Exercise 4.3: Natural and Discrete Sampling

From LNTwww

For natural and discrete sampling

Ideal sampling can be described in time domain by multiplying the analog source signal  $q(t)$  by a  Dirac comb  $p_δ(t)$  :

$$ q_{\rm A}(t) = p_{\delta}(t) \cdot q(t) \hspace{0.05cm}.$$

Dirac impulses  – infinitely narrow and infinitely high –  and accordingly also the  "Dirac comb"  $p_δ(t)$  cannot be realized in practice,  however. 

Here we must assume instead the  "rectangular pulse comb"  $p_{\rm R}(t)$  where the following relation holds:

$$ p_{\rm R}(t) = \left [ \frac{1}{T_{\rm A}} \cdot p_{\rm \delta}(t) \right ]\star g_{\rm R}(t)\hspace{0.4cm}\text{with}\hspace{0.4cm} g_{\rm R}(t) = \left\{ \begin{array}{l} 1 \\ 1/2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} < T_{\rm R}/2\hspace{0.05cm}, \\ {\hspace{0.04cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} = T_{\rm R}/2\hspace{0.05cm}, \\ {\hspace{0.005cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} > T_{\rm R}/2\hspace{0.05cm}. \\ \end{array}$$

The duration  $T_{\rm R}$  of a rectangular pulse  $g_{\rm R}(t)$  should be (significantly) smaller than the distance $T_{\rm A}$ of two samples.

In the diagram this ratio is chosen with  $T_{\rm R}/T_{\rm A} = 0.5$  very large to make the difference between  "natural sampling"  and  "discrete sampling"  especially clear:

  • In natural sampling,  the sampled signal  $q_{\rm A}(t)$  is equal to the product of the rectangular pulse comb  $p_{\rm R}(t)$  and the analog source signal  $q(t)$:
$$q_{\rm A}(t) = p_{\rm R}(t) \cdot q(t) = \left [ \frac{1}{T_{\rm A}} \cdot p_{\rm \delta}(t) \star g_{\rm R}(t)\right ]\cdot q(t)\hspace{0.05cm}.$$
  • In contrast,  the corresponding equation for discrete sampling is:
$$ q_{\rm A}(t) = \left [ \frac{1}{T_{\rm A}} \cdot p_{\rm \delta}(t) \cdot q(t)\right ]\star g_{\rm R}(t)\hspace{0.05cm}.$$

In the graph,  these signals are sketched in blue  (natural sampling)  and green  (discrete sampling)  respectively.

For signal reconstruction,  a rectangular low-pass filter  $H(f)$  with cutoff frequency  $f_{\rm G} = f_{\rm A}/2$  and gain  $T_{\rm A}/T_{\rm R}$  is used in the passband:

$$H(f) = \left\{ \begin{array}{l} T_{\rm A}/T_{\rm R} \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < f_{\rm A}/2}\hspace{0.05cm}, \\ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| > f_{\rm A}/2}\hspace{0.05cm}. \\ \end{array}$$


  • The exercise belongs to the chapter  "Pulse Code Modulation".
  • Reference is made in particular to the page  "Natural and discrete sampling".
  • The sampled source signal is denoted by  $q_{\rm A}(t)$  and its spectral function by  $Q_{\rm A}(f)$.
  • Sampling is always performed at  $ν \cdot T_{\rm A}$.



Let  $T_{\rm R}/T_{\rm A} = 0.5$.  For this,  give the normalized spectrum  $G_{\rm R}(f)/T_{\rm A}$.  What spectral value occurs at  $f = 0$ ?

$G_{\rm R}(f=0)/T_{\rm A} \ = \ $


What is the spectrum  $Q_{\rm A}(f)$  in natural sampling?  Suggestions:

It holds  $Q_{\rm A}(f) = P_{\rm δ}(f) ∗ Q(f)$.
It holds  $Q_{\rm A}(f) = \big[{\rm δ}(f) \cdot (G_{\rm R}(f)/T_{\rm A})\big] ∗ Q(f)$.
It holds  $Q_{\rm A}(f) = \big[P_{\rm δ}(f) ∗ Q(f)\big] \cdot (G_{\rm R}(f)/T_{\rm A})$.


For natural sampling:  Is the specified low-pass suitable for interpolation?



What is the spectrum  $Q_{\rm A}(f)$  for discrete sampling?  Suggestions:

It holds  $Q_{\rm A}(f) = P_{\rm δ}(f) ∗ Q(f)$.
It holds  $Q_{\rm A}(f) = \big[{\rm δ}(f) \cdot (G_{\rm R}(f)/T_{\rm A})\big] ∗ Q(f)$.
It holds  $Q_{\rm A}(f) = \big[P_{\rm δ}(f) ∗ Q(f)\big] \cdot (G_{\rm R}(f)/T_{\rm A})$.


For discrete sampling:  Is the specified low-pass suitable for interpolation?



(1)  The spectrum of the rectangular pulse  $g_{\rm R}(t)$  with amplitude  $1$  and duration  $T_{\rm R}$  is:

$$ G_{\rm R}(f) = T_{\rm R} \cdot {\rm sinc}(f T_{\rm R}) \hspace{0.3cm} {\rm with}\hspace{0.3cm} {\rm sinc}(x) = \sin(\pi x)/(\pi x)\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{G_{\rm R}(f)}{T_{\rm A}} = \frac{T_{\rm R}}{T_{\rm A}} \cdot {\rm sinc}(f T_{\rm R})\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{G_{\rm R}(f = 0)}{T_{\rm A}} =\frac{T_{\rm R}}{T_{\rm A}}\hspace{0.15cm}\underline { = 0.5} \hspace{0.05cm}.$$

(2)  The correct solution is the  second suggested solution:

  • From the given equation in the time domain,  the convolution theorem gives:
$$q_{\rm A}(t) = \left [ \frac{1}{T_{\rm A}} \cdot p_{\rm \delta}(t) \star g_{\rm R}(t)\right ]\cdot q(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}Q_{\rm A}(f) = \left [ \frac{1}{T_{\rm A}}\cdot P_{\rm \delta}(f) \cdot G_{\rm R}(f) \right ] \star Q(f) = \left [ P_{\rm \delta}(f) \cdot \frac{G_{\rm R}(f)}{{T_{\rm A}}} \right ] \star Q(f) \hspace{0.05cm}.$$
  • The first proposed solution is valid only for ideal sampling   (with a Dirac comb)   and the last one for discrete sampling.

(3)  The answer is  YES:

  • Starting from the result of the subtask  (2)  using the spectral function of the Dirac comb,  we obtain.
$$Q_{\rm A}(f) = \left [ P_{\rm \delta}(f) \cdot \frac{G_{\rm R}(f)}{{T_{\rm A}}} \right ] \star Q(f)= \left [ \frac{G_{\rm R}(f)}{{T_{\rm A}}} \cdot \sum_{\mu = -\infty}^{+\infty} \delta(f - \mu \cdot f_{\rm A})\right ] \star Q(f) \hspace{0.05cm}.$$
  • When the sampling theorem is satisfied and the low-pass filter is correct: 
    From the infinite convolution products  only the convolution product with  $μ = 0$  lie in the passband.
  • Taking into account the gain factor  $T_{\rm A}/T_{\rm R}$,  we thus obtain for the spectrum at the filter output:
$$V(f) = \frac{T_{\rm A}}{T_{\rm R}} \cdot \left [ \frac{G_{\rm R}(f = 0)}{{T_{\rm A}}} \cdot \delta(f )\right ] \star Q(f)= Q(f) \hspace{0.05cm}.$$

(4)  The  last suggested solution  is correct.

  • Shifting the factor  $1/T_{\rm A}$  to the rectangular pulse,  we obtain with discrete sampling using the convolution theorem:
$$ q_{\rm A}(t) = \big [ p_{\rm \delta}(t)\cdot q(t) \big ] \star \frac{g_{\rm R}(t)}{T_{\rm A}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}Q_{\rm A}(f)= \big [ P_{\rm \delta}(f)\star Q(f) \big ] \cdot \frac{G_{\rm R}(f)}{T_{\rm A}}\hspace{0.05cm}.$$

(5)  The answer is  NO:

  • The weighting function  $G_{\rm R}(f)$  now involves the inner kernel  $(μ = 0)$  of the convolution product.
  • All other terms  $(μ ≠ 0)$  are eliminated by the low-pass filter. 
  • One obtains here in the relevant range  $|f| < f_{\rm A}/2$:
$$V(f) = \frac{T_{\rm A}}{T_{\rm R}} \cdot \frac{G_{\rm R}(f )}{{T_{\rm A}}} \cdot Q(f) = 2 \cdot 0.5 \cdot {\rm sinc}(f T_{\rm R})\cdot Q(f) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}V(f) = Q(f) \cdot {\rm sinc}(f T_{\rm R})\hspace{0.05cm}.$$
  • If no additional equalization is provided here, the higher frequencies are attenuated according to the  $\rm sinc$ function.
  • The highest  signal frequency  $(f = f_{\rm A}/2)$  is attenuated the most here:
$$V(f = f_{\rm A}/2) = Q( f_{\rm A}/2) \cdot {\rm sinc}( \frac{T_{\rm R}}{2 \cdot T_{\rm A}})= Q( f_{\rm A}/2) \cdot \frac{\sin(\pi/4)}{\pi/4}\approx 0.9 \cdot Q( f_{\rm A}/2) \hspace{0.05cm}.$$