Exercise 2.4Z: Error Probabilities for the Octal System

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:

The exercise belongs to the chapter "Basics of Coded Transmission".

Reference is also made to the chapter "Redundancy-Free Coding" .

Questions

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

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

$p_{\rm S} \ = \ $

$\ \%$

$p_{\rm B} \ = \ $

$\ \%$

$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:

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