Aufgaben:Exercise 4.7Z: About the Water Filling Algorithm - Revision history

Guenter at 13:10, 7 October 2021

2021-10-07T13:10:32Z

Guenter at 14:21, 4 October 2021

2021-10-04T14:21:40Z

Guenter at 13:33, 4 October 2021

2021-10-04T13:33:11Z

Noah at 11:58, 4 October 2021

2021-10-04T11:58:16Z

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Noah at 12:31, 1 October 2021

2021-10-01T12:31:53Z

Guenter at 15:45, 6 September 2021

2021-09-06T15:45:53Z

Guenter at 09:29, 16 August 2021

2021-08-16T09:29:21Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.7Z: Zum Water–Filling–Algorithmus to Aufgaben:Exercise 4.7Z: About the Water Filling Algorithm

2021-08-16T09:24:48Z

Guenter moved page Aufgabe 4.7Z: Zum Water–Filling–Algorithmus to Exercise 4.7Z: About the Water Filling Algorithm

Javier: Text replacement - "„" to """

2021-05-28T14:38:32Z

Text replacement - "„" to """

Javier: Text replacement - "”" to """

2021-05-28T14:32:53Z

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@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Inf_T_4_2_S4d_v2.png|right|frame|Water-filling principle&nbsp; $(K = 4)$]]
+[[File:EN_Inf_Z_4_7.png|right|frame|Water-filling principle&nbsp; $(K = 4)$]]
 We consider&nbsp; $K$&nbsp; parallel Gaussian channels&nbsp; $\rm (AWGN)$&nbsp; with different interference powers&nbsp; $\sigma_k^2$,&nbsp; where&nbsp;  $1 \le k \le K$&nbsp;.&nbsp; The graph illustrates this constellation using&nbsp; $K = 4$&nbsp; as an example.

@@ Line 75: / Line 75: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  <u>Proposed solutions 2 and 3</u> are correct:
+'''(1)'''&nbsp;  <u>Proposed solutions 2 and 3</u>&nbsp; are correct:
-*According to the explanations in the theory section, "Water&ndash;Filling" &nbsp; &#8658; &nbsp; <u>Proposal 2</u> is to be applied when unequal conditions exist.
+*According to the explanations in the theory section&nbsp; "Water&ndash;Filling" &nbsp; &#8658; &nbsp; <u>Proposal 2</u>&nbsp; is to be applied when unequal conditions exist.
-* However, <u>solution proposal 3</u> is also correct: &nbsp; If the channels are equally good, there is nothing to prevent all&nbsp; $K$&nbsp; channels from being filled with the same power &nbsp; &#8658; &nbsp; &nbsp;$P_1 = P_2 =$&nbsp; ...&nbsp; $= P_K = P_X/K$&nbsp;.
+* However,&nbsp; <u>solution proposal 3</u>&nbsp; is also correct: &nbsp; If the channels are equally good, there is nothing to prevent all&nbsp; $K$&nbsp; channels from being filled with the same power &nbsp; &#8658; &nbsp; &nbsp;$P_1 = P_2 =$&nbsp; ...&nbsp; $= P_K = P_X/K$.
@@ Line 89: / Line 89: @@
-[[File:EN_Inf_Z_4_7b.png|right|frame|Best possible distribution of transmission power&nbsp; $P_X = 10$]]
+[[File:EN_Inf_Z_4_7b.png|right|frame|Best possible distribution of transmit power&nbsp; $P_X = 10$]]
 '''(3)'''&nbsp;  According to the adjacent sketch, the following must apply:
 :$$P_2 = P_1 - (\sigma_2^2 - \sigma_1^2) = P_1 -3\hspace{0.3cm}\text{wobei }\hspace{0.3cm}P_1 + P_2 =  P_X = 10$$
@@ Line 98: / Line 98: @@
 \underline{P_1 = 6.5}\hspace{0.05cm},
 \hspace{0.3cm}\underline{P_2 = 3.5}\hspace{0.05cm}.$$
@@ Line 110: / Line 112: @@
-'''(5)'''&nbsp;  For&nbsp; $P_X = 3$&nbsp;, with the same power splitting&nbsp; $(P_1 = P_2 =1.5)$, we obtain:
+'''(5)'''&nbsp;  For&nbsp; $P_X = 3$, &nbsp; with the same power splitting&nbsp; $(P_1 = P_2 =1.5)$,&nbsp; we obtain:
 :$$I(X_1, X_2\hspace{0.05cm};\hspace{0.05cm}Y_1,  Y_2) ={1}/{2} \cdot  {\rm log}_2\hspace{0.05cm}\left ( 1 + \frac{1.5}{1} \right )
 +{1}/{2} \cdot  {\rm log}_2\hspace{0.05cm}\left ( 1 + \frac{1.5}{4} \right )
@@ Line 119: / Line 121: @@
 [[File:EN_Inf_Z_4_7e.png|right|frame|Best possible distribution of the transmit power&nbsp; $P_X = 3$]]
-'''(6)'''&nbsp;  According to the water&ndash;filling algorithm, the total available transmit power&nbsp; $P_X = 3$&nbsp; is now fully allocated to the first channel:
+'''(6)'''&nbsp;  According to the water&ndash;filling algorithm, the total  transmission power&nbsp; $P_X = 3$&nbsp; is now fully allocated to the first channel:
 :$${P_1 = 3}\hspace{0.05cm},
 \hspace{0.3cm}{P_2 = 0}\hspace{0.05cm}.$$
 *This gives for the channel capacity:
@@ Line 130: / Line 135: @@
 Further notes:
-*While for&nbsp; $P_X = 10$&nbsp; the difference between even and best power splitting was only&nbsp; $0.03$&nbsp; bit, for&nbsp; $P_X = 3$&nbsp; the difference is larger,nbsp;  $0.109$&nbsp; bit.
+*While for&nbsp; $P_X = 10$&nbsp; the difference between even and best power splitting was only&nbsp; $0.03$&nbsp; bit, for&nbsp; $P_X = 3$&nbsp; the difference is larger:&nbsp;  $0.109$&nbsp; bit.
 *For even larger&nbsp; $P_X > 10$&nbsp;, the difference between even and best power splitting becomes even smaller.

@@ Line 4: / Line 4: @@
 [[File:EN_Inf_T_4_2_S4d_v2.png|right|frame|Water-filling principle&nbsp; $(K = 4)$]]
-We consider&nbsp; $K$&nbsp; parallel Gaussian channels&nbsp; (AWGN)&nbsp;with different interference powers&nbsp; $\sigma_k^2$, where&nbsp;  $1 \le k \le K$&nbsp;.&nbsp; The graph illustrates this constellation using&nbsp; $K = 4$ as an example.
+We consider&nbsp; $K$&nbsp; parallel Gaussian channels&nbsp; $\rm (AWGN)$&nbsp; with different interference powers&nbsp; $\sigma_k^2$,&nbsp; where&nbsp;  $1 \le k \le K$&nbsp;.&nbsp; The graph illustrates this constellation using&nbsp; $K = 4$&nbsp; as an example.
-The transmit power in the individual channels is denoted by &nbsp;$P_k$&nbsp; , the sum of which must not exceed the specified value&nbsp;$P_X$&nbsp;:
+The transmission power in the individual channels is denoted by &nbsp;$P_k$,&nbsp; the sum of which must not exceed the specified value&nbsp;$P_X$&nbsp;:
 :$$P_1 +\text{...}\hspace{0.05cm}+ P_K = \hspace{0.1cm} \sum_{k= 1}^K
   \hspace{0.1cm}{\rm E} \left [ X_k^2\right ] \le P_{X} \hspace{0.05cm}.$$
@@ Line 12: / Line 12: @@
 :$$I(X_1,\text{...} \hspace{0.05cm}, X_K\hspace{0.05cm};\hspace{0.05cm}Y_1, \text{...}\hspace{0.05cm}, Y_K)
 =  1/2 \cdot \sum_{k= 1}^K  \hspace{0.1cm} {\rm log}_2 \hspace{0.1cm} ( 1 + \frac{P_k}{\sigma_k^2})\hspace{0.05cm},\hspace{0.5cm}
-{\rm Ergebnis\hspace{0.15cm} in \hspace{0.15cm} bit}
+{\rm Result\hspace{0.15cm} in \hspace{0.15cm} \text{"bit"} }
 \hspace{0.05cm}.$$
-The maximum for this is the channel capacity of the total system, where maximization refers to the division of the total power &nbsp;$P_X$&nbsp; among the individual channels:
+The maximum for this is the&nbsp; '''channel capacity of the total system''',&nbsp; where the maximization refers to the division of the total power &nbsp;$P_X$&nbsp; among the individual channels:
-:$$C_K(P_X) = \max_{P_k\hspace{0.05cm},\hspace{0.15cm}{\rm mit} \hspace{0.15cm}P_1 + ... \hspace{0.05cm}+ P_K = P_X} \hspace{-0.5cm} I(X_1, \text{...} \hspace{0.05cm}, X_K\hspace{0.05cm};\hspace{0.05cm}Y_1, \text{...}\hspace{0.05cm}, Y_K) \hspace{0.05cm}.$$
+:$$C_K(P_X) = \max_{P_k\hspace{0.05cm},\hspace{0.15cm}{\rm with} \hspace{0.15cm}P_1 + ... \hspace{0.05cm}+ P_K = P_X} \hspace{-0.5cm} I(X_1, \text{...} \hspace{0.05cm}, X_K\hspace{0.05cm};\hspace{0.05cm}Y_1, \text{...}\hspace{0.05cm}, Y_K) \hspace{0.05cm}.$$
-This maximization can be done with the water&ndash;filling algorithm shown in the above graph for&nbsp; $K = 4$&nbsp;.
+This maximization can be done with the&nbsp; "water&ndash;filling algorithm"&nbsp; shown in the above graph for&nbsp; $K = 4$.
 In the present exercise, this algorithm is to be applied, assuming the following:
 * Two parallel Gaussian channels &nbsp; &#8658; &nbsp; $K = 2$,
-* Normalized noise powers &nbsp; $\sigma_1^2 = 1$&nbsp; und&nbsp; $\sigma_2^2 = 4$,
+* normalized noise powers &nbsp; $\sigma_1^2 = 1$ &nbsp; and &nbsp; $\sigma_2^2 = 4$,
-* Normalized transmit powers &nbsp; $P_X = 10$&nbsp; and &nbsp; $P_X = 3$ respectively.
+* normalized transmission powers &nbsp; $P_X = 10$ &nbsp; and &nbsp; $P_X = 3$ respectively.
@@ Line 31: / Line 31: @@
 *The exercise belongs to the chapter&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang|AWGN channel capacity with continuous value input]].
 *Reference is made in particular to the page&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang#Parallel_Gaussian_channels|Parallel Gaussian Channels]].
-*Since the results are to be given in "bit", the logarithm to base&nbsp; $2$&nbsp; is used in the equations: &nbsp; $\log_2$.
+*Since the results are to be given in&nbsp; "bit",&nbsp; the logarithm to base&nbsp; $2$&nbsp; is used in the equations: &nbsp; $\log_2$.
@@ Line 40: / Line 40: @@
 {Which power allocation strategies are useful?
 |type="[]"}
-- A very noisy channel&nbsp; $k$&nbsp; $($with large noise power&nbsp;  $\sigma_k^2)$&nbsp; hould be allocated a large effective power&nbsp; $P_k$&nbsp;.
+- A very noisy channel&nbsp; $k$&nbsp; $($with large noise power&nbsp;  $\sigma_k^2)$&nbsp; should be allocated a large effective power&nbsp; $P_k$.
-+ A very noisy channel&nbsp; $k$&nbsp; $($with large noise power&nbsp;  $\sigma_k^2)$&nbsp;  hould be allocated a large effective power&nbsp; $P_k$&nbsp;.
++ A very noisy channel&nbsp; $k$&nbsp; $($with large noise power&nbsp;  $\sigma_k^2)$&nbsp;  should be assigned only a small useful power&nbsp; $P_k$.
-+ For equally good channels &nbsp; &#8658; &nbsp; $\sigma_1^2 = \text{...} = \sigma_K^2 = \sigma_N^2$&nbsp; $k$&nbsp; the power&nbsp; $P_k$&nbsp; gshould be evenly distributed.
++ For&nbsp; $K$&nbsp; equally good channels &nbsp; &#8658; &nbsp; $\sigma_1^2 = \text{...} = \sigma_K^2 = \sigma_N^2$&nbsp; the power&nbsp; $P_k$&nbsp; should be evenly distributed.
-{What is the mutual information obtained by distributing the transmit power &nbsp;$P_X = 10$&nbsp; equally to both channels &nbsp;  $(P_1= P_2 = 5)$?
+{What is the mutual information obtained by distributing the transmission power &nbsp;$P_X = 10$&nbsp; equally to both channels &nbsp;  $(P_1= P_2 = 5)$?
 |type="{}"}
 $I(X_1, X_2; Y_1, Y_2) \ = \ $ { 1.877 3% } $\ \rm bit$
-{Let&nbsp; $P_X = P_1 + P_2 = 10$ be further valid.&nbsp; Which optimal power distribution results according to the water&ndash;filling algorithm?
+{Let&nbsp; $P_X = P_1 + P_2 = 10$&nbsp; be further valid.&nbsp; Which optimal power distribution results according to the&nbsp; "water&ndash;filling algorithm"?
 |type="{}"}
 $P_1  \ = \ $ { 6.5 3% }

← Older revision		Revision as of 12:31, 1 October 2021
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−	{{quiz-Header\|Buchseite=~~Informationstheorie~~/~~AWGN–Kanalkapazität bei wertkontinuierlichem Eingang~~	+	{{quiz-Header\|Buchseite=Information_Theory/AWGN_Channel_Capacity_for_Continuous_Input
	}}		}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID2903__Inf_T_4_2_S4d.png|right|frame|Water–Filling–Prinzip&nbsp; $(K = 4)$]]
+[[File:EN_Inf_T_4_2_S4d_v2.png|right|frame|Water–Filling–Prinzip&nbsp; $(K = 4)$]]
 Wir betrachten&nbsp; $K$&nbsp; parallele Gaußsche Kanäle&nbsp; (AWGN)&nbsp; mit unterschiedlichen Störleistungen&nbsp; $\sigma_k^2$, wobei&nbsp;  $1 \le k \le K$&nbsp; gelten soll.&nbsp; Die Grafik verdeutlicht diese Konstellation am Beispiel&nbsp; $K = 4$.
@@ Line 90: / Line 90: @@
-[[File:P_ID2906__Inf_Z_4_7b_neu.png|right|frame|Bestmögliche Aufteilung der Sendeleistung&nbsp; $P_X = 10$]]
+[[File:EN_Inf_Z_4_7b.png|right|frame|Bestmögliche Aufteilung der Sendeleistung&nbsp; $P_X = 10$]]
 '''(3)'''&nbsp;  Entsprechend nebenstehender Skizze muss gelten:
 :$$P_2 = P_1 - (\sigma_2^2 - \sigma_1^2) = P_1 -3\hspace{0.3cm}\text{wobei }\hspace{0.3cm}P_1 + P_2 =  P_X = 10$$
@@ Line 119: / Line 119: @@
-[[File:P_ID2907__Inf_Z_4_7e_neu.png|right|frame|Bestmögliche Aufteilung der Sendeleistung&nbsp; $P_X = 3$]]
+[[File:EN_Inf_Z_4_7e.png|right|frame|Bestmögliche Aufteilung der Sendeleistung&nbsp; $P_X = 3$]]
 '''(6)'''&nbsp;  Entsprechend dem Water&ndash;Filling&ndash;Algorithmus wird die gesamte zur Verfügung stehende Sendeleistung&nbsp; $P_X = 3$&nbsp; nun vollständig dem ersten Kanal zugewiesen:
 :$${P_1 = 3}\hspace{0.05cm},

← Older revision		Revision as of 09:29, 16 August 2021
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−	[[Category:Information Theory: Exercises\|^4.2 AWGN ~~& kontinuierlicher Eingang~~^]]	+	[[Category:Information Theory: Exercises\|^4.2 AWGN and Value-Continuous Input^]]

← Older revision	Revision as of 09:24, 16 August 2021
(No difference)

@@ Line 32: / Line 32: @@
 *Die Aufgabe gehört zum  Kapitel&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang|AWGN–Kanalkapazität bei wertkontinuierlichem Eingang]].
 *Bezug genommen wird insbesondere auf die Seite&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang#Parallele_Gau.C3.9Fsche_Kan.C3.A4le|Parallele Gaußkanäle]].
-*Da die Ergebnisse in &bdquo;bit" angegeben werden sollen, wird in den Gleichungen  der Logarithmus zur Basis&nbsp; $2$&nbsp; verwendet: &nbsp; $\log_2$.
+*Da die Ergebnisse in "bit" angegeben werden sollen, wird in den Gleichungen  der Logarithmus zur Basis&nbsp; $2$&nbsp; verwendet: &nbsp; $\log_2$.
@@ Line 77: / Line 77: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 2 und 3</u>:
-*Nach den Ausführungen im  Theorieteil ist &bdquo;Water&ndash;Filling" &nbsp; &#8658; &nbsp; <u>Vorschlag 2</u> anzuwenden, wenn ungleiche Bedingungen vorliegen.
+*Nach den Ausführungen im  Theorieteil ist "Water&ndash;Filling" &nbsp; &#8658; &nbsp; <u>Vorschlag 2</u> anzuwenden, wenn ungleiche Bedingungen vorliegen.
 * Der <u>Lösungsvorschlag 3</u> ist aber ebenfalls richtig: &nbsp; Bei gleich guten Kanälen spricht nichts dagegen, alle&nbsp; $K$&nbsp; Kanäle mit gleicher Leistung &nbsp; &#8658; &nbsp; &nbsp;$P_1 = P_2 =$&nbsp; ...&nbsp; $= P_K = P_X/K$&nbsp; zu versorgen.