Nyquist rate of 2cos100πt·cos300πt

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

$$\begin{aligned} x(t) &= 2\cos(100\pi t)\cos(300\pi t) \\ &= \cos(400\pi t) + \cos(200\pi t) \end{aligned}$$

Step 2: frequencies ($f = \omega/2\pi$):

Highest frequency $f_m = 200$ Hz.

Step 3: Nyquist rate and interval.

$$\begin{aligned} f_N &= 2f_m = 2\times200 = 400\ \text{Hz} \\ T_N &= \frac{1}{f_N} = \frac{1}{400} = 2.5\ \text{ms} \end{aligned}$$

Answer: Nyquist rate $= 400$ samples/s ($\omega_N = 800\pi$ rad/s); Nyquist interval $= 2.5$ ms.

Note: the highest frequency is 200 Hz, not 150 Hz; multiplying two sinusoids creates the sum frequency.