Difference between revisions of "Aufgaben:Exercise 4.4Z: Pointer Diagram for SSB-AM"

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[[File:P_ID732__Sig_Z_4_4_neu.png|right|frame|Given spectrum  $S_+(f)$]]
+
[[File:P_ID732__Sig_Z_4_4_neu.png|right|frame|Given analytical spectrum  $S_+(f)$]]
 +
 
 
The analytical signal  $s_+(t)$  with the line spectrum  
 
The analytical signal  $s_+(t)$  with the line spectrum  
 
:$$S_{+}(f) =  {\rm 1 \hspace{0.05cm} V} \cdot\delta (f - f_{\rm
 
:$$S_{+}(f) =  {\rm 1 \hspace{0.05cm} V} \cdot\delta (f - f_{\rm
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f_{\rm 60})$$  
 
f_{\rm 60})$$  
 
is to be considered.
 
is to be considered.
Here  $f_{50}$  and  $f_{60}$  are abbreviations for the frequencies  $50 \ \text{kHz}$  and  $60 \ \text{kHz}$, respectively..
+
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&nbsp; [[Modulation_Methods/Einseitenbandmodulation|single sideband amplitude modulation]]&nbsp; (SSB-AM) of a sinusoidal message signal&nbsp; $($Frequenz&nbsp; $f_{\rm N} = 10 \ \text{kHz})$&nbsp; with a cosinusoidal carrier signal&nbsp; $(f_{\rm T} = 50 \ \text{kHz})$&nbsp;, whereby only the <u>upper sideband</u> is transmitted (''USB-Modulation'').
 
 
 
However, the analytical signal could also result from a&nbsp; ''LSB-Modulation''&nbsp; modulation of the same sinusoidal signal if a sinusoidal carrier signal with carrier frequency&nbsp; $f_{\rm T} = 60 \ \text{kHz}$&nbsp; is used.
 
 
 
 
 
  
 +
This analytical signal could occur, for example, with the&nbsp; [[Modulation_Methods/Einseitenbandmodulation|Single Sideband Amplitude Modulation]]&nbsp; $\text{(SSB&ndash;AM)}$&nbsp; of a sinusoidal message signal&nbsp; $($Frequenz&nbsp; $f_{\rm N} = 10 \ \text{kHz})$&nbsp; with a cosinusoidal carrier signal&nbsp; $(f_{\rm T} = 50 \ \text{kHz})$&nbsp;, whereby only the upper sideband is transmitted &nbsp; &rArr; &nbsp; $\text{Upper Sideband Modulation}$.
  
 +
However, the analytical signal could also result from a&nbsp; $\text{Lower Sideband Modulation}$&nbsp; of the same sinusoidal signal if a sinusoidal carrier with frequency&nbsp; $f_{\rm T} = 60 \ \text{kHz}$&nbsp; is used.
  
  
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''Hints:''  
 
''Hints:''  
*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function|Analytical Signal and Its Spectral Function]].
+
*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function|Analytical Signal and its Spectral Function]].
 
   
 
   
*You can check your solution with the interaction module&nbsp; [[Applets:Physikalisches_Signal_%26_Analytisches_Signal|Physikalisches Signal & Analytisches Signal]]&nbsp;.
+
*You can check your solution with the interaction module&nbsp; [[Applets:Physical_Signal_%26_Analytic_Signal|Physical and Analytical Signal]].
 +
 
  
  
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<quiz display=simple>
 
<quiz display=simple>
{Give the analytical signal&nbsp; $s_+(t)$&nbsp; as a formula. What value results at the starting time&nbsp; $t = 0$?
+
{Give the analytical signal&nbsp; $s_+(t)$&nbsp; as a formula.&nbsp; What value results at the starting time&nbsp; $t = 0$?
 
|type="{}"}
 
|type="{}"}
 
$\text{Re}[s_+(t = 0)]\ = \ $  { 1 3% } &nbsp;$\text{V}$
 
$\text{Re}[s_+(t = 0)]\ = \ $  { 1 3% } &nbsp;$\text{V}$
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{What is the maximum value of&nbsp; $|s_+(t)|$&nbsp;? At what time&nbsp; $t_2$&nbsp; is this maximum value reached for the first time?
+
{What is the maximum value of&nbsp; $|s_+(t)|$?&nbsp; At what time&nbsp; $t_2$&nbsp; is this maximum value reached for the first time?
 
|type="{}"}
 
|type="{}"}
 
$|s_+(t)|_{\rm max}\ = \ ${ 2 3% } &nbsp;$\text{V}$
 
$|s_+(t)|_{\rm max}\ = \ ${ 2 3% } &nbsp;$\text{V}$
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60}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({
 
60}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({
 
\omega_{\rm 60}\hspace{0.05cm} t }).$$
 
\omega_{\rm 60}\hspace{0.05cm} t }).$$
The real part of this describes the actual, physical signal:
+
The real part of this describes the actual physical signal:
 
:$$s(t) = {\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm
 
:$$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({
 
50}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({
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Correct is the <u>proposed solution 3</u>:
 
Correct is the <u>proposed solution 3</u>:
*Considering the&nbsp; $50 \ \text{kHz-Cosinussignals}$&nbsp; cosine signal alone, the first zero crossing would occur at&nbsp; $t_1 = T_0/4$&nbsp; , i.e. after&nbsp; $5 \ {\rm &micro; s}$, where&nbsp; $T_0 = 1/f_{50} = 20 \ {\rm &micro; s}$&nbsp; denotes the period duration of this signal.  
+
*Considering the&nbsp; $50 \ \text{kHz}$&nbsp; cosine signal alone, the first zero crossing would occur at&nbsp; $t_1 = T_0/4$&nbsp; , i.e. after&nbsp; $5 \ {\rm &micro; s}$, where&nbsp; $T_0 = 1/f_{50} = 20 \ {\rm &micro; s}$&nbsp; denotes the period duration of this signal.  
 
*The sinusoidal signal with the frequency&nbsp; $60 \ \text{kHz}$&nbsp; is positive during the entire first half-wave&nbsp; $(0 \, \text{...} \, 8.33\ {\rm &micro; s})$&nbsp;.
 
*The sinusoidal signal with the frequency&nbsp; $60 \ \text{kHz}$&nbsp; is positive during the entire first half-wave&nbsp; $(0 \, \text{...} \, 8.33\ {\rm &micro; s})$&nbsp;.
 
*Due to the plus sign, the first zero crossing of&nbsp; $s(t) \ \Rightarrow \ t_1 > 5\ {\rm &micro; s}$ is delayed.  
 
*Due to the plus sign, the first zero crossing of&nbsp; $s(t) \ \Rightarrow \ t_1 > 5\ {\rm &micro; s}$ is delayed.  
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'''(3)'''&nbsp;  The maximum value of&nbsp; $|s_+(t)|$&nbsp; 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.&nbsp; $\underline {2\ \text{ V}}$.
+
'''(3)'''&nbsp;  The maximum value of&nbsp; $|s_+(t)|$&nbsp; 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.&nbsp; $\underline {2\ \text{V}}$.
  
This case is reached for the first time when the faster pointer with angular velocity&nbsp; $\omega_{60}$&nbsp; has caught up its "lag" of&nbsp; $90^{\circ} \; (\pi /2)$&nbsp; with the slower pointer&nbsp; ($\omega_{50}$)&nbsp;:
+
This case is reached for the first time when the faster pointer with circular velocity&nbsp; $\omega_{60}$&nbsp; has caught up its "lag" of&nbsp; $90^{\circ} \; (\pi /2)$&nbsp; with the slower pointer&nbsp; ($\omega_{50}$)&nbsp;:
 
:$$\omega_{\rm 60} \cdot  t_2 - \omega_{\rm
 
:$$\omega_{\rm 60} \cdot  t_2 - \omega_{\rm
 
50}\cdot t_2 = \frac{\pi}{2} \hspace{0.3cm}
 
50}\cdot t_2 = \frac{\pi}{2} \hspace{0.3cm}
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f_{\rm 50})} =  \frac{1}{4
 
f_{\rm 50})} =  \frac{1}{4
 
\cdot(f_{\rm 60}- f_{\rm 50})}\hspace{0.15 cm}\underline{= {\rm 25 \hspace{0.05cm} {\rm &micro; s}}}.$$
 
\cdot(f_{\rm 60}- f_{\rm 50})}\hspace{0.15 cm}\underline{= {\rm 25 \hspace{0.05cm} {\rm &micro; s}}}.$$
*At this point, the two pointers have made&nbsp; $5/4$&nbsp; bzw.&nbsp; $6/4$&nbsp; revolutions respectively and both point in the direction of the imaginary axis (see right graph).
+
*At this point, the two pointers have made&nbsp; $5/4$&nbsp; and&nbsp; $6/4$&nbsp; rotations respectively and both point in the direction of the imaginary axis (see right graph).
*The actual, physical signal&nbsp; $s(t)$ –  i.e. the real part of&nbsp; $s_+(t)$ – is therefore zero at this moment.
+
*The actual physical signal&nbsp; $s(t)$ –  i.e. the real part of&nbsp; $s_+(t)$ – is therefore zero at this moment.
  
  
  
 
'''(4)'''&nbsp;  The condition for&nbsp; $|s_+(t_3)| = 0$&nbsp; is that there is a phase offset of&nbsp; $180^\circ$&nbsp; between the two equally long pointers so that they cancel each other out.
 
'''(4)'''&nbsp;  The condition for&nbsp; $|s_+(t_3)| = 0$&nbsp; is that there is a phase offset of&nbsp; $180^\circ$&nbsp; between the two equally long pointers so that they cancel each other out.
*This further means that the faster pointer has rotated&nbsp; $3\pi /2$&nbsp; further than the&nbsp; $50 \ \text{kHz-component}$.  
+
*This further means that the faster pointer has rotated&nbsp; $3\pi /2$&nbsp; further than the&nbsp; $50 \ \text{kHz}$&nbsp; component.  
  
*Analogous to the sample solution of sub-task&nbsp; '''(3)'''&nbsp;, the following therefore applies:
+
*Analogous to the sample solution of sub-task&nbsp; '''(3)'''&nbsp;, therefore the following applies:
 
:$$t_3  = \frac{3\pi/2}{2\pi (f_{\rm 60}- f_{\rm 50})} \hspace{0.15 cm}\underline{=
 
:$$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 &micro; s}}}.$$
 
  {\rm 75 \hspace{0.05cm} {\rm &micro; s}}}.$$
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__NOEDITSECTION__
 
__NOEDITSECTION__
[[Category:Exercises for Signal Representation|^4.2 Analytical Signal and Its Spectral Function^]]
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[[Category:Signal Representation: Exercises|^4.2 Analytical Signal and its Spectral Function^]]

Revision as of 15:37, 7 May 2021

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:
$$t_3 = \frac{3\pi/2}{2\pi (f_{\rm 60}- f_{\rm 50})} \hspace{0.15 cm}\underline{= {\rm 75 \hspace{0.05cm} {\rm µ s}}}.$$