Difference between revisions of "Aufgaben:Exercise 2.1Z: About the Equivalent Bitrate"

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{{quiz-Header|Buchseite=Digitalsignalübertragung/Grundlagen der codierten Übertragung
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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Basics_of_Coded_Transmission
 
}}
 
}}
  
  
  
[[File:P_ID1309__Dig_Z_2_1.png|right|frame|Quellen- und Codersignal]]
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[[File:P_ID1309__Dig_Z_2_1.png|right|frame|Source signal (top) and encoder signal (bottom)]]
Die obere Darstellung zeigt das Quellensignal $q(t)$ einer redundanzfreien Binärquelle mit Bitdauer $T_{q}$ und Bitrate $R_{q}$. Die beiden Signalparameter $T_{q}$ und $R_{q}$ können der Skizze entnommen werden.
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The upper diagram shows the source signal  $q(t)$  of a redundancy-free binary source with bit duration  $T_{q}$  and  bit rate $R_{q}$. The two signal parameters  $T_{q}$  and  $R_{q}$  can be taken from the sketch.
  
Dieses Binärsignal wird symbolweise codiert und ergibt das unten gezeichnete Codersignal $c(t)$. Alle möglichen Codesymbole kommen in dem dargestellten Signalausschnitt der Dauer $6 / \rm \mu s$ vor.
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*This binary signal is coded symbol-by-symbol and results in the encoder signal  $c(t)$ drawn below.  
Mit der Stufenzahl $M_{c}$ und der Symboldauer $T_{c}$ kann man die äquivalente Bitrate – oder den Informationsfluss – des Codersignals angeben:
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*All possible encoder symbols occur in the signal section of duration  $6 \ \rm µ s$  shown.
 +
*The level number  $M_{c}$  and the symbol duration  $T_{c}$  can be used to specify the equivalent bit rate of the encoder signal:
 
:$$R_c = \frac{{\rm log_2} (M_c)}{T_c} \hspace{0.05cm}.$$
 
:$$R_c = \frac{{\rm log_2} (M_c)}{T_c} \hspace{0.05cm}.$$
Daraus erhält man die relative Redundanz des Codes, wenn man wie hier davon ausgeht, dass die Quelle selbst redundanzfrei ist:
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From this,  one obtains the relative redundancy of the code if one assumes,  as here,  that the source itself is redundancy-free:
 
:$$r_c = \frac{R_c - R_q}{R_c}\hspace{0.05cm}.$$
 
:$$r_c = \frac{R_c - R_q}{R_c}\hspace{0.05cm}.$$
  
  
''Hinweis:''
 
  
 +
Notes:
 +
*The exercise belongs to the chapter   [[Digital_Signal_Transmission/Basics_of_Coded_Transmission|"Basics of Coded Transmission"]].
 +
 +
*The transmission code considered here is the second order bipolar code,  but this is not important for the solution of this exercise.
  
Die Aufgabe bezieht sich auf  [[Digitalsignalübertragung/Grundlagen_der_codierten_Übertragung|Grundlagen der codierten Übertragung]] dieses Buches. Bei dem hier betrachteten Übertragungscode handelt es sich um den Bipolarcode zweiter Ordnung, was jedoch für die Lösung dieser Aufgabe nicht von Bedeutung ist.
 
  
===Fragebogen===
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 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Multiple-Choice Frage
 
|type="[]"}
 
- Falsch
 
+ Richtig
 
  
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{Specify bit duration &nbsp;$(T_{q})$&nbsp; and bit rate &nbsp;$(R_{q})$&nbsp; of the source.
 +
|type="{}"}
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$T_{q}  \ = \ $ { 0.5 3% } $\ \rm &micro; s $
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$R_{q}  \ = \ $ { 2 3% } $\ \rm Mbit/s $
 +
 +
{What are the symbol duration &nbsp;$(T_{c})$&nbsp; and level number &nbsp;$(M_{c})$&nbsp; of the encoder signal?
 +
|type="{}"}
 +
$T_{c} \ = \ $ { 0.5 3% } $\ \rm &micro; s $
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$M_{c} \ = \ $ { 3 3% }
  
{Input-Box Frage
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{What is the equivalent bit rate &nbsp;$R_{c}$&nbsp; of the encoder signal?
 
|type="{}"}
 
|type="{}"}
$\alpha$ = { 0.3 }
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$R_{c}  \ = \ $ { 3.17 3% } $\ \rm Mbit/s $
  
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{Specify the relative redundancy of the code.
 +
|type="{}"}
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$r_{c} \ = \ $ { 36.9 3% } $\ \% $
  
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp;
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'''(1)'''&nbsp; The bit duration&nbsp; $T_{q} = \underline{0.5\ \rm &micro; s}$&nbsp; can be taken from the graphic.
'''(2)'''&nbsp;
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*Since the source is binary and redundancy-free,&nbsp; the following applies to the bit rate of the source:
'''(3)'''&nbsp;
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:$$R_{q}= 1/T_{q}\ \underline{= 2\ \rm Mbit/s}.$$
'''(4)'''&nbsp;
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'''(5)'''&nbsp;
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'''(6)'''&nbsp;
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'''(2)'''&nbsp; For symbol-wise coding,&nbsp; $T_{c} = T_{q}$&nbsp; always applies.
 +
*Thus,&nbsp; in the present example,&nbsp; $T_{c}\ \underline{ = 0.5\ \rm &micro; s}$ is valid.
 +
*The level number&nbsp; $M_{c}\ \underline{ = 3}$&nbsp; can be read from the sketch below.
 +
 
 +
 
 +
'''(3)'''&nbsp; The symbol rate of the encoder signal is&nbsp; $2 \cdot 10^{6}$&nbsp; ternary symbols per second.
 +
*For the equivalent bit rate,&nbsp; the following applies:
 +
:$$R_c = \frac{{\rm log_2} (M_c)}{T_c} = \frac{{\rm log_2}(3)}{0.5\,\,{\rm \mu s}} = \frac{{\rm lg} (3)}{{\rm lg} (2) \cdot 0.5\,\,{\rm \mu s}}= \frac{1.585\,\,{\rm (bit)}}{0.5\,\,{\rm \mu s}}\hspace{0.15cm} \underline {\approx 3.17\,\,{\rm Mbit/s}} \hspace{0.05cm}.$$
 +
 
 +
 
 +
'''(4)'''&nbsp; For relative code redundancy,&nbsp; when the source is redundancy-free,&nbsp; the general rule is:
 +
:$$ r_c = \frac{R_c - R_q}{R_c} = 1- \frac{R_q}{R_c}= 1- \frac{T_c}{T_q \cdot {\rm log_2} (M_c)}\hspace{0.05cm}.$$
 +
*In the case of the second order biploar code considered here,&nbsp; with parameters&nbsp; $T_{c} = T_{q}$&nbsp; and&nbsp; $M_{c} = 3$,&nbsp; the following holds:
 +
:$$r_c = 1- \frac{1}{{\rm log_2} (3)}\hspace{0.15cm}\underline {\approx 36.9 \% }\hspace{0.05cm}.$$
 +
 
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
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[[Category:Aufgaben zu Digitalsignalübertragung|^2.1 Codierte Übertragung - Grundlagen^]]
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[[Category:Digital Signal Transmission: Exercises|^2.1 Basics of Coded Transmission^]]

Latest revision as of 15:54, 3 June 2022


Source signal (top) and encoder signal (bottom)

The upper diagram shows the source signal  $q(t)$  of a redundancy-free binary source with bit duration  $T_{q}$  and  bit rate $R_{q}$. The two signal parameters  $T_{q}$  and  $R_{q}$  can be taken from the sketch.

  • This binary signal is coded symbol-by-symbol and results in the encoder signal  $c(t)$ drawn below.
  • All possible encoder symbols occur in the signal section of duration  $6 \ \rm µ s$  shown.
  • The level number  $M_{c}$  and the symbol duration  $T_{c}$  can be used to specify the equivalent bit rate of the encoder signal:
$$R_c = \frac{{\rm log_2} (M_c)}{T_c} \hspace{0.05cm}.$$

From this,  one obtains the relative redundancy of the code if one assumes,  as here,  that the source itself is redundancy-free:

$$r_c = \frac{R_c - R_q}{R_c}\hspace{0.05cm}.$$


Notes:

  • The transmission code considered here is the second order bipolar code,  but this is not important for the solution of this exercise.


Questions

1

Specify bit duration  $(T_{q})$  and bit rate  $(R_{q})$  of the source.

$T_{q} \ = \ $

$\ \rm µ s $
$R_{q} \ = \ $

$\ \rm Mbit/s $

2

What are the symbol duration  $(T_{c})$  and level number  $(M_{c})$  of the encoder signal?

$T_{c} \ = \ $

$\ \rm µ s $
$M_{c} \ = \ $

3

What is the equivalent bit rate  $R_{c}$  of the encoder signal?

$R_{c} \ = \ $

$\ \rm Mbit/s $

4

Specify the relative redundancy of the code.

$r_{c} \ = \ $

$\ \% $


Solution

(1)  The bit duration  $T_{q} = \underline{0.5\ \rm µ s}$  can be taken from the graphic.

  • Since the source is binary and redundancy-free,  the following applies to the bit rate of the source:
$$R_{q}= 1/T_{q}\ \underline{= 2\ \rm Mbit/s}.$$


(2)  For symbol-wise coding,  $T_{c} = T_{q}$  always applies.

  • Thus,  in the present example,  $T_{c}\ \underline{ = 0.5\ \rm µ s}$ is valid.
  • The level number  $M_{c}\ \underline{ = 3}$  can be read from the sketch below.


(3)  The symbol rate of the encoder signal is  $2 \cdot 10^{6}$  ternary symbols per second.

  • For the equivalent bit rate,  the following applies:
$$R_c = \frac{{\rm log_2} (M_c)}{T_c} = \frac{{\rm log_2}(3)}{0.5\,\,{\rm \mu s}} = \frac{{\rm lg} (3)}{{\rm lg} (2) \cdot 0.5\,\,{\rm \mu s}}= \frac{1.585\,\,{\rm (bit)}}{0.5\,\,{\rm \mu s}}\hspace{0.15cm} \underline {\approx 3.17\,\,{\rm Mbit/s}} \hspace{0.05cm}.$$


(4)  For relative code redundancy,  when the source is redundancy-free,  the general rule is:

$$ r_c = \frac{R_c - R_q}{R_c} = 1- \frac{R_q}{R_c}= 1- \frac{T_c}{T_q \cdot {\rm log_2} (M_c)}\hspace{0.05cm}.$$
  • In the case of the second order biploar code considered here,  with parameters  $T_{c} = T_{q}$  and  $M_{c} = 3$,  the following holds:
$$r_c = 1- \frac{1}{{\rm log_2} (3)}\hspace{0.15cm}\underline {\approx 36.9 \% }\hspace{0.05cm}.$$