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 Modulationsverfahren
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{{quiz-Header|Buchseite=Modulation_Methods/Linear_Digital_Modulation
 
}}
 
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[[File:|right|]]
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[[File:P_ID1702__Mod_Z_4_6.png|right|frame|Specified transmitted signals]]
 +
The figure shows,&nbsp; starting from the same source signal &nbsp;$q(t)$,&nbsp; the transmitted signals with
  
 +
* [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying| Amplitude Shift Keying]]&nbsp; $\rm (ASK)$,
 +
* [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying| Binary Phase Shift Keying ]]&nbsp; $\rm (BPSK)$,
 +
* [[Modulation_Methods/Linear_Digital_Modulation#DPSK_.E2.80.93_Differential_Phase_Shift_Keying| Differential Phase Shift Keying ]]&nbsp; $\rm (DPSK)$.
  
===Fragebogen===
+
 
 +
The transmitted signals are generally designated here as &nbsp;$s_1(t)$, &nbsp;$s_2(t)$&nbsp; and &nbsp;$s_3(t)$.&nbsp; The assignment to the given modulation methods is to be done by you.
 +
 
 +
In addition,&nbsp; the respective average energy per bit  &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; in "Ws" is to be specified for all signals,&nbsp; whereby the following assumptions can be made:
 +
* The&nbsp; (maximum)&nbsp; envelope of all carrier frequency modulated signals is &nbsp;$s_0 = 2\ \rm V$.
 +
* The bit rate of the redundancy-free source signal is &nbsp;$R_{\rm B} = 1 \ \rm Mbit/s$.
 +
* The modulators operate with a working resistance of &nbsp;$R = 50 \ \rm  Ω$.
 +
 
 +
 
 +
For example,&nbsp; for (bipolar) baseband transmission with symbol duration &nbsp;$T_{\rm } = 1/R_{\rm }$&nbsp; would be:
 +
:$$ E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{50 \,{\rm V/A}}= 8 \cdot 10^{-8} \,{\rm Ws}= 0.08 \,\,{\rm &micro; Ws}.$$
 +
 
 +
 
 +
 
 +
 
 +
 
 +
Notes:
 +
*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Linear_Digital_Modulation|Linear Digital Modulation]].
 +
*However, reference is also made to the chapter&nbsp;  [[Digital_Signal_Transmission/Grundlagen_der_codierten_Übertragung|Basics of Coded Transmission]]&nbsp;  in the book "Digital Signal Transmission".
 +
*The powers are given in  &nbsp;$\rm V^2$;&nbsp; they thus refer to the reference resistor &nbsp;$R = 1 \ \rm \Omega$.
 +
 +
 
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Multiple-Choice Frage
+
{Which signal describes ASK?
|type="[]"}
+
|type="()"}
- Falsch
+
- $s_1(t)$,
+ Richtig
+
- $s_2(t)$,
 +
+ $s_3(t)$.
  
 +
{What is the average energy per bit  &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; for ASK?
 +
|type="{}"}
 +
$E_{\rm B} \ = \ $ { 0.02 3% } $\ \rm &micro; Ws$
  
{Input-Box Frage
+
{Which signal describes BPSK?
 +
|type="()"}
 +
+ $s_1(t),$
 +
- $s_2(t),$
 +
- $s_3(t).$
 +
 
 +
{What is the average energy per bit  &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; for BPSK?
 
|type="{}"}
 
|type="{}"}
$\alpha$ = { 0.3 }
+
$E_{\rm B} \ = \ $ { 0.04 3% } $\ \rm &micro; Ws$
 +
 
 +
{Which signal describes DPSK?
 +
|type="()"}
 +
- $s_1(t),$
 +
+ $s_2(t),$
 +
- $s_3(t)$.
 +
 
 +
{What is the average energy per bit  &nbsp; &rArr; &nbsp; $E_{\rm B}$&nbsp; for DPSK?
 +
|type="{}"}
 +
$E_{\rm B} \ = \ $ { 0.04 3% } $\ \rm &micro; Ws$
 +
 
  
  
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</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.'''
+
'''(1)'''&nbsp;  <u>Solution 3</u>&nbsp; is correct:
'''2.'''
+
*The ASK signal results from the multiplication of the here sinusoidal carrier signal&nbsp; $z(t)$&nbsp; with the unipolar source signal&nbsp; $q(t)$.
'''3.'''
+
*It is obvious that&nbsp; $s_3(t)$&nbsp; describes such an ASK signal.
'''4.'''
+
*The unipolar amplitude coefficients of the source signal are&nbsp; $1,\ 1,\ 0,\ 1,\ 0,\ 1,\ 1$.
'''5.'''
+
 
'''6.'''
+
 
'''7.'''
+
 
 +
'''(2)'''&nbsp;  Compared to the bipolar baseband transmission,&nbsp; the following changes can be seen in ASK:
 +
* The energy is halved because of multiplication by the sinusoidal signal.
 +
* 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.
 +
 
 +
 
 +
This gives:
 +
:$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{4 \cdot R} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{4 \cdot 50 \,{\rm V/A}}= 2 \cdot 10^{-8} \,{\rm Ws}\hspace{0.15cm}\underline {= 0.02 \,\,{\rm &micro; Ws}}.$$
 +
 
 +
 
 +
 
 +
'''(3)'''&nbsp;  <u>Solution 1</u>&nbsp; is correct:
 +
*Phase jumps are typical for BPSK.
 +
*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;  Of the changes mentioned under&nbsp; '''(2)'''&nbsp; compared to baseband transmission,&nbsp; only the first one is applicable for BPSK.&nbsp; Thus:
 +
:$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm &micro; Ws}}.$$
 +
 
 +
 
 +
 
 +
'''(5)'''&nbsp;  As can already be assumed,&nbsp; the correct answer is&nbsp; $s_2(t)$ &nbsp; ⇒ &nbsp; <u>solution 2</u>:
 +
 
 +
*The DPSK modulator operates as follows,&nbsp; assuming&nbsp; $m_0 = -1$:&nbsp;
 +
:$$ m_0 = -1, \hspace{0.1cm}a_1 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_1 = -1,$$
 +
:$$m_1 = -1, \hspace{0.1cm}a_2 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_2 = -1,$$
 +
:$$m_2 = -1, \hspace{0.1cm}a_3 = -1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_3 = +1,$$
 +
:$$m_3 = +1, \hspace{0.1cm}a_4 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_4 = +1,$$
 +
:$$m_4 = +1, \hspace{0.1cm}a_5 = -1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_5 = -1,$$
 +
:$$m_5 = -1, \hspace{0.1cm}a_6 = +1 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}m_6 = -1, \,\,{\rm etc.}$$
 +
 
 +
 
 +
 
 +
'''(6)'''&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.
 +
*It follows: &nbsp; DPSK has exactly the same signal energy as BPSK:
 +
:$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm &micro; Ws}}.$$
 +
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Modulationsverfahren|^4.2 Lineare digitale Modulationsverfahren^]]
+
[[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:
$$E_{\rm B} = \frac {s_0^2 \cdot T_{\rm B} }{2 \cdot R} \hspace{0.15cm}\underline {= 0.04 \,\,{\rm µ Ws}}.$$