In Figure P7 .25 is a sampler, followed by an ideallowpass filter, for reconstruction of

Chapter 7, Problem 7.25

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In Figure P7 .25 is a sampler, followed by an ideallowpass filter, for reconstruction of x(t) from its samples xp(t). From the sampling theorem, we know that if \(\omega_{s}=2 \pi / T\) is greater than twice the highest frequency present in x(t) and \(\omega_{c}=\omega_{s} / 2\), then the reconstructed signal xr(t) will exactly equal x(t). If this condition on the bandwidth of x(t) is violated, then xr(t) will not equal x(t). We seek to show in this problem that if \(\omega_{c}=\omega_{s} / 2\), then for any choice of T, xr(t) and x(t) will always be equal at the sampling instants; that is,

\(x_{r}(k T)=x(k T), k=0, \pm 1, \pm 2, \ldots\)

To obtain this result, consider eq. (7.11), which expresses Xr(t) in terms of the samples of

\(x_{r}(t)=\sum_{n=-\infty}^{\infty} x(n T) T \frac{\omega_{c}}{\pi} \frac{\sin \left[\omega_{c}(t-n T)\right]}{\omega_{c}(t-n T)}\)

With \(\omega_{c}=\omega_{s} / 2\), this becomes \(x_{r}(t)=\sum_{n=-\infty}^{\infty} x(n T) \frac{\sin \left[\frac{\pi}{T}(t-n T)\right]}{\frac{\pi}{T}(t-n T)}\)   (P7.25-l)

By considering the values of a for which \([\sin (\alpha)] / \alpha=0\), show from eq. (P7.25-l) that, without any restrictions on x(t), xr(kT) = x(kT) for any integer value of k.