Let's say I know the Fourier transform of a function that is $0$ outside some interval, for example a triangle wave. How can this be used to find the Fourier series of the related periodic function, in this case a train of triangle waves?

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    $\begingroup$ We have the following result: "If $f(t)$ is an integrable function with fourier transform $F(\omega )$ and $T$ is a positive constant,then $\sum _{i=-\infty }^{\infty } f(t+\text{iT})$ is also an integrable function with nth fourier coefficent $\frac{1}{T}F\left(\frac{n}{T}\right)$. Note that the new function is $T$ periodic and i guess is your "the related periodic" function $\endgroup$ – bleh Dec 12 '11 at 16:42

Another(slightly informal but handy) way of seeing it, using tempered distributions:

If $f(t)$ is the triangle function with Fourier transform $F(\omega)$, the train of triangles can be expressed as a convolution $f_T(t)= f(t) \star d_T(t)$ where $d_T(t) = \sum_k \delta(t-kT)$ is an impulse train (or Dirac comb). And the Fourier transform of $d_T(t)$, $D_T(\omega)$ is another impulse train (separated in frequency by 1/T).

Hence, the Fourier transform of $f_T(t)$ is given by $F(\omega) D_T(\omega)$, i.e., a train of Dirac deltas weighted by the values of the original Fourier transform at uniformly sampled frequencies.

Finally, recall that the (coefficients of the) Fourier series a periodic signal are given by the weights of those Dirac deltas.

In short: the Fourier series of the train of triangles is obtained by sampling the original Fourier transform at equally spaced (1/T) frequencies.


If $f$ is your function defined on an interval and $\hat f$ the Fourier transform, then $\hat f$ is the function describing the coefficients which appear in the Fourier series representation of $f$. Think of the Fourier transform as projecting the signal $f$ onto the basis of functions consisting of sines and cosines. You then reconstruct $f$ by taking weighted sums of the basis functions (according to the weights prescribed by the Fourier transform). Note that this new function will be a periodic extension of $f$ to the whole real line.

Wikipedia contains a worked example for a saw tooth wave.

  • $\begingroup$ That Wikipedia article doesn't even seem to mention the Fourier transform of a single period -- am I missing something? $\endgroup$ – Bjorn Dec 12 '11 at 16:14
  • $\begingroup$ @Bjorn if you look at the computation of the coefficients, they are integrating over the interval $[-\pi,\pi]$. So you could pretend that the original function was only supported in this interval (and zero elsewhere); you'd get the same integral. $\endgroup$ – dls Dec 12 '11 at 16:33

You can easily get the Fourier coefficients for the periodic version of the function by sampling the Fourier transform of one period. This process depends on how you periodically extend the function, here is one example:

Let $f: {\mathbb R} \to {\mathbb R}$ with, say, $f(x)=0$ for $x \notin [0,\pi]$ and you suppose we know $\hat{f}$. Let $g$ be the periodic extension of $f$ to all of ${\mathbb R}$ defined by first reflecting $f$ about $x=0$ and then extending periodically, hence $g$ is an even function. You could also do an odd reflection here if you wanted, say, to have an odd periodic function and then make modifications below.

Then you want the Fourier coefficients of $g$, which, since $g$ is even in this case, are given by

$$ a_n = \frac{1}{\pi} \int_{-\pi}^\pi g(x) \cos(nx) dx = \frac{2}{\pi} \int_0^\pi f(x)\cos(nx)dx.$$

Note now that (depending on which convention you use for the Fourier transform), you have

$$ \hat{f}(\omega) = \int_{-\infty}^\infty f(x) e^{-i\omega x} dx = \int_0^\pi f(x)\cos(\omega x) + if(x)\sin(\omega x) dx.$$

By looking at the real part of $\hat{f}$ we have

$${\rm Re}(\hat{f}(\omega)) = \int_0^\pi f(x)\cos(\omega x) dx.$$

Equating this with $a_n$, we see that the Fourier series coefficients are given by

$$ a_n = \frac{\pi}{2} {\rm Re}(\hat{f}(n)).$$


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