Difference between revisions of "Aufgaben:Exercise 2.4Z: Error Probabilities for the Octal System"

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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Redundancy-Free_Coding
 
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[[File:|right|]]
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[[File:EN_Dig_A_2_6.png|right|frame|Octal&nbsp; "random coding"&nbsp; and&nbsp; Gray coding]]
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A digital system with &nbsp;$M = 8$&nbsp; amplitude levels&nbsp; ("octal system")&nbsp; is considered,&nbsp; whose &nbsp;$M – 1 = 7$&nbsp; decision thresholds lie exactly at the respective interval centers.
  
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Each of the equally probable amplitude coefficients &nbsp;$a_{\mu}$&nbsp; with &nbsp;$1 ≤ \mu ≤ 8$&nbsp; can be falsified only into the immediate neighbor coefficients &nbsp;$a_{\mu–1}$&nbsp; and &nbsp;$a_{\mu+1}$,&nbsp; respectively,&nbsp; and in both directions with the same probability &nbsp;$p = 0.01$.&nbsp; Here are some examples:
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*$a_5$&nbsp; passes into coefficient $a_4$ with probability &nbsp;$p = 0.01$&nbsp; and into coefficient &nbsp;$a_6$ with the same probability &nbsp;$p = 0.01$.&nbsp;
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*$a_8$&nbsp; is falsified with probability &nbsp;$p = 0.01$&nbsp; into coefficient &nbsp;$a_7$.&nbsp; No falsification is possible in the other direction.
  
===Fragebogen===
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The mapping of each three binary source symbols into an octal amplitude coefficient happens alternatively according to
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*the second column in the given table,&nbsp; which was generated&nbsp; "randomly"&nbsp; - without strategy,
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*the Gray coding,&nbsp; which is only incompletely indicated in column 3 and is still to be supplemented.
 +
 
 +
 
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The Gray code is given for &nbsp;$M = 4$.&nbsp; For &nbsp;$M = 8$&nbsp; the last two binary characters are to be mirrored at the dashed line.&nbsp; For the first four amplitude coefficients a&nbsp; $\rm L$&nbsp; is to be added at the first place,&nbsp; for &nbsp;$a_{5}, ..., a_{8}$&nbsp; the binary symbol&nbsp; $\rm H$.
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For the two mappings&nbsp; "Random"&nbsp; and&nbsp; "Gray"&nbsp; are to be calculated:
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*the&nbsp; "symbol error probability" &nbsp;$p_{\rm S}$,&nbsp; which is the same in both cases;&nbsp; $p_{\rm S}$&nbsp; indicates the average falsifcation probability of an amplitude coefficient &nbsp;$a_{\mu}$;&nbsp;
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*the&nbsp; "bit error probability" &nbsp;$p_{\rm B}$&nbsp; related to the (decoded) binary symbols.
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Notes:
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*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Grundlagen_der_codierten_Übertragung|"Basics of Coded Transmission"]].
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*Reference is also made to the chapter&nbsp; [[Digital_Signal_Transmission/Redundanzfreie_Codierung|"Redundancy-Free Coding"]] .
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===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Multiple-Choice Frage
 
|type="[]"}
 
- Falsch
 
+ Richtig
 
  
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{To which amplitude coefficient &nbsp;$a_{ \mu}$&nbsp; do the binary sequences &nbsp;$\rm {LHH}$&nbsp; and &nbsp;$\rm {HLL}$ correspond in the Gray code? <br>Please enter index &nbsp;$ \mu$&nbsp;  &nbsp;$(1 <  \mu < 8)$.
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|type="{}"}
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$ \rm {LHH}\text{:}\hspace{0.4cm}  \mu  \ = \ $ { 3 3% }
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$ \rm {HLL}\text{:}\hspace{0.45cm}  \mu  \ = \ $ { 8 3% }
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{Calculate the symbol error probability &nbsp;$p_{\rm S}$.
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|type="{}"}
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$p_{\rm S} \ = \ $ { 1.75 3% } $\ \%$
  
{Input-Box Frage
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{Calculate the bit error probability &nbsp;$p_{\rm B}$&nbsp; for the&nbsp; <u>Gray code</u>.
 
|type="{}"}
 
|type="{}"}
$\alpha$ = { 0.3 }
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$p_{\rm B} \ = \ $ { 0.583 3% } $\ \%$
  
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{Calculate the bit error probability &nbsp;$p_{\rm B}$&nbsp; for the&nbsp; <u>random code</u>.
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|type="{}"}
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$p_{\rm B} \ = \ $ { 0.714 3% } $\ \%$
  
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp;
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'''(1)'''&nbsp; According to the description on the specification page
'''(2)'''&nbsp;
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*$\rm LHH$&nbsp; for the amplitude coefficient&nbsp; $a_{3}$  &nbsp; &rArr; &nbsp; $\underline{\mu =3}$.
'''(3)'''&nbsp;
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*$\rm HLL$&nbsp; for the amplitude coefficient&nbsp; $a_{8}$  &nbsp; &rArr; &nbsp; $\underline{\mu =8}$.
'''(4)'''&nbsp;
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'''(5)'''&nbsp;
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'''(6)'''&nbsp;
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'''(2)'''&nbsp; The outer coefficients&nbsp; $(a_{1}$&nbsp; and&nbsp; $a_{8})$&nbsp; are each falsified with probability&nbsp; $p = 1 \%$,&nbsp; the&nbsp; $M – 2 = 6$&nbsp; inner ones with twice the probability&nbsp; $(2p= 2 \%)$.&nbsp; By averaging,&nbsp; we obtain:
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:$$p_{\rm S} = \frac{2 \cdot 1 + 6 \cdot 2} { 8} \cdot p\hspace{0.15cm}\underline { = 1.75 \,\%} \hspace{0.05cm}.$$
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'''(3)'''&nbsp; Each transmission error&nbsp; (symbol error)&nbsp; results in exactly one bit error in Gray code.&nbsp; However,&nbsp; since each octal symbol contains three binary characters,&nbsp; the following applies
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:$$p_{\rm B} ={p_{\rm S}}/ { 3}\hspace{0.15cm}\underline { = 0.583 \,\%} \hspace{0.05cm}.$$
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 +
 
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'''(4)'''&nbsp; Of the total of seven possible transitions&nbsp; (each in both directions)&nbsp; lead to
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*one error: &nbsp; &nbsp; $\rm HLH \ \Leftrightarrow \ LLH$,
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*two errors: &nbsp; &nbsp;&nbsp; $\rm HLL \ \Leftrightarrow \ HHH$, &nbsp;  &nbsp; $\rm LLL \ \Leftrightarrow \ LHH$, &nbsp;  &nbsp; $\rm HHL \ \Leftrightarrow \ HLH$, &nbsp;  &nbsp; $\rm LLH \ \Leftrightarrow \ LHL$,
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*three errors: &nbsp; &nbsp;&nbsp; $\rm HHH \ \Leftrightarrow \ LLL$, &nbsp;  &nbsp; $\rm LHH \ \Leftrightarrow \ HHL$.
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It follows that:
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:$$p_{\rm B} = \frac{p} { 3} \cdot \frac{1 + 4 \cdot 2 + 2 \cdot 3} { 7} = \frac{15} { 21} \cdot p \hspace{0.15cm}\underline { = 0.714 \,\%} \hspace{0.05cm}.$$
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{{ML-Fuß}}
 
{{ML-Fuß}}
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[[Category:Aufgaben zu Digitalsignalübertragung|^2.2 Redundanzfreie Codierung^]]
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[[Category:Digital Signal Transmission: Exercises|^2.2 Redundancy-Free Coding^]]

Latest revision as of 09:19, 24 May 2022

Octal  "random coding"  and  Gray coding

A digital system with  $M = 8$  amplitude levels  ("octal system")  is considered,  whose  $M – 1 = 7$  decision thresholds lie exactly at the respective interval centers.

Each of the equally probable amplitude coefficients  $a_{\mu}$  with  $1 ≤ \mu ≤ 8$  can be falsified only into the immediate neighbor coefficients  $a_{\mu–1}$  and  $a_{\mu+1}$,  respectively,  and in both directions with the same probability  $p = 0.01$.  Here are some examples:

  • $a_5$  passes into coefficient $a_4$ with probability  $p = 0.01$  and into coefficient  $a_6$ with the same probability  $p = 0.01$. 
  • $a_8$  is falsified with probability  $p = 0.01$  into coefficient  $a_7$.  No falsification is possible in the other direction.


The mapping of each three binary source symbols into an octal amplitude coefficient happens alternatively according to

  • the second column in the given table,  which was generated  "randomly"  - without strategy,
  • the Gray coding,  which is only incompletely indicated in column 3 and is still to be supplemented.


The Gray code is given for  $M = 4$.  For  $M = 8$  the last two binary characters are to be mirrored at the dashed line.  For the first four amplitude coefficients a  $\rm L$  is to be added at the first place,  for  $a_{5}, ..., a_{8}$  the binary symbol  $\rm H$.

For the two mappings  "Random"  and  "Gray"  are to be calculated:

  • the  "symbol error probability"  $p_{\rm S}$,  which is the same in both cases;  $p_{\rm S}$  indicates the average falsifcation probability of an amplitude coefficient  $a_{\mu}$; 
  • the  "bit error probability"  $p_{\rm B}$  related to the (decoded) binary symbols.



Notes:



Questions

1

To which amplitude coefficient  $a_{ \mu}$  do the binary sequences  $\rm {LHH}$  and  $\rm {HLL}$ correspond in the Gray code?
Please enter index  $ \mu$   $(1 < \mu < 8)$.

$ \rm {LHH}\text{:}\hspace{0.4cm} \mu \ = \ $

$ \rm {HLL}\text{:}\hspace{0.45cm} \mu \ = \ $

2

Calculate the symbol error probability  $p_{\rm S}$.

$p_{\rm S} \ = \ $

$\ \%$

3

Calculate the bit error probability  $p_{\rm B}$  for the  Gray code.

$p_{\rm B} \ = \ $

$\ \%$

4

Calculate the bit error probability  $p_{\rm B}$  for the  random code.

$p_{\rm B} \ = \ $

$\ \%$


Solution

(1)  According to the description on the specification page

  • $\rm LHH$  for the amplitude coefficient  $a_{3}$   ⇒   $\underline{\mu =3}$.
  • $\rm HLL$  for the amplitude coefficient  $a_{8}$   ⇒   $\underline{\mu =8}$.


(2)  The outer coefficients  $(a_{1}$  and  $a_{8})$  are each falsified with probability  $p = 1 \%$,  the  $M – 2 = 6$  inner ones with twice the probability  $(2p= 2 \%)$.  By averaging,  we obtain:

$$p_{\rm S} = \frac{2 \cdot 1 + 6 \cdot 2} { 8} \cdot p\hspace{0.15cm}\underline { = 1.75 \,\%} \hspace{0.05cm}.$$


(3)  Each transmission error  (symbol error)  results in exactly one bit error in Gray code.  However,  since each octal symbol contains three binary characters,  the following applies

$$p_{\rm B} ={p_{\rm S}}/ { 3}\hspace{0.15cm}\underline { = 0.583 \,\%} \hspace{0.05cm}.$$


(4)  Of the total of seven possible transitions  (each in both directions)  lead to

  • one error:     $\rm HLH \ \Leftrightarrow \ LLH$,
  • two errors:      $\rm HLL \ \Leftrightarrow \ HHH$,     $\rm LLL \ \Leftrightarrow \ LHH$,     $\rm HHL \ \Leftrightarrow \ HLH$,     $\rm LLH \ \Leftrightarrow \ LHL$,
  • three errors:      $\rm HHH \ \Leftrightarrow \ LLL$,     $\rm LHH \ \Leftrightarrow \ HHL$.


It follows that:

$$p_{\rm B} = \frac{p} { 3} \cdot \frac{1 + 4 \cdot 2 + 2 \cdot 3} { 7} = \frac{15} { 21} \cdot p \hspace{0.15cm}\underline { = 0.714 \,\%} \hspace{0.05cm}.$$