Aufgaben:Exercise 2.11Z: Erasure Channel for Symbols

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Erasure channel for symbols:  "m–BEC"

The channel model  "Binary Erasure Channel"  $\rm (BEC)$ describes a bit-level erasure channel:

• A binary symbol  $(0$  or  $1)$  is correctly transmitted with probability  $1 - \lambda$  and marked as an erasure  "$\rm E$"  with probability  $\lambda$.

• In contrast to the  "Binary Symmetric Channel"  $\rm (BSC)$,  falsifications  $(0 → 1, \ 1 → 0)$  cannot occur here.

A Reed–Solomon code is based on a Galois field   ${\rm GF}(2^m)$   with integer  $m$.  Thus,  each code symbol   $c$   can be represented by  $m$  bits.  If one wants to apply the BEC model here,  one has to modify it to the  "$m$–BEC"  model  as shown in the diagram below for  $m = 2$:

• All code symbols   $($in binary representation  $00, \ 01, \ 10, \ 11)$   are transmitted correctly with probability  $1 - \lambda_2$ .

• Thus,  the probability of an erased symbol  is  $\lambda_2$.

• Note that already a single erased bit leads to the erased received symbol  $y = \rm E$ .

Hints:

• This exercise belongs to the chapter  "Reed–Solomon Decoding at the Erasure Channel".

• For a code based on  ${\rm GF}(2^m)$  the outlined  $2$–BEC model is to be extended to  $m$–BEC.

• The erasure probability of this model is then denoted by  $\lambda_m$.

• For the first subtasks,  use always  $\lambda = 0.2$  for the erasure probability of the basic model according to the upper graph.

Questions

1

$\lambda = 0.2$  is valid.  With what probabilities do the possible received values occur in the  BEC basic model ?

${\rm Pr}(y = 0) \ = \ $ $\ \%$

${\rm Pr}(y = {\rm E}) \ = \ $ $\ \%$

${\rm Pr}(y = {\rm 1}) \ = \ $ $\ \%$

2

$\lambda = 0.2$  is valid.  What is the erasure probability  $\lambda_2$  at symbol level  $(2$–BEC model$)$  when the Reed-Solomon code is based on  $\rm GF(2^2)$?

$\lambda_2 \ = \ $ $\ \%$

3

$\lambda = 0.2$  is valid.  What is the erasure probability  $\lambda_m$,  when the  $m$–BEC model  is fitted to the  $\rm RSC \, (255, \, 223, \, 33)_{256}$?

$\lambda_m \ = \ $ $\ \%$

4

What is the maximum allowed erasure probability  $\lambda$  of the  "BEC basic model"  for  $\lambda_m ≤ 0.2$  to hold?

${\rm Max} \ \big[\lambda\big ] \ = \ $ $\ \%$

5

What is the probability of receiving the  "zero symbol"  in this case?

${\rm Pr}(y_{\rm bin} = 00000000) \ = \ $ $\ \%$

Solution

Solution

(1)  Due to the symmetry of the given BEC model (bit-level erasure channel), the erasure probability is:

$$\ {\rm Pr}(y = {\rm E}) = \lambda \ \underline{= 20\%}.$$

Since code symbols $0$ and $1$ are equally likely, we obtain ${\rm Pr}(y = 0) \ \underline{= 40\%}$ and ${\rm Pr}(y = 1) \ \underline{= 40\%}$ for their probabilities.

(2)  Without limiting generality, we assume the code symbol $c_{\rm binary} = $ "$00$" to solve this exercise.

• According to the 2 BEC model, the receive symbol $y_{\rm binary}$ can then be either "$00$" or canceled $(\rm E)$ and it holds:

$${\rm Pr}(y_{\rm bin} = "00"\hspace{0.05cm} |\hspace{0.05cm} c_{\rm bin} = "00") \hspace{-0.15cm} \ = \ \hspace{-0.15cm} ( 1 - \lambda)^2 = 0.8^2 = 0.64 = 1 - \lambda_2\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \lambda_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 - ( 1 - \lambda)^2 \hspace{0.15cm}\underline{= 36\%}\hspace{0.05cm}. $$

• This assumes that an erasure is avoided only if neither of the two bits has been erased.

(3)  The $\rm RSC \, (255, \, 223, \, 33)_{256}$ is based on the Galois field $\rm GF(256) = GF(2^8) \ \Rightarrow \ \it m = \rm 8$. The result of subtask (2) must now be adapted to this case. For the $8$ BEC holds:

$$1 - \lambda_8 = ( 1 - \lambda)^8 = 0.8^8 \approx 0.168 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \lambda_m = \lambda_8 \hspace{0.15cm}\underline{\approx 83.2\%}\hspace{0.05cm}. $$

(4)  From the condition $\lambda_m ≤ 0.2$ it follows directly $1 - \lambda_m ≥ 0.8$. From this follows further:

$$( 1 - \lambda)^8 \ge 0.8 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 1 - \lambda \ge 0.8^{0.125} \approx 0.9725 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\lambda \hspace{0.15cm} \underline{\le 2.75\%}\hspace{0.05cm}.$$

(5)  Here one proceeds as follows:

• With $\lambda = 0.0275 \Rightarrow \ \lambda_m = 0.2$, $20\%$ of the receiving symbols are erasures.
• The $2^8 = 256$ receiving symbols ($00000000$   ...   $11111111$) are all equally likely. It follows:

$${\rm Pr}(y_{\rm bin} = 00000000) = \hspace{0.1cm}\text{...} \hspace{0.1cm}= {\rm Pr}(y_{\rm bin} = 11111111)= \frac{0.8}{256} \hspace{0.15cm}\underline{= 0.3125\%}\hspace{0.05cm}.$$