Difference between revisions of "Aufgaben:Exercise 1.15: Distance Spectra of HC (7, 4, 3) and HC (8, 4, 4)"

Latest revision as of 16:14, 8 October 2022

Code tables of the $(7, 4, 3)$ Hamming code and the $(8, 4, 4)$ extension

We consider as in the "Exercise 1.9"

the $(7, 4, 3)$ Hamming code and

the extended $(8, 4, 4)$ Hamming code.

The graphic shows the corresponding code tables. In "Exercise 1.12", the syndrome decoding of these two codes has already been covered. In this exercise, the differences regarding the distance spectrum $\{W_{i}\}$ shall now be worked out. For the indexing variable $i = 0, \ \text{...} \ , n$:

The integer $W_{i}$ specifies the number of code words $\underline{x}$ with the "Hamming weight" $\underline{w}_{\rm H}( \underline{x} ) = i$.

For the linear code considered here, $W_{i}$ simultaneously describes the number of code words with the "Hamming distance" $i$ from the all-zero word.

Often one assigns to the number set $\{W_i\}$ a pseudo-function called "weight enumerator function" :

$$\left \{ \hspace{0.05cm} W_i \hspace{0.05cm} \right \} \hspace{0.3cm} \Leftrightarrow \hspace{0.3cm} W(X) = \sum_{i=0 }^{n} W_i \cdot X^{i} = W_0 + W_1 \cdot X + W_2 \cdot X^{2} + ... \hspace{0.05cm} + W_n \cdot X^{n}\hspace{0.05cm}.$$

Bhattacharyya has used the pseudo-function $W(X)$ to specify a channel-independent (upper) bound on the block error probability:

$${\rm Pr(block\:error)} \le{\rm Pr(Bhattacharyya)} = W(\beta) -1 \hspace{0.05cm}.$$

The so-called "Bhattacharyya parameter" is given as follows:

$$\beta = \left\{ \begin{array}{c} \lambda \\ \\ 2 \cdot \sqrt{\varepsilon \cdot (1- \varepsilon)}\\ \\ {\rm e}^{- R \hspace{0.05cm}\cdot \hspace{0.05cm}E_{\rm B}/N_0} \end{array} \right.\quad \begin{array}{*{1}c} {\rm for\hspace{0.15cm} the \hspace{0.15cm}BEC\:model},\\ \\ {\rm for\hspace{0.15cm} the \hspace{0.15cm}BSC\:model}, \\ \\{\rm for\hspace{0.15cm} the \hspace{0.15cm}AWGN\:model}. \end{array}$$

It should be noted that the "Bhattacharyya Bound" is generally very pessimistic. The actual block error probability is often significantly lower.

Hints:

This exercise refers to the chapter "Bounds for Block Error Probability".

A similar topic is covered in "Exercise 1.14" and in "Exercise 1.16".

The channels to be considered are:
- the "BSC model" ("Binary Symmetric Channel"),
- the "BEC model" ("Binary Erasure Channel"),
- the "AWGN channel model".

Questions

$W_{0} \ = \ $

$W_{3} \ = \ $

$W_{4} \ = \ $

$W_{7} \ = \ $

${\rm Pr(Bhattacharyya)} \ = \ $

$\ \%$

${\rm Pr(Bhattacharyya)} \ = \ $

$\ \%$

$\lambda \ = \ $

$10 · \lg {E_{\rm B}/N_0} \ = \ $

$ \ \rm dB$

$10 · \lg {E_{\rm B}/N_0} \ = \ $

$ \ \rm dB$

Solution

(1) By analyzing all the code words of the $(7, 4, 3)$ Hamming code, we see that.

$W_{0} \ \underline{ = \ 1}$ ⇒ code word does not contain a "one" (the zero-word),
$W_{3} \ \underline{ = \ 7}$ ⇒ cod words contain three "ones",
$W_{4} \ \underline{ = \ 7}$ ⇒ code words contain four "ones",
$W_{7} \ \underline{ = \ 1}$ ⇒ code word consists of only "ones.

$W_{i}$ simultaneously specifies the number of code words that differ from the zero-word in $i \ \rm bit$.

(2) The Bhattacharyya bound reads:

$${\rm Pr(block\:error)} \le{\rm Pr(Bhattacharyya)} = W(\beta) -1 \hspace{0.05cm}.$$

The weight function is defined by the subtask (1):

$$W(X) = 1+ 7 \cdot X^{3} + 7 \cdot X^{4} + X^{7}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr(Bhattacharyya)} = 7 \cdot \beta^{3} + 7 \cdot \beta^{4} + \beta^{7} \hspace{0.05cm}.$$

For the Bhattacharyya parameter of the BSC model:

$$\beta = 2 \cdot \sqrt{\varepsilon \cdot (1- \varepsilon)} = 2 \cdot \sqrt{0.01 \cdot 0.99} = 0.199\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr(Bhattacharyya)} = 7 \cdot 0.199^{3} + 7 \cdot 0.199^{4} + 0.199^{7} \hspace{0.15cm} \underline{ \approx 6.6\%} \hspace{0.05cm}.$$

A comparison with the actual block error probability as calculated in "Exercise 1.12",

$${\rm Pr(block\:error)} \approx 21 \cdot \varepsilon^2 = 2.1 \cdot 10^{-3} \hspace{0.05cm},$$

shows that Bhattacharyya provides only a rough bound. In the present case, this bound is more than a factor of $30$ higher than the actual value.

(3) From the code table of the extended $(8, 4, 4)$ Hamming code, the following results are obtained:

$$W(X) = 1+ 14 \cdot X^{4} + X^{8}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr(Bhattacharyya)} = 14 \cdot \beta^{4} + \beta^{8} = 14 \cdot 0.199^{4} + 0.199^{8} \hspace{0.15cm} \underline{ \approx 2.2\%} \hspace{0.05cm}.$$

(4) The equation for the Bhattacharyya parameter is:

$$\beta = \left\{ \begin{array}{c} \lambda \\ \\ 2 \cdot \sqrt{ \varepsilon \cdot (1- \varepsilon)}\\ \\ {\rm e}^{- R \cdot E_{\rm B}/N_0} \end{array} \right.\quad \begin{array}{*{1}c} {\rm for\hspace{0.15cm} the \hspace{0.15cm}BEC model},\\ \\ {\rm for\hspace{0.15cm} the \hspace{0.15cm}BSC model}, \\ \\{\rm for\hspace{0.15cm} the \hspace{0.15cm}AWGN model}. \end{array}$$

With the BEC model, exactly the same bound is obtained when the erasure probability is $\lambda = \beta \ \underline{= 0.199}$.

(5) According to the above equation must apply:

$$\beta = {\rm e}^{- R \hspace{0.05cm}\cdot \hspace{0.05cm} E_{\rm B}/N_0} = 0.199 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} R \cdot E_{\rm B}/N_0 = 10^{0.199} = 1.58 \hspace{0.05cm}.$$

The code rate of the extended $(8, 4, 4)$ Hamming code is $R = 0.5$:

$$E_{\rm B}/N_0 = 3.16 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 \hspace{0.15cm} \underline{\approx 5\,{\rm dB}} \hspace{0.05cm}.$$

(6) Using the code rate $R = 4/7$ of the $(7, 4, 3)$ Hamming code, we obtain:

$$E_{\rm B}/N_0 = 7/4 \cdot 1.58 = 2.765 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 \hspace{0.15cm} \underline{\approx 4.417\,{\rm dB}} \hspace{0.05cm}.$$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Kanalcodierung/Schranken für die Blockfehlerwahrscheinlichkeit}}
+{{quiz-Header|Buchseite=Channel_Coding/Limits_for_Block_Error_Probability}}
-[[File:P_ID2407__KC_A_1_9.png|right|frame|Codetabellen des $(7, 4)$–Hamming–Codes und der $(8, 4)$–Erweiterung]]
+[[File:EN_KC_A_1_9_neu.png|right|frame|Code tables of the&nbsp; $(7, 4, 3)$ Hamming code and the&nbsp; $(8, 4, 4)$ extension]]
-Wir betrachten wie in [[Aufgaben:1.09_Erweiterter_Hamming–Code|Aufgabe 1.9]]
+We consider as in the&nbsp; [[Aufgaben:Exercise_1.09:_Extended_Hamming_Code|"Exercise 1.9"]]
-*den $(7, 4, 3)$–Hamming–Code und
+*the&nbsp; $(7, 4, 3)$&nbsp; Hamming code and
-*den erweiterten $(8, 4, 4)$–Hamming–Code.
+*the extended&nbsp; $(8, 4, 4)$&nbsp; Hamming code.
-Die Grafik zeigt die zugehörigen Codetabellen. In der [[Aufgaben:1.12_Hard_/_Soft_Decision|Aufgabe 1.12]] wurde schon die Syndromdecodierung dieser beiden Codes behandelt.
-In dieser Aufgabe sollen die Unterschiede hinsichtlich des Distanzspektrums $\{W_{i}\}$ herausgearbeitet werden. Für die Laufvariable gilt $i = 0, \ ... \ , n:$
-*Die Integerzahl $W_{i}$ gibt die Zahl der Codeworte $\underline{x}$ mit dem [[Kanalcodierung/Zielsetzung_der_Kanalcodierung#Einige_wichtige_Definitionen_zur_Blockcodierung|Hamming–Gewicht]] $\underline{w}_{\rm H}( \underline{x} ) = i$ an.
+The graphic shows the corresponding code tables.&nbsp; In&nbsp; [[Aufgaben:Exercise_1.12:_Hard_Decision_vs._Soft_Decision|"Exercise 1.12"]],&nbsp; the syndrome decoding of these two codes has already been covered.&nbsp; In this exercise,&nbsp; the differences regarding the distance spectrum&nbsp; $\{W_{i}\}$&nbsp; shall now be worked out.&nbsp; For the indexing variable&nbsp; $i = 0, \ \text{...} \ , n$:
-*Bei den hier betrachteten linearen Code bescheibt $W_{i}$ gleichzeitig die Anzahl der Codeworte mit der [[Kanalcodierung/Schranken_für_die_Blockfehlerwahrscheinlichkeit#Distanzspektrum_eines_linearen_Codes|Hamming–Distanz]] $i$ vom Nullwort.
-*Häufig weist man der Zahlenmenge $\{W_i\}$ einer Pseudo–Funktion zu, die man [[Kanalcodierung/Schranken_für_die_Blockfehlerwahrscheinlichkeit#Distanzspektrum_eines_linearen_Codes|Gewichtsfunktion]] (englisch: ''Weight Enumerator Function'', WEF) nennt:
+*The integer&nbsp; $W_{i}$&nbsp; specifies the number of code words&nbsp; $\underline{x}$&nbsp; with the&nbsp; [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"Hamming weight"]]&nbsp; $\underline{w}_{\rm H}( \underline{x} ) = i$.
+*For the linear code considered here,&nbsp; $W_{i}$&nbsp; simultaneously describes the number of code words with the&nbsp; [[Channel_Coding/Limits_for_Block_Error_Probability#Distance_spectrum_of_a_linear_code|"Hamming distance"]]&nbsp; $i$&nbsp; from the all-zero word.
+*Often one assigns to the number set&nbsp; $\{W_i\}$&nbsp; a pseudo-function called&nbsp; [[Channel_Coding/Limits_for_Block_Error_Probability#Distance_spectrum_of_a_linear_code|"weight enumerator function"]]&nbsp;:
 :$$\left \{ \hspace{0.05cm} W_i \hspace{0.05cm} \right \} \hspace{0.3cm} \Leftrightarrow \hspace{0.3cm} W(X) = \sum_{i=0 }^{n} W_i \cdot X^{i} = W_0 + W_1 \cdot X + W_2 \cdot X^{2} + ... \hspace{0.05cm} + W_n \cdot X^{n}\hspace{0.05cm}.$$
-Bhattacharyya hat die Pseudo–Funktion $W(X;)$ verwendet, um eine kanalunabhängige (obere) Schranke für die Blockfehlerwahrscheinlichkeit anzugeben:
+Bhattacharyya has used the pseudo-function&nbsp; $W(X)$&nbsp; to specify a channel-independent&nbsp; (upper)&nbsp; bound on the block error probability:
+:$${\rm Pr(block\:error)} \le{\rm Pr(Bhattacharyya)} = W(\beta) -1 \hspace{0.05cm}.$$
+The so-called&nbsp; "Bhattacharyya parameter"&nbsp; is given as follows:
+:$$\beta = \left\{ \begin{array}{c} \lambda \\ \\ 2 \cdot \sqrt{\varepsilon \cdot (1- \varepsilon)}\\ \\ {\rm e}^{- R \hspace{0.05cm}\cdot \hspace{0.05cm}E_{\rm B}/N_0} \end{array} \right.\quad \begin{array}{*{1}c} {\rm for\hspace{0.15cm} the \hspace{0.15cm}BEC\:model},\\ \\ {\rm for\hspace{0.15cm} the \hspace{0.15cm}BSC\:model}, \\ \\{\rm for\hspace{0.15cm} the \hspace{0.15cm}AWGN\:model}. \end{array}$$
+It should be noted that the&nbsp; "Bhattacharyya Bound"&nbsp; is generally very pessimistic.&nbsp; The actual block error probability is often significantly lower.
-:$${\rm Pr(Blockfehler)} \le{\rm Pr(Bhattacharyya)} = W(\beta) -1 \hspace{0.05cm}.$$
+Hints:
+*This exercise refers to the chapter&nbsp; [[Channel_Coding/Limits_for_Block_Error_Probability|"Bounds for Block Error Probability"]].
-Der so genannte ''Bhattacharyya–Parameter'' ist dabei wie folgt gegeben:
+*A similar topic is covered in&nbsp; [[Aufgaben:Exercise_1.14:_Bhattacharyya_Bound_for_BEC|"Exercise 1.14"]]&nbsp; and in&nbsp; [[Aufgaben:Exercise_1.16:_Block_Error_Probability_Bounds_for_AWGN|"Exercise 1.16"]].
+* The channels to be considered are:
+** the&nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Symmetric_Channel_.E2.80.93_BSC|"BSC model"]]&nbsp; ("Binary Symmetric Channel"),
+** the&nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_.E2.80.93_BEC|"BEC model"]]&nbsp; ("Binary Erasure Channel"),
+** the&nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#AWGN_channel_at_Binary_Input|"AWGN channel model"]].
-:$$\beta = \left\{ \begin{array}{c} \lambda \\ \\ 2 \cdot \sqrt{\varepsilon \cdot (1- \varepsilon)}\\ \\ {\rm exp}[- R \cdot E_{\rm B}/N_0] \end{array} \right.\quad \begin{array}{*{1}c} {\rm f\ddot{u}r\hspace{0.15cm} das \hspace{0.15cm}BEC-Modell},\\ \\ {\rm f\ddot{u}r\hspace{0.15cm} das \hspace{0.15cm}BSC-Modell}, \\ \\{\rm f\ddot{u}r\hspace{0.15cm} das \hspace{0.15cm}AWGN-Modell}. \end{array}$$
-''Hinweise:''
-* Die Aufgabe bezieht sich auf Kapitel [[Kanalcodierung/Schranken_für_die_Blockfehlerwahrscheinlichkeit|Schranken für die Blockfehlerwahrscheinlichkeit]], ebenso wie [[Aufgaben:1.14_Bhattacharyya–Schranke_für_BEC|Aufgabe 1.14]] und [[Aufgaben:1.16_Schranken_für_AWGN|Aufgabe 1.16]].
-* Als Kanäle sollen betrachtet werden:
-** das [[Kanalcodierung/Kanalmodelle_und_Entscheiderstrukturen#Binary_Symmetric_Channel_.E2.80.93_BSC|BSC–Modell]] (''Binary Symmetric Channel''),
-** das [[Kanalcodierung/Kanalmodelle_und_Entscheiderstrukturen#Binary_Erasure_Channel_.E2.80.93_BEC|BEC–Modell]] (''Binary Erasure Channel''),
-** das [[Kanalcodierung/Kanalmodelle_und_Entscheiderstrukturen#AWGN.E2.80.93Kanal_bei_bin.C3.A4rem_Eingang|AWGN–Kanalmodell]].
-* Anzumerken ist, dass die Bhattacharyya–Schranke im allgemeinen sehr pessimistisch ist. Die tatsächliche Blockfehlerwahrscheinlichkeit liegt oft deutlich darunter.
-* Sollte die Eingabe des Zahlenwertes &bdquo;0&rdquo; erforderlich sein, so geben Sie bitte &bdquo;0.&rdquo; ein.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Geben Sie das Distanzspektrum des $(7, 4, 3)$–Hamming–Codes an.
+{Specify the distance spectrum of the&nbsp; $(7, 4, 3)$ Hamming code.
 |type="{}"}
-$( 7, 4, 3)–{\rm Code} \text{:} \hspace{0.2cm} W_{0} \ = \ ${ 1 3% }
+$W_{0} \ = \ ${ 1 }
-$W_{3} \  = \ ${ 7 3% }
+$W_{3} \  = \ ${ 7 }
-$W_{4} \ = \ ${ 7 3% }
+$W_{4} \ = \ ${ 7 }
-$W_{7} \ = \ ${ 1 3% }
+$W_{7} \ = \ ${ 1 }
-{Wie lautet die Bhattacharyya–Schranke für das BSC–Modell mit $\varepsilon = 0.01?$
+{What is the Bhattacharyya Bound for the&nbsp; $(7, 4, 3)$&nbsp; Hamming code and the BSC model with&nbsp; $\varepsilon = 0.01$?
 |type="{}"}
-$(7, 4, 3)–{\rm Code} \text{:} \hspace{0.2cm} {\rm Pr(Bhattacharyya)} \ = \ $ { 0.666 3% }
+${\rm Pr(Bhattacharyya)} \ = \ $ { 6.6 3% } $\ \%$
-{Wie lautet bei gleichem Kanal die Schranke des erweiterten Codes?
+{Given the same channel,&nbsp; what is the bound of the extended&nbsp; $(8, 4, 4)$&nbsp; Hamming code?
 |type="{}"}
-$(8, 4, 4)–{\rm Code} \text{:} \hspace{0.2cm} {\rm Pr(Bhattacharyya)} \ = \ ${ 0.022 3% }
+${\rm Pr(Bhattacharyya)} \ = \ ${ 2.2 3% } $\ \%$
-{Mit welchem BEC–Parameter $\lambda$ erhält man die genau gleichen Schranken?
+{With which BEC parameter&nbsp; $\lambda$&nbsp; do you get the exact same barriers?
 |type="{}"}
-$\lambda \ = \ $ { 0 3% }
+$\lambda \ = \ $ { 0.199 3% }
-{Betrachten wir nun das AWGN–Modell. Bestimmen Sie $E_{\rm B} / N_{0}$ in $\rm dB$ derart, dass sich für den $(8, 4, 4)$–Code die gleiche Bhattacharyya–Schranke ergibt.
+{We continue to consider the extended&nbsp; $(8, 4, 4)$&nbsp; Hamming code,&nbsp; but now the AWGN model.
+<br>Determine&nbsp; $E_{\rm B} / N_{0}$&nbsp; (in dB)&nbsp; such that the same Bhattacharyya Bound results.
 |type="{}"}
-$(8, 4, 4)–{\rm Code} \text{:} \hspace{0.2cm} 10 · \lg {E_{\rm B}/N_0} \ = \ $ { 5 3% }$ \ \rm dB$
+$10 · \lg {E_{\rm B}/N_0} \ = \ $ { 5 3% }$ \ \rm dB$
-{Ermitteln Sie nun den AWGN–Parameter für den (7, 4, 3)–Hamming–Code.
+{Now determine the AWGN parameter&nbsp; $(10 · \lg {E_{\rm B}/N_0})$&nbsp; for the&nbsp; $(7, 4, 3)$&nbsp; Hamming code.
 |type="{}"}
-$\ (7, 4, 3)–{\rm Code} \text{:} \hspace{0.2cm} 10 · \lg {E_{\rm B}/N_0} \ = \ $ { 4.417 3% }$ \ \rm dB$
+$10 · \lg {E_{\rm B}/N_0} \ = \ $ { 4.417 3% }$ \ \rm dB$
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Durch die Analyse aller Codeworte des $(7, 4, 3)$–Hamming–Codes erkennt man, dass
+'''(1)'''&nbsp; By analyzing all the code words of the&nbsp; $(7, 4, 3)$&nbsp; Hamming code,&nbsp; we see that.
-*$W_{0} \ \underline{ = \ 1}$ Codewort keine Eins beinhaltet (das Nullwort),
+*$W_{0} \ \underline{ = \ 1}$ &nbsp; &rArr; &nbsp; code word does not contain a&nbsp; "one"&nbsp; (the zero-word),
-*$W_{3} \ \underline{ = \ 7}$ Codeworte drei Einsen beinhalten,
+*$W_{3} \ \underline{ = \ 7}$ &nbsp; &rArr; &nbsp; cod words contain three&nbsp; "ones",
-*$W_{4} \ \underline{ = \ 7}$ Codeworte vier Einsen beinhalten,
+*$W_{4} \ \underline{ = \ 7}$ &nbsp; &rArr; &nbsp; code words contain four&nbsp; "ones",
-*$W_{7} \ \underline{ = \ 1}$ Codewort nur aus Einsen besteht.
+*$W_{7} \ \underline{ = \ 1}$ &nbsp; &rArr; &nbsp; code word consists of only&nbsp; "ones.
-$W_{i}$ gibt gleichzeitig die Anzahl der Codeworte an, die sich vom Nullwort in $i \ \rm Bit$ unterscheiden.
+$W_{i}$&nbsp; simultaneously specifies the number of code words that differ from the zero-word in&nbsp; $i \ \rm bit$.
-'''(2)'''&nbsp; Die Bhattacharyya–Schranke lautet:
+'''(2)'''&nbsp; The Bhattacharyya bound reads:
-:$${\rm Pr(Blockfehler)} \le{\rm Pr(Bhattacharyya)} = W(\beta) -1 \hspace{0.05cm}.$$
+:$${\rm Pr(block\:error)} \le{\rm Pr(Bhattacharyya)} = W(\beta) -1 \hspace{0.05cm}.$$
-Die Gewichtsfunktion ist durch die Teilaufgabe (1) festgelegt:
+*The weight function is defined by the subtask&nbsp; '''(1)''':
-:$$W(X) = 1+ 7 \cdot X^{3} + 7 \cdot X^{4} + X^{7}$$
+:$$W(X) = 1+ 7 \cdot X^{3} + 7 \cdot X^{4} + X^{7}\hspace{0.3cm}
+\Rightarrow \hspace{0.3cm} {\rm Pr(Bhattacharyya)} = 7 \cdot \beta^{3} + 7 \cdot \beta^{4} + \beta^{7} \hspace{0.05cm}.$$
+*For the Bhattacharyya parameter of the BSC model:
-:$$\Rightarrow \hspace{0.3cm} {\rm Pr(Bhattacharyya)} = 7 \cdot \beta^{3} + 7 \cdot \beta^{4} + \beta^{7} \hspace{0.05cm}.$$
+:$$\beta = 2 \cdot \sqrt{\varepsilon \cdot (1- \varepsilon)} = 2 \cdot \sqrt{0.01 \cdot 0.99} = 0.199\hspace{0.3cm}
+   \Rightarrow \hspace{0.3cm} {\rm Pr(Bhattacharyya)} = 7 \cdot 0.199^{3} + 7 \cdot 0.199^{4} + 0.199^{7} \hspace{0.15cm} \underline{ \approx 6.6\%} \hspace{0.05cm}.$$
-Für den Bhattacharyya–Parameter des BSC–Modells gilt:
+*A comparison with the actual block error probability as calculated in&nbsp; [[Aufgaben:Exercise_1.12:_Hard_Decision_vs._Soft_Decision|"Exercise 1.12"]],
-:$$\beta = 2 \cdot \sqrt{\varepsilon \cdot (1- \varepsilon)} = 2 \cdot \sqrt{0.01 \cdot 0.99} = 0.199$$
+:$${\rm Pr(block\:error)} \approx 21 \cdot \varepsilon^2 = 2.1 \cdot 10^{-3} \hspace{0.05cm},$$
-:$$ \Rightarrow \hspace{0.3cm} {\rm Pr(Bhattacharyya)} = 7 \cdot 0.199^{3} + 7 \cdot 0.199^{4} + 0.199^{7} \hspace{0.15cm} \underline{ \approx 0.066} \hspace{0.05cm}.$$
-Ein Vergleich mit der tatsächlichen Blockfehlerwahrscheinlichkeit, wie in [[Aufgaben:1.12_Hard_/_Soft_Decision|Aufgabe 1.12]] berechnet:
+:shows that Bhattacharyya provides only a rough bound.&nbsp; In the present case,&nbsp; this bound is more than a factor of&nbsp; $30$&nbsp; higher than the actual value.
-:$$W(X) = 1+ 14 \cdot X^{4} + X^{8}$$
-:$$\Rightarrow \hspace{0.3cm} {\rm Pr(Bhattacharyya)} = 14 \cdot \beta^{4} + \beta^{8} = 14 \cdot 0.199^{4} + 0.199^{8} \hspace{0.15cm} \underline{ \approx 0.022} \hspace{0.05cm}.$$
-zeigt, dass Bhattacharyya nur eine äußerst grobe Schranke bereitstellt. Im vorliegenden Fall liegt diese Schranke um mehr als den Faktor $30$ über dem tatsächlichen Wert.
+'''(3)'''&nbsp; From the code table of the extended&nbsp; $(8, 4, 4)$&nbsp; Hamming code,&nbsp; the following results are obtained:
-'''(3)'''&nbsp; Aus der Codetabelle des $(8, 4, 4)$–Codes erhält man folgende Ergebnisse:
+:$$W(X) = 1+ 14 \cdot X^{4} + X^{8}\hspace{0.3cm}
+\Rightarrow \hspace{0.3cm} {\rm Pr(Bhattacharyya)} = 14 \cdot \beta^{4} + \beta^{8} = 14 \cdot 0.199^{4} + 0.199^{8} \hspace{0.15cm} \underline{ \approx 2.2\%} \hspace{0.05cm}.$$
-:$${\rm Pr(Blockfehler)} \approx 21 \cdot \varepsilon^2 = 2.1 \cdot 10^{-3} \hspace{0.05cm},$$
-'''(4)'''&nbsp; Die Gleichung für den Bhattacharyya–Parameter lautet:
-:$$\beta = \left\{ \begin{array}{c} \lambda \\ \\ 2 \cdot \sqrt{ \varepsilon \cdot (1- \varepsilon)}\\ \\ {\rm exp}[- R \cdot E_{\rm B}/N_0] \end{array} \right.\quad \begin{array}{*{1}c} {\rm f\ddot{u}r\hspace{0.15cm} das \hspace{0.15cm}BEC-Modell},\\ \\ {\rm f\ddot{u}r\hspace{0.15cm} das \hspace{0.15cm}BSC-Modell}, \\ \\{\rm f\ddot{u}r\hspace{0.15cm} das \hspace{0.15cm}AWGN-Modell}. \end{array}$$
+'''(4)'''&nbsp; The equation for the Bhattacharyya parameter is:
-Mit dem BEC–Modell ergibt sich genau die gleiche Schranke, wenn die Auslöschungswahrscheinlichkeit $\lambda = \beta \ \underline{= 0.199}$ beträgt.
+:$$\beta = \left\{ \begin{array}{c} \lambda \\ \\ 2 \cdot \sqrt{ \varepsilon \cdot (1- \varepsilon)}\\ \\ {\rm e}^{- R \cdot E_{\rm B}/N_0} \end{array} \right.\quad \begin{array}{*{1}c} {\rm for\hspace{0.15cm} the \hspace{0.15cm}BEC model},\\ \\ {\rm for\hspace{0.15cm} the \hspace{0.15cm}BSC model}, \\ \\{\rm for\hspace{0.15cm} the \hspace{0.15cm}AWGN model}. \end{array}$$
+With the BEC model,&nbsp; exactly the same bound is obtained when the erasure probability is&nbsp; $\lambda = \beta \ \underline{= 0.199}$.
-'''(5)'''&nbsp; Entsprechend obiger Gleichung muss gelten:
+'''(5)'''&nbsp; According to the above equation must apply:
-:$$\beta = {\rm exp}[- R \cdot E_{\rm B}/N_0] = 0.199 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} R \cdot E_{\rm B}/N_0 = 10^{0.199} = 1.58 \hspace{0.05cm}.$$
+:$$\beta = {\rm e}^{- R \hspace{0.05cm}\cdot \hspace{0.05cm} E_{\rm B}/N_0} = 0.199 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} R \cdot E_{\rm B}/N_0 = 10^{0.199} = 1.58 \hspace{0.05cm}.$$
-Die Coderate des $(8, 4, 4)$–Codes ist $R = 0.5$:
+*The code rate of the extended&nbsp; $(8, 4, 4)$&nbsp; Hamming code is&nbsp; $R = 0.5$:
 :$$E_{\rm B}/N_0 = 3.16 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 \hspace{0.15cm} \underline{\approx 5\,{\rm dB}} \hspace{0.05cm}.$$
-'''(6)'''&nbsp; Mit der Coderate $R = 4/7$ erhält man:
+'''(6)'''&nbsp; Using the code rate&nbsp; $R = 4/7$&nbsp; of the&nbsp; $(7, 4, 3)$&nbsp; Hamming code,&nbsp; we obtain:
 :$$E_{\rm B}/N_0 = 7/4 \cdot 1.58 = 2.765 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 \hspace{0.15cm} \underline{\approx 4.417\,{\rm dB}} \hspace{0.05cm}.$$
@@ Line 132: / Line 148: @@
-[[Category:Aufgaben zu  Kanalcodierung|^1.6 Schranken für die Blockfehlerwahrscheinlichkeit
+[[Category:Channel Coding: Exercises|^1.6 Error Probability Bounds^]]
-^]]