# Fourier Transform (train of impulses)

Perhaps this question has occurred somewhere before, but I learn more if I ask the question myself. We have to determine the Fourier transform of the following function: $$f(t) = \sum_{n=0}^{10} \frac{1}{3^n} \delta(t-5n)$$, but I am not sure how to go about this because I am not so sure what I have in front of me. I also cannot find a question similar to this one with a detailed explanation. To some extent, I understand that it is just a discrete function, and that for every value of $$n$$, there arises a peak everytime $$t=5n$$, with the corresponding "amplitude" $$a$$ but the problem occurs up when I need to actually transform it.

First, the general form of a similar function is given as $$f(t) = \sum_{-\infty}^\infty s[n] \delta(t-n)$$, which then gives that the discrete-time Fourier transform is given by $$\mathcal{F}\{f(t)\}(\omega) = \sum_{-\infty}^{\infty} s[n] e^{-i \omega n}$$. In this case $$s[n]=3^{-n}$$, I guess? But then I still can't really imagine, in either case, what that delta function actually does with the entire transformation and to the summation. And then how I should work with that, in general.

A hint is given that we can make use of the fact that $$\sum_{n=0}^N a r^{-n} = a \frac{1-r^{N+1}}{1-r}$$, but even then I get confused because the delta function also depends on $$n$$ right? so $$a$$ is also dependend of $$n$$ right?

• My brain is even more fogged after writing this Commented Apr 11 at 12:47
• So, after much thought, I may have an idea. Since the peak of $f(t)$, only occurs when $t=5n$, I would say the the period is equal to 5. Which makes $s[n] = 3^{-5n}$ instead of $s[n] = 3^{-n}$. Is this true? Commented Apr 11 at 13:31

You can use the fact that the Fourier transform of $$\delta(t-t_{0})$$ is given by $$e^{-i\omega t_{0}}$$ and that it's a linear integral transformation, hence your sum of deltas will be the sum of Fourier transform of each delta.
Putting these together you get $$\mathcal{F}\left(\sum_{n=0}^{10}3^{-n}\delta(t-5n)\right) = \sum_{n=0}^{10}3^{-n}\mathcal{F}(\delta(t-5n)) = \sum_{n=0}^{10}3^{-n}e^{-i5n\omega} = \frac{\left(\frac{1}{3e^{i5\omega}}\right)^{11}-1}{\frac{1}{3e^{i5\omega}}-1}.$$
I don't know what's troubling you exactly. I'm guessing this is what you'd do, if not tell me since I'm yet to study the Fourier Transform: \begin{aligned} \mathscr F_t\{f(t)\}(\omega)&=\int_{-\infty}^\infty\sum_{n=0}^{10}\frac{1}{3^n}\delta(t-5n)e^{-i\omega t}\,\mathrm dt\\ &=\sum_{n=0}^{10}\frac{1}{3^n}\int_{-\infty}^\infty\delta(t-5n)e^{- i\omega t}\,\mathrm dt\\ &=\sum_{n=0}^{10}\frac{e^{-5 i\omega n}}{3^n}=\sum_{n=0}^{10}\left(\frac{e^{-5 i\omega}}{3}\right)^n\\ &=\dfrac{1-\left(\dfrac{e^{-5 i\omega}}{3}\right)^{11}}{1-\dfrac{e^{-5 i\omega}}{3}}=\dfrac{3^{11}-e^{-55 i\omega}}{3^{10}(1-e^{-5 i\omega})} \end{aligned}
• Thanks! By the way, I think you mean $e^{-2 \pi i f t}$? Because $\omega = 2\pi f$ Commented Apr 11 at 13:55
• And if the integral were not finite, what would change? My simple brain would say that that summation still does not depend on $t$, so can be seen as "constant". Commented Apr 11 at 13:58