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I look at the equations for DFS (Discrete Fourier Series) and DFT (Discrete Fourier transform) and the only difference I notice is that one has a squiggle above the letter and the other doesn't. The equations for x[n] and X[k] are exactly the same except for the presence or absence of a squiggle.

Example:

DFS: $\tilde{x}[n]$ / $\tilde{X}[k]$

compared to:

DTF: $x[n]$ / $X[k]$

What's the difference between "Discrete Fourier Series" and "Discrete Fourier Transform"?

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  • $\begingroup$ I;m going to take a guess that DFS sequence $\tilde{x}[n]$ includes only the first N samples of sequence x[n], and they are repeated over and over ad infinitum... and the DFT is all the samples of x[n] assuming they are periodic with period N?? what's the difference? $\endgroup$
    – MrCasuality
    Jan 23, 2019 at 14:08
  • $\begingroup$ @MrCausality Your guess is probably correct. Then the definitions coincide with DTFT (=Discrere time fourier transform) and DFT (=Discrete fourier transform), which are Fourier transform and series for discrete signals respectively. $\endgroup$
    – Dole
    Jan 23, 2019 at 14:10
  • $\begingroup$ The main difference between them is DFT=N.(DFS). $\endgroup$
    – Ch.Siva Ram Kishore
    Jan 23, 2019 at 14:12
  • $\begingroup$ maybe with DFT you can have statistical variation in the periodic x[n] signal as long as it approaches the same periodic sequence as n approaches infinity... whereas DFS is exactly the same sequence without statistical variations?? I;m just taking a guess because the DFS/DFT description is from a deterministic DSP book...which is missing a little bit of the theory on statistical side... $\endgroup$
    – MrCasuality
    Jan 23, 2019 at 14:16
  • $\begingroup$ a common statistical variation being additive gaussian white noise (AGWN) which averages itself out as n approaches infinity... $\endgroup$
    – MrCasuality
    Jan 23, 2019 at 14:29

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They are the same thing. Just the same.

But for some reasons, we want to call the first period of the DFS sequences $\tilde{x}[n]$ and $\tilde{X}[k]$, as the new DFT sequences $x[n]$, $X[k]$.

They are both periodic. The periodicity of DFS is natural extension of periodicity of complex exponential and periodicity of time sequence, but the periodicity of DFT is also forced by the modulus operator.

Be careful! DFT is also linked with finite length aperiodic sequences $x[n]$ and their DTFT via the frequency sampling relation; i.e., $X[k] = X(e^{j \frac{2\pi}{N}k})$.

Because of this latter relation of the DFT $X[k]$ to the DTFT $X(e^{j \omega})$ of aperiodic (and finite length) sequences $x[n]$, in order to remember and imply the inherent periodicity in DFT (due to its definition through DFS) we shall enforce the modulus operator into the arguments of DFT sequences.

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