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?