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