Aufgaben:Exercise 1.13Z: Binary Erasure Channel Decoding again - Revision history

Hwang at 17:00, 1 November 2022

2022-11-01T17:00:04Z

Guenter at 16:11, 21 July 2022

2022-07-21T16:11:23Z

Guenter at 15:58, 21 July 2022

2022-07-21T15:58:16Z

Noah at 22:00, 16 July 2022

2022-07-16T22:00:34Z

Noah at 21:59, 16 July 2022

2022-07-16T21:59:56Z

Guenter at 10:02, 11 May 2022

2022-05-11T10:02:06Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.13Z: Nochmals BEC–Decodierung to Aufgaben:Exercise 1.13Z: Binary Erasure Channel Decoding again

2022-05-11T09:57:19Z

Guenter moved page Aufgabe 1.13Z: Nochmals BEC–Decodierung to Exercise 1.13Z: Binary Erasure Channel Decoding again

Javier: Text replacement - "„" to """

2021-05-28T14:21:25Z

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2021-05-28T14:19:05Z

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Javier: Text replacement - "Category:Aufgaben zu Kanalcodierung" to "Category:Channel Coding: Exercises"

2021-03-23T12:52:08Z

Text replacement - "Category:Aufgaben zu Kanalcodierung" to "Category:Channel Coding: Exercises"

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 '''(3)'''&nbsp;  If no more than&nbsp; $e_{\rm max} = d_{\rm min}- 1 \ \ \underline{ = 2}$&nbsp; bits are erased,&nbsp; decoding is possible with certainty.
-*Each codeword differs from every other in at least three bit positions.
+*Each code word differs from every other in at least three bit positions.
 *With only two erasures,&nbsp; therefore,&nbsp; the code word can be reconstructed in any case.

@@ Line 50: / Line 50: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The $(7, 4, 3)$ Hamming code is considered here. Accordingly, the minimum distance is $d_{\rm min} \ \underline{= 3}$.
+'''(1)'''&nbsp; The&nbsp; $(7, 4, 3)$&nbsp; Hamming code is considered here.&nbsp; Accordingly,&nbsp; the minimum distance is&nbsp; $d_{\rm min} \ \underline{= 3}$.
-'''(5)'''&nbsp;  Correct are the <u>suggested solutions 1 and 2</u>.
+* $\underline{y}_{\rm D} = (1, 0, {\rm E},  {\rm E},  {\rm E},  {\rm E}, 0)$&nbsp; cannot be decoded because less than&nbsp; $k = 4$&nbsp; bits&nbsp; (number of information bits)&nbsp; arrive.
-* $\underline{y}_{\rm D} = (1, 0, {\rm E},  {\rm E},  {\rm E},  {\rm E}, 0)$&nbsp; cannot be decoded because less than&nbsp; $k = 4$&nbsp; bits (number of information bits) arrive.
+*$\underline{y}_{\rm C} = ( {\rm E},  {\rm E},  {\rm E}, 1, 0, 1, 0)$&nbsp; is not decodable because&nbsp;  $\underline{x} = (0, 1, 1, 1, 0, 1, 0)$&nbsp; and &nbsp; $\underline{x} = (1, 0, 0, 1, 0, 1, 0)$&nbsp; are possible outcomes.
-*Also &nbsp;$\underline{y}_{\rm C} = ( {\rm E},  {\rm E},  {\rm E}, 1, 0, 1, 0)$&nbsp; is not decodable because&nbsp;  $\underline{x} = (0, 1, 1, 1, 0, 1, 0)$&nbsp; and &nbsp; $\underline{x} = (1, 0, 0, 1, 0, 1, 0)$&nbsp; are possible outcomes.
+*$\underline{y}_{\rm B} = ( {\rm E},  {\rm E}, 0,  {\rm E}, 0, 1, 0)$&nbsp; is decodable,&nbsp; since of the 16 possible code words only&nbsp; $\underline{x} = (1, 0, 0, 1, 0, 1, 0)$&nbsp; matches&nbsp; $\underline{y}_{\rm B}$&nbsp; in positions 3, 5, 6, 7.
-*$\underline{y}_{\rm B} = ( {\rm E},  {\rm E}, 0,  {\rm E}, 0, 1, 0)$&nbsp; is decodable, since of the 16 possible codewords only&nbsp; $\underline{x} = (1, 0, 0, 1, 0, 1, 0)$&nbsp; matches&nbsp; $\underline{y}_{\rm B}$&nbsp; in positions 3, 5, 6, 7.
+*$\underline{y}_{\rm A} = (1, 0, 0, 1, {\rm E}, {\rm E}, {\rm E})$&nbsp; is decodable.&nbsp; Only the&nbsp; $m = 3$&nbsp; parity bits are missing.&nbsp; Thus,&nbsp; the information word&nbsp; $\underline{u} = (1, 0, 0, 1)$&nbsp; is also fixed&nbsp; (systematic code).
 {{ML-Fuß}}

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Decoding_of_Linear_Block_Codes}}
-[[File:P_ID2541__KC_Z_1_13.png|right|frame|Code table of the $\rm HC (7, 4, 3)$]]
+[[File:P_ID2541__KC_Z_1_13.png|right|frame|Code table of the&nbsp; $\rm HC (7, 4, 3)$]]
-We consider again as in the&nbsp; [[Aufgaben:Exercise_1.13:_Binary_Erasure_Channel_Decoding|"Exercise 1. 13"]]&nbsp; the decoding of a&nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Hamming_Codes|"Hamming Codes"]]&nbsp; after transmission over an erasure channel &nbsp; ⇒ &nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_. E2.80.93_BEC|"Binary Erasure Channel"]]&nbsp; (abbreviated BEC).
+We consider as in the&nbsp; [[Aufgaben:Exercise_1.13:_Binary_Erasure_Channel_Decoding|"Exercise 1. 13"]]&nbsp; the decoding of a&nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Hamming_Codes|"Hamming Codes"]]&nbsp; after transmission over an erasure channel &nbsp; ⇒ &nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_. E2.80.93_BEC|"Binary Erasure Channel"]]&nbsp; $\rm (BEC)$.
 The&nbsp; $(7, 4, 3)$-Hamming code is fully described by the adjacent code table&nbsp; $\underline{u}_{i} → \underline{x}_{i}$&nbsp; which can be used to find all solutions.
@@ Line 15: / Line 12: @@
 Hints:
 * This exercise belongs to the chapter&nbsp; [[Channel_Coding/Decodierung_linearer_Blockcodes|"Decoding of Linear Block Codes"]].
-* In contrast to&nbsp; [[Aufgaben:Exercise_1.13:_Binary_Erasure_Channel_Decoding|"Exercise 1.13"]]&nbsp; here the solution is not to be found formally, but intuitively.
@@ Line 30: / Line 28: @@
 - NO.
-{Up to how many erasures (maximum number:&nbsp; $e_{\rm max})$&nbsp; is successful decoding guaranteed?
+{Up to how many erasures&nbsp; $($maximum number:&nbsp; $e_{\rm max})$&nbsp; is successful decoding guaranteed?
 |type="{}"}
 $\ e_{\rm max} \ = \ $ { 2 }
-{What is the sent information word&nbsp; $\underline{u}$&nbsp; for&nbsp; $\underline{y} = (1, 0, {\rm E}, {\rm E}, 0, 1, 0)$?
+{The received word is&nbsp; $\underline{y} = (1, 0, {\rm E}, {\rm E}, 0, 1, 0)$.&nbsp; What is the sent information word&nbsp; $\underline{u}$?
 |type="()"}
 - $\underline{u} = (1, 0, 0, 0),$
-+ $\underline{u}= (1, 0, 0, 1),$
++ $\underline{u} = (1, 0, 0, 1),$
 - $\underline{u} = (1, 0, 1, 0),$
 - $\underline{u} = (1, 0, 1, 1).$

@@ Line 3: / Line 3: @@
 [[File:P_ID2541__KC_Z_1_13.png|right|frame|Code table of the $\rm HC (7, 4, 3)$]]
-We consider again as in the&nbsp; [[Aufgaben:Exercise_1.13:_Binary_Erasure_Channel_Decoding|Task 1. 13]]&nbsp; the decoding of a&nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Hamming_Codes|Hamming Codes]]&nbsp; after transmission over an erasure channel &nbsp; ⇒ &nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_. E2.80.93_BEC|Binary Erasure Channel]]&nbsp; (abbreviated BEC).
+We consider again as in the&nbsp; [[Aufgaben:Exercise_1.13:_Binary_Erasure_Channel_Decoding|"Exercise 1. 13"]]&nbsp; the decoding of a&nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Hamming_Codes|"Hamming Codes"]]&nbsp; after transmission over an erasure channel &nbsp; ⇒ &nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_. E2.80.93_BEC|"Binary Erasure Channel"]]&nbsp; (abbreviated BEC).
 The&nbsp; $(7, 4, 3)$-Hamming code is fully described by the adjacent code table&nbsp; $\underline{u}_{i} → \underline{x}_{i}$&nbsp; which can be used to find all solutions.
@@ Line 14: / Line 14: @@
 Hints:
-* This exercise belongs to the chapter&nbsp; [[Channel_Coding/Decodierung_linearer_Blockcodes|Decoding of Linear Block Codes]].
+* This exercise belongs to the chapter&nbsp; [[Channel_Coding/Decodierung_linearer_Blockcodes|"Decoding of Linear Block Codes"]].
-* In contrast to&nbsp; [[Aufgaben:Exercise_1.13:_Binary_Erasure_Channel_Decoding|Exercise 1.13]]&nbsp; here the solution is not to be found formally, but intuitively.
+* In contrast to&nbsp; [[Aufgaben:Exercise_1.13:_Binary_Erasure_Channel_Decoding|"Exercise 1.13"]]&nbsp; here the solution is not to be found formally, but intuitively.

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Kanalcodierung/Decodierung linearer Blockcodes}}
+{{quiz-Header|Buchseite=Channel_Coding/Decoding_of_Linear_Block_Codes}}
-[[File:P_ID2541__KC_Z_1_13.png|right|frame|Codetabelle des&nbsp; $\rm HC \ (7, 4, 3)$]]
+[[File:P_ID2541__KC_Z_1_13.png|right|frame|Code table of the $\rm HC (7, 4, 3)$]]
-Wir betrachten wieder wie in der&nbsp; [[Aufgaben:Aufgabe_1.13:_Decodierung_beim_binären_Auslöschungskanal_(BEC)|Aufgabe 1.13]]&nbsp; die Decodierung eines&nbsp; [[Channel_Coding/Beispiele_binärer_Blockcodes#Hamming.E2.80.93Codes|Hamming–Codes]]&nbsp; nach der Übertragung über einen Auslöschungskanal &nbsp; ⇒ &nbsp; [[Channel_Coding/Kanalmodelle_und_Entscheiderstrukturen#Binary_Erasure_Channel_.E2.80.93_BEC|Binary Erasure Channel]]&nbsp; (abgekürzt BEC).
+We consider again as in the&nbsp; [[Aufgaben:Exercise_1.13:_Binary_Erasure_Channel_Decoding|Task 1. 13]]&nbsp; the decoding of a&nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Hamming_Codes|Hamming Codes]]&nbsp; after transmission over an erasure channel &nbsp; ⇒ &nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_. E2.80.93_BEC|Binary Erasure Channel]]&nbsp; (abbreviated BEC).
-Der&nbsp; $(7, 4, 3)$–Hamming–Code wird durch die nebenstehende Codetabelle&nbsp; $\underline{u}_{i} → \underline{x}_{i}$&nbsp; vollständig beschrieben, anhand derer alle Lösungen gefunden werden können.
+The&nbsp; $(7, 4, 3)$-Hamming code is fully described by the adjacent code table&nbsp; $\underline{u}_{i} → \underline{x}_{i}$&nbsp; which can be used to find all solutions.
@@ Line 13: / Line 13: @@
-''Hinweise'' :
+Hints:
-* Die Aufgabe bezieht sich auf das Kapitel&nbsp; [[Channel_Coding/Decodierung_linearer_Blockcodes|Decodierung linearer Blockcodes]].
+* This exercise belongs to the chapter&nbsp; [[Channel_Coding/Decodierung_linearer_Blockcodes|Decoding of Linear Block Codes]].
-* Im Gegensatz zur&nbsp;  [[Aufgaben:Aufgabe_1.13:_Decodierung_beim_binären_Auslöschungskanal_(BEC)|Aufgabe 1.13]]&nbsp; soll hier die Lösung nicht formal, sondern intuitiv gefunden werden.
+* In contrast to&nbsp; [[Aufgaben:Exercise_1.13:_Binary_Erasure_Channel_Decoding|Exercise 1.13]]&nbsp; here the solution is not to be found formally, but intuitively.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-Wie groß ist die minimale Distanz&nbsp; $\ d_{\rm min}$&nbsp; des vorliegenden Codes?
+What is the minimum distance&nbsp; $\ d_{\rm min}$&nbsp; of the present code?
 |type="{}"}
 $\ d_{\rm min} \ = \ $ { 3 }
-{Ist der Code systematisch?
+{Is the code systematic?
 |type="()"}
-+ JA.
++ YES.
-- NEIN.
+- NO.
-{Bis zu wie vielen Auslöschungen ("Erasures"; &nbsp; maximale Anzahl:&nbsp; $e_{\rm max})$&nbsp; ist eine erfolgreiche Decodierung gewährleistet?
+{Up to how many erasures (maximum number:&nbsp; $e_{\rm max})$&nbsp; is successful decoding guaranteed?
 |type="{}"}
 $\ e_{\rm max} \ = \ $ { 2 }
-{Wie lautet das gesendete Informationswort&nbsp; $\underline{u}$&nbsp; für&nbsp; $\underline{y} = (1, 0, {\rm E}, {\rm E}, 0, 1, 0)$?
+{What is the sent information word&nbsp; $\underline{u}$&nbsp; for&nbsp; $\underline{y} = (1, 0, {\rm E}, {\rm E}, 0, 1, 0)$?
 |type="()"}
 - $\underline{u} = (1, 0, 0, 0),$
@@ Line 41: / Line 41: @@
 - $\underline{u} = (1, 0, 1, 1).$
-{Welche der nachfolgenden Empfangsworte können decodiert werden?
+{Which of the following received words can be decoded?
 |type="[]"}
 + $\underline{y}_{\rm A }= (1, 0, 0, 1, {\rm E}, {\rm E}, {\rm E}),$
@@ Line 50: / Line 50: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Betrachtet wird hier der $(7, 4, 3)$–Hamming–Code. Dementsprechend ist die minimale Distanz $d_{\rm min} \ \underline{= 3}$.
+'''(1)'''&nbsp; The $(7, 4, 3)$ Hamming code is considered here. Accordingly, the minimum distance is $d_{\rm min} \ \underline{= 3}$.
-'''(2)'''&nbsp; Die ersten $k = 4$ Bit eines jeden Codewortes $\underline{x}$ stimmen mit dem Informationswort $\underline{u}$ überein. Richtig ist somit <u>JA</u>.
+'''(2)'''&nbsp; The first $k = 4$ bits of each codeword $\underline{x}$ match the information word $\underline{u}$. Correct is therefore <u>YES</u>.
-'''(3)'''&nbsp;  Werden nicht mehr als $e_{\rm max} = d_{\rm min} – 1 \underline{ = 2}$ Bit ausgelöscht,so ist eine Decodierung mit Sicherheit möglich.
+'''(3)'''&nbsp;  If no more than $e_{\rm max} = d_{\rm min} - 1 \underline{ = 2}$ bits are erased,decoding is possible with certainty.
-*Jedes Codewort unterscheidet sich von jedem anderen in mindestens drei Bitpositionen.
+*Each codeword differs from every other in at least three bit positions.
-*Bei nur zwei Auslöschungen kann deshalb das Codewort in jedem Fall rekonstruiert werden.
+*With only two erasures, therefore, the codeword can be reconstructed in any case.
-'''(4)'''&nbsp;  In der Codetabelle findet man ein einziges Codewort, das mit „$10$” beginnt und mit „$010$” endet, nämlich $\underline{x} = (1, 0, 0, 1, 0, 1, 0)$.
+'''(4)'''&nbsp;  In the code table, one finds a single codeword starting with "$10$" and ending with "$010$", namely $\underline{x} = (1, 0, 0, 1, 0, 1, 0)$.
-Da es sich um einen systematischen Code handelt, beschreiben die ersten $k = 4$ Bit das Informationswort $\underline{u} = (1, 0, 0, 1)$ &nbsp; ⇒&nbsp;  <u>Antwort 2</u>.
+Since this is a systematic code, the first $k = 4$ bits describe the information word $\underline{u} = (1, 0, 0, 1)$ &nbsp; ⇒&nbsp; <u>answer 2</u>.
-'''(5)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 1 und 2</u>.
+'''(5)'''&nbsp;  Correct are the <u>suggested solutions 1 and 2</u>.
-* $\underline{y}_{\rm D} = (1, 0, {\rm E},  {\rm E},  {\rm E},  {\rm E}, 0)$&nbsp; kann nicht decodiert werden, da weniger als&nbsp; $k = 4$&nbsp; Bit (Anzahl der Informationsbit) ankommen.
+* $\underline{y}_{\rm D} = (1, 0, {\rm E},  {\rm E},  {\rm E},  {\rm E}, 0)$&nbsp; cannot be decoded because less than&nbsp; $k = 4$&nbsp; bits (number of information bits) arrive.
-*Auch &nbsp;$\underline{y}_{\rm C} = ( {\rm E},  {\rm E},  {\rm E}, 1, 0, 1, 0)$&nbsp; kann nicht decodierbar, da&nbsp;  $\underline{x} = (0, 1, 1, 1, 0, 1, 0)$&nbsp; und &nbsp; $\underline{x} = (1, 0, 0, 1, 0, 1, 0)$&nbsp; als mögliches Ergebnis in Frage kommen.
+*Also &nbsp;$\underline{y}_{\rm C} = ( {\rm E},  {\rm E},  {\rm E}, 1, 0, 1, 0)$&nbsp; is not decodable because&nbsp;  $\underline{x} = (0, 1, 1, 1, 0, 1, 0)$&nbsp; and &nbsp; $\underline{x} = (1, 0, 0, 1, 0, 1, 0)$&nbsp; are possible outcomes.
-*$\underline{y}_{\rm B} = ( {\rm E},  {\rm E}, 0,  {\rm E}, 0, 1, 0)$&nbsp; ist decodierbar, da von den 16 möglichen Codeworten nur&nbsp; $\underline{x} = (1, 0, 0, 1, 0, 1, 0)$&nbsp; mit&nbsp; $\underline{y}_{\rm B}$&nbsp; in den Positionen 3, 5, 6, 7 übereinstimmt.
+*$\underline{y}_{\rm B} = ( {\rm E},  {\rm E}, 0,  {\rm E}, 0, 1, 0)$&nbsp; is decodable, since of the 16 possible codewords only&nbsp; $\underline{x} = (1, 0, 0, 1, 0, 1, 0)$&nbsp; matches&nbsp; $\underline{y}_{\rm B}$&nbsp; in positions 3, 5, 6, 7.
-*$\underline{y}_{\rm A} = (1, 0, 0, 1, {\rm E}, {\rm E}, {\rm E})$ ist decodierbar. Es fehlen nur die $m = 3$ Prüfbit. Damit liegt das Informationswort $\underline{u} = (1, 0, 0, 1)$ ebenfalls fest (systematischer Code).
+*$\underline{y}_{\rm A} = (1, 0, 0, 1, {\rm E}, {\rm E}, {\rm E})$ is decodable. Only the $m = 3$ parity bits are missing. Thus, the information word $\underline{u} = (1, 0, 0, 1)$ is also fixed (systematic code).
 {{ML-Fuß}}

← Older revision		Revision as of 10:02, 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


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← Older revision	Revision as of 09:57, 11 May 2022
(No difference)

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 - NEIN.
-{Bis zu wie vielen Auslöschungen (&bdquo;Erasures&rdquo;; &nbsp; maximale Anzahl:&nbsp; $e_{\rm max})$&nbsp; ist eine erfolgreiche Decodierung gewährleistet?
+{Bis zu wie vielen Auslöschungen (&bdquo;Erasures"; &nbsp; maximale Anzahl:&nbsp; $e_{\rm max})$&nbsp; ist eine erfolgreiche Decodierung gewährleistet?
 |type="{}"}
 $\ e_{\rm max} \ = \ $ { 2 }

← Older revision		Revision as of 12:52, 23 March 2021
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−	[[Category:~~Aufgaben zu Kanalcodierung~~\|^1.5 Decodierung linearer Blockcodes	+	[[Category:Channel Coding: Exercises\|^1.5 Decodierung linearer Blockcodes


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