Aufgaben:Exercise 2.15: Block Error Probability with AWGN - Revision history

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	the calculation of the block error probability in  [[Channel_Coding/~~Error_Probability_and_Areas_of_Application~~#Block_error_probability_for_RSC_and_BDD\|"Bounded Distance Decoding"]]  $\rm (BDD)$  shall be shown.  The corresponding equation is:		the calculation of the block error probability in  [[Channel_Coding/Error_Probability_and_Application_Areas#Block_error_probability_for_RSC_and_BDD\|"Bounded Distance Decoding"]]  $\rm (BDD)$  shall be shown.  The corresponding equation is:
	:$${\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) =\sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$		:$${\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) =\sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$

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	*But since in the considered code the symbols come from  $\rm GF(2^3)$,  the BSC model with parameter  $\varepsilon$  must also still be adapted to the task.		*But since in the considered code the symbols come from  $\rm GF(2^3)$,  the BSC model with parameter  $\varepsilon$  must also still be adapted to the task.

	*For the falsification probability of the  [[Channel_Coding/~~Error_Probability_and_Areas_of_Application~~#Block_error_probability_for_RSC_and_BDD\|"''m''– BSC"  model]]  applies:		*For the falsification probability of the  [[Channel_Coding/Error_Probability_and_Application_Areas#Block_error_probability_for_RSC_and_BDD\|"''m''– BSC"  model]]  applies:
	:$$\varepsilon_{\rm S} = 1 - (1 - \varepsilon)^m\hspace{0.05cm}.$$		:$$\varepsilon_{\rm S} = 1 - (1 - \varepsilon)^m\hspace{0.05cm}.$$
	:Here it is to be set  $m = 3$  $($three bits per code symbol$)$.		:Here it is to be set  $m = 3$  $($three bits per code symbol$)$.
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	[[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.~~15_Pr(υ_≠_u)_vs._EB/N0~~]]		[[de:Aufgaben:Aufgabe 2.15: RS-Blockfehlerwahrscheinlichkeit bei AWGN]]

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	the calculation of the block error probability in  [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD\|"Bounded Distance Decoding"]]  $\rm (BDD)$  shall be shown.  The corresponding equation is:		the calculation of the block error probability in  [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD\|"Bounded Distance Decoding"]]  $\rm (BDD)$  shall be shown.  The corresponding equation is:
	:$${\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) =		:$${\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) =\sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$
	\sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$

	⇒   The calculation is performed for the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|"AWGN channel"		⇒   The calculation is performed for the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|"AWGN channel"
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	*For the falsification probability of the  [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD\|"''m''– BSC"  model]]  applies:		*For the falsification probability of the  [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD\|"''m''– BSC"  model]]  applies:
	:$$\varepsilon_{\rm S} = 1 - (1 - \varepsilon)^m		:$$\varepsilon_{\rm S} = 1 - (1 - \varepsilon)^m\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	:Here it is to be set  $m = 3$  $($three bits per code symbol$)$.		:Here it is to be set  $m = 3$  $($three bits per code symbol$)$.

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	⇒   For some  $E_{\rm B}/N_0$ values the results are entered in the table above.  The two rows with yellow background are briefly explained here:		⇒   For some  $E_{\rm B}/N_0$ values the results are entered in the table above.  The two rows with yellow background are briefly explained here:
	* For  $10 \cdot \lg {E_{\rm B}/N_0} = 4 \ \rm dB$  we get  $\varepsilon \approx {\rm Q}(1.47) \approx 0.071$  and  $\varepsilon_{\rm S} \approx 0.2$.   The block error probability here can most easily be calculated using the complement:		* For  $10 \cdot \lg {E_{\rm B}/N_0} = 4 \ \rm dB$  we get  $\varepsilon \approx {\rm Q}(1.47) \approx 0.071$  and  $\varepsilon_{\rm S} \approx 0.2$.   The block error probability here can most easily be calculated using the complement:
	:$${\rm Pr(block\:error)} = 1 - \left [ {7 \choose 0} \cdot 0.8^7 + {7 \choose 1} \cdot 0.2 \cdot 0.8^6 + {7 \choose 2} \cdot 0.2^2 \cdot 0.8^5\right ]		:$${\rm Pr(block\:error)} = 1 - \left [ {7 \choose 0} \cdot 0.8^7 + {7 \choose 1} \cdot 0.2 \cdot 0.8^6 + {7 \choose 2} \cdot 0.2^2 \cdot 0.8^5\right ]\approx 0.148 \hspace{0.05cm}.$$
	\approx 0.148 \hspace{0.05cm}.$$

	* For  $10 \cdot \lg {E_{\rm B}/N_0} = 12 \ \rm dB$  one gets  $\varepsilon \approx 1.2 \cdot 10^{-4}$  and  $\varepsilon_{\rm S} \approx 3.5 \cdot 10^{-4}$.  With this very small falsification probability,  the  $f = 3$  term dominates,  and we obtain:		* For  $10 \cdot \lg {E_{\rm B}/N_0} = 12 \ \rm dB$  one gets  $\varepsilon \approx 1.2 \cdot 10^{-4}$  and  $\varepsilon_{\rm S} \approx 3.5 \cdot 10^{-4}$.  With this very small falsification probability,  the  $f = 3$  term dominates,  and we obtain:
	:$${\rm Pr(block\:error)} \approx {7 \choose 3} \cdot (3.5 \cdot 10^{-4})^3 \cdot (1- 3.5 \cdot 10^{-4})^4		:$${\rm Pr(block\:error)} \approx {7 \choose 3} \cdot (3.5 \cdot 10^{-4})^3 \cdot (1- 3.5 \cdot 10^{-4})^4\approx 1.63 \cdot 10^{-9} \hspace{0.05cm}.$$
	\approx 1.63 \cdot 10^{-9} \hspace{0.05cm}.$$

	⇒   You are to calculate the block error probabilities for the rows highlighted in red   $(10 \cdot \lg {E_{\rm B}/N_0} = 5 \ \rm dB, \ 8 \rm dB$,  $10 \ \rm dB)$.		⇒   You are to calculate the block error probabilities for the rows highlighted in red   $(10 \cdot \lg {E_{\rm B}/N_0} = 5 \ \rm dB, \ 8 \rm dB$,  $10 \ \rm dB)$.
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	'''(1)'''  From the table on the information page,  the BSC parameter  $\varepsilon = 0.0505$  can be read.		'''(1)'''  From the table on the information page,  the BSC parameter  $\varepsilon = 0.0505$  can be read.
	*This gives  $\varepsilon_{\rm S}$  for the symbol error probability with  $m = 3$:		*This gives  $\varepsilon_{\rm S}$  for the symbol error probability with  $m = 3$:
	:$$1 - \varepsilon_{\rm S} = (1 - 0.0505)^3 \approx 0.856		:$$1 - \varepsilon_{\rm S} = (1 - 0.0505)^3 \approx 0.856\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\varepsilon_{\rm S} \approx 0.144\hspace{0.05cm}.$$
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	\varepsilon_{\rm S} \approx 0.144
	\hspace{0.05cm}.$$

	*The fastest way to calculate the block error probability is here to use the formula		*The fastest way to calculate the block error probability is here to use the formula
	:$${\rm Pr(block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 - {\rm Pr}(f=0) - {\rm Pr}(f=1) - {\rm Pr}(f=2) = 1 - 1 \cdot 0.856^7 -		:$${\rm Pr(block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 - {\rm Pr}(f=0) - {\rm Pr}(f=1) - {\rm Pr}(f=2) = 1 - 1 \cdot 0.856^7 -7 \cdot 0.144^1 \cdot 0.856^6 - 21 \cdot 0.144^2 \cdot 0.856^5$$
	7 \cdot 0.144^1 \cdot 0.856^6 - 21 \cdot 0.144^2 \cdot 0.856^5$$		:$$\Rightarrow \hspace{0.3cm} {\rm Pr(block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Pr}(\underline{v} \ne \underline{u}) =1 - 0.3368 - 0.3965 - 0.2001 \hspace{0.15cm} \underline{=0.0666}\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm} {\rm Pr(block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Pr}(\underline{v} \ne \underline{u}) =1 - 0.3368 - 0.3965 - 0.2001 \hspace{0.15cm} \underline{=0.0666}
	\hspace{0.05cm}.$$


	'''(2)'''  Following the same calculation procedure as in subtask  '''(1)''',  we obtain with  $\varepsilon_{\rm S} \approx 0.03 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.97$:		'''(2)'''  Following the same calculation procedure as in subtask  '''(1)''',  we obtain with  $\varepsilon_{\rm S} \approx 0.03 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.97$:
	:$${\rm Pr(block\:error)}		:$${\rm Pr(block\:error)}\hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 \hspace{-0.05cm}-\hspace{-0.05cm} 1 \cdot 0.97^7 \hspace{-0.05cm}-\hspace{-0.05cm}7 \cdot 0.03^1 \cdot 0.97^6 \hspace{-0.05cm}-\hspace{-0.05cm} 21 \cdot 0.03^2 \cdot 0.97^5 =1 \hspace{-0.05cm}-\hspace{-0.05cm} 0.8080 \hspace{-0.05cm}-\hspace{-0.05cm} 0.1749\hspace{-0.05cm}-\hspace{-0.05cm} 0.0162= 1 \hspace{-0.05cm}-\hspace{-0.05cm} 0.9991 = 9 \cdot 10^{-4}\hspace{0.05cm}.$$
	\hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 \hspace{-0.05cm}-\hspace{-0.05cm} 1 \cdot 0.97^7 \hspace{-0.05cm}-\hspace{-0.05cm}
	7 \cdot 0.03^1 \cdot 0.97^6 \hspace{-0.05cm}-\hspace{-0.05cm} 21 \cdot 0.03^2 \cdot 0.97^5 =1 \hspace{-0.05cm}-\hspace{-0.05cm} 0.8080 \hspace{-0.05cm}-\hspace{-0.05cm} 0.1749\hspace{-0.05cm}-\hspace{-0.05cm} 0.0162= 1 \hspace{-0.05cm}-\hspace{-0.05cm} 0.9991 = 9 \cdot 10^{-4}
	\hspace{0.05cm}.$$

	*You can see that here the difference between two numbers of almost the same size must be formed,  so that the result could be affected by an error.		*You can see that here the difference between two numbers of almost the same size must be formed,  so that the result could be affected by an error.

	*Therefore we still calculate the following quantities:		*Therefore we still calculate the following quantities:
	:$${\rm Pr}(f=3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}		:$${\rm Pr}(f=3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{7 \choose 3} \cdot \varepsilon_{\rm S}^3 \cdot (1 - \varepsilon_{\rm S})^4 = 35 \cdot 0.03^3 \cdot 0.97^4 = 8.366 \cdot 10^{-4}\hspace{0.05cm},$$
	{7 \choose 3} \cdot \varepsilon_{\rm S}^3 \cdot (1 - \varepsilon_{\rm S})^4 = 35 \cdot 0.03^3 \cdot 0.97^4 = 8.366 \cdot 10^{-4}\hspace{0.05cm},$$		:$${\rm Pr}(f=4) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{7 \choose 4} \cdot \varepsilon_{\rm S}^4 \cdot (1 - \varepsilon_{\rm S})^3 = 35 \cdot 0.03^4 \cdot 0.97^3 = 0.259 \cdot 10^{-4}\hspace{0.05cm},$$
	:$${\rm Pr}(f=4) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}		:$${\rm Pr}(f=5) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{7 \choose 5} \cdot \varepsilon_{\rm S}^5 \cdot (1 - \varepsilon_{\rm S})^2 = 21 \cdot 0.03^5 \cdot 0.97^2 = 0.005 \cdot 10^{-4}$$
	{7 \choose 4} \cdot \varepsilon_{\rm S}^4 \cdot (1 - \varepsilon_{\rm S})^3 = 35 \cdot 0.03^4 \cdot 0.97^3 = 0.259 \cdot 10^{-4}\hspace{0.05cm},$$
	:$${\rm Pr}(f=5) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}
	{7 \choose 5} \cdot \varepsilon_{\rm S}^5 \cdot (1 - \varepsilon_{\rm S})^2 = 21 \cdot 0.03^5 \cdot 0.97^2 = 0.005 \cdot 10^{-4}$$
	:$$\Rightarrow \hspace{0.3cm} {\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) \approx {\rm Pr}(f=3) + {\rm Pr}(f=4) + {\rm Pr}(f=5) \hspace{0.15cm} \underline{=8.63 \cdot 10^{-4}} \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} {\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) \approx {\rm Pr}(f=3) + {\rm Pr}(f=4) + {\rm Pr}(f=5) \hspace{0.15cm} \underline{=8.63 \cdot 10^{-4}} \hspace{0.05cm}.$$

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	'''(3)'''  Here  $\varepsilon_{\rm S} = 0.005 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.995$  is already given in the table.		'''(3)'''  Here  $\varepsilon_{\rm S} = 0.005 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.995$  is already given in the table.
	*The  (by far)  dominant term in the calculation of the block error probability is  ${\rm Pr}(f = 3)$:		*The  (by far)  dominant term in the calculation of the block error probability is  ${\rm Pr}(f = 3)$:
	:$${\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) \approx {\rm Pr}(f=3) = {7 \choose 3} \cdot 0.005^3 \cdot 0.995^4		:$${\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) \approx {\rm Pr}(f=3) = {7 \choose 3} \cdot 0.005^3 \cdot 0.995^4\hspace{0.15cm} \underline{\approx 4.3 \cdot 10^{-6}} \hspace{0.05cm}.$$
	\hspace{0.15cm} \underline{\approx 4.3 \cdot 10^{-6}} \hspace{0.05cm}.$$


	'''(4)'''  For the BSC parameter  $\varepsilon$  holds with  $\varepsilon_{\rm S} = 0.1$:		'''(4)'''  For the BSC parameter  $\varepsilon$  holds with  $\varepsilon_{\rm S} = 0.1$:
	:$$\varepsilon = 1 -(1 - \varepsilon_{\rm S})^{1/3} = 1 - 0.9^{1/3} \approx 0.0345		:$$\varepsilon = 1 -(1 - \varepsilon_{\rm S})^{1/3} = 1 - 0.9^{1/3} \approx 0.0345\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*The relation between  $\varepsilon$  and  $E_{\rm B}/N_0$  is:		*The relation between  $\varepsilon$  and  $E_{\rm B}/N_0$  is:
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	*The inverse  $x = {\rm Q}^{-1}(0.0345)$  is obtained with the applet  [[Applets:QFunction\|"Complementary Gaussian Error Functions"]]  to  $x = 1.82$.  This further gives:		*The inverse  $x = {\rm Q}^{-1}(0.0345)$  is obtained with the applet  [[Applets:QFunction\|"Complementary Gaussian Error Functions"]]  to  $x = 1.82$.  This further gives:
	:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{1.82^2}{2R \cdot 3/7} \approx 3.864		:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{1.82^2}{2R \cdot 3/7} \approx 3.864\hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.15cm}(E_{\rm B}/N_0)\hspace{0.15cm} \underline{\approx 5.87 \,\, {\rm dB}} \hspace{0.05cm}. $$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	10 \cdot {\rm lg}\hspace{0.15cm}(E_{\rm B}/N_0)
	\hspace{0.15cm} \underline{\approx 5.87 \,\, {\rm dB}} \hspace{0.05cm}. $$


	'''(5)'''  After the same calculation one obtains		'''(5)'''  After the same calculation one obtains
	* for  $\varepsilon_{\rm S} = 10^{-2} \ \Rightarrow \ \varepsilon \approx 0.33 \cdot 10^{-2} \ \Rightarrow \ x = {\rm Q}^{-1}(\varepsilon) = 2.71$		* for  $\varepsilon_{\rm S} = 10^{-2} \ \Rightarrow \ \varepsilon \approx 0.33 \cdot 10^{-2} \ \Rightarrow \ x = {\rm Q}^{-1}(\varepsilon) = 2.71$
	:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{2.71^2}{2R \cdot 3/7} \approx 8.568		:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{2.71^2}{2R \cdot 3/7} \approx 8.568\hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.15cm}(E_{\rm B}/N_0)\hspace{0.15cm} \underline{\approx 9.32 \,\, {\rm dB}} \hspace{0.05cm}, $$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	10 \cdot {\rm lg}\hspace{0.15cm}(E_{\rm B}/N_0)
	\hspace{0.15cm} \underline{\approx 9.32 \,\, {\rm dB}} \hspace{0.05cm}, $$

	* for $\varepsilon_{\rm S} = 10^{-3} \ \Rightarrow \ \varepsilon \approx 0.33 \cdot 10^{-3} \ \Rightarrow \ x = {\rm Q}^{-1}(\varepsilon) = 3.4$:		* for $\varepsilon_{\rm S} = 10^{-3} \ \Rightarrow \ \varepsilon \approx 0.33 \cdot 10^{-3} \ \Rightarrow \ x = {\rm Q}^{-1}(\varepsilon) = 3.4$:
	:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{3.4^2}{2R \cdot 3/7} \approx 13.487		:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{3.4^2}{2R \cdot 3/7} \approx 13.487\hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.15cm}(E_{\rm B}/N_0)\hspace{0.15cm} \underline{\approx 11.3 \,\, {\rm dB}} \hspace{0.05cm}. $$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	10 \cdot {\rm lg}\hspace{0.15cm}(E_{\rm B}/N_0)
	\hspace{0.15cm} \underline{\approx 11.3 \,\, {\rm dB}} \hspace{0.05cm}. $$

	[[File:EN_KC_A_2_15e.ng.png\|right\|frame\|Results for $\rm RSC \, (7, \, 3, \, 5)_8$ decoding]]		[[File:EN_KC_A_2_15e.ng.png\|right\|frame\|Results for $\rm RSC \, (7, \, 3, \, 5)_8$ decoding]]

Maintenance script: Add German interlanguage link

2026-03-16T12:31:03Z

Add German interlanguage link

← Older revision		Revision as of 14:31, 16 March 2026
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	[[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:2.15_Pr(υ_≠_u)_vs._EB/N0]]

Guenter at 16:12, 13 March 2023

2023-03-13T16:12:30Z

← Older revision		Revision as of 18:12, 13 March 2023
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	\hspace{0.15cm} \underline{\approx 11.3 \,\, {\rm dB}} \hspace{0.05cm}. $$		\hspace{0.15cm} \underline{\approx 11.3 \,\, {\rm dB}} \hspace{0.05cm}. $$

	[[File:EN_KC_A_2_15e.ng.png\|right\|frame\|Results for $\rm RSC \, (7, \, 3, \, 5)_8$ decoding ~~'''Korrektur''' ohne Codierung~~]]		[[File:EN_KC_A_2_15e.ng.png\|right\|frame\|Results for $\rm RSC \, (7, \, 3, \, 5)_8$ decoding]]

	The graph shows		The graph shows

Guenter at 14:47, 8 November 2022

2022-11-08T14:47:09Z

← Older revision		Revision as of 16:47, 8 November 2022
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	\hspace{0.15cm} \underline{\approx 11.3 \,\, {\rm dB}} \hspace{0.05cm}. $$		\hspace{0.15cm} \underline{\approx 11.3 \,\, {\rm dB}} \hspace{0.05cm}. $$

	[[File:~~P_ID2572__KC_A_2_15e_neu~~.png\|right\|frame\|Results for $\rm RSC \, (7, \, 3, \, 5)_8$ decoding '''Korrektur''' ohne Codierung]]		[[File:EN_KC_A_2_15e.ng.png\|right\|frame\|Results for $\rm RSC \, (7, \, 3, \, 5)_8$ decoding '''Korrektur''' ohne Codierung]]

	The graph shows		The graph shows

Guenter at 15:45, 1 November 2022

2022-11-01T15:45:54Z

← Older revision		Revision as of 17:45, 1 November 2022
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From the table on the information page, the BSC parameter $\varepsilon = 0.0505$ can be read.		'''(1)'''  From the table on the information page,  the BSC parameter  $\varepsilon = 0.0505$  can be read.
	*This gives $\varepsilon_{\rm S}$ for the symbol ~~corruption~~ probability with $m = 3$:		*This gives  $\varepsilon_{\rm S}$  for the symbol error probability with  $m = 3$:
	:$$1 - \varepsilon_{\rm S} = (1 - 0.0505)^3 \approx 0.856		:$$1 - \varepsilon_{\rm S} = (1 - 0.0505)^3 \approx 0.856
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}		\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*The fastest way to calculate the block error probability here is to use the formula		*The fastest way to calculate the block error probability is here to use the formula
	:$${\rm Pr(~~Block~~\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 - {\rm Pr}(f=0) - {\rm Pr}(f=1) - {\rm Pr}(f=2) = 1 - 1 \cdot 0.856^7 -		:$${\rm Pr(block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 - {\rm Pr}(f=0) - {\rm Pr}(f=1) - {\rm Pr}(f=2) = 1 - 1 \cdot 0.856^7 -
	7 \cdot 0.144^1 \cdot 0.856^6 - 21 \cdot 0.144^2 \cdot 0.856^5$$		7 \cdot 0.144^1 \cdot 0.856^6 - 21 \cdot 0.144^2 \cdot 0.856^5$$
	:$$\Rightarrow \hspace{0.3cm} {\rm Pr(~~Block~~\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Pr}(\underline{v} \ne \underline{u}) =1 - 0.3368 - 0.3965 - 0.2001 \hspace{0.15cm} \underline{=0.0666}		:$$\Rightarrow \hspace{0.3cm} {\rm Pr(block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Pr}(\underline{v} \ne \underline{u}) =1 - 0.3368 - 0.3965 - 0.2001 \hspace{0.15cm} \underline{=0.0666}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$


	'''(2)'''  Following the same calculation procedure as in subtask '''(1)''', ~~the following is obtained~~ with $\varepsilon_{\rm S} \approx 0.03 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.97$:		'''(2)'''  Following the same calculation procedure as in subtask  '''(1)''',  we obtain with  $\varepsilon_{\rm S} \approx 0.03 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.97$:
	:$${\rm Pr(~~Block~~\:error)}		:$${\rm Pr(block\:error)}
	\hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 \hspace{-0.05cm}-\hspace{-0.05cm} 1 \cdot 0.97^7 \hspace{-0.05cm}-\hspace{-0.05cm}		\hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 \hspace{-0.05cm}-\hspace{-0.05cm} 1 \cdot 0.97^7 \hspace{-0.05cm}-\hspace{-0.05cm}
	7 \cdot 0.03^1 \cdot 0.97^6 \hspace{-0.05cm}-\hspace{-0.05cm} 21 \cdot 0.03^2 \cdot 0.97^5 =1 \hspace{-0.05cm}-\hspace{-0.05cm} 0.8080 \hspace{-0.05cm}-\hspace{-0.05cm} 0.1749\hspace{-0.05cm}-\hspace{-0.05cm} 0.0162= 1 \hspace{-0.05cm}-\hspace{-0.05cm} 0.9991 = 9 \cdot 10^{-4}		7 \cdot 0.03^1 \cdot 0.97^6 \hspace{-0.05cm}-\hspace{-0.05cm} 21 \cdot 0.03^2 \cdot 0.97^5 =1 \hspace{-0.05cm}-\hspace{-0.05cm} 0.8080 \hspace{-0.05cm}-\hspace{-0.05cm} 0.1749\hspace{-0.05cm}-\hspace{-0.05cm} 0.0162= 1 \hspace{-0.05cm}-\hspace{-0.05cm} 0.9991 = 9 \cdot 10^{-4}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*You can see that here the difference between two numbers of almost the same size must be formed, so that the result could be affected by an error.		*You can see that here the difference between two numbers of almost the same size must be formed,  so that the result could be affected by an error.

	*Therefore we still calculate the following quantities:		*Therefore we still calculate the following quantities:
	:$${\rm Pr}(f=3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}		:$${\rm Pr}(f=3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}
Line 111:		Line 112:
	:$${\rm Pr}(f=5) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}		:$${\rm Pr}(f=5) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}
	{7 \choose 5} \cdot \varepsilon_{\rm S}^5 \cdot (1 - \varepsilon_{\rm S})^2 = 21 \cdot 0.03^5 \cdot 0.97^2 = 0.005 \cdot 10^{-4}$$		{7 \choose 5} \cdot \varepsilon_{\rm S}^5 \cdot (1 - \varepsilon_{\rm S})^2 = 21 \cdot 0.03^5 \cdot 0.97^2 = 0.005 \cdot 10^{-4}$$
	:$$\Rightarrow \hspace{0.3cm} {\rm Pr(~~Blockfehler~~)} = {\rm Pr}(\underline{v} \ne \underline{u}) \approx {\rm Pr}(f=3) + {\rm Pr}(f=4) + {\rm Pr}(f=5) \hspace{0.15cm} \underline{=8.63 \cdot 10^{-4}} \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} {\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) \approx {\rm Pr}(f=3) + {\rm Pr}(f=4) + {\rm Pr}(f=5) \hspace{0.15cm} \underline{=8.63 \cdot 10^{-4}} \hspace{0.05cm}.$$

	*The terms for $f = 6$ and $f = 7$ can be omitted here. They do not provide a relevant contribution.		*The terms for  $f = 6$  and  $f = 7$  can be omitted here.  They do not provide a relevant contribution.




	'''(3)'''  Here $\varepsilon_{\rm S} = 0.005 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.995$ is already given in the table.		'''(3)'''  Here  $\varepsilon_{\rm S} = 0.005 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.995$  is already given in the table.
	*The (by far) dominant term in the calculation of the block error probability is ${\rm Pr}(f = 3)$:.		*The  (by far)  dominant term in the calculation of the block error probability is  ${\rm Pr}(f = 3)$:
	:$${\rm Pr(~~Blockfehler~~)} = {\rm Pr}(\underline{v} \ne \underline{u}) \approx {\rm Pr}(f=3) = {7 \choose 3} \cdot 0.005^3 \cdot 0.995^4		:$${\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) \approx {\rm Pr}(f=3) = {7 \choose 3} \cdot 0.005^3 \cdot 0.995^4
	\hspace{0.15cm} \underline{\approx 4.3 \cdot 10^{-6}} \hspace{0.05cm}.$$		\hspace{0.15cm} \underline{\approx 4.3 \cdot 10^{-6}} \hspace{0.05cm}.$$


	'''(4)'''  For the BSC parameter $\varepsilon$ holds with $\varepsilon_{\rm S} = 0.1$:		'''(4)'''  For the BSC parameter  $\varepsilon$  holds with  $\varepsilon_{\rm S} = 0.1$:
	:$$\varepsilon = 1 -(1 - \varepsilon_{\rm S})^{1/3} = 1 - 0.9^{1/3} \approx 0.0345		:$$\varepsilon = 1 -(1 - \varepsilon_{\rm S})^{1/3} = 1 - 0.9^{1/3} \approx 0.0345
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*The relation between $\varepsilon$ and $E_{\rm B}/N_0$ is:		*The relation between  $\varepsilon$  and  $E_{\rm B}/N_0$  is:
	:$$\varepsilon = {\rm Q}(x)\hspace{0.05cm}, \hspace{0.5cm} x = \sqrt{2 \cdot R \cdot E_{\rm B}/N_0}\hspace{0.05cm}.$$		:$$\varepsilon = {\rm Q}(x)\hspace{0.05cm}, \hspace{0.5cm} x = \sqrt{2 \cdot R \cdot E_{\rm B}/N_0}\hspace{0.05cm}.$$

	*The inverse $x = {\rm Q}^{-1}(0.0345)$ is obtained with the applet [[Applets:QFunction\|"Complementary Gaussian Error Functions"]] to $x = 1.82$.		*The inverse  $x = {\rm Q}^{-1}(0.0345)$  is obtained with the applet  [[Applets:QFunction\|"Complementary Gaussian Error Functions"]]  to  $x = 1.82$.  This further gives:
	*This further gives:
	:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{1.82^2}{2R \cdot 3/7} \approx 3.864		:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{1.82^2}{2R \cdot 3/7} \approx 3.864
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}		\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
Line 140:		Line 140:

	'''(5)'''  After the same calculation one obtains		'''(5)'''  After the same calculation one obtains
	* ~~für~~ $\varepsilon_{\rm S} = 10^{-2} \ \Rightarrow \ \varepsilon \approx 0.33 \cdot 10^{-2} \ \Rightarrow \ x = {\rm Q}^{-1}(\varepsilon) = 2.71$		* for  $\varepsilon_{\rm S} = 10^{-2} \ \Rightarrow \ \varepsilon \approx 0.33 \cdot 10^{-2} \ \Rightarrow \ x = {\rm Q}^{-1}(\varepsilon) = 2.71$
	:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{2.71^2}{2R \cdot 3/7} \approx 8.568		:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{2.71^2}{2R \cdot 3/7} \approx 8.568
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}		\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
Line 152:		Line 152:
	\hspace{0.15cm} \underline{\approx 11.3 \,\, {\rm dB}} \hspace{0.05cm}. $$		\hspace{0.15cm} \underline{\approx 11.3 \,\, {\rm dB}} \hspace{0.05cm}. $$

	[[File:P_ID2572__KC_A_2_15e_neu.png\|right\|frame\|Results for $rm RSC \, (7, \, 3, \, 5)_8$ decoding]]		[[File:P_ID2572__KC_A_2_15e_neu.png\|right\|frame\|Results for $\rm RSC \, (7, \, 3, \, 5)_8$ decoding '''Korrektur''' ohne Codierung]]
	~~<br><br><br><br><br><br><br><br><br><br>~~
	The graph shows the course of the block error probability as a function of $10 \cdot \lg {E_{\rm B}/N_0}$ ~~as well as~~ the completely filled result table.		The graph shows
			*the course of the block error probability as function of $10 \cdot \lg {E_{\rm B}/N_0}$,

			* and the completely filled result table.

	~~One can see the clearly less favorable (asymptotic) behavior of this short (green) code $\rm RSC \, (7, \, 5, \, 3)_8$ compared to the (red) comparison code $\rm RSC \, (255, \, 223, \, 33)_8$:~~

			One can see the clearly less favorable  (asymptotic)  behavior of the  (green)  code $\rm RSC \, (7, \, 5, \, 3)_8$ compared to the  (red)  comparison code $\rm RSC \, (255, \, 223, \, 33)_8$:

	*For abscissa values smaller than $10 \ \rm dB$ the result is even worse than without coding.		# For abscissa values smaller than  $10 \ \rm dB$  the result is even worse than without coding.
	*Therefore it should be pointed out again that ~~this~~ $\rm RSC \, (7, \, 3, \, 5)_8$ has little practical meaning.		# It should be pointed out again that the  $\rm RSC \, (7, \, 3, \, 5)_8$  has only little practical meaning.
	*It was chosen for this exercise only to be able to demonstrate with reasonable effort the calculation of the block error probability for "Bounded Distance Decoding" (BDD).		# It was chosen for this exercise only to be able to demonstrate with reasonable effort the calculation of the block error probability for  "Bounded Distance Decoding"  $\rm (BDD)$.
	{{ML-Fuß}}		{{ML-Fuß}}

	[[Category:Channel Coding: Exercises\|^2.6 Block Error Probability of RS Codes^]]		[[Category:Channel Coding: Exercises\|^2.6 Block Error Probability of RS Codes^]]

Guenter at 15:23, 1 November 2022

2022-11-01T15:23:38Z

← Older revision		Revision as of 17:23, 1 November 2022
Line 51:		Line 51:
	* We refer you here to the two interactive HTML5/JavaScript applets		* We refer you here to the two interactive HTML5/JavaScript applets
	:*[[Applets:Complementary_Gaussian_Error_Functions\|"Complementary Gaussian error functions"]],  and		:*[[Applets:Complementary_Gaussian_Error_Functions\|"Complementary Gaussian error functions"]],  and

	:*[[Applets:Binomial_and_Poisson_Distribution_(Applet)\|"Binomial and Poisson Distribution"]].		:*[[Applets:Binomial_and_Poisson_Distribution_(Applet)\|"Binomial and Poisson Distribution"]].

Guenter at 15:22, 1 November 2022

2022-11-01T15:22:11Z

← Older revision		Revision as of 17:22, 1 November 2022
Line 49:		Line 49:
	* The exercise belongs to the chapter  [[Channel_Coding/Error_Probability_and_Areas_of_Application\| "Error Probability and Application Areas"]].		* The exercise belongs to the chapter  [[Channel_Coding/Error_Probability_and_Areas_of_Application\| "Error Probability and Application Areas"]].

	* We refer you here to the two interactive applets		* We refer you here to the two interactive HTML5/JavaScript applets
	:*[[Applets:Complementary_Gaussian_Error_Functions\|"Complementary Gaussian error functions"]],  and		:*[[Applets:Complementary_Gaussian_Error_Functions\|"Complementary Gaussian error functions"]],  and

Guenter at 15:17, 1 November 2022

2022-11-01T15:17:20Z

← Older revision		Revision as of 17:17, 1 November 2022
Line 2:		Line 2:

	[[File:P_ID2571__KC_A_2_15neu.png\|right\|frame\|Incomplete table of results]]		[[File:P_ID2571__KC_A_2_15neu.png\|right\|frame\|Incomplete table of results]]
	Using the example of  $\rm RSC \, (7, \, 3, \, 5)_8$  with ~~the~~ parameters		Using the example of  $\rm RSC \, (7, \, 3, \, 5)_8$  with parameters
	* $n = 7$  (number of code symbols),		* $n = 7$  $($number of code symbols$)$,
	* $~~k =3$  (number of information symbols)~~,
	* $t = 2$  (correction capability).

			* $k =3$  $($number of information symbols$)$,

	the calculation of the block error probability in  [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD\|"Bounded Distance Decoding"]]  (BDD) shall be shown. The corresponding equation is:		* $t = 2$  $($correction capability$)$.
	:$${\rm Pr(~~Block~~\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) =

			the calculation of the block error probability in  [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD\|"Bounded Distance Decoding"]]  $\rm (BDD)$  shall be shown.  The corresponding equation is:
			:$${\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) =
	\sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$		\sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$

	The calculation is performed for the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|"AWGN channel"		⇒   The calculation is performed for the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|"AWGN channel"
	]] characterized by the parameter  $E_{\rm B}/N_0$~~ .~~		]]  characterized by the parameter  $E_{\rm B}/N_0$:

	*The quotient  $E_{\rm B}/{N_0}$  can be expressed by the relation		*The quotient  $E_{\rm B}/{N_0}$  can be expressed by the relation
	:$$\varepsilon = {\rm Q} \big (\sqrt{{2 \cdot R \cdot E_{\rm B}}/{N_0}} \big ) $$		:$$\varepsilon = {\rm Q} \big (\sqrt{{2 \cdot R \cdot E_{\rm B}}/{N_0}} \big ) $$
	into the  [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Symmetric_Channel_.E2.80. 93_BSC\|"BSC model"]]  where  $R$  denotes the code rate  $($here:  $R = 3/7)$  and  ${\rm Q}(x)$  indicates the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|"complementary Gaussian error integral"]]~~ ~~.		:into the  [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Symmetric_Channel_.E2.80. 93_BSC\|"BSC model"]]  where  $R$  denotes the code rate  $($here:  $R = 3/7)$  and  ${\rm Q}(x)$  indicates the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|"complementary Gaussian error integral"]].
	*But since in the considered code the symbols come from  $\rm GF(2^3)$ , the BSC model with parameter  $\varepsilon$  must also still be adapted to the task.
	*For the ~~corruption~~ probability of the  [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD\|"$m$ BSC model"]]  applies~~, where here  $m = 3$~~  ~~is to be set (three bits per code symbol):~~		*But since in the considered code the symbols come from  $\rm GF(2^3)$,  the BSC model with parameter  $\varepsilon$  must also still be adapted to the task.

			*For the falsification probability of the  [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD\|"''m''– BSC"  model]]  applies:
	:$$\varepsilon_{\rm S} = 1 - (1 - \varepsilon)^m		:$$\varepsilon_{\rm S} = 1 - (1 - \varepsilon)^m
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
			:Here it is to be set  $m = 3$  $($three bits per code symbol$)$.


	For some  $E_{\rm B}/N_0$ values the results are entered in the table above. The two rows with yellow background are briefly explained here:		⇒   For some  $E_{\rm B}/N_0$ values the results are entered in the table above.  The two rows with yellow background are briefly explained here:
	* For  $10 \cdot \lg {E_{\rm B}/N_0} = 4 \ \rm dB$  we get  $\varepsilon \approx {\rm Q}(1.47) \approx 0.071$  and  $\varepsilon_{\rm S} \approx 0.2$. The block error probability here can most easily be calculated using the complement:		* For  $10 \cdot \lg {E_{\rm B}/N_0} = 4 \ \rm dB$  we get  $\varepsilon \approx {\rm Q}(1.47) \approx 0.071$  and  $\varepsilon_{\rm S} \approx 0.2$.   The block error probability here can most easily be calculated using the complement:
	:$${\rm Pr(~~Block~~\:error)} = 1 - \left [ {7 \choose 0} \cdot 0.8^7 + {7 \choose 1} \cdot 0.2 \cdot 0.8^6 + {7 \choose 2} \cdot 0.2^2 \cdot 0.8^5\right ]		:$${\rm Pr(block\:error)} = 1 - \left [ {7 \choose 0} \cdot 0.8^7 + {7 \choose 1} \cdot 0.2 \cdot 0.8^6 + {7 \choose 2} \cdot 0.2^2 \cdot 0.8^5\right ]
	\approx 0.148 \hspace{0.05cm}.$$		\approx 0.148 \hspace{0.05cm}.$$
	* For  $10 \cdot \lg {E_{\rm B}/N_0} = 12 \ \rm dB$  one gets  $\varepsilon \approx 1.2 \cdot 10^{-4}$  and  $\varepsilon_{\rm S} \approx 3.5 \cdot 10^{-4}$. With this very small ~~corruption~~ probability, the  $f = 3$ term dominates, and we obtain:
	:$${\rm Pr(~~Block~~\:error)} \approx {7 \choose 3} \cdot (3.5 \cdot 10^{-4})^3 \cdot (1- 3.5 \cdot 10^{-4})^4		* For  $10 \cdot \lg {E_{\rm B}/N_0} = 12 \ \rm dB$  one gets  $\varepsilon \approx 1.2 \cdot 10^{-4}$  and  $\varepsilon_{\rm S} \approx 3.5 \cdot 10^{-4}$.  With this very small falsification probability,  the  $f = 3$  term dominates,  and we obtain:
			:$${\rm Pr(block\:error)} \approx {7 \choose 3} \cdot (3.5 \cdot 10^{-4})^3 \cdot (1- 3.5 \cdot 10^{-4})^4
	\approx 1.63 \cdot 10^{-9} \hspace{0.05cm}.$$		\approx 1.63 \cdot 10^{-9} \hspace{0.05cm}.$$

	*You are to calculate the block error probabilities for the rows highlighted in red  $(10 \cdot \lg {E_{\rm B}/N_0} = 5 \ \rm dB, \ 8 \rm dB$,  $10 \ \rm dB)$~~ ~~ .		⇒   You are to calculate the block error probabilities for the rows highlighted in red   $(10 \cdot \lg {E_{\rm B}/N_0} = 5 \ \rm dB, \ 8 \rm dB$,  $10 \ \rm dB)$.
	*The rows with blue background show some results of  [[Aufgaben:Exercise_2.15Z:_Block_Error_Probability_once_more\|"Exercise 2.15Z"]]. There  ${\rm Pr}(\underline{v} ≠ \underline{u})$  is calculated for  $\varepsilon_{\rm S} = 10\%,  \ 1\%$  $0.1\%$.		:*The rows with blue background show some results of  [[Aufgaben:Exercise_2.15Z:_Block_Error_Probability_once_more\|"Exercise 2.15Z"]].  There  ${\rm Pr}(\underline{v} ≠ \underline{u})$  is calculated for  $\varepsilon_{\rm S} = 10\%,  \ 1\%$  $0.1\%$.
	*In subtasks '''(4)''' and '''(5)''' you are to establish the relationship between the ~~size~~  $\varepsilon_{\rm S}$  and the AWGN parameter  $E_{\rm B}/N_0$  thus completing the above table.		:*In subtasks '''(4)''' and '''(5)''' you are to establish the relationship between the quantity  $\varepsilon_{\rm S}$  and the AWGN parameter  $E_{\rm B}/N_0$  thus completing the above table.



Line 41:		Line 46:


			<u>Hints:</u>
			* The exercise belongs to the chapter  [[Channel_Coding/Error_Probability_and_Areas_of_Application\| "Error Probability and Application Areas"]].

			* We refer you here to the two interactive applets
			:*[[Applets:Complementary_Gaussian_Error_Functions\|"Complementary Gaussian error functions"]],  and

	~~Hints~~:		:*[[Applets:Binomial_and_Poisson_Distribution_(Applet)\|"Binomial and Poisson Distribution"]].
	* ~~The exercise belongs to the chapter  [[Channel_Coding/Error_Probability_and_Areas_of_Application\| "Error Probability and Application Areas"]].~~
	* We refer you here to the two interactive applets
	~~::[[Applets:Complementary_Gaussian_Error_Functions\|"Complementary Gaussian error functions"]]  and ~~
	::[[Applets:Binomial_and_Poisson_Distribution_(Applet)\|"Binomial and Poisson Distribution"]].


Line 57:		Line 62:
	{What is the block error probability for  $10 \cdot \lg {E_{\rm B}/N_0} \hspace{0.15cm}\underline{= 5 \ \rm dB}$?		{What is the block error probability for  $10 \cdot \lg {E_{\rm B}/N_0} \hspace{0.15cm}\underline{= 5 \ \rm dB}$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr(~~Block~~\:error)} \ = \ ${ 6.66 3% } $\ \cdot 10^{-2}$		${\rm Pr(block\:error)} \ = \ ${ 6.66 3% } $\ \cdot 10^{-2}$

	{What is the block error probability for  $10 \cdot \lg {E_{\rm B}/N_0} \hspace{0.15cm}\underline{= 8 \ \rm dB}$?		{What is the block error probability for  $10 \cdot \lg {E_{\rm B}/N_0} \hspace{0.15cm}\underline{= 8 \ \rm dB}$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr(~~Block~~\:error)} \ = \ ${ 8.63 3% } $\ \cdot 10^{-4}$		${\rm Pr(block\:error)} \ = \ ${ 8.63 3% } $\ \cdot 10^{-4}$

	{What is the block error probability for  $10 \cdot \lg {E_{\rm B}/N_0}\hspace{0.15cm}\underline{ = 10 \ \rm dB}$?		{What is the block error probability for  $10 \cdot \lg {E_{\rm B}/N_0}\hspace{0.15cm}\underline{ = 10 \ \rm dB}$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr(~~Block~~\:error)} \ = \ ${ 4.3 3% } $\ \cdot 10^{-6}$		${\rm Pr(block\:error)} \ = \ ${ 4.3 3% } $\ \cdot 10^{-6}$

	{How is  $\varepsilon_{\rm S} = 0.1$  related to  $10 \cdot \lg {E_{\rm B}/N_0}$ ?   ''Note:''  Use the given applet to calculate  ${\rm Q}(x)$.		{How is  $\varepsilon_{\rm S} = 0.1$  related to  $10 \cdot \lg {E_{\rm B}/N_0}$ ?   <u>Note:</u>  Use the given applet to calculate  ${\rm Q}(x)$.
	\|type="{}"}		\|type="{}"}
	$\varepsilon_{\rm S} = 10^{-1} \text{:} \hspace{0.4cm} 10 \cdot \lg {E_{\rm B}/N_0} \ = \ ${ 5.87 3% } $\ \rm dB$		$\varepsilon_{\rm S} = 10^{-1} \text{:} \hspace{0.4cm} 10 \cdot \lg {E_{\rm B}/N_0} \ = \ ${ 5.87 3% } $\ \rm dB$

	{Find also the  $E_{\rm B}/N_0$ values  $($in  $\rm dB)$  for  $\varepsilon_{\rm S} = 0.01$  and  $\varepsilon_{\rm S} = 0.001$. Complete the table...		{Find also the  $E_{\rm B}/N_0$ values  $($in  $\rm dB)$  for  $\varepsilon_{\rm S} = 0.01$  and  $\varepsilon_{\rm S} = 0.001$. Complete the table.
	\|type="{}"}		\|type="{}"}
	$\varepsilon_{\rm S} = 10^{-2} \text{:} \hspace{0.4cm} 10 \cdot \lg {E_{\rm B}/N_0} \ = \ $ { 9.32 3% } $\ \rm dB$		$\varepsilon_{\rm S} = 10^{-2} \text{:} \hspace{0.4cm} 10 \cdot \lg {E_{\rm B}/N_0} \ = \ $ { 9.32 3% } $\ \rm dB$

Noah at 20:12, 14 September 2022

2022-09-14T20:12:45Z

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