Prove sampling theorem; aliasing effect

Statement

A continuous-time signal $x(t)$ band-limited to $\omega_M$ ($X(j\omega) = 0$ for $|\omega| > \omega_M$) is uniquely determined by its samples $x(nT)$ if the sampling frequency $\omega_s = 2\pi/T > 2\omega_M$.

Proof. Let $x(t)$ be band-limited: $X(j\omega) = 0$ for $|\omega| > \omega_M$.

Step 1 – sampling model. Sample with the impulse train $p(t) = \sum_n \delta(t-nT)$:

$$x_p(t) = x(t)p(t) = \sum_{n=-\infty}^{\infty} x(nT)\,\delta(t-nT)$$

Step 2 – transform of $p(t)$. $p(t)$ is periodic with period $T$; its Fourier series coefficients are all $1/T$, so

$$P(j\omega) = \frac{2\pi}{T}\sum_{k=-\infty}^{\infty}\delta(\omega - k\omega_s), \qquad \omega_s = \frac{2\pi}{T}$$

Step 3 – multiplication property.

$$\begin{aligned} X_p(j\omega) &= \frac{1}{2\pi}X(j\omega) * P(j\omega) \\ &= \frac{1}{T}\sum_{k=-\infty}^{\infty} X\big(j(\omega - k\omega_s)\big) \end{aligned}$$

Step 4 – no-overlap condition. The $k=0$ copy lies in $[-\omega_M, \omega_M]$ and the $k=1$ copy starts at $\omega_s - \omega_M$. They do not overlap if

$$\omega_s - \omega_M > \omega_M \;\Rightarrow\; \omega_s > 2\omega_M$$

Step 5 – recovery. Pass $x_p(t)$ through an ideal LPF $H(j\omega) = T$ for $|\omega| < \omega_c$, 0 otherwise, with $\omega_M < \omega_c < \omega_s - \omega_M$. Only the $k=0$ term survives:

$$X_r(j\omega) = H(j\omega)X_p(j\omega) = T\cdot\frac{1}{T}X(j\omega) = X(j\omega)$$

Hence $x_r(t) = x(t)$: the samples determine $x(t)$ uniquely. $\blacksquare$

Aliasing effect

If $\omega_s < 2\omega_M$, the $k=1$ copy in $X_p(j\omega)$ starts at $\omega_s - \omega_M$, which is less than $\omega_M$, so neighbouring copies overlap. In the overlap region the spectral values add, and a component at $\omega_0 > \omega_s/2$ appears at $\omega_s - \omega_0$. The LPF then outputs a distorted signal. This is aliasing.

Example: 700 Hz sampled at 1000 Hz gives the same samples as 300 Hz.

It is avoided by sampling above the Nyquist rate and using an anti-aliasing LPF before sampling.