Difference between revisions of "Aufgaben:Exercise 4.7Z: Signal Shapes for ASK, BPSK and DPSK"

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{{quiz-Header|Buchseite=Modulationsverfahren/Lineare digitale Modulation
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{{quiz-Header|Buchseite=Modulation_Methods/Linear_Digital_Modulation
 
}}
 
}}
  
[[File:P_ID1702__Mod_Z_4_6.png|right|frame|Vorgegebene Sendesignale]]
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[[File:P_ID1702__Mod_Z_4_6.png|right|frame|Specified transmitted signals]]
Die Abbildung zeigt jeweils ausgehend vom gleichen Quellensignal  $q(t)$  die Sendesignale bei
+
The figure shows,  starting from the same source signal  $q(t)$,  the transmitted signals with
  
* [[Modulation_Methods/Lineare_digitale_Modulationsverfahren#ASK_.E2.80.93_Amplitude_Shift_Keying| Amplitude Shift Keying]]  (ASK),  
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* [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying| Amplitude Shift Keying]]  $\rm (ASK)$,  
* [[Modulation_Methods/Lineare_digitale_Modulationsverfahren#BPSK_.E2.80.93_Binary_Phase_Shift_Keying| Binary Phase Shift Keying ]]  (BPSK),
+
* [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying| Binary Phase Shift Keying ]]  $\rm (BPSK)$,
* [[Modulation_Methods/Lineare_digitale_Modulationsverfahren#DPSK_.E2.80.93_Differential_Phase_Shift_Keying| Differential Phase Shift Keying ]]  (DPSK).
+
* [[Modulation_Methods/Linear_Digital_Modulation#DPSK_.E2.80.93_Differential_Phase_Shift_Keying| Differential Phase Shift Keying ]]  $\rm (DPSK)$.
  
  
Die Sendesignale sind hier allgemein mit  $s_1(t)$,  $s_2(t)$  und  $s_3(t)$  bezeichnet.  Die Zuordnung zu den vorgegebenen Modulationsverfahren soll von Ihnen vorgenommen werden.
+
The transmitted signals are generally designated here as  $s_1(t)$,  $s_2(t)$  and  $s_3(t)$.  The assignment to the given modulation methods is to be done by you.
  
Außerdem soll für alle Signale die jeweilige mittlere Energie pro Bit   ⇒   $E_{\rm B}$  in „Ws” angegeben werden, wobei folgende Annahmen getroffen werden können:  
+
In addition,  the respective average energy per bit   ⇒   $E_{\rm B}$  in "Ws" is to be specified for all signals,  whereby the following assumptions can be made:
* Die (maximale) Hüllkurve aller trägerfrequenzmodulierten Signale ist  $s_0 = 2\ \rm V$.
+
* The  (maximum)  envelope of all carrier frequency modulated signals is  $s_0 = 2\ \rm V$.
* Die Bitrate des redundanzfreien Quellensignals beträgt  $R_{\rm B} = 1 \ \rm Mbit/s$.
+
* The bit rate of the redundancy-free source signal is  $R_{\rm B} = 1 \ \rm Mbit/s$.
* Die Modulatoren arbeiten mit einem Arbeitswiderstand von  $R = 50 \ \rm  Ω$.
+
* The modulators operate with a working resistance of  $R = 50 \ \rm  Ω$.
  
  
Beispielsweise würde bei (bipolarer) Basisbandübertragung mit der Symboldauer  $T_{\rm } = 1/R_{\rm }$  gelten:
+
For example,  for (bipolar) baseband transmission with symbol duration  $T_{\rm } = 1/R_{\rm }$  would be:
 
:$$ E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{50 \,{\rm V/A}}= 8 \cdot 10^{-8} \,{\rm Ws}= 0.08 \,\,{\rm µ Ws}.$$
 
:$$ E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{50 \,{\rm V/A}}= 8 \cdot 10^{-8} \,{\rm Ws}= 0.08 \,\,{\rm µ Ws}.$$
  
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+
Notes:  
 
+
*The exercise belongs to the chapter  [[Modulation_Methods/Linear_Digital_Modulation|Linear Digital Modulation]].
 
+
*However, reference is also made to the chapter   [[Digital_Signal_Transmission/Grundlagen_der_codierten_Übertragung|Basics of Coded Transmission]]   in the book "Digital Signal Transmission".
 
+
*The powers are given in   $\rm V^2$;  they thus refer to the reference resistor  $R = 1 \ \rm \Omega$.
''Hinweise:''
 
*Die Aufgabe gehört zum  Kapitel  [[Modulation_Methods/Lineare_digitale_Modulation|Lineare digitale Modulation]].
 
*Bezug genommen wird aber auch auf das Kapitel   [[Digitalsignalübertragung/Grundlagen_der_codierten_Übertragung|Grundlagen der codierten Übertragung]]   im Buch „Digitalsignalübertragung”.
 
*Die Leistungen sind in   $\rm V^2$  anzugeben;  sie beziehen sich somit auf den Bezugswiderstand  $R = 1 \ \rm \Omega$.
 
 
   
 
   
  
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welcher Signalverlauf beschreibt die ASK?
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{Which signal describes ASK?
 
|type="()"}
 
|type="()"}
 
- $s_1(t)$,
 
- $s_1(t)$,
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+ $s_3(t)$.
 
+ $s_3(t)$.
  
{Welche mittlere Energie pro Bit &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; ergibt sich bei der ASK?
+
{What is the average energy per bit &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; for ASK?
 
|type="{}"}
 
|type="{}"}
 
  $E_{\rm B} \ = \ $ { 0.02 3% } $\ \rm &micro; Ws$
 
  $E_{\rm B} \ = \ $ { 0.02 3% } $\ \rm &micro; Ws$
  
{Welcher Signalverlauf beschreibt die BPSK?
+
{Which signal describes BPSK?
 
|type="()"}
 
|type="()"}
 
+ $s_1(t),$
 
+ $s_1(t),$
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- $s_3(t).$
 
- $s_3(t).$
  
{Welche mittlere Energie pro Bit &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; ergibt sich bei der BPSK?
+
{What is the average energy per bit &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; for BPSK?
 
|type="{}"}
 
|type="{}"}
 
$E_{\rm B} \ = \ $ { 0.04 3% } $\ \rm &micro; Ws$
 
$E_{\rm B} \ = \ $ { 0.04 3% } $\ \rm &micro; Ws$
  
{Welcher Signalverlauf beschreibt die DPSK?
+
{Which signal describes DPSK?
 
|type="()"}
 
|type="()"}
 
- $s_1(t),$
 
- $s_1(t),$
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- $s_3(t)$.
 
- $s_3(t)$.
  
{Welche mittlere Energie pro Bit &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; ergibt sich bei der DPSK?
+
{What is the average energy per bit &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; for DPSK?
 
|type="{}"}
 
|type="{}"}
 
$E_{\rm B} \ = \ $ { 0.04 3% } $\ \rm &micro; Ws$
 
$E_{\rm B} \ = \ $ { 0.04 3% } $\ \rm &micro; Ws$
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</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 3</u>:
+
'''(1)'''&nbsp;  <u>Solution 3</u>&nbsp; is correct:
*Das ASK–Signal ergibt sich aus der Multiplikation des hier sinusförmigen Trägersignals&nbsp; $z(t)$&nbsp; mit dem unipolaren Quellensignal&nbsp; $q(t)$.  
+
*The ASK signal results from the multiplication of the here sinusoidal carrier signal&nbsp; $z(t)$&nbsp; with the unipolar source signal&nbsp; $q(t)$.  
*Es ist offensichtlich, dass&nbsp; $s_3(t)$&nbsp; ein solches ASK–Signal beschreibt.
+
*It is obvious that&nbsp; $s_3(t)$&nbsp; describes such an ASK signal.
*Die unipolaren Amplitudenkoeffizienten des Quellensignals lauten&nbsp; $1,\ 1,\ 0,\ 1,\ 0,\ 1,\ 1$.
+
*The unipolar amplitude coefficients of the source signal are&nbsp; $1,\ 1,\ 0,\ 1,\ 0,\ 1,\ 1$.
  
  
  
'''(2)'''&nbsp;  Gegenüber der bipolaren Basisbandübertragung sind bei der ASK folgende Änderungen zu erkennen:
+
'''(2)'''&nbsp;  Compared to the bipolar baseband transmission,&nbsp; the following changes can be seen in ASK:
* Die Energie wird wegen der Multiplikation mit dem Sinussignal halbiert.
+
* The energy is halved because of multiplication by the sinusoidal signal.
* Da&nbsp; $q(t)$&nbsp; als redundanzfrei vorausgesetzt wird, gilt in der Hälfte der Zeit&nbsp; $s_3(t) = 0$, wodurch die Energie nochmals halbiert wird.
+
* Since&nbsp; $q(t)$&nbsp; is assumed to be redundancy-free, in half the time&nbsp; $s_3(t) = 0$,&nbsp; which halves the energy again.
  
  
Damit ergibt sich:
+
This gives:
 
:$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{4 \cdot R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{4 \cdot 50 \,{\rm V/A}}= 2 \cdot 10^{-8} \,{\rm Ws}\hspace{0.15cm}\underline {= 0.02 \,\,{\rm &micro; Ws}}.$$
 
:$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{4 \cdot R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{4 \cdot 50 \,{\rm V/A}}= 2 \cdot 10^{-8} \,{\rm Ws}\hspace{0.15cm}\underline {= 0.02 \,\,{\rm &micro; Ws}}.$$
  
  
  
'''(3)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 1</u>:
+
'''(3)'''&nbsp;  <u>Solution 1</u>&nbsp; is correct:
*Typisch für die BPSK sind Phasensprünge.  
+
*Phase jumps are typical for BPSK.
*Da stets das gleiche Quellensignal vorausgesetzt wurde, treten diese Phasensprünge genau dann auf, wenn es im ASK–Signal&nbsp; $s_3(t)$&nbsp; einen Symbolwechsel gibt.
+
*Since the same source signal was always assumed,&nbsp; these phase jumps occur exactly when there is a symbol change in the ASK signal&nbsp; $s_3(t)$.&nbsp;  
  
  
  
'''(4)'''&nbsp;  Von der unter&nbsp; '''(2)'''&nbsp; genannten Veränderung gegenüber der Basisbandübertragung ist bei BPSK nur die erste zutreffend.&nbsp; Damit gilt:
+
'''(4)'''&nbsp;  Of the changes mentioned under&nbsp; '''(2)'''&nbsp; compared to baseband transmission,&nbsp; only the first one is applicable for BPSK.&nbsp; Thus:
 
:$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm &micro; Ws}}.$$
 
:$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm &micro; Ws}}.$$
  
  
  
'''(5)'''&nbsp;  Wie bereits zu vermuten ist, lautet die richtige Antwort&nbsp; $s_2(t)$ &nbsp; ⇒ &nbsp; <u>Lösungsvorschlag 2</u>:  
+
'''(5)'''&nbsp;  As can already be assumed,&nbsp; the correct answer is&nbsp; $s_2(t)$ &nbsp; ⇒ &nbsp; <u>solution 2</u>:  
  
*Der DPSK–Modulator arbeitet wie folgt, wobei&nbsp; $m_0 = -1$&nbsp; vorausgesetzt wird:
+
*The DPSK modulator operates as follows,&nbsp; assuming&nbsp; $m_0 = -1$:&nbsp;  
 
:$$ m_0 = -1, \hspace{0.1cm}a_1 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_1 = -1,$$
 
:$$ m_0 = -1, \hspace{0.1cm}a_1 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_1 = -1,$$
 
:$$m_1 = -1, \hspace{0.1cm}a_2 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_2 = -1,$$  
 
:$$m_1 = -1, \hspace{0.1cm}a_2 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_2 = -1,$$  
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:$$m_3 = +1, \hspace{0.1cm}a_4 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_4 = +1,$$
 
:$$m_3 = +1, \hspace{0.1cm}a_4 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_4 = +1,$$
 
:$$m_4 = +1, \hspace{0.1cm}a_5 = -1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_5 = -1,$$  
 
:$$m_4 = +1, \hspace{0.1cm}a_5 = -1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_5 = -1,$$  
:$$m_5 = -1, \hspace{0.1cm}a_6 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_6 = -1, \,\,{\rm usw.}$$
+
:$$m_5 = -1, \hspace{0.1cm}a_6 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_6 = -1, \,\,{\rm etc.}$$
  
  
  
'''(6)'''&nbsp;  Ein Vergleich der beiden Signale&nbsp; $s_1(t)$&nbsp; und&nbsp; $s_2(t)$&nbsp; zeigt, dass sich hinsichtlich der Signalenergie nichts ändert.  
+
'''(6)'''&nbsp;  A comparison of the two signals&nbsp; $s_1(t)$&nbsp; and&nbsp; $s_2(t)$&nbsp; shows that nothing changes with respect to the signal energy.
*Daraus folgt: &nbsp; Die DPSK weist die genau gleiche Signalenergie auf wie die BPSK:
+
*It follows: &nbsp; DPSK has exactly the same signal energy as BPSK:
 
:$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm &micro; Ws}}.$$
 
:$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm &micro; Ws}}.$$
  
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[[Category:Aufgaben zu Modulationsverfahren|^4.2 Lineare digitale Modulation^]]
+
[[Category:Modulation Methods: Exercises|^4.2 Linear Digital Modulation^]]

Latest revision as of 14:06, 15 April 2022

Specified transmitted signals

The figure shows,  starting from the same source signal  $q(t)$,  the transmitted signals with


The transmitted signals are generally designated here as  $s_1(t)$,  $s_2(t)$  and  $s_3(t)$.  The assignment to the given modulation methods is to be done by you.

In addition,  the respective average energy per bit   ⇒   $E_{\rm B}$  in "Ws" is to be specified for all signals,  whereby the following assumptions can be made:

  • The  (maximum)  envelope of all carrier frequency modulated signals is  $s_0 = 2\ \rm V$.
  • The bit rate of the redundancy-free source signal is  $R_{\rm B} = 1 \ \rm Mbit/s$.
  • The modulators operate with a working resistance of  $R = 50 \ \rm Ω$.


For example,  for (bipolar) baseband transmission with symbol duration  $T_{\rm } = 1/R_{\rm }$  would be:

$$ E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{50 \,{\rm V/A}}= 8 \cdot 10^{-8} \,{\rm Ws}= 0.08 \,\,{\rm µ Ws}.$$



Notes:

  • The exercise belongs to the chapter  Linear Digital Modulation.
  • However, reference is also made to the chapter  Basics of Coded Transmission  in the book "Digital Signal Transmission".
  • The powers are given in  $\rm V^2$;  they thus refer to the reference resistor  $R = 1 \ \rm \Omega$.



Questions

1

Which signal describes ASK?

$s_1(t)$,
$s_2(t)$,
$s_3(t)$.

2

What is the average energy per bit   ⇒   $E_{\rm B}$  for ASK?

$E_{\rm B} \ = \ $

$\ \rm µ Ws$

3

Which signal describes BPSK?

$s_1(t),$
$s_2(t),$
$s_3(t).$

4

What is the average energy per bit   ⇒   $E_{\rm B}$  for BPSK?

$E_{\rm B} \ = \ $

$\ \rm µ Ws$

5

Which signal describes DPSK?

$s_1(t),$
$s_2(t),$
$s_3(t)$.

6

What is the average energy per bit   ⇒   $E_{\rm B}$  for DPSK?

$E_{\rm B} \ = \ $

$\ \rm µ Ws$


Solution

(1)  Solution 3  is correct:

  • The ASK signal results from the multiplication of the here sinusoidal carrier signal  $z(t)$  with the unipolar source signal  $q(t)$.
  • It is obvious that  $s_3(t)$  describes such an ASK signal.
  • The unipolar amplitude coefficients of the source signal are  $1,\ 1,\ 0,\ 1,\ 0,\ 1,\ 1$.


(2)  Compared to the bipolar baseband transmission,  the following changes can be seen in ASK:

  • The energy is halved because of multiplication by the sinusoidal signal.
  • Since  $q(t)$  is assumed to be redundancy-free, in half the time  $s_3(t) = 0$,  which halves the energy again.


This gives:

$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{4 \cdot R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{4 \cdot 50 \,{\rm V/A}}= 2 \cdot 10^{-8} \,{\rm Ws}\hspace{0.15cm}\underline {= 0.02 \,\,{\rm µ Ws}}.$$


(3)  Solution 1  is correct:

  • Phase jumps are typical for BPSK.
  • Since the same source signal was always assumed,  these phase jumps occur exactly when there is a symbol change in the ASK signal  $s_3(t)$. 


(4)  Of the changes mentioned under  (2)  compared to baseband transmission,  only the first one is applicable for BPSK.  Thus:

$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm µ Ws}}.$$


(5)  As can already be assumed,  the correct answer is  $s_2(t)$   ⇒   solution 2:

  • The DPSK modulator operates as follows,  assuming  $m_0 = -1$: 
$$ m_0 = -1, \hspace{0.1cm}a_1 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_1 = -1,$$
$$m_1 = -1, \hspace{0.1cm}a_2 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_2 = -1,$$
$$m_2 = -1, \hspace{0.1cm}a_3 = -1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_3 = +1,$$
$$m_3 = +1, \hspace{0.1cm}a_4 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_4 = +1,$$
$$m_4 = +1, \hspace{0.1cm}a_5 = -1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_5 = -1,$$
$$m_5 = -1, \hspace{0.1cm}a_6 = +1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}m_6 = -1, \,\,{\rm etc.}$$


(6)  A comparison of the two signals  $s_1(t)$  and  $s_2(t)$  shows that nothing changes with respect to the signal energy.

  • It follows:   DPSK has exactly the same signal energy as BPSK:
$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm µ Ws}}.$$