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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.