make this work well. The smooth pairs are not only smooth; the smoothness of one function implies the rapid decay of the other, whereas these functions don't decay at all. You need a bit of sophisticated theory to even make these Fourier transforms well-defined because the naive integrals don't converge, so it's unlikely that you can approximate them well by a finite sum.

share|cite|improve this answer
I am still wondering that, since these Fourier transforms are still "well defined" in some sense, shouldn't a dense sampling of N>>1 cycles of f(x) give a good approximation of g(k) for k<klim, with klim determined by the density of the sampling? Naively I expect klim ~ 1/delta_x, but in reality klim is much smaller than that... You are right that the actual klim should have to do with the converging property of the Fourier integral. Is there perhaps a way to quantify this? – user1342516 May 10 '12 at 11:51
@joriki Do you have a solution to the numerical evaluation of the Fourier transform with oscillating function? Thank you – ecook Mar 21 '15 at 14:03

The "established rule" you're looking for is related to the Nyquist-Shannon sampling theorem, so I'd suggest researching that term. Non-rigorously, you have to sample things at least twice per up/down wiggle in order to not lose information about the function. This kind of makes sense: you have to catch every peak and trough, or you're not getting the whole picture. You have to sample twice per period of the highest frequency present in the function. If you calculate the FFT of such a properly sampled sequence, you'll see that the FFT well approximates the Fourier transform.

Unfortunately, your function is of infinite support in both the $x$ and $k$ domains: it does not have a highest frequency, so the Nyquist theorem does not strictly apply. However, we can still non-rigorously apply it. The results depend on where you truncate $f(x)$, since you can't feed infinite samples into an FFT.

Let's calculate the "instantaneous" frequency of $f(x)$. It is given by the derivative of the cosine argument: $$ k_\text{inst}(x) = \frac{d}{dx}x^2 = 2x $$ At the endpoints of a finite domain of truncation $x\in(-x_c,x_c)$, the highest frequency present is $$ k_\text{inst-max}=2x_c, $$ and the Nyquist rate is double that $$ k_\text{Nyq}\approx 4x_c $$ To convert to a "length" per sample (where "length" is really your units of measure for $x$) we take the reciprocal and multiply by $2\pi$ $$ \nu_s=\frac{2\pi}{4x_c}=\frac{\pi}{2x_c} $$ The sampling length should be at least that; in practice you take some fraction of that that length for finer resolution (oversampling). This is Delta_x in your parlance. Here's an example with $x_c=5$ and no oversampling ($\nu_s$ as above):

a plot

You can see that the samples just barely capture the peaks and valleys of the function most of the time. For better resolution, up the oversampling; here's an example with the sampling length halved.

A figure

As you can see, this does a better job of capturing the behavior of the function. Finally, you'll only see agreement in the $k$ domain with the expected Fourier transform over the limited set of frequencies that are present in your truncation of $f$, and even then you'll see better and better agreement at higher and higher oversampling.

Finally, there are some other things you can do like windowing the function before taking the FFT to smooth out issues in the $k$ domain, so I would suggest looking into window functions on Wikipedia.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.