Aufgaben:Exercise 4.9: Higher-Level Modulation - Revision history

Guenter at 14:28, 4 November 2021

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Guenter at 14:24, 4 November 2021

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Guenter at 14:03, 4 November 2021

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Noah at 15:41, 2 November 2021

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Noah at 15:40, 2 November 2021

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Noah at 15:40, 2 November 2021

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Noah at 13:16, 12 October 2021

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Guenter at 13:13, 7 October 2021

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Noah at 11:58, 4 October 2021

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Guenter at 15:30, 27 September 2021

2021-09-27T15:30:24Z

@@ Line 104: / Line 104: @@
 In the graph, the two&nbsp; "8&ndash;ASK"nbsp; systems are also drawn as dots according to propositions 4 and 5.
-* The purple dot is above the &nbsp;$C_{\rm 8&ndash;ASK}$ curve &nbsp; &#8658; &nbsp; $R = 2.5$ and $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$ are not enough to decode the 8&ndash;ASK without errors &nbsp; &#8658; &nbsp; $R > C{\rm 8&ndash;ASK}$ &nbsp; &#8658; &nbsp; the channel coding theorem is not satisfied &nbsp; &#8658; &nbsp; answer 4 is wrong.
+* The purple dot is above the &nbsp;$C_{\rm 8&ndash;ASK}$&nbsp; curve &nbsp; &#8658; &nbsp; $R = 2.5$ and $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$ are not enough to decode the&nbsp; "8&ndash;ASK"&nbsp; without errors &nbsp; &#8658; &nbsp; $R > C_{\rm 8&ndash;ASK}$ &nbsp; &#8658; &nbsp; channel coding theorem is not satisfied &nbsp; &#8658; &nbsp; answer 4 is wrong.
-* However, if we reduce the code rate to $R = 2 < C_{\rm 8&ndash;ASK}$  according to the yellow dot for the same $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$, the channel coding theorem is satisfied  &nbsp; &#8658; &nbsp; Answer 5 is correct.
+* However, if we reduce the code rate to&nbsp; $R = 2 < C_{\rm 8&ndash;ASK}$&nbsp;  for the same&nbsp; $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$&nbsp; according to the yellow dot,&nbsp; the channel coding theorem is satisfied  &nbsp; &#8658; &nbsp; answer 5 is correct.
 {{ML-Fuß}}

@@ Line 74: / Line 74: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Proposition 2</u> is correct, as shown by the calculation for &nbsp;$10 \cdot \lg (E_{\rm S}/{N_0}) = 15 \ \rm dB$ &nbsp;&nbsp;&#8658; &nbsp; $E_{\rm S}/{N_0} = 31.62$ zeigt:
+'''(1)'''&nbsp; <u>Proposition 2</u>&nbsp; is correct,&nbsp; as shown by the calculation for &nbsp;$10 \cdot \lg (E_{\rm S}/{N_0}) = 15 \ \rm dB$ &nbsp;&nbsp;&#8658; &nbsp; $E_{\rm S}/{N_0} = 31.62$:
-:$$C_2(15\hspace{0.1cm}{\rm dB}) = {1}/{2} \cdot  {\rm log}_2 \hspace{0.1cm} ( 1 +   2 \cdot 31.62 ) = {1}/{2} \cdot  {\rm log}_2 \hspace{0.1cm} ( 64.25 ) \approx 3\,{\rm bit/Kanalzugriff}\hspace{0.05cm}. $$
+:$$C_2(15\hspace{0.1cm}{\rm dB}) = {1}/{2} \cdot  {\rm log}_2 \hspace{0.1cm} ( 1 +   2 \cdot 31.62 ) = {1}/{2} \cdot  {\rm log}_2 \hspace{0.1cm} ( 64.25 ) \approx 3\,{\rm bit/use}\hspace{0.05cm}. $$
 *The other two proposed solutions provide the following numerical values:
-:$$C_3(15\hspace{0.1cm}{\rm dB}) \  =  \  {\rm log}_2 \hspace{0.1cm} ( 1 +   31.62 ) \approx 5.03\,{\rm bit/Kanalzugriff}\hspace{0.05cm},$$
+:$$C_3(15\hspace{0.1cm}{\rm dB}) \  =  \  {\rm log}_2 \hspace{0.1cm} ( 1 +   31.62 ) \approx 5.03\,{\rm bit/use}\hspace{0.05cm},$$
-:$$ C_1(15\hspace{0.1cm}{\rm dB}) \  =  \   C_3/2 \approx 2.51\,{\rm bit/Kanalzugriff}\hspace{0.05cm}.$$
+:$$ C_1(15\hspace{0.1cm}{\rm dB}) \  =  \   C_3/2 \approx 2.51\,{\rm bit/use}\hspace{0.05cm}.$$
-*The proposed solution 3 corresponds to the case of ''two independent Gaussian channels'' with half transmit power per channel.
+*The proposed solution 3 corresponds to the case of&nbsp; "two independent Gaussian channels"&nbsp; with half transmission power per channel.
+'''(2)'''&nbsp;<u>Proposed solutions 1, 2 and 4</u>&nbsp; are correct:
-'''(2)'''&nbsp;<u>Proposed solutions 1, 2 and 4</u> are correct:
+*If one would replace &nbsp;$E_{\rm S}$&nbsp; by &nbsp;$E_{\rm B}$,&nbsp; then the statement 3 would be also correct.
-*If one would replace &nbsp;$E_{\rm S}$&nbsp; by &nbsp;$E_{\rm B}$&nbsp;, then also the statement 3 would be correct.
+*For &nbsp;$E_{\rm B}/{N_0} < \ln (2)$ &nbsp; &rArr;  &nbsp; $C_{\rm Gaussian} &equiv; 0$&nbsp; is valid,&nbsp;  and therefore also &nbsp;$C_{\rm BPSK} &equiv; 0$.
 [[File:EN_Inf_A_4_9e_v2.png|right|frame|Channel capacity limits for <br>BPSK, 4–ASK and 8–ASK]]
-'''(4)'''&nbsp; <u>Proposed solutions 1, 2 and 5</u> are correct:
+<br><br>
-*From the brown curve, one can see the correctness of the first two statements.
+'''(4)'''&nbsp; <u>Proposed solutions 1, 2 and 5</u>&nbsp; are correct:
-*The 8&ndash;PSK with I&ndash; and Q&ndash;components &ndash; i.e. with&nbsp; $K = 2$&nbsp; dimensions &ndash; lies slightly above the brown curve for small &nbsp;$E_{\rm S}/{N_0}$ values &nbsp; &rArr; &nbsp; the answer 3 is incorrect.
+*From the brown curve,&nbsp; one can see the correctness of the first two statements.
-In the graph, the two 8&ndash;ASK&ndash;systems are also drawn as dots according to propositions 4 and 5.
+In the graph, the two&nbsp; "8&ndash;ASK"nbsp; systems are also drawn as dots according to propositions 4 and 5.
-* The purple dot is above the &nbsp;$C_{\rm 8&ndash;ASK}$ curve &nbsp; &#8658; &nbsp; $R = 2.5$ and $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$ are not enough to decode the 8&ndash;ASK without errors &nbsp; &#8658; &nbsp; $R > C$ &nbsp; &#8658; &nbsp; the channel coding theorem is not satisfied &nbsp; &#8658; &nbsp; answer 4 is wrong.
+* The purple dot is above the &nbsp;$C_{\rm 8&ndash;ASK}$ curve &nbsp; &#8658; &nbsp; $R = 2.5$ and $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$ are not enough to decode the 8&ndash;ASK without errors &nbsp; &#8658; &nbsp; $R > C{\rm 8&ndash;ASK}$ &nbsp; &#8658; &nbsp; the channel coding theorem is not satisfied &nbsp; &#8658; &nbsp; answer 4 is wrong.
 * However, if we reduce the code rate to $R = 2 < C_{\rm 8&ndash;ASK}$  according to the yellow dot for the same $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$, the channel coding theorem is satisfied  &nbsp; &#8658; &nbsp; Answer 5 is correct.

@@ Line 5: / Line 5: @@
 [[File:EN_Inf_A_4_9.png|right|frame|Some channel capacity curves]]
 The graph shows AWGN channel capacity curves over the&nbsp; $10 \cdot \lg (E_{\rm S}/{N_0})$:
-* $C_\text{Gauß}$:&nbsp; &nbsp; Shannon's boundary curve,
+* $C_\text{Gaussian}$:&nbsp; &nbsp; Shannon's boundary curve,
-* $C_\text{BPSK}$:&nbsp; &nbsp; valid for ''Binary Phase Shift Keying''.
+* $C_\text{BPSK}$:&nbsp; &nbsp; valid for&nbsp; "Binary Phase Shift Keying".
@@ Line 19: / Line 16: @@
 Hints:
 *The task belongs to the chapter&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertdiskretem_Eingang|AWGN channel capacity with discrete value input]].
-*Reference is made in particular to the page&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertdiskretem_Eingang#The_channel_capacity_.7F.27.22.60UNIQ-MathJax81-QINU.60.22.27.7F_as_a_function_of_.7F.27.22.60UNIQ-MathJax82-QINU.60.22.27.7F|The channel capacity $C$ as a function of $E_{\rm S}/{N_0}$]].
+*Reference is made in particular to the page&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertdiskretem_Eingang#The_channel_capacity_.7F.27.22.60UNIQ-MathJax81-QINU.60.22.27.7F_as_a_function_of_.7F.27.22.60UNIQ-MathJax82-QINU.60.22.27.7F|Channel capacity&nbsp; $C$&nbsp; as a function of&nbsp; $E_{\rm S}/{N_0}$]].
-*Since the results are to be given in "bit", "log" &nbsp;&#8658;&nbsp; "log<sub>2</sub>" is used in the equations.
+*Since the results are to be given in&nbsp; "bit" &nbsp; &rArr; &nbsp;  "log" &nbsp;&#8658;&nbsp; "log<sub>2</sub>" is used in the equations.
-*The modulation methods mentioned in the questions are described in terms of their signal space constellation:
+*The modulation methods mentioned in the questions are described in terms of their signal space constellation&nbsp; <br>(see lower graph).
 [[File:EN_Inf_A_4_9_Zusatz.png|right|frame|Proposed signal space constellations]]
-''Notes on nomenclature:'':
+'''Notes on nomenclature:'''
-*In the literature,"BPSK" is sometimes also referred to as "2&ndash;ASK"
+*In the literature,&nbsp; "BPSK" is sometimes also referred to as&nbsp; "2&ndash;ASK":
 :$$x &#8712; X = \{+1,\ -1\}.$$
-*In contrast, in our learning tutorial $\rm LNTwww$ we understand as "ASK" the unipolar case
+*In contrast,&nbsp; in our learning tutorial we understand as&nbsp; "ASK"&nbsp; the unipolar case:
 :$$x &#8712; X = \{0,\ 1 \}.$$
 *Therefore, according to our nomenclature:
-:$$C_\text{AK} < C_\text{BPSK}$$
+:$$C_\text{ASK} < C_\text{BPSK}$$
-This fact is irrelevant for the solution of the present problem.
+But:&nbsp; This fact is irrelevant for the solution of the present problem.
@@ Line 40: / Line 37: @@
 <quiz display=simple>
-{What equation underlies the Shannon boundary curve &nbsp;$C_{\rm Gauß}$&nbsp; ?
+{What equation underlies Shannon's boundary curve &nbsp;$C_{\rm Gaussian}$?
 |type="[]"}
-- &nbsp; $C_{\rm Gauß}  = C_1= {1}/{2} \cdot  {\rm log}_2 \hspace{0.1cm} ( 1 + E_{\rm S}/{N_0})$ ,
+- &nbsp; $C_{\rm Gaussian}  = C_1= {1}/{2} \cdot  {\rm log}_2 \hspace{0.1cm} ( 1 + E_{\rm S}/{N_0})$ ,
-+ &nbsp; $C_{\rm Gauß}  = C_2= {1}/{2} \cdot  {\rm log}_2 \hspace{0.1cm} ( 1 + 2 \cdot E_{\rm S}/{N_0})$ ,
++ &nbsp; $C_{\rm Gaussian}  = C_2= {1}/{2} \cdot  {\rm log}_2 \hspace{0.1cm} ( 1 + 2 \cdot E_{\rm S}/{N_0})$ ,
-- &nbsp; $C_{\rm Gauß}  = C_3=   {\rm log}_2 \hspace{0.1cm} ( 1 + E_{\rm S}/{N_0})$ .
+- &nbsp; $C_{\rm Gaussian}  = C_3=   {\rm log}_2 \hspace{0.1cm} ( 1 + E_{\rm S}/{N_0})$ .
-{Which statements are true for the green curve &nbsp;$C_{\rm BPSK}$&nbsp;?
+{Which statements are true for the green curve &nbsp;$(C_{\rm BPSK})$?
 |type="[]"}
 + $C_{\rm BPSK}$&nbsp; cannot be given in closed form.
 + $C_{\rm BPSK}$&nbsp; is greater than zero if &nbsp;$E_{\rm S}/{N_0} > 0$&nbsp; is assumed.
-- For &nbsp;$E_{\rm S}/{N_0} < \ln (2)$&nbsp;,&nbsp;$C_{\rm BPSK} &equiv; 0$.
+- For &nbsp;$E_{\rm S}/{N_0} < \ln (2)$ &nbsp; &rArr; &nbsp; $C_{\rm BPSK} &equiv; 0$.
-+ In the whole range &nbsp;$C_{\rm BPSK} < C_{\rm Gauß} $ is valid.
++ In the whole range &nbsp;$C_{\rm BPSK} < C_{\rm Gaussian} $&nbsp; is valid.
-{Which statements are true for the red curve &nbsp;$C_{\rm red}$&nbsp;?
+{Which statements are true for the red curve &nbsp;$(C_{\rm red})$?
 |type="[]"}
-- For the associated random variable &nbsp;$X$&nbsp; gilt &nbsp;$M_X = |X| = 2$.
+- For the associated random variable &nbsp;$X$&nbsp; holds &nbsp;$M_X = |X| = 2$.
-+ For the associated random variable &nbsp;$X$&nbsp; gilt &nbsp;$M_X = |X| = 4$.
++ For the associated random variable &nbsp;$X$&nbsp; holds &nbsp;$M_X = |X| = 4$.
-+ $C_{\rm red}$&nbsp; is simultaneously the channel capacity of the 4&ndash;ASK.
++ $C_{\rm red}$&nbsp; is simultaneously the channel capacity of the&nbsp; "4&ndash;ASK".
-- $C_{\rm red}$&nbsp; is simultaneously the channel capacity of the 4&ndash;QAM.
+- $C_{\rm red}$&nbsp; is simultaneously the channel capacity of the&nbsp; "4&ndash;QAM".
 + For all &nbsp;$E_{\rm S}/{N_0} > 0$&nbsp; &nbsp;$C_{\rm red}$&nbsp; is between "green" and "brown".
-{Which statements are true for the brown curve &nbsp;$C_{\rm brown}$&nbsp; ? &nbsp;($p_{\rm B}$: &nbsp; bit error probability)
+{Which statements are true for the brown curve &nbsp;$(C_{\rm brown})$? <br>Note:&nbsp; $p_{\rm B}$&nbsp; denotes the bit error probability here.
 |type="[]"}
-+ For the associated random variable &nbsp;$X$&nbsp;, &nbsp;$M_X = |X| = 8$.
++ For the associated random variable &nbsp;$X$ &nbsp; &rArr; &nbsp; $M_X = |X| = 8$.
-+ $C_{\rm brown}$&nbsp; is simultaneously the channel capacity of the 8&ndash;ASK.
++ $C_{\rm brown}$&nbsp; is simultaneously the channel capacity of the&nbsp; "8&ndash;ASK".
-- $C_{\rm brown}$&nbsp; is simultaneously the channel capacity of the 8&ndash;PSK.
+- $C_{\rm brown}$&nbsp; is simultaneously the channel capacity of the&nbsp; "8&ndash;PSK".
-- $p_{\rm B} &equiv; 0$&nbsp; is possible with 8&ndash;ASK,&nbsp; $R = 2.5$&nbsp; and&nbsp; $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$&nbsp;.
+- $p_{\rm B} &equiv; 0$&nbsp; is possible with&nbsp; "8&ndash;ASK",&nbsp; $R = 2.5$&nbsp; and&nbsp; $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$.
-+ $p_{\rm B} &equiv; 0$&nbsp; is possible with 8&ndash;ASK,&nbsp; $R = 2$&nbsp; and&nbsp; $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$&nbsp;.
++ $p_{\rm B} &equiv; 0$&nbsp; is possible with&nbsp; "8&ndash;ASK",&nbsp; $R = 2.0$&nbsp; and&nbsp; $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$&nbsp;.

@@ Line 9: / Line 9: @@
-The two other curves &nbsp;$C_\text{rot}$&nbsp; and &nbsp;$C_\text{braun}$&nbsp; should be analyzed and assigned to possible modulation schemes in subtasks&nbsp; '''(3)'''&nbsp; and&nbsp; '''(4)'''&nbsp;.
+The two other curves &nbsp;$C_\text{red}$&nbsp; and &nbsp;$C_\text{brown}$&nbsp; should be analyzed and assigned to possible modulation schemes in subtasks&nbsp; '''(3)'''&nbsp; and&nbsp; '''(4)'''&nbsp;.
@@ Line 54: / Line 54: @@
 + In the whole range &nbsp;$C_{\rm BPSK} < C_{\rm Gauß} $ is valid.
-{Which statements are true for the red curve &nbsp;$C_{\rm rot}$&nbsp;?
+{Which statements are true for the red curve &nbsp;$C_{\rm red}$&nbsp;?
 |type="[]"}
 - For the associated random variable &nbsp;$X$&nbsp; gilt &nbsp;$M_X = |X| = 2$.
 + For the associated random variable &nbsp;$X$&nbsp; gilt &nbsp;$M_X = |X| = 4$.
-+ $C_{\rm rot}$&nbsp; is simultaneously the channel capacity of the 4&ndash;ASK.
++ $C_{\rm red}$&nbsp; is simultaneously the channel capacity of the 4&ndash;ASK.
-- $C_{\rm rot}$&nbsp; is simultaneously the channel capacity of the 4&ndash;QAM.
+- $C_{\rm red}$&nbsp; is simultaneously the channel capacity of the 4&ndash;QAM.
-+ For all &nbsp;$E_{\rm S}/{N_0} > 0$&nbsp; &nbsp;$C_{\rm rot}$&nbsp; is between "grün" and "braun".
++ For all &nbsp;$E_{\rm S}/{N_0} > 0$&nbsp; &nbsp;$C_{\rm red}$&nbsp; is between "green" and "brown".
-{Which statements are true for the brown curve &nbsp;$C_{\rm braun}$&nbsp; ? &nbsp;($p_{\rm B}$: &nbsp; bit error probability)
+{Which statements are true for the brown curve &nbsp;$C_{\rm brown}$&nbsp; ? &nbsp;($p_{\rm B}$: &nbsp; bit error probability)
 |type="[]"}
 + For the associated random variable &nbsp;$X$&nbsp;, &nbsp;$M_X = |X| = 8$.
-+ $C_{\rm braun}$&nbsp; is simultaneously the channel capacity of the 8&ndash;ASK.
++ $C_{\rm brown}$&nbsp; is simultaneously the channel capacity of the 8&ndash;ASK.
-- $C_{\rm braun}$&nbsp; is simultaneously the channel capacity of the 8&ndash;PSK.
+- $C_{\rm brown}$&nbsp; is simultaneously the channel capacity of the 8&ndash;PSK.
 - $p_{\rm B} &equiv; 0$&nbsp; is possible with 8&ndash;ASK,&nbsp; $R = 2.5$&nbsp; and&nbsp; $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$&nbsp;.
 + $p_{\rm B} &equiv; 0$&nbsp; is possible with 8&ndash;ASK,&nbsp; $R = 2$&nbsp; and&nbsp; $10 \cdot \lg (E_{\rm S}/{N_0}) = 10 \ \rm dB$&nbsp;.
@@ Line 95: / Line 95: @@
 '''(3)'''&nbsp; <u>Statements 2, 3 and 5</u> are correct::
-*The red curve &nbsp;$C_{\rm rot}$&nbsp; is always above &nbsp;$C_{\rm BPSK}$&nbsp;, but below &nbsp;$C_{\rm braun}$&nbsp; and the Shannon boundary curve &nbsp;$C_{\rm Gauß}$.
+*The red curve &nbsp;$C_{\rm red}$&nbsp; is always above &nbsp;$C_{\rm BPSK}$&nbsp;, but below &nbsp;$C_{\rm brown}$&nbsp; and the Shannon boundary curve &nbsp;$C_{\rm Gauß}$.
 *The statements also hold if for certain &nbsp;$E_{\rm S}/{N_0}$ values curves are indistinguishable within the character precision.
-*From the limit &nbsp;$C_{\rm rot}= 2 \ \rm bit/channel use$&nbsp; for &nbsp;$E_{\rm S}/{N_0}  &#8594; &#8734;$&nbsp;, the symbol range &nbsp;$M_X = |X| = 4$.
+*From the limit &nbsp;$C_{\rm red}= 2 \ \rm bit/channel use$&nbsp; for &nbsp;$E_{\rm S}/{N_0}  &#8594; &#8734;$&nbsp;, the symbol range &nbsp;$M_X = |X| = 4$.
 *Thus, the red curve describes the 4&ndash;ASK.&nbsp; $M_X = |X| = 2$&nbsp; would apply to the BPSK.
-*The 4&ndash;QAM leads exactly to the same final value "2 bit/channel use".&nbsp; For small &nbsp;$E_{\rm S}/{N_0}$ values, however, the channel capacity &nbsp;$C_{\rm 4&ndash;QAM}$&nbsp; is above the red curve, since &nbsp;$C_{\rm rot}$&nbsp; is bounded by the Gaussian boundary curve &nbsp;$C_2$&nbsp;, but $C_{\rm 4&ndash;QAM}$&nbsp; is bounded by &nbsp;$C_3$.
+*The 4&ndash;QAM leads exactly to the same final value "2 bit/channel use".&nbsp; For small &nbsp;$E_{\rm S}/{N_0}$ values, however, the channel capacity &nbsp;$C_{\rm 4&ndash;QAM}$&nbsp; is above the red curve, since &nbsp;$C_{\rm red}$&nbsp; is bounded by the Gaussian boundary curve &nbsp;$C_2$&nbsp;, but $C_{\rm 4&ndash;QAM}$&nbsp; is bounded by &nbsp;$C_3$.

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Inf_A_4_9.png|right|frame|Some channel capacitance curves]]
+[[File:EN_Inf_A_4_9.png|right|frame|Some channel capacity curves]]
 The graph shows AWGN channel capacity curves over the&nbsp; $10 \cdot \lg (E_{\rm S}/{N_0})$:
 * $C_\text{Gauß}$:&nbsp; &nbsp; Shannon's boundary curve,

@@ Line 108: / Line 108: @@
-[[File:EN_Inf_A_4_9e.png|right|frame|Kanalkapazitätsgrenzen für <br>BPSK, 4–ASK und 8–ASK]]
+[[File:EN_Inf_A_4_9e_v2.png|right|frame|Kanalkapazitätsgrenzen für <br>BPSK, 4–ASK und 8–ASK]]
 '''(4)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1, 2 und 5</u>:
 *Aus dem braunen Kurvenverlauf erkennt man die Richtigkeit der beiden ersten Aussagen.

← Older revision		Revision as of 11:58, 4 October 2021
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−	{{quiz-Header\|Buchseite=~~Informationstheorie~~/~~AWGN–Kanalkapazität bei wertdiskretem Eingang~~	+	{{quiz-Header\|Buchseite=Information_Theory/AWGN_Channel_Capacity_for_Discrete_Input
	}}		}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID2952__Inf_A_4_9.png|right|frame|Einige Kanalkapazitätskurven]]
+[[File:EN_Inf_A_4_9.png|right|frame|Einige Kanalkapazitätskurven]]
 Die Grafik zeigt  AWGN&ndash;Kanalkapazitätskurven über der Abszisse&nbsp; $10 \cdot \lg (E_{\rm S}/{N_0})$:
 * $C_\text{Gauß}$:&nbsp; &nbsp; Shannonsche Grenzkurve,
@@ Line 25: / Line 25: @@
-[[File:P_ID2953__Inf_A_4_9_Zusatz.png|right|frame|Vorgeschlagene Signalraumkonstellationen]]
+[[File:EN_Inf_A_4_9_Zusatz.png|right|frame|Vorgeschlagene Signalraumkonstellationen]]
 ''Anmerkungen zur Nomenklatur'':
@@ Line 108: / Line 108: @@
-[[File:P_ID2954__Inf_A_4_9e.png|right|frame|Kanalkapazitätsgrenzen für <br>BPSK, 4–ASK und 8–ASK]]
+[[File:EN_Inf_A_4_9e.png|right|frame|Kanalkapazitätsgrenzen für <br>BPSK, 4–ASK und 8–ASK]]
 '''(4)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1, 2 und 5</u>:
 *Aus dem braunen Kurvenverlauf erkennt man die Richtigkeit der beiden ersten Aussagen.