Prove sampling theorem; Nyquist of product

Sampling theorem

Statement: A signal $x(t)$ band-limited to $\omega_m$ ($X(j\omega) = 0$ for $|\omega| > \omega_m$) is uniquely determined by its samples $x(nT)$ if the sampling frequency $\omega_s = \frac{2\pi}{T} \ge 2\omega_m$.

Proof: Model sampling as multiplication by an impulse train:

$$x_p(t) = x(t)\,p(t), \qquad p(t) = \sum_{n=-\infty}^{\infty}\delta(t - nT)$$

The FT of the periodic impulse train is $P(j\omega) = \frac{2\pi}{T}\sum_k\delta(\omega - k\omega_s)$. Multiplication in time is convolution in frequency:

$$\begin{aligned} X_p(j\omega) &= \frac{1}{2\pi}X(j\omega) * P(j\omega) \\ &= \frac{1}{T}\sum_{k=-\infty}^{\infty}X(j(\omega - k\omega_s)) \end{aligned}$$

So the sampled spectrum is the original spectrum repeated every $\omega_s$. The copy at $k=0$ spans $-\omega_m$ to $\omega_m$, and the next starts at $\omega_s - \omega_m$. They do not overlap if $\omega_s - \omega_m \ge \omega_m$, i.e. $\omega_s \ge 2\omega_m$. Then an ideal low-pass filter with gain $T$ and cutoff $\omega_m \le \omega_c \le \omega_s - \omega_m$ passes only the $k=0$ copy, giving back $X(j\omega)$ and hence $x(t)$ exactly. If $\omega_s < 2\omega_m$, the copies overlap (aliasing) and recovery is impossible. Hence proved.

Nyquist rate of $x(t) = \frac{1}{2\pi}\cos(4000\pi t)\cos(1000\pi t)$

$$x(t) = \frac{1}{4\pi}\left[\cos(5000\pi t) + \cos(3000\pi t)\right]$$

Frequencies: $\frac{5000\pi}{2\pi} = 2500$ Hz and $\frac{3000\pi}{2\pi} = 1500$ Hz, so $f_m = 2500$ Hz.

$$f_N = 2f_m = 5000\ \text{Hz}, \qquad T_N = \frac{1}{5000} = 0.2\ \text{ms}$$

Answer: Nyquist rate $= 5000$ samples/s (5 kHz); Nyquist interval $= 0.2$ ms.