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:
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$\begingroup$ is that the Kronecker delta you mean? are $n$ and $N$ integers? $\endgroup$ – robert bristow-johnson Apr 5 '15 at 14:07
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$\begingroup$ yes sir, its kronecker delta $\endgroup$ – minimal risk Apr 5 '15 at 14:33
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$\begingroup$ What happens if you apply the formula? (Read: please show us your work and where you're stuck). $\endgroup$ – Matt L. Apr 5 '15 at 14:37
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$\begingroup$ 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 $\endgroup$ – minimal risk Apr 5 '15 at 14:58
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$\sum\limits_{r=-\infty}^{\infty}\delta[n-rN]$ 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 $a_k = 1/N$.
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$\begingroup$ How do you cancel the sum that is within the outer sum? and get it to be delta[n]? except that I pretty much get it so thank you. $\endgroup$ – minimal risk Apr 5 '15 at 16:39
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$\begingroup$ 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. $\endgroup$ – Eloi Marin Apr 5 '15 at 17:14
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$\begingroup$ Thank you so much Eloi! I find this forum very helpful for us the engineering students, where its hard to come by a community for engineering topics. $\endgroup$ – minimal risk Apr 6 '15 at 6:09
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$\begingroup$ I'm glad I could help you! Indeed this site is a nice place to learn, asking but also answering questions (I'm also a student of DSP right now). If you found my answer useful, you can vote or mark it as "accepted"! $\endgroup$ – Eloi Marin Apr 6 '15 at 7:14
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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.
