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For a bandlimited signal $x(t)$ that is reconstructed after Nyquist-sampling at intervals $T$,

$$x(t) = \sum_{k=-\infty}^{\infty} {x[k]\textrm{sinc}\left(\frac{t-kT}{T}\right)}$$

where $x[k]$ = $x(kT)$. For a non-band limited signal $x(t)$, the reconstructed signal will not be same as $x(t)$. Let me call that $p(t)$. Writing the same formula again, so

$$p(t) = \sum_{k=-\infty}^{\infty} {x[k]\textrm{ sinc}\left(\frac{t-kT}{T}\right)}$$

I am trying to see if there is an alternate way of expressing $p(t)$ in terms of $x(t)$ WITHOUT INVOLVING AN INFINITE SUM. It need not be simpler looking, but just wondering if there are other ways of expressing it.

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  • $\begingroup$ I think you mean $x[kT]$ and $(t-kT)$ in your equations. The way it's currently written doesn't make any sense. $\endgroup$ – Hilmar Mar 25 '16 at 11:48
  • $\begingroup$ @Hilmar, I agree, I edited the question. $\endgroup$ – MBaz Mar 25 '16 at 13:48
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It seems to me that the only way is to cheat and hide the sum behind a variable. For example, say that $X_T(f)$ is the Fourier transform of

$$x_T(t) = T \sum_{n=-\infty}^\infty x(nT)\delta(t-nT).$$

One can show that

$$ \begin{align} X_T(f) &= \mathcal F \lbrace x_T(t) \rbrace \\ &= \sum_{k=-\infty}^\infty X\left(f - \frac{k}{T}\right) \\ \end{align} $$

where $ X(f) = \mathcal F \lbrace x(t) \rbrace .$ Then,

$$p(t)=\mathcal F^{-1} \lbrace H(f)X_T(f) \rbrace$$

where $H(f)$ is a brick-wall low-pass filter with cutoff frequency $f_c=1/(2T)$:

$$ H(f) = \begin{cases} 1 \quad \text{for }|f|<f_c \\ 0 \quad \text{for }|f|>f_c \\ \end{cases} $$

and $\mathcal F^{-1}$ represents the inverse Fourier transform. This is cheating, though, since in general $X_T(f)$ involves an infinite sum.

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