Aufgaben:Exercise 1.13: Binary Erasure Channel Decoding - 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.~~13_BEC–Decodierung~~]]		[[de:Aufgaben:Aufgabe 1.13: Decodierung beim binären Auslöschungskanal (BEC)]]

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	The equation of determination is thus:		The equation of determination is thus:
	:$${ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T}= { \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T}\hspace{0.3cm}		:$${ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T}= { \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \begin{pmatrix} 1 &0\\ 1 &0\\ 1 &1 \end{pmatrix} \cdot \begin{pmatrix} z_2 \\ z_7 \end{pmatrix} = \begin{pmatrix} 1 &1 &0 &1 &0\\ 0 &1 &1 &0 &1\\ 1 &0 &1 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm} \begin{pmatrix} 1 &0\\ 1 &0\\ 1 &1 \end{pmatrix} \cdot \begin{pmatrix} z_2 \\ z_7 \end{pmatrix} = \begin{pmatrix} 1 &1 &0 &1 &0\\ 0 &1 &1 &0 &1\\ 1 &0 &1 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \hspace{0.05cm}.$$

	This results in three equations for the two unknowns  $z_{2}$  and  $z_{7}$:		This results in three equations for the two unknowns  $z_{2}$  and  $z_{7}$:
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	'''(3)'''  One obtains after some matrix multiplications:		'''(3)'''  One obtains after some matrix multiplications:
	:$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T} = \begin{pmatrix} 1 &1 &1 &0\\ 0 &1 &1 &1\\ 1 &1 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \hspace{0.05cm},\hspace{1cm}		:$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T} = \begin{pmatrix} 1 &1 &1 &0\\ 0 &1 &1 &1\\ 1 &1 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \hspace{0.05cm},\hspace{1cm}{ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T} = \begin{pmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} z_5 \\ z_6 \\ z_7 \end{pmatrix} = \begin{pmatrix} z_5 \\ z_6 \\ z_7 \end{pmatrix} \hspace{0.05cm}.$$
	{ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T} = \begin{pmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} z_5 \\ z_6 \\ z_7 \end{pmatrix} = \begin{pmatrix} z_5 \\ z_6 \\ z_7 \end{pmatrix} \hspace{0.05cm}.$$

	*By equating it follows  $z_{5} = 0, \ z_{6} = 0, \ z_{7} = 1$   ⇒   <u>Suggested solution 2</u>.		*By equating it follows  $z_{5} = 0, \ z_{6} = 0, \ z_{7} = 1$   ⇒   <u>Suggested solution 2</u>.
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	'''(5)'''  Analogous to subtask  '''(3)''',  we obtain:		'''(5)'''  Analogous to subtask  '''(3)''',  we obtain:

	:$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T} = \begin{pmatrix} 1 &1 &1 &1\\ 0 &1 &1 &0\\ 1 &1 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \hspace{0.05cm},\hspace{1cm}		:$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T} = \begin{pmatrix} 1 &1 &1 &1\\ 0 &1 &1 &0\\ 1 &1 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \hspace{0.05cm},\hspace{1cm}{ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T} = \begin{pmatrix} 0 &0 &0\\ 1 &1 &0\\ 1 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} z_4 \\ z_6 \\ z_7 \end{pmatrix} = \begin{pmatrix} 0 \\ z_4 + z_6 \\ z_4 + z_7 \end{pmatrix} \hspace{0.05cm}.$$
	{ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T} = \begin{pmatrix} 0 &0 &0\\ 1 &1 &0\\ 1 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} z_4 \\ z_6 \\ z_7 \end{pmatrix} = \begin{pmatrix} 0 \\ z_4 + z_6 \\ z_4 + z_7 \end{pmatrix} \hspace{0.05cm}.$$

	*If we now set the two column vectors equal,  we obtain only two equations for the three unknowns   ⇒   <u>Suggested solution 4</u>.  Or in other words:		*If we now set the two column vectors equal,  we obtain only two equations for the three unknowns   ⇒   <u>Suggested solution 4</u>.  Or in other words:

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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:1.13_BEC–Decodierung]]

Hwang at 16:02, 23 January 2023

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	- $\underline{z} = (1, 0, 0, 1, 0, 0, 1).$		- $\underline{z} = (1, 0, 0, 1, 0, 0, 1).$

	{What are the consequences of the red entries for  $\boldsymbol{\rm H}_{\rm K}$  and  $z_{\rm K}$  (see graphic on the information ~~page~~)?		{What are the consequences of the red entries for  $\boldsymbol{\rm H}_{\rm K}$  and  $z_{\rm K}$  (see graphic in the information section)?
	\|type="[]"}		\|type="[]"}
	+ The erasure vector is  $\underline{z}_{\rm E} = (z_{5}, z_{6}, z_{7}).$		+ The erasure vector is  $\underline{z}_{\rm E} = (z_{5}, z_{6}, z_{7}).$
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	- For the present  $\underline{y}$   no unique decoding is possible.		- For the present  $\underline{y}$   no unique decoding is possible.

	{What are the consequences of the green entries for  $\boldsymbol{\rm H}_{\rm K}$  and  $z_{\rm K}$  (see graphic on the information ~~page~~)?		{What are the consequences of the green entries for  $\boldsymbol{\rm H}_{\rm K}$  and  $z_{\rm K}$  (see graphic in the information section)?
	\|type="[]"}		\|type="[]"}
	+ The received word is  $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E}).$		+ The received word is  $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E}).$

Hwang at 16:59, 1 November 2022

2022-11-01T16:59:37Z

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	* This exercise belongs to the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\|"Decoding of Linear Block Codes"]].		* This exercise belongs to the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\|"Decoding of Linear Block Codes"]].

	* The algorithm for mapping the received word  $\underline{y}$  to the correct ~~codeword~~  $\underline{z} = \underline{x}$   is described in detail in the  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"theory section"]] .		* The algorithm for mapping the received word  $\underline{y}$  to the correct code word  $\underline{z} = \underline{x}$   is described in detail in the  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"theory section"]] .

	* We would like to remind again that in BEC decoding we use the first decoder block  $\underline{y} → \underline{z}$  as  "code word finder"  since wrong decisions are excluded here.  Each received word is decoded correctly,  or it may not be decoded at all.		* We would like to remind again that in BEC decoding we use the first decoder block  $\underline{y} → \underline{z}$  as  "code word finder"  since wrong decisions are excluded here.  Each received word is decoded correctly,  or it may not be decoded at all.
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	*In contrast   $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E})$  according to subtask  '''(4)'''  could not be decoded. In the code table, besides  $(1, 1, 0, 1, 0, 0, 1)$  with  $(1, 1, 0, 0, 0, 1, 0)$  one can recognize another code word  (highlighted in green),  which becomes the received word  $\underline{y}$  due to the  $n_{\rm E} = 3$  erasures concerning bit 4, 6 and 7.		*In contrast   $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E})$  according to subtask  '''(4)'''  could not be decoded. In the code table, besides  $(1, 1, 0, 1, 0, 0, 1)$  with  $(1, 1, 0, 0, 0, 1, 0)$  one can recognize another code word  (highlighted in green),  which becomes the received word  $\underline{y}$  due to the  $n_{\rm E} = 3$  erasures concerning bit 4, 6 and 7.

	*This case,  when the  $n_{\rm E} = d_{\rm min}$  extinctions affect exactly the  $d_{\rm min}$  different bits of two ~~codewords~~, leads to a matrix  $\mathbf{H}_{\rm E}$  with rank smaller  $d_{\rm min}$.		*This case,  when the  $n_{\rm E} = d_{\rm min}$  extinctions affect exactly the  $d_{\rm min}$  different bits of two code words, leads to a matrix  $\mathbf{H}_{\rm E}$  with rank smaller  $d_{\rm min}$.

	*If  $n_{\rm E} > d_{\rm min}$, the number  $n - n_{\rm E}$  of bits not erased is smaller than the number  $k$  of information bits.  In this case,  of course,  the code word cannot be decoded.		*If  $n_{\rm E} > d_{\rm min}$, the number  $n - n_{\rm E}$  of bits not erased is smaller than the number  $k$  of information bits.  In this case,  of course,  the code word cannot be decoded.

Guenter at 12:26, 28 September 2022

2022-09-28T12:26:58Z

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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:EN_KC_A_1_13.png\|right\|frame\|Decoding at the BEC]		[[File:EN_KC_A_1_13.png\|right\|frame\|Decoding at the BEC]]

	We assume here the model in section  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"Decoding at the Binary Erasure Channel"]]  (BEC configuration highlighted in green):		We assume here the model in section  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"Decoding at the Binary Erasure Channel"]]  (BEC configuration highlighted in green):

Guenter at 14:28, 17 August 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:~~KC_A_1_13_neu~~.png\|right\|frame\|Decoding at the BEC ~~'''Korrektur''']~~]		[[File:EN_KC_A_1_13.png\|right\|frame\|Decoding at the BEC]

	We assume here the model in section  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"Decoding at the Binary Erasure Channel"]]  (BEC configuration highlighted in green):		We assume here the model in section  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"Decoding at the Binary Erasure Channel"]]  (BEC configuration highlighted in green):

Guenter at 15:47, 21 July 2022

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	*This case,  when the  $n_{\rm E} = d_{\rm min}$  extinctions affect exactly the  $d_{\rm min}$  different bits of two codewords, leads to a matrix  $\mathbf{H}_{\rm E}$  with rank smaller  $d_{\rm min}$.		*This case,  when the  $n_{\rm E} = d_{\rm min}$  extinctions affect exactly the  $d_{\rm min}$  different bits of two codewords, leads to a matrix  $\mathbf{H}_{\rm E}$  with rank smaller  $d_{\rm min}$.

	*If  $~~\boldsymbol{\rm H}_~~{\rm E} > d_{\rm min}$, the number  $n - n_{\rm E}$  of bits not erased is smaller than the number  $k$  of information bits.  In this case,  of course,  the code word cannot be decoded.		*If  $n_{\rm E} > d_{\rm min}$, the number  $n - n_{\rm E}$  of bits not erased is smaller than the number  $k$  of information bits.  In this case,  of course,  the code word cannot be decoded.

Guenter at 14:53, 21 July 2022

2022-07-21T14:53:07Z

Show changes

Guenter at 13:05, 21 July 2022

2022-07-21T13:05:18Z

← Older revision		Revision as of 15:05, 21 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:KC_A_1_13_neu.png\|right\|frame\|Decoding at the BEC]]		[[File:KC_A_1_13_neu.png\|right\|frame\|Decoding at the BEC '''Korrektur''']]

	We assume here the model in section  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"Decoding at the Binary Erasure Channel"]]  (BEC configuration highlighted in green):		We assume here the model in section  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"Decoding at the Binary Erasure Channel"]]  (BEC configuration highlighted in green):

	*Each information word  $\underline{u}$  is encoded blockwise and yields the ~~codeword~~  $\underline{x}$. Let the block code be linear and completely given by its parity-check matrix  $\boldsymbol{\rm H}$.		*Each information word  $\underline{u}$  is encoded blockwise and yields the code word  $\underline{x}$.  Let the block code be linear and completely given by its parity-check matrix  $\boldsymbol{\rm H}$.

	*During transmission  $n_{\rm E}$  bits of the ~~codeword~~ are erased   ⇒  [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_.E2.80.93_BEC\|"Binary Erasure Channel"]]  (BEC). The ~~codeword~~  $\underline{x}$  thus becomes the received word  $\underline{y}$.		*During transmission  $n_{\rm E}$  bits of the code word are erased   ⇒  [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_.E2.80.93_BEC\|"Binary Erasure Channel"]]  $\rm (BEC)$.  The code word  $\underline{x}$  thus becomes the received word  $\underline{y}$.

	*If the number  $n_{\rm E}$  of erasures is less than the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding\|"minimum distance"]]  $d_{\rm min}$  of the code, it is possible to reconstruct from  $\underline{y}$  the ~~codeword~~  $\underline{z} = \underline{x}$  without error, and thus the correct information word  $\underline{v} = \underline{u}$ is also obtained.		*If the number  $n_{\rm E}$  of erasures is less than the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding\|"minimum distance"]]  $d_{\rm min}$  of the code,  it is possible to reconstruct from  $\underline{y}$  the code word  $\underline{z} = \underline{x}$  without error,  and thus the correct information word  $\underline{v} = \underline{u}$  is also obtained.


	For exercise description, let us now consider the Hamming ~~codeword~~  $\underline{x} = (0, 1, 0, 1, 1, 0, 0)$  and the received word  $\underline{y} = (0, 1, {\rm E} , {\rm E}, 1, 0, 0).$		For exercise description,  let us now consider the Hamming code word  $\underline{x} = (0, 1, 0, 1, 1, 0, 0)$  and the received word  $\underline{y} = (0, 1, {\rm E} , {\rm E}, 1, 0, 0).$

	*The third and fourth bit were thus erased by the channel.		*The third and fourth bit were thus erased by the channel.  The  "code word finder"  thus has the task to determine the vector  $z_{\rm E} = (z_{3}, z_{4})$  with  $z_{3}, \ z_{4} \in \{0, 1\}$  to be determined.  This is done according to the equation
	*The ~~codeword~~ finder thus has the ~~exercise~~ to determine the vector  $z_{\rm E} = (z_{3}, z_{4})$ with $z_{3}, \ z_{4} \in \{0, 1\}$  to be determined. This is done according to the equation

	:$${ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T}= { \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T}\hspace{0.05cm}.$$		:$${ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T}= { \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T}\hspace{0.05cm}.$$
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	:$$\underline{z}_{\rm K} = (0, 1, 1, 0, 0)\hspace{0.05cm},\hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm K} = \begin{pmatrix} 1 &1 &1 &0 &0\\ 0 &1 &0 &1 &0\\ 1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}, \hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm E} = \begin{pmatrix} 1 &0\\ 1 &1\\ 0 &1 \end{pmatrix} \hspace{0.05cm}.$$		:$$\underline{z}_{\rm K} = (0, 1, 1, 0, 0)\hspace{0.05cm},\hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm K} = \begin{pmatrix} 1 &1 &1 &0 &0\\ 0 &1 &0 &1 &0\\ 1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}, \hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm E} = \begin{pmatrix} 1 &0\\ 1 &1\\ 0 &1 \end{pmatrix} \hspace{0.05cm}.$$

	*This equation gives two equations ~~of determination~~ for the bits to be determined, whose solution leads to the result  $z_{3} = 0$  and  $z_{4} = 1$~~ ~~.		*This equation gives two equations for the bits to be determined,  whose solution leads to the result  $z_{3} = 0$  and  $z_{4} = 1$.





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	Hints:		Hints:
	* This exercise belongs to the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\|"Decoding of Linear Block Codes"]].		* This exercise belongs to the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\|"Decoding of Linear Block Codes"]].
	* The algorithm for mapping the received word  $\underline{y}$  to the correct codeword  $\underline{z} = \underline{x}$  is described in detail in the  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"~~Theory~~ section"]] .
	* We would like to remind again that in BEC decoding we use the first decoder block  $\underline{y} → \underline{z}$  as  ~~''codeword~~ finder''  since wrong decisions are excluded here. Each received word is decoded correctly, or it may not be decoded at all.		* The algorithm for mapping the received word  $\underline{y}$  to the correct codeword  $\underline{z} = \underline{x}$   is described in detail in the  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"theory section"]] .
	*In the BSC model, on the other hand, decoding errors cannot be avoided. Accordingly, we refer to the corresponding block there as  ''code word estimator''.
			* We would like to remind again that in BEC decoding we use the first decoder block  $\underline{y} → \underline{z}$  as  "code word finder"  since wrong decisions are excluded here.  Each received word is decoded correctly,  or it may not be decoded at all.

			*In the BSC model,  on the other hand,  decoding errors cannot be avoided.  Accordingly,  we refer to the corresponding block there as  "code word estimator".


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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{  $\underline{y} = (1, {\rm E}, 0, 1, 0, 0, {\rm E})$ was received. Which sequence does the ~~codeword~~ finder decide to use?		{  $\underline{y} = (1, {\rm E}, 0, 1, 0, 0, {\rm E})$  was received.  Which sequence does the code word finder decide to use?
	\|type="()"}		\|type="()"}
	- $\underline{z} = (1, 0, 0, 1, 0, 0, 0),$		- $\underline{z} = (1, 0, 0, 1, 0, 0, 0),$
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	+ The erasure vector is  $\underline{z}_{\rm E} = (z_{5}, z_{6}, z_{7}).$		+ The erasure vector is  $\underline{z}_{\rm E} = (z_{5}, z_{6}, z_{7}).$
	+ The received word is  $\underline{y} = (1, 1, 0, 1, {\rm E}, {\rm E}, {\rm E}).$		+ The received word is  $\underline{y} = (1, 1, 0, 1, {\rm E}, {\rm E}, {\rm E}).$
	- $\boldsymbol{\rm H}_{\rm E}$  is a  $2 \times 3$ matrix.		- $\boldsymbol{\rm H}_{\rm E}$  is a  $2 \times 3$  matrix.
	+ $\boldsymbol{\rm H}_{\rm E}$  is a  $3 \times 3$ matrix.		+ $\boldsymbol{\rm H}_{\rm E}$  is a  $3 \times 3$  matrix.

	{Now apply  $\underline{y} = (1, 1, 0, 1, {\rm E}, {\rm E}, {\rm E}).$ Which ~~codeword~~ is selected?		{Now apply  $\underline{y} = (1, 1, 0, 1, {\rm E}, {\rm E}, {\rm E}).$  Which code word is selected?
	\|type="()"}		\|type="()"}
	- $\underline{z} = (1, 1, 0, 1, 1, 1, 0),$		- $\underline{z} = (1, 1, 0, 1, 1, 1, 0),$
	+ $\underline{z} = (1, 1, 0, 1, 0, 0, 1),$		+ $\underline{z} = (1, 1, 0, 1, 0, 0, 1),$
	- $\underline{z} = (1, 1, 0, 0, 0, 1, 0).$		- $\underline{z} = (1, 1, 0, 0, 0, 1, 0).$
	- For the present  $\underline{y}$  no unique decoding is possible.		- For the present  $\underline{y}$   no unique decoding is possible.

	{What are the consequences of the green entries for  $\boldsymbol{\rm H}_{\rm K}$  and  $z_{\rm K}$  (see graphic on the information page)?		{What are the consequences of the green entries for  $\boldsymbol{\rm H}_{\rm K}$  and  $z_{\rm K}$  (see graphic on the information page)?
	\|type="[]"}		\|type="[]"}
	+ The received word is  $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E}).$		+ The received word is  $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E}).$
	- $\boldsymbol{\rm H}_{\rm K}$  differs from ~~subquestion~~ '''(2)''' in the last row.		- $\boldsymbol{\rm H}_{\rm K}$  differs from subtask  '''(2)'''  in the last row.
	+ $\boldsymbol{\rm H}_{\rm K}$  differs from ~~subquestion~~ '''(2)''' in the last column.		+ $\boldsymbol{\rm H}_{\rm K}$  differs from subtask  '''(2)'''  in the last column.

	{Now apply  $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E}).$ Which ~~codeword~~ is selected?		{Now apply  $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E}).$  Which code word is selected?
	\|type="()"}		\|type="()"}
	- $\underline{z} = (1, 1, 0, 1, 1, 1, 0),$		- $\underline{z} = (1, 1, 0, 1, 1, 1, 0),$
	- $\underline{z} = (1, 1, 0, 1, 0, 0, 1),$		- $\underline{z} = (1, 1, 0, 1, 0, 0, 1),$
	- $\underline{z} = (1, 1, 0, 0, 0, 1, 0).$		- $\underline{z} = (1, 1, 0, 0, 0, 1, 0).$
	+ For the present  $\underline{y}$  no unique decoding is possible.		+ For the present  $\underline{y}$   no unique decoding is possible.

	{Which statements result for the correction capability at the BEC? $n_{\rm E}$  indicates in the following the number of erasures.		{Which statements result for the correction capability at the BEC?  $n_{\rm E}$  indicates in the following the number of erasures.
	\|type="[]"}		\|type="[]"}
	+ For  $n_{\rm E} < d_{\rm min}$  unique decoding is always possible.		+ For  $n_{\rm E} < d_{\rm min}$  unique decoding is always possible.

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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:EN_KC_A_1_13.png\|right\|frame\|Decoding at the BEC]		[[File:EN_KC_A_1_13.png\|right\|frame\|Decoding at the BEC]]

	We assume here the model in section  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"Decoding at the Binary Erasure Channel"]]  (BEC configuration highlighted in green):		We assume here the model in section  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"Decoding at the Binary Erasure Channel"]]  (BEC configuration highlighted in green):

← Older revision		Revision as of 16:28, 17 August 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:~~KC_A_1_13_neu~~.png\|right\|frame\|Decoding at the BEC ~~'''Korrektur''']~~]		[[File:EN_KC_A_1_13.png\|right\|frame\|Decoding at the BEC]

	We assume here the model in section  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"Decoding at the Binary Erasure Channel"]]  (BEC configuration highlighted in green):		We assume here the model in section  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel\|"Decoding at the Binary Erasure Channel"]]  (BEC configuration highlighted in green):