# What's the difference between “Discrete Fourier Series” and “Discrete Fourier Transform”? [duplicate]

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"?

• 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? – MrCasuality Jan 23 at 14:08
• @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. – Dole Jan 23 at 14:10
• The main difference between them is DFT=N.(DFS). – Ch.Siva Ram Kishore Jan 23 at 14:12
• 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... – MrCasuality Jan 23 at 14:16
• a common statistical variation being additive gaussian white noise (AGWN) which averages itself out as n approaches infinity... – MrCasuality Jan 23 at 14:29

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 $$x[n]$$, $$X[k]$$.
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.