Difference between revisions of "Aufgaben:Exercise 1.09: Extended Hamming Code"

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[[File:EN_KC_A_1_9.png|right|frame|$\text{(7, 4)}$&ndash;Hamming–Code (gelb hinterlegt) <br>und&nbsp; $\text{(8, 4)}$&ndash;Erweiterung  (grün)]]
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[[File:EN_KC_A_1_9.png|right|frame|$\text{(7, 4)}$ Hamming code (yellow background) <br>and&nbsp; $\text{(8, 4)}$ extension (green).]]
  
Es sollen zwei Codes miteinander verglichen werden, deren Codetabellen rechts angegeben sind.  
+
Two codes are to be compared, whose code tables are given on the right.  
*Die ersten vier Bit eines jeden Codewortes&nbsp; $\underline{x}$&nbsp; sind gleich dem jeweiligen Informationswort&nbsp; $\underline{u}$&nbsp; (schwarze Schrift).  
+
*The first four bits of each code word&nbsp; $\underline{x}$&nbsp; are equal to the respective information word&nbsp; $\underline{u}$&nbsp; (black font).  
*Danach folgen&nbsp; $m = n- k$&nbsp; Prüfbit (rote Schrift).
+
*Then follow&nbsp; $m = n- k$&nbsp; parity bit (red font).
  
  
Der systematische&nbsp; $\text{(7, 4)}$–Hamming–Code wurde bereits in&nbsp; [[Aufgaben:1.6_Zum_(7,_4)–Hamming–Code|Aufgabe 1.6]]&nbsp; sowie&nbsp; [[Aufgaben:Aufgabe_1.07:_Prüf-_und_Generatormatrix_des_HC_(7,_4,_3)|Aufgabe 1.7]]&nbsp; behandelt. Prüfmatrix und Generatormatrix dieses Codes sind wie folgt gegeben:
+
The systematic&nbsp; $\text{(7, 4)}$-Hamming code has already been discussed in&nbsp; [[Aufgaben:Exercise_1.6:_(7,_4)_Hamming_Code|Exercise 1.6]]&nbsp; and&nbsp; [[Aufgaben:Exercise_1.6:_(7,_4)_Hamming_Code|Exercise 1.7]]&nbsp;. The parity-check matrix and generator matrix of this code are given as follows:
 
:$${ \boldsymbol{\rm H}}_1 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix}\hspace{0.05cm},$$
 
:$${ \boldsymbol{\rm H}}_1 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix}\hspace{0.05cm},$$
  
 
:$${ \boldsymbol{\rm G}}_1 = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$
 
:$${ \boldsymbol{\rm G}}_1 = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$
  
Im weiteren Verlauf der Aufgabe wird dieser (gelb hinterlegte) Code&nbsp; $\mathcal{C}_{1}$&nbsp; genannt.
+
In the further course of the exercise this (yellow highlighted) code is called&nbsp; $\mathcal{C}_{1}$&nbsp;.
  
  
Die rechte Spalte in obiger Tabelle gibt einen Blockcode mit den Parametern&nbsp; $n = 8$&nbsp; und&nbsp; $k = 4$&nbsp; an, der in der Literatur meist als „Erweiterter Hamming–Code” bezeichnet wird. Wir nennen diesen (grün hinterlegten) Code im Folgenden&nbsp; $\mathcal{C}_{2}$&nbsp; und bezeichnen dessen Prüfmatrix mit&nbsp; ${ \boldsymbol{\rm H}}_{2}$&nbsp; und die dazugehörige Generatormatrix mit&nbsp; ${ \boldsymbol{\rm G}}_{2}$ .
+
The right column in the above table specifies a block code with parameters&nbsp; $n = 8$&nbsp; and&nbsp; $k = 4$&nbsp;, usually referred to in the literature as the "Extended Hamming Code". We refer to this code (highlighted in green) in the following&nbsp; $\mathcal{C}_{2}$&nbsp; and denote its parity-check matrix by&nbsp; ${ \boldsymbol{\rm H}}_{2}$&nbsp; and the corresponding generator matrix by&nbsp; ${ \boldsymbol{\rm G}}_{2}$ .
  
Die Fragen zu dieser Aufgabe beziehen sich auf
+
The questions for this exercise are related to
*die&nbsp; [[Channel_Coding/Zielsetzung_der_Kanalcodierung#Einige_wichtige_Definitionen_zur_Blockcodierung|Coderate]],
+
*the&nbsp; [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|code rate]],
*die&nbsp; [[Channel_Coding/Zielsetzung_der_Kanalcodierung#Einige_wichtige_Definitionen_zur_Blockcodierung|minimale Distanz]]&nbsp; zwischen zwei Codeworten,
+
*the&nbsp; [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|minimum distance]]&nbsp; between two codewords,
*die&nbsp; [[Channel_Coding/Allgemeine_Beschreibung_linearer_Blockcodes#Codefestlegung_durch_die_Pr.C3.BCfmatrix|Prüfmatrix]]&nbsp; und die&nbsp; [[Channel_Coding/Allgemeine_Beschreibung_linearer_Blockcodes#Codefestlegung_durch_die_Generatormatrix|Generatormatrix]]&nbsp; des erweiterten&nbsp; $\text{(8, 4)}$–Hamming–Codes.
+
*the&nbsp; [[Channel_Coding/General_Description_of_Linear_Block_Codes#Code_definition_by_the_parity-check_matrix|parity-check matrix]]&nbsp; and the&nbsp; [[Channel_Coding/General_Description_of_Linear_Block_Codes#Code_definition_by_the_generator_matrix|generator matrix]]&nbsp; of the extended&nbsp; $\text{(8, 4)}$ Hamming code.
  
  
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''Hinweise'':  
+
Hints:  
  
*Die Aufgabe gehört zum Kapitel&nbsp; [[Channel_Coding/Allgemeine_Beschreibung_linearer_Blockcodes|Allgemeine Beschreibung linearer Blockcodes]].
+
*This exercise belongs to the chapter&nbsp; [[Channel_Coding/General_Description_of_Linear_Block_Codes|General Description of Linear Block Codes]].
*Beachten Sie bei der Lösung, dass&nbsp; $\mathcal{C}_{1}$&nbsp; und&nbsp; $\mathcal{C}_{2}$&nbsp; jeweils&nbsp; [[Channel_Coding/Allgemeine_Beschreibung_linearer_Blockcodes#Systematische_Codes|systematische Codes]]&nbsp; sind.  
+
*Note in the solution that&nbsp; $\mathcal{C}_{1}$&nbsp; and&nbsp; $\mathcal{C}_{2}$&nbsp; are each&nbsp; [[Channel_Coding/General_Description_of_Linear_Block_Codes#Systematic_Codes|Systematic Codes]]&nbsp;.  
*Die folgende&nbsp; [[Aufgaben:Aufgabe_1.09Z:_Erweiterung_und/oder_Punktierung|Aufgabe 1.9Z]]&nbsp; behandelt die Erweiterung von Codes in etwas allgemeinerer Form.
+
*The following&nbsp; [[Aufgaben:Exercise_1.09Z:_Extension_and/or_Puncturing|Exercise 1.9Z]]&nbsp; deals with the extension of codes in somewhat more general terms.
 
   
 
   
  
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
  
{Geben Sie die Coderaten von&nbsp; $\mathcal{C}_{1}$&nbsp; und&nbsp; $\mathcal{C}_{2}$&nbsp; an.
+
{Specify the code rates of&nbsp; $\mathcal{C}_{1}$&nbsp; and&nbsp; $\mathcal{C}_{2}$&nbsp;.
 
|type="{}"}
 
|type="{}"}
 
$\mathcal{C}_{1}\text{:}\hspace{0.4cm}R \ = \ $ { 0.571 3% }
 
$\mathcal{C}_{1}\text{:}\hspace{0.4cm}R \ = \ $ { 0.571 3% }
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{Geben Sie die minimalen Distanzen von&nbsp; $\mathcal{C}_{1}$&nbsp; und&nbsp; $\mathcal{C}_{2}$&nbsp; an.
+
{Give the minimum distances of&nbsp; $\mathcal{C}_{1}$&nbsp; and&nbsp; $\mathcal{C}_{2}$&nbsp; .
 
|type="{}"}
 
|type="{}"}
 
$\mathcal{C}_{1}\text{:}\hspace{0.4cm}d_{\rm min} \ = \ $ { 3 }
 
$\mathcal{C}_{1}\text{:}\hspace{0.4cm}d_{\rm min} \ = \ $ { 3 }
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{Welches Format besitzt die Prüfmatrix&nbsp; $\boldsymbol{\rm H}_{2}$&nbsp; von&nbsp; $\mathcal{C}_{2}$?
+
{What is the format of the parity-check matrix&nbsp; $\boldsymbol{\rm H}_{2}$&nbsp; of&nbsp; $\mathcal{C}_{2}$?
 
|type="{}"}
 
|type="{}"}
 
${\rm Spaltenzahl} \ = \ $ { 8 }
 
${\rm Spaltenzahl} \ = \ $ { 8 }
 
${\rm Zeilenzahl} \ = \ $    { 4 }
 
${\rm Zeilenzahl} \ = \ $    { 4 }
  
{Leiten Sie aus der Codetabelle die Gleichung für das Codebit&nbsp; $x_ {8} (= p_{4})$&nbsp; ab. Welche Angabe ist richtig?
+
{Derive the equation for the code bit&nbsp; $x_ {8} (= p_{4})$&nbsp; from the code table. Which specification is correct?
 
|type="()"}
 
|type="()"}
  
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+ $x_{8} = x_{1}⊕x_{2}⊕x_{3}⊕x_{4}⊕x_{5}⊕x_{6}⊕x_{7}.$
 
+ $x_{8} = x_{1}⊕x_{2}⊕x_{3}⊕x_{4}⊕x_{5}⊕x_{6}⊕x_{7}.$
  
{Welche Aussagen gelten für&nbsp; ${ \boldsymbol{\rm H}}_{2}$? ''Hinweis:'' &nbsp;Richtig sind 3 von 4 Antworten.
+
{Which statements are true for&nbsp; ${ \boldsymbol{\rm H}}_{2}$? ''Hint:'' &nbsp;Correct are 3 out of 4 answers.
 
|type="[]"}
 
|type="[]"}
  
+ Die Zeile 1 lautet:  &nbsp; $1 1 0 1 1 0 0 0$.
+
+ Row 1 reads:  &nbsp; $1 1 0 1 1 0 0 0$.
+ Die Zeile 2 lautet:  &nbsp; $0 1 1 1 0 1 0 0$.
+
+ Row 2 reads:  &nbsp; $0 1 1 1 0 1 0 0$.
- Die Zeile 3 lautet:  &nbsp;  $0 0 0 0 1 1 1 1$.
+
- Row 3 reads:  &nbsp;  $0 0 0 0 1 1 1 1$.
+ Die Zeile 4 lautet:  &nbsp;  $1 1 1 1 1 1 1 1$.
+
+ Row 4 reads:  &nbsp;  $1 1 1 1 1 1 1 1$.
  
{Welche Umformung ist für die letzte Zeile von&nbsp; ${ \boldsymbol{\rm H}}_{2}$&nbsp; zulässig?
+
{Which transformation is allowed for the last row of&nbsp; ${ \boldsymbol{\rm H}}_{2}$&nbsp;?
 
|type="()"}
 
|type="()"}
 
- $1 1 1 1 1 1 1 1  →  0 0 0 0 0 0 0 0,$
 
- $1 1 1 1 1 1 1 1  →  0 0 0 0 0 0 0 0,$
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- $1 1 1 1 1 1 1 1  →  0 0 1 0 1 0 0 0.$
 
- $1 1 1 1 1 1 1 1  →  0 0 1 0 1 0 0 0.$
  
{Geben Sie die zugehörige Generatormatrix&nbsp; ${ \boldsymbol{\rm G}}_{2}$&nbsp; an. Welche Aussagen treffen zu?
+
{Give the corresponding generator matrix&nbsp; ${ \boldsymbol{\rm G}}_{2}$&nbsp;. Which statements are true?
 
|type="[]"}
 
|type="[]"}
- ${ \boldsymbol{\rm G}}_{2}$&nbsp; hat gleiches Format wie die Matrix&nbsp; ${ \boldsymbol{\rm G}}_{1}$ des&nbsp; $\text{(7, 4)}$–Codes.
+
- ${ \boldsymbol{\rm G}}_{2}$&nbsp; has same format as matrix&nbsp; ${ \boldsymbol{\rm G}}_{1}$ of a&nbsp; $\text{(7, 4)}$ code.
+ ${ \boldsymbol{\rm G}}_{2}$&nbsp; beginnt wie ${ \boldsymbol{\rm G}}_{1}$&nbsp; mit einer Diagonalmatrix&nbsp; ${ \boldsymbol{\rm I}}_{4}$ .
+
+ ${ \boldsymbol{\rm G}}_{2}$&nbsp; starts like ${ \boldsymbol{\rm G}}_{1}$&nbsp; with a diagonal matrix&nbsp; ${ \boldsymbol{\rm I}}_{4}$ .
+ ${ \boldsymbol{\rm G}}_{2}$&nbsp; hat im betrachteten Beispiel das gleiche Format wie&nbsp; ${ \boldsymbol{\rm H}}_{2}$ .
+
+ ${ \boldsymbol{\rm G}}_{2}$&nbsp; in the considered example has the same format as&nbsp; ${ \boldsymbol{\rm H}}_{2}$ .
  
  
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</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Die entsprechende Gleichung für die Coderate lautet in beiden Fällen $R = k/n\text{:}$
+
'''(1)'''&nbsp; The corresponding equation for the code rate in both cases is $R = k/n\text{:}$
 
*$\mathcal{C}_{1} \text{:} \  n = 7, k = 4\  ⇒ \ R = 4/7  \underline {= 0.571},$
 
*$\mathcal{C}_{1} \text{:} \  n = 7, k = 4\  ⇒ \ R = 4/7  \underline {= 0.571},$
 
*$\mathcal{C}_{2} \text{:} \ n = 8, k = 4 \ ⇒ \ R = 4/8  \underline { =0.5}.$
 
*$\mathcal{C}_{2} \text{:} \ n = 8, k = 4 \ ⇒ \ R = 4/8  \underline { =0.5}.$
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'''(2)'''&nbsp; Die minimale Distanz des (7, 4, 3)–Hamming–Codes $\mathcal{C}_{1}$ beträgt $d_{\rm min} \underline{= 3}$, was allein schon aus der Namensgebung ablesbar ist.  
+
'''(2)'''&nbsp; The minimum distance of the (7, 4, 3)-Hamming code $\mathcal{C}_{1}$ is $d_{\rm min} \underline{= 3}$, which can be read from the naming alone.  
  
*Aus der Tabelle auf der Angabenseite ist ersichtlich, dass für den erweiterten Hamming–Code $d_{\rm min} \underline{= 4}$ gilt.
+
*From the table on the information page, it can be seen that for the extended Hamming code $d_{\rm min} \underline{= 4}$ holds.
* $\mathcal{C}_{2}$ bezeichnet man deshalb in der Literatur auch als einen (8, 4, 4)–Blockcode.
+
* $\mathcal{C}_{2}$ is therefore also called a (8, 4, 4) block code in the literature.
  
  
  
'''(3)'''&nbsp; Die Prüfmatrix ${ \boldsymbol{\rm H}}$ besteht im Allgemeinen aus $n$ Spalten und $m = n k$ Zeilen, wobei $m$ die Anzahl der Prüfgleichungen angibt.  
+
'''(3)'''&nbsp; The parity-check matrix ${ \boldsymbol{\rm H}}$ generally consists of $n$ columns and $m = n - k$ rows, where $m$ indicates the number of parity-check equations.  
*Beim (7, 4, 3)–Hamming–Code ist ${ \boldsymbol{\rm H}}$ eine 3 × 7–Matrix.  
+
*For the (7, 4, 3)-Hamming code, ${ \boldsymbol{\rm H}}$ is a 3 × 7 matrix.  
*Für den erweiterten Hamming–Code &nbsp; ⇒ &nbsp; Code $\mathcal{C}_{2}$ gilt demgegenüber $\underline{n = 8}$ (Spaltenzahl) und $\underline{m = 4}$ (Zeilenzahl).
+
*For the extended Hamming code &nbsp; ⇒ &nbsp; code $\mathcal{C}_{2}$, on the other hand, $\underline{n = 8}$ (column number) and $\underline{m = 4}$ (row number) holds.
  
  
  
'''(4)'''&nbsp; Aus der Codetabelle auf der Angabenseite erkennt man, dass allein <u>Antwort 3</u> richtig ist.  
+
'''(4)'''&nbsp; From the code table on the information page you can see that only <u>answer 3</u> is correct.  
*Das Prüfbit $p_{4}$ ist so zu bestimmen, dass die Modulo–2–Summe über alle Bits des Codewortes den Wert $0$ ergibt.
+
*The parity bit $p_{4}$ is to be determined in such a way that the modulo 2 sum over all bits of the code word results in the value $0$.
  
  
  
'''(5)'''&nbsp; Anzumerken ist zunächst, dass die Angabe der Prüfmatrix nie eindeutig ist, schon allein deshalb, weil die Reihenfolge der Prüfgleichungen vertauschbar ist.  
+
'''(5)'''&nbsp; It should first be noted that the specification of the parity-check matrix is never unambiguous, if only because the order of the parity-check equations is interchangeable.  
*Unter Berücksichtigung des Hinweises, dass nur eine der vorgegebenen Zeilen falsch ist, ist ${ \boldsymbol{\rm H}}_{2}$ allerdings eindeutig bestimmt:
+
*However, considering that only one of the given rows is wrong, ${ \boldsymbol{\rm H}}_{2}$ is uniquely determined:
  
 
:$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
 
:$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
  
*Richtig sind also die <u>Aussagen 1, 2 und 4</u>. Die Zeilen dieser Prüfmatrix stehen in dieser Reihenfolge für die vier Prüfgleichungen:
+
*Correct are therefore the <u>statements 1, 2 and 4</u>. The rows of this parity-check matrix represent the four parity-check equations in this order:
  
 
:$$ x_1\oplus  x_2 \oplus x_4 \oplus x_5 = 0 \hspace{0.05cm},$$
 
:$$ x_1\oplus  x_2 \oplus x_4 \oplus x_5 = 0 \hspace{0.05cm},$$
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'''(6)'''&nbsp;  Richtig ist die <u>Antwort 2</u>:  
+
'''(6)'''&nbsp;  Correct is the <u>answer 2</u>:  
*Zu diesem Ergebnis kommt man, wenn man die letzte Zeile durch die Modulo–2–Summe über alle vier Zeilen ersetzt, was erlaubt ist.  
+
*This result is obtained by replacing the last row with the modulo 2 sum over all four rows, which is allowed.  
*Der Vorschlag 1 stellt keine Prüfgleichung dar.  
+
*Proposition 1 does not represent a parity-check equation.  
*Der Vorschlag 3 steht für die Prüfgleichung $x_{3}⊕x_{5} = 0$, was auch nicht den Gegebenheiten entspricht.
+
*Proposal 3 represents the parity-check equation $x_{3}⊕x_{5} = 0$, which also does not correspond to the facts.
  
  
Entsprechend dem richtigen Lösungsvorschlag 2 wird dagegen die Prüfgleichung
+
According to the correct solution suggestion 2, on the other hand, the parity-check equation becomes
 
:$$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0$$
 
:$$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0$$
durch folgende neue Prüfgleichung ersetzt:
+
replaced by the following new parity-check equation:
 
:$$x_1 \oplus x_2 \oplus x_3 \oplus x_8 = 0 \hspace{0.05cm}.$$
 
:$$x_1 \oplus x_2 \oplus x_3 \oplus x_8 = 0 \hspace{0.05cm}.$$
Die modifizierte Prüfmatrix lautet nun:
+
The modified parity-check matrix is now:
 
:$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &0 &0 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$
 
:$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &0 &0 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$
  
  
  
'''(7)'''&nbsp;  Nach dieser Matrixmanipulation liegt ${ \boldsymbol{\rm H}}_{2}$ in der für systematische Codes typischen Form vor:
+
'''(7)'''&nbsp;  After this matrix manipulation, ${ \boldsymbol{\rm H}}_{2}$ is in the form typical for systematic codes:
  
 
:$${ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_m \right)\hspace{0.3cm} \Rightarrow\hspace{0.3cm} m = 4 {\rm :}\hspace{0.3cm}{ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_4 \right) \hspace{0.05cm}.$$
 
:$${ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_m \right)\hspace{0.3cm} \Rightarrow\hspace{0.3cm} m = 4 {\rm :}\hspace{0.3cm}{ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_4 \right) \hspace{0.05cm}.$$
  
Damit lautet die Generatormatrix:
+
Thus, the generator matrix is:
 
:$${ \boldsymbol{\rm G_{2}}} =\left({ \boldsymbol{\rm I}}_4 \: ; \: { \boldsymbol{\rm P}}\right) = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1 &1\\ 0 &1 &0 &0 &1 &1 &0 &1\\ 0 &0 &1 &0 &0 &1 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$
 
:$${ \boldsymbol{\rm G_{2}}} =\left({ \boldsymbol{\rm I}}_4 \: ; \: { \boldsymbol{\rm P}}\right) = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1 &1\\ 0 &1 &0 &0 &1 &1 &0 &1\\ 0 &0 &1 &0 &0 &1 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$
  
Richtig sind also die <u>Aussagen 2 und 3</u>:
+
So the <u>statements 2 and 3</u> are correct:
* ${ \boldsymbol{\rm G}}_{2}$ beginnt wie ${ \boldsymbol{\rm G}}_{1}$ (siehe Angabenblatt) mit einer Diagonalmatrix ${ \boldsymbol{\rm I}}_{4}$ , hat aber im Gegensatz zu ${ \boldsymbol{\rm G}}_{1}$ nun 8 Spalten.  
+
* ${ \boldsymbol{\rm G}}_{2}$ starts like ${ \boldsymbol{\rm G}}_{1}$ (see specification sheet) with a diagonal matrix ${ \boldsymbol{\rm I}}_{4}$ , but unlike ${ \boldsymbol{\rm G}}_{1}$ now has 8 columns.  
*Im vorliegenden Fall $n = 8, k = 4 \ ⇒ \ m = 4$ sind sowohl ${ \boldsymbol{\rm G}}_{2}$ als auch ${ \boldsymbol{\rm H}}_{2}$ jeweils 4×8–Matrizen.
+
*In the present case $n = 8, k = 4 \ ⇒ \ m = 4$ both ${ \boldsymbol{\rm G}}_{2}$ and ${ \boldsymbol{\rm H}}_{2}$ are 4×8 matrices respectively.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  

Revision as of 17:06, 6 July 2022

$\text{(7, 4)}$ Hamming code (yellow background)
and  $\text{(8, 4)}$ extension (green).

Two codes are to be compared, whose code tables are given on the right.

  • The first four bits of each code word  $\underline{x}$  are equal to the respective information word  $\underline{u}$  (black font).
  • Then follow  $m = n- k$  parity bit (red font).


The systematic  $\text{(7, 4)}$-Hamming code has already been discussed in  Exercise 1.6  and  Exercise 1.7 . The parity-check matrix and generator matrix of this code are given as follows:

$${ \boldsymbol{\rm H}}_1 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix}\hspace{0.05cm},$$
$${ \boldsymbol{\rm G}}_1 = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$

In the further course of the exercise this (yellow highlighted) code is called  $\mathcal{C}_{1}$ .


The right column in the above table specifies a block code with parameters  $n = 8$  and  $k = 4$ , usually referred to in the literature as the "Extended Hamming Code". We refer to this code (highlighted in green) in the following  $\mathcal{C}_{2}$  and denote its parity-check matrix by  ${ \boldsymbol{\rm H}}_{2}$  and the corresponding generator matrix by  ${ \boldsymbol{\rm G}}_{2}$ .

The questions for this exercise are related to




Hints:



Questions

1

Specify the code rates of  $\mathcal{C}_{1}$  and  $\mathcal{C}_{2}$ .

$\mathcal{C}_{1}\text{:}\hspace{0.4cm}R \ = \ $

$\mathcal{C}_{2}\text{:}\hspace{0.4cm}R \ = \ $

2

Give the minimum distances of  $\mathcal{C}_{1}$  and  $\mathcal{C}_{2}$  .

$\mathcal{C}_{1}\text{:}\hspace{0.4cm}d_{\rm min} \ = \ $

$\mathcal{C}_{2}\text{:}\hspace{0.4cm}d_{\rm min} \ = \ $

3

What is the format of the parity-check matrix  $\boldsymbol{\rm H}_{2}$  of  $\mathcal{C}_{2}$?

${\rm Spaltenzahl} \ = \ $

${\rm Zeilenzahl} \ = \ $

4

Derive the equation for the code bit  $x_ {8} (= p_{4})$  from the code table. Which specification is correct?

$x_{8} = 0.$
$x_{8} = x_{1}⊕x_{2}⊕x_{4}⊕x_{5}.$
$x_{8} = x_{1}⊕x_{2}⊕x_{3}⊕x_{4}⊕x_{5}⊕x_{6}⊕x_{7}.$

5

Which statements are true for  ${ \boldsymbol{\rm H}}_{2}$? Hint:  Correct are 3 out of 4 answers.

Row 1 reads:   $1 1 0 1 1 0 0 0$.
Row 2 reads:   $0 1 1 1 0 1 0 0$.
Row 3 reads:   $0 0 0 0 1 1 1 1$.
Row 4 reads:   $1 1 1 1 1 1 1 1$.

6

Which transformation is allowed for the last row of  ${ \boldsymbol{\rm H}}_{2}$ ?

$1 1 1 1 1 1 1 1 → 0 0 0 0 0 0 0 0,$
$1 1 1 1 1 1 1 1 → 1 1 1 0 0 0 0 1,$
$1 1 1 1 1 1 1 1 → 0 0 1 0 1 0 0 0.$

7

Give the corresponding generator matrix  ${ \boldsymbol{\rm G}}_{2}$ . Which statements are true?

${ \boldsymbol{\rm G}}_{2}$  has same format as matrix  ${ \boldsymbol{\rm G}}_{1}$ of a  $\text{(7, 4)}$ code.
${ \boldsymbol{\rm G}}_{2}$  starts like ${ \boldsymbol{\rm G}}_{1}$  with a diagonal matrix  ${ \boldsymbol{\rm I}}_{4}$ .
${ \boldsymbol{\rm G}}_{2}$  in the considered example has the same format as  ${ \boldsymbol{\rm H}}_{2}$ .


Solution

(1)  The corresponding equation for the code rate in both cases is $R = k/n\text{:}$

  • $\mathcal{C}_{1} \text{:} \ n = 7, k = 4\ ⇒ \ R = 4/7 \underline {= 0.571},$
  • $\mathcal{C}_{2} \text{:} \ n = 8, k = 4 \ ⇒ \ R = 4/8 \underline { =0.5}.$


(2)  The minimum distance of the (7, 4, 3)-Hamming code $\mathcal{C}_{1}$ is $d_{\rm min} \underline{= 3}$, which can be read from the naming alone.

  • From the table on the information page, it can be seen that for the extended Hamming code $d_{\rm min} \underline{= 4}$ holds.
  • $\mathcal{C}_{2}$ is therefore also called a (8, 4, 4) block code in the literature.


(3)  The parity-check matrix ${ \boldsymbol{\rm H}}$ generally consists of $n$ columns and $m = n - k$ rows, where $m$ indicates the number of parity-check equations.

  • For the (7, 4, 3)-Hamming code, ${ \boldsymbol{\rm H}}$ is a 3 × 7 matrix.
  • For the extended Hamming code   ⇒   code $\mathcal{C}_{2}$, on the other hand, $\underline{n = 8}$ (column number) and $\underline{m = 4}$ (row number) holds.


(4)  From the code table on the information page you can see that only answer 3 is correct.

  • The parity bit $p_{4}$ is to be determined in such a way that the modulo 2 sum over all bits of the code word results in the value $0$.


(5)  It should first be noted that the specification of the parity-check matrix is never unambiguous, if only because the order of the parity-check equations is interchangeable.

  • However, considering that only one of the given rows is wrong, ${ \boldsymbol{\rm H}}_{2}$ is uniquely determined:
$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
  • Correct are therefore the statements 1, 2 and 4. The rows of this parity-check matrix represent the four parity-check equations in this order:
$$ x_1\oplus x_2 \oplus x_4 \oplus x_5 = 0 \hspace{0.05cm},$$
$$x_2 \oplus x_3 \oplus x_4 \oplus x_6 = 0 \hspace{0.05cm},$$
$$ x_1 \oplus x_3 \oplus x_4 \oplus x_7 = 0 \hspace{0.05cm},$$
$$ x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0 \hspace{0.05cm}.$$


(6)  Correct is the answer 2:

  • This result is obtained by replacing the last row with the modulo 2 sum over all four rows, which is allowed.
  • Proposition 1 does not represent a parity-check equation.
  • Proposal 3 represents the parity-check equation $x_{3}⊕x_{5} = 0$, which also does not correspond to the facts.


According to the correct solution suggestion 2, on the other hand, the parity-check equation becomes

$$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0$$

replaced by the following new parity-check equation:

$$x_1 \oplus x_2 \oplus x_3 \oplus x_8 = 0 \hspace{0.05cm}.$$

The modified parity-check matrix is now:

$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &0 &0 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$


(7)  After this matrix manipulation, ${ \boldsymbol{\rm H}}_{2}$ is in the form typical for systematic codes:

$${ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_m \right)\hspace{0.3cm} \Rightarrow\hspace{0.3cm} m = 4 {\rm :}\hspace{0.3cm}{ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_4 \right) \hspace{0.05cm}.$$

Thus, the generator matrix is:

$${ \boldsymbol{\rm G_{2}}} =\left({ \boldsymbol{\rm I}}_4 \: ; \: { \boldsymbol{\rm P}}\right) = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1 &1\\ 0 &1 &0 &0 &1 &1 &0 &1\\ 0 &0 &1 &0 &0 &1 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$

So the statements 2 and 3 are correct:

  • ${ \boldsymbol{\rm G}}_{2}$ starts like ${ \boldsymbol{\rm G}}_{1}$ (see specification sheet) with a diagonal matrix ${ \boldsymbol{\rm I}}_{4}$ , but unlike ${ \boldsymbol{\rm G}}_{1}$ now has 8 columns.
  • In the present case $n = 8, k = 4 \ ⇒ \ m = 4$ both ${ \boldsymbol{\rm G}}_{2}$ and ${ \boldsymbol{\rm H}}_{2}$ are 4×8 matrices respectively.