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.  
 
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.  
  

Revision as of 17:03, 9 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 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:

  • $a_5$  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.


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 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 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.




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

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


(2)  The outer coefficients ($a_{1}$ and $a_{8}$) are each distorted 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:     HLH $\Leftrightarrow$ LLH,
  • two errors:      HLL $\Leftrightarrow$ HHH, LLL $\Leftrightarrow$ LHH, HHL $\Leftrightarrow$ HLH, LLH $\Leftrightarrow$ LHL,
  • three errors:       HHH $\Leftrightarrow$ LLL, LHH $\Leftrightarrow$ HHL.


It follows that:

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