Sub-sampling; sampling sinc(5πt) at 10 Hz

Sub-sampling theorem

A band-pass signal occupying $f_L$ to $f_H$ (bandwidth $B = f_H - f_L$) can be sampled at a rate much lower than $2f_H$, roughly $2B$, without loss of information. The valid rates are

$$\frac{2f_H}{k} \le f_s \le \frac{2f_L}{k - 1}, \qquad 1 \le k \le \left\lfloor\frac{f_H}{B}\right\rfloor$$

Here aliasing is used on purpose: the band is shifted down to baseband without overlap. This is used in IF-sampling receivers.

Numerical: $x(t) = \text{sinc}(5\pi t)$, $f_s = 10$ Hz

Using $\text{sinc}(\theta) = \sin\theta/\theta$:

$$x(t) = \frac{\sin 5\pi t}{5\pi t} \;\Longleftrightarrow\; X(f) = \frac{1}{5}\,\text{rect}\!\left(\frac{f}{5}\right)$$

so $X(f) = 0.2$ for $|f| < 2.5$ Hz and zero elsewhere. Bandwidth $W = 2.5$ Hz, Nyquist rate $= 5$ Hz.

i) Sampled signal ($T_s = 0.1$ s): $x(nT_s) = \dfrac{\sin(n\pi/2)}{n\pi/2}$

                            1
                            ^
                     0.637  |  0.637
                       ^    |    ^
 0.127                 |    |    |                 0.127
   ^                   |    |    |                   ^
---+----+----+----+----+----+----+----+----+----+----+-------> t (s)
             |                             |
             v                             v
          -0.212                        -0.212
 -0.5      -0.3      -0.1   0   0.1       0.3       0.5

Impulses occur every 0.1 s, with heights following the sinc envelope; the samples at $t = \pm 0.2, \pm 0.4$ s are zero.

ii) Spectrum of the sampled signal

$$X_s(f) = f_s\sum_n X(f - nf_s) = 2\sum_n \text{rect}\!\left(\frac{f - 10n}{5}\right)$$

Rectangles of height $10 \times 0.2 = 2$ and width 5 Hz, centred at $0, \pm 10, \pm 20, \pm 30$ Hz:

  2 __     ___     ___     ___     ___     ___     __
    | |   |   |   |   |   |   |   |   |   |   |   | |
----+-+---+---+---+---+---+---+---+---+---+---+---+-+--> f (Hz)
   -30     -20     -10      0      10      20      30

Occupied bands for $|f| < 30$ Hz: $|f| < 2.5$, $7.5$–$12.5$, $17.5$–$22.5$ and $27.5$–$30$ Hz (and their mirror images at negative frequencies).

iii) Recovery Yes. $f_s = 10$ Hz is greater than the Nyquist rate $2W = 5$ Hz, so the copies are separated by 5 Hz gaps (2.5 to 7.5 Hz). An ideal LPF with gain $1/10$ and cut-off anywhere between 2.5 and 7.5 Hz (e.g. 5 Hz) recovers $x(t)$ exactly.