I have a question that is similar to this one but slightly different.

If I have discrete signal $$s(t) = \sum_k n_k \delta(t-kT_0),\quad k=0,1,\dotsc,$$ where $n_k$ are just some scalar numbers. What is the Fourier transform of $s(t)$? I think it should be some kind of a convolution $$S(f) = G(f)\star\sum_m\delta(f-m/T_0),\quad m=0,1,\dotsc,$$ but what is $G(f)$? Is there an analytical expression for it in terms of $n_k$?


1 Answer 1


The Fourier transform of $\delta(t)$ is $1$, so the FT of $\delta(t-a)$ is $e^{-ia\omega}$. By linearity,

$$S(\omega) = \sum_k n_k e^{-ikT_0\omega}.$$

Unless $n_k$ have a simple dependence on $k$, there is probably no closed formula for the sum.

Note: I'm assuming $\hat f(\omega) = \int_{-\infty}^\infty e^{-i\omega t}f(t)\,dt$. Other scalings of the Fourier transform are also common.

  • $\begingroup$ Yes this makes sense. So $S(\omega)$ is indeed periodic in frequency ($\Omega = 1/T_0$), and that's why I thought it could be expressed using a convolution. $\endgroup$
    – LWZ
    Commented Jun 5, 2013 at 21:25

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