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

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	[[File:P_ID2905__Inf_A_4_7_neu.png\|right\|frame\|Signal space points in digital modulation]]		[[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  $Y = X + N$  was given in the  [[Information_Theory/~~AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang~~#~~Channel_capacity_of_the_AWGN_channels~~\|theory section]]  as follows <br>(with the additional unit  "bit"):		The channel capacity of the AWGN channel with the indicator  $Y = X + N$  was given in the  [[Information_Theory/AWGN_Channel_Capacity_for_Continuous-Valued_Input#Channel_capacity_of_the_AWGN_channel\|theory section]]  as follows <br>(with the additional unit  "bit"):
	:$$C_{\rm AWGN}(P_X,\ P_N) = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + {P_X}/{P_N} \right )\hspace{0.05cm}.$$		:$$C_{\rm AWGN}(P_X,\ P_N) = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + {P_X}/{P_N} \right )\hspace{0.05cm}.$$
	The quantities used have the following meaning:		The quantities used have the following meaning:
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	Hints:		Hints:
	*The exercise belongs to the chapter  [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang\|AWGN channel capacity with continuous value input]].		*The exercise belongs to the chapter  [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang\|AWGN channel capacity with continuous value input]].
	*Reference is made in particular to the page  [[Information_Theory/~~AWGN_Channel_Capacity_for_Continuous_Input~~#Parallel_Gaussian_channels\|Parallel Gaussian Channels]].		*Reference is made in particular to the page  [[Information_Theory/AWGN_Channel_Capacity_for_Continuous-Valued_Input#Parallel_Gaussian_channels\|Parallel Gaussian Channels]].
	*Since the results are to be given in "bit", wird "log<sub>2</sub>" is used.		*Since the results are to be given in "bit", wird "log<sub>2</sub>" is used.

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	[[Category:Information Theory: Exercises\|^4.2 AWGN and Value-Continuous Input^]]		[[Category:Information Theory: Exercises\|^4.2 AWGN and Value-Continuous Input^]]
	[[de:Aufgaben:4.~~07_K_parallele_Gaußkanäle~~]]		[[de:Aufgaben:Aufgabe 4.7: Mehrere parallele Gaußkanäle]]

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	:$$C_{\rm nat}(\xi, K) ={K}/{2} \cdot {\rm ln}\hspace{0.05cm}\left ( 1 + {\xi}/{K} \right )\hspace{0.05cm}.$$		:$$C_{\rm nat}(\xi, K) ={K}/{2} \cdot {\rm ln}\hspace{0.05cm}\left ( 1 + {\xi}/{K} \right )\hspace{0.05cm}.$$
	*Then,  for large values of  $K$  i.e., for small values of the quotient  $\varepsilon =\xi/K$  holds:		*Then,  for large values of  $K$  i.e., for small values of the quotient  $\varepsilon =\xi/K$  holds:
	:$${\rm ln}\hspace{0.05cm}\left ( 1 + \varepsilon \right )=		:$${\rm ln}\hspace{0.05cm}\left ( 1 + \varepsilon \right )=\varepsilon - \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{3} - ...\hspace{0.3cm}\Rightarrow \hspace{0.3cm}C_{\rm nat}(\xi, K) = \frac{K}{2} \cdot \left [ \frac{\xi}{K} - \frac{\xi^2}{2K^2} +\frac{\xi^3}{3K^3} - \text{...} \right ]$$
	\varepsilon - \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{3} - ...		:$$\hspace{0.3cm}\Rightarrow \hspace{0.3cm}C_{\rm bit}(\xi, K) = \frac{\xi}{2 \cdot {\rm ln}\hspace{0.1cm}(2)} \cdot \left [ 1 - \frac{\xi}{2K} +\frac{\xi^2}{3K^2} -\frac{\xi^3}{4K^3} +\frac{\xi^4}{5K^4} - \text{...} \right ] \hspace{0.05cm}.$$
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	C_{\rm nat}(\xi, K) = \frac{K}{2} \cdot \left [ \frac{\xi}{K} - \frac{\xi^2}{2K^2} +
	\frac{\xi^3}{3K^3} - \text{...} \right ]$$
	:$$\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	C_{\rm bit}(\xi, K) = \frac{\xi}{2 \cdot {\rm ln}\hspace{0.1cm}(2)} \cdot \left [ 1 - \frac{\xi}{2K} +
	\frac{\xi^2}{3K^2} -\frac{\xi^3}{4K^3} +
	\frac{\xi^4}{5K^4} - \text{...} \right ] \hspace{0.05cm}.$$
	* For  $K → ∞$ , the proposed value is:		* For  $K → ∞$ , the proposed value is:
	:$$C_{\rm bit}(\xi, K \rightarrow\infty) = \frac{\xi}{2 \cdot {\rm ln}\hspace{0.1cm}(2)} =		:$$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}.$$
	\frac{P_X/P_N}{2 \cdot {\rm ln}\hspace{0.1cm}(2)} \hspace{0.05cm}.$$
	*For smaller values of  $K$,  the result is always a smaller  $C$–value, since		*For smaller values of  $K$,  the result is always a smaller  $C$–value, since
	:$$\frac{\xi}{2K} > \frac{\xi^2}{3K^2}\hspace{0.05cm}, \hspace{0.5cm}		:$$\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.}$$
	\frac{\xi^3}{4K^3} > \frac{\xi^4}{5K^4} \hspace{0.05cm}, \hspace{0.5cm} {\rm usw.}$$

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

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	[[Category:Information Theory: Exercises\|^4.2 AWGN and Value-Continuous Input^]]		[[Category:Information Theory: Exercises\|^4.2 AWGN and Value-Continuous Input^]]
			[[de:Aufgaben:4.07_K_parallele_Gaußkanäle]]

Guenter at 12:33, 4 October 2021

2021-10-04T12:33:08Z

Guenter at 12:23, 4 October 2021

2021-10-04T12:23:55Z

← Older revision		Revision as of 14:23, 4 October 2021
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	* [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying\|Amplitude Shift Keying]]  (ASK),		* [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying\|Amplitude Shift Keying]]  (ASK),
	* [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying\|Binary Phase Shift Keying]]  (BPSK),		* [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying\|Binary Phase Shift Keying]]  (BPSK),
	* [[Modulation_Methods/Further_AM_Variants#~~Quadratur.E2.80.93Amplitudenmodulation_~~.28QAM.29\|Quadrature Amplitude Modulation]]  (here:   4-QAM),		* [[Modulation_Methods/Further_AM_Variants#Quadrature_Amplitude_Modulation_.28QAM.29\|Quadrature Amplitude Modulation]]  (here:   4-QAM),
	*[[Examples_of_Communication_Systems/Further_Developments_of_the_GSM#Enhanced_Data_Rates_for_GSM_Evolution\|Phase Shift Keying]]  (here:   8–PSK for GSM Evolution),		*[[Examples_of_Communication_Systems/Further_Developments_of_the_GSM#Enhanced_Data_Rates_for_GSM_Evolution\|Phase Shift Keying]]  (here:   8–PSK for GSM Evolution),
	* [[Modulation_Methods/Quadrature_Amplitude_Modulation#Other_signal_space_constellations\|Combined ASK/PSK Modulation]]  (here:   16-ASK/PSK).		* [[Modulation_Methods/Quadrature_Amplitude_Modulation#Other_signal_space_constellations\|Combined ASK/PSK Modulation]]  (here:   16-ASK/PSK).

Guenter at 12:21, 4 October 2021

2021-10-04T12:21:01Z

← Older revision		Revision as of 14:21, 4 October 2021
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	[[File:P_ID2905__Inf_A_4_7_neu.png\|right\|frame\|Signal space points in digital modulation]]		[[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 $Y = X + N$  was given in the  [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang#Channel_capacity_of_the_AWGN_channels\|theory section]]  as follows <br>(with the additional unit "bit"):		The channel capacity of the AWGN channel with the indicator  $Y = X + N$  was given in the  [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang#Channel_capacity_of_the_AWGN_channels\|theory section]]  as follows <br>(with the additional unit  "bit"):
	:$$C_{\rm AWGN}(P_X,\ P_N) = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + {P_X}/{P_N} \right )\hspace{0.05cm}.$$		:$$C_{\rm AWGN}(P_X,\ P_N) = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + {P_X}/{P_N} \right )\hspace{0.05cm}.$$
	The quantities used have the following meaning:		The quantities used have the following meaning:
	* $P_X$  is the ~~transmit~~ power   ⇒   variance of the random ~~variablee~~  $X$,		* $P_X$  is the transmission power   ⇒   variance of the random variable  $X$,
	* $P_N$  is the noise power   ⇒   variance of the random variable  $N$.		* $P_N$  is the noise power   ⇒   variance of the random variable  $N$.

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	{What is the channel capacity  $C_K$  for  $K$  equal channels  (each with the noise power  $P_N$  and the ~~transmit~~ power  $P_X(K)$?		{What is the channel capacity  $C_K$  for  $K$  equal channels,  each with the noise power  $P_N$  and the transmission power  $P_X(K)$?
	\|type="()"}		\|type="()"}
	-   $C_K = K/2 \cdot \log_2 \ \big[1 + P_X/P_N \big]$.		-   $C_K = K/2 \cdot \log_2 \ \big[1 + P_X/P_N \big]$.
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	{Is there a (theoretical) optimum with respect to the number ~~of channels~~  $K$ ?		{Is there a  (theoretical)  optimum with respect to the number  $K$  of channels?
	\|type="[]"}		\|type="[]"}
	- Yes:   The largest channel capacity results for  $K = 2$.		- Yes:   The largest channel capacity results for  $K = 2$.
	- Yes:   The largest channel capacity results for  $K = 4$.		- Yes:   The largest channel capacity results for  $K = 4$.
	+ No:   The larger  $K$, the larger the channel capacity.		+ No:   The larger  $K$, the larger the channel capacity.
	+ The limit value for  $K \to \infty$  (in bit)  is  $C_K = P_X/P_N/2/\ln (2)$  in "bit".		+ The limit value for  $K \to \infty$  (in bit)  is  $C_K = P_X/P_N/2/\ln (2)$  in  "bit".

Noah at 10:33, 4 October 2021

2021-10-04T10:33:30Z

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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.

Noah at 10:33, 4 October 2021

2021-10-04T10:33:11Z

Show changes

Noah at 12:31, 1 October 2021

2021-10-01T12:31:46Z

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	{{quiz-Header\|Buchseite=~~Informationstheorie~~/~~AWGN–Kanalkapazität bei wertkontinuierlichem Eingang~~		{{quiz-Header\|Buchseite=Information_Theory/AWGN_Channel_Capacity_for_Continuous_Input
	}}		}}

Guenter at 09:28, 16 August 2021

2021-08-16T09:28:39Z

← Older revision		Revision as of 11:28, 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 14:23, 4 October 2021
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	* [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying\|Amplitude Shift Keying]]  (ASK),		* [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying\|Amplitude Shift Keying]]  (ASK),
	* [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying\|Binary Phase Shift Keying]]  (BPSK),		* [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying\|Binary Phase Shift Keying]]  (BPSK),
	* [[Modulation_Methods/Further_AM_Variants#~~Quadratur.E2.80.93Amplitudenmodulation_~~.28QAM.29\|Quadrature Amplitude Modulation]]  (here:   4-QAM),		* [[Modulation_Methods/Further_AM_Variants#Quadrature_Amplitude_Modulation_.28QAM.29\|Quadrature Amplitude Modulation]]  (here:   4-QAM),
	*[[Examples_of_Communication_Systems/Further_Developments_of_the_GSM#Enhanced_Data_Rates_for_GSM_Evolution\|Phase Shift Keying]]  (here:   8–PSK for GSM Evolution),		*[[Examples_of_Communication_Systems/Further_Developments_of_the_GSM#Enhanced_Data_Rates_for_GSM_Evolution\|Phase Shift Keying]]  (here:   8–PSK for GSM Evolution),
	* [[Modulation_Methods/Quadrature_Amplitude_Modulation#Other_signal_space_constellations\|Combined ASK/PSK Modulation]]  (here:   16-ASK/PSK).		* [[Modulation_Methods/Quadrature_Amplitude_Modulation#Other_signal_space_constellations\|Combined ASK/PSK Modulation]]  (here:   16-ASK/PSK).

← Older revision		Revision as of 14:33, 4 October 2021
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The parameter  $K$  is equal to the dimension of the signal space representation:		'''(1)'''  The parameter  $K$  is equal to the dimension of the signal space representation:
	* For <u>ASK and BPSK</u>,  $\underline{K=1}$.		* For  <u>ASK and BPSK</u>,  $\underline{K=1}$.
	* For <u> constellations 3 to 5</u>,however,  $\underline{K=2}$  (orthogonal modulation with cosine and sine).		* For  <u> constellations 3 to 5</u>,  however,  $\underline{K=2}$  (orthogonal modulation with cosine and sine).



	'''(2)'''  Correct is the <u>proposed solution 2</u>:		'''(2)'''  Correct is the <u>proposed solution 2</u>:
	*For each of the channels  $(1 ≤ k ≤ K)$ , the channel capacitance is   $C = 1/2 \cdot \log_2 \ \big[1 + (P_X/K) /P_N) \big]$.  The total capacitance is then larger by a factor of  $K$ :		*For each of the channels  $(1 ≤ k ≤ K)$,  the channel capacitance is   $C = 1/2 \cdot \log_2 \ \big[1 + (P_X/K) /P_N) \big]$.
			*The total capacitance is then larger by a factor of  $K$ :
	:$$C_K(P_X) = \sum_{k= 1}^K \hspace{0.1cm}C_k = \frac{K}{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + \frac{P_X}{K \cdot P_N} \right )\hspace{0.05cm}.$$		:$$C_K(P_X) = \sum_{k= 1}^K \hspace{0.1cm}C_k = \frac{K}{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + \frac{P_X}{K \cdot P_N} \right )\hspace{0.05cm}.$$
	*The proposed solution 1 is too positive. This would apply when limiting the total power to  $K · P_X$ .		*The proposed solution 1 is too positive.  This would apply when limiting the total power to  $K · P_X$ .
	*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,  which is obviously not true.


	'''(3)'''  The table shows the results for  $K = 1$,  $K = 2$  and  $K = 4$  , and various signa–to–noise power ratios  $\xi = P_X/P_N$. For  $\xi = P_X/P_N = 15$  (highlighted column), the result is:		[[File:P_ID2902__Inf_A_4_7c.png\|right\|frame\|Channel capacity  $C_K$  of  $K$  parallel Gaussian channels for different  $\xi = P_X/P_N$]]
	~~[[File:P_ID2902__Inf_A_4_7c.png\|right\|frame\|Channel capacity  $C_K$  of  $K$  parallel Gaussian channels for different  $\xi = P_X/P_N$]]~~		'''(3)'''  The table shows the results for  $K = 1$,  $K = 2$  and  $K = 4$,  and various signa–to–noise power ratios  $\xi = P_X/P_N$. <br>For  $\xi = P_X/P_N = 15$  (highlighted column),  the result is:

	* $K=1$:   $C_K = 1/2 · \log_2 \ (16)\hspace{0.05cm}\underline{ = 2.000}$ bit,		* $K=1$:   $C_K = 1/2 · \log_2 \ (16)\hspace{0.05cm}\underline{ = 2.000}$ bit,
	* $K=2$:   $C_K = 1/2 · \log_2 \ (8.5)\hspace{0.05cm}\underline{ = 3.087}$ bit,		* $K=2$:   $C_K = 1/2 · \log_2 \ (8.5)\hspace{0.05cm}\underline{ = 3.087}$ bit,
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	*We now write the channel capacity using the natural logarithm and the abbreviation  $\xi = P_X/P_N$:		*We now write the channel capacity using the natural logarithm and the abbreviation  $\xi = P_X/P_N$:
	:$$C_{\rm nat}(\xi, K) ={K}/{2} \cdot {\rm ln}\hspace{0.05cm}\left ( 1 + {\xi}/{K} \right )\hspace{0.05cm}.$$		:$$C_{\rm nat}(\xi, K) ={K}/{2} \cdot {\rm ln}\hspace{0.05cm}\left ( 1 + {\xi}/{K} \right )\hspace{0.05cm}.$$
	*Then, for large values of  $K$i.e., for small values of the quotient  $\varepsilon =\xi/K$ ~~jholds~~:		*Then,  for large values of  $K$  i.e., for small values of the quotient  $\varepsilon =\xi/K$  holds:
	:$${\rm ln}\hspace{0.05cm}\left ( 1 + \varepsilon \right )=		:$${\rm ln}\hspace{0.05cm}\left ( 1 + \varepsilon \right )=
	\varepsilon - \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{3} - ...		\varepsilon - \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{3} - ...
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	:$$C_{\rm bit}(\xi, K \rightarrow\infty) = \frac{\xi}{2 \cdot {\rm ln}\hspace{0.1cm}(2)} =		:$$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}.$$		\frac{P_X/P_N}{2 \cdot {\rm ln}\hspace{0.1cm}(2)} \hspace{0.05cm}.$$
	*For smaller values of  $K$  , the result is always a ~~smalle~~  $C$–value, since		*For smaller values of  $K$,  the result is always a smaller  $C$–value, since
	:$$\frac{\xi}{2K} > \frac{\xi^2}{3K^2}\hspace{0.05cm}, \hspace{0.5cm}		:$$\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.}$$		\frac{\xi^3}{4K^3} > \frac{\xi^4}{5K^4} \hspace{0.05cm}, \hspace{0.5cm} {\rm usw.}$$

	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 away from the theoretical maximum  $($for $K → ∞)$  for large   $\xi$–values.