# Relation between Fourier Series & Fourier transform [duplicate]

So i was just revising some basic DSP concepts. Just wanted to verify this fact.

Fourier series represents a periodic signal $\hat{x}(t)$ with period P as a countably infinite sum of sinusoids of frequency $0$, $\frac{1}{P},\frac{2}{P},\frac{3}{P}...$. This converges to the signal in the interval, $-\frac{P}{2} < t < +\frac{P}{2}$, and if the time domain signal is periodic, then over the whole time domain.

Fourier Transform is sorta like a limit of the fourier series where P goes to $\infty$.

So i know that the fourier transform of $\operatorname{rect}(t)$ is $\operatorname{sinc}(f)$ ( ignoring the scaling factors ) . And that the fourier series of a $\operatorname{rect}()$ is given by http://mathworld.wolfram.com/FourierSeriesSquareWave.html ( which is also a $\operatorname{sinc}()$ in the frequency domain ) .

I just wanted to confirm the following

If I sample the $\operatorname{sinc}()$ i obtain from the fourier transform of a $\operatorname{rect}()$, and use those values to reconstruct a fourier series, will i end up getting a square wave ?

• Yes, that is correct (ignoring scaling factors, as you say). – MBaz Feb 18 '16 at 2:05
• there is a small problem directly comparing the periodic extension of $\operatorname{rect}(t)$ and the Wolfram square wave page you cite and that is that the periodic extension of $\operatorname{rect}(t)$ must be an even-symmetry function with a DC component (because it toggles between 0 and 1) with only $\cos()$ terms in the Fourier Series and the Wolfram page is an odd-symmetry square wave, so it has only $\sin()$ terms in the Fourier Series and has no DC component. – robert bristow-johnson Feb 18 '16 at 3:03
• @robertbristow-johnson I just realized that, i just wanted to get my general understanding right without figuring out the mathematical details like phase/time shifts and scalings etc. – Abhinav Vishak Iyappan Feb 18 '16 at 14:07

So the Fourier Series is an example of sampling in the frequency domain causing periodicity in the time domain. The corresponding example (using duality of the Fourier Transform) is the so-called DTFT (Discrete-Time Fourier Transform) $X(e^{j\omega})$ of a sampled sequence $x[n]$ which is naturally periodic with period $2 \pi$. And normally we consider $X(e^{j\omega})$ with $-\pi < \omega \le +\pi$ .
• I try to apply this uniform sampling argument to the CTFT $X(\omega)$ of a pure sinusoid $x(t)=\cos(\omega t)$, but I think I cannot remember the consequence of sampling an impulse function which is to be used in order to represent CTFT of the $x(t)$ at least convergent informally via generalized distributions of $\delta(t)$ – Fat32 Feb 18 '16 at 13:40
• you can't really sample $\delta(t)$ or $\delta(f)$. (and those functions don't really exist in reality anyway.) if you were attempting to sample the Fourier Transform of $x(t) = \cos(2 pi f_0 t)$ (which is $X(f) = \frac{1}{2}\left(\delta(f+f_0)+\delta(f+f_0)\right)$, what you would be attempting is to periodically extend $x(t)$ (which was previously periodic with a likely different period). if the sampling "period" in the frequency domain was not exactly $f_0$, then you get zero, which means periodically extending $\cos(2 \pi f_0 t)$ by some other period than $\frac{1}{f_0}$ adds to zero. – robert bristow-johnson Feb 18 '16 at 16:14