Solution:

Assume an analog waveform $\mathrm{x}(\mathrm{t})$, as shown in Fig. (a), with a Fourier transform, X(f), which is zero outside the interval $\left(-f_m\ltf\ltf_m\right)$, as shown in Fig. (b). The sampling of $\mathrm{x}(\mathrm{t})$ can be viewed as the product of x(t) with a train of unit impulses functions, $x_s(t)$, shown in Fig. (c), and defined as follows:

$ x_\delta(t)=\sum_{n=-\infty}^{\infty} \delta\left(t-n T_s\right)\\ $

Let us choose $T_s=\frac{1}{2 f_m}$, so that the Nyquist rate is just satisfied.

Using the shifting property of the impulse function the $x_{\mathrm{s}}(t)$, shown in Fig. (e), can be given by,

$ x_s(t)=x(t) x_\delta(t)=\sum_{n=-\infty}^{\infty} x(t) \delta\left(t-n T_s\right)\\ $

$ =\sum_{n=-\infty}^{\infty} x\left(n T_s\right) \delta\left(t-n T_s\right)\\ $

Using the frequency convolution property of Fourier transform, the time product $x(t) x_\delta(t)$ transform to the frequency domain convolution $X(f) \otimes X_\delta(f)$, where $X_\delta(f)$ is the Fourier transform of $x_\delta(t)$ and given by,

$ X_\delta(f)=\frac{1}{T_s} \sum_{n=-\infty}^{\infty} \delta\left(f-n f_s\right)\\ $

The convolution with an impulse function simply shifts the original function, as follows:

$ X(f) \otimes \delta\left(f-n f_s\right)=X\left(f-n f_s\right)\\ $

The Fourier transform of the sampled waveform, $X_s(f)$, can be given by:

$ \begin{aligned} X_s(f)=X(f) \otimes X_\delta(f) &=X(f) \otimes\left[\frac{1}{T_s} \sum_{n=-\infty}^{\infty} \delta\left(f-n f_s\right)\right] \\ &=\frac{1}{T_s} \sum_{n=-\infty}^{\infty} X\left(f-n f_s\right) \end{aligned} $