Sampling theorem via impulse train sampling

Nyquist sampling theorem: a signal band-limited to $\omega_M$ is completely determined by its samples $x(nT)$ if $\omega_s = 2\pi/T > 2\omega_M$. The minimum rate $2\omega_M$ is the Nyquist rate.

Frequency-domain analysis of impulse-train sampling

Let $x(t)$ be band-limited to $\omega_M$. Impulse-train sampling multiplies it by a periodic impulse train of period $T$:

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

The Fourier transform of the impulse train is also an impulse train, with spacing $\omega_s = 2\pi/T$:

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

Multiplication in time is convolution in frequency:

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

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

X(jw):            /\
                 /  \
           -----/----\------> w
              -wM    wM

Xp(jw), ws > 2wM (no overlap):
    /\        /\        /\
   /  \      /  \      /  \
--/----\----/----\----/----\--> w
    -ws        0        ws
       gap between copies = ws - 2wM

Conclusion from the spectrum

• The copy centred at $k=0$ occupies $-\omega_M$ to $\omega_M$; the next copy starts at $\omega_s - \omega_M$.
• The copies do not overlap only if $\omega_s - \omega_M > \omega_M$, i.e. $\omega_s > 2\omega_M$. This is the Nyquist condition.
• When it holds, an ideal low-pass filter with gain $T$ and cutoff $\omega_c$ such that $\omega_M < \omega_c < \omega_s - \omega_M$ passes only the $k=0$ copy, giving back $X(j\omega)$ exactly. Hence $x(t)$ is fully recovered.
• If $\omega_s < 2\omega_M$, the copies overlap (aliasing) and $x(t)$ cannot be recovered.