# Sampling + Bandpass

A one dimensional function $f(x)=2\cos \pi \xi_0 x$ is sampled at a rate $\xi_s$, which is just above $\xi_0$. Commom reconstrucion filter such as the zero-or-first order-hold circuits have a passband greater than $\pm \xi_s/2$ with the first zero crossings at $\pm \xi_s$, as shown is attach image. Show that reconstructed funcion is of the form:

$$g(x)=2(a+b\cos 2\pi \xi_s x )\cos \pi \xi_0 x + 2b\sin2\pi\xi_s x \sin \pi \xi_0 x,$$

where $a:=H(\xi_0/2)\xi_s$ e $b:=H(\xi_s-\xi_0/2)\xi_s.$

How I will be able to start the solution the this question?, I don't get see how is a filter? Why say bandpass (The figure like a low pass filter)? ... (I get the fourier transform of $f$, this is $\delta(w-\xi_0/2) + \delta(w+\xi_0/2)$)

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 I recommend you ask the moderators to migrate this question to the signal processing site dsp.SE – Dilip Sarwate Dec 16 '12 at 3:55