Prove sampling theorem for low pass signals

Statement (low-pass sampling theorem): If a continuous-time signal $x(t)$ is band-limited with $X(j\omega) = 0$ for $|\omega| > \omega_M$, then $x(t)$ is uniquely determined by its samples $x(nT)$, $n = 0, \pm1, \dots$, provided

$$\omega_s = \frac{2\pi}{T} > 2\omega_M$$

$2\omega_M$ is the Nyquist rate.

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. $\blacksquare$

X(jw)          Xp(jw), ws > 2wM          LPF H(jw)
  /\         /\      /\      /\         ______
 /  \       /  \    /  \    /  \       |      |
-wM wM   -ws        0       ws       -wc     wc

In time domain the recovered signal is the sinc interpolation $x(t) = \sum_n x(nT)\,\dfrac{\sin(\omega_s(t-nT)/2)}{\omega_s(t-nT)/2}$ (with $\omega_c = \omega_s/2$).