Nyquist-Kotelnikov sampling theorem; aliasing and aperture

Nyquist–Kotelnikov sampling theorem

A signal $g(t)$ band-limited to $W$ Hz is completely described by its instantaneous values taken at uniform intervals $T_s \le \dfrac{1}{2W}$, and can be exactly reconstructed from them. The minimum rate $f_s = 2W$ is the Nyquist rate. (Kotelnikov proved it independently in 1933; it is also called the Kotelnikov theorem or the Whittaker–Kotelnikov–Shannon (WKS) theorem.)

Time-domain analysis

Ideal sampling multiplies $g(t)$ by a periodic impulse train:

$$g_\delta(t) = g(t)\sum_n\delta(t - nT_s) = \sum_n g(nT_s)\,\delta(t - nT_s)$$

Reconstruction by an ideal LPF of bandwidth $W$ (with $f_s = 2W$ and $\text{sinc}(x) = \sin x/x$, $h(t) = \text{sinc}(2\pi Wt)$) gives

$$g(t) = \sum_{n=-\infty}^{\infty} g(nT_s)\,\text{sinc}\big(2\pi W(t - nT_s)\big)$$

At $t = kT_s$ only the $k$-th sinc is nonzero (all the others are at their zeros), so the curve passes exactly through every sample, and between samples the sincs interpolate the signal.

g(t)      .-~~-.
        .'  |   '.
     _.' |  |  |  '._
     |   |  |  |  |  |   samples every Ts
   -2Ts -Ts 0  Ts 2Ts

Frequency-domain analysis

The impulse train has Fourier series $\delta_{T_s}(t) = f_s\sum_n e^{j2\pi nf_st}$, so

$$G_\delta(f) = f_s\sum_{n=-\infty}^{\infty}G(f - nf_s)$$

     /\        /\        /\
 ___/  \______/  \______/  \___ f
   -fs   -W  0  W     fs

• Copies of $G(f)$ sit at $0, \pm f_s, \pm 2f_s, \ldots$
• They do not overlap if $f_s - W \ge W$, i.e. $f_s \ge 2W$.
• An ideal LPF $H(f) = T_s$ for $|f| < W$ then gives $G(f)$ back.
• If $f_s < 2W$ the copies overlap and $G(f)$ is lost.

Aliasing and aperture effect

• Aliasing: overlap of spectral copies when $f_s < 2W$; high frequencies appear as false low frequencies. It is prevented by an anti-aliasing filter and $f_s > 2W$.
• Aperture effect: in flat-top sampling with pulse width $\tau$, the spectrum is multiplied by $H(f) = \tau\,\text{sinc}(\pi f\tau)$, which attenuates high frequencies. It is corrected with an equalizer $1/H(f)$ or by keeping $\tau$ small.