Prove sampling theorem; aliasing and aperture effect

Sampling theorem

A band-limited signal $g(t)$ with no frequency components above $W$ Hz is uniquely determined by its samples taken at uniform intervals $T_s \le 1/(2W)$, and can be recovered from them exactly by an ideal low-pass filter.

Proof

Step 1: spectrum of the sampled signal. Ideal sampling:

$$g_\delta(t) = g(t)\,\delta_{T_s}(t) = \sum_n g(nT_s)\delta(t - nT_s)$$

The periodic impulse train has Fourier series $\delta_{T_s}(t) = \dfrac{1}{T_s}\sum_n e^{j2\pi nf_st}$. Hence

$$g_\delta(t) = f_s\sum_n g(t)e^{j2\pi nf_st} \;\Longleftrightarrow\; G_\delta(f) = f_s\sum_{n=-\infty}^{\infty}G(f - nf_s)$$

The spectrum is $G(f)$ repeated at every multiple of $f_s$.

Step 2: no overlap. Each copy occupies $nf_s - W$ to $nf_s + W$. Copies $n = 0$ and $n = 1$ do not overlap if $W \le f_s - W$, i.e. $f_s \ge 2W$.

Step 3: recovery. Pass $g_\delta(t)$ through an ideal LPF $H(f) = T_s\,\text{rect}(f/2W)$. Only the $n = 0$ term survives: $G_\delta(f)H(f) = G(f)$. In the time domain, with $f_s = 2W$ and $\text{sinc}(x) = \sin x/x$, $h(t) = \text{sinc}(2\pi Wt)$ and

$$g(t) = \sum_n g(nT_s)\,h(t - nT_s) = \sum_n g\!\left(\frac{n}{2W}\right)\text{sinc}(2\pi Wt - n\pi)$$

Thus $g(t)$ is fully recovered from its samples, which proves the theorem.

 Gd(f):  /\      /\      /\
    ____/  \____/  \____/  \____
       -fs      0      fs
            |<-LPF->|   (needs fs >= 2W)

Aliasing effect

When $f_s < 2W$ the spectral copies overlap; components above $f_s/2$ fold back and appear as false low frequencies, and the signal cannot be recovered. Aliasing is prevented by an anti-aliasing LPF and by sampling above $2W$.

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 the high message frequencies. This distortion is the aperture effect. It is reduced by a small $\tau$ and corrected by an equalizer $1/H(f)$.