Aufgaben:Exercise 1.11Z: Syndrome Decoding again - Revision history

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	[[Category:Channel Coding: Exercises\|^1.5 Linear Block Code Decoding^]]		[[Category:Channel Coding: Exercises\|^1.5 Linear Block Code Decoding^]]
	[[de:Aufgaben:1.~~11Z_Nochmals_Syndromdecodierung~~]]		[[de:Aufgaben:Aufgabe 1.11Z: Nochmals Syndromdecodierung]]

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	[[Category:Channel Coding: Exercises\|^1.5 Linear Block Code Decoding^]]		[[Category:Channel Coding: Exercises\|^1.5 Linear Block Code Decoding^]]
			[[de:Aufgaben:1.11Z_Nochmals_Syndromdecodierung]]

Guenter at 11:24, 20 July 2022

2022-07-20T11:24:26Z

← Older revision		Revision as of 13:24, 20 July 2022
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The answer is <u>YES</u>, as can be seen from the given parity-check matrix $\mathbf{H}$:		'''(1)'''  The answer is  <u>YES</u>,  as can be seen from the given parity-check matrix  $\mathbf{H}$:
	*This contains a $3×3$ diagonal matrix at the end.		*This contains a  $3×3$  diagonal matrix at the end.
	*The code words are therefore:		*The code words are therefore:

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	'''(2)'''  With this received vector $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$ all parity-check equations are satisfied:		'''(2)'''  With this received vector  $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$,  all parity-check equations are satisfied:

	:$$u_1 \oplus u_2 \oplus u_4 \oplus p_1 = 1 \oplus 0 \oplus 1 \oplus 0 = 0 \hspace{0.05cm},$$		:$$u_1 \oplus u_2 \oplus u_4 \oplus p_1 = 1 \oplus 0 \oplus 1 \oplus 0 = 0 \hspace{0.05cm},$$
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	:$$u_1 \oplus u_3 \oplus u_4 \oplus p_3 = 1 \oplus 0 \oplus 1 \oplus 0 = 0 \hspace{0.05cm}.$$		:$$u_1 \oplus u_3 \oplus u_4 \oplus p_3 = 1 \oplus 0 \oplus 1 \oplus 0 = 0 \hspace{0.05cm}.$$

	Accordingly, the correct answer is <u>YES</u>.		Accordingly,  the correct answer is  <u>YES</u>.


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	'''(3)'''  It holds $\underline{s} = \underline{y} · \boldsymbol{\rm H}^{\rm T}$:		'''(3)'''  It holds $\underline{s} = \underline{y} · \boldsymbol{\rm H}^{\rm T}$:

	:$$ \underline{s} = \begin{pmatrix} 1 &0 &0 &1 &0 &1 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 &0 &1\\ 1 &1 &0\\ 0 &1 &1\\ 1 &1 &1\\ 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} = \begin{pmatrix} 0 &0 &0 \end{pmatrix} = \underline{s}_0 \hspace{0.2cm} \Rightarrow\hspace{0.2cm} \hspace{0.15cm} \underline{ \rm ~~Antwort~~ \hspace{0.15cm}1} \hspace{0.05cm}.$$		:$$ \underline{s} = \begin{pmatrix} 1 &0 &0 &1 &0 &1 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 &0 &1\\ 1 &1 &0\\ 0 &1 &1\\ 1 &1 &1\\ 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} = \begin{pmatrix} 0 &0 &0 \end{pmatrix} = \underline{s}_0 \hspace{0.2cm} \Rightarrow\hspace{0.2cm} \hspace{0.15cm} \underline{ \rm Answer \hspace{0.15cm}1} \hspace{0.05cm}.$$



	'''(4)'''  One could now for each $\underline{y}$ check the equation $\underline{y} · \boldsymbol{\rm H}^{\rm T} = (0, 0, 0)$. Now here the result shall be obtained in a different way:		'''(4)'''  One could now for each  $\underline{y}$  check the equation  $\underline{y} · \boldsymbol{\rm H}^{\rm T} = (0, 0, 0)$.  Now here the result shall be obtained in a different way:

	*$\underline{y}= (1, 1, 0, 1, 0, 1, 0)$ differs from $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$ in bit $u_{2}$, which is used only in the first two parity-check equations, but not in the last ⇒ $\underline{s} = \underline{s}_{6} = (1, 1, 0)$.		*$\underline{y}= (1, 1, 0, 1, 0, 1, 0)$  differs from  $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$  in bit  $u_{2}$,  which is used only in the first two parity-check equations,  but not in the last ⇒ $\underline{s} = \underline{s}_{6} = (1, 1, 0)$.

	*Applying the parity-check equations to $\underline{y} = (0, 1, 0, 1, 0, 0, 1)$, we obtain $\underline{s} = \underline{s}_{0} = (0, 0, 0)$, as evidenced by the following calculation:		*Applying the parity-check equations to  $\underline{y} = (0, 1, 0, 1, 0, 0, 1)$,  we obtain $\underline{s} = \underline{s}_{0} = (0, 0, 0)$, as evidenced by the following calculation:

	:$$u_1 \oplus u_2 \oplus u_4 \oplus p_1 = 0 \oplus 1 \oplus 1 \oplus 0 = 0 \hspace{0.05cm},$$		:$$u_1 \oplus u_2 \oplus u_4 \oplus p_1 = 0 \oplus 1 \oplus 1 \oplus 0 = 0 \hspace{0.05cm},$$
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	:$$u_1 \oplus u_3 \oplus u_4 \oplus p_3 = 0 \oplus 0 \oplus 1 \oplus 1 = 0 \hspace{0.05cm}.$$		:$$u_1 \oplus u_3 \oplus u_4 \oplus p_3 = 0 \oplus 0 \oplus 1 \oplus 1 = 0 \hspace{0.05cm}.$$

	*The same result is obtained with the received vector $\underline{y} = (0, 1, 1, 0, 1, 0, 1),$ which differs from the vector $(1, 0, 0, 1, 0, 1, 0)$ in all seven bit positions:		*The same result is obtained with the received vector  $\underline{y} = (0, 1, 1, 0, 1, 0, 1),$  which differs from the vector  $(1, 0, 0, 1, 0, 1, 0)$  in all seven bit positions:

	:$$u_1 \oplus u_2 \oplus u_4 \oplus p_1 = 0 \oplus 1 \oplus 0 \oplus 1 = 0 \hspace{0.05cm},$$		:$$u_1 \oplus u_2 \oplus u_4 \oplus p_1 = 0 \oplus 1 \oplus 0 \oplus 1 = 0 \hspace{0.05cm},$$
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	:$$u_1 \oplus u_3 \oplus u_4 \oplus p_3 = 0 \oplus 1 \oplus 0 \oplus 1 = 0 \hspace{0.05cm}.$$		:$$u_1 \oplus u_3 \oplus u_4 \oplus p_3 = 0 \oplus 1 \oplus 0 \oplus 1 = 0 \hspace{0.05cm}.$$

	So the correct answers are <u>answers 2 and 3</u>.		So the correct answers are  <u>answers 2 and 3</u>.

	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 11:02, 20 July 2022

2022-07-20T11:02:01Z

← Older revision		Revision as of 13:02, 20 July 2022
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	- No.		- No.

	{~~Received~~  $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$. Is this a valid ~~codeword~~?		{With received vector  $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$.  Is this a valid code word?
	\|type="()"}		\|type="()"}
	+ Yes,		+ Yes,
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	- $\underline{s} = \underline{s}_{7} = (1, 1, 1).$		- $\underline{s} = \underline{s}_{7} = (1, 1, 1).$

	{Which received words lead to the same syndrome as in subtask '''(3)'''?		{Which received words lead to the same syndrome as in subtask  '''(3)'''?
	\|type="[]"}		\|type="[]"}
	- $\underline{y} = (1, 1, 0, 1, 0, 1, 0),$		- $\underline{y} = (1, 1, 0, 1, 0, 1, 0),$

Guenter at 15:39, 19 July 2022

2022-07-19T15:39:43Z

← Older revision		Revision as of 17:39, 19 July 2022
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	{{quiz-Header\|Buchseite=Channel_Coding/Decoding_of_Linear_Block_Codes}}		{{quiz-Header\|Buchseite=Channel_Coding/Decoding_of_Linear_Block_Codes}}

	[[File:P_ID2399__KC_Z_1_10.png\|right\|frame\|~~Diagram:~~ Parity-check ~~equations~~]]		[[File:P_ID2399__KC_Z_1_10.png\|right\|frame\|Parity-check chart]]

	The same constellation is considered as in  [[Aufgaben:Exercise_1.11:_Syndrome_Decoding\|"Exercise 1.11"]]~~, namely the~~ decoding of a  $(7, 4, 3)$ Hamming code with the parity-check matrix		The same constellation is considered as in  [[Aufgaben:Exercise_1.11:_Syndrome_Decoding\|"Exercise 1.11"]]:  The decoding of a  $(7, 4, 3)$  Hamming code with the parity-check matrix

	:$${ \boldsymbol{\rm H}}_{\rm } = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$		:$${ \boldsymbol{\rm H}}_{\rm } = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$

	Accordingly, the generator ~~polynomial~~ is:		Accordingly,  the generator matrix is:

	:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$		:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$
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	:$$\underline{s} = \underline{y} \cdot { \boldsymbol{\rm H}}^{\rm T} \in {\rm GF}(2^m) \hspace{0.05cm}.$$		:$$\underline{s} = \underline{y} \cdot { \boldsymbol{\rm H}}^{\rm T} \in {\rm GF}(2^m) \hspace{0.05cm}.$$

	With this result, any single error in the ~~codeword~~ can be corrected in the Hamming code under consideration.		With this result,  any single error in the code word can be corrected in the Hamming code under consideration.
	*In the error-free case  $\underline{s} = \underline{s}_{0} = (0, 0, 0)$.		*In the error-free case  $\underline{s} = \underline{s}_{0} = (0, 0, 0)$.
	*But even three transmission errors may result in  $\underline{s}_{0} = (0, 0, 0)$ , so these errors remain undetected.
			*But even three transmission errors may result in  $\underline{s}_{0} = (0, 0, 0)$,  so these errors remain undetected.


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	Hints:		Hints:
	* The exercise belongs to the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\|"Decoding of Linear Block Codes"]].		* The exercise belongs to the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\|"Decoding of Linear Block Codes"]].
	* For more information on syndrome decoding, see the specification sheet for  [[Aufgaben:Exercise_1.11:_Syndrome_Decoding\|"Exercise 1.11"]].
			* For more information on syndrome decoding,  see the specification sheet for  [[Aufgaben:Exercise_1.11:_Syndrome_Decoding\|"Exercise 1.11"]].

	* The graph illustrates the three parity-check equations corresponding to the parity-check matrix:		* The graph illustrates the three parity-check equations corresponding to the parity-check matrix:
	**first row: red circle,		**first row:   red circle,
	**second row: green circle,		**second row:   green circle,
	**third row: blue circle.		**third row:   blue circle.

Noah at 15:27, 14 July 2022

2022-07-14T15:27:10Z

Noah at 15:26, 14 July 2022

2022-07-14T15:26:42Z

Show changes

Noah at 21:55, 13 July 2022

2022-07-13T21:55:23Z

← Older revision		Revision as of 23:55, 13 July 2022
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	{{quiz-Header\|Buchseite=~~Kanalcodierung~~/~~Decodierung linearer Blockcodes~~}}		{{quiz-Header\|Buchseite=Channel_Coding/Decoding_of_Linear_Block_Codes}}

	[[File:P_ID2399__KC_Z_1_10.png\|right\|frame\|~~Schaubild~~: ~~Prüfgleichungen~~]]		[[File:P_ID2399__KC_Z_1_10.png\|right\|frame\|Diagram: Parity-check equations]]

	~~Betrachtet wird die gleiche Konstellation wie~~ in  [[Aufgaben:1.~~11_Syndromdecodierung~~\|~~Aufgabe~~ 1.11]], ~~nämlich die Decodierung eines~~  $(7, 4, 3)$~~–Hamming–Codes mit der Prüfmatrix~~		The same constellation is considered as in  [[Aufgaben:Exercise_1.11:_Syndrome_Decoding\|Exercise 1.11]], namely the decoding of a  $(7, 4, 3)$ Hamming code with the parity-check matrix

	:$${ \boldsymbol{\rm H}}_{\rm } = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$		:$${ \boldsymbol{\rm H}}_{\rm } = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$

	~~Dementsprechend lautet das Generatorpolynom~~:		Accordingly, the generator polynomial is:

	:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$		:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$

Guenter at 09:59, 11 May 2022

2022-05-11T09:59:40Z

← Older revision		Revision as of 11:59, 11 May 2022
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	[[Category:Channel Coding: Exercises\|^1.5 ~~Decodierung linearer Blockcodes~~^]]		[[Category:Channel Coding: Exercises\|^1.5 Linear Block Code Decoding^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 1.11Z: Nochmals Syndromdecodierung to Aufgaben:Exercise 1.11Z: Syndrome Decoding again

2022-05-11T09:53:03Z

Guenter moved page Aufgaben:Aufgabe 1.11Z: Nochmals Syndromdecodierung to Aufgaben:Exercise 1.11Z: Syndrome Decoding again

← Older revision		Revision as of 11:53, 11 May 2022
(No difference)

← Older revision		Revision as of 17:27, 14 July 2022
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	[[File:P_ID2399__KC_Z_1_10.png\|right\|frame\|Diagram: Parity-check equations]]		[[File:P_ID2399__KC_Z_1_10.png\|right\|frame\|Diagram: Parity-check equations]]

	The same constellation is considered as in  [[Aufgaben:Exercise_1.11:_Syndrome_Decoding\|Exercise 1.11]], namely the decoding of a  $(7, 4, 3)$ Hamming code with the parity-check matrix		The same constellation is considered as in  [[Aufgaben:Exercise_1.11:_Syndrome_Decoding\|"Exercise 1.11"]], namely the decoding of a  $(7, 4, 3)$ Hamming code with the parity-check matrix

	:$${ \boldsymbol{\rm H}}_{\rm } = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$		:$${ \boldsymbol{\rm H}}_{\rm } = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$
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	:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$		:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$

	In  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Principle_of_syndrome_decoding\|~~Syndrome~~ decoding]]  one forms from the received vector  $\underline{y}$  the syndrome  $\underline{s}$  according to the equation		In  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Principle_of_syndrome_decoding\|"syndrome decoding"]]  one forms from the received vector  $\underline{y}$  the syndrome  $\underline{s}$  according to the equation

	:$$\underline{s} = \underline{y} \cdot { \boldsymbol{\rm H}}^{\rm T} \in {\rm GF}(2^m) \hspace{0.05cm}.$$		:$$\underline{s} = \underline{y} \cdot { \boldsymbol{\rm H}}^{\rm T} \in {\rm GF}(2^m) \hspace{0.05cm}.$$
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	Hints:		Hints:
	* The exercise belongs to the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\|Decoding of Linear Block Codes]].		* The exercise belongs to the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\|"Decoding of Linear Block Codes"]].
	* For more information on syndrome decoding, see the specification sheet for  [[Aufgaben:Exercise_1.11:_Syndrome_Decoding\|Exercise 1.11]].		* For more information on syndrome decoding, see the specification sheet for  [[Aufgaben:Exercise_1.11:_Syndrome_Decoding\|"Exercise 1.11"]].
	* The graph illustrates the three parity-check equations corresponding to the parity-check matrix:		* The graph illustrates the three parity-check equations corresponding to the parity-check matrix:
	**first row: red circle,		**first row: red circle,