Aufgaben:Exercise 1.17: About the Channel Coding Theorem - Revision history

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	[[Category:Channel Coding: Exercises\|^1.7 Channel Capacity^]]		[[Category:Channel Coding: Exercises\|^1.7 Channel Capacity^]]
	[[de:Aufgaben:1.~~17_Coderate_vs._EB/N0~~]]		[[de:Aufgaben:Aufgabe 1.17: Zum Kanalcodierungstheorem]]

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	'''(4)'''  Now,  according to the given equation:		'''(4)'''  Now,  according to the given equation:
	:$$1 - {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} x} = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x = \frac{-{\rm ln}(0.5)}{-0.4} = 1.73\hspace{0.3cm}		:$$1 - {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} x} = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x = \frac{-{\rm ln}(0.5)}{-0.4} = 1.73\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 1.73 - 1.6 = 0.13 \,{\rm dB}\hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 1.73 - 1.6 = 0.13 \,{\rm dB}\hspace{0.05cm}.$$

	* $10 · \lg {E_{\rm B}/N_0}$  could thus be decreased by  $3 \ \rm dB - 0.13 \ dB = 2.87 \ dB$,  i.e. by a factor  $A = 10^{0.287}\hspace{0.15cm} \underline{= 1.94} \hspace{0.05cm}.$		* $10 · \lg {E_{\rm B}/N_0}$  could thus be decreased by  $3 \ \rm dB - 0.13 \ dB = 2.87 \ dB$,  i.e. by a factor  $A = 10^{0.287}\hspace{0.15cm} \underline{= 1.94} \hspace{0.05cm}.$

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	[[Category:Channel Coding: Exercises\|^1.7 Channel Capacity^]]		[[Category:Channel Coding: Exercises\|^1.7 Channel Capacity^]]
			[[de:Aufgaben:1.17_Coderate_vs._EB/N0]]

Guenter at 14:48, 28 September 2022

2022-09-28T14:48:36Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Correct are <u>all proposed solutions</u>:		'''(1)'''  Correct are  <u>all proposed solutions</u>:
	*From the graph you can already see that the rate of  $\rm Z$  is greater than the rate of  $\rm Y$  and the rate of  $\rm Y$  is greater than the rate of  $\rm X$.		*From the graph you can already see that the rate of  $\rm Z$  is greater than the rate of  $\rm Y$  and the rate of  $\rm Y$  is greater than the rate of  $\rm X$.
	*The actual rates of these three systems are $R_{\rm X} = 4/7 = 0.571$, $R_{\rm Y} = 11/15 = 0.733$, and $R_{\rm Z} = 26/31 = 0.839$.
	*In addition, since the $(31, 26, 3)$ Hamming code   ⇒   code  $\rm Z$  has the largest codeword length $n$, it requires $n$ despite a larger code rate  $R$  for  ${\rm BER} = 10^{-5}$  a smaller  $E_{\rm B}/N_{0}$  than the other two Hamming codes.		*The actual rates of these three systems are  $R_{\rm X} = 4/7 = 0.571$,  $R_{\rm Y} = 11/15 = 0.733$,  and $R_{\rm Z} = 26/31 = 0.839$.

			*In addition,  since the  $(31, 26, 3)$  Hamming code   ⇒   code  $\rm Z$  has the largest code word length  $n$,  it requires  $n$  despite a larger code rate  $R$  for  ${\rm BER} = 10^{-5}$  a smaller  $E_{\rm B}/N_{0}$  than the other two Hamming codes.


	'''(2)'''  Correct is <u>the answer 2</u>:
	*For a smaller bit error rate, one always needs a larger $E_{\rm B}/N_{0}$.		'''(2)'''  Correct is  <u>the answer 2</u>:
	*There is no vertical shift, because even with $\rm BER = 10^{-10}$ there is no change in the code rates.		*For a smaller bit error rate,  one always needs a larger  $E_{\rm B}/N_{0}$.

			*There is no vertical shift,  because even with  $\rm BER = 10^{-10}$  there is no change in the code rates.



	'''(3)'''  For the logarithmized AWGN parameter $10 · \lg {E_{\rm B}/N_0} = 3 \ {\rm dB}$, we obtain the auxiliary quantity $x = 1.6 + 3 = 4.6.$ This gives:		'''(3)'''  For the logarithmized AWGN parameter  $10 · \lg {E_{\rm B}/N_0} = 3 \ {\rm dB}$,  we obtain the auxiliary quantity  $x = 1.6 + 3 = 4.6.$  This gives:

	:$$R_{\rm max} = C (x = 4.6)= 1 - {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} 4.6} \hspace{0.15cm} \underline{= 0.84} \hspace{0.05cm}.$$		:$$R_{\rm max} = C (x = 4.6)= 1 - {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} 4.6} \hspace{0.15cm} \underline{= 0.84} \hspace{0.05cm}.$$
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	'''(4)'''  Now, according to the given equation:		'''(4)'''  Now,  according to the given equation:
	:$$1 - {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} x} = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x = \frac{-{\rm ln}(0.5)}{-0.4} = 1.73\hspace{0.3cm}		:$$1 - {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} x} = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x = \frac{-{\rm ln}(0.5)}{-0.4} = 1.73\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 1.73 - 1.6 = 0.13 \,{\rm dB}\hspace{0.05cm}.$$		\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 1.73 - 1.6 = 0.13 \,{\rm dB}\hspace{0.05cm}.$$

	* $10 · \lg {E_{\rm B}/N_0}$ could thus be decreased by $3 \ \rm dB - 0.13 \ dB = 2.87 \ dB$, i.e. by a factor $A = 10^{0.287}\hspace{0.15cm} \underline{= 1.94} \hspace{0.05cm}.$		* $10 · \lg {E_{\rm B}/N_0}$  could thus be decreased by  $3 \ \rm dB - 0.13 \ dB = 2.87 \ dB$,  i.e. by a factor  $A = 10^{0.287}\hspace{0.15cm} \underline{= 1.94} \hspace{0.05cm}.$
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 14:44, 28 September 2022

2022-09-28T14:44:59Z

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	The graph shows the maximum allowable code rate  $R < C$  according to Shannon's  [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding#Channel_coding_theorem_and_channel_capacity\|"channel coding theorem"]]:		The graph shows the maximum allowable code rate  $R < C$  according to Shannon's  [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding#Channel_coding_theorem_and_channel_capacity\|"channel coding theorem"]]:

	*The green ~~limit~~ curve indicates the channel capacity  $C$  for the AWGN channel assuming a binary input signal ("BPSK").		*The green curve indicates the channel capacity  $C$  for the AWGN channel assuming a binary input signal  ("BPSK").

	*In the  [[Aufgaben:Exercise_1.17Z:_BPSK_Channel_Capacity\|"Exercise 1.17Z"]]  a simple approximation is given for this. With the second abscissa		*In the  [[Aufgaben:Exercise_1.17Z:_BPSK_Channel_Capacity\|Exercise 1.17Z]]  a simple approximation is given for this.  With the second abscissa

	:$$x = \frac {1.6\,{\rm dB} + 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 }{1\,{\rm dB}}$$		:$$x = \frac {1.6\,{\rm dB} + 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 }{1\,{\rm dB}}$$
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	:$$C \approx \hspace{0.15cm} \left\{ \begin{array}{c} 1 - {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} x} \\ \\ 0 \end{array} \right.\quad \begin{array}{*{1}c} {\rm for\hspace{0.15cm}} x > 0, \\ \\{\rm for\hspace{0.15cm}} x < 0. \end{array}$$		:$$C \approx \hspace{0.15cm} \left\{ \begin{array}{c} 1 - {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} x} \\ \\ 0 \end{array} \right.\quad \begin{array}{*{1}c} {\rm for\hspace{0.15cm}} x > 0, \\ \\{\rm for\hspace{0.15cm}} x < 0. \end{array}$$

	*If  $R < C$ ~~holds~~, ~~then~~ a code can be found which leads to the error probability "zero" for infinitely long blocks  $(n → ∞)$ . What this code looks like is not determined by the channel coding theorem and does not matter for this exercise.		*For  $R < C$,  a code can be found which leads to the error probability  "zero"  for infinitely long blocks  $(n → ∞)$.

			*What this code looks like is not determined by the channel coding theorem and does not matter for this exercise.


	Drawn into the graph as dots are the characteristics of established coding systems:		Drawn into the graph as dots are the characteristics of established coding systems:
	*The dots  $\rm X$,  $\rm Y$~~  and~~  $\rm Z$  mark three Hamming codes of different code lengths, namely with  $n = 7$,  $n = 15$  and  $n = 31$.		*The dots  $\rm X$,  $\rm Y$,  $\rm Z$  mark three Hamming codes of different code lengths, namely with  $n = 7$,  $n = 15$  and  $n = 31$.
	*The coding system  $\rm W$  is characterized by the parameters  $R = 0.5$  and  $10 \ - \lg {E_{\rm B}/N_0} = 3 {\rm dB}$ .
			*The coding system  $\rm W$  is characterized by the parameters  $R = 0.5$  and  $10 \ \cdot \lg {E_{\rm B}/N_0} = 3 {\rm dB}$ .




			Hints:
			*This exercise belongs to the topic of chapter  [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding\|"Information Theoretical Limits of Channel Coding"]].

			*The information theoretical limit  "channel capacity"  refers to the error probability  $\rm BER =0$.

			*The plotted points of real transmission systems,  on the other hand,  result under the assumption  $\rm BER = 10^{-5}$.
	~~Hints:~~
	*This exercise belongs to the topic of chapter  [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding\|"Information Theoretical Limits of Channel Coding"]].
	*The information theoretical limit "channel capacity" refers to the error probability  $\rm BER =0$.
	*The plotted points of real transmission systems, on the other hand, result under the assumption  $\rm BER = 10^{-5}$.


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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Which of the points belong to which Hamming code? ''Note:''   The graph was created for  $\rm BER = 10^{-5}$ .		{Which of the points belong to which Hamming code? Note:   The graph was created for  $\rm BER = 10^{-5}$ .
	\|type="[]"}		\|type="[]"}
	+ $\rm X$  denotes the  $(7, 4, 3)$ Hamming code.		+ $\rm X$  denotes the  $(7, 4, 3)$ Hamming code.

Guenter at 12:38, 28 September 2022

2022-09-28T12:38:33Z

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	}}		}}

	[[File:EN_KC_A_1_17.png\|right\|frame\|~~Kanalkapazität~~ (~~grün~~) ~~und Coderaten~~ (~~rote Punkte~~) ~~einiger etablierter Systeme~~]]		[[File:EN_KC_A_1_17.png\|right\|frame\|Channel capacity  $($green$)$  and code rates  $($red dots$)$  of some established systems]]

	The graph shows the maximum allowable code rate  $R < C$  according to Shannon's  [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding#Channel_coding_theorem_and_channel_capacity\|"channel coding theorem"]]:		The graph shows the maximum allowable code rate  $R < C$  according to Shannon's  [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding#Channel_coding_theorem_and_channel_capacity\|"channel coding theorem"]]:
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	\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 1.73 - 1.6 = 0.13 \,{\rm dB}\hspace{0.05cm}.$$		\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 1.73 - 1.6 = 0.13 \,{\rm dB}\hspace{0.05cm}.$$

	$10 · \lg {E_{\rm B}/N_0}$ could thus be decreased by $3 \ \rm dB - 0.13 \ dB = 2.87 \ dB$, i.e. by a factor $A = 10^{0.287}\hspace{0.15cm} \underline{= 1.94} \hspace{0.05cm}.$		* $10 · \lg {E_{\rm B}/N_0}$ could thus be decreased by $3 \ \rm dB - 0.13 \ dB = 2.87 \ dB$, i.e. by a factor $A = 10^{0.287}\hspace{0.15cm} \underline{= 1.94} \hspace{0.05cm}.$
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 14:37, 17 August 2022

2022-08-17T14:37:16Z

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	}}		}}

	[[File:~~P_ID2412__KC_A_1_16~~.png\|right\|frame\|Kanalkapazität (grün) und Coderaten (rote Punkte) einiger etablierter Systeme]]		[[File:EN_KC_A_1_17.png\|right\|frame\|Kanalkapazität (grün) und Coderaten (rote Punkte) einiger etablierter Systeme]]

	The graph shows the maximum allowable code rate  $R < C$  according to Shannon's  [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding#Channel_coding_theorem_and_channel_capacity\|"channel coding theorem"]]:		The graph shows the maximum allowable code rate  $R < C$  according to Shannon's  [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding#Channel_coding_theorem_and_channel_capacity\|"channel coding theorem"]]:

Noah at 15:14, 16 August 2022

2022-08-16T15:14:04Z

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	Hints:		Hints:
	*This exercise belongs to the topic of chapter  [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding\|"~~information theoretical limits~~ of ~~channel coding~~"]].		*This exercise belongs to the topic of chapter  [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding\|"Information Theoretical Limits of Channel Coding"]].
	*The information theoretical limit "channel capacity" refers to the error probability  $\rm BER =0$.		*The information theoretical limit "channel capacity" refers to the error probability  $\rm BER =0$.
	*The plotted points of real transmission systems, on the other hand, result under the assumption  $\rm BER = 10^{-5}$.		*The plotted points of real transmission systems, on the other hand, result under the assumption  $\rm BER = 10^{-5}$.

Noah at 15:04, 16 August 2022

2022-08-16T15:04:46Z

Show changes

Noah at 21:21, 14 August 2022

2022-08-14T21:21:57Z

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	{{quiz-Header\|Buchseite=~~Kanalcodierung~~/~~Informationstheoretische Grenzen der Kanalcodierung~~		{{quiz-Header\|Buchseite=Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding