Aufgaben:Exercise 4.12: Regular and Irregular Tanner Graph - Revision history

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	[[Category:Channel Coding: Exercises\|^4.4 Low–density Parity–check Codes^]]		[[Category:Channel Coding: Exercises\|^4.4 Low–density Parity–check Codes^]]
	[[de:Aufgaben:4.~~12_Regulärer/irregulärer_Tanner–Graph~~]]		[[de:Aufgaben:Aufgabe 4.12: Regulärer und irregulärer Tanner–Graph]]

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	*By adding the row  "$0 \ 0 \ 0 \ 1 \ 1 \ 1$"  to the  $\mathbf{H}_{\rm A}$  matrix,  one obtains:		*By adding the row  "$0 \ 0 \ 0 \ 1 \ 1 \ 1$"  to the  $\mathbf{H}_{\rm A}$  matrix,  one obtains:
	:$${ \boldsymbol{\rm H}}_{\rm B} =		:$${ \boldsymbol{\rm H}}_{\rm B} =\begin{pmatrix}1 &1 &0 &1 &0 &0\\1 &0 &1 &0 &1 &0\\0 &1 &1 &0 &0 &1\\0 &0 &0 &1 &1 &1\end{pmatrix}\hspace{0.05cm}.$$
	\begin{pmatrix}
	1 &1 &0 &1 &0 &0\\
	1 &0 &1 &0 &1 &0\\
	0 &1 &1 &0 &0 &1\\
	0 &0 &0 &1 &1 &1
	\end{pmatrix}\hspace{0.05cm}.$$

	:The modifications are marked in red in the adjacent graphic.		:The modifications are marked in red in the adjacent graphic.

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	* For the rate of the regular code $\rm B$  to be constructed,  the following lower bound then applies:		* For the rate of the regular code $\rm B$  to be constructed,  the following lower bound then applies:
	:$$R \ge 1 - \frac{w_{\rm C}}{w_{\rm R}}		:$$R \ge 1 - \frac{w_{\rm C}}{w_{\rm R}}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


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	*Thus,  this exercise is based on the following parity-check equation:		*Thus,  this exercise is based on the following parity-check equation:
	:$${ \boldsymbol{\rm H}}_{\rm A} =		:$${ \boldsymbol{\rm H}}_{\rm A} =\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}.$$
	\begin{pmatrix}
	1 &1 &0 &1 &0 &0\\
	1 &0 &1 &0 &1 &0\\
	0 &1 &1 &0 &0 &1
	\end{pmatrix}\hspace{0.05cm}.$$

	*In the diagram,  the parity-check equations are illustrated as red  (row 1),  green  (row 2)  resp. blue  (row 3)  groupings.		*In the diagram,  the parity-check equations are illustrated as red  (row 1),  green  (row 2)  resp. blue  (row 3)  groupings.
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	*The Hamming weights of the rows and columns,  respectively,  are $w_{\rm R} = 3$  and  $w_{\rm C} = 2$.		*The Hamming weights of the rows and columns,  respectively,  are $w_{\rm R} = 3$  and  $w_{\rm C} = 2$.
	*This gives as lower bound for the code rate:		*This gives as lower bound for the code rate:
	:$$R \ge 1 - \frac{w_{\rm C}}{w_{\rm R}}		:$$R \ge 1 - \frac{w_{\rm C}}{w_{\rm R}}= 1 - {2}/{3} = 1/3\hspace{0.05cm}.$$
	= 1 - {2}/{3} = 1/3
	\hspace{0.05cm}.$$
	*The  $\mathbf{H}$  manipulation does not change the generator matrix  $\mathbf{G}$.		*The  $\mathbf{H}$  manipulation does not change the generator matrix  $\mathbf{G}$.

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	[[Category:Channel Coding: Exercises\|^4.4 Low–density Parity–check Codes^]]		[[Category:Channel Coding: Exercises\|^4.4 Low–density Parity–check Codes^]]
			[[de:Aufgaben:4.12_Regulärer/irregulärer_Tanner–Graph]]

Guenter at 15:39, 17 December 2022

2022-12-17T15:39:22Z

← Older revision		Revision as of 17:39, 17 December 2022
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	===Solutiojn===		===Solutiojn===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The number of $\mathbf{H}_{\rm A}$&~~ndash~~;rows is equal to the number of ~~<i>~~check nodes~~</i>~~ $C_j$ in the Tanner graph  ⇒  $\underline{m = 3}$, and the number $\underline{n = 6}$ of ~~<i>~~variable nodes~~</i>~~ $V_i$ is equal to the column number.		'''(1)'''  The number of  $\mathbf{H}_{\rm A}$  rows is equal to the number of check nodes  $C_j$  in the Tanner graph   ⇒   $\underline{m = 3}$,  <br>and the number  $\underline{n = 6}$  of variable nodes  $V_i$  is equal to the column number.


			[[File:P_ID3073__KC_A_4_12c_v1.png\|right\|frame\|Underlying parity-check equations]]

	'''(2)'''  Correct are <u>answers 1 and 3</u> in contrast to statement 2:		'''(2)'''  Correct are the  <u>answers 1 and 3</u>  in contrast to statement 2:
	*The second $\mathbf{H}_{\rm A}$&~~ndash~~;row is rather "$1 \ 0 \ 1 \ 0 \ 1 \ 0$". ~~Thus, this exercise is based on the following parity-check equation:~~		*The second  $\mathbf{H}_{\rm A}$  row is rather  "$1 \ 0 \ 1 \ 0 \ 1 \ 0$".

	~~[[File:P_ID3073__KC_A_4_12c_v1.png\|right\|frame\|Underlying parity-check equations]]~~

			*Thus,  this exercise is based on the following parity-check equation:
	:$${ \boldsymbol{\rm H}}_{\rm A} =		:$${ \boldsymbol{\rm H}}_{\rm A} =
	\begin{pmatrix}		\begin{pmatrix}
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	\end{pmatrix}\hspace{0.05cm}.$$		\end{pmatrix}\hspace{0.05cm}.$$

	*In the diagram, the parity-check equations are illustrated as red (row 1), green (row 2)~~, and~~ blue (row 3) groupings~~, respectively~~.		*In the diagram,  the parity-check equations are illustrated as red  (row 1),  green  (row 2)  resp. blue  (row 3)  groupings.



	'''(3)'''  Correct are <u>solutions 1 and 3</u>:		'''(3)'''  Correct are the  <u>solutions 1 and 3</u>:
	* The $\mathbf{H}$ matrix ends with a $3 &times 3$ diagonal matrix   ⇒   systematic code.		* The  $\mathbf{H}$  matrix ends with a  $3 &times 3$  diagonal matrix   ⇒   systematic code.
	* Thus, the Hamming weights of the last three columns are $w_{\rm S}(4) = w_{\rm S}(5) = w_{\rm S}(6) = 1$.
	* For the first three columns, $w_{\rm S}(1) = w_{\rm S}(2) = w_{\rm S}(3) = 2$   ⇒   irregular code.
	* The three matrix rows are linearly independent. Thus $k = n - m = 6 - 3 = 3$ and $R = k/n = 1/2$ holds.

			* Thus,  the Hamming weights of the last three columns are  $w_{\rm C}(4) = w_{\rm C}(5) = w_{\rm C}(6) = 1$.

			* For the first three columns,  $w_{\rm C}(1) = w_{\rm C}(2) = w_{\rm C}(3) = 2$   ⇒   irregular code.

			* The three matrix rows are linearly independent.

	[[File: P_ID3074__KC_A_4_12d_v1.png\|right\|frame\|Modified Tanner graph for code $\rm B$]]		*Thus  $k = n - m = 6 - 3 = 3$  and  $R = k/n = 1/2$  holds.
	'''(4)'''  Correct is the <u>proposed solution 1</u>:
	*Looking at the previous Tanner&~~ndash~~;~~s graph,~~ one can see the correctness of proposed solution 1.
	*By adding the row "$0 \ 0 \ 0 \ 1 \ 1 \ 1$" to the $\mathbf{H}_{\rm A}$ matrix, one obtains:
			[[File: P_ID3074__KC_A_4_12d_v1.png\|right\|frame\|Modified Tanner graph for code  $\rm B$]]
			'''(4)'''  Correct is the  <u>proposed solution 1</u>:
			*Looking at the previous Tanner graph,  one can see the correctness of the proposed solution 1.

			*By adding the row  "$0 \ 0 \ 0 \ 1 \ 1 \ 1$"  to the  $\mathbf{H}_{\rm A}$  matrix,  one obtains:
	:$${ \boldsymbol{\rm H}}_{\rm B} =		:$${ \boldsymbol{\rm H}}_{\rm B} =
	\begin{pmatrix}		\begin{pmatrix}
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	\end{pmatrix}\hspace{0.05cm}.$$		\end{pmatrix}\hspace{0.05cm}.$$

	The modifications are marked in red in the adjacent graphic.		:The modifications are marked in red in the adjacent graphic.

	~~Due to the newly added check node $C_4$ and the connections with $V_4, \ V_5$ and $V_6$, there are now~~
	* from all variable nodes $V_i$ two lines, and
	* from all Check Nodes $C_j$ uniformly four.

			*Due to the newly added check node  $C_4$  and the connections with  $V_4, \ V_5$ and $V_6$,  there are now
			:* from all variable nodes  $V_i$  two lines,  and

	This is the condition for the code $\rm B$ to be regular.		:* from all check nodes  $C_j$  uniformly four.
			*This is the condition for the code  $\rm B$  to be regular.



	'''(5)'''  Correct are the <u>solutions 2 and 3</u>:		'''(5)'''  Correct are the  <u>solutions 2 and 3</u>:
	*The construction in subtask (4) yields a regular code.		*The construction in subtask  '''(4)'''  yields a regular code.
	*The Hamming weights of the rows and columns, respectively, are $w_{\rm Z} = 3$ and $w_{\rm S} = 2$.
			*The Hamming weights of the rows and columns,  respectively,  are $w_{\rm R} = 3$  and  $w_{\rm C} = 2$.
	*This gives as lower bound for the code rate:		*This gives as lower bound for the code rate:
	:$$R \ge 1 - \frac{w_{\rm S}}{w_{\rm Z}}		:$$R \ge 1 - \frac{w_{\rm C}}{w_{\rm R}}
	= 1 - {2}/{3} = 1/3		= 1 - {2}/{3} = 1/3
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*The $\mathbf{H}$ manipulation does not change the generator matrix $\mathbf{G}$.		*The  $\mathbf{H}$  manipulation does not change the generator matrix  $\mathbf{G}$.
	*The same code is still sent with code rate $R = 1/2$.
			*The same code is still sent with code rate  $R = 1/2$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 15:23, 17 December 2022

2022-12-17T15:23:32Z

← Older revision		Revision as of 17:23, 17 December 2022
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	{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Low-Density_Parity_Check_Codes}}		{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Low-Density_Parity_Check_Codes}}

	[[File:P_ID3072__KC_A_4_12_v1.png\|right\|frame\|~~Given~~ Tanner graph for code  $\rm A$]]		[[File:P_ID3072__KC_A_4_12_v1.png\|right\|frame\|Tanner graph for code  $\rm A$]]
	~~Dargestellt ist ein~~ Tanner~~–Graph eines Codes~~  $\rm A$  ~~mit~~		Shown is a Tanner graph of a code  $\rm A$  with
	* the  ~~<i>~~variable nodes~~</i>  (abbreviated VNs)~~  $V_1, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \ V_6$, where  $V_i$  denotes the  $i$th code word bit (whether information~~–~~ or parity bit) and corresponds to the  $i$th ~~column of the parity-check matrix;~~		* the  variable nodes  $V_1, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \ V_6$, . where  $V_i$  denotes the  $i$<sup>th</sup>  code word bit  $($whether information bit or parity bit$)$  and corresponds to the  $i$<sup>th</sup>  column of the parity-check matrix;
	* the  check nodes</i>  ~~(abbreviated CNs)  $C_1, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \ C_3$, which represent the rows~~ of ~~the  $\mathbf{H}_{\rm A}$ matrix and hence~~ the parity-check ~~equations.~~

			* the  check nodes</i>  $C_1, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \ C_3$,  which represent the rows of the  $\mathbf{H}_{\rm A}$  matrix and hence the parity-check equations.

	An edge between  $V_i$  and  $C_j$  indicates that the  $i$th code word symbol is involved in the  $j$th parity-check equation. In this case, the element  $h_{j,\hspace{0.05cm}i}$  of the parity-check matrix is equal  $1$.

			An edge between  $V_i$  and  $C_j$  indicates that the  $i$<sup>th</sup>  code word symbol is involved in the  $j$<sup>th</sup>  parity-check equation.  In this case,  the element  $h_{j,\hspace{0.05cm}i}$  of the parity-check matrix is equal  $1$.

	In ~~the~~ exercise, the relation between the above Tanner~~–~~graph  $($valid for the code  $\rm A)$  and the matrix  $\mathbf{H}_{\rm A}$  shall be given. In addition, the Tanner–graph to a parity-check matrix  $\mathbf{H}_{\rm B}$  shall be set up, resulting from  $\mathbf{H}_{\rm A}$  adding another row. This is to be determined so that the associated code  $\rm B$  is regular. This means:		In this exercise, the relation between the above Tanner graph  $($valid for the code  $\rm A)$  and the matrix  $\mathbf{H}_{\rm A}$  shall be given.
	* From all  <i>variable nodes</i>  $V_i$  $($with  ~~$1 ≤ i ≤ n)$  go off equal numbers of edges, likewise from all  <i>check nodes</i>  $C_j$  $($with  $1 ≤ j ≤ m)$.~~
	* The Hamming weights of all rows of  $\mathbf{H}_{\rm B}$  are each said to be equal  $(w_{\rm Z})$, as are the Hamming–weights of all columns  $(w_{\rm S})$.
	* For the rate of the regular code to be constructed $\rm B$  the following lower bound then applies:
	~~:$$R \ge 1 - \frac{w_{\rm S}}{w_{\rm Z}}~~
	~~\hspace{0.05cm}.$$~~

			In addition,  the Tanner graph to a parity-check matrix  $\mathbf{H}_{\rm B}$  shall be set up,  resulting from  $\mathbf{H}_{\rm A}$  adding another row.  This is to be determined so that the associated code  $\rm B$  is regular.  This means:
			* From all  variable nodes  $V_i$  $($with  $1 ≤ i ≤ n)$  go off an equal number of edges,  <br>likewise from all  check nodes  $C_j$  $($with  $1 ≤ j ≤ m)$.

			* The Hamming weights of all rows of  $\mathbf{H}_{\rm B}$  are each said to be equal  $(w_{\rm R})$,  as are the Hamming weights of all columns  $(w_{\rm C})$.

			* For the rate of the regular code $\rm B$  to be constructed,  the following lower bound then applies:
			:$$R \ge 1 - \frac{w_{\rm C}}{w_{\rm R}}
			\hspace{0.05cm}.$$




			<u>Hints:</u>
			*This exercise belongs to the chapter  [[Channel_Coding/The_Basics_of_Low-Density_Parity_Check_Codes\|"Basics of Low–density Parity–check Codes"]].

	~~Hints:~~		*Reference is made in particular to the section  [[Channel_Coding/The_Basics_of_Low-Density_Parity_Check_Codes#Two-part_LDPC_graph_representation_-_Tanner_graph\|"Two-part LDPC graph representation – Tanner graph"]].
	*This exercise belongs to the chapter  [[Channel_Coding/The_Basics_of_Low-Density_Parity_Check_Codes\|"Basics of Low–density Parity–check Codes"]].
	*Reference is made in particular to the ~~page~~  [[Channel_Coding/The_Basics_of_Low-Density_Parity_Check_Codes#Two-part_LDPC_graph_representation_-_Tanner_graph\|"Two-part LDPC graph representation – Tanner graph"]].


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	{Which statements are true based on the Tanner graph?		{Which statements are true based on the Tanner graph?
	\|type="[]"}		\|type="[]"}
	+ The row 1 of  $\mathbf{H}_{\rm A}$&~~ndash~~;matrix is  "$1 \ 1 \ 0 \ 1 \ 0 \ 0$".		+ The row 1 of the  $\mathbf{H}_{\rm A}$  matrix is  "$1 \ 1 \ 0 \ 1 \ 0 \ 0$".
	- The row 2 of  $\mathbf{H}_{\rm A}$&~~ndash~~;matrix is  "$1 \ 0 \ 1 \ 0 \ 0 \ 1$".		- The row 2 of the  $\mathbf{H}_{\rm A}$  matrix is  "$1 \ 0 \ 1 \ 0 \ 0 \ 1$".
	+ The row 3 of  $\mathbf{H}_{\rm A}$&~~ndash~~;matrix is  "$0 \ 1 \ 1 \ 0 \ 0 \ 1$".		+ The row 3 of the  $\mathbf{H}_{\rm A}$  matrix is  "$0 \ 1 \ 1 \ 0 \ 0 \ 1$".

	{What are the properties of ~~the~~ code  $\rm A$ ?		{What are the properties of code  $\rm A$ ?
	\|type="[]"}		\|type="[]"}
	+ The code is systematic.		+ The code is systematic.
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	- The code rate is  $R = 1/3$.		- The code rate is  $R = 1/3$.

	{The matrix  $\mathbf{H}_{\rm B}$  is obtained from  $\mathbf{H}_{\rm A}$  by adding one more row. By which fourth row does a regular code  $\rm B$ result?		{The matrix  $\mathbf{H}_{\rm B}$  is obtained from  $\mathbf{H}_{\rm A}$  by adding one more row.  By which fourth row does a regular code  $\rm B$  result?
	\|type="[]"}		\|type="[]"}
	+By adding "$0 \ 0 \ 0 \ 1 \ 1 \ 1$".		+By adding "$0 \ 0 \ 0 \ 1 \ 1 \ 1$".
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	- By adding any other row.		- By adding any other row.

	{What are the properties of ~~the~~ code  $\rm B$~~ ~~?		{What are the properties of code  $\rm B$?
	\|type="[]"}		\|type="[]"}
	- The code is systematic.		- The code is systematic.

Noah at 20:21, 10 December 2022

2022-12-10T20:21:06Z

Show changes

Guenter: Guenter moved page Aufgaben:Aufgabe 4.12: Regulärer und irregulärer Tanner–Graph to Aufgaben:Exercise 4.12: Regular and Irregular Tanner Graph

2022-11-22T13:00:36Z

Guenter moved page Aufgaben:Aufgabe 4.12: Regulärer und irregulärer Tanner–Graph to Aufgaben:Exercise 4.12: Regular and Irregular Tanner Graph

← Older revision	Revision as of 15:00, 22 November 2022
(No difference)

Javier: Text replacement - "„" to """

2021-05-28T14:36:04Z

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← Older revision		Revision as of 16:36, 28 May 2021
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	{Welche Aussagen sind aufgrund des Tanner–Graphen zutreffend?		{Welche Aussagen sind aufgrund des Tanner–Graphen zutreffend?
	\|type="[]"}		\|type="[]"}
	+ Die Zeile 1 der  $\mathbf{H}_{\rm A}$–Matrix ist  ~~&bdquo;~~$1 \ 1 \ 0 \ 1 \ 0 \ 0$".		+ Die Zeile 1 der  $\mathbf{H}_{\rm A}$–Matrix ist  "$1 \ 1 \ 0 \ 1 \ 0 \ 0$".
	- Die Zeile 2 der  $\mathbf{H}_{\rm A}$–Matrix ist  ~~&bdquo;~~$1 \ 0 \ 1 \ 0 \ 0 \ 1$".		- Die Zeile 2 der  $\mathbf{H}_{\rm A}$–Matrix ist  "$1 \ 0 \ 1 \ 0 \ 0 \ 1$".
	+ Die Zeile 3 der  $\mathbf{H}_{\rm A}$–Matrix ist  ~~&bdquo;~~$0 \ 1 \ 1 \ 0 \ 0 \ 1$".		+ Die Zeile 3 der  $\mathbf{H}_{\rm A}$–Matrix ist  "$0 \ 1 \ 1 \ 0 \ 0 \ 1$".

	{Welche Eigenschaften weist der Code  $\rm A$  auf?		{Welche Eigenschaften weist der Code  $\rm A$  auf?
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	{Die Matrix  $\mathbf{H}_{\rm B}$  ergibt sich aus  $\mathbf{H}_{\rm A}$  durch Hinzufügen einer weiteren Zeile. Durch welche vierte Zeile ergibt sich ein regulärer Code  $\rm B$?		{Die Matrix  $\mathbf{H}_{\rm B}$  ergibt sich aus  $\mathbf{H}_{\rm A}$  durch Hinzufügen einer weiteren Zeile. Durch welche vierte Zeile ergibt sich ein regulärer Code  $\rm B$?
	\|type="[]"}		\|type="[]"}
	+ Durch Hinzufügen von ~~&bdquo;~~$0 \ 0 \ 0 \ 1 \ 1 \ 1$".		+ Durch Hinzufügen von "$0 \ 0 \ 0 \ 1 \ 1 \ 1$".
	- Durch Hinzufügen von ~~&bdquo;~~$1 \ 1 \ 1 \ 1 \ 1 \ 1$".		- Durch Hinzufügen von "$1 \ 1 \ 1 \ 1 \ 1 \ 1$".
	- Durch Hinzufügen irgend einer anderen Zeile.		- Durch Hinzufügen irgend einer anderen Zeile.

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	'''(2)'''  Richtig sind die <u>Antworten 1 und 3</u> im Gegensatz zur Aussage 2:		'''(2)'''  Richtig sind die <u>Antworten 1 und 3</u> im Gegensatz zur Aussage 2:
	*Die zweite $\mathbf{H}_{\rm A}$–Zeile lautet vielmehr ~~&bdquo;~~$1 \ 0 \ 1 \ 0 \ 1 \ 0$". Somit liegt dieser Aufgabe die folgende Prüfgleichung zugrunde:		*Die zweite $\mathbf{H}_{\rm A}$–Zeile lautet vielmehr "$1 \ 0 \ 1 \ 0 \ 1 \ 0$". Somit liegt dieser Aufgabe die folgende Prüfgleichung zugrunde:

	[[File:P_ID3073__KC_A_4_12c_v1.png\|right\|frame\|Zugrunde liegende Prüfgleichungen]]		[[File:P_ID3073__KC_A_4_12c_v1.png\|right\|frame\|Zugrunde liegende Prüfgleichungen]]
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	'''(4)'''  Richtig ist der <u>Lösungsvorschlag 1</u>:		'''(4)'''  Richtig ist der <u>Lösungsvorschlag 1</u>:
	*Betrachtet man den bisherigen Tanner–Graphen, so erkennt man die Richtigkeit von Lösungsvorschlag 1.		*Betrachtet man den bisherigen Tanner–Graphen, so erkennt man die Richtigkeit von Lösungsvorschlag 1.
	*Durch Hinzufügen der Zeile ~~&bdquo;~~$0 \ 0 \ 0 \ 1 \ 1 \ 1$" zur $\mathbf{H}_{\rm A}$–Matrix erhält man:		*Durch Hinzufügen der Zeile "$0 \ 0 \ 0 \ 1 \ 1 \ 1$" zur $\mathbf{H}_{\rm A}$–Matrix erhält man:
	:$${ \boldsymbol{\rm H}}_{\rm B} =		:$${ \boldsymbol{\rm H}}_{\rm B} =
	\begin{pmatrix}		\begin{pmatrix}

Javier: Text replacement - "”" to """

2021-05-28T14:31:27Z

Text replacement - "”" to """

← Older revision		Revision as of 16:31, 28 May 2021
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	{Welche Aussagen sind aufgrund des Tanner–Graphen zutreffend?		{Welche Aussagen sind aufgrund des Tanner–Graphen zutreffend?
	\|type="[]"}		\|type="[]"}
	+ Die Zeile 1 der  $\mathbf{H}_{\rm A}$–Matrix ist  &bdquo;$1 \ 1 \ 0 \ 1 \ 0 \ 0$~~”~~.		+ Die Zeile 1 der  $\mathbf{H}_{\rm A}$–Matrix ist  &bdquo;$1 \ 1 \ 0 \ 1 \ 0 \ 0$".
	- Die Zeile 2 der  $\mathbf{H}_{\rm A}$–Matrix ist  &bdquo;$1 \ 0 \ 1 \ 0 \ 0 \ 1$~~”~~.		- Die Zeile 2 der  $\mathbf{H}_{\rm A}$–Matrix ist  &bdquo;$1 \ 0 \ 1 \ 0 \ 0 \ 1$".
	+ Die Zeile 3 der  $\mathbf{H}_{\rm A}$–Matrix ist  &bdquo;$0 \ 1 \ 1 \ 0 \ 0 \ 1$~~”~~.		+ Die Zeile 3 der  $\mathbf{H}_{\rm A}$–Matrix ist  &bdquo;$0 \ 1 \ 1 \ 0 \ 0 \ 1$".

	{Welche Eigenschaften weist der Code  $\rm A$  auf?		{Welche Eigenschaften weist der Code  $\rm A$  auf?
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	{Die Matrix  $\mathbf{H}_{\rm B}$  ergibt sich aus  $\mathbf{H}_{\rm A}$  durch Hinzufügen einer weiteren Zeile. Durch welche vierte Zeile ergibt sich ein regulärer Code  $\rm B$?		{Die Matrix  $\mathbf{H}_{\rm B}$  ergibt sich aus  $\mathbf{H}_{\rm A}$  durch Hinzufügen einer weiteren Zeile. Durch welche vierte Zeile ergibt sich ein regulärer Code  $\rm B$?
	\|type="[]"}		\|type="[]"}
	+ Durch Hinzufügen von &bdquo;$0 \ 0 \ 0 \ 1 \ 1 \ 1$~~”~~.		+ Durch Hinzufügen von &bdquo;$0 \ 0 \ 0 \ 1 \ 1 \ 1$".
	- Durch Hinzufügen von &bdquo;$1 \ 1 \ 1 \ 1 \ 1 \ 1$~~”~~.		- Durch Hinzufügen von &bdquo;$1 \ 1 \ 1 \ 1 \ 1 \ 1$".
	- Durch Hinzufügen irgend einer anderen Zeile.		- Durch Hinzufügen irgend einer anderen Zeile.

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	'''(2)'''  Richtig sind die <u>Antworten 1 und 3</u> im Gegensatz zur Aussage 2:		'''(2)'''  Richtig sind die <u>Antworten 1 und 3</u> im Gegensatz zur Aussage 2:
	*Die zweite $\mathbf{H}_{\rm A}$–Zeile lautet vielmehr &bdquo;$1 \ 0 \ 1 \ 0 \ 1 \ 0$~~”~~. Somit liegt dieser Aufgabe die folgende Prüfgleichung zugrunde:		*Die zweite $\mathbf{H}_{\rm A}$–Zeile lautet vielmehr &bdquo;$1 \ 0 \ 1 \ 0 \ 1 \ 0$". Somit liegt dieser Aufgabe die folgende Prüfgleichung zugrunde:

	[[File:P_ID3073__KC_A_4_12c_v1.png\|right\|frame\|Zugrunde liegende Prüfgleichungen]]		[[File:P_ID3073__KC_A_4_12c_v1.png\|right\|frame\|Zugrunde liegende Prüfgleichungen]]
Line 99:		Line 99:
	'''(4)'''  Richtig ist der <u>Lösungsvorschlag 1</u>:		'''(4)'''  Richtig ist der <u>Lösungsvorschlag 1</u>:
	*Betrachtet man den bisherigen Tanner–Graphen, so erkennt man die Richtigkeit von Lösungsvorschlag 1.		*Betrachtet man den bisherigen Tanner–Graphen, so erkennt man die Richtigkeit von Lösungsvorschlag 1.
	*Durch Hinzufügen der Zeile &bdquo;$0 \ 0 \ 0 \ 1 \ 1 \ 1$~~”~~ zur $\mathbf{H}_{\rm A}$–Matrix erhält man:		*Durch Hinzufügen der Zeile &bdquo;$0 \ 0 \ 0 \ 1 \ 1 \ 1$" zur $\mathbf{H}_{\rm A}$–Matrix erhält man:
	:$${ \boldsymbol{\rm H}}_{\rm B} =		:$${ \boldsymbol{\rm H}}_{\rm B} =
	\begin{pmatrix}		\begin{pmatrix}