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

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[[File:P_ID1326__Dig_Z_2_4.png|right|frame|Octal "random coding" and gray coding]]
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[[File:EN_Dig_A_2_6.png|right|frame|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.  
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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 distorted 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:
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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  $p = 0.01$.   
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*$a_5$  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 distorted with probability  $p = 0.01$  into coefficient  $a_7$.  No distortion is possible in the other direction.
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*$a_8$  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 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,
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*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 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 '''L''' is to be added at the first place, for  $a_{5}, ..., a_{8}$  the binary symbol '''H'''.
 
  
For the two mappings "Random" and "Gray" are to be calculated:
+
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  $\rm H$.
*the ''symbol error probability''  $p_{\rm S}$, which is the same in both cases; $p_{\rm S}$  indicates the average distortion probability of an amplitude coefficient  $a_{\mu}$;   
 
*the ''bit error probability''  $p_{\rm B}$  related to the (decoded) binary symbols.
 
  
 +
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.
  
  
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 +
Notes:
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*The exercise belongs to the chapter   [[Digital_Signal_Transmission/Grundlagen_der_codierten_Übertragung|"Basics of Coded Transmission"]].
  
''Notes:''
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*Reference is also made to the chapter  [[Digital_Signal_Transmission/Redundanzfreie_Codierung|"Redundancy-Free Coding"]] .
*The exercise belongs to the chapter   [[Digital_Signal_Transmission/Grundlagen_der_codierten_Übertragung|Basics of Coded Transmission]].
 
*Reference is also made to the chapter  [[Digital_Signal_Transmission/Redundanzfreie_Codierung|Redundancy-Free Coding]] .
 
 
   
 
   
  
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<quiz display=simple>
 
<quiz display=simple>
  
{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)$.
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{To which amplitude coefficient &nbsp;$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)$.
 
|type="{}"}
 
|type="{}"}
 
$ \rm {LHH}\text{:}\hspace{0.4cm}  \mu  \ = \ $ { 3 3% }  
 
$ \rm {LHH}\text{:}\hspace{0.4cm}  \mu  \ = \ $ { 3 3% }  
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$p_{\rm S} \ = \ $ { 1.75 3% } $\ \%$
 
$p_{\rm S} \ = \ $ { 1.75 3% } $\ \%$
  
{Calculate the bit error probability &nbsp;$p_{\rm B}$&nbsp; for the <u>gray code</u>.
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{Calculate the bit error probability &nbsp;$p_{\rm B}$&nbsp; for the&nbsp; <u>Gray code</u>.
 
|type="{}"}
 
|type="{}"}
 
$p_{\rm B} \ = \ $ { 0.583 3% } $\ \%$
 
$p_{\rm B} \ = \ $ { 0.583 3% } $\ \%$
  
{Calculate the bit error probability &nbsp;$p_{\rm B}$&nbsp; for the <u>random code</u>.
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{Calculate the bit error probability &nbsp;$p_{\rm B}$&nbsp; for the&nbsp; <u>random code</u>.
 
|type="{}"}
 
|type="{}"}
 
$p_{\rm B} \ = \ $ { 0.714 3% } $\ \%$
 
$p_{\rm B} \ = \ $ { 0.714 3% } $\ \%$
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{{ML-Kopf}}
 
{{ML-Kopf}}
 
'''(1)'''&nbsp;  According to the description on the specification page
 
'''(1)'''&nbsp;  According to the description on the specification page
*"LHH" for the amplitude coefficient $a_{3}$  &nbsp; &rArr; &nbsp; $\underline{\mu =3}$.  
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*$\rm LHH$&nbsp; for the amplitude coefficient&nbsp; $a_{3}$  &nbsp; &rArr; &nbsp; $\underline{\mu =3}$.  
*"HLL" for the amplitude coefficient $a_{8}$  &nbsp; &rArr; &nbsp; $\underline{\mu =8}$.  
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*$\rm HLL$&nbsp; for the amplitude coefficient&nbsp; $a_{8}$  &nbsp; &rArr; &nbsp; $\underline{\mu =8}$.  
  
  
'''(2)'''&nbsp; The outer coefficients ($a_{1}$ and $a_{8}$) are each distorted with probability $p = 1 \%$, <br>the $M – 2 = 6$ inner ones with twice the probability $(2p= 2 \%)$. By averaging, we obtain:
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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:
 
:$$p_{\rm S} = \frac{2 \cdot 1 + 6 \cdot 2} { 8} \cdot p\hspace{0.15cm}\underline { = 1.75 \,\%} \hspace{0.05cm}.$$
 
:$$p_{\rm S} = \frac{2 \cdot 1 + 6 \cdot 2} { 8} \cdot p\hspace{0.15cm}\underline { = 1.75 \,\%} \hspace{0.05cm}.$$
  
'''(3)'''&nbsp; 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
+
 
 +
'''(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
 
:$$p_{\rm B} ={p_{\rm S}}/ { 3}\hspace{0.15cm}\underline { = 0.583 \,\%} \hspace{0.05cm}.$$
 
:$$p_{\rm B} ={p_{\rm S}}/ { 3}\hspace{0.15cm}\underline { = 0.583 \,\%} \hspace{0.05cm}.$$
  
'''(4)'''&nbsp; Of the total of seven possible transitions (each in both directions) lead to
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*one error: &nbsp; &nbsp; '''HLH''' $\Leftrightarrow$ '''LLH''',
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'''(4)'''&nbsp; Of the total of seven possible transitions&nbsp; (each in both directions)&nbsp; lead to
*two errors: &nbsp; &nbsp;&nbsp; '''HLL''' $\Leftrightarrow$ '''HHH''', '''LLL''' $\Leftrightarrow$ '''LHH''', '''HHL''' $\Leftrightarrow$ '''HLH''', '''LLH''' $\Leftrightarrow$ '''LHL''',
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*one error: &nbsp; &nbsp; $\rm HLH \ \Leftrightarrow \ LLH$,
*three errors: &nbsp; &nbsp; &nbsp; '''HHH''' $\Leftrightarrow$ '''LLL''', '''LHH''' $\Leftrightarrow$ '''HHL'''.
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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; $\rm LLH \ \Leftrightarrow \ LHL$,  
 +
*three errors: &nbsp; &nbsp;&nbsp; $\rm HHH \ \Leftrightarrow \ LLL$, &nbsp;  &nbsp; $\rm LHH \ \Leftrightarrow \ HHL$.
  
  

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:

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