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

Guenter at 14:48, 28 September 2022

2022-09-28T14:48:36Z

Guenter at 14:44, 28 September 2022

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Guenter at 12:38, 28 September 2022

2022-09-28T12:38:33Z

Guenter at 14:37, 17 August 2022

2022-08-17T14:37:16Z

Noah at 15:14, 16 August 2022

2022-08-16T15:14:04Z

Noah at 15:04, 16 August 2022

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Noah at 21:21, 14 August 2022

2022-08-14T21:21:57Z

Guenter at 11:21, 11 May 2022

2022-05-11T11:21:55Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.17: Zum Kanalcodierungstheorem to Aufgaben:Exercise 1.17: About the Channel Coding Theorem

2022-05-11T10:15:20Z

Guenter moved page Aufgabe 1.17: Zum Kanalcodierungstheorem to Exercise 1.17: About the Channel Coding Theorem

Javier: Text replacement - "„" to """

2021-05-28T14:26:38Z

Text replacement - "„" to """

@@ Line 67: / Line 67: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Correct are <u>all proposed solutions</u>:
+'''(1)'''&nbsp; Correct are&nbsp; <u>all proposed solutions</u>:
 *From the graph you can already see that the rate of &nbsp;$\rm Z$&nbsp; is greater than the rate of &nbsp;$\rm Y$&nbsp; and the rate of &nbsp;$\rm Y$&nbsp; is greater than the rate of &nbsp;$\rm X$.
-*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 &nbsp; ⇒ &nbsp; code &nbsp;$\rm Z$&nbsp; has the largest codeword length $n$, it requires $n$ despite a larger code rate&nbsp; $R$&nbsp; for&nbsp; ${\rm BER} = 10^{-5}$&nbsp; a smaller&nbsp; $E_{\rm B}/N_{0}$&nbsp; than the other two Hamming codes.
+*The actual rates of these three systems are&nbsp; $R_{\rm X} = 4/7 = 0.571$,&nbsp; $R_{\rm Y} = 11/15 = 0.733$,&nbsp; and $R_{\rm Z} = 26/31 = 0.839$.
 '''(2)'''&nbsp; Correct is <u>the answer 2</u>:
-*For a smaller bit error rate, one always needs a larger $E_{\rm B}/N_{0}$.
+'''(2)'''&nbsp; Correct is&nbsp; <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,&nbsp; one always needs a larger&nbsp; $E_{\rm B}/N_{0}$.
-'''(3)'''&nbsp; 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)'''&nbsp; For the logarithmized AWGN parameter&nbsp; $10 · \lg {E_{\rm B}/N_0} = 3 \ {\rm dB}$,&nbsp; we obtain the auxiliary quantity&nbsp; $x = 1.6 + 3 = 4.6.$&nbsp; This gives:
 :$$R_{\rm max} = C (x = 4.6)= 1 - {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} 4.6} \hspace{0.15cm} \underline{= 0.84} \hspace{0.05cm}.$$
@@ Line 86: / Line 89: @@
-'''(4)'''&nbsp; Now, according to the given equation:
+'''(4)'''&nbsp; Now, &nbsp;according to the given equation:
 :$$1 -  {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} x} = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x = \frac{-{\rm ln}(0.5)}{-0.4} = 1.73\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 = 1.73 - 1.6 = 0.13 \,{\rm dB}\hspace{0.05cm}.$$
-* $10 · \lg {E_{\rm B}/N_0}$ could thus be decreased by $3 \ \rm dB - 0.13 \ dB = 2.87 \ dB$, i.e. by a factor $A = 10^{0.287}\hspace{0.15cm} \underline{= 1.94} \hspace{0.05cm}.$
+* $10 · \lg {E_{\rm B}/N_0}$&nbsp; could thus be decreased by&nbsp; $3 \ \rm dB - 0.13 \ dB = 2.87 \ dB$,&nbsp; i.e. by a factor&nbsp; $A = 10^{0.287}\hspace{0.15cm} \underline{= 1.94} \hspace{0.05cm}.$
 {{ML-Fuß}}

@@ Line 9: / Line 9: @@
 The graph shows the maximum allowable code rate&nbsp; $R < C$&nbsp; according to Shannon's&nbsp; [[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&nbsp; $C$&nbsp; for the AWGN channel assuming a binary input signal ("BPSK").
+*The green curve indicates the channel capacity&nbsp; $C$&nbsp; for the AWGN channel assuming a binary input signal&nbsp; ("BPSK").
-*In the&nbsp; [[Aufgaben:Exercise_1.17Z:_BPSK_Channel_Capacity|"Exercise 1.17Z"]]&nbsp; a simple approximation is given for this. With the second abscissa
+*In the&nbsp; [[Aufgaben:Exercise_1.17Z:_BPSK_Channel_Capacity|Exercise 1.17Z]]&nbsp; a simple approximation is given for this.&nbsp; With the second abscissa
 :$$x = \frac {1.6\,{\rm dB} + 10 \cdot {\rm lg} \hspace{0.1cm} E_{\rm B}/N_0 }{1\,{\rm dB}}$$
@@ Line 19: / Line 19: @@
 :$$C \approx \hspace{0.15cm} \left\{ \begin{array}{c} 1 - {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} x} \\ \\ 0 \end{array} \right.\quad \begin{array}{*{1}c} {\rm for\hspace{0.15cm}} x > 0, \\ \\{\rm for\hspace{0.15cm}} x < 0. \end{array}$$
-*If&nbsp; $R < C$ holds, then a code can be found which leads to the error probability "zero" for infinitely long blocks&nbsp; $(n → ∞)$&nbsp;. What this code looks like is not determined by the channel coding theorem and does not matter for this exercise.
+*For&nbsp; $R < C$,&nbsp; a code can be found which leads to the error probability&nbsp; "zero"&nbsp; for infinitely long blocks&nbsp; $(n → ∞)$.&nbsp;
 Drawn into the graph as dots are the characteristics of established coding systems:
-*The dots&nbsp; $\rm X$,&nbsp; $\rm Y$&nbsp; and&nbsp; $\rm Z$&nbsp; mark three Hamming codes of different code lengths, namely with&nbsp; $n = 7$,&nbsp; $n = 15$&nbsp; and&nbsp; $n = 31$.
+*The dots&nbsp; $\rm X$,&nbsp; $\rm Y$,&nbsp; $\rm Z$&nbsp; mark three Hamming codes of different code lengths, namely with&nbsp; $n = 7$,&nbsp; $n = 15$&nbsp; and&nbsp; $n = 31$.
-*The coding system&nbsp; $\rm W$&nbsp; is characterized by the parameters&nbsp; $R = 0.5$&nbsp; and&nbsp; $10 \ - \lg {E_{\rm B}/N_0} = 3 {\rm dB}$&nbsp;.
+*The plotted points of real transmission systems,&nbsp; on the other hand,&nbsp; result under the assumption&nbsp; $\rm BER = 10^{-5}$.
@@ Line 42: / Line 44: @@
 ===Questions===
 <quiz display=simple>
-{Which of the points belong to which Hamming code? ''Note:'' &nbsp; The graph was created for&nbsp; $\rm BER = 10^{-5}$&nbsp;.
+{Which of the points belong to which Hamming code? Note: &nbsp; The graph was created for&nbsp; $\rm BER = 10^{-5}$&nbsp;.
 |type="[]"}
 + $\rm X$&nbsp; denotes the&nbsp; $(7, 4, 3)$ Hamming code.

@@ Line 5: / Line 5: @@
 }}
-[[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&nbsp; $($green$)$&nbsp; and code rates&nbsp; $($red dots$)$&nbsp; of some established systems]]
 The graph shows the maximum allowable code rate&nbsp; $R < C$&nbsp; according to Shannon's&nbsp; [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding#Channel_coding_theorem_and_channel_capacity|"channel coding theorem"]]:
@@ Line 88: / Line 88: @@
 \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}}

@@ Line 5: / Line 5: @@
 }}
-[[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&nbsp; $R < C$&nbsp; according to Shannon's&nbsp; [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding#Channel_coding_theorem_and_channel_capacity|"channel coding theorem"]]:

@@ Line 33: / Line 33: @@
 Hints:
-*This exercise belongs to the topic of chapter&nbsp; [[Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding|"information theoretical limits of channel coding"]].
+*This exercise belongs to the topic of chapter&nbsp; [[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&nbsp; $\rm BER =0$.
 *The plotted points of real transmission systems, on the other hand, result under the assumption&nbsp; $\rm BER = 10^{-5}$.

← Older revision		Revision as of 21:21, 14 August 2022
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−	{{quiz-Header\|Buchseite=~~Kanalcodierung~~/~~Informationstheoretische Grenzen der Kanalcodierung~~	+	{{quiz-Header\|Buchseite=Channel_Coding/Information_Theoretical_Limits_of_Channel_Coding

← Older revision		Revision as of 11:21, 11 May 2022
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−	[[Category:Channel Coding: Exercises\|^1.7 ~~Informationstheoretische Grenzen~~^]]	+	[[Category:Channel Coding: Exercises\|^1.7 Channel Capacity^]]

← Older revision	Revision as of 10:15, 11 May 2022
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← Older revision		Revision as of 14:26, 28 May 2021
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	:$$C \approx \hspace{0.15cm} \left\{ \begin{array}{c} 1 - {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} x} \\ \\ 0 \end{array} \right.\quad \begin{array}{*{1}c} {\rm f\ddot{u}r\hspace{0.15cm}} x > 0, \\ \\{\rm f\ddot{u}r\hspace{0.15cm}} x < 0. \end{array}$$		:$$C \approx \hspace{0.15cm} \left\{ \begin{array}{c} 1 - {\rm e}^{- 0.4 \hspace{0.05cm} \cdot \hspace{0.05cm} x} \\ \\ 0 \end{array} \right.\quad \begin{array}{*{1}c} {\rm f\ddot{u}r\hspace{0.15cm}} x > 0, \\ \\{\rm f\ddot{u}r\hspace{0.15cm}} x < 0. \end{array}$$

−	*Gilt  $R < C$, so kann ein Code gefunden werden, der bei unendlich langen Blöcken  $(n → ∞)$  zur Fehlerwahrscheinlichkeit ~~&bdquo;~~Null" führt. Wie dieser Code aussieht, ist durch das Kanalcodierungstheorem nicht festgelegt und spielt für diese Aufgabe auch keine Rolle.	+	*Gilt  $R < C$, so kann ein Code gefunden werden, der bei unendlich langen Blöcken  $(n → ∞)$  zur Fehlerwahrscheinlichkeit "Null" führt. Wie dieser Code aussieht, ist durch das Kanalcodierungstheorem nicht festgelegt und spielt für diese Aufgabe auch keine Rolle.