Use Fourier transform to find Fourier series coeficcients

Question

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 $x(t)$ .

Answer 1

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 $$ \mathcal{F}\{x(t)\} = \sum_{n=-\infty}^\infty X_n \mathcal{F}\{e^{i 2 \pi n/T_0}\} = \sum_{n=-\infty}^\infty X_n \delta(\omega - 2 \pi n/T_0) $$

Answer 2

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)

Answer 3

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.