My problem is related to the periodicity of DFT. Having the following expression

$$ Y_{k}=\sum_{n=0}^{2N-1}e^{-j\frac{2\pi mk}{2N}} $$

I can easly find that the upper function is $2N$ periodic. So if $k \in[0,1,..,2NK]$ I would get $K$ concatenated versions of original signal between $0$ ans $2N$. If I truncate the original signal (which is $2N$ long) in the time domain by a window which is $W$ times smaller I will obtain the following expression

$$ Y_{k}=\sum_{m=0}^{2N/W-1}e^{-j\frac{2\pi mk}{2N}}=\frac{\sin{\left(\frac{\pi k}{W}\right)}}{\sin{\left(\frac{\pi k}{2N}\right)}} $$

for which it still holds $2N$ periodicity assuming that $\frac{2N}{W}$ is an integer number.

  • My question is what happens if $\frac{2N}{W}$ is not an integer?
  • How will this influences the periodicity?

Since I have to take integer number of time samples I assume that $\frac{2N}{W}$ should be floored or ceiled and in that case I would get \begin{equation} Y_{k}=\sum_{m=0}^{\text{ceil}(2N/W-1)}e^{-j\frac{2\pi mk}{2N}}=\frac{\sin{\left(\frac{\pi k}{2N}\text{ceil}\left(\frac{2N}{W}\right)\right)}}{\sin{\left(\frac{\pi k}{2N}\right)}} \end{equation}

  • Is this function still $2N$ periodic? Because If I evaluated for $k \in[0,1,..,2NK$] I will get $K$ copies of the original signal but they will be somehow scaled, so not completely identical. Identical copies I only get if $\frac{2N}{W}$ is an integer.
  • Could somehow provide me explanation for that?

1 Answer 1


As you have correctly observed, $2N/W$ must be an integer, because the window can only have an integer number of samples. Furthermore, regardless of the upper summation limit,

$$Y_k=\sum_{m=0}^Ke^{-j\frac{2\pi mk}{2N}}$$

is always $2N$-periodic because

$$Y_{k+2N}=\sum_{m=0}^Ke^{-j\frac{2\pi m(k+2N)}{2N}}=\sum_{m=0}^Ke^{-j\frac{2\pi mk}{2N}e^{-j2\pi m}}=\sum_{m=0}^Ke^{-j\frac{2\pi mk}{2N}}=Y_k$$


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