Sampling theorem; Nyquist of 1 + cos + sin

Sampling theorem

Sampling theorem: A continuous-time signal $x(t)$ that is band-limited to $\omega_M$ (that is, $X(j\omega)=0$ for $|\omega|>\omega_M$) is uniquely determined by its samples $x(nT)$ if the sampling frequency satisfies

$$\omega_s = \frac{2\pi}{T} > 2\omega_M \quad\text{or}\quad f_s > 2f_M$$

The minimum rate $2f_M$ is called the Nyquist rate, and $T_N = 1/(2f_M)$ is the Nyquist interval.

The original signal is recovered by passing the samples through an ideal low-pass filter of cutoff between $\omega_M$ and $\omega_s-\omega_M$ and gain $T$.

Nyquist rate of $x(t) = 1 + \cos(2000\pi t) + \sin(4000\pi t)$

Write each term as $\cos(2\pi f t)$ or $\sin(2\pi f t)$ and read its frequency:

The highest frequency present is $f_M = 2000$ Hz, so

$$\begin{aligned} f_N &= 2 f_M = 2 \times 2000 = 4000\ \text{Hz} \\ \omega_N &= 2\pi f_N = 8000\pi\ \text{rad/s} \end{aligned}$$

Answer: Nyquist rate = 4000 Hz (4000 samples/s, or $8000\pi$ rad/s).