Aufgaben:Exercise 1.2: A Simple Binary Channel Code - Revision history

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	[[Category:Channel Coding: Exercises\|^1.1 Objective of Channel Coding^]]		[[Category:Channel Coding: Exercises\|^1.1 Objective of Channel Coding^]]
	[[de:Aufgaben:1.~~2_Einfacher_binärer_Kanalcode~~]]		[[de:Aufgaben:Aufgabe 1.2: Einfacher binärer Kanalcode]]

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	[[Category:Channel Coding: Exercises\|^1.1 Objective of Channel Coding^]]		[[Category:Channel Coding: Exercises\|^1.1 Objective of Channel Coding^]]
			[[de:Aufgaben:1.2_Einfacher_binärer_Kanalcode]]

Guenter at 14:09, 17 August 2022

2022-08-17T14:09:04Z

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	[[File:~~EN_KC_A_1_2~~.png\|right\|frame\|Clarification ~~of channel~~ coding and decoding ~~'''Korrektur'''~~]]		[[File:EN_KC_A_1_2_bitte.png\|right\|frame\|Clarification:  Channel coding and decoding ]]
	The graph illustrates the channel coding  $\mathcal{C}$  considered here:		The graph illustrates the channel coding  $\mathcal{C}$  considered here:
	*There are four possible information blocks  $\underline{u} = (u_{1},\ u_{2},\ \text{...}\hspace{0.05cm} ,\ u_{k})$.		*There are four possible information blocks  $\underline{u} = (u_{1},\ u_{2},\ \text{...}\hspace{0.05cm} ,\ u_{k})$.

Guenter at 12:34, 6 June 2022

2022-06-06T12:34:02Z

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	'''(4)'''  Correct is  <u>Yes</u>:		'''(4)'''  Correct is  <u>Yes</u>:  A systematic code is characterized by the fact that in each case the first  $k$  bits of the code words are identical to the information block.
	*A systematic code is characterized by the fact that in each case the first  $k$  bits of the code words are identical to the information block.

Guenter at 12:31, 6 June 2022

2022-06-06T12:31:51Z

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	*Each information block  $\underline{u}$  is uniquely assigned  (recognizable by the same color)  to the code word  $\underline{x}= (x_{1},\ x_{2},\ \text{...}\hspace{0.05cm} ,\ x_{n})$ .		*Each information block  $\underline{u}$  is uniquely assigned  (recognizable by the same color)  to the code word  $\underline{x}= (x_{1},\ x_{2},\ \text{...}\hspace{0.05cm} ,\ x_{n})$ .

	*Due to decoding errors  $(0 → 1, \ 1 → 0)$  there are more than 4,  namely 16 different ~~receiving~~ words  $\underline{y} = (y_{1},\ y_{2},\ \text{...} \hspace{0.05cm} ,\ y_{n})$.		*Due to decoding errors  $(0 → 1, \ 1 → 0)$  there are more than 4,  namely 16 different receive words  $\underline{y} = (y_{1},\ y_{2},\ \text{...} \hspace{0.05cm} ,\ y_{n})$.


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	'''(2)'''  Each code word  $\underline{x}$  is uniquely associated with an information block  $\underline{u}$.		'''(2)'''  Each code word  $\underline{x}$  is uniquely associated with an information block  $\underline{u}$.
	*By corruptions of a single of the total  $n$  bits of a code word  $\underline{x}$  the ~~receiving~~ words  $\underline{y}$  result.		*By corruptions of a single of the total  $n$  bits of a code word  $\underline{x}$  the receive words  $\underline{y}$  result.
	*From the number  $(16 = 2^4$)  of possible ~~receiving~~ words follows  $\underline{ n = 4}$.		*From the number  $(16 = 2^4$)  of possible receive words follows  $\underline{ n = 4}$.

Guenter at 12:29, 6 June 2022

2022-06-06T12:29:37Z

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	$ d_{\rm H} \ (\underline{x}_1,\ \underline{x}_2) \ = \ $ { 4 }		$ d_{\rm H} \ (\underline{x}_1,\ \underline{x}_2) \ = \ $ { 4 }

	{What is the minimum Hamming distance of the code  $\mathcal{C}$  under consideration$?		{What is the minimum Hamming distance of the code  $\mathcal{C}$  under consideration?
	\|type="{}"}		\|type="{}"}
	$ d_{\rm min} (\mathcal{C}) \ = \ $ { 2 }		$ d_{\rm min} (\mathcal{C}) \ = \ $ { 2 }
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The code size here is given to $\|\mathcal{C}\| = 4$.		'''(1)'''  The code size here is given to  $\|\mathcal{C}\| = 4$.
	*In general, $\|\mathcal{C}\|= 2^k$.		*In general,  $\|\mathcal{C}\|= 2^k$.
	*From this follows $\underline{ k = 2}$.		*From this follows  $\underline{ k = 2}$.



	'''(2)'''  Each ~~codeword~~ $\underline{x}$ is uniquely associated with an information block $\underline{u}$.		'''(2)'''  Each code word  $\underline{x}$  is uniquely associated with an information block  $\underline{u}$.
	*By corruptions of single of the total $n$ bits of a ~~codeword~~ $\underline{x}$ the ~~receive~~ words $\underline{y}$ result.		*By corruptions of a single of the total  $n$  bits of a code word  $\underline{x}$  the receiving words  $\underline{y}$  result.
	*From the number $(16 = 2^4$) of possible ~~receive~~ words follows $\underline{ n = 4}$.		*From the number  $(16 = 2^4$)  of possible receiving words follows  $\underline{ n = 4}$.



	'''(3)'''  The code rate is by definition $R = k/n$.		'''(3)'''  The code rate is by definition  $R = k/n$.  Using the above results,  we obtain $\underline{R = 0.5}$.
	*Using the above results, we obtain $\underline{R = 0.5}$.


			'''(4)'''  Correct is  <u>Yes</u>:
			*A systematic code is characterized by the fact that in each case the first  $k$  bits of the code words are identical to the information block.

	~~'''(4)'''  Correct <u>Yes</u>:~~
	*A systematic code is characterized by the fact that in each case the first $k$ bits of the code words are identical to the information block.


			'''(5)'''  The Hamming weight of a binary code is equal to the algebraic sum  $x_1 + x_2 + \text{...} + x_n$  over all code word elements.  Thus:
	'''(5)'''  The Hamming weight of a binary code is equal to the algebraic sum $x_1 + x_2 + \text{...} + x_n$ over all code word elements. Thus:
	:$$w_{\rm H}(\underline{x}_0) \hspace{0.15cm} \underline {= 0} \hspace{0.05cm}, \hspace{0.3cm}w_{\rm H}(\underline{x}_1) \hspace{0.15cm} \underline {= 2} \hspace{0.05cm}, \hspace{0.3cm} w_{\rm H}(\underline{x}_2) \hspace{0.15cm} \underline {= 2}\hspace{0.05cm}, \hspace{0.3cm}w_{\rm H}(\underline{x}_3) \hspace{0.15cm} \underline {= 4}\hspace{0.05cm}.$$		:$$w_{\rm H}(\underline{x}_0) \hspace{0.15cm} \underline {= 0} \hspace{0.05cm}, \hspace{0.3cm}w_{\rm H}(\underline{x}_1) \hspace{0.15cm} \underline {= 2} \hspace{0.05cm}, \hspace{0.3cm} w_{\rm H}(\underline{x}_2) \hspace{0.15cm} \underline {= 2}\hspace{0.05cm}, \hspace{0.3cm}w_{\rm H}(\underline{x}_3) \hspace{0.15cm} \underline {= 4}\hspace{0.05cm}.$$


	'''(6)'''  The Hamming distance between two ~~codewords~~ here can only take the values $2$ and $4$:		'''(6)'''  The Hamming distance between two code words here can only take the values  $2$  and  $4$:
	:$$d_{\rm H}(\underline{x}_0, \hspace{0.05cm}\underline{x}_1) \hspace{0.15cm} \underline {= 2}\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{x}_0, \hspace{0.05cm}\underline{x}_2) = 2\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{x}_0, \hspace{0.05cm}\underline{x}_3) \hspace{0.15cm} \underline {= 4}\hspace{0.05cm},$$		:$$d_{\rm H}(\underline{x}_0, \hspace{0.05cm}\underline{x}_1) \hspace{0.15cm} \underline {= 2}\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{x}_0, \hspace{0.05cm}\underline{x}_2) = 2\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{x}_0, \hspace{0.05cm}\underline{x}_3) \hspace{0.15cm} \underline {= 4}\hspace{0.05cm},$$

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	'''(7)'''  From the result of subtask (6) it follows $d_{\rm min}(\mathcal{C}) \hspace{0.15cm}\underline{= 2}$.		'''(7)'''  From the result of subtask  '''(6)'''  it follows  $d_{\rm min}(\mathcal{C}) \hspace{0.15cm}\underline{= 2}$.
	*Generally, for this quantity:		*Generally,  for this quantity:
	:$$d_{\rm min}(\mathcal{C}) = \min_{\substack{\underline{x},\hspace{0.05cm}\underline{x}' \hspace{0.05cm}\in \hspace{0.05cm} \mathcal{C} \ {\underline{x}} \hspace{0.05cm}\ne \hspace{0.05cm} \underline{x}'}}\hspace{0.1cm}d_{\rm H}(\underline{x}, \hspace{0.05cm}\underline{x}')\hspace{0.05cm}.$$		:$$d_{\rm min}(\mathcal{C}) = \min_{\substack{\underline{x},\hspace{0.05cm}\underline{x}' \hspace{0.05cm}\in \hspace{0.05cm} \mathcal{C} \ {\underline{x}} \hspace{0.05cm}\ne \hspace{0.05cm} \underline{x}'}}\hspace{0.1cm}d_{\rm H}(\underline{x}, \hspace{0.05cm}\underline{x}')\hspace{0.05cm}.$$

Guenter at 12:21, 6 June 2022

2022-06-06T12:21:49Z

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	}}		}}

	[[File:EN_KC_A_1_2.png\|right\|frame\|Clarification of channel ~~(de)~~coding]]		[[File:EN_KC_A_1_2.png\|right\|frame\|Clarification of channel coding and decoding '''Korrektur''']]
	The graph illustrates the channel coding ~~considered here~~  $\mathcal{C}$:		The graph illustrates the channel coding  $\mathcal{C}$  considered here:
	*There are four possible information blocks  $\underline{u} = (u_{1}, u_{2}, \text{...}\hspace{0.05cm} , u_{k~~})$.~~		*There are four possible information blocks  $\underline{u} = (u_{1},\ u_{2},\ \text{...}\hspace{0.05cm} ,\ u_{k})$.
	*Each information block  $\underline{u}$  is uniquely assigned (recognizable by the same color) to the codeword  $\underline{x}= (x_{1}, x_{2}, \text{...}\hspace{0.05cm} , x_{n})$ .
	*Due to decoding errors  $(0 → 1, \ 1 → 0)$  there are more than 4, namely 16 different receive words  $\underline{y} = (y_{1}, y_{2}, \text{...} \hspace{0.05cm} , y_{n})$.

			*Each information block  $\underline{u}$  is uniquely assigned  (recognizable by the same color)  to the code word  $\underline{x}= (x_{1},\ x_{2},\ \text{...}\hspace{0.05cm} ,\ x_{n})$ .

	~~From subtask '''(4)''' we consider the following mapping:~~		*Due to decoding errors  $(0 → 1, \ 1 → 0)$  there are more than 4,  namely 16 different receiving words  $\underline{y} = (y_{1},\ y_{2},\ \text{...} \hspace{0.05cm} ,\ y_{n})$.
	:$$~~\underline{u_0} =~~ (0, 0) \~~leftrightarrow (0, 0, 0,~~ 0) ~~= \underline{x_0}\hspace{0.05cm}~~,$$
	:$$\underline{~~u_1~~} = (~~0, 1) \leftrightarrow (0, 1, 0,~~ 1~~) = \underline{x_1}\hspace{0.05cm~~},$$
	~~:$$~~\~~underline~~{~~u_2~~} ~~= (1~~, 0) \~~leftrightarrow (1, 0, 1, 0) =~~ \~~underline~~{~~x_2~~}\hspace{0.05cm},$$
	~~:$$~~\~~underline~~{~~u_3~~} ~~= (1, 1~~) ~~\leftrightarrow (1, 1, 1, 1) = \underline{x_3}\hspace{0~~.~~05cm}.$$~~


			From subtask  '''(4)'''  we consider the following mapping:
			:$$\underline{u_0} = (0,\ 0) \leftrightarrow (0,\ 0,\ 0,\ 0) = \underline{x_0}\hspace{0.05cm},$$
			:$$\underline{u_1} = (0,\ 1) \leftrightarrow (0,\ 1,\ 0,\ 1) = \underline{x_1}\hspace{0.05cm},$$
			:$$\underline{u_2} = (1,\ 0) \leftrightarrow (1,\ 0,\ 1,\ 0) = \underline{x_2}\hspace{0.05cm},$$
			:$$\underline{u_3} = (1,\ 1) \leftrightarrow (1,\ 1,\ 1,\ 1) = \underline{x_3}\hspace{0.05cm}.$$



			Hints:
			*This exercise belongs to the chapter  [[Channel_Coding/Objective_of_Channel_Coding\|"Objective of Channel Coding"]]

	~~Hints:~~
	*This exercise belongs to the chapter  [[Channel_Coding/Objective_of_Channel_Coding\|Obective of Channel Coding]]
	*Reference is made in particular to the pages		*Reference is made in particular to the pages
	::[[Channel_Coding/Objective_of_Channel_Coding#Block_diagram_and_requirements\|Block diagram and requirements]],		:*[[Channel_Coding/Objective_of_Channel_Coding#Block_diagram_and_requirements\|"Block diagram and requirements"]],
	::[[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding\|Important definitions for block coding]].		:*[[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding\|"Important definitions for block coding"]].


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	{What is the ~~codeword~~ length $n$?		{What is the code word length  $n$?
	\|type="{}"}		\|type="{}"}
	$n \ = \ $ { 4 }		$n \ = \ $ { 4 }
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	{Indicate the Hamming distances between the following code words.		{Indicate the Hamming distances between the following code words.
	\|type="{}"}		\|type="{}"}
	$ d_{\rm H} \ (\underline{x}_0, \underline{x}_1) \ = \ $ { 2 }		$ d_{\rm H} \ (\underline{x}_0,\ \underline{x}_1) \ = \ $ { 2 }
	$ d_{\rm H} \ (\underline{x}_0, \underline{x}_3) \ = \ $ { 4 }		$ d_{\rm H} \ (\underline{x}_0,\ \underline{x}_3) \ = \ $ { 4 }
	$ d_{\rm H} \ (\underline{x}_1, \underline{x}_2) \ = \ $ { 4 }		$ d_{\rm H} \ (\underline{x}_1,\ \underline{x}_2) \ = \ $ { 4 }

	{What is the minimum Hamming distance of the code ~~under consideration~~  $\mathcal{C}$?		{What is the minimum Hamming distance of the code  $\mathcal{C}$  under consideration$?
	\|type="{}"}		\|type="{}"}
	$ d_{\rm min} (\mathcal{C}) \ = \ $ { 2 }		$ d_{\rm min} (\mathcal{C}) \ = \ $ { 2 }

Noah at 16:06, 4 June 2022

2022-06-04T16:06:20Z

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Guenter at 15:25, 2 June 2022

2022-06-02T15:25:14Z

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	[[File:~~P_ID2381__KC_A_1_2~~.png\|right\|frame\|Verdeutlichung der Kanal(de)codierung]]		[[File:EN_KC_A_1_2.png\|right\|frame\|Verdeutlichung der Kanal(de)codierung]]
	Die Grafik verdeutlicht die hier betrachtete Kanalcodierung  $\mathcal{C}$:		Die Grafik verdeutlicht die hier betrachtete Kanalcodierung  $\mathcal{C}$:
	*Es gibt vier mögliche Informationsblöcke  $\underline{u} = (u_{1}, u_{2}, \text{...}\hspace{0.05cm} , u_{k})$.		*Es gibt vier mögliche Informationsblöcke  $\underline{u} = (u_{1}, u_{2}, \text{...}\hspace{0.05cm} , u_{k})$.

Guenter at 09:10, 11 May 2022

2022-05-11T09:10:44Z

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	[[Category:Channel Coding: Exercises\|^1.1 ~~Zielsetzung der Kanalcodierung~~^]]		[[Category:Channel Coding: Exercises\|^1.1 Objective of Channel Coding^]]