Aufgaben:Exercise 2.12Z: Reed-Solomon Syndrome Calculation - Revision history

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	[[Category:Channel Coding: Exercises\|^2.5 Reed-Solomon Error Correction^]]		[[Category:Channel Coding: Exercises\|^2.5 Reed-Solomon Error Correction^]]
	[[de:Aufgaben:2.~~12Z_Reed–Solomon–Syndromberechnung~~]]		[[de:Aufgaben:Aufgabe 2.12Z: Reed–Solomon–Syndromberechnung]]

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 *Given is the parity-check matrix&nbsp; $\mathbf{H}$&nbsp; of the considered code and its transpose:
-:$${ \boldsymbol{\rm H}} =
+:$${ \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}\end{pmatrix} \hspace{0.05cm},\hspace{0.4cm}{ \boldsymbol{\rm H}}^{\rm T} =\begin{pmatrix}1 & 1 & 1 \\\alpha^1 & \alpha^2 & \alpha^3 \\\alpha^2 & \alpha^4 & \alpha^6 \\\alpha^3 & \alpha^6 & \alpha^2 \\\alpha^4 & \alpha^1 & \alpha^{5} \\\alpha^5 & \alpha^{3} & \alpha^{1} \\\alpha^6 & \alpha^{5} & \alpha^{4}\end{pmatrix} \hspace{0.05cm}.$$
@@ Line 72: / Line 57: @@
 [[File:EN_KC_Z_2_5_neu.png|right|frame|Conversion tables for the Galois field&nbsp; $\rm GF(2^3)$]]
 '''(1)'''&nbsp; The corresponding equation for syndrome calculation is:
-:$$\underline {s} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (s_0, s_1, s_2) =
+:$$\underline {s} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (s_0, s_1, s_2) =\begin{pmatrix}\alpha,0, \alpha^3,0, 1, \alpha,0\end{pmatrix}\cdot\begin{pmatrix}1 & 1 & 1 \\\alpha^1 & \alpha^2 & \alpha^3 \\\alpha^2 & \alpha^4 & \alpha^6 \\\alpha^3 & \alpha^6 & \alpha^2 \\\alpha^4 & \alpha^1 & \alpha^{5} \\\alpha^5 & \alpha^{3} & \alpha^{1} \\\alpha^6 & \alpha^{5} & \alpha^{4}\end{pmatrix}\hspace{0.05cm}.$$
 *The first element results in

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	*The first element results in		*The first element results in
	:$$s_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^2 + 1 \cdot \alpha^4 + \alpha \cdot \alpha^5=		:$$s_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^2 + 1 \cdot \alpha^4 + \alpha \cdot \alpha^5=\alpha + \alpha^5 + \alpha^4+ \alpha^6$$
	\alpha + \alpha^5 + \alpha^4+ \alpha^6$$
	:$$\Rightarrow\hspace{0.3cm} s_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (\alpha) + (\alpha^2 + \alpha+ 1)+ (\alpha^2 + \alpha) + + (\alpha^2 + 1) = \alpha^2 + \alpha = \alpha^4\hspace{0.05cm}.$$		:$$\Rightarrow\hspace{0.3cm} s_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (\alpha) + (\alpha^2 + \alpha+ 1)+ (\alpha^2 + \alpha) + + (\alpha^2 + 1) = \alpha^2 + \alpha = \alpha^4\hspace{0.05cm}.$$

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	'''(2)'''  Correspondingly,  for the second syndrome element,  the  <u>proposed solution 2</u>  applies accordingly:		'''(2)'''  Correspondingly,  for the second syndrome element,  the  <u>proposed solution 2</u>  applies accordingly:
	:$$s_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^4 + 1 \cdot \alpha^1 + \alpha \cdot \alpha^3=		:$$s_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^4 + 1 \cdot \alpha^1 + \alpha \cdot \alpha^3=\alpha + \alpha^7 + \alpha+ \alpha^4= 1 + \alpha^4 = \alpha^2 + \alpha + 1 = \alpha^5\hspace{0.05cm}.$$
	\alpha + \alpha^7 + \alpha+ \alpha^4= 1 + \alpha^4 = \alpha^2 + \alpha + 1 = \alpha^5
	\hspace{0.05cm}.$$


	'''(3)'''  To calculate $s_2$,  the received word must be multiplied by the last matrix column:		'''(3)'''  To calculate $s_2$,  the received word must be multiplied by the last matrix column:
	:$$s_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^6 + 1 \cdot \alpha^5 + \alpha \cdot \alpha^1=		:$$s_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^6 + 1 \cdot \alpha^5 + \alpha \cdot \alpha^1=\alpha + \alpha^2 + \alpha^5 + \alpha^2=\alpha^5 + \alpha = (\alpha^2 + \alpha + 1) + \alpha = \alpha^2 + 1 = \alpha^5\hspace{0.05cm}.$$
	\alpha + \alpha^2 + \alpha^5 + \alpha^2=\alpha^5 + \alpha = (\alpha^2 + \alpha + 1) + \alpha = \alpha^2 + 1 = \alpha^5
	\hspace{0.05cm}.$$

	*Correct is the  <u>proposed solution 3</u>.		*Correct is the  <u>proposed solution 3</u>.

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	[[Category:Channel Coding: Exercises\|^2.5 Reed-Solomon Error Correction^]]		[[Category:Channel Coding: Exercises\|^2.5 Reed-Solomon Error Correction^]]
			[[de:Aufgaben:2.12Z_Reed–Solomon–Syndromberechnung]]

Hwang at 15:30, 23 January 2023

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	Given is the received word   $\underline{y} = (\alpha, \, 0, \, \alpha^3, \, 0, \, 1, \, \alpha, \, 0)$.		Given is the received word   $\underline{y} = (\alpha, \, 0, \, \alpha^3, \, 0, \, 1, \, \alpha, \, 0)$.

	*Based on the syndrome   $\underline {s} = (s_0, s_1, s_2) = \underline {y} \cdot { \boldsymbol{\rm H }}^{\rm T}$   it is to check whether individual symbols of the received vector  $\underline{y}$  were ~~corrupted~~ during transmission.		*Based on the syndrome   $\underline {s} = (s_0, s_1, s_2) = \underline {y} \cdot { \boldsymbol{\rm H }}^{\rm T}$   it is to check whether individual symbols of the received vector  $\underline{y}$  were falsified during transmission.

	*Given is the parity-check matrix  $\mathbf{H}$  of the considered code and its transpose:		*Given is the parity-check matrix  $\mathbf{H}$  of the considered code and its transpose:

Guenter at 16:18, 30 October 2022

2022-10-30T16:18:02Z

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	* Since the received word can actually be decoded according to the specification   ⇒   $\underline{r = 1}$.		* Since the received word can actually be decoded according to the specification   ⇒   $\underline{r = 1}$.

			*Without this specification  "The received word can be decoded",  this subtask would not be solvable.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 15:53, 30 October 2022

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:~~P_ID2560__KC_T_2_5_Darstellung~~.png\|right\|frame\|Conversion tables for the Galois field $\rm GF(2^3)$]]		[[File:EN_KC_Z_2_5_neu.png\|right\|frame\|Conversion tables for the Galois field  $\rm GF(2^3)$]]
	'''(1)'''  The corresponding equation for syndrome calculation is:		'''(1)'''  The corresponding equation for syndrome calculation is:
	:$$\underline {s} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (s_0, s_1, s_2) =		:$$\underline {s} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (s_0, s_1, s_2) =
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	The first element results in		*The first element results in
	:$$s_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^2 + 1 \cdot \alpha^4 + \alpha \cdot \alpha^5=		:$$s_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^2 + 1 \cdot \alpha^4 + \alpha \cdot \alpha^5=
	\alpha + \alpha^5 + \alpha^4+ \alpha^6$$		\alpha + \alpha^5 + \alpha^4+ \alpha^6$$
	:$$\Rightarrow\hspace{0.3cm} s_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (\alpha) + (\alpha^2 + \alpha+ 1)+ (\alpha^2 + \alpha) + + (\alpha^2 + 1) = \alpha^2 + \alpha = \alpha^4\hspace{0.05cm}.$$		:$$\Rightarrow\hspace{0.3cm} s_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (\alpha) + (\alpha^2 + \alpha+ 1)+ (\alpha^2 + \alpha) + + (\alpha^2 + 1) = \alpha^2 + \alpha = \alpha^4\hspace{0.05cm}.$$

	Correct is the <u>proposed solution 1</u>.		*Correct is the  <u>proposed solution 1</u>.


	'''(2)'''  Correspondingly, for the second syndrome element, the <u>proposed solution 2</u> applies accordingly:		'''(2)'''  Correspondingly,  for the second syndrome element,  the  <u>proposed solution 2</u>  applies accordingly:
	:$$s_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^4 + 1 \cdot \alpha^1 + \alpha \cdot \alpha^3=		:$$s_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^4 + 1 \cdot \alpha^1 + \alpha \cdot \alpha^3=
	\alpha + \alpha^7 + \alpha+ \alpha^4= 1 + \alpha^4 = \alpha^2 + \alpha + 1 = \alpha^5		\alpha + \alpha^7 + \alpha+ \alpha^4= 1 + \alpha^4 = \alpha^2 + \alpha + 1 = \alpha^5
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	'''(3)'''  To calculate $s_2$ must be multiplied by the last matrix column:		'''(3)'''  To calculate $s_2$,  the received word must be multiplied by the last matrix column:
	:$$s_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^6 + 1 \cdot \alpha^5 + \alpha \cdot \alpha^1=		:$$s_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^6 + 1 \cdot \alpha^5 + \alpha \cdot \alpha^1=
	\alpha + \alpha^2 + \alpha^5 + \alpha^2=\alpha^5 + \alpha = (\alpha^2 + \alpha + 1) + \alpha = \alpha^2 + 1 = \alpha^5		\alpha + \alpha^2 + \alpha^5 + \alpha^2=\alpha^5 + \alpha = (\alpha^2 + \alpha + 1) + \alpha = \alpha^2 + 1 = \alpha^5
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	Correct is the <u>proposed solution 3</u>.		*Correct is the  <u>proposed solution 3</u>.


	'''(4)'''  Due to the calculated syndrome $\underline{s} = (\alpha^4, \, \alpha^5, \, \alpha^6) ≠ 0$ the received word contains at least one symbol error   ⇒   $r > 0$.		'''(4)'''  Due to the calculated syndrome   $\underline{s} = (\alpha^4, \, \alpha^5, \, \alpha^6) ≠ 0$,  the received word contains at least one symbol error   ⇒   $r > 0$.
	*The present Reed–Solomon–code $(7, \, 4, \, 4)_8 \ \Rightarrow \ d_{\rm min} = 4$ cannot correct more than $t = &lfloor;d_{\rm min}/2&rfloor; = 1$ errors.		*The present Reed–Solomon–code  $(7, \, 4, \, 4)_8 \ \Rightarrow \ d_{\rm min} = 4$  cannot correct more than  $t = &lfloor;d_{\rm min}/2&rfloor; = 1$  errors.
	* Since the received word can actually be decoded according to the specification,  $\underline{r = 1}$ ~~holds~~.
			* Since the received word can actually be decoded according to the specification   ⇒   $\underline{r = 1}$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 15:43, 30 October 2022

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	# coefficient vector representation.		# coefficient vector representation.

	* Given the received word   $\underline{y} = (\alpha, \, 0, \, \alpha^3, \, 0, \, 1, \, \alpha, \, 0)$.
			Given is the received word   $\underline{y} = (\alpha, \, 0, \, \alpha^3, \, 0, \, 1, \, \alpha, \, 0)$.

	*Based on the syndrome   $\underline {s} = (s_0, s_1, s_2) = \underline {y} \cdot { \boldsymbol{\rm H }}^{\rm T}$   it is to check whether individual symbols of the received vector  $\underline{y}$  were corrupted during transmission.		*Based on the syndrome   $\underline {s} = (s_0, s_1, s_2) = \underline {y} \cdot { \boldsymbol{\rm H }}^{\rm T}$   it is to check whether individual symbols of the received vector  $\underline{y}$  were corrupted during transmission.
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	Hints:  This exercise refers to the section  [[Channel_Coding/Error_Correction_According_to_Reed-Solomon_Coding#Step_.28A.29:_Evaluation_of_the_syndrome_in_BDD\| "Step (A): Evaluation of the syndrome in BDD"]]  of the chapter  "Error ~~coding~~ according to Reed–Solomon ~~coding~~".		Hints:  This exercise refers to the section  [[Channel_Coding/Error_Correction_According_to_Reed-Solomon_Coding#Step_.28A.29:_Evaluation_of_the_syndrome_in_BDD\| "Step (A): Evaluation of the syndrome in BDD"]]  of the chapter  "Error Correction according to Reed–Solomon Coding".


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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{~~Received was~~  $\underline{y} = (\alpha, \, 0, \, \alpha^3, \, 0, \, 1, \, \alpha, \, 0)$. Specify the first element of the syndrome  $\underline{s} = (s_0, \, s_1, \, s_2)$~~ ~~.		{It holds for the weceived word:   $\underline{y} = (\alpha, \, 0, \, \alpha^3, \, 0, \, 1, \, \alpha, \, 0)$. Specify the first element of the syndrome  $\underline{s} = (s_0, \, s_1, \, s_2)$.
	\|type="()"}		\|type="()"}
	+ $s_0 = \alpha^4$,		+ $s_0 = \alpha^4$,
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	- $s_2 = 0, \, 1, \, \alpha, \, \alpha^2$ oder $\alpha^3$.		- $s_2 = 0, \, 1, \, \alpha, \, \alpha^2$ oder $\alpha^3$.

	{~~Know~~ that the received word ~~at hand~~  $\underline{y}$  can be decoded correctly. How many symbol errors does the received word contain?		{Known is that the received word   $\underline{y}$   can be decoded correctly.  How many symbol errors does the received word contain?
	\|type="{}"}		\|type="{}"}
	$r \ = \ ${ 1 3% }		$r \ = \ ${ 1 3% }

Guenter at 15:35, 30 October 2022

2022-10-30T15:35:50Z

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	{{quiz-Header\|Buchseite=Channel_Coding/Error_Correction_According_to_Reed-Solomon_Coding}}		{{quiz-Header\|Buchseite=Channel_Coding/Error_Correction_According_to_Reed-Solomon_Coding}}

	[[File:~~P_ID2559__KC_T_2_5_Darstellung~~.png\|right\|frame\|~~Conversion table for the Galois field ~~ $\rm GF(2^3)$]]		[[File:EN_KC_Z_2_5_neu.png\|right\|frame\|$\rm GF(2^3)$  representation as powers, polynomials, vectors]]
	As in the  [[Aufgaben:Exercise_2.12:_Decoding_at_RSC_(7,_4,_4)_to_Base_8\|"Exercise 2. 12"]]  we consider the Reed–Solomon code  $(7, \, 4, \, 4)_8$ based on the Galois field  ${\rm GF}(q)$  with  $q = 8 = 2^3$ . The graph shows the corresponding conversion table.		As in the  [[Aufgaben:Exercise_2.12:_Decoding_at_RSC_(7,_4,_4)_to_Base_8\|"Exercise 2. 12"]]  we consider the Reed–Solomon code  $(7, \, 4, \, 4)_8$  based on the Galois field  ${\rm GF}(q)$  with   $q = 8 = 2^3$.   The graph shows the corresponding conversion table.

	Given are the possible code symbols in exponent representation $($~~potencies~~ of  $\alpha)$  ~~and in polynomial– and~~ coefficient vector representation.		Given are the possible code symbols in
			# exponent representation  $($powers of  $\alpha)$
			# polynomialrepresentation,
			# coefficient vector representation.

	Given the received word  $\underline{y} = (\alpha, \, 0, \, \alpha^3, \, 0, \, 1, \, \alpha, \, 0)$. ~~Based on the syndrome~~		* Given the received word   $\underline{y} = (\alpha, \, 0, \, \alpha^3, \, 0, \, 1, \, \alpha, \, 0)$.
	~~:$$\underline {s} = (s_0, s_1, s_2) = \underline {y} \cdot { \boldsymbol{\rm H }}^{\rm T}$$~~

	is to check whether individual symbols of the received vector  $\underline{y}$  were corrupted during transmission. Given is the parity-check matrix  $\mathbf{H}$  of the considered code and its transpose:		*Based on the syndrome   $\underline {s} = (s_0, s_1, s_2) = \underline {y} \cdot { \boldsymbol{\rm H }}^{\rm T}$   it is to check whether individual symbols of the received vector  $\underline{y}$  were corrupted during transmission.

			*Given is the parity-check matrix  $\mathbf{H}$  of the considered code and its transpose:
	:$${ \boldsymbol{\rm H}} =		:$${ \boldsymbol{\rm H}} =
	\begin{pmatrix}		\begin{pmatrix}
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			Hints:  This exercise refers to the section  [[Channel_Coding/Error_Correction_According_to_Reed-Solomon_Coding#Step_.28A.29:_Evaluation_of_the_syndrome_in_BDD\| "Step (A): Evaluation of the syndrome in BDD"]]  of the chapter  "Error coding according to Reed–Solomon coding".




	Hints:
	* This exercise refers to the ~~page~~  [[Channel_Coding/Error_Correction_According_to_Reed-Solomon_Coding#Step_.28A.29:_Evaluation_of_the_syndrome_in_BDD\| "Step (A): Evaluation of the syndrome in BDD"]]  of the chapter "Error coding according to Reed–Solomon coding".

Noah at 16:58, 12 September 2022

2022-09-12T16:58:54Z

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	{{quiz-Header\|Buchseite=~~Kanalcodierung~~/~~Fehlerkorrektur nach Reed–Solomon–Codierung~~}}		{{quiz-Header\|Buchseite=Channel_Coding/Error_Correction_According_to_Reed-Solomon_Coding}}

	[[File:P_ID2559__KC_T_2_5_Darstellung.png\|right\|frame\|Conversion table for the Galois field  $\rm GF(2^3)$]]		[[File:P_ID2559__KC_T_2_5_Darstellung.png\|right\|frame\|Conversion table for the Galois field  $\rm GF(2^3)$]]
	~~Wie~~ in ~~der~~  [[Aufgaben:Exercise_2.12:_Decoding_at_RSC_(7,_4,_4)_to_Base_8\|"~~Aufgabe~~ 2.12"]]  ~~betrachten wir den~~ Reed–Solomon~~–Code~~  $(7, \, 4, \, 4)_8$~~, der auf dem Galoisfeld~~  ${\rm GF}(q)$  ~~mit~~  $q = 8 = 2^3$  ~~basiert~~. ~~Die Grafik zeigt die zugehörige Umrechnungstabelle~~.		As in the  [[Aufgaben:Exercise_2.12:_Decoding_at_RSC_(7,_4,_4)_to_Base_8\|"Exercise 2. 12"]]  we consider the Reed–Solomon code  $(7, \, 4, \, 4)_8$ based on the Galois field  ${\rm GF}(q)$  with  $q = 8 = 2^3$ . The graph shows the corresponding conversion table.

	~~Gegeben sind die möglichen Codesymbole~~ in ~~Exponentendarstellung~~ $($~~Potenzen von~~  $\alpha)$  ~~sowie~~ in ~~Polynom~~– ~~und Koeffizientenvektordarstellung~~.		Given are the possible code symbols in exponent representation $($potencies of  $\alpha)$  and in polynomial– and coefficient vector representation.

	~~Vorgegeben ist das Empfangswort~~  $\underline{y} = (\alpha, \, 0, \, \alpha^3, \, 0, \, 1, \, \alpha, \, 0)$. ~~Anhand des Syndroms~~		Given the received word  $\underline{y} = (\alpha, \, 0, \, \alpha^3, \, 0, \, 1, \, \alpha, \, 0)$. Based on the syndrome
	:$$\underline {s} = (s_0, s_1, s_2) = \underline {y} \cdot { \boldsymbol{\rm H }}^{\rm T}$$		:$$\underline {s} = (s_0, s_1, s_2) = \underline {y} \cdot { \boldsymbol{\rm H }}^{\rm T}$$

	~~soll überprüft werden, ob einzelne Symbole des Empfangsvektors~~  $\underline{y}$  ~~bei der Übertragung verfälscht wurden~~. ~~Gegeben ist hierzu die Prüfmatrix~~  $\mathbf{H}$  ~~des betrachteten Codes und deren Transponierte~~:		is to check whether individual symbols of the received vector  $\underline{y}$  were corrupted during transmission. Given is the parity-check matrix  $\mathbf{H}$  of the considered code and its transpose:
	:$${ \boldsymbol{\rm H}} =		:$${ \boldsymbol{\rm H}} =
	\begin{pmatrix}		\begin{pmatrix}
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	~~''Hinweis~~:''		Hints:
	* ~~Die Aufgabe bezieht sich auf die Seite~~  [[Channel_Coding/~~Fehlerkorrektur_nach_Reed%E2%80%93Solomon%E2%80%93Codierung~~#~~Schritt_~~.28A.29:~~_Auswertung_des_Syndroms_beim_BDD~~\| ~~Schritt  $\rm~~ (A)$: ~~Auswertung des Syndroms beim~~ BDD]]  ~~des Kapitels~~ "~~Fehlercodierung nach~~ Reed–Solomon~~–Codierung~~".		* This exercise refers to the page  [[Channel_Coding/Error_Correction_According_to_Reed-Solomon_Coding#Step_.28A.29:_Evaluation_of_the_syndrome_in_BDD\| "Step (A): Evaluation of the syndrome in BDD"]]  of the chapter "Error coding according to Reed–Solomon coding".


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	===~~Fragebogen~~===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{~~Empfangen wurde~~  $\underline{y} = (\alpha, \, 0, \, \alpha^3, \, 0, \, 1, \, \alpha, \, 0)$. ~~Geben Sie das erste Element des Syndroms~~  $\underline{s} = (s_0, \, s_1, \, s_2)$  an.		{Received was  $\underline{y} = (\alpha, \, 0, \, \alpha^3, \, 0, \, 1, \, \alpha, \, 0)$. Specify the first element of the syndrome  $\underline{s} = (s_0, \, s_1, \, s_2)$ .
	\|type="()"}		\|type="()"}
	+ $s_0 = \alpha^4$,		+ $s_0 = \alpha^4$,
	- $s_0 = \alpha^5$,		- $s_0 = \alpha^5$,
	- $s_0 = \alpha^6$,		- $s_0 = \alpha^6$,
	- $s_0 = 0, \, 1, \, \alpha, \, \alpha^2$  ~~oder~~  $\alpha^3$.		- $s_0 = 0, \, 1, \, \alpha, \, \alpha^2$  or  $\alpha^3$.

	{~~Wie lautet bei gleichem Empfangswort das zweite Syndromelement~~?		{What is the second syndrome element for the same received word?
	\|type="()"}		\|type="()"}
	- $s_1 = \alpha^4$,		- $s_1 = \alpha^4$,
	+ $s_1 = \alpha^5$,		+ $s_1 = \alpha^5$,
	- $s_1 = \alpha^6$,		- $s_1 = \alpha^6$,
	- $s_1 = 0, \, 1, \, \alpha, \, \alpha^2$  ~~oder~~  $\alpha^3$.		- $s_1 = 0, \, 1, \, \alpha, \, \alpha^2$  or  $\alpha^3$.

	{~~Wie lautet bei gleichem Empfangswort das dritte Syndromelement~~?		{What is the third syndrome element for the same received word?
	\|type="()"}		\|type="()"}
	- $s_2 = \alpha^4$,		- $s_2 = \alpha^4$,
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	- $s_2 = 0, \, 1, \, \alpha, \, \alpha^2$ oder $\alpha^3$.		- $s_2 = 0, \, 1, \, \alpha, \, \alpha^2$ oder $\alpha^3$.

	{~~Bekannt ist, dass das vorliegende Empfangswort~~  $\underline{y}$  ~~richtig decodiert werden kann~~. ~~Wieviele Symbolfehler beinhaltet das Empfangswort~~?		{Know that the received word at hand  $\underline{y}$  can be decoded correctly. How many symbol errors does the received word contain?
	\|type="{}"}		\|type="{}"}
	$r \ = \ ${ 1 3% }		$r \ = \ ${ 1 3% }
	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
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	[[File:P_ID2560__KC_T_2_5_Darstellung.png\|right\|frame\|~~Umrechnungstabellen für das Galoisfeld~~ $\rm GF(2^3)$]]		[[File:P_ID2560__KC_T_2_5_Darstellung.png\|right\|frame\|Conversion tables for the Galois field $\rm GF(2^3)$]]
	'''(1)'''  ~~Die entsprechende Gleichung zur Syndromberechnung lautet~~:		'''(1)'''  The corresponding equation for syndrome calculation is:
	:$$\underline {s} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (s_0, s_1, s_2) =		:$$\underline {s} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (s_0, s_1, s_2) =
	\begin{pmatrix}		\begin{pmatrix}
Line 88:		Line 88:
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	~~Das erste Element ergibt sich zu~~		The first element results in
	:$$s_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^2 + 1 \cdot \alpha^4 + \alpha \cdot \alpha^5=		:$$s_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^2 + 1 \cdot \alpha^4 + \alpha \cdot \alpha^5=
	\alpha + \alpha^5 + \alpha^4+ \alpha^6$$		\alpha + \alpha^5 + \alpha^4+ \alpha^6$$
	:$$\Rightarrow\hspace{0.3cm} s_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (\alpha) + (\alpha^2 + \alpha+ 1)+ (\alpha^2 + \alpha) + + (\alpha^2 + 1) = \alpha^2 + \alpha = \alpha^4\hspace{0.05cm}.$$		:$$\Rightarrow\hspace{0.3cm} s_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (\alpha) + (\alpha^2 + \alpha+ 1)+ (\alpha^2 + \alpha) + + (\alpha^2 + 1) = \alpha^2 + \alpha = \alpha^4\hspace{0.05cm}.$$

	~~Richtig ist der~~ <u>~~Lösungsvorschlag~~ 1</u>.		Correct is the <u>proposed solution 1</u>.


	'''(2)'''  ~~Entsprechend gilt für das zweite Syndromelement entsprechend dem der~~ <u>~~Lösungsvorschlag~~ 2</u>:		'''(2)'''  Correspondingly, for the second syndrome element, the <u>proposed solution 2</u> applies accordingly:
	:$$s_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^4 + 1 \cdot \alpha^1 + \alpha \cdot \alpha^3=		:$$s_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^4 + 1 \cdot \alpha^1 + \alpha \cdot \alpha^3=
	\alpha + \alpha^7 + \alpha+ \alpha^4= 1 + \alpha^4 = \alpha^2 + \alpha + 1 = \alpha^5		\alpha + \alpha^7 + \alpha+ \alpha^4= 1 + \alpha^4 = \alpha^2 + \alpha + 1 = \alpha^5
Line 102:		Line 102:


	'''(3)'''  ~~Zur Berechnung von~~ $s_2$ ~~muss mit der letzten Matrixspalte multipliziert werden~~:		'''(3)'''  To calculate $s_2$ must be multiplied by the last matrix column:
	:$$s_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^6 + 1 \cdot \alpha^5 + \alpha \cdot \alpha^1=		:$$s_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot 1 + \alpha^3 \cdot \alpha^6 + 1 \cdot \alpha^5 + \alpha \cdot \alpha^1=
	\alpha + \alpha^2 + \alpha^5 + \alpha^2=\alpha^5 + \alpha = (\alpha^2 + \alpha + 1) + \alpha = \alpha^2 + 1 = \alpha^5		\alpha + \alpha^2 + \alpha^5 + \alpha^2=\alpha^5 + \alpha = (\alpha^2 + \alpha + 1) + \alpha = \alpha^2 + 1 = \alpha^5
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	~~Richtig ist der~~ <u>~~Lösungsvorschlag~~ 3</u>.		Correct is the <u>proposed solution 3</u>.


	'''(4)'''  ~~Aufgrund des errechneten Syndroms~~ $\underline{s} = (\alpha^4, \, \alpha^5, \, \alpha^6) ≠ 0$ ~~beinhaltet das Empfangswort mindestens einen Symbolfehler~~   ⇒   $r > 0$.		'''(4)'''  Due to the calculated syndrome $\underline{s} = (\alpha^4, \, \alpha^5, \, \alpha^6) ≠ 0$ the received word contains at least one symbol error   ⇒   $r > 0$.
	*~~Der vorliegende~~ Reed–Solomon–~~Code~~ $(7, \, 4, \, 4)_8 \ \Rightarrow \ d_{\rm min} = 4$ ~~kann nicht mehr als~~ $t = &lfloor;d_{\rm min}/2&rfloor; = 1$ ~~Fehler korrigieren~~.		*The present Reed–Solomon–code $(7, \, 4, \, 4)_8 \ \Rightarrow \ d_{\rm min} = 4$ cannot correct more than $t = &lfloor;d_{\rm min}/2&rfloor; = 1$ errors.
	* ~~Da das Empfangswort gemäß der Angabe tatsächlich decodiert werden kann~~, ~~gilt~~  $\underline{r = 1}$.		* Since the received word can actually be decoded according to the specification,  $\underline{r = 1}$ holds.
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