Aufgaben:Exercise 4.7: Several Parallel Gaussian Channels - Revision history

Guenter at 12:33, 4 October 2021

2021-10-04T12:33:08Z

Guenter at 12:23, 4 October 2021

2021-10-04T12:23:55Z

Guenter at 12:21, 4 October 2021

2021-10-04T12:21:01Z

Noah at 10:33, 4 October 2021

2021-10-04T10:33:30Z

Noah at 10:33, 4 October 2021

2021-10-04T10:33:11Z

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Noah at 12:31, 1 October 2021

2021-10-01T12:31:46Z

Guenter at 09:28, 16 August 2021

2021-08-16T09:28:39Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.7: Mehrere parallele Gaußkanäle to Aufgaben:Exercise 4.7: Several Parallel Gaussian Channels

2021-08-16T09:23:52Z

Guenter moved page Aufgabe 4.7: Mehrere parallele Gaußkanäle to Exercise 4.7: Several Parallel Gaussian Channels

Javier: Text replacement - "”" to """

2021-05-28T14:41:12Z

Text replacement - "”" to """

Javier: Text replacement - "„" to """

2021-05-28T14:37:18Z

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@@ Line 91: / Line 91: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; The parameter&nbsp; $K$&nbsp; is equal to the dimension of the signal space representation:
-* For <u>ASK and BPSK</u>,&nbsp; $\underline{K=1}$.
+* For&nbsp; <u>ASK and BPSK</u>,&nbsp; $\underline{K=1}$.
-* For <u> constellations 3 to 5</u>,however,&nbsp; $\underline{K=2}$&nbsp; (orthogonal modulation with cosine and sine).
+* For&nbsp; <u> constellations 3 to 5</u>,&nbsp; however,&nbsp; $\underline{K=2}$&nbsp; (orthogonal modulation with cosine and sine).
 '''(2)'''&nbsp; Correct is the <u>proposed solution 2</u>:
-*For each of the channels&nbsp; $(1 &#8804; k &#8804; K)$&nbsp;, the channel capacitance is &nbsp; $C = 1/2 \cdot \log_2 \ \big[1 + (P_X/K) /P_N) \big]$.&nbsp; The total capacitance is then larger by a factor of&nbsp; $K$&nbsp;:
+*For each of the channels&nbsp; $(1 &#8804; k &#8804; K)$,&nbsp; the channel capacitance is &nbsp; $C = 1/2 \cdot \log_2 \ \big[1 + (P_X/K) /P_N) \big]$.&nbsp;
 :$$C_K(P_X)  = \sum_{k= 1}^K \hspace{0.1cm}C_k = \frac{K}{2} \cdot  {\rm log}_2\hspace{0.05cm}\left ( 1 + \frac{P_X}{K \cdot P_N} \right )\hspace{0.05cm}.$$
-*The proposed solution 1 is too positive. This would apply when limiting the total power to&nbsp; $K &middot; P_X$&nbsp;.
+*The proposed solution 1 is too positive.&nbsp; This would apply when limiting the total power to&nbsp; $K &middot; P_X$&nbsp;.
-*Proposition 3 would imply that no capacity increase is achieved by using multiple independent channels, which is obviously not true.
+*Proposition 3 would imply that no capacity increase is achieved by using multiple independent channels,&nbsp; which is obviously not true.
-'''(3)'''&nbsp; The table shows the results for&nbsp; $K = 1$,&nbsp; $K = 2$&nbsp; and&nbsp; $K = 4$&nbsp; , and various signa&ndash;to&ndash;noise power ratios&nbsp; $\xi = P_X/P_N$. For&nbsp; $\xi = P_X/P_N = 15$&nbsp; (highlighted column), the result is:
+[[File:P_ID2902__Inf_A_4_7c.png|right|frame|Channel capacity&nbsp; $C_K$&nbsp; of&nbsp; $K$&nbsp; parallel Gaussian channels for different&nbsp; $\xi = P_X/P_N$]]
-[[File:P_ID2902__Inf_A_4_7c.png|right|frame|Channel capacity&nbsp; $C_K$&nbsp; of&nbsp; $K$&nbsp; parallel Gaussian channels for different&nbsp; $\xi = P_X/P_N$]]
+'''(3)'''&nbsp; The table shows the results for&nbsp; $K = 1$,&nbsp; $K = 2$&nbsp; and&nbsp; $K = 4$,&nbsp; and various signa&ndash;to&ndash;noise power ratios&nbsp; $\xi = P_X/P_N$. <br>For&nbsp; $\xi = P_X/P_N = 15$&nbsp; (highlighted column),&nbsp; the result is:
 *  $K=1$:&nbsp;&nbsp; $C_K = 1/2 &middot; \log_2 \ (16)\hspace{0.05cm}\underline{ = 2.000}$ bit,
 * $K=2$:&nbsp;&nbsp; $C_K = 1/2 &middot; \log_2 \ (8.5)\hspace{0.05cm}\underline{ = 3.087}$ bit,
@@ Line 113: / Line 115: @@
 *We now write the channel capacity using the natural logarithm and the abbreviation&nbsp; $\xi = P_X/P_N$:
 :$$C_{\rm nat}(\xi, K)  ={K}/{2} \cdot  {\rm ln}\hspace{0.05cm}\left ( 1 + {\xi}/{K} \right )\hspace{0.05cm}.$$
-*Then, for large values of&nbsp; $K$i.e., for small values of the quotient&nbsp; $\varepsilon =\xi/K$&nbsp;jholds:
+*Then,&nbsp; for large values of&nbsp; $K$&nbsp; i.e., for small values of the quotient&nbsp; $\varepsilon =\xi/K$&nbsp; holds:
 :$${\rm ln}\hspace{0.05cm}\left ( 1 + \varepsilon \right )=
 \varepsilon - \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{3} - ...
@@ Line 126: / Line 128: @@
 :$$C_{\rm bit}(\xi, K \rightarrow\infty)  = \frac{\xi}{2 \cdot {\rm ln}\hspace{0.1cm}(2)} =
 \frac{P_X/P_N}{2 \cdot {\rm ln}\hspace{0.1cm}(2)} \hspace{0.05cm}.$$
-*For smaller values of&nbsp; $K$&nbsp; , the result is always a smalle&nbsp; $C$&ndash;value, since
+*For smaller values of&nbsp; $K$,&nbsp; the result is always a smaller&nbsp; $C$&ndash;value, since
 :$$\frac{\xi}{2K} > \frac{\xi^2}{3K^2}\hspace{0.05cm}, \hspace{0.5cm}
 \frac{\xi^3}{4K^3} > \frac{\xi^4}{5K^4}  \hspace{0.05cm}, \hspace{0.5cm}  {\rm usw.}$$
-The last row of the table shows: &nbsp; With&nbsp;  $K = 4$&nbsp;one is still far from the theoretical maximum&nbsp; $($for $K &#8594; &#8734;)$&nbsp;for large &nbsp; $\xi$&ndash;values.
+The last row of the table shows: &nbsp; With&nbsp;  $K = 4$&nbsp; one is still far away from the theoretical maximum&nbsp; $($for $K &#8594; &#8734;)$&nbsp; for large &nbsp; $\xi$&ndash;values.

@@ Line 22: / Line 22: @@
 * [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying|Amplitude Shift Keying]]&nbsp; (ASK),
 * [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying|Binary Phase Shift Keying]]&nbsp; (BPSK),
-* [[Modulation_Methods/Further_AM_Variants#Quadratur.E2.80.93Amplitudenmodulation_.28QAM.29|Quadrature Amplitude Modulation]]&nbsp; (here: &nbsp; 4-QAM),
+* [[Modulation_Methods/Further_AM_Variants#Quadrature_Amplitude_Modulation_.28QAM.29|Quadrature Amplitude Modulation]]&nbsp; (here: &nbsp; 4-QAM),
 *[[Examples_of_Communication_Systems/Further_Developments_of_the_GSM#Enhanced_Data_Rates_for_GSM_Evolution|Phase Shift Keying]]&nbsp; (here: &nbsp; 8&ndash;PSK for GSM Evolution),
 * [[Modulation_Methods/Quadrature_Amplitude_Modulation#Other_signal_space_constellations|Combined ASK/PSK Modulation]]&nbsp; (here: &nbsp; 16-ASK/PSK).

@@ Line 4: / Line 4: @@
 [[File:P_ID2905__Inf_A_4_7_neu.png|right|frame|Signal space points in digital modulation]]
-The channel capacity of the AWGN channel with the indicator $Y = X + N$&nbsp; was given in the&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang#Channel_capacity_of_the_AWGN_channels|theory section]]&nbsp; as follows <br>(with the additional unit "bit"):
+The channel capacity of the AWGN channel with the indicator&nbsp; $Y = X + N$&nbsp; was given in the&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang#Channel_capacity_of_the_AWGN_channels|theory section]]&nbsp; as follows <br>(with the additional unit&nbsp; "bit"):
 :$$C_{\rm AWGN}(P_X,\ P_N) = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + {P_X}/{P_N} \right )\hspace{0.05cm}.$$
 The quantities used have the following meaning:
-* $P_X$&nbsp; is the transmit power &nbsp; &#8658; &nbsp; variance of the random variablee&nbsp; $X$,
+* $P_X$&nbsp; is the transmission power &nbsp; &#8658; &nbsp; variance of the random variable&nbsp; $X$,
 * $P_N$&nbsp; is the noise power &nbsp; &#8658; &nbsp; variance of the random variable&nbsp; $N$.
@@ Line 59: / Line 59: @@
-{What is the channel capacity &nbsp;$C_K$&nbsp; for&nbsp; $K$&nbsp; equal channels&nbsp; (each with the noise power  &nbsp;$P_N$&nbsp; and the transmit power &nbsp;$P_X(K)$?
+{What is the channel capacity &nbsp;$C_K$&nbsp; for&nbsp; $K$&nbsp; equal channels,&nbsp; each with the noise power  &nbsp;$P_N$&nbsp; and the transmission power &nbsp;$P_X(K)$?
 |type="()"}
 -  &nbsp; $C_K = K/2 \cdot  \log_2 \ \big[1 + P_X/P_N \big]$.
@@ Line 76: / Line 76: @@
-{Is there a (theoretical) optimum with respect to the number of channels&nbsp; $K$&nbsp;?
+{Is there a&nbsp; (theoretical)&nbsp; optimum with respect to the number&nbsp; $K$&nbsp; of channels?
 |type="[]"}
 - Yes: &nbsp; The largest channel capacity results for &nbsp;$K = 2$.
 - Yes: &nbsp; The largest channel capacity results for &nbsp;$K = 4$.
 + No: &nbsp;  The larger&nbsp; $K$, the larger the channel capacity.
-+ The limit value for&nbsp; $K \to \infty$&nbsp; (in bit)&nbsp; is &nbsp;$C_K = P_X/P_N/2/\ln (2)$&nbsp; in "bit".
++ The limit value for&nbsp; $K \to \infty$&nbsp; (in bit)&nbsp; is &nbsp;$C_K = P_X/P_N/2/\ln (2)$&nbsp; in&nbsp; "bit".

← Older revision		Revision as of 10:33, 4 October 2021
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	\frac{\xi^3}{4K^3} > \frac{\xi^4}{5K^4} \hspace{0.05cm}, \hspace{0.5cm} {\rm usw.}$$		\frac{\xi^3}{4K^3} > \frac{\xi^4}{5K^4} \hspace{0.05cm}, \hspace{0.5cm} {\rm usw.}$$

−	The last row of the table shows   With  $K = 4$ one is still far from the theoretical maximum  $($for $K → ∞)$ for large   $\xi$–values.	+	The last row of the table shows:   With  $K = 4$ one is still far from the theoretical maximum  $($for $K → ∞)$ for large   $\xi$–values.

← 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
	}}		}}

← Older revision		Revision as of 09:28, 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 09:23, 16 August 2021
(No difference)

@@ Line 4: / Line 4: @@
 [[File:P_ID2905__Inf_A_4_7_neu.png|right|frame|Signalraumpunkte bei digitaler Modulation]]
-Die Kanalkapazität des AWGN&ndash;Kanals mit dem Kennzeichen $Y = X + N$&nbsp; wurde im&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang#Kanalkapazit.C3.A4t_des_AWGN.E2.80.93Kanals|Theorieteil]]&nbsp; wie folgt angegeben <br>(mit der Zusatz&ndash;Einheit "bit&rdquo;):
+Die Kanalkapazität des AWGN&ndash;Kanals mit dem Kennzeichen $Y = X + N$&nbsp; wurde im&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang#Kanalkapazit.C3.A4t_des_AWGN.E2.80.93Kanals|Theorieteil]]&nbsp; wie folgt angegeben <br>(mit der Zusatz&ndash;Einheit "bit"):
 :$$C_{\rm AWGN}(P_X,\ P_N) = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + {P_X}/{P_N} \right )\hspace{0.05cm}.$$
 Die verwendeten Größen haben folgende Bedeutung:
@@ Line 40: / Line 40: @@
 *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 "bit&rdquo; angegeben werden sollen, wird  "log<sub>2</sub>&rdquo; verwendet.
+*Da die Ergebnisse in "bit" angegeben werden sollen, wird  "log<sub>2</sub>" verwendet.
@@ Line 81: / Line 81: @@
 - Ja: &nbsp; Die größte Kanalkapazität ergibt sich für &nbsp;$K = 4$.
 + Nein: &nbsp;  Je größer&nbsp; $K$, desto größer ist die Kanalkapazität.
-+ Der Grenzwert für&nbsp; $K \to \infty$&nbsp; (in bit)&nbsp; ist &nbsp;$C_K = P_X/P_N/2/\ln (2)$&nbsp; in "bit&rdquo;.
++ Der Grenzwert für&nbsp; $K \to \infty$&nbsp; (in bit)&nbsp; ist &nbsp;$C_K = P_X/P_N/2/\ln (2)$&nbsp; in "bit".