# Exercise 4.2Z: About the Sampling Theorem

Harmonic oscillations of different phase

The  sampling theorem  states that the sampling frequency  $f_{\rm A} = 1/T_{\rm A}$  must be at least twice as large as the largest frequency  $f_\text {N, max}$  contained in the source signal  $q(t)$:

$$f_{\rm A} \ge 2 \cdot f_{\rm N,\hspace{0.05cm}max}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} T_{\rm A} \le \frac{1}{2 \cdot f_{\rm N, \hspace{0.05cm}max}}\hspace{0.05cm}.$$

If this condition is met,  then at the receiver the message signal can be passed through a rectangular  (ideal)  low-pass filter with frequency response

$$H(f) = \left\{ \begin{array}{l} 1 \\ 1/2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < f_{\rm G},} \\ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| = f_{\rm G},} \\ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| > f_{\rm G}} \\ \end{array}$$

can be completely reconstructed, that is, it is then  $v(t) = q(t)$.

• The cutoff frequency  $f_{\rm G}$  is to be chosen equal to half the sampling frequency.
• The equal sign is generally valid only if the spectrum  $Q(f)$  does not contain a discrete spectral line at frequency  $f_\text {N, max}$.

In this exercise,  three different source signals are considered,  each of which can be expressed as a harmonic oscillation

$$q(t) = A \cdot \cos (2 \pi \cdot f_{\rm N} \cdot t - \varphi)$$

with amplitude  $A = 1\ \rm V$  and frequency  $f_{\rm N}= 5 \ \rm kHz$.  For the spectral function  $Q(f)$  of all represented time signals generally holds:

$$Q(f) = \frac{A}{2} \cdot \delta (f- f_{\rm N}) \cdot {\rm e}^{-{\rm j}\hspace{0.04cm}\cdot \hspace{0.04cm}\varphi}+ \frac{A}{2} \cdot \delta (f+ f_{\rm N}) \cdot {\rm e}^{+{\rm j}\hspace{0.04cm}\cdot \hspace{0.04cm}\varphi}\hspace{0.05cm}.$$

The oscillations sketched in the graph differ only by the phase  $φ$:

• $φ_1 = 0$   ⇒   cosine signal  $q_1(t)$,
• $φ_2 = π/2 \ (= 90^\circ)$   ⇒   sinusoidal signal  $q_2(t)$,
• $φ_3 = π/4 \ (= 45^\circ)$   ⇒   signal  $q_3(t)$.

Hints:

• The exercise belongs to the chapter  "Pulse Code Modulation".
• Reference is made in particular to the page  "Sampling and Signal Reconstruction".
• 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  $ν \cdot T_{\rm A}$.

### Questions

1

Which statements are valid with  $f_{\rm A} = 11\ \rm kHz$?

 The sampling theorem is always satisfied. All signals can be reconstructed by a low-pass filter. It is always true:  $Q_{\rm A}(f = 5 \ {\rm kHz}) = Q(f = 5 \ \rm kHz)$.

2

What sampling distance results with  $f_{\rm A} = 10\ \rm kHz$?

 $T_{\rm A} \ = \$ $\ \rm ms$

3

Which statements are valid for the signal  $q_1(t)$  and  $f_{\rm A} = 10\ \rm kHz$?

 It holds  $Q_{\rm A}(f = 5 \ {\rm kHz)} = Q_1(f = 5 \ \rm kHz)$. A complete signal reconstruction is possible   ⇒   $v_1(t) = q_1(t)$. The reconstructed signal is  $v_1(t) \equiv 0$.

4

What statements hold for the signal  $q_2(t)$  and  $f_{\rm A} = 10\ \rm kHz$?

 It holds  $Q_{\rm A}(f = 5 \ {\rm kHz)} = Q_2(f = 5 \ \rm kHz)$. A complete signal reconstruction is possible   ⇒   $v_2(t) = q_2(t)$. The reconstructed signal is  $v_2(t) \equiv 0$.

5

What statements hold for the signal  $q_3(t)$ and $f_{\rm A} = 10\ \rm kHz$?

 It holds  $Q_{\rm A}(f = 5 \ {\rm kHz)} = Q_3(f = 5 \ \rm kHz)$. A complete signal reconstruction is possible   ⇒   $v_3(t) = q_3(t)$. The reconstructed signal is  $v_3(t) \equiv 0$.

### Solution

#### Solution

(1)  All statements  are true:

Spectral function of the sampled signal
• The sampling theorem is satisfied by  $f_{\rm A} = 11 \ \rm kHz > 2 \cdot 5 \ \rm kHz$  so that a complete signal reconstruction is always possible.
• The spectrum  $Q_{\rm A}(f)$  results from  $Q(f)$  by periodic continuation at the respective frequency spacing  $f_{\rm A}$,  which is generally illustrated in the graph.
• By a rectangular low-pass with  $f_{\rm G} = f_{\rm A}/2 = 5.5 \ \rm kHz$  the original spectrum  $Q(f)$ is obtained.

The shift by

• $f_{\rm A} = 11 \ \rm kHz$  yields the lines at  $+6 \ \rm kHz$  and  $+16 \ \rm kHz$,
• $-f_{\rm A} = -11 \ \rm kHz$  yields the lines at  $-6 \ \rm kHz$  and  $-16 \ \rm kHz$,
• $2 - f_{\rm A} = 22 \ \rm kHz$  yields the lines at  $+17 \ \rm kHz$  and  $+27 \ \rm kHz$,
• $-2 - f_{\rm A}= -22 \ \rm kHz$  yields the lines at  $-17 \ \rm kHz$, $-27 \ \rm kHz$.

(2)  The sampling distance is equal to the reciprocal of the sampling frequency:

$$T_{\rm A} = {1}/{f_{\rm A} }\hspace{0.15cm}\underline { = 0.1\,{\rm ms}} \hspace{0.05cm}.$$

(3)  The correct solution is  suggestion 2:

Spectral function of the sampled cosine signal
• For the cosinusoidal signal,  according to this graph with  $f_{\rm A} = 10 \rm \ kHz$:  All spectral lines of  $Q_{\rm A}(f)$:  are real.
• The periodization of  $Q(f)$  with  $f_{\rm A} = 10 \rm \ kHz$  leads to a Dirac comb with spectral lines at  $±f_{\rm N}$,  $±f_{\rm N}± f_{\rm A}$,  $±f_{\rm N}± 2f_{\rm A}$, . ..
• Through the superpositions,  all Dirac functions have weight  $A$,  while the spectral lines of  $Q(f)$  are weighted only by  $A/2$  each.
• Because  $H(f = f_{\rm N}) = H(f = f_{\rm G}) = 0.5$  the spectrum  $V_1(f)$  after the low-pass is identical to  $Q_1(f)$   ⇒   $v_1(t) = q_1(t)$.
• In the time domain, the signal reconstruction can be thought of as follows:   The samples of  $q_1(t)$  lie exactly at the signal maxima and minima.
• The lowpass filter forms the cosine signal with correct amplitude, frequency and phase.

Sampled sine signal

(4)  Correct is  suggested solution 2:

• All sampled values of  $q_2(t)$  now lie exactly at the zero crossings of the sinusoidal signal,  which means that here  $q_{\rm A}(t) \equiv 0$  holds.  However,  this naturally also gives  $v_2(t) \equiv 0$.
• In the spectral domain,  the result can be derived using the graph for subtask  (1)
⇒  $Q(f)$  is purely imaginary and the imaginary parts at  $±f_{\rm N}$  have different signs.
• Thus,  one positive and one negative part cancel each other in periodization
⇒   $Q_{\rm A}(f) \equiv 0$   ⇒   $V_2(f) \equiv 0$.

Sampled harmonic oscillation with phase  $φ_3 = π/4$

(5)  None of the given solutions  is correct:

• If in the graph for the subtask  (1)  the sampling frequency  $f_{\rm A} = 11 \ \rm kHz$  is replaced by  $f_{\rm A} = 10 \ \rm kHz$,  the real parts add up,  but the imaginary parts cancel out.
• This means that now  $Q_{\rm A}(f)$  and  $V_3(f)$  are real spectra.  This further means:
• The phase information is lost  $(φ = 0)$  and the output signal  $v_3(t)$  is a cosine signal.
• $q_3(t)$  and  $v_3(t)$  thus differ in both amplitude and phase.  Only the frequency remains the same.

The graph shows

• turquoise the signal $q_3(t)$  and its samples  (circles),  and
• red dashed the output signal  $v_3(t)$  of the low-pass.

You can see that the low-pass gives exactly the result you would probably choose if you were to draw a curve through the samples  (circles).