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

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Noah at 20:12, 14 September 2022

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Guenter at 15:54, 4 September 2022

2022-09-04T15:54:41Z

Guenter at 15:53, 4 September 2022

2022-09-04T15:53:44Z

Guenter: Guenter moved page Aufgaben:Aufgabe 2.15: RS-Blockfehlerwahrscheinlichkeit bei AWGN to Aufgaben:Exercise 2.15: Block Error Probability with AWGN

2022-09-04T15:40:23Z

Guenter moved page Aufgabe 2.15: RS-Blockfehlerwahrscheinlichkeit bei AWGN to Exercise 2.15: Block Error Probability with AWGN

@@ Line 152: / Line 152: @@
 \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

@@ Line 152: / Line 152: @@
 \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

@@ Line 83: / Line 83: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; From the table on the information page, the BSC parameter $\varepsilon = 0.0505$ can be read.
+'''(1)'''&nbsp; From the table on the information page,&nbsp;  the BSC parameter&nbsp; $\varepsilon = 0.0505$&nbsp; can be read.
-*This gives $\varepsilon_{\rm S}$ for the symbol corruption probability with $m = 3$:
+*This gives&nbsp; $\varepsilon_{\rm S}$&nbsp; for the symbol error probability with&nbsp; $m = 3$:
 :$$1 - \varepsilon_{\rm S} = (1 - 0.0505)^3 \approx 0.856
 \hspace{0.3cm}\Rightarrow  \hspace{0.3cm}
@@ Line 90: / Line 90: @@
 \hspace{0.05cm}.$$
-*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 -
 \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}
+:$$\Rightarrow \hspace{0.3cm} {\rm Pr(block\:error)}   \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Pr}(\underline{v} \ne \underline{u}) =1 - 0.3368 - 0.3965 - 0.2001 \hspace{0.15cm} \underline{=0.0666}
 \hspace{0.05cm}.$$
-'''(2)'''&nbsp; 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)'''&nbsp; Following the same calculation procedure as in subtask&nbsp; '''(1)''',&nbsp; we obtain with&nbsp; $\varepsilon_{\rm S} \approx 0.03 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.97$:
-:$${\rm Pr(Block\:error)}
+:$${\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}
 \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}.$$
-*You can see that here the difference between two numbers of almost the same size must be formed, so that the result could be affected by an error.
+*You can see that here the difference between two numbers of almost the same size must be formed,&nbsp; so that the result could be affected by an error.
 *Therefore we still calculate the following quantities:
 :$${\rm Pr}(f=3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}
@@ Line 111: / Line 112: @@
 :$${\rm Pr}(f=5) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}
 {7 \choose 5} \cdot \varepsilon_{\rm S}^5 \cdot (1 - \varepsilon_{\rm S})^2 = 21 \cdot 0.03^5 \cdot 0.97^2 = 0.005 \cdot 10^{-4}$$
-:$$\Rightarrow  \hspace{0.3cm} {\rm Pr(Blockfehler)} = {\rm Pr}(\underline{v} \ne \underline{u})   \approx {\rm Pr}(f=3) +  {\rm Pr}(f=4) + {\rm Pr}(f=5)  \hspace{0.15cm} \underline{=8.63 \cdot 10^{-4}} \hspace{0.05cm}.$$
+:$$\Rightarrow  \hspace{0.3cm} {\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u})   \approx {\rm Pr}(f=3) +  {\rm Pr}(f=4) + {\rm Pr}(f=5)  \hspace{0.15cm} \underline{=8.63 \cdot 10^{-4}} \hspace{0.05cm}.$$
-*The terms for $f = 6$ and $f = 7$ can be omitted here. They do not provide a relevant contribution.
+*The terms for&nbsp; $f = 6$&nbsp; and&nbsp; $f = 7$&nbsp; can be omitted here.&nbsp; They do not provide a relevant contribution.
-'''(3)'''&nbsp; Here $\varepsilon_{\rm S} = 0.005 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.995$ is already given in the table.
+'''(3)'''&nbsp; Here&nbsp; $\varepsilon_{\rm S} = 0.005 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.995$&nbsp; 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&nbsp; (by far)&nbsp; dominant term in the calculation of the block error probability is&nbsp; ${\rm Pr}(f = 3)$:
-:$${\rm Pr(Blockfehler)} = {\rm Pr}(\underline{v} \ne \underline{u})  \approx {\rm Pr}(f=3) = {7 \choose 3} \cdot 0.005^3 \cdot 0.995^4
+:$${\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}.$$
-'''(4)'''&nbsp; For the BSC parameter $\varepsilon$ holds with $\varepsilon_{\rm S} = 0.1$:
+'''(4)'''&nbsp; For the BSC parameter&nbsp; $\varepsilon$&nbsp; holds with&nbsp; $\varepsilon_{\rm S} = 0.1$:
 :$$\varepsilon = 1 -(1 - \varepsilon_{\rm S})^{1/3} = 1 - 0.9^{1/3} \approx 0.0345
 \hspace{0.05cm}.$$
-*The relation between $\varepsilon$ and $E_{\rm B}/N_0$ is:
+*The relation between&nbsp; $\varepsilon$&nbsp; and&nbsp; $E_{\rm B}/N_0$&nbsp; is:
 :$$\varepsilon = {\rm Q}(x)\hspace{0.05cm}, \hspace{0.5cm} x = \sqrt{2 \cdot R \cdot E_{\rm B}/N_0}\hspace{0.05cm}.$$
-*The inverse $x = {\rm Q}^{-1}(0.0345)$ is obtained with the applet [[Applets:QFunction|"Complementary Gaussian Error Functions"]] to $x = 1.82$.
+*The inverse&nbsp; $x = {\rm Q}^{-1}(0.0345)$&nbsp; is obtained with the applet&nbsp; [[Applets:QFunction|"Complementary Gaussian Error Functions"]]&nbsp; to&nbsp; $x = 1.82$.&nbsp; This further gives:
 :$$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}
@@ Line 140: / Line 140: @@
 '''(5)'''&nbsp; 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&nbsp; $\varepsilon_{\rm S} = 10^{-2} \ \Rightarrow \ \varepsilon \approx 0.33 \cdot 10^{-2} \ \Rightarrow \ x = {\rm Q}^{-1}(\varepsilon) = 2.71$
 :$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{2.71^2}{2R \cdot 3/7} \approx 8.568
 \hspace{0.3cm}  \Rightarrow  \hspace{0.3cm}
@@ Line 152: / Line 152: @@
 \hspace{0.15cm} \underline{\approx 11.3 \,\, {\rm dB}} \hspace{0.05cm}. $$
-[[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
-*For abscissa values smaller than $10 \ \rm dB$ the result is even worse than without coding.
+# For abscissa values smaller than&nbsp; $10 \ \rm dB$&nbsp; 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&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$&nbsp; 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&nbsp; "Bounded Distance Decoding"&nbsp; $\rm (BDD)$.
 {{ML-Fuß}}
 [[Category:Channel Coding: Exercises|^2.6 Block Error Probability of RS Codes^]]

@@ Line 51: / Line 51: @@
 * We refer you here to the two interactive HTML5/JavaScript applets&nbsp;
 :*[[Applets:Complementary_Gaussian_Error_Functions|"Complementary Gaussian error functions"]],&nbsp; and&nbsp;
 :*[[Applets:Binomial_and_Poisson_Distribution_(Applet)|"Binomial and Poisson Distribution"]].

@@ Line 49: / Line 49: @@
 * The exercise belongs to the chapter&nbsp; [[Channel_Coding/Error_Probability_and_Areas_of_Application| "Error Probability and Application Areas"]].
-* We refer you here to the two interactive applets&nbsp;
+* We refer you here to the two interactive HTML5/JavaScript applets&nbsp;
 :*[[Applets:Complementary_Gaussian_Error_Functions|"Complementary Gaussian error functions"]],&nbsp; and&nbsp;

@@ Line 2: / Line 2: @@
 [[File:P_ID2571__KC_A_2_15neu.png|right|frame|Incomplete table of results]]
-Using the example of&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$&nbsp; with the parameters
+Using the example of&nbsp; $\rm RSC \, (7, \, 3, \, 5)_8$&nbsp; with parameters
-* $n = 7$&nbsp; (number of code symbols),
+* $n = 7$&nbsp; $($number of code symbols$)$,
-the calculation of the block error probability in&nbsp; [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD|"Bounded Distance Decoding"]]&nbsp; (BDD) shall be shown. The corresponding equation is:
+* $t = 2$&nbsp; $($correction capability$)$.
-:$${\rm Pr(Block\:error)}  = {\rm Pr}(\underline{v} \ne \underline{u}) =
 \sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$
-The calculation is performed for the&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|"AWGN channel"
+&rArr; &nbsp; The calculation is performed for the&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|"AWGN channel"
-]] characterized by the parameter&nbsp; $E_{\rm B}/N_0$&nbsp;.
+]]&nbsp; characterized by the parameter&nbsp; $E_{\rm B}/N_0$:
 *The quotient&nbsp; $E_{\rm B}/{N_0}$&nbsp; can be expressed by the relation
 :$$\varepsilon = {\rm Q} \big (\sqrt{{2 \cdot R \cdot E_{\rm B}}/{N_0}} \big ) $$
-into the&nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Symmetric_Channel_.E2.80. 93_BSC|"BSC model"]]&nbsp; where&nbsp; $R$&nbsp; denotes the code rate&nbsp; $($here: &nbsp;$R = 3/7)$&nbsp; and&nbsp; ${\rm Q}(x)$&nbsp; indicates the&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|"complementary Gaussian error integral"]]&nbsp;.
+:into the&nbsp; [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Symmetric_Channel_.E2.80. 93_BSC|"BSC model"]]&nbsp; where&nbsp; $R$&nbsp; denotes the code rate&nbsp; $($here: &nbsp;$R = 3/7)$&nbsp; and&nbsp; ${\rm Q}(x)$&nbsp; indicates the&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|"complementary Gaussian error integral"]].
-*But since in the considered code the symbols come from&nbsp; $\rm GF(2^3)$&nbsp;, the BSC model with parameter&nbsp; $\varepsilon$&nbsp; must also still be adapted to the task.
-*For the corruption probability of the&nbsp; [[Channel_Coding/Error_Probability_and_Areas_of_Application#Block_error_probability_for_RSC_and_BDD|"$m$ BSC model"]]&nbsp; applies, where here&nbsp; $m = 3$&nbsp; is to be set (three bits per code symbol):
+*But since in the considered code the symbols come from&nbsp; $\rm GF(2^3)$,&nbsp; the BSC model with parameter&nbsp; $\varepsilon$&nbsp; must also still be adapted to the task.
 :$$\varepsilon_{\rm S} = 1 - (1 - \varepsilon)^m
 \hspace{0.05cm}.$$
-For some&nbsp; $E_{\rm B}/N_0$ values the results are entered in the table above. The two rows with yellow background are briefly explained here:
+&rArr; &nbsp; For some&nbsp; $E_{\rm B}/N_0$ values the results are entered in the table above.&nbsp; The two rows with yellow background are briefly explained here:
-* For&nbsp; $10 \cdot \lg {E_{\rm B}/N_0} = 4 \ \rm dB$&nbsp; we get&nbsp; $\varepsilon \approx {\rm Q}(1.47) \approx 0.071$&nbsp; and&nbsp; $\varepsilon_{\rm S} \approx 0.2$. The block error probability here can most easily be calculated using the complement:
+* For&nbsp; $10 \cdot \lg {E_{\rm B}/N_0} = 4 \ \rm dB$&nbsp; we get&nbsp; $\varepsilon \approx {\rm Q}(1.47) \approx 0.071$&nbsp; and&nbsp; $\varepsilon_{\rm S} \approx 0.2$. &nbsp; 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}.$$
-* For&nbsp; $10 \cdot \lg {E_{\rm B}/N_0} = 12 \ \rm dB$&nbsp; one gets&nbsp; $\varepsilon \approx 1.2 \cdot 10^{-4}$&nbsp; and&nbsp; $\varepsilon_{\rm S} \approx 3.5 \cdot 10^{-4}$. With this very small corruption probability, the&nbsp; $f = 3$ term dominates, and we obtain:
-:$${\rm Pr(Block\:error)}  \approx  {7 \choose 3} \cdot (3.5 \cdot 10^{-4})^3 \cdot (1- 3.5 \cdot 10^{-4})^4
+* For&nbsp; $10 \cdot \lg {E_{\rm B}/N_0} = 12 \ \rm dB$&nbsp; one gets&nbsp; $\varepsilon \approx 1.2 \cdot 10^{-4}$&nbsp; and&nbsp; $\varepsilon_{\rm S} \approx 3.5 \cdot 10^{-4}$.&nbsp; With this very small falsification probability,&nbsp; the&nbsp; $f = 3$&nbsp; term dominates,&nbsp; and we obtain:
 \approx 1.63 \cdot 10^{-9}  \hspace{0.05cm}.$$
-*You are to calculate the block error probabilities for the rows highlighted in red&nbsp; $(10 \cdot \lg {E_{\rm B}/N_0} = 5 \ \rm dB, \ 8 \rm dB$,&nbsp; $10 \ \rm dB)$&nbsp; .
+&rArr; &nbsp; You are to calculate the block error probabilities for the rows highlighted in red &nbsp; $(10 \cdot \lg {E_{\rm B}/N_0} = 5 \ \rm dB, \ 8 \rm dB$,&nbsp; $10 \ \rm dB)$.
-*The rows with blue background show some results of&nbsp; [[Aufgaben:Exercise_2.15Z:_Block_Error_Probability_once_more|"Exercise 2.15Z"]]. There&nbsp; ${\rm Pr}(\underline{v} &ne; \underline{u})$&nbsp; is calculated for&nbsp; $\varepsilon_{\rm S} = 10\%,&nbsp; \ 1\%$&nbsp; $0.1\%$.
+:*The rows with blue background show some results of&nbsp; [[Aufgaben:Exercise_2.15Z:_Block_Error_Probability_once_more|"Exercise 2.15Z"]].&nbsp; There&nbsp; ${\rm Pr}(\underline{v} &ne; \underline{u})$&nbsp; is calculated for&nbsp; $\varepsilon_{\rm S} = 10\%,&nbsp; \ 1\%$&nbsp; $0.1\%$.
-*In subtasks '''(4)''' and '''(5)''' you are to establish the relationship between the size&nbsp; $\varepsilon_{\rm S}$&nbsp; and the AWGN parameter&nbsp; $E_{\rm B}/N_0$&nbsp; thus completing the above table.
+:*In subtasks '''(4)''' and '''(5)''' you are to establish the relationship between the quantity&nbsp; $\varepsilon_{\rm S}$&nbsp; and the AWGN parameter&nbsp; $E_{\rm B}/N_0$&nbsp; thus completing the above table.
@@ Line 41: / Line 46: @@
-Hints:
+:*[[Applets:Binomial_and_Poisson_Distribution_(Applet)|"Binomial and Poisson Distribution"]].
@@ Line 57: / Line 62: @@
 {What is the block error probability for&nbsp;  $10 \cdot \lg {E_{\rm B}/N_0} \hspace{0.15cm}\underline{= 5 \ \rm dB}$?
 |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&nbsp; $10 \cdot \lg {E_{\rm B}/N_0} \hspace{0.15cm}\underline{= 8 \ \rm dB}$?
 |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&nbsp; $10 \cdot \lg {E_{\rm B}/N_0}\hspace{0.15cm}\underline{ = 10 \ \rm dB}$?
 |type="{}"}
-${\rm Pr(Block\:error)} \ = \ ${ 4.3 3% } $\ \cdot 10^{-6}$
+${\rm Pr(block\:error)} \ = \ ${ 4.3 3% } $\ \cdot 10^{-6}$
-{How is&nbsp; $\varepsilon_{\rm S} = 0.1$&nbsp; related to&nbsp; $10 \cdot \lg {E_{\rm B}/N_0}$&nbsp;? &nbsp; ''Note:'' &nbsp;Use the given applet to calculate&nbsp; ${\rm Q}(x)$.
+{How is&nbsp; $\varepsilon_{\rm S} = 0.1$&nbsp; related to&nbsp; $10 \cdot \lg {E_{\rm B}/N_0}$&nbsp;? &nbsp; <u>Note:</u> &nbsp;Use the given applet to calculate&nbsp; ${\rm Q}(x)$.
 |type="{}"}
 $\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&nbsp; $E_{\rm B}/N_0$ values&nbsp; $($in&nbsp; $\rm dB)$&nbsp; for&nbsp; $\varepsilon_{\rm S} = 0.01$&nbsp; and&nbsp; $\varepsilon_{\rm S} = 0.001$. Complete the table...
+{Find also the&nbsp; $E_{\rm B}/N_0$ values&nbsp; $($in&nbsp; $\rm dB)$&nbsp; for&nbsp; $\varepsilon_{\rm S} = 0.01$&nbsp; and&nbsp; $\varepsilon_{\rm S} = 0.001$. Complete the table.
 |type="{}"}
 $\varepsilon_{\rm S} = 10^{-2} \text{:} \hspace{0.4cm} 10 \cdot \lg {E_{\rm B}/N_0} \ = \ $ { 9.32 3% } $\ \rm dB$

← Older revision		Revision as of 15:54, 4 September 2022
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−	[[Category:Channel Coding: Exercises\|^2.6 Block Error Probability of ~~Reed-Solomon~~ Codes^]]	+	[[Category:Channel Coding: Exercises\|^2.6 Block Error Probability of RS Codes^]]

← Older revision		Revision as of 15:53, 4 September 2022
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−	[[Category:Channel Coding: Exercises\|^2.6 ~~RSC–Fehlerwahrscheinlichkeit~~^]]	+	[[Category:Channel Coding: Exercises\|^2.6 Block Error Probability of Reed-Solomon Codes^]]

← Older revision	Revision as of 15:40, 4 September 2022
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