# Why are these functions periodic?

Say I have some discrete function whose DFT is always 0 at even indices and arbitrary values at odd indices. I noticed that this class of functions is always periodic.

For instance, if $$F = [0.5, 0, 0.7, 0, 0.9, 0, 1.2, 0, 1.7, 0, -1, 0, i, 0 ,2i, 0]$$

Let $f = \operatorname{ifft} [F]$ where $\operatorname{ifft}$ is the inverse fast fourier transform. The magnitude of $f$, $|f|$, then looks like the following:

I noticed you end up with some sort of periodic function no matter what the values at the odd indices of the frequency spectrum are. I'm not too sure why and was hoping to get some insight into this.

Note how in the image above the waveform between 1 and 7 is identical to the the waveform between 9 and 15. I'm specifically talking about this sort of behaviour, I know that finite frequency spectrums result in periodic waveforms.

• That is because every even-th term of your data is zero. Jan 5, 2018 at 2:24
• @SangchulLee Well, yes, but why does that result in the periodicity? Jan 5, 2018 at 2:25
• @SangchulLee Perfect, could you please post that as an answer so I can accept it? Jan 5, 2018 at 2:37
• Glad it helped :) Jan 5, 2018 at 2:46

Consider data $(x_0, x_1, \cdots, x_{2N-1})$ of length $2N$ and assume that $x_{2k-1} = 0$ for all $k = 1, \cdots, N$. Then
$$f_k = \sum_{n=0}^{2N-1} x_{n} e^{-i\pi k n/N} = \sum_{n=0}^{N-1} x_{2n} e^{-2i\pi k n/N}.$$
Notice that the last sum is exactly the same as the DFT of data $(x_0, x_2, \cdots, x_{2(N-1)})$ of length $N$, hence $f$ also has period $N$.
• Yes, but I meant why are there multiple periods in the resulting waveform? Look at how the waveform is identical at $n$ between 1 and 7 and between 9 and 15. Jan 5, 2018 at 2:18