Aufgaben:Exercise 1.12: Hard Decision vs. Soft Decision - Revision history

Guenter at 15:00, 30 September 2022

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Guenter at 10:00, 11 May 2022

2022-05-11T10:00:08Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.12: Hard Decision vs. Soft Decision to Aufgaben:Exercise 1.12: Hard Decision vs. Soft Decision

2022-05-11T09:53:44Z

Guenter moved page Aufgabe 1.12: Hard Decision vs. Soft Decision to Exercise 1.12: Hard Decision vs. Soft Decision

Javier: Text replacement - "”" to """

2021-05-28T14:27:42Z

Text replacement - "”" to """

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Kanalcodierung/Decodierung linearer Blockcodes}}
-[[File:EN_KC_A_1_12.png|right|frame|Block error rate of the&nbsp; $\rm HC(7, \, 4, \, 3)$&nbsp; for <br>"Hard Decision" and "Soft Decision" '''Korrektur''']]
+[[File:EN_KC_A_1_12.png|right|frame|Block error rate of the&nbsp; $\rm HC(7, \, 4, \, 3)$&nbsp; for <br>"Hard Decision" and "Soft Decision"]]
 The figure shows the block error probability for the&nbsp; [[Channel_Coding/General_Description_of_Linear_Block_Codes#Some_properties_of_the_.287.2C_4.2C_3.29_Hamming_code|"$(7, \, 4, \, 3)$ Hamming code"]],&nbsp; where two variants are considered for the receiver:

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Kanalcodierung/Decodierung linearer Blockcodes}}
-[[File:P_ID2408__KC_A_1_12.png|right|frame|Block error rate of the&nbsp; $\rm HC(7, \, 4, \, 3)$&nbsp; for <br>"Hard Decision" and "Soft Decision" '''Korrektur''']]
+[[File:EN_KC_A_1_12.png|right|frame|Block error rate of the&nbsp; $\rm HC(7, \, 4, \, 3)$&nbsp; for <br>"Hard Decision" and "Soft Decision" '''Korrektur''']]
 The figure shows the block error probability for the&nbsp; [[Channel_Coding/General_Description_of_Linear_Block_Codes#Some_properties_of_the_.287.2C_4.2C_3.29_Hamming_code|"$(7, \, 4, \, 3)$ Hamming code"]],&nbsp; where two variants are considered for the receiver:

@@ Line 114: / Line 114: @@
 :$$10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 20 \cdot {\rm lg} \hspace{0.1cm}(2.33) - 10 \cdot {\rm lg} \hspace{0.1cm} (8/7) = 7.35\,{\rm dB} - 0.58\,{\rm dB}\hspace{0.15cm}\underline{\approx 6.77\,{\rm dB}}\hspace{0.05cm}.$$
+* Analogously,&nbsp; for&nbsp; $\varepsilon = 0.001 \ ⇒ \ {\rm Q}^{-1}(\varepsilon) = 3.09$:
 :$$10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 20 \cdot {\rm lg} \hspace{0.1cm}(3.09) - 0.58\,{\rm dB}\hspace{0.15cm}\underline{\approx 9.22\,{\rm dB}}\hspace{0.05cm}.$$

@@ Line 70: / Line 70: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Every Hamming code is perfect and has the minimum distance $d_{\rm min} = 3$.
+'''(1)'''&nbsp; Every Hamming code is perfect and has the minimum distance&nbsp; $d_{\rm min} = 3$.
-*Therefore, one bit error in the codeword can be corrected, while two bit errors will always cause the codeword to fail &nbsp; ⇒ &nbsp; parameter $t = 1$.
+*Therefore,&nbsp; one bit error in the code word can be corrected,&nbsp; while two bit errors will always cause the code word to fail &nbsp; ⇒ &nbsp; parameter&nbsp; $t = 1$.
-*Thus, for the block error probability:
+*Thus,&nbsp; for the block error probability:
 :$${\rm Pr(block\:error)} \hspace{-0.15cm}\ = \ \hspace{-0.15cm} 1 - {\rm Pr(no\hspace{0.15cm} block\:error)} - {\rm Pr(one\hspace{0.15cm} block\:error)} = 1 - (1 - \varepsilon)^7 - 7 \cdot \varepsilon \cdot (1 - \varepsilon)^6 \hspace{0.05cm}.$$
@@ Line 79: / Line 79: @@
-'''(2)'''&nbsp; Each $(n, \, k, \, 3)$ Hamming code can correct only one bit error. Thus, for the BSC channel, the general rule is with codeword length $n$:
+'''(2)'''&nbsp; Each&nbsp; $(n, \, k, \, 3)$&nbsp; Hamming code can correct only one bit error.&nbsp; Thus,&nbsp; for the BSC channel,&nbsp; the general rule is with code word length&nbsp; $n$:
 :$${\rm Pr(block\:error)} = 1 - \text{Pr(no bit error)} - \text{Pr(one bit error)}  = 1 - (1 - \varepsilon)^n - n \cdot \varepsilon \cdot (1 - \varepsilon)^{n-1}.$$
-*After series expansion of &nbsp; $(1 - \varepsilon)^n$ &nbsp; or of &nbsp; $n \cdot \varepsilon \cdot (1 - \varepsilon)^{n-1}$ &nbsp; we obtain from this:
+*After series expansion of &nbsp; "$(1 - \varepsilon)^n$" &nbsp; or of &nbsp; "$n \cdot \varepsilon \cdot (1 - \varepsilon)^{n-1}$" &nbsp; we obtain from this:
 :$${\rm Pr(block\:error)} = 1 - \left [ 1 - {n \choose 1}\cdot \varepsilon + {n \choose 2}\cdot \varepsilon^2 - \hspace{0.05cm}\text{...} \hspace{0.05cm} \right ] -\left [ n \cdot \varepsilon \cdot \left ( 1 - {{n-1} \choose 1}\cdot \varepsilon + {{n-1} \choose 2}\cdot \varepsilon^2 - \hspace{0.05cm}\text{...} \hspace{0.05cm}\right ) \right ] \hspace{0.05cm}.$$
-*If we neglect all terms with $\varepsilon^3, \ \varepsilon^4, \ \text{...}$&nbsp; we finally get:
+*If we neglect all terms with&nbsp; $\varepsilon^3, \ \varepsilon^4, \ \text{...}$&nbsp; we finally get:
 :$${\rm Pr(block\:error)} \hspace{-0.15cm}\ = \ \hspace{-0.15cm} n \cdot \varepsilon - {n \choose 2}\cdot \varepsilon^2 - n \cdot \varepsilon + n \cdot \varepsilon {{n-1} \choose 1}\cdot \varepsilon + \hspace{0.05cm}\text{...}\hspace{0.05cm} = -n/2 \cdot (n-1)\cdot \varepsilon^2 + n \cdot (n-1)\cdot \varepsilon^2 = n \cdot (n-1)/2 \cdot \varepsilon^2 \hspace{0.05cm}.$$
-The correct solution is <u>suggestion 1</u>.
+The correct solution is&nbsp; <u>suggestion 1</u>.
-For the $(7, \, 4, \, 3)$ Hamming code, it follows:
+For the&nbsp; $(7, \, 4, \, 3)$&nbsp; Hamming code,&nbsp; it follows:
 :$${\rm Pr(block\:error)} \le \left\{ \begin{array}{c} 2.03 \cdot 10^{-3}\\ 2.09 \cdot 10^{-5} \end{array} \right.\quad \begin{array}{*{1}c} {\rm for}\hspace{0.15cm} \varepsilon = 10^{-2} \\ {\rm for}\hspace{0.15cm} \varepsilon = 10^{-3} \\ \end{array} \hspace{0.05cm}.$$
-*By comparison with the result of the subtask (1) one recognizes the validity of this approximation.
+*By comparison with the result of the subtask&nbsp; '''(1)'''&nbsp; one recognizes the validity of this approximation.&nbsp; The smaller the BSC falsification probability $\varepsilon$,&nbsp; the better it is.
-'''(3)'''&nbsp; The results of subtask (2) can be summarized as follows:
+'''(3)'''&nbsp; The results of subtask&nbsp; '''(2)'''&nbsp; can be summarized as follows:
 :$${\rm Pr(block\:error)} = \left\{ \begin{array}{l} 3 \cdot \varepsilon^2 \\ 21 \cdot \varepsilon^2\\ 105 \cdot \varepsilon^2\\ \end{array} \right.\quad \begin{array}{*{1}l} {\rm for}\hspace{0.15cm} n = 3 \\ {\rm for}\hspace{0.15cm} n = 7 \\ {\rm for}\hspace{0.15cm} n = 15 \\ \end{array} \hspace{0.05cm}.$$
-*Right is <u>answer 1</u>.
+*Right is <u>answer 1</u>.&nbsp; Of course, the Hamming code with the lowest rate&nbsp; $R = 1/3$,&nbsp; i.e. with the greatest relative redundancy,&nbsp; has the lowest block error probability.
-'''(4)'''&nbsp;  At Hard Decision, with the complementary Gaussian error function ${\rm Q}(x)$:
+'''(4)'''&nbsp;  At Hard Decision,&nbsp; with the complementary Gaussian error function&nbsp; ${\rm Q}(x)$:
 :$$\varepsilon = {\rm Q}\left ( \sqrt{2 \cdot R \cdot E_{\rm B}/N_0} \right )\hspace{0.3cm} \Rightarrow \hspace{0.3cm} E_{\rm B}/N_0 = \frac{[{\rm Q}^{-1}(\varepsilon)]^2}{2R}\hspace{0.3cm}
   \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 20 \cdot {\rm lg} \hspace{0.1cm}[{\rm Q}^{-1}(\varepsilon)] - 10 \cdot {\rm lg} \hspace{0.1cm} (2R) \hspace{0.05cm}.$$
-*From this we obtain with $\varepsilon = 0.01 \ ⇒ \ {\rm Q}^{–1}(\varepsilon) = 2.33$:
+*From this we obtain with&nbsp; $\varepsilon = 0.01 \ ⇒ \ {\rm Q}^{–1}(\varepsilon) = 2.33$:
 :$$10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 20 \cdot {\rm lg} \hspace{0.1cm}(2.33) - 10 \cdot {\rm lg} \hspace{0.1cm} (8/7) = 7.35\,{\rm dB} - 0.58\,{\rm dB}\hspace{0.15cm}\underline{\approx 6.77\,{\rm dB}}\hspace{0.05cm}.$$
@@ Line 121: / Line 119: @@
-'''(5)'''&nbsp; Here we refer to the block error probability $10^{–5}$.
+'''(5)'''&nbsp; Here we refer to the block error probability&nbsp; $10^{–5}$.
-*According to the result of subtask (2), the BSC corruption probability must then not be greater than
+*According to the result of subtask&nbsp; '''(2)''',&nbsp; the BSC falsification probability must then not be greater than
 :$$\varepsilon = \sqrt{{10^{-5}}/{21}} = 6.9 \cdot 10^{-4} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Q}^{-1}(\varepsilon) = 3.2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 9.52\,{\rm dB}\hspace{0.05cm}.$$
-*For''Soft Decision'' according to specification $8 \ {\rm dB} suffices \ ⇒ \ 10 · \lg {G_{\rm SD}} \ \underline{= 1.52 \ {\rm dB}}$.
+*For&nbsp; "Soft Decision"&nbsp; according to the specification graph:&nbsp; $10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 8\,{\rm dB}\hspace{0.05cm}$ &nbsp; &rArr; &nbsp;  $10 · \lg {G_{\rm SD}} \ \underline{= 1.52 \ {\rm dB}}$.
 {{ML-Fuß}}

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Kanalcodierung/Decodierung linearer Blockcodes}}
-[[File:P_ID2408__KC_A_1_12.png|right|frame|Block error rate of the&nbsp; $\rm HC(7, \, 4, \, 3)$&nbsp; for <br>"Hard Decision" and "Soft Decision"]]
+[[File:P_ID2408__KC_A_1_12.png|right|frame|Block error rate of the&nbsp; $\rm HC(7, \, 4, \, 3)$&nbsp; for <br>"Hard Decision" and "Soft Decision" '''Korrektur''']]
-The figure shows the block error probability for the&nbsp; [[Channel_Coding/General_Description_of_Linear_Block_Codes#Some_properties_of_the_.287.2C_4.2C_3.29_Hamming_code|$(7, \, 4, \, 3)$ Hamming code]], where two variants are considered for the receiver:
+The figure shows the block error probability for the&nbsp; [[Channel_Coding/General_Description_of_Linear_Block_Codes#Some_properties_of_the_.287.2C_4.2C_3.29_Hamming_code|"$(7, \, 4, \, 3)$ Hamming code"]],&nbsp; where two variants are considered for the receiver:
-*Maximum likelihood detection with hard decisions $($"Hard Decision", &nbsp;$\rm HD)$, which in the present case (perfect code) can also be realized by syndrome decoding, yields the red curve (with circle marker).
+*Maximum likelihood decoding with hard decisions&nbsp; $($"Hard Decision", &nbsp;$\rm HD)$,&nbsp; which in the present case&nbsp; $($perfect code$)$&nbsp; can also be realized by syndrome decoding,&nbsp; yields the red curve&nbsp; $($with circle marker$)$.
-*The channel can be replaced by the BSC model in a simplified way for "Hard Decision". The relationship between the BSC parameter&nbsp; $\varepsilon$&nbsp; and the AWGN quotient&nbsp; $E_{\rm B}/N_{0}$&nbsp; (used in the graph) is given as follows:
+*The channel can be replaced by the BSC model in a simplified way for&nbsp; "Hard Decision".&nbsp; The relationship between the BSC parameter&nbsp; $\varepsilon$&nbsp; and the AWGN quotient&nbsp; $E_{\rm B}/N_{0}$&nbsp; $($used in the graph$)$ is given as follows:
 :$$\varepsilon = {\rm Q}\left ( \sqrt{2 \cdot R \cdot E_{\rm B}/N_0} \right ) \hspace{0.05cm}.$$
-:Here denotes&nbsp; ${\rm Q}(x)$&nbsp; the ''complementary Gaussian error function'' and&nbsp; $R$&nbsp; the code rate.
+:Here&nbsp; ${\rm Q}(x)$&nbsp;  denotes the&nbsp; "complementary Gaussian error function"&nbsp; and&nbsp; $R$&nbsp; the code rate.
-*The green curve (with crosses) shows the block error probability for "soft" decisions $($"Soft Decision", &nbsp;$\rm SD)$. This function curve cannot be given in closed-mathematical form. The curve drawn in the graph is an upper bound given in [Fri96]:
+*The green curve&nbsp; $($with crosses$)$ shows the block error probability for soft decisions&nbsp; $($"Soft Decision", &nbsp;$\rm SD)$.&nbsp; This curve cannot be given in closed-mathematical form.&nbsp; The curve drawn in the graph is an upper bound given in&nbsp; [Fri96]:
 :$$ {\rm Pr(block\:error)} \hspace{-0.15cm}\ \le \ \hspace{-0.15cm} 7 \cdot {\rm Q}\left ( \sqrt{ 3 \cdot \frac{2 \cdot R \cdot E_{\rm B}}{N_0}} \right )+7 \cdot {\rm Q}\left ( \sqrt{ 4 \cdot \frac{2 \cdot R \cdot E_{\rm B}}{N_0}} \right ) + {\rm Q}\left ( \sqrt{ 7 \cdot \frac{2 \cdot R \cdot E_{\rm B}}{N_0}} \right ) \hspace{0.05cm}.$$
-:*The respective first factor in the argument of the&nbsp; $\rm Q$ function indicates the possible Hamming distances: &nbsp; $i = 3, \, 4, \, 7$.
+:#The respective first factor in the arguments of the&nbsp; $\rm Q$&ndash;function indicates Hamming distances &nbsp; &rArr; &nbsp; $i = 3, \, 4, \, 7$.
-:*The prefactors take into account multiplicities&nbsp; $W_{3} = W_{4} = 7$&nbsp; and&nbsp; $W_{7} = 1$. &nbsp; $R = 4/7$&nbsp; describes the code rate.
+:#The prefactors take into account&nbsp; "multiplicities"&nbsp; $W_{3} = W_{4} = 7$&nbsp; and&nbsp; $W_{7} = 1$. &nbsp;
-:*For&nbsp; $10 · \lg {E_{\rm B}/N_0} > 8  \ \rm dB$&nbsp; ist&nbsp; $\rm Pr(block\:error) < 10^{–5}$.
+:#$R = 4/7$&nbsp; describes the code rate.
@@ Line 25: / Line 26: @@
-Hints:
+*For the subtasks&nbsp; '''(1)'''&nbsp; to&nbsp; '''(4)''',&nbsp; "Hard Decision" is always assumed.
@@ Line 39: / Line 41: @@
 ===Questions===
 <quiz display=simple>
-{What is the block error probability of the&nbsp; $(7, \, 4, \, 3)$ Hamming code in "Hard Decision"?
+{What is the block error probability of the&nbsp; $(7, \, 4, \, 3)$ Hamming code in&nbsp; "Hard Decision"?
 |type="{}"}
 $\varepsilon = 10^{-2} \text{:} \hspace{0.4cm} {\rm Pr(block\:error)} \ = \ $ { 2.03 3% } $\ \cdot 10^{-3} $
 $\varepsilon = 10^{-3} \text{:} \hspace{0.4cm} {\rm Pr(block\:error)} \ = \ $ { 0.021 3% } $\ \cdot 10^{-3} $
-{How can we approximate the block error probability of a Hamming code, assuming "hard decision"?
+{How can we approximate the block error probability of a Hamming code,&nbsp; assuming "Hard Decision"?
 |type="()"}
 + ${\rm Pr(block\:error)} = n · (n–1)/2 · \varepsilon^2.$
@@ Line 52: / Line 54: @@
 {Which Hamming code has the smallest block error probability with constant BSC parameter&nbsp; $\varepsilon$?
 |type="()"}
-+ the Hamming code&nbsp; $(3, \, 1, \, 3)$, identical to the "repetition code"&nbsp; $\rm RC \ (3, \, 1, \, 3)$,
++ the Hamming code&nbsp; $(3, \, 1, \, 3)$,&nbsp; identical to the&nbsp; "repetition code"&nbsp; $\rm RC \ (3, \, 1, \, 3)$,
 - the Hamming code&nbsp; $(7, \, 4, \, 3)$,
 - the Hamming code&nbsp; $(15, \, 11, \, 3)$.
@@ Line 61: / Line 63: @@
 $\varepsilon = 10^{-3} \text{:} \hspace{0.4cm} 10 · \lg {E_{\rm B}/N_0} \ = \ $ { 9.22 3% } $\ \rm dB$
-{What is the gain (in dB) to be achieved by "Soft Decision" if the block error probability is not to exceed&nbsp; $10^{-5}$&nbsp; ?
+{What is the gain&nbsp; $($in dB$)$&nbsp; to be achieved by&nbsp; "Soft Decision"&nbsp; if the block error probability is not to exceed&nbsp; $10^{-5}$ ?
 |type="{}"}
 $\ 10 · \lg \, {G_{\rm SD}} \ = \ $ { 1.52 3% } $ \ \rm dB$

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← Older revision	Revision as of 09:53, 11 May 2022
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@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Kanalcodierung/Decodierung linearer Blockcodes}}
-[[File:P_ID2408__KC_A_1_12.png|right|frame|Blockfehlerrate des&nbsp; $\rm HC(7, \, 4, \, 3)$&nbsp; bei <br>"Hard Decision&rdquo; und "Soft Decision&rdquo;]]
+[[File:P_ID2408__KC_A_1_12.png|right|frame|Blockfehlerrate des&nbsp; $\rm HC(7, \, 4, \, 3)$&nbsp; bei <br>"Hard Decision" und "Soft Decision"]]
 Die Abbildung zeigt die Blockfehlerwahrscheinlichkeit für den&nbsp; [[Channel_Coding/Allgemeine_Beschreibung_linearer_Blockcodes#Einige_Eigenschaften_des_.287.2C_4.2C_3.29.E2.80.93Hamming.E2.80.93Codes|$(7, \, 4, \, 3)$–Hamming–Code]], wobei für den Empfänger zwei Varianten berücksichtigt sind:
-*Bei Maximum–Likelihood–Detektion mit harten Entscheidungen $($"Hard Decision&rdquo;, &nbsp;$\rm HD)$, die im vorliegenden Fall (perfekter Code) auch durch Syndromdecodierung realisiert werden kann, ergibt sich die rote Kurve (mit Kreismarkierung).
+*Bei Maximum–Likelihood–Detektion mit harten Entscheidungen $($"Hard Decision", &nbsp;$\rm HD)$, die im vorliegenden Fall (perfekter Code) auch durch Syndromdecodierung realisiert werden kann, ergibt sich die rote Kurve (mit Kreismarkierung).
-*Der Kanal kann bei "Hard Decision&rdquo; vereinfacht durch das BSC–Modell ersetzt werden. Der Zusammenhang zwischen dem BSC–Parameter&nbsp; $\varepsilon$&nbsp; und dem AWGN–Quotienten&nbsp; $E_{\rm B}/N_{0}$&nbsp; (in der Grafik verwendet) ist wie folgt gegeben:
+*Der Kanal kann bei "Hard Decision" vereinfacht durch das BSC–Modell ersetzt werden. Der Zusammenhang zwischen dem BSC–Parameter&nbsp; $\varepsilon$&nbsp; und dem AWGN–Quotienten&nbsp; $E_{\rm B}/N_{0}$&nbsp; (in der Grafik verwendet) ist wie folgt gegeben:
 :$$\varepsilon = {\rm Q}\left ( \sqrt{2 \cdot R \cdot E_{\rm B}/N_0} \right ) \hspace{0.05cm}.$$
@@ Line 13: / Line 13: @@
 :Hier bezeichnet&nbsp; ${\rm Q}(x)$&nbsp; die ''komplementäre Gaußsche Fehlerfunktion'' und&nbsp; $R$&nbsp; die Coderate.
-*Die grüne Kurve (mit Kreuzen) zeigt die Blockfehlerwahrscheinlichkeit bei „weichen” Entscheidungen $($"Soft Decision&rdquo;, &nbsp;$\rm SD)$. Dieser Funktionsverlauf lässt sich nicht in geschlossen–mathematischer Form angeben. Die in der Grafik eingezeichnete Kurve ist eine in [Fri96] angegebene obere Schranke:
+*Die grüne Kurve (mit Kreuzen) zeigt die Blockfehlerwahrscheinlichkeit bei „weichen” Entscheidungen $($"Soft Decision", &nbsp;$\rm SD)$. Dieser Funktionsverlauf lässt sich nicht in geschlossen–mathematischer Form angeben. Die in der Grafik eingezeichnete Kurve ist eine in [Fri96] angegebene obere Schranke:
 :$$ {\rm Pr(Blockfehler)} \hspace{-0.15cm}\ \le \ \hspace{-0.15cm} 7 \cdot {\rm Q}\left ( \sqrt{ 3 \cdot \frac{2 \cdot R \cdot E_{\rm B}}{N_0}} \right )+7 \cdot {\rm Q}\left ( \sqrt{ 4 \cdot \frac{2 \cdot R \cdot E_{\rm B}}{N_0}} \right ) + {\rm Q}\left ( \sqrt{ 7 \cdot \frac{2 \cdot R \cdot E_{\rm B}}{N_0}} \right ) \hspace{0.05cm}.$$
@@ Line 29: / Line 29: @@
 ''Hinweise:''
 * Die Aufgabe bezieht sich auf das Kapitel&nbsp; [[Channel_Coding/Decodierung_linearer_Blockcodes|Decodierung linearer Blockcodes]].
-* Die oben zitierte Literaturstelle&nbsp; [Fri96]&nbsp; verweist auf das Buch <br>"Friedrichs, B.: Kanalcodierung – Grundlagen und Anwendungen in modernen Kommunikationssystemen. Berlin – Heidelberg: Springer, 1996&rdquo;.
+* Die oben zitierte Literaturstelle&nbsp; [Fri96]&nbsp; verweist auf das Buch <br>"Friedrichs, B.: Kanalcodierung – Grundlagen und Anwendungen in modernen Kommunikationssystemen. Berlin – Heidelberg: Springer, 1996".
 * Verwenden Sie für numerische Ergebnisse das Berechnungsmodul&nbsp;  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|Komplementäre Gaußsche Fehlerfunktionen]].
-*Für die Teilaufgaben '''(1)''' bis '''(4)''' wird stets von  "Hard Decision&rdquo; ausgegangen.
+*Für die Teilaufgaben '''(1)''' bis '''(4)''' wird stets von  "Hard Decision" ausgegangen.
@@ Line 39: / Line 39: @@
 ===Fragebogen===
 <quiz display=simple>
-{Welche Blockfehlerwahrscheinlichkeit besitzt der&nbsp; $(7, \, 4, \, 3)$–Hamming–Code bei "Hard Decision&rdquo;?
+{Welche Blockfehlerwahrscheinlichkeit besitzt der&nbsp; $(7, \, 4, \, 3)$–Hamming–Code bei "Hard Decision"?
 |type="{}"}
 $\varepsilon = 10^{-2} \text{:} \hspace{0.4cm} {\rm Pr(Blockfehler)} \ = \ $ { 2.03 3% } $\ \cdot 10^{-3} $
 $\varepsilon = 10^{-3} \text{:} \hspace{0.4cm} {\rm Pr(Blockfehler)} \ = \ $ { 0.021 3% } $\ \cdot 10^{-3} $
-{Wie kann man die Blockfehlerwahrscheinlichkeit eines Hamming–Codes annähern, "Hard Decision&rdquo;  vorausgesetzt?
+{Wie kann man die Blockfehlerwahrscheinlichkeit eines Hamming–Codes annähern, "Hard Decision"  vorausgesetzt?
 |type="()"}
 + ${\rm Pr(Blockfehler)} = n · (n–1)/2 · \varepsilon^2.$
@@ Line 52: / Line 52: @@
 {Welcher Hamming–Code besitzt die kleinste Blockfehlerwahrscheinlichkeit bei konstantem BSC–Parameter&nbsp; $\varepsilon$?
 |type="()"}
-+ der Hamming–Code&nbsp; $(3, \, 1, \, 3)$, identisch mit dem "Repetition Code&rdquo;&nbsp; $\rm RC \ (3, \, 1, \, 3)$,
++ der Hamming–Code&nbsp; $(3, \, 1, \, 3)$, identisch mit dem "Repetition Code"&nbsp; $\rm RC \ (3, \, 1, \, 3)$,
 - der Hamming–Code&nbsp; $(7, \, 4, \, 3)$,
 - der Hamming–Code&nbsp; $(15, \, 11, \, 3)$.
@@ Line 61: / Line 61: @@
 $\varepsilon = 10^{-3} \text{:} \hspace{0.4cm} 10 · \lg {E_{\rm B}/N_0} \ = \ $ { 9.22 3% } $\ \rm dB$
-{Welcher Gewinn (in dB) ist durch "Soft Decision&rdquo; zu erzielen, wenn die Blockfehlerwahrscheinlichkeit den Wert&nbsp; $10^{–5}$&nbsp; nicht überschreiten soll?
+{Welcher Gewinn (in dB) ist durch "Soft Decision" zu erzielen, wenn die Blockfehlerwahrscheinlichkeit den Wert&nbsp; $10^{–5}$&nbsp; nicht überschreiten soll?
 |type="{}"}
 $\ 10 · \lg \, {G_{\rm SD}} \ = \ $ { 1.52 3% } $ \ \rm dB$