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

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{{quiz-Header|Buchseite=*Buch*/*Kapitel*
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{{quiz-Header|Buchseite=Signal_Representation/Analytical_Signal_and_Its_Spectral_Function
 
}}
 
}}
  
[[File:*|250px|right|*]]
+
[[File:P_ID718__Sig_A_4_4.png|250px|right|frame|Spectrum of the analytical signal]]
  
 +
We assume a cosine-shaped source signal&nbsp; $q(t)$&nbsp; with
 +
*amplitude&nbsp; $A_{\rm N} = 0.8 \ \text{V}$&nbsp;  and
 +
*frequency&nbsp; $f_{\rm N}= 10 \ \text{kHz}$.
  
===Fragebogen===
+
 
 +
The frequency conversion is done by means of&nbsp; [[Modulation_Methods/Zweiseitenband-Amplitudenmodulation#ZSB-Amplitudenmodulation_mit_Tr.C3.A4ger|"Double-Sideband Amplitude Modulation with Carrier"]].
 +
 
 +
The modulated signal&nbsp; $s(t)$&nbsp; is with the (normalised) carrier&nbsp; $z(t) = \text{cos}(\omega_{\rm T} \cdot t)$&nbsp; and the DC component&nbsp; $q_0 = 1 \ \text{V}$:
 +
 +
:$$\begin{align*} s(t) & = \left(q_0 + q(t)\right) \cdot z(t)= \left({\rm 1 \hspace{0.05cm}
 +
V}  + {\rm 0.8 \hspace{0.05cm}V}\cdot {\cos} ( \omega_{\rm N}\cdot  t)\right)
 +
\cdot {\cos} ( \omega_{\rm T}\cdot  t) = \\ & =  q_0 \cdot {\cos} ( \omega_{\rm T}\cdot t) +
 +
{A_{\rm N}}/{2} \cdot {\cos} ( (\omega_{\rm T}+ \omega_{\rm N}) \cdot t)
 +
+  {A_{\rm N}}/{2} \cdot {\cos} ( (\omega_{\rm T}- \omega_{\rm N}) \cdot t).\end{align*}$$
 +
 
 +
The first term describes the carrier, the second term the so-called upper sideband&nbsp; $\rm (OSB)$&nbsp;  and the last term the lower sideband&nbsp; $\rm (USB)$.
 +
 
 +
The sketch shows the spectrum&nbsp; $S_+(f)$&nbsp; of the corresponding analytical signal for&nbsp; $f_{\rm T} = 50 \ \text{kHz}$. You can see
 +
*the carrier (red),
 +
*the upper sideband (blue),&nbsp; and
 +
*the lower sideband (grün).
 +
 
 +
 
 +
In subtask&nbsp; '''(5)'''&nbsp; the magnitude of&nbsp; $s_+(t)$&nbsp; is asked for.&nbsp; This is the length of the resulting pointer.
 +
 
 +
 
 +
 
 +
 
 +
 
 +
''Hints:''
 +
*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function|Analytical Signal and its Spectral Function]].
 +
*The interactive applet&nbsp; [[Applets:Physical_Signal_%26_Analytic_Signal|Physical and Analytical Signal]]&nbsp; illustrates the topic covered here.
 +
*In this task we use the following nomenclature because of the German original:
 +
#The index&nbsp; $\rm N$&nbsp; stands for "source signal" &nbsp; &rArr; &nbsp; (German:&nbsp; "Nachrichtenignal").
 +
#The index&nbsp; $\rm T$&nbsp; stands for&nbsp; "carrier" &nbsp; &rArr; &nbsp; (German:&nbsp; "Trägersignal").
 +
#$\rm OSB$&nbsp; denotes the&nbsp; "upper sideband" &nbsp; &rArr; &nbsp; (German:&nbsp; "oberes Seitenband").
 +
#$\rm USB$&nbsp; denotes the&nbsp; "lower sideband" &nbsp; &rArr; &nbsp; (German:&nbsp; "unteres Seitenband").
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Multiple-Choice Frage
+
{What is the analytical signal&nbsp; $s_+(t)$.&nbsp; What is its magnitude at time&nbsp; $t  = 0$?
 +
|type="{}"}
 +
$\text{Re}[s_+(t=0)]\ = \ $  { 1.8 3% } &nbsp;$\text{V}$
 +
$\text{Im}[s_+(t=0)]\ = \ $ { 0. } &nbsp;$\text{V}$
 +
 
 +
{Which of the following statements are true?
 
|type="[]"}
 
|type="[]"}
- Falsch
+
+ $s_+(t)$&nbsp; results from&nbsp; $s(t)$, if&nbsp; $\cos(\text{...})$&nbsp; is replaced by&nbsp; ${\rm e}^{{\rm j}(\text{...})}$&nbsp;.
+ Richtig
+
- If&nbsp; $s(t)$&nbsp; is an even time function,&nbsp; $s_+(t)$&nbsp; is purely real.
 +
- At no time does the imaginary part of&nbsp; $s_+(t)$ disappear.
 +
 
  
 +
{What is the value of the analytical signal at time&nbsp; $t = 5 \ {\rm &micro;}\text{s}$?
 +
|type="{}"}
 +
$\text{Re}[s_+(t=5  \ {\rm &micro;} \text{s})]\ = \ $ { 0. } &nbsp;$\text{V}$
 +
$\text{Im}[s_+(t=5 \ {\rm &micro;} \text{s})]\ = \ $ { 1.761 3% } &nbsp;$\text{V}$
  
{Input-Box Frage
+
{What is the value of&nbsp; $s_+(t)$&nbsp; at time&nbsp; $t = 20 \ {\rm &micro;}\text{s}$?
 
|type="{}"}
 
|type="{}"}
<math> \alpha = </math> { 0.3 _5 }
+
$\text{Re}[s_+(t=20 \ {\rm &micro;} \text{s})]\ = \ $ { 1.236 3% } &nbsp;$\text{V}$
 +
$\text{Im}[s_+(t=20 \ {\rm &micro;} \text{s})]\ = \ $ { 0. } &nbsp;$\text{V}$
  
 +
{What is the smallest possible pointer length?&nbsp; At what time &nbsp; $t_{\text{min}}$&nbsp; does this value occur for the first time?
 +
|type="{}"}
 +
$|s_+(t)|_{\text{min}}\ = \ $ { 0.2 3% } &nbsp;$\text{V}$
 +
$t_{\text{min}}\ = \ $ { 50 3% } &nbsp;${\rm &micro;} \text{s}$
  
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
 
 +
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.'''  Antwort 1
+
'''(1)'''&nbsp;  By inverse Fourier transform of&nbsp; $S_+(f)$&nbsp; taking into account the&nbsp; "Shifting Theorem":
 +
 
 +
:$$s_{+}(t) = {\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm
 +
j}\hspace{0.05cm} \omega_{\rm 50}\hspace{0.05cm} t } + {\rm 0.4
 +
\hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm}
 +
\omega_{\rm 60} \hspace{0.05cm} t }+ {\rm 0.4 \hspace{0.05cm} V}
 +
\cdot {\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm
 +
40}\hspace{0.05cm} t }.$$
 +
 
 +
[[File:EN_Sig_A_4_4_ML.png|center|frame|Three different analytical signals]]
 +
 
 +
*The expression describes the sum of three pointers rotating at different circular velocities.
 +
*In the above equation, for example,&nbsp;  $\omega_{60} = 2\pi (f_{\rm T} + f_{\rm N}) = 2\pi \cdot 60 \ \text{kHz}$.  
 +
*At time&nbsp; $t = 0$&nbsp; all three pointers point in the direction of the real axis (see left graph).
 +
*One obtains the <u>real value</u>&nbsp; $s_+(t = 0) \;\underline{=  1.8 \ \text{V}}$.
 +
 
 +
<br clear=all>
 +
'''(2)'''&nbsp; The <u>first statement</u> is correct and results from the&nbsp; "Hilbert transform".&nbsp; On the other hand, the next two statements are'nt correct:
 +
*$s_+(t)$&nbsp; is always a complex time function with exception of the limiting case&nbsp; $s(t) \equiv 0$.
 +
*However, every complex function also has purely real values at some points in time.
 +
*The&nbsp; "pointer group"&nbsp; always rotates in a mathematically positive direction.
 +
*If the sum vector crosses the real axis, the imaginary part disappears at this point and&nbsp; $s_+(t)$&nbsp; is purely real.
 +
 
 +
 
 +
 
 +
'''(3)'''&nbsp;  The period duration of the carrier signal is&nbsp; $T_0 = 1/f_T = 20 \ {\rm &micro;} \text{s}$.
 +
*After&nbsp; $t = 5 \ {\rm &micro;} \text{s}$&nbsp;  (see middle graph) the carrier has thus rotated by&nbsp; $90^{\circ}$.
 +
*The blue pointer&nbsp; $\rm (OSB)$&nbsp; rotates&nbsp; $20\%$&nbsp; faster, the green one&nbsp; $\rm (USB)$&nbsp; $20\%$&nbsp; slower than the red rotary pointer (carrier signal):
 +
 +
:$$s_{+}({\rm 5 \hspace{0.05cm} {\rm &micro;}  s})  =  {\rm 1
 +
\hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 2 \pi
 +
\hspace{0.03cm} \cdot \hspace{0.08cm}50 \hspace{0.03cm} \cdot
 +
\hspace{0.08cm}0.005 } + {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm
 +
e}^{{\rm j}\hspace{0.05cm} 2 \pi \hspace{0.03cm} \cdot
 +
\hspace{0.08cm}60 \hspace{0.03cm} \cdot \hspace{0.08cm}0.005
 +
}+ {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm
 +
j}\hspace{0.05cm} 2 \pi \hspace{0.03cm} \cdot \hspace{0.08cm}40
 +
\hspace{0.03cm} \cdot \hspace{0.08cm}0.005 } =  {\rm 1
 +
\hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 90^\circ
 +
}+ {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm
 +
j}\hspace{0.05cm} 108^\circ }+{\rm 0.4 \hspace{0.05cm} V} \cdot
 +
{\rm e}^{{\rm j}\hspace{0.05cm} 72^\circ }.$$
 +
 
 +
*Thus, the angles travelled in&nbsp; $ 5 \ {\rm &micro;} \text{s}$&nbsp; by OSB and USB are&nbsp; $108^{\circ}$&nbsp; and&nbsp; $72^{\circ}$ respectively.
 +
*Since at this time the real parts of OSB and USB compensate,&nbsp; $s_+(t=5  \ {\rm &micro;}  \text{s})$&nbsp; is <u>purely imaginary</u> and we obtain:
 +
 +
:$${\rm Im}\left[s_{+}(t = {\rm 5 \hspace{0.05cm} {\rm &micro;}  s})\right] =
 +
{\rm 1 \hspace{0.05cm} V} + 2 \cdot {\rm 0.4 \hspace{0.05cm}
 +
V}\cdot \cos (18^\circ ) \hspace{0.15 cm}\underline{= {\rm 1.761 \hspace{0.05cm} V}}.$$
 +
 
 +
 
 +
 
 +
'''(4)'''&nbsp;  After one rotation of the red carrier, i.e. at time $t$ = $T_0 = 20 \ {\rm &micro;} \text{s}$, the blue pointer has already covered&nbsp; $72^{\circ}$&nbsp; more and the green pointer correspondingly&nbsp;  $72^{\circ}$&nbsp; less.&nbsp; The sum of the three pointers is again <u>real</u> and results in accordance with the graph on the right:
 +
 +
:$${\rm Re}\left[s_{+}({\rm 20 \hspace{0.05cm} {\rm &micro;}  s})\right] =
 +
{\rm 1 \hspace{0.05cm} V} + 2 \cdot {\rm 0.4 \hspace{0.05cm}
 +
V}\cdot \cos (72^\circ ) \hspace{0.15 cm}\underline{= {\rm 1.236 \hspace{0.05cm} V}}.$$
 +
 
 +
 
 +
'''(5)'''&nbsp; The magnitude is minimum when the pointers of the two sidebands are offset from the carrier by&nbsp; $180^{\circ}$&nbsp;.&nbsp; It follows:
 +
 +
:$$|s_{+}(t)|_{\rm min} = {\rm 1 \hspace{0.05cm} V} - 2 \cdot {\rm
 +
0.4 \hspace{0.05cm} V} \hspace{0.15 cm}\underline{= {\rm 0.2 \hspace{0.05cm} V}}.$$
 +
 
 +
Within one period&nbsp; $T_0$&nbsp; of the carrier, a phase offset of&nbsp; $\pm72^{\circ}$&nbsp; occurs with respect to the pointers of the two sidebands.&nbsp; From this follows:
 +
:$$t_{\text{min}} = 180^{\circ}/72^{\circ} \cdot T_0 = 2.5 \cdot T_0  \;\underline{= 50 \ {\rm &micro;} \text{s}}.$$
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
 
__NOEDITSECTION__
 
__NOEDITSECTION__
[[Category:Aufgaben zu Signaldarstellung|^4. Bandpassartige Signale^]]
+
[[Category:Signal Representation: Exercises|^4.2 Analytical Signal and its Spectral Function^]]

Latest revision as of 15:29, 7 May 2021

Spectrum of the analytical signal

We assume a cosine-shaped source signal  $q(t)$  with

  • amplitude  $A_{\rm N} = 0.8 \ \text{V}$  and
  • frequency  $f_{\rm N}= 10 \ \text{kHz}$.


The frequency conversion is done by means of  "Double-Sideband Amplitude Modulation with Carrier".

The modulated signal  $s(t)$  is with the (normalised) carrier  $z(t) = \text{cos}(\omega_{\rm T} \cdot t)$  and the DC component  $q_0 = 1 \ \text{V}$:

$$\begin{align*} s(t) & = \left(q_0 + q(t)\right) \cdot z(t)= \left({\rm 1 \hspace{0.05cm} V} + {\rm 0.8 \hspace{0.05cm}V}\cdot {\cos} ( \omega_{\rm N}\cdot t)\right) \cdot {\cos} ( \omega_{\rm T}\cdot t) = \\ & = q_0 \cdot {\cos} ( \omega_{\rm T}\cdot t) + {A_{\rm N}}/{2} \cdot {\cos} ( (\omega_{\rm T}+ \omega_{\rm N}) \cdot t) + {A_{\rm N}}/{2} \cdot {\cos} ( (\omega_{\rm T}- \omega_{\rm N}) \cdot t).\end{align*}$$

The first term describes the carrier, the second term the so-called upper sideband  $\rm (OSB)$  and the last term the lower sideband  $\rm (USB)$.

The sketch shows the spectrum  $S_+(f)$  of the corresponding analytical signal for  $f_{\rm T} = 50 \ \text{kHz}$. You can see

  • the carrier (red),
  • the upper sideband (blue),  and
  • the lower sideband (grün).


In subtask  (5)  the magnitude of  $s_+(t)$  is asked for.  This is the length of the resulting pointer.



Hints:

  1. The index  $\rm N$  stands for "source signal"   ⇒   (German:  "Nachrichtenignal").
  2. The index  $\rm T$  stands for  "carrier"   ⇒   (German:  "Trägersignal").
  3. $\rm OSB$  denotes the  "upper sideband"   ⇒   (German:  "oberes Seitenband").
  4. $\rm USB$  denotes the  "lower sideband"   ⇒   (German:  "unteres Seitenband").

Questions

1

What is the analytical signal  $s_+(t)$.  What is its magnitude at time  $t = 0$?

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

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

 $\text{V}$

2

Which of the following statements are true?

$s_+(t)$  results from  $s(t)$, if  $\cos(\text{...})$  is replaced by  ${\rm e}^{{\rm j}(\text{...})}$ .
If  $s(t)$  is an even time function,  $s_+(t)$  is purely real.
At no time does the imaginary part of  $s_+(t)$ disappear.

3

What is the value of the analytical signal at time  $t = 5 \ {\rm µ}\text{s}$?

$\text{Re}[s_+(t=5 \ {\rm µ} \text{s})]\ = \ $

 $\text{V}$
$\text{Im}[s_+(t=5 \ {\rm µ} \text{s})]\ = \ $

 $\text{V}$

4

What is the value of  $s_+(t)$  at time  $t = 20 \ {\rm µ}\text{s}$?

$\text{Re}[s_+(t=20 \ {\rm µ} \text{s})]\ = \ $

 $\text{V}$
$\text{Im}[s_+(t=20 \ {\rm µ} \text{s})]\ = \ $

 $\text{V}$

5

What is the smallest possible pointer length?  At what time   $t_{\text{min}}$  does this value occur for the first time?

$|s_+(t)|_{\text{min}}\ = \ $

 $\text{V}$
$t_{\text{min}}\ = \ $

 ${\rm µ} \text{s}$


Solution

(1)  By inverse Fourier transform of  $S_+(f)$  taking into account the  "Shifting Theorem":

$$s_{+}(t) = {\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm 50}\hspace{0.05cm} t } + {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm 60} \hspace{0.05cm} t }+ {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm 40}\hspace{0.05cm} t }.$$
Three different analytical signals
  • The expression describes the sum of three pointers rotating at different circular velocities.
  • In the above equation, for example,  $\omega_{60} = 2\pi (f_{\rm T} + f_{\rm N}) = 2\pi \cdot 60 \ \text{kHz}$.
  • At time  $t = 0$  all three pointers point in the direction of the real axis (see left graph).
  • One obtains the real value  $s_+(t = 0) \;\underline{= 1.8 \ \text{V}}$.


(2)  The first statement is correct and results from the  "Hilbert transform".  On the other hand, the next two statements are'nt correct:

  • $s_+(t)$  is always a complex time function with exception of the limiting case  $s(t) \equiv 0$.
  • However, every complex function also has purely real values at some points in time.
  • The  "pointer group"  always rotates in a mathematically positive direction.
  • If the sum vector crosses the real axis, the imaginary part disappears at this point and  $s_+(t)$  is purely real.


(3)  The period duration of the carrier signal is  $T_0 = 1/f_T = 20 \ {\rm µ} \text{s}$.

  • After  $t = 5 \ {\rm µ} \text{s}$  (see middle graph) the carrier has thus rotated by  $90^{\circ}$.
  • The blue pointer  $\rm (OSB)$  rotates  $20\%$  faster, the green one  $\rm (USB)$  $20\%$  slower than the red rotary pointer (carrier signal):
$$s_{+}({\rm 5 \hspace{0.05cm} {\rm µ} s}) = {\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 2 \pi \hspace{0.03cm} \cdot \hspace{0.08cm}50 \hspace{0.03cm} \cdot \hspace{0.08cm}0.005 } + {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 2 \pi \hspace{0.03cm} \cdot \hspace{0.08cm}60 \hspace{0.03cm} \cdot \hspace{0.08cm}0.005 }+ {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 2 \pi \hspace{0.03cm} \cdot \hspace{0.08cm}40 \hspace{0.03cm} \cdot \hspace{0.08cm}0.005 } = {\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 90^\circ }+ {\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 108^\circ }+{\rm 0.4 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} 72^\circ }.$$
  • Thus, the angles travelled in  $ 5 \ {\rm µ} \text{s}$  by OSB and USB are  $108^{\circ}$  and  $72^{\circ}$ respectively.
  • Since at this time the real parts of OSB and USB compensate,  $s_+(t=5 \ {\rm µ} \text{s})$  is purely imaginary and we obtain:
$${\rm Im}\left[s_{+}(t = {\rm 5 \hspace{0.05cm} {\rm µ} s})\right] = {\rm 1 \hspace{0.05cm} V} + 2 \cdot {\rm 0.4 \hspace{0.05cm} V}\cdot \cos (18^\circ ) \hspace{0.15 cm}\underline{= {\rm 1.761 \hspace{0.05cm} V}}.$$


(4)  After one rotation of the red carrier, i.e. at time $t$ = $T_0 = 20 \ {\rm µ} \text{s}$, the blue pointer has already covered  $72^{\circ}$  more and the green pointer correspondingly  $72^{\circ}$  less.  The sum of the three pointers is again real and results in accordance with the graph on the right:

$${\rm Re}\left[s_{+}({\rm 20 \hspace{0.05cm} {\rm µ} s})\right] = {\rm 1 \hspace{0.05cm} V} + 2 \cdot {\rm 0.4 \hspace{0.05cm} V}\cdot \cos (72^\circ ) \hspace{0.15 cm}\underline{= {\rm 1.236 \hspace{0.05cm} V}}.$$


(5)  The magnitude is minimum when the pointers of the two sidebands are offset from the carrier by  $180^{\circ}$ .  It follows:

$$|s_{+}(t)|_{\rm min} = {\rm 1 \hspace{0.05cm} V} - 2 \cdot {\rm 0.4 \hspace{0.05cm} V} \hspace{0.15 cm}\underline{= {\rm 0.2 \hspace{0.05cm} V}}.$$

Within one period  $T_0$  of the carrier, a phase offset of  $\pm72^{\circ}$  occurs with respect to the pointers of the two sidebands.  From this follows:

$$t_{\text{min}} = 180^{\circ}/72^{\circ} \cdot T_0 = 2.5 \cdot T_0 \;\underline{= 50 \ {\rm µ} \text{s}}.$$