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I have trouble trying to work out a fourier transform of sampling function. I also looked at other questions on math stackexchange but they just don't make much sense to me.

\begin{equation} f(t)=\sum_{j=0}^{\frac{T}{\Delta t}-1} \delta(t-j\Delta t) \text{ where, } T=n\Delta t \end{equation}

I'm not even sure how to approach this problem - I'm not sure how the delta function inside of the sum works out to be when you apply the fourier transform. I think that you should get something like this - but also, I'm not really sure.

\begin{equation} f(\omega)=\sum_{j=0}^{\Delta t-1} \delta(\omega-j\frac{T}{\Delta t}) \end{equation}

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  • $\begingroup$ The Fourier transform is linear, so applying to a sum of things just means: apply it to each thing and sum the results. Do you know or can you calculate the Fourier transform of a single delta function? $\endgroup$ – NickD Apr 11 '18 at 16:24
  • $\begingroup$ Well I know that fourier transformation of delta function $\delta(x-x_0) = e^{-2\pi ikx_0}$.... so in this case, I would get $\delta(t-j\Delta t)=e^{-2 \pi ik(j\Delta t)}$ ... am I right here? $\endgroup$ – Rigel Apr 11 '18 at 20:23

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