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

Guenter at 11:24, 20 July 2022

2022-07-20T11:24:26Z

Guenter at 11:02, 20 July 2022

2022-07-20T11:02:01Z

Guenter at 15:39, 19 July 2022

2022-07-19T15:39:43Z

Noah at 15:27, 14 July 2022

2022-07-14T15:27:10Z

Noah at 15:26, 14 July 2022

2022-07-14T15:26:42Z

Noah at 21:55, 13 July 2022

2022-07-13T21:55:23Z

Guenter at 09:59, 11 May 2022

2022-05-11T09:59:40Z

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 Aufgabe 1.11Z: Nochmals Syndromdecodierung to Exercise 1.11Z: Syndrome Decoding again

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

2021-03-23T12:48:54Z

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

Javier: Text replacement - "[[Kanalcodierung" to "[[Channel_Coding"

2020-07-09T12:55:39Z

Text replacement - "[[Kanalcodierung" to "[[Channel_Coding"

@@ Line 66: / Line 66: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The answer is <u>YES</u>, as can be seen from the given parity-check matrix $\mathbf{H}$:
+'''(1)'''&nbsp; The answer is&nbsp; <u>YES</u>,&nbsp; as can be seen from the given parity-check matrix&nbsp; $\mathbf{H}$:
-*This contains a $3×3$ diagonal matrix at the end.
+*This contains a&nbsp; $3×3$&nbsp; diagonal matrix at the end.
 *The code words are therefore:
@@ Line 74: / Line 74: @@
-'''(2)'''&nbsp; With this received vector $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$ all parity-check equations are satisfied:
+'''(2)'''&nbsp; With this received vector&nbsp; $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$,&nbsp; all parity-check equations are satisfied:
 :$$u_1 \oplus u_2 \oplus u_4 \oplus p_1 = 1 \oplus 0 \oplus 1 \oplus 0 = 0 \hspace{0.05cm},$$
@@ Line 80: / Line 80: @@
 :$$u_1 \oplus u_3 \oplus u_4 \oplus p_3 = 1 \oplus 0 \oplus 1 \oplus 0 = 0 \hspace{0.05cm}.$$
-Accordingly, the correct answer is <u>YES</u>.
+Accordingly,&nbsp; the correct answer is&nbsp; <u>YES</u>.
@@ Line 86: / Line 86: @@
 '''(3)'''&nbsp;   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)'''&nbsp; 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)'''&nbsp; One could now for each&nbsp; $\underline{y}$&nbsp; check the equation&nbsp; $\underline{y} · \boldsymbol{\rm H}^{\rm T} = (0, 0, 0)$.&nbsp; 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  ⇒  $\underline{s} = \underline{s}_{6} = (1, 1, 0)$.
+*$\underline{y}= (1, 1, 0, 1, 0, 1, 0)$&nbsp; differs from&nbsp; $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$&nbsp; in bit&nbsp; $u_{2}$,&nbsp; which is used only in the first two parity-check equations,&nbsp; 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&nbsp;  $\underline{y} = (0, 1, 0, 1, 0, 0, 1)$,&nbsp; we obtain $\underline{s} = \underline{s}_{0} = (0, 0, 0)$, as evidenced by the following calculation:
 :$$u_1 \oplus u_2 \oplus u_4 \oplus p_1 = 0 \oplus 1 \oplus 1 \oplus 0 = 0 \hspace{0.05cm},$$
@@ Line 100: / Line 100: @@
 :$$u_1 \oplus u_3 \oplus u_4 \oplus p_3 = 0 \oplus 0 \oplus 1 \oplus 1 = 0 \hspace{0.05cm}.$$
-*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&nbsp; $\underline{y} = (0, 1, 1, 0, 1, 0, 1),$&nbsp; which differs from the vector&nbsp; $(1, 0, 0, 1, 0, 1, 0)$&nbsp; in all seven bit positions:
 :$$u_1 \oplus u_2 \oplus u_4 \oplus p_1 = 0 \oplus 1 \oplus 0 \oplus 1 = 0 \hspace{0.05cm},$$
@@ Line 106: / Line 106: @@
 :$$u_1 \oplus u_3 \oplus u_4 \oplus p_3 = 0 \oplus 1 \oplus 0 \oplus 1 = 0 \hspace{0.05cm}.$$
-So the correct answers are <u>answers 2 and 3</u>.
+So the correct answers are&nbsp; <u>answers 2 and 3</u>.
 {{ML-Fuß}}

@@ Line 46: / Line 46: @@
 - No.
-{Received&nbsp; $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$. Is this a valid codeword?
+{With received vector&nbsp; $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$.&nbsp; Is this a valid code word?
 |type="()"}
 + Yes,
@@ Line 57: / Line 57: @@
 - $\underline{s} = \underline{s}_{7} = (1, 1, 1).$
-{Which received words lead to the same syndrome as in subtask '''(3)'''?
+{Which received words lead to the same syndrome as in subtask&nbsp; '''(3)'''?
 |type="[]"}
 -  $\underline{y} = (1, 1, 0, 1, 0, 1, 0),$

@@ Line 1: / Line 1: @@
 {{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&nbsp; [[Aufgaben:Exercise_1.11:_Syndrome_Decoding|"Exercise 1.11"]], namely the decoding of a&nbsp; $(7, 4, 3)$ Hamming code with the parity-check matrix
+The same constellation is considered as in&nbsp; [[Aufgaben:Exercise_1.11:_Syndrome_Decoding|"Exercise 1.11"]]:&nbsp; The decoding of a&nbsp; $(7, 4, 3)$&nbsp; Hamming code with the parity-check matrix
 :$${ \boldsymbol{\rm H}}_{\rm } = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$
-Accordingly, the generator polynomial is:
+Accordingly,&nbsp; the generator matrix is:
 :$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$
@@ Line 15: / Line 15: @@
 :$$\underline{s} = \underline{y} \cdot { \boldsymbol{\rm H}}^{\rm T} \in {\rm GF}(2^m) \hspace{0.05cm}.$$
-With this result, any single error in the codeword can be corrected in the Hamming code under consideration.
+With this result,&nbsp; any single error in the code word can be corrected in the Hamming code under consideration.
 *In the error-free case&nbsp; $\underline{s} = \underline{s}_{0} = (0, 0, 0)$.
-*But even three transmission errors may result in&nbsp; $\underline{s}_{0} = (0, 0, 0)$&nbsp;, so these errors remain undetected.
@@ Line 26: / Line 27: @@
 Hints:
 * The exercise belongs to the chapter&nbsp; [[Channel_Coding/Decoding_of_Linear_Block_Codes|"Decoding of Linear Block Codes"]].
 * For more information on syndrome decoding, see the specification sheet for&nbsp; [[Aufgaben:Exercise_1.11:_Syndrome_Decoding|"Exercise 1.11"]].
 * The graph illustrates the three parity-check equations corresponding to the parity-check matrix:
-**first row: red circle,
+**first row: &nbsp; red circle,
-**second row: green circle,
+**second row: &nbsp; green circle,
-**third row: blue circle.
+**third row: &nbsp; blue circle.

@@ Line 3: / Line 3: @@
 [[File:P_ID2399__KC_Z_1_10.png|right|frame|Diagram: Parity-check equations]]
-The same constellation is considered as in&nbsp; [[Aufgaben:Exercise_1.11:_Syndrome_Decoding|Exercise 1.11]], namely the decoding of a&nbsp; $(7, 4, 3)$ Hamming code with the parity-check matrix
+The same constellation is considered as in&nbsp; [[Aufgaben:Exercise_1.11:_Syndrome_Decoding|"Exercise 1.11"]], namely the decoding of a&nbsp; $(7, 4, 3)$ Hamming code with the parity-check matrix
 :$${ \boldsymbol{\rm H}}_{\rm } = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$
@@ Line 11: / Line 11: @@
 :$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$
-In&nbsp; [[Channel_Coding/Decoding_of_Linear_Block_Codes#Principle_of_syndrome_decoding|Syndrome decoding]]&nbsp; one forms from the received vector&nbsp; $\underline{y}$&nbsp; the syndrome&nbsp; $\underline{s}$&nbsp; according to the equation
+In&nbsp; [[Channel_Coding/Decoding_of_Linear_Block_Codes#Principle_of_syndrome_decoding|"syndrome decoding"]]&nbsp; one forms from the received vector&nbsp; $\underline{y}$&nbsp; the syndrome&nbsp; $\underline{s}$&nbsp; according to the equation
 :$$\underline{s} = \underline{y} \cdot { \boldsymbol{\rm H}}^{\rm T} \in {\rm GF}(2^m) \hspace{0.05cm}.$$
@@ Line 26: / Line 26: @@
 Hints:
-* The exercise belongs to the chapter&nbsp; [[Channel_Coding/Decoding_of_Linear_Block_Codes|Decoding of Linear Block Codes]].
+* The exercise belongs to the chapter&nbsp; [[Channel_Coding/Decoding_of_Linear_Block_Codes|"Decoding of Linear Block Codes"]].
-* For more information on syndrome decoding, see the specification sheet for&nbsp; [[Aufgaben:Exercise_1.11:_Syndrome_Decoding|Exercise 1.11]].
+* For more information on syndrome decoding, see the specification sheet for&nbsp; [[Aufgaben:Exercise_1.11:_Syndrome_Decoding|"Exercise 1.11"]].
 * The graph illustrates the three parity-check equations corresponding to the parity-check matrix:
 **first row: red circle,

@@ Line 11: / Line 11: @@
 :$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$
-Bei der&nbsp; [[Channel_Coding/Decodierung_linearer_Blockcodes#Prinzip_der_Syndromdecodierung|Syndromdecodierung]]&nbsp; bildet man aus dem Empfangsvektor&nbsp; $\underline{y}$&nbsp; das Syndrom&nbsp; $\underline{s}$&nbsp; entsprechend der Gleichung
+In&nbsp; [[Channel_Coding/Decoding_of_Linear_Block_Codes#Principle_of_syndrome_decoding|Syndrome decoding]]&nbsp; one forms from the received vector&nbsp; $\underline{y}$&nbsp; the syndrome&nbsp; $\underline{s}$&nbsp; according to the equation
 :$$\underline{s} = \underline{y} \cdot { \boldsymbol{\rm H}}^{\rm T} \in {\rm GF}(2^m) \hspace{0.05cm}.$$
-Mit diesem Ergebnis lässt sich beim betrachteten Hamming–Code ein jeder Einzelfehler im Codewort korrigieren.
+With this result, any single error in the codeword can be corrected in the Hamming code under consideration.
-*Im fehlerfreien Fall gilt&nbsp; $\underline{s} = \underline{s}_{0} = (0, 0, 0)$.
+*In the error-free case&nbsp; $\underline{s} = \underline{s}_{0} = (0, 0, 0)$.
-*Aber auch bei drei Übertragungsfehlern kann sich unter Umständen&nbsp; $\underline{s}_{0} = (0, 0, 0)$&nbsp; ergeben, so dass diese Fehler unerkannt bleiben.
+*But even three transmission errors may result in&nbsp; $\underline{s}_{0} = (0, 0, 0)$&nbsp;, so these errors remain undetected.
@@ Line 25: / Line 25: @@
-''Hinweise:''
+Hints:
-* Die Aufgabe gehört zum Kapitel&nbsp; [[Channel_Coding/Decodierung_linearer_Blockcodes|Decodierung linearer Blockcodes]].
+* The exercise belongs to the chapter&nbsp; [[Channel_Coding/Decoding_of_Linear_Block_Codes|Decoding of Linear Block Codes]].
-* Weitere Informationen zur Syndromdecodierung finden Sie im Angabenblatt zur&nbsp; [[Aufgaben:1.11_Syndromdecodierung|Aufgabe 1.11]].
+* For more information on syndrome decoding, see the specification sheet for&nbsp; [[Aufgaben:Exercise_1.11:_Syndrome_Decoding|Exercise 1.11]].
-* Die Grafik verdeutlicht die drei Prüfgleichungen entsprechend der Prüfmatrix:
+* The graph illustrates the three parity-check equations corresponding to the parity-check matrix:
-**erste Zeile: rote Gruppierung,
+**first row: red circle,
-**zweite Zeile: grüne Gruppierung,
+**second row: green circle,
-**dritte Zeile: blaue Gruppierung.
+**third row: blue circle.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Handelt es sich um einen systematischen Code?
+{Is it a systematic code?
 |type="()"}
-+ Ja,
++ Yes,
-- Nein.
+- No.
-{Empfangen wurde&nbsp; $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$. Ist dies ein gültiges Codewort?
+{Received&nbsp; $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$. Is this a valid codeword?
 |type="()"}
-+ Ja,
++ Yes,
-- Nein.
+- No.
-{Welches Syndrom ergibt sich mit diesem Empfangswort?
+{What syndrome results with this received word?
 |type="()"}
 + $\underline{s} = \underline{s}_{0} = (0, 0, 0),$
@@ Line 54: / Line 54: @@
 - $\underline{s} = \underline{s}_{7} = (1, 1, 1).$
-{Welche Empfangsworte führen zum gleichen Syndrom wie in Teilaufgabe '''(3)'''?
+{Which received words lead to the same syndrome as in subtask '''(3)'''?
 |type="[]"}
 -  $\underline{y} = (1, 1, 0, 1, 0, 1, 0),$
@@ Line 61: / Line 61: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die Antwort ist <u>JA</u>, wie man aus der vorgegebenen Prüfmatrix $\mathbf{H}$ erkennt:
+'''(1)'''&nbsp; The answer is <u>YES</u>, as can be seen from the given parity-check matrix $\mathbf{H}$:
-*Diese beinhaltet am Ende eine $3×3$–Diagonalmatrix.
+*This contains a $3×3$ diagonal matrix at the end.
-*Die Codeworte lauten demzufolge:
+*The code words are therefore:
 :$$ \underline{x} = ( x_1, x_2, x_3, x_4, x_5, x_6, x_7) = ( u_1, u_2, u_3, u_4, p_1, p_2, p_{3}) \hspace{0.05cm}.$$
@@ Line 71: / Line 71: @@
-'''(2)'''&nbsp; Mit diesem Empfangsvektor $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$ werden alle Prüfgleichungen erfüllt:
+'''(2)'''&nbsp; With this received vector $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$ all parity-check equations are satisfied:
 :$$u_1 \oplus u_2 \oplus u_4 \oplus p_1 = 1 \oplus 0 \oplus 1 \oplus 0 = 0 \hspace{0.05cm},$$
@@ Line 77: / Line 77: @@
 :$$u_1 \oplus u_3 \oplus u_4 \oplus p_3 = 1 \oplus 0 \oplus 1 \oplus 0 = 0 \hspace{0.05cm}.$$
-Richtig ist dementsprechend die Antwort <u>JA</u>.
+Accordingly, the correct answer is <u>YES</u>.
-'''(3)'''&nbsp;   Es gilt $\underline{s} = \underline{y} · \boldsymbol{\rm H}^{\rm T}$:
+'''(3)'''&nbsp;   It holds $\underline{s} = \underline{y} · \boldsymbol{\rm H}^{\rm T}$:
 :$$ \underline{s} = \begin{pmatrix} 1 &0 &0 &1 &0 &1 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 &0 &1\\ 1 &1 &0\\ 0 &1 &1\\ 1 &1 &1\\ 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} = \begin{pmatrix} 0 &0 &0 \end{pmatrix} = \underline{s}_0 \hspace{0.2cm} \Rightarrow\hspace{0.2cm} \hspace{0.15cm} \underline{ \rm Antwort \hspace{0.15cm}1} \hspace{0.05cm}.$$
@@ Line 87: / Line 87: @@
-'''(4)'''&nbsp; Man könnte nun für jedes $\underline{y}$ die Gleichung $\underline{y} · \boldsymbol{\rm H}^{\rm T} = (0, 0, 0)$ überprüfen. Hier soll nun das Ergebnis auf anderem Wege gewonnen werden:
+'''(4)'''&nbsp; 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:
-*$\underline{y}= (1, 1, 0, 1, 0, 1, 0)$ unterscheidet sich von $\underline{y} = (1, 0, 0, 1, 0, 1, 0)$ im Bit $u_{2}$, das nur in den beiden ersten Prüfgleichungen verwendet wird, nicht jedoch in der letzten  ⇒  $\underline{s} = \underline{s}_{6} = (1, 1, 0)$.
+*$\underline{y}= (1, 1, 0, 1, 0, 1, 0)$ 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  ⇒  $\underline{s} = \underline{s}_{6} = (1, 1, 0)$.
-*Wendet man die Prüfgleichungen auf $\underline{y} = (0, 1, 0, 1, 0, 0, 1)$ an, so erhält man $\underline{s} = \underline{s}_{0} = (0, 0, 0)$, wie die folgende Rechnung belegt:
+*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:
 :$$u_1 \oplus u_2 \oplus u_4 \oplus p_1 = 0 \oplus 1 \oplus 1 \oplus 0 = 0 \hspace{0.05cm},$$
@@ Line 97: / Line 97: @@
 :$$u_1 \oplus u_3 \oplus u_4 \oplus p_3 = 0 \oplus 0 \oplus 1 \oplus 1 = 0 \hspace{0.05cm}.$$
-*Zum gleichen Ergebnis kommt man mit dem Empfangsvektor $\underline{y}  = (0, 1, 1, 0, 1, 0, 1),$ der sich vom Vektor $(1, 0, 0, 1, 0, 1, 0)$ in allen sieben Bitpositionen unterscheidet:
+*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:
 :$$u_1 \oplus u_2 \oplus u_4 \oplus p_1 = 0 \oplus 1 \oplus 0 \oplus 1 = 0 \hspace{0.05cm},$$
@@ Line 103: / Line 103: @@
 :$$u_1 \oplus u_3 \oplus u_4 \oplus p_3 = 0 \oplus 1 \oplus 0 \oplus 1 = 0 \hspace{0.05cm}.$$
-Richtig sind also die <u>Antworten 2 und 3</u>.
+So the correct answers are <u>answers 2 and 3</u>.
 {{ML-Fuß}}

← Older revision		Revision as of 09:59, 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^]]

← Older revision	Revision as of 09:53, 11 May 2022
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← Older revision		Revision as of 12:48, 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^]]