Aufgaben:Exercise 2.13: Decoding at RSC (7, 3, 5) to Base 8

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ELP allocation schemes for  $r = 1, \ r = 2, \ r = 3$

In the  "Exercise 2.12"  we used the so-called Petersen algorithm

• for error correction respective

• for decoding

the Reed–Solomon code  $(7, \, 4, \, 4)_8$  which can due to the minimum distance  $d_{\rm min} = 4$  $(t = 1)$ only correct one symbol error.

In this exercise we now consider the  ${\rm RSC} \, (7, \, 3, \, 5)_8 \ \Rightarrow \ d_{\rm min} = 5 \ \Rightarrow \ t = 2$,  whose parity-check matrix is as follows:

$${ \boldsymbol{\rm H}} = \begin{pmatrix} 1 & \alpha^1 & \alpha^2 & \alpha^3 & \alpha^4 & \alpha^5 & \alpha^6\\ 1 & \alpha^2 & \alpha^4 & \alpha^6 & \alpha^1 & \alpha^{3} & \alpha^{5}\\ 1 & \alpha^3 & \alpha^6 & \alpha^2 & \alpha^{5} & \alpha^{1} & \alpha^{4}\\ 1 & \alpha^4 & \alpha^1 & \alpha^{5} & \alpha^{2} & \alpha^{6} & \alpha^{3} \end{pmatrix} .$$

The received word under consideration   $\underline{y} = (\alpha^2, \, \alpha^3, \, \alpha, \, \alpha^5, \, \alpha^4, \, \alpha^2, \, 1)$   results in the syndrome 

$$\underline{s} = \underline{y} \cdot \mathbf{H}^{\rm T} = (0, \, 1, \, \alpha^5, \, \alpha^2).$$

The further decoding procedure is done according to the following theory pages:

• Step  $\rm (B)$:  "Set up/evaluate the ELP coefficient vector";   "ELP"  stands for  "Error Locator Polynomial".

• Step  $\rm (C)$:  "Localization of error positions",

• Step  $\rm (D)$:  "Final error correction".

Hints:

• This exercise belongs to the chapter  "Error correction According to Reed–Solomon Coding".

• In the above graphic you can see the assignments of the  "ELP coefficients"  ${\it \underline{\Lambda}}_l$  assuming that there are symbol errors in the received words $r = 1, \ r = 2$  respectively.  $r = 3$ .

Questions

1

Which allocation schemes might be relevant for this exercise?

	The schema highlighted in blue  $(r = 1)$.
	The schema highlighted in red  $(r = 2)$.
	The scheme highlighted in green  $(r = 3)$.

2

Can the syndrome   $\underline{s} = (0, \, 1, \, \alpha^5, \, \alpha^2)$   be caused by one symbol error?

	YES.
	NO.

3

Can the syndrome   $\underline{s} = (0, \, 1, \, \alpha^5, \, \alpha^2)$   be caused by two symbol errors?

	YES.
	NO.

4

Which symbols of the code word have been falsified?

	Symbol 0,
	Symbol 1,
	Symbol 2,
	Symbol 3,
	Symbol 4,
	Symbol 5,
	Symbol 6.

5

What is the error vector  $\underline{e}$?  Give also the decoding result   $\underline{z} = \underline{y} + \underline{e}$ .

	$\underline{e} = (1, \, 0, \, 0, \, 0, \, 0, \, 0, \, \alpha^6)$,
	$\underline{e} = (\alpha^6, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1)$,
	$\underline{e} = (0, \, 0, \, 1, \, \alpha^6, \, 0, \, 0, \, 0)$,
	$\underline{e} = (0, \, 0, \, \alpha^6, \, 1, \, 0, \, 0, \, 0)$.

Solution

Solution

(1)  Correct are the  proposed solutions 1 and 2:

• The ${\rm RSC} \, (7, \, 3, \, 5)_8$  can correct up to  $t = 2$  symbol errors.

• The actual symbol error count  $r$  must not be larger.

Conversion tables for  $\rm GF(2^3)$

(2)  Assuming  $r = 1$,  the  $n-k-1$  governing equations  ${\it \underline{\Lambda}}_l \cdot \underline{s}^{\rm T} = 0$  for  $\lambda_0$  are as follows :

$$\lambda_0 \cdot 0 \hspace{-0.15cm} \ + \ \hspace{-0.15cm} 1 = 0 \hspace{0.75cm} \Rightarrow \hspace{0.3cm} \lambda_0 \hspace{0.15cm} {\rm unspecified} \hspace{0.05cm},$$

$$\lambda_0 \cdot 1 \hspace{-0.15cm} \ + \ \hspace{-0.15cm} \alpha^5 = 0 \hspace{0.55cm} \Rightarrow \hspace{0.3cm} \lambda_0 = \alpha^5 \hspace{0.05cm},$$

$$\lambda_0 \cdot \alpha^5 \hspace{-0.15cm} \ + \ \hspace{-0.15cm} \alpha^2 = 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \lambda_0 = \alpha^{-3} = \alpha^4 \hspace{0.05cm}.$$

• The assumption  $r = 1$  would be justified only if the same  $\lambda_0$  value resulted from all these three equations.

• This is not the case here   ⇒   Answer NO.

(3)  Assuming the occupation for  $r = 2$,  we obtain two equations of determination for  $\lambda_0$  and  $\lambda_1$:

$$\lambda_0 \cdot 0 \hspace{-0.15cm} \ + \ \hspace{-0.15cm} \lambda_1 \cdot 1 +\alpha^5 = 0 \hspace{0.38cm} \Rightarrow \hspace{0.3cm} \lambda_1 = \alpha^5 \hspace{0.05cm},$$

$$\lambda_0 \cdot 1 \hspace{-0.15cm} \ + \ \hspace{-0.15cm} \lambda_1 \cdot \alpha^5 +\alpha^2 = 0 \hspace{0.15cm} \Rightarrow \hspace{0.3cm} \lambda_0 = \alpha^{5+5} +\alpha^2 = \alpha^{3} +\alpha^2 =\alpha^{5} \hspace{0.05cm}.$$

• The system of equations can be solved under the assumption  $r = 2$   ⇒   Answer YES.

• The results  $\lambda_0 = \lambda_1 = \alpha^5$  obtained here are processed in the next subtask.

(4)  With the result  $\lambda_0 = \lambda_1 = \alpha^5$,  the error locator polynomial  $($"key equation"$)$  is:

$${\it \Lambda}(x)=x \cdot \big ({\it \lambda}_0 + {\it \lambda}_1 \cdot x + x^2 \big ) =x \cdot \big (\alpha^5 + \alpha^5 \cdot x + x^2 )\hspace{0.05cm}.$$

• This function has zeros for  $x = \alpha^2$  and  $x = \alpha^3$:

$${\it \Lambda}(x = \alpha^2 )\hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha^2 \cdot \big ( \alpha^5 + \alpha^7 + \alpha^4 \big ) = \alpha^2 \cdot \big ( \alpha^5 + 1 + \alpha^4 \big )= 0\hspace{0.05cm},$$

$${\it \Lambda}(x = \alpha^3 )\hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha^3 \cdot \big ( \alpha^5 + \alpha^8 + \alpha^6 \big ) = \alpha^3 \cdot \big ( \alpha^5 + \alpha + \alpha^6 \big )= 0\hspace{0.05cm}.$$

• Falsified are consequently the symbols at positions  2  and  3   ⇒   Solutions 3 and 4.

(5)  After the result of the subtask  (4)  only the last two proposed solutions come into question:

$$\underline{e} = (0, \, 0, \, e_2, \, e_3, \, 0, \, 0, \, 0).$$

• The approach is therefore accordingly   $\underline{e} \cdot \mathbf{H}^{\rm T} = \underline{s}$:

$$(0, 0, e_2, e_3, 0, 0, 0) \cdot \begin{pmatrix} 1 & 1 & 1 & 1\\ \alpha^1 & \alpha^2 & \alpha^3 & \alpha^4\\ \alpha^2 & \alpha^4 & \alpha^6 & \alpha^1\\ \alpha^3 & \alpha^6 & \alpha^2 & \alpha^5\\ \alpha^4 & \alpha^1 & \alpha^{5} & \alpha^2\\ \alpha^5 & \alpha^{3} & \alpha^{1} & \alpha^6\\ \alpha^6 & \alpha^{5} & \alpha^{4} & \alpha^3 \end{pmatrix} \hspace{0.15cm}\stackrel{!}{=} \hspace{0.15cm} (0, 1, \alpha^5, \alpha^2) \hspace{0.05cm}. $$

$$\Rightarrow \hspace{0.3cm} e_2 \cdot \alpha^{2} \hspace{-0.15cm} \ + \ \hspace{-0.15cm} e_3 \cdot \alpha^{3} = 0 \hspace{0.05cm}, \hspace{0.8cm} e_2 \cdot \alpha^{4} + e_6 \cdot \alpha^{3} = 1\hspace{0.05cm},\hspace{0.8cm} e_2 \cdot \alpha^{6} \hspace{-0.15cm} \ + \ \hspace{-0.15cm} e_3 \cdot \alpha^{2} = \alpha^{5} \hspace{0.05cm}, \hspace{0.6cm} e_2 \cdot \alpha^{1} + e_6 \cdot \alpha^{5} = \alpha^{2} \hspace{0.05cm}. $$

• All four equations are satisfied with  $e_2 = 1$  as well as  $e_3 = \alpha^6$   ⇒   Proposed solution 3:

$$\underline {e} = (0, \hspace{0.05cm}0, \hspace{0.05cm}1, \hspace{0.05cm}\alpha^6, \hspace{0.05cm}0, \hspace{0.05cm}0, \hspace{0.05cm}0) $$

• With   $\alpha + 1 = \alpha^3$  and  $\alpha^5 + \alpha^6 = \alpha$   we get from the given received word  $\underline{y} = (\alpha^2, \, \alpha^3, \, \alpha, \, \alpha^5, \, \alpha^4, \, \alpha^2, \, 1)$  to the decoding result

$$\underline {z} = (\alpha^2, \hspace{0.05cm}\alpha^3, \hspace{0.05cm}\alpha^3, \hspace{0.05cm}\alpha, \hspace{0.05cm}\alpha^4, \hspace{0.05cm}\alpha^2, \hspace{0.05cm}1) \hspace{0.05cm}. $$

• In the  "Exercise 2.7"  it was shown that this is a valid code word of  ${\rm RSC} \, (7, \, 3, \, 5)_8$.

• The corresponding information word is  $\underline{u} = (\alpha^4, \, 1, \, \alpha^3)$.