A signal processing snippet. Let $latex x(t)$ be a bandlimited signal restricted to the interval $latex [T_1, T_2]$, so that in particular
$latex x(t) = \displaystyle{\sum_{n=0}^{N-1} x(n \Delta t) \text{sinc}\left(\frac{t – n\Delta t}{\Delta t}\right)} 1_{T_1 \leq t \leq T_2}$
Here, as opposed to previous entries, we have defined $latex \text{sinc}(t) = \sin(\pi t)/(\pi t)$ (I have had a change of heart). Then the Fourier transform of $latex x(t)$ is
$latex X(f) = \displaystyle{\sum_{n=0}^{N-1}} x(n\Delta t) e^{-i 2\pi f n \Delta t} R(n \Delta t – T_2, n \Delta t – T_1, f – 1/2\Delta t, f + 1/2 \Delta t) \,\Delta t$
where we define
$latex R(t_1, t_2, f_1, f_2) = \dfrac{(\text{Ei}(i 2\pi f_2 t_2) – \text{Ei}(i 2\pi f_2 t_1)) – (\text{Ei}(i 2\pi f_1 t_2) – \text{Ei}(i 2\pi f_1 t_1))}{i 2\pi}$
and $latex \text{Ei}$ is the exponential integral, which for imaginary arguments is
$latex \text{Ei}(it) = i \dfrac{\pi}{2} -\displaystyle{\int_{t}^{\infty}} \dfrac{e^{i…
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