Nyquist rate of (1/2π)cos50πt·cos500πt

Nyquist sampling theorem

A continuous-time signal band-limited to $f_m$ Hz can be exactly recovered from its uniformly spaced samples if the sampling rate is at least twice the highest frequency, $f_s \ge 2f_m$. The minimum rate $2f_m$ is the Nyquist rate, and $1/(2f_m)$ is the Nyquist interval. Sampling below this rate causes aliasing.

Nyquist rate of $x(t) = \frac{1}{2\pi}\cos(50\pi t)\cos(500\pi t)$

Using $\cos A\cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$:

$$\begin{aligned} x(t) &= \frac{1}{2\pi}\cdot\frac{1}{2}\left[\cos(550\pi t) + \cos(450\pi t)\right] \\ &= \frac{1}{4\pi}\cos(550\pi t) + \frac{1}{4\pi}\cos(450\pi t) \end{aligned}$$

Highest frequency $f_m = 275$ Hz.

$$\begin{aligned} f_N &= 2f_m = 2\times275 = 550\ \text{Hz} \\ T_N &= \frac{1}{550} = 1.82\ \text{ms} \end{aligned}$$

Answer: Nyquist rate $= 550$ samples/s ($\omega_N = 1100\pi$ rad/s); Nyquist interval $\approx 1.82$ ms.