Minimum sampling rate (sampling theorem)

Statement (sampling theorem): if $x(t)$ has no frequency components above $\omega_m$, it can be recovered completely from its samples only if the sampling frequency satisfies $\omega_s \ge 2\omega_m$ (strictly, $\omega_s > 2\omega_m$). The minimum rate $2\omega_m$ is the Nyquist rate.

Proof

The ideal sampler output is

$$x^*(t) = x(t)\sum_{k=-\infty}^{\infty}\delta(t-kT)$$

The impulse train is periodic with period $T$, so expand it as a Fourier series ($\omega_s = 2\pi/T$):

$$\sum_{k}\delta(t-kT) = \frac{1}{T}\sum_{n=-\infty}^{\infty} e^{jn\omega_s t}$$

Then

$$x^*(t) = \frac{1}{T}\sum_{n=-\infty}^{\infty} x(t)\,e^{jn\omega_s t}$$

Using the frequency-shift property $\mathcal{F}[x(t)e^{jn\omega_s t}] = X(j(\omega - n\omega_s))$:

$$X^*(j\omega) = \frac{1}{T}\sum_{n=-\infty}^{\infty} X(j(\omega + n\omega_s))$$

So the spectrum of the sampled signal is the original spectrum (scaled by $1/T$) repeated at every multiple of $\omega_s$.

Condition for no overlap

 ws > 2wm: no overlap
   /\        /\        /\
 -/--\------/--\------/--\--> w
 -ws   -wm  0  wm   ws
          |<-LPF->|

 ws < 2wm: overlap (aliasing)
   /\  /\  /\
 -/--\/--\/--\--> w

The primary band occupies $-\omega_m$ to $\omega_m$. The first copy occupies $\omega_s-\omega_m$ to $\omega_s+\omega_m$. They do not overlap if

$$\omega_s - \omega_m \ge \omega_m \quad\Rightarrow\quad \omega_s \ge 2\omega_m$$

Under this condition, an ideal low-pass filter with gain $T$ and cut-off $\omega_s/2$ removes all copies and returns $X(j\omega)$, i.e. $x(t)$ is recovered exactly. If $\omega_s < 2\omega_m$, the bands overlap (aliasing): high-frequency components appear as low frequencies, and no filter can separate them.

Hence the minimum sampling rate is $\omega_s = 2\omega_m$.

Example: a signal limited to 50 Hz must be sampled at more than 100 Hz. In practice control systems sample much faster (often 10 times the closed-loop bandwidth).