How to derive the discrete delta function from geometric sum of complex sinusoids?

Question

Using this in the context of Fourier transforms. This should probably be an easy derivation for you guys, but I forget how to derive it.

$$ \sum_{n=0}^{L-1}e^{-j\frac{2\pi k}{L}n} = \frac{1-e^{-j2\pi k}}{1-e^{\frac{-j2\pi k}{L}}} $$

which somehow can also be reduced to $$L\delta [k] $$

Write the thing inside the sum as $u^n$, where $u = \exp(-2j\pi k/L)$. Then it's just a sum of a geometric series. — John Hughes, Dec 11 '18 at 22:53
Gah, yes I see it now. Any hints about the reduction to the discrete delta function? — user50420, Dec 11 '18 at 22:56
Nope --- none whatsoever. if you actually want people to address that as well, it'd be useful to say so in your question rather than just adding it as an after-the-fact note. — John Hughes, Dec 11 '18 at 23:00
true, I made the update. Thanks for helping with the first half, though. — user50420, Dec 11 '18 at 23:13
Is the term inside the summation $e^{-j \frac{2 \pi k n}{L}}$? (i.e. missing $n$?) — Godfather, Dec 11 '18 at 23:15

score 2 · Answer 1 · answered Dec 11 '18 at 23:21

When $k=0$, the summation is simply $L$.

For $k \neq 0$, write the numerator as: $e^{-j \pi k} \left( e^{j \pi k} - e^{-j \pi k} \right)$ = $e^{-j \pi k} \times 2j \sin(k \pi)$, which is $0$ for $k \neq 0$, because $\sin(k \pi)=0$.

Thus, you have a summation which is $L$ for $k=0$ and $0$ for $k \neq 0$. This can be expressed as $L \delta[k]$.

Also, the denominator should read $1- e^{-j \frac{2 \pi k}{L}}$.

How to derive the discrete delta function from geometric sum of complex sinusoids?

1 Answers1