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Consider the following analog signal

$$x(t) = 2 \sin(100\pi t)$$

The signal $x(t)$ is sampled with a sampling rate $F_s=50\textrm{Hz}$. Determine the discrete time signal. Plot the discrete time signal.

Also determine the total number of samples.

I don't understand how to approach this question.

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$x(t)$ is a model for an analog signal. After sampling, the real $t$ variable is turned into an integer variable $k$, corresponding to a regular splitting of the real time interval, starting from some origin time $t_0$, evenly spaced by a sampling period $T_s$ being the inverse of the sampling frequency: $T_s = \tfrac{1}{F_s}$.

A discrete signal is thus:

$$ x[k] = x(t_0+k.T_s) = 2\sin(100\pi (t_0+k.T_s)) = 2\sin(100\pi t_0+k.100\pi/50)) $$

Thus, since $$2\sin(100\pi t_0+k.100\pi/50)) = 2\sin(100\pi t_0+2k\pi))= 2\sin(100\pi t_0))$$

the discretized signal is determined by a constant value, depending only on the initial sampling time $t_0$. You can plot if with many plotting software (even Excel).

The total number of samples is ambiguous. Since $k$ take any integer value, the number of discrete samples is infinite. Since it is constant, $x[n]$ takes only one value.

Explanation: you start from a continuous sine, and only get a constant signal. This is an illustration of the Nyquist-Shannon sampling theorem (or aliasing). If the sampling frequency is not consistent with signal maximal frequency, the discrete signal might be a poor version of the analog signal.

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