Nyquist rate of (1/2π)cos4000πt·cos1000πt

Sampling theorem

A band-limited signal with no frequency components above $f_m$ Hz can be completely represented by, and exactly recovered from, its uniformly spaced samples if the sampling frequency satisfies $f_s \ge 2f_m$. The minimum rate $2f_m$ is the Nyquist rate; recovery is done with an ideal low-pass filter.

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

Step 1: convert the product to a sum using $\cos A\cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$:

$$\begin{aligned} x(t) &= \frac{1}{2\pi}\cdot\frac{1}{2}\left[\cos(5000\pi t) + \cos(3000\pi t)\right] \\ &= \frac{1}{4\pi}\cos(5000\pi t) + \frac{1}{4\pi}\cos(3000\pi t) \end{aligned}$$

Step 2: frequencies.

Highest frequency $f_m = 2500$ Hz.

Step 3: Nyquist rate and interval.

$$\begin{aligned} f_N &= 2f_m = 2\times2500 = 5000\ \text{Hz} \\ T_N &= \frac{1}{f_N} = \frac{1}{5000} = 0.2\ \text{ms} \end{aligned}$$

Answer: Nyquist rate $= 5000$ samples/s (5 kHz, $\omega_N = 10000\pi$ rad/s); Nyquist interval $= 0.2$ ms $= 200\ \mu$s.

Note: the highest frequency is not 2000 Hz (from $\cos 4000\pi t$ alone); multiplication creates the sum frequency 2500 Hz.