Aufgaben:Exercise 1.09Z: Extension and/or Puncturing - Revision history

Guenter at 14:37, 11 July 2022

2022-07-11T14:37:13Z

Guenter at 14:16, 11 July 2022

2022-07-11T14:16:34Z

Noah at 21:56, 7 July 2022

2022-07-07T21:56:17Z

Show changes

Guenter at 14:46, 14 June 2022

2022-06-14T14:46:57Z

Guenter at 09:49, 11 May 2022

2022-05-11T09:49:56Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.09Z: Erweiterung und/oder Punktierung to Aufgaben:Exercise 1.09Z: Extension and/or Puncturing

2022-05-11T09:42:35Z

Guenter moved page Aufgabe 1.09Z: Erweiterung und/oder Punktierung to Exercise 1.09Z: Extension and/or Puncturing

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

2021-03-23T12:50:31Z

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

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

2020-07-09T12:46:05Z

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

Tasnad: Text replacement - "[File:" to "[File:"

2020-05-26T14:09:43Z

Text replacement - "[File:" to "[File:"

Guenter at 16:27, 9 May 2019

2019-05-09T16:27:17Z

@@ Line 82: / Line 82: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The rate of the $(5, \, 2)$ code is $R = 2/5 \ \underline{ = 0.4}$.
+'''(1)'''&nbsp; The rate of the&nbsp; $(5, \, 2)$&nbsp; code is&nbsp; $R = 2/5 \ \underline{ = 0.4}$.
-*From the given code, we further recognize the minimum distance $d_{\rm min} \ \underline{ = 3}$.
+*From the given code,&nbsp; we further recognize the minimum distance&nbsp; $d_{\rm min} \ \underline{ = 3}$.
-'''(2)'''&nbsp; When extending from the $(5, \, 2)$ code to the $(6, \, 2)$ code, another parity bit is added.
+'''(2)'''&nbsp; When extending from the&nbsp; $(5, \, 2)$&nbsp; code to the&nbsp; $(6, \, 2)$&nbsp; code,&nbsp; another parity bit is added.
-*The codeword thus has the form
+*The code word thus has the form
 :$$\underline{x} = ( x_1, x_2, x_3, x_4, x_5, x_6) = ( u_1, u_2, p_1, p_2, p_{3}, p_4) \hspace{0.05cm}.$$
@@ Line 96: / Line 96: @@
 :$$p_4 = x_6 = x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \hspace{0.05cm}.$$
-*That is, the new parity bit $p_{4}$ is chosen to result in an even number of ones in each codeword &nbsp;⇒&nbsp; <u>Answer 2</u>.
+*That is,&nbsp; the new parity bit&nbsp; $p_{4}$&nbsp; is chosen to result in an even number of ones in each code word &nbsp; ⇒ &nbsp; <u>Answer 2</u>.
 *Solving this task with the parity-check matrix, we get
 :$${ \boldsymbol{\rm H}}_{(6,\hspace{0.05cm} 2)} = \begin{pmatrix} 1 &0 &1 &0 &0 &0\\ 1 &1 &0 &1 &0 &0\\ 0 &1 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 \end{pmatrix}	\hspace{0.3cm} \Rightarrow\hspace{0.3cm} { \boldsymbol{\rm H}}_{{\rm (6,\hspace{0.05cm} 2)\hspace{0.05cm}sys}} = \begin{pmatrix} 1 &0 &1 &0 &0 &0\\ 1 &1 &0 &1 &0 &0\\ 0 &1 &0 &0 &1 &0\\ 1 &1 &0 &0 &0 &1 \end{pmatrix}\hspace{0.3cm}
 \Rightarrow\hspace{0.3cm} { \boldsymbol{\rm G}}_{{\rm (6,\hspace{0.05cm} 2)\hspace{0.05cm}sys}} = \begin{pmatrix} 1 &0 &1 &1 &0 &1\\ 0 &1 &0 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$
-*The two rows of the generator matrix $\boldsymbol{\rm G}$ give two of the four codewords, the modulo $2$ sum gives the third, and finally the all zero word has to be considered.
+*The two rows of the generator matrix&nbsp; $\boldsymbol{\rm G}$&nbsp; give two of the four code words,&nbsp; the modulo 2 sum gives the third,&nbsp; and finally the all zero word has to be considered.
-'''(3)'''&nbsp; After extension from the $(5, \, 2)$ code to the $(6, \, 2)$ code.
+'''(3)'''&nbsp; After extension from the&nbsp; $(5, \, 2)$&nbsp; code to the&nbsp; $(6, \, 2)$&nbsp; code.
-*decreases the rate from $R = 2/5$ to $R = 2/6 \ \underline{= 0.333}$,
+*decreases the rate from&nbsp; $R = 2/5$&nbsp; to&nbsp; $R = 2/6 \ \underline{= 0.333}$,
 '''(4)'''&nbsp; Using the same procedure as in (3), we obtain
 :$${ \boldsymbol{\rm H}}_{(7,\hspace{0.05cm} 2)} \hspace{-0.05cm}=\hspace{-0.05cm} \begin{pmatrix} 1 &0 &1 &0 &0 &0 &0\\ 1 &1 &0 &1 &0 &0 &0\\ 0 &1 &0 &0 &1 &0 &0\\ 1 &1 &0 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 &1 \end{pmatrix}	\hspace{0.15cm} \Rightarrow\hspace{0.15cm} { \boldsymbol{\rm H}}_{{\rm (7,\hspace{0.05cm} 2)\hspace{0.05cm}sys}} \hspace{-0.05cm}=\hspace{-0.05cm} \begin{pmatrix} 1 &0 &1 &0 &0 &0 &0\\ 1 &1 &0 &1 &0 &0 &0\\ 0 &1 &0 &0 &1 &0 &0\\ 1 &1 &0 &0 &0 &1 &0\\ 0 &0 &0 &0 &0 &0 &1 \end{pmatrix}\hspace{0.15cm} \Rightarrow\hspace{0.15cm} { \boldsymbol{\rm G}}_{{\rm (6,\hspace{0.05cm} 2)\hspace{0.05cm}sys}} \hspace{-0.05cm}=\hspace{-0.05cm} \begin{pmatrix} 1 &0 &1 &1 &0 &1 &0 \\ 0 &1 &0 &1 &1 &1 &0 \end{pmatrix}\hspace{0.05cm}.$$
-⇒&nbsp; <u>Both answers</u> are correct.
+⇒&nbsp; <u>Both answers</u>&nbsp; are correct.
-'''(5)'''&nbsp; The rate is now $R = 2/7 = \underline{0.266}$.
+'''(5)'''&nbsp; The code rate is now&nbsp; $R = 2/7 \ \underline{=0.266}$.
-*The minimum distance is still $d_{\rm min} \ \underline{= 4}$ , as can be seen from the $(7, \, 2)$ code words:
+*The minimum distance is still&nbsp; $d_{\rm min} \ \underline{= 4}$,&nbsp; as can be seen from the&nbsp; $(7, \, 2)$&nbsp; code words:
-:$$\mathcal{C} = \{ (0, 0, 0, 0, 0, 0, 0), \hspace{0.1cm}(0, 1, 0, 1, 1, 1, 0), \hspace{0.1cm}(1, 0, 1, 1, 0, 1, 0), \hspace{0.1cm}(1, 1, 1, 0, 1, 0, 0) \}\hspace{0.05cm}.$$
+:$$\mathcal{C} = \{ (0, 0, 0, 0, 0, 0, 0), \hspace{0.3cm}(0, 1, 0, 1, 1, 1, 0), \hspace{0.3cm}(1, 0, 1, 1, 0, 1, 0), \hspace{0.3cm}(1, 1, 1, 0, 1, 0, 0) \}\hspace{0.05cm}.$$
-<u>In general:</u> &nbsp; If the minimum distance of a code is even, it cannot be increased by extension.
+<u>In general:</u> &nbsp; If the minimum distance of a code is even,&nbsp; it cannot be increased by extension.
-'''(6)'''&nbsp; Correct are the <u>statements 1 and 2</u>:
+'''(6)'''&nbsp; Correct are the&nbsp; <u>statements 1 and 2</u>:
-*By crossing out the last row and the last column, we obtain for parity-check matrix and generator matrix, respectively (each in systematic form):
+*By crossing out the last row and the last column,&nbsp; we obtain for parity-check matrix and generator matrix,&nbsp; respectively (each in systematic form):
 :$${ \boldsymbol{\rm H}}_{(4,\hspace{0.05cm} 2)} = \begin{pmatrix} 1 &0 &1 &0 \\ 1 &1 &0 &1 \end{pmatrix}	\hspace{0.3cm} \Rightarrow\hspace{0.3cm} { \boldsymbol{\rm G}}_{{\rm (4,\hspace{0.05cm} 2)}} = \begin{pmatrix} 1 &0 &1 &1 \\ 0 &1 &0 &1 \end{pmatrix}\hspace{0.05cm}.$$
-*From the generator matrix we get the mentioned codewords $(1, 0, 1, 1), \, (0, 1, 0, 1), \, (1, 1, 1, 0)$ as row sum as well as the null word $(0, 0, 0, 0)$. The minimum distance of this code is $d_{\rm min}= 2$, which is smaller than the minimum distance $d_{\rm min}= 3$ of the $(5, \, 2)$ code.
+*From the generator matrix we get the mentioned code words&nbsp; $(1, 0, 1, 1), \, (0, 1, 0, 1), \, (1, 1, 1, 0)$&nbsp; as row sum as well as the null word&nbsp; $(0, 0, 0, 0)$.&nbsp;
-<u>In general:</u> &nbsp; Puncturing makes $d_{\rm min}$ smaller by $1$ (if it was even before) or it stays the same. This can be illustrated by generating the $(3, \, 2)$ block code by another puncturing (of the parity bit $p_{2}$). This code
+*This code&nbsp; $ \mathcal{C} = \{ (0, 0, 0), \hspace{0.3cm}(0, 1, 1), \hspace{0.3cm}(1, 0, 1), \hspace{0.3cm}(1, 1, 0) \}$&nbsp; has the same minimum distance&nbsp; $d_{\rm min}= 2$&nbsp; as the&nbsp; $(4, \, 2)$&nbsp; code.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 [[File:EN_KC_Z_1_9.png|right|frame|Extension and puncturing]]
-Often you know a code that seems to be suitable for an application, but its code rate does not exactly match the specifications.
+Often you know a code that seems to be suitable for an application,&nbsp; but its code rate does not exactly match the specifications.
-There are several possibilities for rate adaptation
+There are several possibilities for rate adaptation:
 '''Extension''':
-<br>Starting from the&nbsp; $(n, \, k)$ code whose parity-check matrix&nbsp; $\mathbf{H}$&nbsp; is given, one obtains a&nbsp; $(n+1, \, k)$ code by extending the parity-check matrix by one row and one column and adding zeros and ones to the new matrix elements according to the upper graph. One adds a new parity bit
+<br>Starting from the&nbsp; $(n, \, k)$&nbsp; code whose parity-check matrix&nbsp; $\mathbf{H}$&nbsp; is given,&nbsp; one obtains a&nbsp; $(n+1, \, k)$&nbsp; code by extending the parity-check matrix by one row and one column and adding zeros and ones to the new matrix elements according to the upper graph.&nbsp; So,&nbsp; one adds a new parity bit
 :$$x_{n+1} = x_1 \oplus x_2 \oplus ... \hspace{0.05cm} \oplus x_n$$
-and thus a new parity-check equation is added, which is considered in&nbsp; $\mathbf{H}\hspace{0.05cm}'$&nbsp; .
+and thus a new parity-check equation is added,&nbsp; which is considered in&nbsp; $\mathbf{H}\hspace{0.05cm}'$&nbsp; .
 '''Puncturing''':
-<br>According to the figure below, one arrives at a&nbsp; $(n-1, \, k)$ code of larger rate by omitting a parity bit and a parity-check equation, which is equivalent to deleting one row and one column from the parity-check matrix&nbsp; $\mathbf{H}$&nbsp;.
+<br>According to the figure below,&nbsp; one arrives at a&nbsp; $(n-1, \, k)$&nbsp; code of larger rate by omitting a parity bit and a parity-check equation,&nbsp; which is equivalent to deleting one row and one column from the parity-check matrix&nbsp; $\mathbf{H}$&nbsp;.
 '''Shortening''':
-<br>If an information bit is omitted instead of a parity bit, the result is a&nbsp; $(n-1, \, k-1)$ code of smaller rate.
+<br>If an information bit is omitted instead of a parity bit,&nbsp; the result is a&nbsp; $(n-1, \, k-1)$&nbsp; code of smaller rate.
-In this exercise, starting from a&nbsp; $(5, \, 2)$ block code
+In this exercise,&nbsp; starting from a&nbsp; $(5, \, 2)$&nbsp; block code
-:$$\mathcal{C} = \{ (0, 0, 0, 0, 0) \hspace{0.1cm}, (0, 1, 0, 1, 1) \hspace{0.1cm},(1, 0, 1, 1, 0) \hspace{0.1cm},(1, 1, 1, 0, 1) \}$$
+:$$\mathcal{C} = \{ (0, 0, 0, 0, 0), \hspace{0.3cm} (0, 1, 0, 1, 1), \hspace{0.3cm} (1, 0, 1, 1, 0), \hspace{0.3cm} (1, 1, 1, 0, 1) \}$$
-the following codes are constructed and analyzed:
+the following codes are to constructed and analyzed:
-*one&nbsp; $(6, \, 2)$ code by single extension,
+*one&nbsp; $(6, \, 2)$&nbsp; code by single extension,
-*one&nbsp; $(7, \, 2)$ code by extending it again,
+*one&nbsp; $(7, \, 2)$&nbsp; code by extending it again,
-*one&nbsp; $(4, \, 2)$ code by puncturing.
+*one&nbsp; $(4, \, 2)$&nbsp; code by puncturing.
-The parity-check matrix and the generator matrix of the systematic&nbsp; $(5, \, 2)$ code are:
+The parity-check matrix and the generator matrix of the systematic&nbsp; $(5, \, 2)$&nbsp; code are:
-:$${ \boldsymbol{\rm H}}_{(5, 2)} = \begin{pmatrix} 1 &0 &1 &0 &0\\ 1 &1 &0 &1 &0\\ 0 &1 &0 &0 &1 \end{pmatrix}	\hspace{0.3cm} \Leftrightarrow\hspace{0.3cm} { \boldsymbol{\rm G}}_{(5, 2)} = \begin{pmatrix} 1 &0 &1 &1 &0\\ 0 &1 &0 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
+:$${ \boldsymbol{\rm H}}_{(5,\ 2)} = \begin{pmatrix} 1 &0 &1 &0 &0\\ 1 &1 &0 &1 &0\\ 0 &1 &0 &0 &1 \end{pmatrix}	\hspace{0.3cm} \Leftrightarrow\hspace{0.3cm} { \boldsymbol{\rm G}}_{(5,\ 2)} = \begin{pmatrix} 1 &0 &1 &1 &0\\ 0 &1 &0 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
-Hints :
+*In the&nbsp; [[Aufgaben:Exercise_1.09:_Extended_Hamming_Code|"Exercise 1.9"]]&nbsp; it is exemplified how the&nbsp; $(7, \, 4, \, 3)$&nbsp; Hamming code is turned into a&nbsp; $(8, \, 4, \, 4)$&nbsp; code by extension.
@@ Line 49: / Line 48: @@
 ===Questions===
 <quiz display=simple>
-{Specify the characteristics of the given&nbsp; $(5, \, 2)$ code.
+{Specify the characteristics of the given&nbsp; $(5, \, 2)$&nbsp; code.
 |type="{}"}
 $R \ = \ $ { 0.4 3% }
 $d_{\rm min} \ = \ $ {  3 3% }
-{What code words does the $(6, \, 2)$ code have after expansion?
+{What code words does&nbsp; the $(6, \, 2)$&nbsp; code have after expansion?
 |type="[]"}
 - $(0 0 0 0 0 1), \ (0 1 0 1 1 0), \ (1 0 1 1 0 0), \ (1 1 1 0 1 1).$
 + $(0 0 0 0 0 0), \ (0 1 0 1 1 1), \ (1 0 1 1 0 1), \ (1 1 1 0 1 0).$
-{Specify the characteristics of the extended&nbsp; $(6, \, 2)$ code.
+{Specify the characteristics of the extended&nbsp; $(6, \, 2)$&nbsp; code.
 |type="{}"}
 $R \ = \ $ { 0.333 3% }
 $d_{\rm min} \ = \ $ {  4 3% }
-{What is the systematic generator matrix&nbsp; $\boldsymbol{\rm G}$&nbsp; of the&nbsp; $(7, \, 2)$ code?
+{What is the systematic generator matrix&nbsp; $\boldsymbol{\rm G}$&nbsp; of the&nbsp; $(7, \, 2)$&nbsp; code?
 |type="[]"}
 + Row 1 of&nbsp; $\boldsymbol{\rm G} \text{:} \hspace{0.2cm} 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 0.$
 + Row 2 of&nbsp;  $\boldsymbol{\rm G} \text{:} \hspace{0.2cm} 0, \, 1, \, 0, \, 1, \, 1, \, 1, \, 0.$
-{Specify the characteristics of the extended&nbsp; $(7, \, 2)$ code.
+{Specify the characteristics of the extended&nbsp; $(7, \, 2)$&nbsp; code.
 |type="{}"}
 $R \ = \ $ { 0.266 3% }
 $d_{\rm min} \ = \ $ {  4 3% }
-{Which statements are true for the $(4, \, 2)$ code (puncturing the last parity bit)?
+{Which statements are true for the&nbsp; $(4, \, 2)$&nbsp; code&nbsp; (puncturing the last parity bit)?
 |type="[]"}
 + The code rate is now&nbsp; $R = 2/4 = 0.5$.
-+ $C_{(4, 2)} = \{(0, 0, 0, 0),  \, (1, 0, 1, 1), \, (0, 1, 0, 1), \, (1, 1, 1, 0)\}$.
++ $C_{(4,\ 2)} = \{(0, 0, 0, 0),  \, (1, 0, 1, 1), \, (0, 1, 0, 1), \, (1, 1, 1, 0)\}$.
 - The minimum distance remains unchanged from the&nbsp; $(5, \, 2)$ code.
 </quiz>

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Kanalcodierung/Allgemeine Beschreibung linearer Blockcodes}}
-[[File:P_ID2403__KC_Z_1_9.png|right|frame|Zur Erweiterung und Punktierung]]
+[[File:EN_KC_Z_1_9.png|right|frame|Zur Erweiterung und Punktierung]]
 Häufig kennt man einen Code, der für eine Anwendung als geeignet erscheint, dessen Coderate aber nicht exakt mit den Vorgaben übereinstimmt.

← Older revision		Revision as of 09:49, 11 May 2022
Line 144:		Line 144:


−	[[Category:Channel Coding: Exercises\|^1.4 ~~Beschreibung linearer Blockcodes~~^]]	+	[[Category:Channel Coding: Exercises\|^1.4 Linear Block Code Description^]]

← Older revision		Revision as of 12:50, 23 March 2021
Line 144:		Line 144:


−	[[Category:~~Aufgaben zu Kanalcodierung~~\|^1.4 Beschreibung linearer Blockcodes^]]	+	[[Category:Channel Coding: Exercises\|^1.4 Beschreibung linearer Blockcodes^]]

@@ Line 42: / Line 42: @@
 ''Hinweise'' :
-*Die Aufgabe gehört zum Kapitel&nbsp; [[Kanalcodierung/Allgemeine_Beschreibung_linearer_Blockcodes|Allgemeine Beschreibung linearer Blockcodes]].
+*Die Aufgabe gehört zum Kapitel&nbsp; [[Channel_Coding/Allgemeine_Beschreibung_linearer_Blockcodes|Allgemeine Beschreibung linearer Blockcodes]].
 * In der&nbsp; [[Aufgaben:Aufgabe_1.09:_Erweiterter_Hamming–Code|Aufgabe 1.9]]&nbsp; wird beispielhaft gezeigt, wie aus dem&nbsp; $(7, \, 4, \, 3)$–Hamming–Code durch Erweiterung ein&nbsp; $(8, \, 4, \, 4)$–Code entsteht.

← Older revision	Revision as of 14:09, 26 May 2020
(No difference)

@@ Line 83: / Line 83: @@
 ===Musterlösung===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die Rate des $(5, \, 2)$–Codes ist $R = 2/5 \ \underline{ = 0.4}$. Aus dem angegebenen Code erkennt man weiterhin die minimale Distanz $d_{\rm min} \ \underline{ = 3}$.
+'''(1)'''&nbsp; Die Rate des $(5, \, 2)$–Codes ist $R = 2/5 \ \underline{ = 0.4}$.
 '''(2)'''&nbsp; Bei Erweiterung vom $(5, \, 2)$–Code zum $(6, \, 2)$–Code wird ein weiteres Prüfbit hinzugefügt. Das Codewort hat somit die Form
 :$$\underline{x} = ( x_1, x_2, x_3, x_4, x_5, x_6) = ( u_1, u_2, p_1, p_2, p_{3}, p_4) \hspace{0.05cm}.$$
-Für das hinzugekommene Prüfbit muss dabei gelten:
+*Für das hinzugekommene Prüfbit muss dabei gelten:
 :$$p_4 = x_6 = x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \hspace{0.05cm}.$$
-Das heißt: Das neue Prüfbit $p_{4}$ wird so gewählt, dass sich in jedem Codewort eine gerade Anzahl von Einsen ergibt &nbsp;⇒&nbsp; <u>Antwort 2</u>. Löst man diese Aufgabe mit der Prüfmatrix, so erhält man
+*Das heißt: Das neue Prüfbit $p_{4}$ wird so gewählt, dass sich in jedem Codewort eine gerade Anzahl von Einsen ergibt &nbsp;⇒&nbsp; <u>Antwort 2</u>.
 :$${ \boldsymbol{\rm H}}_{(6,\hspace{0.05cm} 2)} = \begin{pmatrix} 1 &0 &1 &0 &0 &0\\ 1 &1 &0 &1 &0 &0\\ 0 &1 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 \end{pmatrix}	\hspace{0.3cm} \Rightarrow\hspace{0.3cm} { \boldsymbol{\rm H}}_{{\rm (6,\hspace{0.05cm} 2)\hspace{0.05cm}sys}} = \begin{pmatrix} 1 &0 &1 &0 &0 &0\\ 1 &1 &0 &1 &0 &0\\ 0 &1 &0 &0 &1 &0\\ 1 &1 &0 &0 &0 &1 \end{pmatrix}\hspace{0.3cm}
 \Rightarrow\hspace{0.3cm} { \boldsymbol{\rm G}}_{{\rm (6,\hspace{0.05cm} 2)\hspace{0.05cm}sys}} = \begin{pmatrix} 1 &0 &1 &1 &0 &1\\ 0 &1 &0 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$
-Die beiden Zeilen der Generatormatrix $\boldsymbol{\rm G}$ ergeben zwei der vier Codeworte, die Modulo–$2$–Summe das dritte und schließlich ist auch noch das Nullwort zu berücksichtigen.
+*Die beiden Zeilen der Generatormatrix $\boldsymbol{\rm G}$ ergeben zwei der vier Codeworte, die Modulo–$2$–Summe das dritte und schließlich ist auch noch das Nullwort zu berücksichtigen.
@@ Line 106: / Line 111: @@
-<u>Allgemein gilt:</u> Erweitert man einen Code, so nimmt die Rate ab und die Minimaldistanz erhöht sich um $1$, falls $d_{\rm min}$ vorher ungerade war.
+<u>Allgemein gilt:</u> &nbsp; Erweitert man einen Code, so nimmt die Rate ab und die Minimaldistanz erhöht sich um $1$, falls $d_{\rm min}$ vorher ungerade war.
@@ Line 115: / Line 121: @@
 '''(5)'''&nbsp; Die Rate beträgt nun $R = 2/7 = \underline{0.266}$. Die Minimaldistanz ist weiterhin $d_{\rm min} \ \underline{= 4}$ , wie man aus den $(7, \, 2)$–Codeworten ablesen kann:
 :$$\mathcal{C} = \{ (0, 0, 0, 0, 0, 0, 0), \hspace{0.1cm}(0, 1, 0, 1, 1, 1, 0), \hspace{0.1cm}(1, 0, 1, 1, 0, 1, 0), \hspace{0.1cm}(1, 1, 1, 0, 1, 0, 0) \}\hspace{0.05cm}.$$
-<u>Allgemein gilt:</u> Ist die Minimaldistanz eines Codes geradzahlig, so kann diese durch Erweiterung nicht vergrößert werden.
+<u>Allgemein gilt:</u> &nbsp; Ist die Minimaldistanz eines Codes geradzahlig, so kann diese durch Erweiterung nicht vergrößert werden.
@@ Line 128: / Line 137: @@
-<u>Allgemein gilt:</u> Durch Punktierung wird $d_{\rm min}$ um $1$ kleiner (wenn sie vorher gerade war) oder sie bleibt gleich. Dies kann man sich verdeutlichen, wenn man durch eine weitere Punktierung (des Prüfbits $p_{2}$) den $(3, \, 2)$–Blockcode generiert. Dieser Code
+<u>Allgemein gilt:</u> &nbsp; Durch Punktierung wird $d_{\rm min}$ um $1$ kleiner (wenn sie vorher gerade war) oder sie bleibt gleich. Dies kann man sich verdeutlichen, wenn man durch eine weitere Punktierung (des Prüfbits $p_{2}$) den $(3, \, 2)$–Blockcode generiert. Dieser Code
 :$$ \mathcal{C} = \{ (0, 0, 0), \hspace{0.1cm}(0, 1, 1), \hspace{0.1cm}(1, 0, 1), \hspace{0.1cm}(1, 1, 0) \}$$
 besitzt die gleiche Minimaldistanz $d_{\rm min}= 2$ wie der $(4, \, 2)$–Code.