Aufgaben:Exercise 2.08: Generator Polynomials for Reed-Solomon: Difference between revisions

Latest revision as of 17:56, 16 March 2026

Four generator matrices, three of which describe Reed-Solomon codes

In the "Exercise 2.7" you should determine the code words of the $\rm RSC \, (7, \, 3, \, 5)_8$ via a polynomial. However, you can also determine the code word $\underline{c}$ from the information word $\underline{u}$ and the generator matrix $\mathbf{G}$ according to the following equation:

$$\underline {c} = \underline {u} \cdot { \boldsymbol{\rm G}}\hspace{0.05cm}.$$

Two of these generator matrices describe the $\rm RSC \, (7, \, 3, \, 5)_8$. In the subtask (1) is explicitly asked which.

Another generator matrix belongs to $\rm RSC \, (7, \, 5, \, 3)_8$, which is considered in subtask (3).

Hints:

This exercise belongs to the chapter "Definition and Properties of Reed-Solomon Codes".

Important information about Reed–Solomon codes can also be found in "Exercise 2.7".

Questions

Solution

(1) Correct are the solutions 2 and 3 ⇒ matrices $\mathbf{G}_{\rm B}$ and $\mathbf{G}_{\rm C}$.

In the matrix $\mathbf{G}_{\rm C}$ the allowed transformations $\alpha^8 = \alpha, \ \alpha^{10} = \alpha^3$ and $\alpha^{12} = \alpha^5$ have already been considered.

The matrix $\mathbf{G}_{\rm A}$ holds for the $(7, \, 5, \, 3)$ Hamming code and $\mathbf{G}_{\rm D}$ belongs to the $\rm RSC \, (7, \, 5, \, 3)_8$. See subtask (3) for more details.

(2) In the $\rm RSC \, (7, \, 3, \, 5)_8$, in each coding step are processed $k = 3$ information symbols,
in coding step 1 according to the specification the symbols $\alpha^4, \ 1$ and $\alpha^3$.

With the generator matrix $\mathbf{G}_{\rm C}$ thus holds:

$$\underline {c} = \underline {u} \cdot { \boldsymbol{\rm G}}_{\rm C} =\begin{pmatrix}\alpha^4 & 1 & \alpha^3\end{pmatrix} \cdot\begin{pmatrix}1 & 1 & 1 & 1 & 1 & 1 & 1\\1 & \alpha^1 & \alpha^2 & \alpha^3 & \alpha^4 & \alpha^5 & \alpha^6\\1 & \alpha^2 & \alpha^4 & \alpha^6 & \alpha^1 & \alpha^{3} & \alpha^{5}\end{pmatrix}\hspace{0.05cm}. $$

$\rm GF(2^3)$ representation as powers, polynomials, vectors

This results according to the adjacent auxiliary table:

$$c_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha^{4}\cdot 1 + 1 \cdot 1 + \alpha^{3}\cdot 1 =(110) + (001) + (011)= (100) = \alpha^{2} \hspace{0.05cm},$$

$$c_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha^{4}\cdot 1 + 1 \cdot \alpha + \alpha^{3}\cdot \alpha^{2}=(110) + (010) + (110) = (011) = \alpha^{3} \hspace{0.05cm},$$

$$c_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha^{4}\cdot 1 + 1 \cdot \alpha^{2} + \alpha^{3}\cdot \alpha^{4}=(110) + (100) + (001) = (011) = \alpha^{3} \hspace{0.05cm},$$

$$c_3 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha^{4}\cdot 1 + 1 \cdot \alpha^{3} + \alpha^{3}\cdot \alpha^{6}=$(110) + (011) + (100) = (001) = 1 \hspace{0.05cm},$$

$$c_4 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha^{4}\cdot 1 + 1 \cdot \alpha^{4} + \alpha^{3}\cdot \alpha^{1}= \alpha^{4} \hspace{0.05cm},$$

$$c_5 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha^{4}\cdot 1 + 1 \cdot \alpha^{5} + \alpha^{3}\cdot \alpha^{3}=(110) + (111) + (101) = (100) = \alpha^{2} \hspace{0.05cm},$$

$$c_6 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha^{4}\cdot 1 + 1 \cdot \alpha^{6} + \alpha^{3}\cdot \alpha^{5}=(\alpha^{2} + \alpha) + (\alpha^2 +1) + \alpha = 1 \hspace{0.05cm}.$$

You get exactly the same result as in subtask (4) of "Exercise 2.7". Correct are the solutions 1 and 2.
So it is not $c_6 = 0$, but $c_6 = 1$.

(3) At the $\rm RSC \, (7, \, 5, \, 3)_8$, the information word $\underline{u} = (u_0, \, u_1, \, u_2, \, u_3, \, u_4)$ must be considered.

With the generator matrix $\mathbf{G}_{\rm D}$ one obtains:

$$\underline {c} = \underline {u} \cdot { \boldsymbol{\rm G}}_{\rm D} =\begin{pmatrix}\alpha^4 & 1 & \alpha^3 & 0 & \alpha^6\end{pmatrix} \cdot\begin{pmatrix}1 & 1 & 1 & 1 & 1 & 1 & 1\\1 & \alpha^1 & \alpha^2 & \alpha^3 & \alpha^4 & \alpha^5 & \alpha^6\\1 & \alpha^2 & \alpha^4 & \alpha^6 & \alpha^1 & \alpha^{3} & \alpha^{5}\\1 & \alpha^3 & \alpha^6 & \alpha^2 & \alpha^5 & \alpha^{1} & \alpha^{4}\\1 & \alpha^4 & \alpha^1 & \alpha^5 & \alpha^2 & \alpha^{6} & \alpha^{3}\end{pmatrix}\hspace{0.05cm}. $$

From this it follows:

$$c_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha^{4}\cdot 1 + 1 \cdot 1 + \alpha^{3}\cdot 1 + 0 \cdot 1 + \alpha^{6}\cdot 1= (110) + (001) + (011) + (000) + (101) = (001) = 1 \hspace{0.05cm},$$

$$c_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \left [ \alpha^{4}\cdot 1 + 1 \cdot \alpha + \alpha^{3}\cdot \alpha^{2} \right ] + 0 \cdot \alpha^{3} + \alpha^{6}\cdot \alpha^{4}= \left [ \alpha^{3} \right ] + \alpha^{3} = 0 \hspace{0.05cm}.$$

This takes into account that the bracket expression $[ \ \text{...} \ ]$ corresponds exactly to the result $c_1$ of subtask (2).

Corresponding is also considered in the following calculations:

$$c_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \left [ \alpha^{3} \right ] + \alpha^{6}\cdot \alpha^{1}=\left [ \alpha^{3} \right ] + \alpha^{7} =(011) + (001) = (010) = \alpha^{1} \hspace{0.05cm},$$

$$c_3 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \left [ 1 \right ] + \alpha^{6}\cdot \alpha^{5}=\left [ 1 \right ] + \alpha^{4}=(001) + (110) = (111) = \alpha^{5} \hspace{0.05cm},$$

$$c_4 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \left [ \alpha^{4} \right ] + \alpha^{6}\cdot \alpha^{2}=\left [ \alpha^{4} \right ] + \alpha^{1} = (110) + (010) = (100)= \alpha^{2} \hspace{0.05cm},$$

$$c_5 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \left [ \alpha^{2} \right ] + \alpha^{6}\cdot \alpha^{6}=\left [ \alpha^{2} \right ] + \alpha^{5} =(100) + (111) = (011) = \alpha^{3} \hspace{0.05cm},$$

$$c_6 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \left [ 1 \right ] + \alpha^{6}\cdot \alpha^{3}= \left [ 1 \right ] + \alpha^{2}= (001) + (100) = (101) = \alpha^{6} \hspace{0.05cm}.$$

This means: All proposed solutions are correct.

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Kanalcodierung/Definition und Eigenschaften von Reed–Solomon–Codes}}
+{{quiz-Header|Buchseite=Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes}}
-[[File:P_ID2525__KC_A_2_8.png|right|frame|Vier Generatormatrizen, drei davon beschreiben Reed–Solomon–Codes]]
+[[File:P_ID2525__KC_A_2_8.png|right|frame|Four generator matrices, three of which describe Reed-Solomon codes]]
-In der [[Aufgaben:2.07_Reed%E2%80%93Solomon%E2%80%93Code_(7,_3,_5)(Base_8)|Aufgabe A2.7]] sollten Sie die Codeworte des $\rm RSC \, (7, \, 3, \, 5)_8$ über ein Polynom ermitteln. Man kann aber das Codewort $\underline{c}$ auch aus dem Informationswort $\underline{u}$ und der Generatormatrix $\mathbf{G}$ gemäß der folgenden Gleichung bestimmen:
+In the&nbsp; [[Aufgaben:Exercise_2.07:_Reed-Solomon_Code_(7,_3,_5)_to_Base_8|"Exercise 2.7"]]&nbsp; you should determine the code words of the&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$&nbsp; via a polynomial.&nbsp; However,&nbsp; you can also determine the code word&nbsp; $\underline{c}$ &nbsp; from the information word&nbsp; $\underline{u}$&nbsp; and the generator matrix&nbsp; $\mathbf{G}$&nbsp; according to the following equation:
-:$$\underline {c} = \underline {u} \cdot { \boldsymbol{\rm G}}
+:$$\underline {c} = \underline {u} \cdot { \boldsymbol{\rm G}}\hspace{0.05cm}.$$
-\hspace{0.05cm}.$$
-Zwei der vorgegebenen Generatormatrizen beschreiben den $\rm RSC \, (7, \, 3, \, 5)_8$. In der Teilaufgabe (1) ist explizit gefragt, welche. Eine weitere Generatormatrix gehört zum $\rm RSC \, (7, \, 5, \, 3)_8$, der in der Teilaufgabe (3) betrachtet wird.
+*Two of these generator matrices describe the&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$.&nbsp; In the subtask&nbsp; '''(1)'''&nbsp; is explicitly asked which.
+*Another generator matrix belongs to&nbsp; $\rm RSC \, (7, \, 5, \, 3)_8$,&nbsp; which is considered in subtask&nbsp; '''(3)'''.
-''Hinweise:''
-* Die Aufgabe gehört zum Themengebiet des Kapitels [[Kanalcodierung/Definition_und_Eigenschaften_von_Reed%E2%80%93Solomon%E2%80%93Codes| Definition und Eigenschaften von Reed&ndash;Solomon&ndash;Codes]].
-* Wichtige Informationen zu den Reed&ndash;Solomon&ndash;Codes finden Sie auch in der Angabe zur [[Aufgaben:2.07_Reed%E2%80%93Solomon%E2%80%93Code_(7,_3,_5)(Base_8)| Aufgabe A2.7]].
+Hints:
+* This exercise belongs to the chapter&nbsp; [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes| "Definition and Properties of Reed-Solomon Codes"]].
-===Fragebogen===
+* Important information about Reed&ndash;Solomon codes can also be found in&nbsp; [[Aufgaben:Exercise_2.07:_Reed-Solomon_Code_(7,_3,_5)_to_Base_8| "Exercise 2.7"]].
+===Questions===
 <quiz display=simple>
-{Welche der Generatorpolynome beschreiben den $\rm RSC \, (7, \, 3, \, 5)_8$?
+{Which of the generator polynomials describe the&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$?
 |type="[]"}
-- $\mathbf{G}_{\rm A}$,
+- the matrix&nbsp; $\mathbf{G}_{\rm A}$,
-+ $\mathbf{G}_{\rm B}$,
++ the matrix&nbsp; $\mathbf{G}_{\rm B}$,
-+ $\mathbf{G}_{\rm C}$,
++ the matrix&nbsp; $\mathbf{G}_{\rm C}$,
-- $\mathbf{G}_{\rm D}$.
+- the matrix&nbsp; $\mathbf{G}_{\rm D}$.
-{Die Informationsfolge beginnt mit $\alpha^4, \, 1, \, \alpha^3, \, 0, \, \alpha^6$. Bestimmen Sie das erste Codewort für den $\rm RSC \, (7, \, 3, \, 5)_8$.
+{The information sequence starts with&nbsp; $\alpha^4, \, 1, \, \alpha^3, \, 0, \, \alpha^6$.&nbsp; Determine the first code word for the&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$.
 |type="[]"}
-+ Es gilt $c_0 = \alpha^2$,
++ It holds&nbsp; $c_0 = \alpha^2$,
-+ Es gilt $c_1 = \alpha^3$,
++ It holds&nbsp; $c_1 = \alpha^3$,
-- Es gilt $c_6 = 0$.
+- It holds&nbsp; $c_6 = 0$.
-{Wie lautet bei gleicher Informationsfolge das Codewort für den $\rm RSC \, (7, \, 5, \, 3)_8$?
+{What is the code word for the&nbsp; $\rm RSC \, (7, \, 5, \, 3)_8$ given the same sequence of information?
 |type="[]"}
-+ Es gilt $c_0 = 1$,
++ It holds&nbsp; $c_0 = 1$,
-+ Es gilt $c_1 = 0$,
++ It holds&nbsp; $c_1 = 0$,
-+ Es gilt $c_6 = \alpha^6$.
++ It holds&nbsp; $c_6 = \alpha^6$.
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 2 und 3</u> &nbsp;&#8658;&nbsp; Matrizen $\mathbf{G}_{\rm B}$ und $\mathbf{G}_{\rm C}$, wobei in der Matrix $\mathbf{G}_{\rm C}$ bereits die erlaubten Umformungen $\alpha^8 = \alpha, \ \alpha^{10} = \alpha^3$ und $\alpha^{12} = \alpha^5$ berücksichtigt wurden. Die Matrix $\mathbf{G}_{\rm A}$ gilt für den $(7, \, 5, \, 3)$&ndash;Hamming&ndash;Code und $\mathbf{G}_{\rm D}$ gehört zum $\rm RSC \, (7, \, 5, \, 3)_8$. Siehe hierzu Teilaufgabe (3).
+'''(1)'''&nbsp; Correct are the&nbsp; <u>solutions 2 and 3</u> &nbsp; &#8658; &nbsp; matrices&nbsp; $\mathbf{G}_{\rm B}$&nbsp; and&nbsp; $\mathbf{G}_{\rm C}$.
+*In the matrix&nbsp; $\mathbf{G}_{\rm C}$&nbsp; the allowed transformations&nbsp; $\alpha^8 = \alpha, \ \alpha^{10} = \alpha^3$&nbsp; and&nbsp; $\alpha^{12} = \alpha^5$&nbsp; have already been considered.
+*The matrix&nbsp; $\mathbf{G}_{\rm A}$&nbsp; holds for the&nbsp; $(7, \, 5, \, 3)$&nbsp; Hamming code and&nbsp; $\mathbf{G}_{\rm D}$&nbsp; belongs to the&nbsp; $\rm RSC \, (7, \, 5, \, 3)_8$.&nbsp; See subtask&nbsp; '''(3)'''&nbsp; for more details.
+'''(2)'''&nbsp; In the&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$,&nbsp;  in each coding step are processed&nbsp;  $k = 3$&nbsp; information symbols, &nbsp; <br>in coding step 1 according to the specification the symbols&nbsp; $\alpha^4, \ 1$ and $\alpha^3$.
-'''(2)'''&nbsp; Beim $\rm RSC \, (7, \, 3, \, 5)_8$ werden in jedem Codierschritt $k = 3$ Informationssymbole verarbeitet, im Codierschritt 1 die Symbole $\alpha^4, \ 1$ und $\alpha^3$. Mit der Generatormatrix $\mathbf{G}_{\rm C}$ gilt somit:
+*With the generator matrix $\mathbf{G}_{\rm C}$ thus holds:
-:$$\underline {c} = \underline {u} \cdot { \boldsymbol{\rm G}}_{\rm C} =
+:$$\underline {c} = \underline {u} \cdot { \boldsymbol{\rm G}}_{\rm C} =\begin{pmatrix}\alpha^4 & 1 & \alpha^3\end{pmatrix} \cdot\begin{pmatrix}1 & 1 & 1 & 1 & 1 & 1 & 1\\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}\end{pmatrix}\hspace{0.05cm}. $$
-\begin{pmatrix}
+[[File:EN_KC_Z_2_5_neu.png|right|frame|$\rm GF(2^3)$&nbsp; representation as powers, polynomials, vectors]]
-		\alpha^4 & 1 & \alpha^3
-\end{pmatrix} \cdot
+*This results according to the adjacent auxiliary table:
-\begin{pmatrix}
+:$$c_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}   \alpha^{4}\cdot 1  + 1 \cdot 1 +  \alpha^{3}\cdot 1 =(110) + (001) + (011)= (100) = \alpha^{2} \hspace{0.05cm},$$
-& 1 & 1 & 1 & 1 & 1 & 1\\
+:$$c_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}   \alpha^{4}\cdot 1 + 1 \cdot \alpha + \alpha^{3}\cdot \alpha^{2}=(110)  + (010) + (110) = (011) = \alpha^{3} \hspace{0.05cm},$$
-& \alpha^1 & \alpha^2 & \alpha^3 & \alpha^4 & \alpha^5 & \alpha^6\\
+:$$c_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}    \alpha^{4}\cdot 1 + 1 \cdot \alpha^{2} + \alpha^{3}\cdot \alpha^{4}=(110) + (100) + (001) = (011) = \alpha^{3} \hspace{0.05cm},$$
-& \alpha^2 & \alpha^4 & \alpha^6 & \alpha^1 & \alpha^{3} & \alpha^{5}
+:$$c_3 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}   \alpha^{4}\cdot 1 + 1 \cdot \alpha^{3} + \alpha^{3}\cdot \alpha^{6}=$(110) + (011) + (100) = (001) = 1 \hspace{0.05cm},$$
-\end{pmatrix}\hspace{0.05cm}. $$
+:$$c_4 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}    \alpha^{4}\cdot 1 + 1 \cdot \alpha^{4} + \alpha^{3}\cdot \alpha^{1}= \alpha^{4} \hspace{0.05cm},$$
+:$$c_5 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}    \alpha^{4}\cdot 1 + 1 \cdot \alpha^{5} + \alpha^{3}\cdot \alpha^{3}=(110) + (111) + (101) =  (100) = \alpha^{2} \hspace{0.05cm},$$
+:$$c_6 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  \alpha^{4}\cdot 1 + 1 \cdot \alpha^{6} + \alpha^{3}\cdot \alpha^{5}=(\alpha^{2} + \alpha) + (\alpha^2 +1) + \alpha = 1 \hspace{0.05cm}.$$
-[[File:P_ID2584__KC_T_2_5_Darstellung.png|right|frame|]] Damit ergibt sich entsprechend der nebenstehenden Tabelle:
+*You get exactly the same result as in subtask&nbsp; '''(4)'''&nbsp; of&nbsp; [[Aufgaben:Exercise_2.07:_Reed-Solomon_Code_(7,_3,_5)_to_Base_8|"Exercise 2.7"]].&nbsp; Correct are the&nbsp; <u>solutions 1 and 2</u>.
-:$$c_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}   \alpha^{4}\cdot 1  + 1 \cdot 1 +  \alpha^{3}\cdot 1 =$$
+*So it is not&nbsp; $c_6 = 0$,&nbsp; but $c_6 = 1$.
-:$$\ = \ \hspace{-0.15cm}
-(110) + (001) + (011)= (100) = \alpha^{2} \hspace{0.05cm},$$
-:$$c_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}   \alpha^{4}\cdot 1 + 1 \cdot \alpha + \alpha^{3}\cdot \alpha^{2}= $$
-:$$\ = \ \hspace{-0.15cm}
- (110)  + (010) + (110) = (011) = \alpha^{3} \hspace{0.05cm},$$
-:$$c_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}    \alpha^{4}\cdot 1 + 1 \cdot \alpha^{2} + \alpha^{3}\cdot \alpha^{4}=$$
-:$$\ = \ \hspace{-0.15cm}
- (110) + (100) + (001) = (011) = \alpha^{3} \hspace{0.05cm},$$
-:$$c_3 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}   \alpha^{4}\cdot 1 + 1 \cdot \alpha^{3} + \alpha^{3}\cdot \alpha^{6}=$$
-:$$\ = \ \hspace{-0.15cm}
- (110) + (011) + (100) = (001) = 1 \hspace{0.05cm},$$
-:$$c_4 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}    \alpha^{4}\cdot 1 + 1 \cdot \alpha^{4} + \alpha^{3}\cdot \alpha^{1}
- = \alpha^{4} \hspace{0.05cm},$$
-:$$c_5 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}    \alpha^{4}\cdot 1 + 1 \cdot \alpha^{5} + \alpha^{3}\cdot \alpha^{3}=$$
-:$$\ = \ \hspace{-0.15cm}
- (110) + (111) + (101) =  (100) = \alpha^{2} \hspace{0.05cm},$$
-:$$c_6 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  \alpha^{4}\cdot 1 + 1 \cdot \alpha^{6} + \alpha^{3}\cdot \alpha^{5}=$$
-:$$\ = \ \hspace{-0.15cm}
- (\alpha^{2} + \alpha) + (\alpha^2 +1) + \alpha = 1 \hspace{0.05cm}.$$
-Man erhält das genau gleiche Ergebnis wie in der Teilaufgabe (4) von [[Aufgaben:2.07_Reed%E2%80%93Solomon%E2%80%93Code_(7,_3,_5)(Base_8)|Aufgabe A2.7]]. Richtig sind die <u>Lösungsvorschläge 1 und 2</u>. Es gilt nicht $c_6 = 0$, sondern $c_6 = 1$.
-'''(3)'''&nbsp; Beim $\rm RSC \, (7, \, 5, \, 3)_8$ ist nun das Informationswort $\underline{u} = (u_0, \, u_1, \, u_2, \, u_3, \, u_4)$ zu berücksichtigen. Mit der Generatormatrix $\mathbf{G}_{\rm D}$ erhält man somit:
+'''(3)'''&nbsp; At the&nbsp; $\rm RSC \, (7, \, 5, \, 3)_8$,&nbsp; the information word&nbsp; $\underline{u} = (u_0, \, u_1, \, u_2, \, u_3, \, u_4)$&nbsp; must be considered.
-:$$\underline {c} = \underline {u} \cdot { \boldsymbol{\rm G}}_{\rm D} =
+*With the generator matrix&nbsp; $\mathbf{G}_{\rm D}$&nbsp; one obtains:
-\begin{pmatrix}
+:$$\underline {c} = \underline {u} \cdot { \boldsymbol{\rm G}}_{\rm D} =\begin{pmatrix}\alpha^4 & 1 & \alpha^3 & 0 & \alpha^6\end{pmatrix} \cdot\begin{pmatrix}1 & 1 & 1 & 1 & 1 & 1 & 1\\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}\hspace{0.05cm}. $$
-		\alpha^4 & 1 & \alpha^3 & 0 & \alpha^6
-\end{pmatrix} \cdot
-\begin{pmatrix}
-& 1 & 1 & 1 & 1 & 1 & 1\\
-& \alpha^1 & \alpha^2 & \alpha^3 & \alpha^4 & \alpha^5 & \alpha^6\\
-& \alpha^2 & \alpha^4 & \alpha^6 & \alpha^1 & \alpha^{3} & \alpha^{5}\\
-& \alpha^3 & \alpha^6 & \alpha^2 & \alpha^5 & \alpha^{1} & \alpha^{4}\\
-& \alpha^4 & \alpha^1 & \alpha^5 & \alpha^2 & \alpha^{6} & \alpha^{3}
-\end{pmatrix}\hspace{0.05cm}. $$
-Daraus folgt:
+*From this it follows:
-:$$c_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}   \alpha^{4}\cdot 1  + 1 \cdot 1 +  \alpha^{3}\cdot 1  + 0 \cdot 1 +  \alpha^{6}\cdot 1=$$
+:$$c_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}   \alpha^{4}\cdot 1  + 1 \cdot 1 +  \alpha^{3}\cdot 1  + 0 \cdot 1 +  \alpha^{6}\cdot 1= (110) + (001) + (011) + (000) +  (101) = (001) = 1 \hspace{0.05cm},$$
-:$$\ = \ \hspace{-0.15cm} (110) + (001) + (011) + (000) +  (101) = (001) = 1 \hspace{0.05cm},$$
 :$$c_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  \left [ \alpha^{4}\cdot 1 + 1 \cdot \alpha + \alpha^{3}\cdot \alpha^{2} \right ] + 0 \cdot \alpha^{3} +  \alpha^{6}\cdot \alpha^{4}=  \left [ \alpha^{3} \right ] + \alpha^{3} = 0 \hspace{0.05cm}.$$
-Hierbei ist berücksichtigt, dass der Klammerausdruck $[ \ ... \ ]$ genau dem Ergebnis $c_1$ der Teilaufgabe (2) entspricht. Entsprechendes wird bei den folgenden Berechnungen ebenfalls berücksichtigt:
+*This takes into account that the bracket expression&nbsp; $[ \ \text{...} \ ]$&nbsp; corresponds exactly to the result&nbsp; $c_1$&nbsp; of subtask&nbsp; '''(2)'''.
-:$$c_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}    \left [ \alpha^{3} \right ] + \alpha^{6}\cdot \alpha^{1}=
-\left [ \alpha^{3} \right ] + \alpha^{7} =
+*Corresponding is also considered in the following calculations:
- (011) + (001) = (010) = \alpha^{1} \hspace{0.05cm},$$
+:$$c_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}    \left [ \alpha^{3} \right ] + \alpha^{6}\cdot \alpha^{1}=\left [ \alpha^{3} \right ] + \alpha^{7} =(011) + (001) = (010) = \alpha^{1} \hspace{0.05cm},$$
-:$$c_3 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}   \left [ 1 \right ] + \alpha^{6}\cdot \alpha^{5}=
+:$$c_3 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}   \left [ 1 \right ] + \alpha^{6}\cdot \alpha^{5}=\left [ 1 \right ] + \alpha^{4}=(001) + (110)  = (111) = \alpha^{5} \hspace{0.05cm},$$
-\left [ 1 \right ] + \alpha^{4}=
+:$$c_4 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}    \left [ \alpha^{4} \right ] + \alpha^{6}\cdot \alpha^{2}=\left [ \alpha^{4} \right ] + \alpha^{1} = (110) + (010) = (100)= \alpha^{2} \hspace{0.05cm},$$
- (001) + (110)  = (111) = \alpha^{5} \hspace{0.05cm},$$
+:$$c_5 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}    \left [ \alpha^{2} \right ] + \alpha^{6}\cdot \alpha^{6}=\left [ \alpha^{2} \right ] + \alpha^{5} =(100) + (111) = (011) = \alpha^{3} \hspace{0.05cm},$$
-:$$c_4 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}    \left [ \alpha^{4} \right ] + \alpha^{6}\cdot \alpha^{2}=
+:$$c_6 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  \left [ 1 \right ] + \alpha^{6}\cdot \alpha^{3}= \left [ 1 \right ] + \alpha^{2}=  (001) + (100) = (101) = \alpha^{6} \hspace{0.05cm}.$$
-\left [ \alpha^{4} \right ] + \alpha^{1} = (110) + (010) = (100)
- = \alpha^{2} \hspace{0.05cm},$$
-:$$c_5 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}    \left [ \alpha^{2} \right ] + \alpha^{6}\cdot \alpha^{6}=
-\left [ \alpha^{2} \right ] + \alpha^{5} =
- (100) + (111) = (011) = \alpha^{3} \hspace{0.05cm},$$
-:$$c_6 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  \left [ 1 \right ] + \alpha^{6}\cdot \alpha^{3}= \left [ 1 \right ] + \alpha^{2}
- =  (001) + (100) = (101) = \alpha^{6} \hspace{0.05cm}.$$
-Das heißt: <u>Alle Lösungsvorschläge</u> sind richtig.
+*This means:&nbsp; <u>All proposed solutions</u>&nbsp; are correct.
 {{ML-Fuß}}
-[[Category:Aufgaben zu  Kanalcodierung|^2.3 Definition und Eigenschaften von Reed–Solomon–Codes^]]
+[[Category:Channel Coding: Exercises|^2.3 Reed–Solomon Codes^]]
+[[de:Aufgaben:Aufgabe 2.08: Generatorpolynome für Reed-Solomon]]

	the matrix $\mathbf{G}_{\rm A}$,
	the matrix $\mathbf{G}_{\rm B}$,
	the matrix $\mathbf{G}_{\rm C}$,
	the matrix $\mathbf{G}_{\rm D}$.

	It holds $c_0 = \alpha^2$,
	It holds $c_1 = \alpha^3$,
	It holds $c_6 = 0$.

	It holds $c_0 = 1$,
	It holds $c_1 = 0$,
	It holds $c_6 = \alpha^6$.