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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 ?

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  • $\begingroup$ Yes, that is correct (ignoring scaling factors, as you say). $\endgroup$
    – MBaz
    Commented Feb 18, 2016 at 2:05
  • $\begingroup$ 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. $\endgroup$
    – robert bristow-johnson
    Commented Feb 18, 2016 at 3:03
  • $\begingroup$ @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. $\endgroup$
    – Abhinav Vishak Iyappan
    Commented Feb 18, 2016 at 14:07

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Regarding reciprocal domains, such as the "time" domain and "frequency" domain, that are related to each other by the Fourier Transform, whenever you uniformly sample in one domain, it causes periodic extension in the other domain. It does that always.

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 $-\pi < \omega \le +\pi$ .

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  • $\begingroup$ 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 $\delta(t)$ $\endgroup$
    – Fat32
    Commented Feb 18, 2016 at 13:40
  • $\begingroup$ 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. $\endgroup$
    – robert bristow-johnson
    Commented Feb 18, 2016 at 16:14

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