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

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:55:07Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:55, 16 March 2026
Line 2:		Line 2:

	[[File:P_ID2574__KC_Z_2_15.png\|right\|frame\|Binominal probabilities]]		[[File:P_ID2574__KC_Z_2_15.png\|right\|frame\|Binominal probabilities]]
	Using a Reed–Solomon code with the correctability  $t$  and  [[Channel_Coding/~~Error_Probability_and_Areas_of_Application~~#Block_error_probability_for_RSC_and_BDD\|"Bounded Distance Decoding"]]  $\rm $  $\rm (BDD)$  is obtained for the block error probability with		Using a Reed–Solomon code with the correctability  $t$  and  [[Channel_Coding/Error_Probability_and_Application_Areas#Block_error_probability_for_RSC_and_BDD\|"Bounded Distance Decoding"]]  $\rm $  $\rm (BDD)$  is obtained for the block error probability with
	* the code word length  $n$  and		* the code word length  $n$  and

Line 98:		Line 98:

	[[Category:Channel Coding: Exercises\|^2.6 Block Error Probability of RS Codes^]]		[[Category:Channel Coding: Exercises\|^2.6 Block Error Probability of RS Codes^]]
	[[de:Aufgaben:2.~~15Z_Nochmals_Pr(υ_≠_u)_für_BDD~~]]		[[de:Aufgaben:Aufgabe 2.15Z: Nochmals RS-Blockfehlerwahrscheinlichkeit]]

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:47:41Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:47, 16 March 2026
Line 7:		Line 7:
	* the symbol falsification probability  $\varepsilon_{\rm S}$:		* the symbol falsification probability  $\varepsilon_{\rm S}$:

	:$${\rm Pr(block\:error)} =		:$${\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}.$$
	\sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$

	In this exercise,  the block error probability for the  $\rm RSC \, (7, \, 3, \, 5)_8$  and various  $\varepsilon_{\rm S}$  values shall be calculated and approximated.		In this exercise,  the block error probability for the  $\rm RSC \, (7, \, 3, \, 5)_8$  and various  $\varepsilon_{\rm S}$  values shall be calculated and approximated.
Line 55:		Line 54:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  For the  $\rm RSC \, (7, \, 3, \, 5)_8$,  because  $d_{\rm min} = 5 \ \Rightarrow \ t = 2$  gives the block error probability:		'''(1)'''  For the  $\rm RSC \, (7, \, 3, \, 5)_8$,  because  $d_{\rm min} = 5 \ \Rightarrow \ t = 2$  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} $$
	\sum_{f = 3}^{7} {7 \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{7-f} $$		:$$\Rightarrow \hspace{0.3cm}{\rm Pr(block\:error)} ={7 \choose 3} \cdot 0.1^3 \cdot 0.9^4 + {7 \choose 4} \cdot 0.1^4 \cdot 0.9^3+ {7 \choose 5} \cdot 0.1^5 \cdot 0.9^2+{7 \choose 6} \cdot 0.1^6 \cdot 0.9+ {7 \choose 7} \cdot 0.1^7\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm}{\rm Pr(block\:error)} ={7 \choose 3} \cdot 0.1^3 \cdot 0.9^4 + {7 \choose 4} \cdot 0.1^4 \cdot 0.9^3
	+ {7 \choose 5} \cdot 0.1^5 \cdot 0.9^2+
	{7 \choose 6} \cdot 0.1^6 \cdot 0.9+ {7 \choose 7} \cdot 0.1^7
	\hspace{0.05cm}.$$

	*According to this calculation,  five terms would have to be considered.  However,  since		*According to this calculation,  five terms would have to be considered.  However,  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$$
	\sum_{f = 0}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} = 1$$

	:is also valid,  the following calculation is faster:		:is also valid,  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}.$$
	+ {7 \choose 2} \cdot 0.1^2 \cdot 0.9^5 \big ] =1 - \big [ 0.4783 + 0.3720 + 0.1240 \big ] \hspace{0.15cm} \underline{= 2.57 \cdot 10^{-2}}
	\hspace{0.05cm}.$$


	'''(2)'''  Analogous to the subtask  '''(1)'''  one obtains here:		'''(2)'''  Analogous to the subtask  '''(1)'''  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}.$$
	+ 21 \cdot 0.01^2 \cdot 0.99^5 \big ] =1 - \big [ 0.9321 + 0.0659 + 0.0020 \big ] \approx 0
	\hspace{0.05cm}.$$

	*This means:   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:   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.

	*The full calculation yields here:		*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 $$
	{7 \choose 3} \cdot 0.01^3 \cdot 0.99^4 + {7 \choose 4} \cdot 0.01^4 \cdot 0.99^3 +		:$$\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}.$$
	{7 \choose 5} \cdot 0.01^5 \cdot 0.99^2+ {7 \choose 6} \cdot 0.01^6 \cdot 0.99+
	{7 \choose 7} \cdot 0.01^7 $$
	:$$\Rightarrow \hspace{0.3cm}{\rm Pr(block\:error)}= 10^{-6} \cdot
	\big [ 33.6209 + 0.3396 + 0.0021 + ... \big ] \hspace{0.15cm} \underline{ \approx 3.396 \cdot 10^{-5}}
	\hspace{0.05cm}.$$



	'''(3)'''  From the sample solution to the subtask  '''(2)'''  the result can be read directly:		'''(3)'''  From the sample solution to the subtask  '''(2)'''  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}.$$
	\hspace{0.05cm}.$$

	*The relative error is about  $-1\%$.		*The relative error is about  $-1\%$.
Line 101:		Line 85:

	'''(4)'''  Restricting to the relevant term  $(f = 3)$,  we get  $\varepsilon_{\rm S} = 0.001$:		'''(4)'''  Restricting to the relevant term  $(f = 3)$,  we get  $\varepsilon_{\rm S} = 0.001$:
	:$${\rm Pr(block\:error)} \approx		:$${\rm Pr(block\:error)} \approx{7 \choose 3} \cdot [10^{-3}]^3 \cdot 0.999^4 \hspace{0.15cm} \underline{ \approx 3.49 \cdot 10^{-8}}\hspace{0.05cm}.$$
	{7 \choose 3} \cdot [10^{-3}]^3 \cdot 0.999^4 \hspace{0.15cm} \underline{ \approx 3.49 \cdot 10^{-8}}
	\hspace{0.05cm}.$$

	*The relative error here is only about  $-0.1\%$.		*The relative error here is only about  $-0.1\%$.
Line 109:		Line 91:

	'''(5)'''  According to the derived approximation,  the following holds for the considered code:		'''(5)'''  According to the derived approximation,  the following holds for the considered code:
	:$${\rm Pr(block\:error)} \approx {7 \choose 3} \cdot {\varepsilon_{\rm S}}^3 = 35 \cdot {\varepsilon_{\rm S}}^3\hspace{0.3cm}		:$${\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}} =\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}.$$
	\Rightarrow \hspace{0.3cm} {\rm Pr(block\:error)} = 10^{-10}: \hspace{0.4cm} {\varepsilon_{\rm S}} =
	\big ( \frac{10^{-10}}{35} \big )^{1/3} = 2.857^{1/3} \cdot 10^{-4}
	\hspace{0.15cm} \underline{ \approx 1.42 \cdot 10^{-4}}\hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

Maintenance script: Add German interlanguage link

2026-03-16T12:31:02Z

Add German interlanguage link

← Older revision		Revision as of 14:31, 16 March 2026
Line 119:		Line 119:

	[[Category:Channel Coding: Exercises\|^2.6 Block Error Probability of RS Codes^]]		[[Category:Channel Coding: Exercises\|^2.6 Block Error Probability of RS Codes^]]
			[[de:Aufgaben:2.15Z_Nochmals_Pr(υ_≠_u)_für_BDD]]

Guenter at 13:37, 2 November 2022

2022-11-02T13:37:26Z

← Older revision		Revision as of 15:37, 2 November 2022
Line 2:		Line 2:

	[[File:P_ID2574__KC_Z_2_15.png\|right\|frame\|Binominal probabilities]]		[[File:P_ID2574__KC_Z_2_15.png\|right\|frame\|Binominal probabilities]]
	Using a Reed–Solomon code with the correctability  $t$  and  [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD\|"Bounded Distance Decoding"]]  $\rm $  $\rm (BDD)$  is obtained with		Using a Reed–Solomon code with the correctability  $t$  and  [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD\|"Bounded Distance Decoding"]]  $\rm $  $\rm (BDD)$  is obtained for the block error probability with
	* the code word length  $n$  and		* the code word length  $n$  and

	* the symbol falsification probability  $\varepsilon_{\rm S}$		* the symbol falsification probability  $\varepsilon_{\rm S}$:


	~~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}.$$		\sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$

Guenter at 16:39, 1 November 2022

2022-11-01T16:39:54Z

← Older revision		Revision as of 18:39, 1 November 2022
Line 56:		Line 56:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  For the $\rm RSC \, (7, \, 3, \, 5)_8$, because $d_{\rm min} = 5 \ \Rightarrow \ t = 2$ gives the block error probability:		'''(1)'''  For the  $\rm RSC \, (7, \, 3, \, 5)_8$,  because  $d_{\rm min} = 5 \ \Rightarrow \ t = 2$  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} $$		\sum_{f = 3}^{7} {7 \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{7-f} $$
	:$$\Rightarrow \hspace{0.3cm}{\rm Pr(~~Block~~\:error)} ={7 \choose 3} \cdot 0.1^3 \cdot 0.9^4 + {7 \choose 4} \cdot 0.1^4 \cdot 0.9^3		:$$\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 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		{7 \choose 6} \cdot 0.1^6 \cdot 0.9+ {7 \choose 7} \cdot 0.1^7
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*According to this calculation, five terms would have to be considered. However, since		*According to this calculation,  five terms would have to be considered.  However,  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$$		\sum_{f = 0}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} = 1$$

	is also valid, the following calculation is faster:		:is also valid,  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}}		+ {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}.$$		\hspace{0.05cm}.$$


	'''(2)'''  Analogous to the subtask '''(1)''' one obtains here:		'''(2)'''  Analogous to the subtask  '''(1)'''  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		+ 21 \cdot 0.01^2 \cdot 0.99^5 \big ] =1 - \big [ 0.9321 + 0.0659 + 0.0020 \big ] \approx 0
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*This means:   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:   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.

	*The full calculation yields here:		*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 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 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 $$		{7 \choose 7} \cdot 0.01^7 $$
	:$$\Rightarrow \hspace{0.3cm}{\rm Pr(~~Block~~\:error)}= 10^{-6} \cdot		:$$\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}}		\big [ 33.6209 + 0.3396 + 0.0021 + ... \big ] \hspace{0.15cm} \underline{ \approx 3.396 \cdot 10^{-5}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
Line 91:		Line 92:


	'''(3)'''  From the sample solution to the subtask '''(2)''' the result can be read directly:		'''(3)'''  From the sample solution to the subtask  '''(2)'''  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}.$$		\hspace{0.05cm}.$$

	*The relative error is about $-1\%$.		*The relative error is about  $-1\%$.
	*The minus sign indicates that this is only an approximation and not a bound:
	*The approximate value is slightly smaller than the actual value.

			*The minus sign indicates that this is only an approximation and not a bound:  The approximate value is slightly smaller than the actual value.


	'''(4)'''  Restricting to the relevant term $(f = 3)$, we get $\varepsilon_{\rm S} = 0.001$:
	:$${\rm Pr(~~Block~~\:error)} \approx		'''(4)'''  Restricting to the relevant term  $(f = 3)$,  we get  $\varepsilon_{\rm S} = 0.001$:
			:$${\rm Pr(block\:error)} \approx
	{7 \choose 3} \cdot [10^{-3}]^3 \cdot 0.999^4 \hspace{0.15cm} \underline{ \approx 3.49 \cdot 10^{-8}}		{7 \choose 3} \cdot [10^{-3}]^3 \cdot 0.999^4 \hspace{0.15cm} \underline{ \approx 3.49 \cdot 10^{-8}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*The relative error here is ~~also~~ only about $-0.1\%$.		*The relative error here is only about  $-0.1\%$.


	'''(5)'''  According to the derived approximation, the following holds for the considered code:		'''(5)'''  According to the derived approximation,  the following holds for the considered code:
	:$${\rm Pr(~~Block~~\:error)} \approx {7 \choose 3} \cdot {\varepsilon_{\rm S}}^3 = 35 \cdot {\varepsilon_{\rm S}}^3\hspace{0.3cm}		:$${\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}		\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}.$$		\hspace{0.15cm} \underline{ \approx 1.42 \cdot 10^{-4}}\hspace{0.05cm}.$$

Guenter at 16:29, 1 November 2022

2022-11-01T16:29:45Z

← Older revision		Revision as of 18:29, 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"]].

Guenter at 16:27, 1 November 2022

2022-11-01T16:27:34Z

← Older revision		Revision as of 18:27, 1 November 2022
Line 2:		Line 2:

	[[File:P_ID2574__KC_Z_2_15.png\|right\|frame\|Binominal probabilities]]		[[File:P_ID2574__KC_Z_2_15.png\|right\|frame\|Binominal probabilities]]
	Using a Reed–Solomon~~–~~code with the correctability  $t$  and  [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD\|"Bounded Distance Decoding"]]  (BDD) is obtained with		Using a Reed–Solomon code with the correctability  $t$  and  [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD\|"Bounded Distance Decoding"]]  $\rm $  $\rm (BDD)$  is obtained with
	* the code word length  $n$  and		* the code word length  $n$  and
	* the symbol ~~corruption~~ probability  $\varepsilon_{\rm S}$
			* the symbol falsification probability  $\varepsilon_{\rm S}$


	for the block error probability:		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}.$$		\sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$

	In this exercise, the block error probability for the  $\rm RSC \, (7, \, 3, \, 5)_8$  and various  $\varepsilon_{\rm S}$ values shall be calculated and approximated.		In this exercise,  the block error probability for the  $\rm RSC \, (7, \, 3, \, 5)_8$  and various  $\varepsilon_{\rm S}$  values shall be calculated and approximated.

	The above equation is reminiscent of the  [[Theory_of_Stochastic_Signals/Binomial_Distribution\|"Biomial Distribution"]]. The graph shows the probabilities of the binomial distribution for parameters  $n = 7$  (codeword length) and  $\varepsilon_{\rm S} = 0.25$  (symbol corruption probability).


			*The above equation is reminiscent of the  [[Theory_of_Stochastic_Signals/Binomial_Distribution\|"Biomial Distribution"]].

			*The graph shows the probabilities of the binomial distribution for parameters  $n = 7$  ("code word length") and  $\varepsilon_{\rm S} = 0.25$  ("symbol falsification probability").


Line 24:		Line 24:
	Hints:		Hints:
	* 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 applet  [[Applets:Binomial-_und_Poissonverteilung_(Applet)\|"Binomial and Poisson Distribution"]]~~ ~~.
			* For checking,  you can use the interactive HTML5/JavaScript applet  [[Applets:Binomial-_und_Poissonverteilung_(Applet)\|"Binomial and Poisson Distribution"]].


Line 34:		Line 35:
	{What is the block error probability for  $\varepsilon_{\rm S} = 10^{-1}$?		{What is the block error probability for  $\varepsilon_{\rm S} = 10^{-1}$?
	\|type="{}"}		\|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  $\varepsilon_{\rm S} =10^{-2}$?		{What is the block error probability for  $\varepsilon_{\rm S} =10^{-2}$?
	\|type="{}"}		\|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  $\varepsilon_{\rm S} =10^{-2}$, considering only the term  $f = t + 1$ ?		{What result is obtained for  $\varepsilon_{\rm S} =10^{-2}$,  considering only the term  $f = t + 1$ ?
	\|type="{}"}		\|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  $\varepsilon_{\rm S} = 10^{-3}$?		{What result is obtained approximately for  $\varepsilon_{\rm S} = 10^{-3}$?
	\|type="{}"}		\|type="{}"}
	$\rm Pr(~~Block~~\:error) \ \approx \ ${ 3.49 3% } $\ \cdot 10^{-8}$		$\rm Pr(block\:error) \ \approx \ ${ 3.49 3% } $\ \cdot 10^{-8}$

	{What  $\varepsilon_{\rm S}$  is needed for the block error probability  $10^{-10}$?		{What  $\varepsilon_{\rm S}$  is needed for the block error probability  $10^{-10}$?

Noah at 20:38, 14 September 2022

2022-09-14T20:38:28Z

← Older revision		Revision as of 22:38, 14 September 2022
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 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 $$		{7 \choose 7} \cdot 0.01^7 $$
	:$$\Rightarrow \hspace{0.3cm}{\rm Pr(~~Blockfehler~~)}= 10^{-6} \cdot		:$$\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}}		\big [ 33.6209 + 0.3396 + 0.0021 + ... \big ] \hspace{0.15cm} \underline{ \approx 3.396 \cdot 10^{-5}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
Line 110:		Line 110:
	'''(5)'''  According to the derived approximation, the following holds for the considered code:		'''(5)'''  According to the derived approximation, the following holds for the considered code:
	:$${\rm Pr(Block\:error)} \approx {7 \choose 3} \cdot {\varepsilon_{\rm S}}^3 = 35 \cdot {\varepsilon_{\rm S}}^3\hspace{0.3cm}		:$${\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}		\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}.$$		\hspace{0.15cm} \underline{ \approx 1.42 \cdot 10^{-4}}\hspace{0.05cm}.$$

Noah at 20:36, 14 September 2022

2022-09-14T20:36:47Z

← Older revision		Revision as of 22:36, 14 September 2022
Line 24:		Line 24:
	Hints:		Hints:
	* 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 applet  [[Applets:~~Binomial_and_Poisson_distribution_~~(Applet)\|"Binomial and Poisson Distribution"]] .		* For checking, you can use the interactive applet  [[Applets:Binomial-_und_Poissonverteilung_(Applet)\|"Binomial and Poisson Distribution"]] .

Noah at 20:36, 14 September 2022

2022-09-14T20:36:27Z

Show changes