Aufgaben:Exercise 1.7: PDF of Rice Fading - Revision history

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	:$$f_a(a) ={a}/{\sigma^2} \cdot {\rm e}^{ -{(a^2+ \|z_0\|^2) }/({2\sigma^2})}\cdot {\rm I}_0 \left [ {a \cdot \|z_0\|}/{\sigma^2} \right ]\hspace{0.05cm}.$$		:$$f_a(a) ={a}/{\sigma^2} \cdot {\rm e}^{ -{(a^2+ \|z_0\|^2) }/({2\sigma^2})}\cdot {\rm I}_0 \left [ {a \cdot \|z_0\|}/{\sigma^2} \right ]\hspace{0.05cm}.$$
	* For example, the modified zeroth order Bessel function returns the following values:		* For example, the modified zeroth order Bessel function returns the following values:
	:$${\rm I }_0 (2) = 2.28\hspace{0.05cm},\hspace{0.2cm}{\rm I }_0 (4) = 11.30\hspace{0.05cm},\hspace{0.2cm}{\rm I }_0 (3) = 67.23		:$${\rm I }_0 (2) = 2.28\hspace{0.05cm},\hspace{0.2cm}{\rm I }_0 (4) = 11.30\hspace{0.05cm},\hspace{0.2cm}{\rm I }_0 (3) = 67.23\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	* The power (noncentral second moment) of the multiplicative factor  $\|z(t)\|$ is		* The power (noncentral second moment) of the multiplicative factor  $\|z(t)\|$ is
	:$${\rm E}\left [ a^2 \right ] = {\rm E}\left [ \|z(t)\|^2 \right ] = 2 \cdot \sigma^2 + \|z_0\|^2		:$${\rm E}\left [ a^2 \right ] = {\rm E}\left [ \|z(t)\|^2 \right ] = 2 \cdot \sigma^2 + \|z_0\|^2\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	* With  $z_0 = 0$,  the Rice fading  becomes Rayleigh fading, which is more critical.  In this case, the probability that  $a$  lies in the yellow-shaded area between  $0$  and  $1$  is		* With  $z_0 = 0$,  the Rice fading  becomes Rayleigh fading, which is more critical.  In this case, the probability that  $a$  lies in the yellow-shaded area between  $0$  and  $1$  is
	:$$ {\rm Pr}(a \le 1) = 1 - {\rm e}^{-0.5/\sigma^2} \approx 0.4		:$$ {\rm Pr}(a \le 1) = 1 - {\rm e}^{-0.5/\sigma^2} \approx 0.4\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	In this task the probability  ${\rm Pr}(a ≤ 1)$  for  $\|z_0\| ≠ 0$  is to be approximated. There are two ways to do this, namely:		In this task the probability  ${\rm Pr}(a ≤ 1)$  for  $\|z_0\| ≠ 0$  is to be approximated. There are two ways to do this, namely:

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	[[Category:Mobile Communications: Exercises\|^1.4 Fading with Direct Path Component^]]		[[Category:Mobile Communications: Exercises\|^1.4 Fading with Direct Path Component^]]
			[[de:Exercises:Exercise_1.7:_PDF_of_Rice_Fading]]

Javier at 08:19, 7 July 2021

2021-07-07T08:19:42Z

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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Calculate some PDF values for  $\|z_0\| = 2$  and  $\sigma = 2$:		{Calculate some PDF values for  $\|z_0\| = 2$  and  $\sigma = 1$:
	\|type="{}"}		\|type="{}"}
	$f_a(a = 1) \ = \ ${ 0.187 3% }		$f_a(a = 1) \ = \ ${ 0.187 3% }
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  With  $\|z_0\| = 2$  and  $\sigma = 2$  the Rice PDF is		'''(1)'''  With  $\|z_0\| = 2$  and  $\sigma = 1$  the Rice PDF is
	:$$f_a(a) = a \cdot {\rm exp} [ -\frac{a^2 + 4}{2}] \cdot {\rm I}_0 (2a)\hspace{0.05cm}.$$		:$$f_a(a) = a \cdot {\rm exp} [ -\frac{a^2 + 4}{2}] \cdot {\rm I}_0 (2a)\hspace{0.05cm}.$$

Javier: Text replacement - "Category:Exercises for Mobile Communications" to "Category:Mobile Communications: Exercises"

2021-03-23T12:37:11Z

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	[[Category:~~Exercises for~~ Mobile Communications\|^1.4 Fading with Direct Path Component^]]		[[Category:Mobile Communications: Exercises\|^1.4 Fading with Direct Path Component^]]

Javier: Text replacement - "===Solutions===" to "===Solution==="

2020-12-15T14:11:11Z

Text replacement - "===Solutions===" to "===Solution==="

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	</quiz>		</quiz>

	===~~Solutions~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  With  $\|z_0\| = 2$  and  $\sigma = 2$  the Rice PDF is		'''(1)'''  With  $\|z_0\| = 2$  and  $\sigma = 2$  the Rice PDF is

Javier: Text replacement - "Mobile Kommunikation/Nichtfrequenzselektives Fading mit Direktkomponente" to "Mobile_Communications/Non-Frequency_Selective_Fading_With_Direct_Component"

2020-12-15T13:58:40Z

Text replacement - "Mobile Kommunikation/Nichtfrequenzselektives Fading mit Direktkomponente" to "Mobile_Communications/Non-Frequency_Selective_Fading_With_Direct_Component"

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	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Nichtfrequenzselektives Fading mit Direktkomponente~~}}		{{quiz-Header\|Buchseite=Mobile_Communications/Non-Frequency_Selective_Fading_With_Direct_Component}}

	[[File:P_ID2133__Mob_A_1_7.png\|right\|frame\| Rice fading for different values of  $\|z_0\|^2$]]		[[File:P_ID2133__Mob_A_1_7.png\|right\|frame\| Rice fading for different values of  $\|z_0\|^2$]]

Guenter at 15:26, 8 December 2020

2020-12-08T15:26:31Z

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	:$${\rm Pr}(a \le 1) \approx {\rm Pr}(g \le -3.37) = {\rm Q}(3.37) { \approx 0.04\,\%} \hspace{0.05cm}.$$		:$${\rm Pr}(a \le 1) \approx {\rm Pr}(g \le -3.37) = {\rm Q}(3.37) { \approx 0.04\,\%} \hspace{0.05cm}.$$

	*If one assumes the equal power Gaussian ~~distributio~~ n (see the note to the last subtask), the mean value is  $m_g = \sqrt{21}\approx 4.58$,  and the probability would then be		*If one assumes the equal power Gaussian distribution  (see the note to the last subtask), the mean value is  $m_g = \sqrt{21}\approx 4.58$,  and the probability would then be
	:$${\rm Pr}(a \le 1) \approx {\rm Q}(3.58) \hspace{0.15cm} \underline{ \approx 0.02\,\%} \hspace{0.05cm}.$$		:$${\rm Pr}(a \le 1) \approx {\rm Q}(3.58) \hspace{0.15cm} \underline{ \approx 0.02\,\%} \hspace{0.05cm}.$$

Guenter at 15:18, 8 December 2020

2020-12-08T15:18:13Z

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	''Notes:''		''Notes:''
	* This task belongs to chapter  [[Mobile_Communications/Non-frequency_selective_fading_with_direct_component\| Non-frequency selective fading with direct component]].		* This task belongs to chapter  [[Mobile_Communications/Non-frequency_selective_fading_with_direct_component\| Non-frequency selective fading with direct component]].
	* For the numerical solutions of the last subtasks, we recommend the ~~interaction module~~  [[Applets:Complementary_Gaussian_Error_Functions\|Complementary Gaussian Error Functions]].		* For the numerical solutions of the last subtasks, we recommend the applet  [[Applets:Complementary_Gaussian_Error_Functions\|Complementary Gaussian Error Functions]].

Javier: Text replacement - "PSD" to "PDS"

2020-08-25T09:57:02Z

Text replacement - "PSD" to "PDS"

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	[[File:P_ID2133__Mob_A_1_7.png\|right\|frame\| Rice fading for different values of  $\|z_0\|^2$]]		[[File:P_ID2133__Mob_A_1_7.png\|right\|frame\| Rice fading for different values of  $\|z_0\|^2$]]
	As you can see in the diagram, we consider the same scenario as in  [[Aufgaben:Exercise_1.6:~~_Autocorrelation_Function_and_PSD_with_Rice_Fading~~\| Exercise 1.6]]:		As you can see in the diagram, we consider the same scenario as in  [[Aufgaben:Exercise_1.6:_Autocorrelation_Function_and_PDS_with_Rice_Fading\| Exercise 1.6]]:
	* Rice fading  with variance of the Gaussian processes   $\sigma^2 = 1$  and parameter  $\|z_0\|$  for the direct path.		* Rice fading  with variance of the Gaussian processes   $\sigma^2 = 1$  and parameter  $\|z_0\|$  for the direct path.
	* Regarding direct path, we are interested in the parameter values  $\|z_0\|^2 = 0, \ 2, \ 4, \ 10, \ 20$  (see graph).		* Regarding direct path, we are interested in the parameter values  $\|z_0\|^2 = 0, \ 2, \ 4, \ 10, \ 20$  (see graph).

Guenter at 10:00, 13 August 2020

2020-08-13T10:00:57Z

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	===Solutions===		===Solutions===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  With $\|z_0\| = 2$ and $\sigma = 2$ the Rice PDF is		'''(1)'''  With  $\|z_0\| = 2$  and  $\sigma = 2$  the Rice PDF is
	$$f_a(a) = a \cdot {\rm exp} [ -\frac{a^2 + 4}{2}] \cdot {\rm I}_0 (2a)\hspace{0.05cm}.$$		:$$f_a(a) = a \cdot {\rm exp} [ -\frac{a^2 + 4}{2}] \cdot {\rm I}_0 (2a)\hspace{0.05cm}.$$

	*This gives the desired values:		*This gives the desired values:
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	'''(2)'''  With the result of the subtask '''(1)'''   ⇒   $f_a(a = 1) = 0.187$ the triangle approximation gives		'''(2)'''  With the result of the subtask  '''(1)'''   ⇒   $f_a(a = 1) = 0.187$  the triangle approximation gives
	$${\rm Pr}(a \le 1) = {1}/{2} \cdot 0.187 \cdot 1\hspace{0.15cm} \underline{ \approx 9.4\,\%} \hspace{0.05cm}.$$		:$${\rm Pr}(a \le 1) = {1}/{2} \cdot 0.187 \cdot 1\hspace{0.15cm} \underline{ \approx 9.4\,\%} \hspace{0.05cm}.$$

	*This result will be a bit too large, because the blue curve is below the connecting line from $(0, 0)$ to $(1, 0.187)$   ⇒   convex curve.		*This result will be a bit too large, because the blue curve is below the connecting line from  $(0, 0)$  to  $(1, 0.187)$   ⇒   convex curve.

	~~'''(3)'''  For the red curve the WDF–value $f_a(a = 1) \approx 0.35$ can be read from the graph:~~
	~~$${\rm Pr}(a \le 1) = \frac{1}{2} \cdot 0.35 \hspace{0.15cm} \underline{ \approx 17.5\,\%} \hspace{0.05cm}.$$~~

	*The actual probability value will be slightly larger because the red curve ~~is concave in~~ the ~~range between~~ $0$ ~~and~~ $1$.		'''(3)'''  For the red curve the PDF value  $f_a(a = 1) \approx 0.35$  can be read from the graph:
			:$${\rm Pr}(a \le 1) = \frac{1}{2} \cdot 0.35 \hspace{0.15cm} \underline{ \approx 17.5\,\%} \hspace{0.05cm}.$$

			*The actual probability value will be slightly larger because the red curve is concave in the range between  $0$  and  $1$.


	'''(4)'''  The Gaussian approximation states that one can approximate the Rice distribution by a Gaussian distribution with mean $\|z_0\| = \sqrt{10} = 3.16$ and standard deviation $\sigma = 1$ if the quotient $\|z_0\|/\sigma$ is sufficiently large. Then we have
	~~$${\rm Pr}(a \le 1) \approx {\rm Pr}(g \le -2.16) = {\rm Q}(2.16) \hspace{0.15cm} \underline{ \approx 1.5\,\%} \hspace{0.05cm}.$$~~

	*Here, $g$ denotes a Gaussian distributed random variable with mean $0$ and standard deviation $\sigma = 1$.		'''(4)'''  The Gaussian approximation states that one can approximate the Rice distribution by a Gaussian distribution with mean  $\|z_0\| = \sqrt{10} = 3.16$  and standard deviation  $\sigma = 1$  if the quotient  $\|z_0\|/\sigma$   is sufficiently large.  Then we have
	*The numerical value was determined with the specified interactive [[Applets:~~QFunction~~\|~~Applet~~]].		:$${\rm Pr}(a \le 1) \approx {\rm Pr}(g \le -2.16) = {\rm Q}(2.16) \hspace{0.15cm} \underline{ \approx 1.5\,\%} \hspace{0.05cm}.$$

			*Here,  $g$  denotes a Gaussian distributed random variable with mean  $0$  and standard deviation  $\sigma = 1$.
			*The numerical value was determined with the specified interactive applet  [[Applets:Complementary_Gaussian_Error_Functions\|Complementary Gaussian Error Functions]].


	<i>Note:</i>   The Gaussian approximation is certainly associated with a certain error here:		<i>Note:</i>   The Gaussian approximation is certainly associated with a certain error here:
	*From the graph you can see that the average value of the green curve is not $a = 3.16$ , but rather $3.31$.		*From the graph you can see that the average value of the green curve is not  $a = 3.16$,  but rather  $3.31$.
	*Then the power of the Gaussian approximation $(3.31^2 + 1^2 = 12)$ is exactly the same as that of the Rice distribution:		*Then the power of the Gaussian approximation  $(3.31^2 + 1^2 = 12)$  is exactly the same as that of the Rice distribution:
	:$$\|z_0\|^2 + 2 \sigma^2= 10 + 2 =12\hspace{0.05cm}.$$		:$$\|z_0\|^2 + 2 \sigma^2= 10 + 2 =12\hspace{0.05cm}.$$


	'''(5)'''  Using the same calculation method, replace the Rice PDF with a Gaussian PDF with mean value $\sqrt{20} \approx $4.47 and standard deviation $\sigma = $1 and you get		'''(5)'''  Using the same calculation method, replace the Rice PDF with a Gaussian PDF with mean value  $\sqrt{20} \approx 4.47$  and standard deviation  $\sigma = 1$  and you get
	$${\rm Pr}(a \le 1) \approx {\rm Pr}(g \le -3.37) = {\rm Q}(3.37) { \approx 0.04\,\%} \hspace{0.05cm}.$$		:$${\rm Pr}(a \le 1) \approx {\rm Pr}(g \le -3.37) = {\rm Q}(3.37) { \approx 0.04\,\%} \hspace{0.05cm}.$$

	*If one assumes the equal power Gaussian ~~distribution~~ (see the note to the last subtask), the mean value is $m_g = \sqrt{21}\approx 4.58$, and the probability would then be		*If one assumes the equal power Gaussian distributio n (see the note to the last subtask), the mean value is  $m_g = \sqrt{21}\approx 4.58$,  and the probability would then be
	$${\rm Pr}(a \le 1) \approx {\rm Q}(3.58) \hspace{0.15cm} \underline{ \approx 0.02\,\%} \hspace{0.05cm}.$$		:$${\rm Pr}(a \le 1) \approx {\rm Q}(3.58) \hspace{0.15cm} \underline{ \approx 0.02\,\%} \hspace{0.05cm}.$$

	{{ML-Fuß}}		{{ML-Fuß}}

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	[[File:P_ID2133__Mob_A_1_7.png\|right\|frame\| Rice fading for different values of  $\|z_0\|^2$]]		[[File:P_ID2133__Mob_A_1_7.png\|right\|frame\| Rice fading for different values of  $\|z_0\|^2$]]
	As you can see in the diagram, we consider the same scenario as in  [[Aufgaben:Exercise_1.6:~~_Autocorrelation_Function_and_PSD_with_Rice_Fading~~\| Exercise 1.6]]:		As you can see in the diagram, we consider the same scenario as in  [[Aufgaben:Exercise_1.6:_Autocorrelation_Function_and_PDS_with_Rice_Fading\| Exercise 1.6]]:
	* Rice fading  with variance of the Gaussian processes   $\sigma^2 = 1$  and parameter  $\|z_0\|$  for the direct path.		* Rice fading  with variance of the Gaussian processes   $\sigma^2 = 1$  and parameter  $\|z_0\|$  for the direct path.
	* Regarding direct path, we are interested in the parameter values  $\|z_0\|^2 = 0, \ 2, \ 4, \ 10, \ 20$  (see graph).		* Regarding direct path, we are interested in the parameter values  $\|z_0\|^2 = 0, \ 2, \ 4, \ 10, \ 20$  (see graph).