Aliasing in a band-limited signal

Let $x(t)$ be band-limited with bandwidth $W$, i.e. $X(j\omega) = 0$ for $|\omega| > W$ (in rad/s). It is sampled with an impulse train of period $T$, sampling frequency $\omega_s = 2\pi/T$.

Spectrum of the sampled signal

$$X_p(j\omega) = \frac{1}{T}\sum_{k=-\infty}^{\infty} X\big(j(\omega - k\omega_s)\big)$$

The baseband copy ($k=0$) occupies $[-W, W]$. The copy for $k=1$ occupies $[\omega_s - W,\ \omega_s + W]$.

Three cases

ws < 2W :
       k=0           k=1
     _______       _______
    /       \     /       \
---/----------\-/----------\---> w
  -W     ws-W  W  ws       ws+W
          |<-->|
         overlap

What aliasing does

• In the overlap band $\omega_s - W < |\omega| < W$, $X_p$ is the sum of the original spectrum and the tail of the neighbouring copy. A component at $\omega_0 > \omega_s/2$ appears at the alias frequency $\omega_s - \omega_0$.
• A low-pass filter cannot separate the two, so the reconstructed signal differs from $x(t)$: high frequencies are lost and false low frequencies appear.
• Example: $W = 2\pi(1000)$, $\omega_s = 2\pi(1500)$. A 900 Hz component appears at $1500 - 900 = 600$ Hz.

Avoiding it: sample at $\omega_s > 2W$ and place an anti-aliasing low-pass filter of cutoff $\omega_s/2$ before the sampler.