Aufgaben:Exercise 2.15Z: Block Error Probability once more - Revision history

Guenter at 13:37, 2 November 2022

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Guenter at 16:39, 1 November 2022

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Guenter at 16:29, 1 November 2022

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Guenter at 16:27, 1 November 2022

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Noah at 20:38, 14 September 2022

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Noah at 20:36, 14 September 2022

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Noah at 20:36, 14 September 2022

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Guenter at 15:57, 4 September 2022

2022-09-04T15:57:36Z

Guenter: Guenter moved page Aufgaben:Aufgabe 2.15Z: Nochmals RS-Blockfehlerwahrscheinlichkeit to Aufgaben:Exercise 2.15Z: Block Error Probability once more

2022-09-04T15:41:12Z

Guenter moved page Aufgabe 2.15Z: Nochmals RS-Blockfehlerwahrscheinlichkeit to Exercise 2.15Z: Block Error Probability once more

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

2021-03-23T12:52:51Z

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

@@ Line 2: / Line 2: @@
 [[File:P_ID2574__KC_Z_2_15.png|right|frame|Binominal probabilities]]
-Using a Reed&ndash;Solomon code with the correctability&nbsp; $t$&nbsp; and&nbsp; [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD|"Bounded Distance Decoding"]]&nbsp; $\rm $&nbsp; $\rm (BDD)$&nbsp; is obtained with
+Using a Reed&ndash;Solomon code with the correctability&nbsp; $t$&nbsp; and&nbsp; [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD|"Bounded Distance Decoding"]]&nbsp; $\rm $&nbsp; $\rm (BDD)$&nbsp; is obtained for the block error probability with
 * the code word length&nbsp; $n$&nbsp; and
-* the symbol falsification probability&nbsp; $\varepsilon_{\rm S}$
+* the symbol falsification probability&nbsp; $\varepsilon_{\rm S}$:
 :$${\rm Pr(block\:error)}  =
 \sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$

@@ Line 56: / Line 56: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; For the $\rm RSC \, (7, \, 3, \, 5)_8$, because $d_{\rm min} = 5 \ \Rightarrow \ t = 2$ gives the block error probability:
+'''(1)'''&nbsp; For the&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$,&nbsp; because&nbsp; $d_{\rm min} = 5 \ \Rightarrow \ t = 2$&nbsp; gives the block error probability:
-:$${\rm Pr(Block\:error)}  \hspace{-0.15cm} \ = \ \hspace{-0.15cm}
+:$${\rm Pr(block\:error)}  \hspace{-0.15cm} \ = \ \hspace{-0.15cm}
 \sum_{f = 3}^{7} {7 \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{7-f} $$
-:$$\Rightarrow \hspace{0.3cm}{\rm Pr(Block\:error)} ={7 \choose 3} \cdot 0.1^3 \cdot 0.9^4 + {7 \choose 4} \cdot 0.1^4 \cdot 0.9^3
+:$$\Rightarrow \hspace{0.3cm}{\rm Pr(block\:error)} ={7 \choose 3} \cdot 0.1^3 \cdot 0.9^4 + {7 \choose 4} \cdot 0.1^4 \cdot 0.9^3
 + {7 \choose 5} \cdot 0.1^5 \cdot 0.9^2+
   {7 \choose 6} \cdot 0.1^6 \cdot 0.9+ {7 \choose 7} \cdot 0.1^7
 \hspace{0.05cm}.$$
-*According to this calculation, five terms would have to be considered. However, since
+*According to this calculation,&nbsp; five terms would have to be considered.&nbsp; However,&nbsp; since
-:$${\rm Pr(Block\:error)}  =
+:$${\rm Pr(block\:error)}  =
 \sum_{f = 0}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} = 1$$
-is also valid, the following calculation is faster:
+:is also valid,&nbsp; the following calculation is faster:
-:$${\rm Pr(Block\:error)} =1 - \big [ {7 \choose 0}  \cdot 0.9^7 + {7 \choose 1} \cdot 0.1 \cdot 0.9^6
+:$${\rm Pr(block\:error)} =1 - \big [ {7 \choose 0}  \cdot 0.9^7 + {7 \choose 1} \cdot 0.1 \cdot 0.9^6
 + {7 \choose 2} \cdot 0.1^2 \cdot 0.9^5 \big ] =1 - \big [ 0.4783 + 0.3720 + 0.1240  \big ] \hspace{0.15cm} \underline{= 2.57 \cdot 10^{-2}}
 \hspace{0.05cm}.$$
-'''(2)'''&nbsp; Analogous to the subtask '''(1)''' one obtains here:
+'''(2)'''&nbsp; Analogous to the subtask&nbsp; '''(1)'''&nbsp; one obtains here:
-:$${\rm Pr(Block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 - \big [ 0.99^7 + 7 \cdot 0.01 \cdot 0.99^6
+:$${\rm Pr(block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 - \big [ 0.99^7 + 7 \cdot 0.01 \cdot 0.99^6
 + 21 \cdot 0.01^2 \cdot 0.99^5 \big ] =1 - \big [ 0.9321 + 0.0659 + 0.0020  \big ] \approx 0
 \hspace{0.05cm}.$$
-*This means: &nbsp; For the probability $\varepsilon_{\rm S} = 0.01$ the simplified calculation is very error-prone, because a value close to $1$ results for the bracket expression.
+*This means: &nbsp; For the probability&nbsp; $\varepsilon_{\rm S} = 0.01$&nbsp; the simplified calculation is very error-prone,&nbsp; because a value close to&nbsp; $1$&nbsp; results for the bracket expression.
 *The full calculation yields here:
-:$${\rm Pr(Block\:error)}  \hspace{-0.15cm} \ = \ \hspace{-0.15cm}
+:$${\rm Pr(block\:error)}  \hspace{-0.15cm} \ = \ \hspace{-0.15cm}
 {7 \choose 3} \cdot 0.01^3 \cdot 0.99^4 + {7 \choose 4} \cdot 0.01^4 \cdot 0.99^3 +
 {7 \choose 5} \cdot 0.01^5 \cdot 0.99^2+  {7 \choose 6} \cdot 0.01^6 \cdot 0.99+
 {7 \choose 7} \cdot 0.01^7 $$
-:$$\Rightarrow \hspace{0.3cm}{\rm Pr(Block\:error)}= 10^{-6} \cdot
+:$$\Rightarrow \hspace{0.3cm}{\rm Pr(block\:error)}= 10^{-6} \cdot
 \big [ 33.6209 +  0.3396 + 0.0021 + ... \big ] \hspace{0.15cm} \underline{ \approx  3.396 \cdot 10^{-5}}
 \hspace{0.05cm}.$$
@@ Line 91: / Line 92: @@
-'''(3)'''&nbsp; From the sample solution to the subtask '''(2)''' the result can be read directly:
+'''(3)'''&nbsp; From the sample solution to the subtask&nbsp; '''(2)'''&nbsp; the result can be read directly:
-:$${\rm Pr(Block\:error)} \hspace{0.15cm}  \underline{ \approx  3.362 \cdot 10^{-5}}
+:$${\rm Pr(block\:error)} \hspace{0.15cm}  \underline{ \approx  3.362 \cdot 10^{-5}}
 \hspace{0.05cm}.$$
-*The relative error is about $-1\%$.
+*The relative error is about&nbsp; $-1\%$.
 '''(4)'''&nbsp; Restricting to the relevant term $(f = 3)$, we get $\varepsilon_{\rm S} = 0.001$:
-:$${\rm Pr(Block\:error)}  \approx
+'''(4)'''&nbsp; Restricting to the relevant term&nbsp; $(f = 3)$,&nbsp; we get&nbsp; $\varepsilon_{\rm S} = 0.001$:
 {7 \choose 3} \cdot [10^{-3}]^3 \cdot 0.999^4  \hspace{0.15cm} \underline{ \approx  3.49 \cdot 10^{-8}}
 \hspace{0.05cm}.$$
-*The relative error here is also only about $-0.1\%$.
+*The relative error here is  only about&nbsp; $-0.1\%$.
-'''(5)'''&nbsp; According to the derived approximation, the following holds for the considered code:
+'''(5)'''&nbsp; According to the derived approximation,&nbsp; the following holds for the considered code:
-:$${\rm Pr(Block\:error)}  \approx {7 \choose 3} \cdot {\varepsilon_{\rm S}}^3 = 35 \cdot {\varepsilon_{\rm S}}^3\hspace{0.3cm}
+:$${\rm Pr(block\:error)}  \approx {7 \choose 3} \cdot {\varepsilon_{\rm S}}^3 = 35 \cdot {\varepsilon_{\rm S}}^3\hspace{0.3cm}
-\Rightarrow  \hspace{0.3cm} {\rm Pr(Block\:error)} = 10^{-10}: \hspace{0.4cm} {\varepsilon_{\rm S}} =
+\Rightarrow  \hspace{0.3cm} {\rm Pr(block\:error)} = 10^{-10}: \hspace{0.4cm} {\varepsilon_{\rm S}} =
 \big ( \frac{10^{-10}}{35} \big )^{1/3} = 2.857^{1/3} \cdot 10^{-4}
 \hspace{0.15cm} \underline{ \approx  1.42 \cdot 10^{-4}}\hspace{0.05cm}.$$

← Older revision		Revision as of 16:29, 1 November 2022
Line 25:		Line 25:
	* This exercise belongs to the chapter  [[Channel_Coding/Error_Probability_and_Areas_of_Application\| "Error Probability and Application Areas"]].		* This exercise belongs to the chapter  [[Channel_Coding/Error_Probability_and_Areas_of_Application\| "Error Probability and Application Areas"]].

−	* For checking,  you can use the interactive HTML5/JavaScript applet  [[Applets:~~Binomial-_und_Poissonverteilung_~~(Applet)\|"Binomial and Poisson Distribution"]].	+	* For checking,  you can use the interactive HTML5/JavaScript applet  [[Applets:Binomial_and_Poisson_Distribution_(Applet)\|"Binomial and Poisson Distribution"]].

@@ Line 2: / Line 2: @@
 [[File:P_ID2574__KC_Z_2_15.png|right|frame|Binominal probabilities]]
-Using a Reed&ndash;Solomon&ndash;code with the correctability&nbsp; $t$&nbsp; and&nbsp; [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD|"Bounded Distance Decoding"]]&nbsp; (BDD) is obtained with
+Using a Reed&ndash;Solomon code with the correctability&nbsp; $t$&nbsp; and&nbsp; [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD|"Bounded Distance Decoding"]]&nbsp; $\rm $&nbsp; $\rm (BDD)$&nbsp; is obtained with
 * the code word length&nbsp; $n$&nbsp; and
-* the symbol corruption probability&nbsp; $\varepsilon_{\rm S}$
 for the block error probability:
-:$${\rm Pr(Block\:error)}  =
+:$${\rm Pr(block\:error)}  =
 \sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$
-In this exercise, the block error probability for the&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$&nbsp; and various&nbsp; $\varepsilon_{\rm S}$ values shall be calculated and approximated.
+In this exercise,&nbsp; the block error probability for the&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$&nbsp; and various&nbsp; $\varepsilon_{\rm S}$&nbsp; values shall be calculated and approximated.
@@ Line 24: / Line 24: @@
 Hints:
 * This exercise belongs to the chapter&nbsp; [[Channel_Coding/Error_Probability_and_Areas_of_Application| "Error Probability and Application Areas"]].
-* For checking, you can use the interactive applet&nbsp; [[Applets:Binomial-_und_Poissonverteilung_(Applet)|"Binomial and Poisson Distribution"]]&nbsp;.
@@ Line 34: / Line 35: @@
 {What is the block error probability for&nbsp; $\varepsilon_{\rm S} = 10^{-1}$?
 |type="{}"}
-$\rm Pr(Block\:error) \ = \ ${ 2.57 3% } $\ \cdot 10^{-2}$
+$\rm Pr(block\:error) \ = \ ${ 2.57 3% } $\ \cdot 10^{-2}$
 {What is the block error probability for&nbsp; $\varepsilon_{\rm S} =10^{-2}$?
 |type="{}"}
-$\rm Pr(Block\:error) \ = \ ${ 3.396 3% } $\ \cdot 10^{-5}$
+$\rm Pr(block\:error) \ = \ ${ 3.396 3% } $\ \cdot 10^{-5}$
-{What result is obtained for&nbsp; $\varepsilon_{\rm S} =10^{-2}$, considering only the term&nbsp; $f = t + 1$&nbsp;?
+{What result is obtained for&nbsp; $\varepsilon_{\rm S} =10^{-2}$,&nbsp; considering only the term&nbsp; $f = t + 1$&nbsp;?
 |type="{}"}
-$\rm Näherung \text{:} \hspace{0.2cm} Pr(Block\:error) \ \approx \ ${ 3.362 3% } $\ \cdot 10^{-5}$
+$\rm Approximation \text{:} \hspace{0.2cm} Pr(block\:error) \ \approx \ ${ 3.362 3% } $\ \cdot 10^{-5}$
 {What result is obtained approximately for&nbsp; $\varepsilon_{\rm S} = 10^{-3}$?
 |type="{}"}
-$\rm Pr(Block\:error) \ \approx \ ${ 3.49 3% } $\ \cdot 10^{-8}$
+$\rm Pr(block\:error) \ \approx \ ${ 3.49 3% } $\ \cdot 10^{-8}$
 {What&nbsp; $\varepsilon_{\rm S}$&nbsp; is needed for the block error probability&nbsp; $10^{-10}$?

@@ Line 84: / Line 84: @@
 {7 \choose 5} \cdot 0.01^5 \cdot 0.99^2+  {7 \choose 6} \cdot 0.01^6 \cdot 0.99+
 {7 \choose 7} \cdot 0.01^7 $$
-:$$\Rightarrow \hspace{0.3cm}{\rm Pr(Blockfehler)}= 10^{-6} \cdot
+:$$\Rightarrow \hspace{0.3cm}{\rm Pr(Block\:error)}= 10^{-6} \cdot
 \big [ 33.6209 +  0.3396 + 0.0021 + ... \big ] \hspace{0.15cm} \underline{ \approx  3.396 \cdot 10^{-5}}
 \hspace{0.05cm}.$$
@@ Line 110: / Line 110: @@
 '''(5)'''&nbsp; According to the derived approximation, the following holds for the considered code:
 :$${\rm Pr(Block\:error)}  \approx {7 \choose 3} \cdot {\varepsilon_{\rm S}}^3 = 35 \cdot {\varepsilon_{\rm S}}^3\hspace{0.3cm}
-\Rightarrow  \hspace{0.3cm} {\rm Pr(Blockfehler)} = 10^{-10}: \hspace{0.4cm} {\varepsilon_{\rm S}} =
+\Rightarrow  \hspace{0.3cm} {\rm Pr(Block\:error)} = 10^{-10}: \hspace{0.4cm} {\varepsilon_{\rm S}} =
 \big ( \frac{10^{-10}}{35} \big )^{1/3} = 2.857^{1/3} \cdot 10^{-4}
 \hspace{0.15cm} \underline{ \approx  1.42 \cdot 10^{-4}}\hspace{0.05cm}.$$

← Older revision		Revision as of 15:57, 4 September 2022
Line 118:		Line 118:


−	[[Category:Channel Coding: Exercises\|^2.6 ~~RSC–Fehlerwahrscheinlichkeit~~^]]	+	[[Category:Channel Coding: Exercises\|^2.6 Block Error Probability of RS Codes^]]

← Older revision		Revision as of 12:52, 23 March 2021
Line 118:		Line 118:


−	[[Category:~~Aufgaben zu Kanalcodierung~~\|^2.6 RSC–Fehlerwahrscheinlichkeit^]]	+	[[Category:Channel Coding: Exercises\|^2.6 RSC–Fehlerwahrscheinlichkeit^]]