I understand that the Fourier Transform can be seen as a generalisation of the Fourier Series, where the period $T_0 \to \infty$ . Now I have encountered this strange question (in an engineering course on signal analysis):

Given a periodic function $x(t)$, find the Fourier Series coefficients $X_n$ by using the Fourier Transform.

What does this mean? How can it be done? As I see it, FS and FT are similar concepts, but they are not the same operation.

For reference, $x(t) = rect(\frac{t-0.25}{0.25}) * \Delta _1 (t)$ but I am seeking an answer in terms of any periodic function $x(t)$ .

  • $\begingroup$ In order to find the Fourier Series, we need to specify the period. In other words, what is $T_0$ in this quetion? $\endgroup$ – Omnomnomnom Mar 10 '14 at 20:08
  • $\begingroup$ What exactly is $\Delta_1(t)$? Is it some sort of sum of delta functions? $\endgroup$ – Omnomnomnom Mar 10 '14 at 20:28
  • $\begingroup$ $\Delta _1 (t)$ is the so-called "Dirac Comb", which is a periodic extension of the "Dirac Delta" or "Impulse" function. This is the notation given by my texbook, "Modern Digital and Analog Communication Systems" by Lathi and Ding. In this notation, the subscript of Delta is the period. Therefore the period of $x(t)$ is 1. $\endgroup$ – rudolfbyker Mar 10 '14 at 20:36
  • $\begingroup$ Using the accepted answer by @Omnomnomnom, the correct answer for the example is: $X_n=\frac{1}{4} sinc(\frac{\pi n}{4}) e^{- \frac{j \pi n}{2}}$ $\endgroup$ – rudolfbyker Mar 10 '14 at 21:19

Suppose we have a function $\tilde x(t)$ that is zero except on the interval $[-T_0/2,T_0/2]$ (on which $\tilde x(t) = x(t)$) and whose Fourier transform is given by $$ \widehat x(\omega) = \int_{-\infty}^\infty \tilde x(t) e^{-i\omega t}dt = \int_{-T_0/2}^{T_0/2} x(t) e^{-i\omega t}dt $$ Using $\widehat x(\omega)$, we would like to find the Fourier series for the $T_0$-periodic function that agrees with $x(t)$ on this interval. We note that the coefficients of the Fourier series for $x$ are given by $$ X_n = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x(t)e^{-i (2 \pi n/T_0) t}\,dt $$ for any integer $n$. Notice the similarity! From here, you can derive $$ X_n = \frac{1}{T_0}\widehat{x}(2 \pi n/T_0) $$ Alternatively, let's say you wanted to look directly at $\mathcal F\{x(t)\}$. Note that $x(t) = \sum_{n = -\infty}^\infty X_n e^{i 2 \pi n/T_0}$. It follows that $$ \mathcal{F}\{x(t)\} = \sum_{n=-\infty}^\infty X_n \mathcal{F}\{e^{i 2 \pi n/T_0}\} = \sum_{n=-\infty}^\infty X_n \delta(\omega - 2 \pi n/T_0) $$

  • $\begingroup$ This is starting to make more sense now! Although there is a slight difference between the original question and this answer which still confuses me: In your answer, x(t) is non-periodic to start with. Am I correct in understanding that your non-periodic x(t) is a "helping" function to be able to do the FT, then you substited the FT into the expression for FS coefficients, which will give a FS which is equal to the periodic extension of x(t) ? Then maybe we should call the non-periodic function something else, like p(t)? Then $x(t)=p(t)*\Delta_1(t)$ and $p(t)=rect(\frac{t-0.25}{0.25})$ ? $\endgroup$ – rudolfbyker Mar 10 '14 at 20:58
  • $\begingroup$ @rudolfbyker You have the right idea. In this case, my "helping function" was $\tilde x(t)$, as opposed to $x(t)$ (sorry if that was unclear). $\endgroup$ – Omnomnomnom Mar 10 '14 at 23:56

Strictly speaking, the fourier transform is not defined for non-zero periodic functions. Such functions aren't in $L_1$ nor in $L_2$, so the fourier transform does not exist. If you attempt to compute it, you run into trouble when doing the integration, since the integral won't converge.

To apply the fourier transform to such functions, you must first generalize it to distributions. Some engineering texts do that, although usually without formal proofs that this is actually sane. You can put this construction on a firm mathematical basis, but it's not trivial. (Though it isn't that hard either - Rudin's book "Functional Analysis" for example includes a quite accessible derivation of the main results)

  • $\begingroup$ I'm having a hard time understanding this... What is $L_1$ and $L_2$ ? $\endgroup$ – rudolfbyker Mar 10 '14 at 20:38
  • $\begingroup$ @fgp You can extend the fourier transform to $\mathcal S'$, wich will allow you to define the FT for periodic functions. Also, the FT on $\mathbb T^n$ is actually also a fourier transform, mapping $2\pi$-periodic functions to a sequence in $\ell^\infty(\mathbb Z)$. $\endgroup$ – AlexR Mar 10 '14 at 20:50
  • $\begingroup$ @AlexR If by $\mathcal{S}'$ you mean the dual of the Schwartz-space, sure, but that entails using distributions, no? In particular, the fourier transform of a periodic function will then be a distribution of the form $\sum_{i\in\mathbb{N}} c_i \delta_i$ I think. Regarding $\mathbb{T}^n$ - calling that transform a fourier transform just dodges the issue. That is (at least in the one-dimensional case) exactly what you usually call a fourier series. $\endgroup$ – fgp Mar 10 '14 at 23:05
  • $\begingroup$ @fgp The terms can be used synonymously when working with $\mathbb T$ and yes, $\mathcal S'$ denotes the space of tempered distributions. $\endgroup$ – AlexR Mar 11 '14 at 1:53
  • $\begingroup$ @AlexR Then I fail to see what your comment actually criticizes. Since the question says "As I see it, FS and FT are similar concepts, but they are not the same operation.", simply answering "they're synonyms" would exactly be helpfull. And I did explain that distributions unify fourier transforms and series, but you, well, need distributions for that. $\endgroup$ – fgp Mar 11 '14 at 2:04

You question is about the relation of the two fourier transforms, on $\mathbb R$ and on $\mathbb T = \mathbb R / 2\pi\mathbb Z$. The first fourier transform can be given for $f\in L^1(\mathbb R)$ as $$\mathcal F f(\xi) = \frac1{\sqrt{2\pi}} \int_{\mathbb R} f(x) e^{-i\xi x}dx$$ While the other maps a $2\pi$-periodic function $g:\mathbb T \to \mathbb R$ by $$\mathcal F g(k) = \frac1{2\pi} \int_{\mathbb T} g(x) e^{-ikx} dx$$ Now the relation asked for is the following identity. Let $g$ be a $2\pi$-periodic function and $f\in\mathcal S(\mathbb R)$ a schwartz-function satisfying $$g(x) = \sum_{k\in\mathbb Z} f(x+2\pi k)$$ Then $$\mathcal F g(k) = \mathcal F f(k) \qquad \forall k\in\mathbb Z$$ Where the first FT is on $\mathbb T$ and the second is on $\mathbb R$.
However, the terms fourier transform and fourier series can be used synonymously if the domain of the fourier transform is clear ($\mathbb T$). Note that other periods than $2\pi$ can be deduced from this and similar equalities hold.


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