# Discrete Fourier series for a sum of deltas

so Im given a discrete sum $$x[n] = \sum\limits_{r=\infty}^{+\infty}\delta[n-rN]$$ how do I calculate its discrete Fourier series coefficients?

Thank you.

edit: this is what i've come to up to now:

• is that the Kronecker delta you mean? are $n$ and $N$ integers? Apr 5, 2015 at 14:07
• yes sir, its kronecker delta Apr 5, 2015 at 14:33
• What happens if you apply the formula? (Read: please show us your work and where you're stuck). Apr 5, 2015 at 14:37
• i added a picture of where Im stuck. As you can see, in line num 3, the sum within a sum of this delta by two indexes feels dependent.. so I dont know exactly how to proceede Apr 5, 2015 at 14:58

$\sum\limits_{r=-\infty}^{\infty}\delta[n-rN]$ is a periodic Kronecker delta with period $N$, but the sum to get the fourier coefficients only "uses" one period $\sum\limits_{n = 0}^{N-1}$ so you don't have to worry about the periodic extension of $\delta$ and take just $x[n] = \delta[n]$ as your function. This is going to cancel out every exponential in the sum except when $n=0$, so the result will be independent of the particular coefficient and be just $a_k = 1/N$.
• The inner sum doesn't affect the outer, so you can remove it. Why doesn't affect? Because it only changes the $rN$ term inside delta's time index, and the only way this could make some difference is when $n = rN$ (the condition for the delta function to be 1) but since the outer sum goes from $n = 0$ to $n = N-1$, $n$ is never going to be as big as $N$, so the $rN$ term won't change anything. Apr 5, 2015 at 17:14
If you draw the sequence $x[n]$ you'll see right away how it works. You have an impulse at $n=0$, at $n=-N$, at $n=N$, etc. But if you sum over only one period, no matter which period you choose, there's always only one impulse in that period. So you're left with only one sum index contributing to the final result. I'm sure you can take it from here.