Aufgaben:Exercise 4.7: Product Code Decoding - Revision history

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	[[Category:Channel Coding: Exercises\|^4.2 About the Product Codes^]]		[[Category:Channel Coding: Exercises\|^4.2 About the Product Codes^]]
	[[de:Aufgaben:4.~~7_Produktcode–Decodierung~~]]		[[de:Aufgaben:Aufgabe 4.7: Decodierung von Produktcodes]]

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	* For the second row results the syndrome		* For the second row results the syndrome
	:$$		:$$\underline{s} = \underline{y}_2 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{ \boldsymbol{\rm H}}_1^{\rm T} = \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot \hspace{-0.05cm}\begin{pmatrix}1 &0 &1 \\1 &1 &0 \\0 &1 &1 \\1 &1 &1 \\1 &0 &0 \\0 &1 &0 \\0 &0 &1\end{pmatrix} \hspace{-0.05cm}=$$
	\underline{s} = \underline{y}_2 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{ \boldsymbol{\rm H}}_1^{\rm T} = \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot \hspace{-0.05cm}\begin{pmatrix}1 &0 &1 \\1 &1 &0 \\0 &1 &1 \\1 &1 &1 \\1 &0 &0 \\0 &1 &0 \\0 &0 &1\end{pmatrix} \hspace{-0.05cm}=$$		:$$\Rightarrow \hspace{0.3cm} \underline{s} =\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )+ \left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}1 \right )=\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right )= \underline{s}_1 \hspace{0.03cm}.$$
	:$$		* According to the upper syndrome table on the information page,  the last bit is incorrectly  "corrected".  Incorrect corrections are entered in red in the upper graphic.* Correspondingly applies to the fourth row::$$\underline{s} = \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0 \right ) \cdot { \boldsymbol{\rm H}}_1^{\rm T} =\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )+ \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )=\left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right )= \underline{s}_4 \hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm} \underline{s} =\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )+ \left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}1 \right )=\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right )= \underline{s}_1 \hspace{0.03cm}.$$		* This causes a miscorrection of bit  '''5'''.:* Vertical decoding of columns 1, 3, 4, 5, 6, and 7 is straightforward because there is at most one error per column, which can be corrected by the truncated Hamming code $\rm (6, \ 3, \ 3)$.:* In column 2, however, there is a miscorrection of the last bit according to the lower syndrome table. With the transpose of the $\rm (6, \ 3, \ 3)$ parity-check matrix $\mathbf{H}_2$ results namely::$$\underline{s}= \underline{y}_{2{\rm S}}\cdot { \boldsymbol{\rm H}}_2^{\rm T}= \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot\begin{pmatrix}1 &1 &0 \\1 &0 &1 \\0 &1 &1 \\1 &0 &0 \\0 &1 &0 \\0 &0 &1\end{pmatrix}=\left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right )+ \left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right )=\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right ) = \underline{s}_1.$$
	* According to the upper syndrome table on the information page,  the last bit is incorrectly  "corrected".  Incorrect corrections are entered in red in the upper graphic.* Correspondingly applies to the fourth row::$$
	\underline{s} = \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0 \right ) \cdot { \boldsymbol{\rm H}}_1^{\rm T} =\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )+ \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )=\left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right )= \underline{s}_4 \hspace{0.05cm}.$$
	* This causes a miscorrection of bit  '''5'''.:* Vertical decoding of columns 1, 3, 4, 5, 6, and 7 is straightforward because there is at most one error per column, which can be corrected by the truncated Hamming code $\rm (6, \ 3, \ 3)$.:* In column 2, however, there is a miscorrection of the last bit according to the lower syndrome table. With the transpose of the $\rm (6, \ 3, \ 3)$ parity-check matrix $\mathbf{H}_2$ results namely::$$
	\underline{s}= \underline{y}_{2{\rm S}}\cdot { \boldsymbol{\rm H}}_2^{\rm T}= \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot\begin{pmatrix}1 &1 &0 \\1 &0 &1 \\0 &1 &1 \\1 &0 &0 \\0 &1 &0 \\0 &0 &1\end{pmatrix}=\left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right )+ \left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right )=\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right ) = \underline{s}_1.$$
	* The second horizontal decoding is problem-free, since now at most one error occurs in each row  ⇒  <u>Solution suggestion 3</u>.		* The second horizontal decoding is problem-free, since now at most one error occurs in each row  ⇒  <u>Solution suggestion 3</u>.

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	* For the second row results the syndrome		* For the second row results the syndrome
	:$$\underline{s} = \underline{y}_2 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{ \boldsymbol{\rm H}}_1^{\rm T} = \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot \hspace{-0.05cm}		:$$
	\begin{pmatrix}		\underline{s} = \underline{y}_2 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{ \boldsymbol{\rm H}}_1^{\rm T} = \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot \hspace{-0.05cm}\begin{pmatrix}1 &0 &1 \\1 &1 &0 \\0 &1 &1 \\1 &1 &1 \\1 &0 &0 \\0 &1 &0 \\0 &0 &1\end{pmatrix} \hspace{-0.05cm}=$$
	1 &0 &1 \\		:$$
	1 &1 &0 \\		\Rightarrow \hspace{0.3cm} \underline{s} =\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )+ \left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}1 \right )=\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right )= \underline{s}_1 \hspace{0.03cm}.$$
	0 &1 &1 \\		* According to the upper syndrome table on the information page,  the last bit is incorrectly  "corrected".  Incorrect corrections are entered in red in the upper graphic.* Correspondingly applies to the fourth row::$$
	1 &1 &1 \\		\underline{s} = \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0 \right ) \cdot { \boldsymbol{\rm H}}_1^{\rm T} =\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )+ \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )=\left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right )= \underline{s}_4 \hspace{0.05cm}.$$
	1 &0 &0 \\		* This causes a miscorrection of bit  '''5'''.:* Vertical decoding of columns 1, 3, 4, 5, 6, and 7 is straightforward because there is at most one error per column, which can be corrected by the truncated Hamming code $\rm (6, \ 3, \ 3)$.:* In column 2, however, there is a miscorrection of the last bit according to the lower syndrome table. With the transpose of the $\rm (6, \ 3, \ 3)$ parity-check matrix $\mathbf{H}_2$ results namely::$$
	0 &1 &0 \\		\underline{s}= \underline{y}_{2{\rm S}}\cdot { \boldsymbol{\rm H}}_2^{\rm T}= \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot\begin{pmatrix}1 &1 &0 \\1 &0 &1 \\0 &1 &1 \\1 &0 &0 \\0 &1 &0 \\0 &0 &1\end{pmatrix}=\left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right )+ \left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right )=\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right ) = \underline{s}_1.$$
	0 &0 &1
	\end{pmatrix} \hspace{-0.05cm}=$$:$$\Rightarrow \hspace{0.3cm} \underline{s} =
	\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )
	+ \left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}1 \right )=
	\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right )
	= \underline{s}_1 \hspace{0.03cm}.$$* According to the upper syndrome table on the information page,  the last bit is incorrectly  "corrected".  Incorrect corrections are entered in red in the upper graphic.* Correspondingly applies to the fourth row::$$\underline{s} = \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0 \right ) \cdot { \boldsymbol{\rm H}}_1^{\rm T} =
	\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )
	+ \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )=
	\left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right )= \underline{s}_4 \hspace{0.05cm}.$$* This causes a miscorrection of bit  '''5'''.:* Vertical decoding of columns 1, 3, 4, 5, 6, and 7 is straightforward because there is at most one error per column, which can be corrected by the truncated Hamming code $\rm (6, \ 3, \ 3)$.:* In column 2, however, there is a miscorrection of the last bit according to the lower syndrome table. With the transpose of the $\rm (6, \ 3, \ 3)$ parity-check matrix $\mathbf{H}_2$ results namely::$$\underline{s}= \underline{y}_{2{\rm S}}\cdot { \boldsymbol{\rm H}}_2^{\rm T}
	= \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot
	\begin{pmatrix}
	1 &1 &0 \\
	1 &0 &1 \\
	0 &1 &1 \\
	1 &0 &0 \\
	0 &1 &0 \\
	0 &0 &1
	\end{pmatrix}=
	\left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right )
	+ \left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right )=
	\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right ) = \underline{s}_1.$$
	* The second horizontal decoding is problem-free, since now at most one error occurs in each row  ⇒  <u>Solution suggestion 3</u>.		* The second horizontal decoding is problem-free, since now at most one error occurs in each row  ⇒  <u>Solution suggestion 3</u>.

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 The parity-check matrices of these component codes are:
-:$${ \boldsymbol{\rm H}}_1
+:$${ \boldsymbol{\rm H}}_1=  \begin{pmatrix}1 &1 &0 &1 &1 &0 &0 \\0 &1 &1 &1 &0 &1 &0 \\1 &0 &1 &1 &0 &0 &1\end{pmatrix} \hspace{0.05cm},\hspace{0.8cm}{ \boldsymbol{\rm H}}_2=  \begin{pmatrix}1 &1 &0  &1 &0 &0 \\1 &0 &1  &0 &1 &0 \\0 &1 &1  &0 &0 &1\end{pmatrix} \hspace{0.05cm}.$$For the sake of completeness,&nbsp; the generator matrices are also given,&nbsp; but they are not needed to solve the exercise::$${ \boldsymbol{\rm G}}_1=  \begin{pmatrix}1 &0 &0 &0 &1 &0 &1 \\0 &1 &0 &0 &1 &1 &0 \\0 &0 &1 &0 &0 &1 &1 \\0 &0 &0 &1 &1 &1 &1\end{pmatrix} \hspace{0.05cm},\hspace{0.8cm}{ \boldsymbol{\rm G}}_2=  \begin{pmatrix}1 &0 &0 &1 &1 &0  \\0 &1 &0 &1 &0 &1  \\0 &0 &1 &0 &1 &1\end{pmatrix} \hspace{0.05cm}.$$
 The&nbsp; "Hard Decision Decoding"&nbsp; of this code is preferably done iteratively,&nbsp; by alternately&nbsp; "syndrome decoding"&nbsp; all rows and then all columns.&nbsp; See section&nbsp; [[Channel_Coding/The_Basics_of_Product_Codes#Iterative_syndrome_decoding_of_product_codes| "Iterative Syndrome Decoding of Product Codes"]].

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	\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )		\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )
	+ \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )=		+ \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )=
	\left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right )= \underline{s}_4 \hspace{0.05cm}.$$		\left ( 1, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right )= \underline{s}_4 \hspace{0.05cm}.$$* This causes a miscorrection of bit  '''5'''.:* Vertical decoding of columns 1, 3, 4, 5, 6, and 7 is straightforward because there is at most one error per column, which can be corrected by the truncated Hamming code $\rm (6, \ 3, \ 3)$.:* In column 2, however, there is a miscorrection of the last bit according to the lower syndrome table. With the transpose of the $\rm (6, \ 3, \ 3)$ parity-check matrix $\mathbf{H}_2$ results namely::$$\underline{s}= \underline{y}_{2{\rm S}}\cdot { \boldsymbol{\rm H}}_2^{\rm T}
	* This causes a miscorrection of bit  '''5'''.

	:* Vertical decoding of columns 1, 3, 4, 5, 6, and 7 is straightforward because there is at most one error per column, which can be corrected by the truncated Hamming code $\rm (6, \ 3, \ 3)$.

	:* In column 2, however, there is a miscorrection of the last bit according to the lower syndrome table. With the transpose of the $\rm (6, \ 3, \ 3)$ parity-check matrix $\mathbf{H}_2$ results namely:
	:$$\underline{s}= \underline{y}_{2{\rm S}}\cdot { \boldsymbol{\rm H}}_2^{\rm T}
	= \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot		= \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot
	\begin{pmatrix}		\begin{pmatrix}

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	1 &0 &1 &0 &1 &0 \\		1 &0 &1 &0 &1 &0 \\
	0 &1 &1 &0 &0 &1		0 &1 &1 &0 &0 &1
	\end{pmatrix} \hspace{0.05cm}.$$		\end{pmatrix} \hspace{0.05cm}.$$For the sake of completeness,  the generator matrices are also given,  but they are not needed to solve the exercise::$${ \boldsymbol{\rm G}}_1

	For the sake of completeness,  the generator matrices are also given,  but they are not needed to solve the exercise:
	:$${ \boldsymbol{\rm G}}_1
	= \begin{pmatrix}		= \begin{pmatrix}
	1 &0 &0 &0 &1 &0 &1 \\		1 &0 &0 &0 &1 &0 &1 \\
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	0 &1 &0 \\		0 &1 &0 \\
	0 &0 &1		0 &0 &1
	\end{pmatrix} \hspace{-0.05cm}=$$		\end{pmatrix} \hspace{-0.05cm}=$$:$$\Rightarrow \hspace{0.3cm} \underline{s} =
	:$$\Rightarrow \hspace{0.3cm} \underline{s} =
	\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )		\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )
	+ \left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}1 \right )=		+ \left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}1 \right )=
	\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right )		\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}1 \right )
	= \underline{s}_1 \hspace{0.03cm}.$$		= \underline{s}_1 \hspace{0.03cm}.$$* According to the upper syndrome table on the information page,  the last bit is incorrectly  "corrected".  Incorrect corrections are entered in red in the upper graphic.* Correspondingly applies to the fourth row::$$\underline{s} = \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0 \right ) \cdot { \boldsymbol{\rm H}}_1^{\rm T} =

	* According to the upper syndrome table on the information page,  the last bit is incorrectly  "corrected".  Incorrect corrections are entered in red in the upper graphic.

	* Correspondingly applies to the fourth row:
	:$$\underline{s} = \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}1, \hspace{0.03cm}0 \right ) \cdot { \boldsymbol{\rm H}}_1^{\rm T} =
	\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )		\left ( 1, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )
	+ \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )=		+ \left ( 0, \hspace{0.03cm} 1, \hspace{0.03cm}0 \right )=

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	[[Category:Channel Coding: Exercises\|^4.2 About the Product Codes^]]		[[Category:Channel Coding: Exercises\|^4.2 About the Product Codes^]]
			[[de:Aufgaben:4.7_Produktcode–Decodierung]]

Guenter at 15:29, 13 March 2023

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	#The corrected code word is then  $\underline{y} + \underline{e}_{\mu}$.		#The corrected code word is then  $\underline{y} + \underline{e}_{\mu}$.

	[[File:EN_KC_A_4_7_Zusatz_v2.png\|right\|frame\|Predefined encoder and receiver matrices ~~'''Korrektur''' encoded~~]]		[[File:EN_KC_A_4_7_Zusatz_v2.png\|right\|frame\|Predefined encoder and receiver matrices]]


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	'''(1)'''  The decoding process of the received matrix  $\mathbf{A}$  is illustrated by the following diagram.		'''(1)'''  The decoding process of the received matrix  $\mathbf{A}$  is illustrated by the following diagram.

	[[File:EN_KC_A_4_7a_v2.png\|right\|frame\|Syndrome decoding of the two-dimensional received matrix $\mathbf{A}$ ~~'''Korrektur''' bereits zweite Grafik falsch. Auch Deutscg~~]]		[[File:EN_KC_A_4_7a_v2.png\|right\|frame\|Syndrome decoding of the two-dimensional received matrix $\mathbf{A}$ ]]

	* The single errors in rows  1, 3, 5 and 6  are detected by  $(7, \ 4, \ 3)$  Hamming code and can be corrected <br> ⇒  green markings in the graphic  "1st iteration horizontal".		* The single errors in rows  1, 3, 5 and 6  are detected by  $(7, \ 4, \ 3)$  Hamming code and can be corrected <br> ⇒  green markings in the graphic  "1st iteration horizontal".

Guenter at 11:52, 21 December 2022

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← Older revision		Revision as of 13:52, 21 December 2022
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	#The corrected code word is then  $\underline{y} + \underline{e}_{\mu}$.		#The corrected code word is then  $\underline{y} + \underline{e}_{\mu}$.

	[[File:~~EN_KC_A_4_7_Zusatz~~.png\|right\|frame\|Predefined encoder and receiver matrices '''Korrektur''' encoded]]		[[File:EN_KC_A_4_7_Zusatz_v2.png\|right\|frame\|Predefined encoder and receiver matrices '''Korrektur''' encoded]]

Guenter at 16:26, 19 December 2022

2022-12-19T16:26:20Z

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	{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}		{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}

	[[File:~~EN_KC_A_4_7_v2~~.png\|right\|frame\|Syndrome tables of the considered component codes  $\mathcal{C}_1$  and  $\mathcal{C}_2$]]		[[File:EN_KC_T_4_2_S2b_v2.png\|right\|frame\|Syndrome tables of the considered component codes  $\mathcal{C}_1$  and  $\mathcal{C}_2$]]
	We consider as in the  [[Aufgaben:Exercise_4.6:_Product_Code_Generation\|$\text{Exercise 4.6}$]]  a product code based on		We consider as in the  [[Aufgaben:Exercise_4.6:_Product_Code_Generation\|$\text{Exercise 4.6}$]]  a product code based on
	* the Hamming code  $\rm HC \ (7, \ 4, \ 3)$   ⇒   $\mathcal{C}_1$,		* the Hamming code  $\rm HC \ (7, \ 4, \ 3)$   ⇒   $\mathcal{C}_1$,