Exercise 4.4Z: Pointer Diagram for SSB-AM

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Given analytical spectrum  $S_+(f)$

The analytical signal  $s_+(t)$  with the line spectrum

$$S_{+}(f) = {\rm 1 \hspace{0.05cm} V} \cdot\delta (f - f_{\rm 50})- {\rm j} \cdot {\rm 1 \hspace{0.05cm} V} \cdot\delta (f - f_{\rm 60})$$

is to be considered. Here  $f_{50}$  and  $f_{60}$  are abbreviations for the frequencies  $50 \ \text{kHz}$  and  $60 \ \text{kHz}$, respectively.

This analytical signal could occur, for example, with the  Single Sideband Amplitude Modulation  $\text{(SSB–AM)}$  of a sinusoidal message signal  $($Frequenz  $f_{\rm N} = 10 \ \text{kHz})$  with a cosinusoidal carrier signal  $(f_{\rm T} = 50 \ \text{kHz})$ , whereby only the upper sideband is transmitted   ⇒   $\text{Upper Sideband Modulation}$.

However, the analytical signal could also result from a  $\text{Lower Sideband Modulation}$  of the same sinusoidal signal if a sinusoidal carrier with frequency  $f_{\rm T} = 60 \ \text{kHz}$  is used.



Hints:



Questions

1

Give the analytical signal  $s_+(t)$  as a formula.  What value results at the starting time  $t = 0$?

$\text{Re}[s_+(t = 0)]\ = \ $

 $\text{V}$
$\text{Im}[s_+(t = 0)]\ = \ $

 $\text{V}$

2

At what time  $t_1$  does the first zero crossing of the physical signal  $s(t)$  occur relative to the first zero crossing of the  $50 \ \text{kHz-cosine signal}$ ?
Note:   The latter is at time  $T_0/4 = 1/(4 \cdot f_{50}) = 5 \ µ \text{s}$.

It is  $t_1 < 5 \ {\rm µ} \text{s}$.
It is  $t_1 = 5 \ {\rm µ}\text{s}$.
It is  $t_1 > 5 \ {\rm µ} \text{s}$.

3

What is the maximum value of  $|s_+(t)|$?  At what time  $t_2$  is this maximum value reached for the first time?

$|s_+(t)|_{\rm max}\ = \ $

 $\text{V}$
$t_2\ = \ $

 ${\rm µ s}$

4

At what time  $t_3$  is the pointer length  $|s_+(t)|$  equal to zero for the first time?

$t_3\ = \ $

 ${\rm µ s}$


Solution

Three different analytical signals

(1)  The analytical signal is generally:

$$s_{+}(t) = {\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm 50}\hspace{0.05cm} t } - {\rm j}\cdot{\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm 60} \hspace{0.05cm} t }.$$

At time  $t = 0$  the complex exponential functions each take the value  $1$  and one obtains (see left graph):

  • $\text{Re}[s_+(t = 0)] \; \underline{= +1\ \text{V}}$,
  • $\text{Im}[s_+(t = 0)]\; \underline{ = \,-\hspace{-0.08cm}1\ \text{V}}$.


(2)  For the analytical signal it can also be written:

$$s_{+}(t) = {\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm 50}\hspace{0.05cm} t }) + {\rm j} \cdot{\rm 1 \hspace{0.05cm} V} \cdot \sin({ \omega_{\rm 50}\hspace{0.05cm} t }) - {\rm j} \cdot {\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm 60}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({ \omega_{\rm 60}\hspace{0.05cm} t }).$$

The real part of this describes the actual physical signal:

$$s(t) = {\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm 50}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({ \omega_{\rm 60}\hspace{0.05cm} t }).$$

Correct is the proposed solution 3:

  • Considering the  $50 \ \text{kHz}$  cosine signal alone, the first zero crossing would occur at  $t_1 = T_0/4$  , i.e. after  $5 \ {\rm µ s}$, where  $T_0 = 1/f_{50} = 20 \ {\rm µ s}$  denotes the period duration of this signal.
  • The sinusoidal signal with the frequency  $60 \ \text{kHz}$  is positive during the entire first half-wave  $(0 \, \text{...} \, 8.33\ {\rm µ s})$ .
  • Due to the plus sign, the first zero crossing of  $s(t) \ \Rightarrow \ t_1 > 5\ {\rm µ s}$ is delayed.
  • The middle graph shows the analytical signal at time  $t = T_0/4$, when the red carrier would have its zero crossing.
  • The zero crossing of the violet cumulative pointer only occurs when it points in the direction of the imaginary axis. Then  $s(t_1) = \text{Re}[s_+(t_1)] = 0$.


(3)  The maximum value of  $|s_+(t)|$  is reached when both pointers point in the same direction. The amount of the sum pointer is then equal to the sum of the two individual pointers; i.e.  $\underline {2\ \text{V}}$.

This case is reached for the first time when the faster pointer with circular velocity  $\omega_{60}$  has caught up its "lag" of  $90^{\circ} \; (\pi /2)$  with the slower pointer  ($\omega_{50}$) :

$$\omega_{\rm 60} \cdot t_2 - \omega_{\rm 50}\cdot t_2 = \frac{\pi}{2} \hspace{0.3cm} \Rightarrow\hspace{0.3cm}t_2 = \frac{\pi/2}{2\pi (f_{\rm 60}- f_{\rm 50})} = \frac{1}{4 \cdot(f_{\rm 60}- f_{\rm 50})}\hspace{0.15 cm}\underline{= {\rm 25 \hspace{0.05cm} {\rm µ s}}}.$$
  • At this point, the two pointers have made  $5/4$  and  $6/4$  rotations respectively and both point in the direction of the imaginary axis (see right graph).
  • The actual physical signal  $s(t)$ – i.e. the real part of  $s_+(t)$ – is therefore zero at this moment.


(4)  The condition for  $|s_+(t_3)| = 0$  is that there is a phase offset of  $180^\circ$  between the two equally long pointers so that they cancel each other out.

  • This further means that the faster pointer has rotated  $3\pi /2$  further than the  $50 \ \text{kHz}$  component.
  • Analogous to the sample solution of sub-task  (3) , therefore the following applies:
$$t_3 = \frac{3\pi/2}{2\pi (f_{\rm 60}- f_{\rm 50})} \hspace{0.15 cm}\underline{= {\rm 75 \hspace{0.05cm} {\rm µ s}}}.$$