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A signal sampled in time domain has replicas of its spectrum in frequency domain. This implies that to transmit a sampled signal we need ifinite bandwidth.How is this possible?

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You are right. If you really want to send the sampled signal $$x_s(t) = {\delta}_{T_s}(t) x_c(t) = ( \sum_k \delta (t- k T_s) ) x_c(t) = \sum_k x_c(k T_s) \delta (t-k T_s)$$ then this weighted impulse train constitudes an infinite amount of (shifted replicas of base period) bandwidth in frequency domain.

Therefore this signal is not transmitted anywhere... Indeed the impulse sampling cannot be realized in practice either. It's just a mathematical model which provides a quantitative description of the conversion process from a continuous time-signal into a discrete-time sequence whose samples have values equal to the continuous-time signal at those sampling impulse instants.

The sampling process is used as the initial stage of conversion from continuous to discrete time domains and it's the sequence of those samples which are processed later on for whatever purpose you have.

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One normally doesn't send sampled signals, but instead sends arrays or vectors of numbers or values representing the sampled signal. Thus, the required information bandwidth is determined by the number of bits per sample and the sample rate.

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This link may be helpful for understanding how a sampled signal is reconstructed:

sampling theorem

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The idealization is that by sampling you get a signal that can change its value instantly (zero between samples, non-zero at sample points), the theory says that implies an infinite bandwidth. So, yes, in the analog domain, ideally, a perfect sampled signal has infinite bandwidth.

However, the process of keeping the samples and discard all the zeros between samples (as many as there are in $|\mathbb{R}| - f_{sampling} \cdot t = \infty$) has the effect of folding the spectrum such that each replica falls over the centered one. This way, in the digital domain, you only have available one replica at the origin.

Now, when going back from digital to analog, one doesn't actually fill in with zeros in between the samples (which would lead to infinite bandwidth), but with some kind of interpolation that smooths the time signal changes. This smooth interpolation lower the needed bandwidth. The ideal case is to output a shifted scaled $sinc(t - nT)$ function at each output sample, since the signal bandwidth will be that of the interpolation function and since the bandwidth of a $sinc(t)$ is bounded, the output bandwidth is bounded.

In practical implementations, however, $sinc(t)$ is not time-bounded, so a common method is to use a constant pulse of sampling interval duration, which is not bounded in frequency (it is a $sinc(\omega)$), but goes to zero very fast as it approach $-\infty$ or $+\infty$, then discard its tails (usually something in the transmission path attenuates these tails) which are negligible. This way, you only have one replica at the origin in the analog domain, with infinite lenght tails which are lost. Anyhow, lost energy is very low.

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