Aufgaben:Exercise 1.3: Rayleigh Fading - Revision history

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	[[Category:Mobile Communications: Exercises\|^1.2 PDF of Rayleigh Fading^]]		[[Category:Mobile Communications: Exercises\|^1.2 PDF of Rayleigh Fading^]]
	[[de:~~Exercises~~:~~Exercise_1~~.3:~~_Rayleigh_Fading~~]]		[[de:Aufgaben:Exercise 1.3: Rayleigh Fading]]

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	* The magnitude  $a(t) = \|z(t)\|$  has a Rayleigh PDF, from which the name "Rayleigh Fading" is derived:		* The magnitude  $a(t) = \|z(t)\|$  has a Rayleigh PDF, from which the name "Rayleigh Fading" is derived:
	:$$f_a(a) =		:$$f_a(a) =\left\{ \begin{array}{c} a/\sigma^2 \cdot {\rm e}^ { -a^2/(2\sigma^2)} \\0 \end{array} \right.\quad\begin{array}{*{1}c} {\rm for}\hspace{0.15cm} a \ge 0\\ {\rm for}\hspace{0.15cm} a < 0 \\ \end{array}\hspace{0.05cm}.$$
	\left\{ \begin{array}{c} a/\sigma^2 \cdot {\rm e}^ { -a^2/(2\sigma^2)} \\
	0 \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm for}\hspace{0.15cm} a \ge 0
	\\ {\rm for}\hspace{0.15cm} a < 0 \\ \end{array}
	\hspace{0.05cm}.$$

	* The squared magnitude  $p(t) = a(t)^2 = \|z(t)\|^2$  is exponentially distributed according to the equation		* The squared magnitude  $p(t) = a(t)^2 = \|z(t)\|^2$  is exponentially distributed according to the equation
	:$$f_p(p) = \left\{ \begin{array}{c} 1/(2\sigma^2) \cdot {\rm e}^ { -p/(2\sigma^2)} \\		:$$f_p(p) = \left\{ \begin{array}{c} 1/(2\sigma^2) \cdot {\rm e}^ { -p/(2\sigma^2)} \\0 \end{array} \right.\quad\begin{array}{*{1}c} {\rm for}\hspace{0.15cm} p \ge 0\\ {\rm for}\hspace{0.15cm} p < 0 \\ \end{array}\hspace{0.05cm}.$$
	0 \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm for}\hspace{0.15cm} p \ge 0
	\\ {\rm for}\hspace{0.15cm} p < 0 \\ \end{array}
	\hspace{0.05cm}.$$

	Measurements have shown that the time intervals with  $a(t) ≤ 1$  $($highlighted in yellow in the graphic$)$ add up to  $\text{59 ms}$  $($intervals highlighted in red$).$  Being the total measurement time  $\text{150 ms}$,  the probability that the magnitude of the Rayleigh fading is less than or equal to  $1$  is		Measurements have shown that the time intervals with  $a(t) ≤ 1$  $($highlighted in yellow in the graphic$)$ add up to  $\text{59 ms}$  $($intervals highlighted in red$).$  Being the total measurement time  $\text{150 ms}$,  the probability that the magnitude of the Rayleigh fading is less than or equal to  $1$  is
	:$${\rm Pr}(a(t) \le 1) = \frac{59\,\,{\,{\rm ms}}}{150\,\,{\rm ms}} = 39.4 \%		:$${\rm Pr}(a(t) \le 1) = \frac{59\,\,{\,{\rm ms}}}{150\,\,{\rm ms}} = 39.4 \%\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	In the lower graph, the value range between  $\text{-3 dB}$  and  $\text{+3 dB}$  of the logarithmic Rayleigh coefficient  $20 \cdot {\rm lg} \ a(t)$  is highlighted in green.  The subtask '''(4)'''  refers to this.		In the lower graph, the value range between  $\text{-3 dB}$  and  $\text{+3 dB}$  of the logarithmic Rayleigh coefficient  $20 \cdot {\rm lg} \ a(t)$  is highlighted in green.  The subtask '''(4)'''  refers to this.
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From  ${\rm Max}[a(t)] = 2$  follows directly:		'''(1)'''  From  ${\rm Max}[a(t)] = 2$  follows directly:
	:$${\rm Max} \left [ 20 \cdot {\rm lg}\hspace{0.15cm}a(t) \right ] = 20 \cdot {\rm lg}\hspace{0.15cm}(2) \hspace{0.15cm} \underline{\approx 6\,\,{\rm dB}}		:$${\rm Max} \left [ 20 \cdot {\rm lg}\hspace{0.15cm}a(t) \right ] = 20 \cdot {\rm lg}\hspace{0.15cm}(2) \hspace{0.15cm} \underline{\approx 6\,\,{\rm dB}}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


	'''(2)'''  The maximum value of the square  $p(t) = a(t)^2$  is		'''(2)'''  The maximum value of the square  $p(t) = a(t)^2$  is
	:$${\rm Max} \left [ p(t) \right ] = {\rm Max} \left [ a(t)^2 \right ] \hspace{0.15cm} \underline{=4}		:$${\rm Max} \left [ p(t) \right ] = {\rm Max} \left [ a(t)^2 \right ] \hspace{0.15cm} \underline{=4}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*The logarithmic representation of the squared magnitude  $p(t)$  is identical to the logarithmic representation of the magnitude  $a(t)$.		*The logarithmic representation of the squared magnitude  $p(t)$  is identical to the logarithmic representation of the magnitude  $a(t)$.
	*Since  $p(t)$  is a power quantity:		*Since  $p(t)$  is a power quantity:
	:$${\rm Max} \left [ p(t) \right ] = {\rm Max} \left [ a(t)^2 \right ] \hspace{0.15cm} \underline{= 4}		:$${\rm Max} \left [ p(t) \right ] = {\rm Max} \left [ a(t)^2 \right ] \hspace{0.15cm} \underline{= 4}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*The maximum value is thus also  $\underline{\approx 6\,\,{\rm dB}}$.		*The maximum value is thus also  $\underline{\approx 6\,\,{\rm dB}}$.
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	'''(3)'''  The condition  $a(t) ≤ 1$  is equivalent to the requirement  $p(t) = a(t)^2 ≤ 1$.		'''(3)'''  The condition  $a(t) ≤ 1$  is equivalent to the requirement  $p(t) = a(t)^2 ≤ 1$.
	*The absolute square is known to be exponentially distributed, and for  $p ≥ 0$  we have		*The absolute square is known to be exponentially distributed, and for  $p ≥ 0$  we have
	:$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm e}^{ -p/(2\sigma^2)}		:$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm e}^{ -p/(2\sigma^2)}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	[[File:EN_Mob_A_1_3_c.png\|right\|frame\|PDF and probability regions ]]		[[File:EN_Mob_A_1_3_c.png\|right\|frame\|PDF and probability regions ]]
	*It follows:		*It follows:

	:$${\rm Pr}(p(t) \le 1) = \frac{1}{2\sigma^2} \cdot \int_{0}^{1}{\rm e}^{ -p/(2\sigma^2)} \hspace{0.15cm}{\rm d}p =		:$${\rm Pr}(p(t) \le 1) = \frac{1}{2\sigma^2} \cdot \int_{0}^{1}{\rm e}^{ -p/(2\sigma^2)} \hspace{0.15cm}{\rm d}p =1 - {\rm e}^{ -1/(2\sigma^2)} = 0.394$$
	1 - {\rm e}^{ -1/(2\sigma^2)} = 0.394$$		:$$\Rightarrow \hspace{0.3cm} {\rm e}^{ -1/(2\sigma^2)} =0.606 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\sigma^2 = \frac{1}{2 \cdot {\rm ln}\hspace{0.1cm}(0.606)} = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\underline{\sigma = 1} \hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm} {\rm e}^{ -1/(2\sigma^2)} =0.606 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	\sigma^2 = \frac{1}{2 \cdot {\rm ln}\hspace{0.1cm}(0.606)} = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	\underline{\sigma = 1} \hspace{0.05cm}.$$

	The graph shows		The graph shows
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	'''(4)'''  From  $10 \cdot {\rm lg} \ p_1 = \ -3 \ \ \rm dB$  follows  $p_1 = 0.5$.  The upper limit of the integration range results from the condition  $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$,  so  $p_2 = 2$.		'''(4)'''  From  $10 \cdot {\rm lg} \ p_1 = \ -3 \ \ \rm dB$  follows  $p_1 = 0.5$.  The upper limit of the integration range results from the condition  $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$,  so  $p_2 = 2$.
	*This gives, according to the above graph:		*This gives, according to the above graph:
	$${\rm Pr}(-3\,\,{\rm dB}\le 10 \cdot {\rm lg}\hspace{0.15cm}p(t) \le +3\,\,{\rm dB}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \int_{0.5}^{2}f_p(p)\hspace{0.15cm}{\rm d}p =		$${\rm Pr}(-3\,\,{\rm dB}\le 10 \cdot {\rm lg}\hspace{0.15cm}p(t) \le +3\,\,{\rm dB}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \int_{0.5}^{2}f_p(p)\hspace{0.15cm}{\rm d}p =\left [ - {\rm e}^{ -{p}/(2\sigma^2)}\hspace{0.15cm} \right ]_{0.5}^{2} ={\rm e}^{-0.25}- {\rm e}^{-1} \approx 0.779 - 0.368 \hspace{0.15cm} \underline{ = 0.411} \hspace{0.05cm}.$$
	\left [ - {\rm e}^{ -{p}/(2\sigma^2)}\hspace{0.15cm} \right ]_{0.5}^{2} ={\rm e}^{-0.25}- {\rm e}^{-1} \approx 0.779 - 0.368 \hspace{0.15cm} \underline{ = 0.411} \hspace{0.05cm}.$$

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	[[Category:Mobile Communications: Exercises\|^1.2 PDF of Rayleigh Fading^]]		[[Category:Mobile Communications: Exercises\|^1.2 PDF of Rayleigh Fading^]]
			[[de:Exercises:Exercise_1.3:_Rayleigh_Fading]]

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	''Notes:''		''Notes:''
	* This task belongs to chapter  [[Mobile_Communications/Probability_density_of_Rayleigh_fading\|Probability density of Rayleigh fading]]  of this book.		* This task belongs to chapter  [[Mobile_Communications/Probability_density_of_Rayleigh_fading\|Probability density of Rayleigh fading]]  of this book.
	* A similar topic is treated with a different approach in chapter  [[Theory_of_Stochastic_Signals/Further_distributions\|Further distributions]]  of the book "Stochastic Signal Theory~~”~~.		* A similar topic is treated with a different approach in chapter  [[Theory_of_Stochastic_Signals/Further_distributions\|Further distributions]]  of the book "Stochastic Signal Theory".
	* To check your results, you can use the interactive applet  [[Applets:PDF,_CDF_and_Moments_of_Special_Distributions\|PDF, CDF and Moments of Special Distributions]]  of the book "Stochastic Signal Theory~~”~~.		* To check your results, you can use the interactive applet  [[Applets:PDF,_CDF_and_Moments_of_Special_Distributions\|PDF, CDF and Moments of Special Distributions]]  of the book "Stochastic Signal Theory".

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	''Notes:''		''Notes:''
	* This task belongs to chapter  [[Mobile_Communications/Probability_density_of_Rayleigh_fading\|Probability density of Rayleigh fading]]  of this book.		* This task belongs to chapter  [[Mobile_Communications/Probability_density_of_Rayleigh_fading\|Probability density of Rayleigh fading]]  of this book.
	* A similar topic is treated with a different approach in chapter  [[Theory_of_Stochastic_Signals/Further_distributions\|Further distributions]]  of the book ~~&bdquo;~~Stochastic Signal Theory”.		* A similar topic is treated with a different approach in chapter  [[Theory_of_Stochastic_Signals/Further_distributions\|Further distributions]]  of the book "Stochastic Signal Theory”.
	* To check your results, you can use the interactive applet  [[Applets:PDF,_CDF_and_Moments_of_Special_Distributions\|PDF, CDF and Moments of Special Distributions]]  of the book ~~&bdquo;~~Stochastic Signal Theory”.		* To check your results, you can use the interactive applet  [[Applets:PDF,_CDF_and_Moments_of_Special_Distributions\|PDF, CDF and Moments of Special Distributions]]  of the book "Stochastic Signal Theory”.

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	[[Category:~~Exercises for~~ Mobile Communications\|^1.2 PDF of Rayleigh Fading^]]		[[Category:Mobile Communications: Exercises\|^1.2 PDF of Rayleigh Fading^]]

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	An example of such a Rayleigh channel occurs in urban mobile communications when narrow-band signals are used with ranges between  $50$  and  $100$  meters.		An example of such a Rayleigh channel occurs in urban mobile communications when narrow-band signals are used with ranges between  $50$  and  $100$  meters.

	Looking at the radio signals  $s(t)$  and  $r(t)$  in the equivalent low pass range  $($that is, around the frequency  $f = 0)$,  the signal transmission is described completely by the equation		Looking at the radio signals  $s(t)$  and  $r(t)$  in the equivalent low-pass range  $($that is, around the frequency  $f = 0)$,  the signal transmission is described completely by the equation
	:$$r(t)= z(t) \cdot s(t)$$		:$$r(t)= z(t) \cdot s(t)$$

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	</quiz>		</quiz>

	===~~Solutions~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From  ${\rm Max}[a(t)] = 2$  follows directly:		'''(1)'''  From  ${\rm Max}[a(t)] = 2$  follows directly:

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	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Wahrscheinlichkeitsdichte des Rayleigh–Fadings~~}}		{{quiz-Header\|Buchseite=Mobile_Communications/Probability_Density_of_Rayleigh_Fading}}

	[[File:P_ID2106__Mob_A_1_3.png\|right\|frame\|Time evolution of Rayleigh fading]]		[[File:P_ID2106__Mob_A_1_3.png\|right\|frame\|Time evolution of Rayleigh fading]]

Guenter at 16:38, 4 December 2020

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	* on the left side the probability  ${\rm Pr}(p(t) ≤ 1)$,		* on the left side the probability  ${\rm Pr}(p(t) ≤ 1)$,
	* on the right side the probability  ${\rm Pr}(0.5 \le p(t) ≤ 2)$.		* on the right side the probability  ${\rm Pr}(0.5 \le p(t) ≤ 2)$.
			<br clear=all>
			'''(4)'''  From  $10 \cdot {\rm lg} \ p_1 = \ -3 \ \ \rm dB$  follows  $p_1 = 0.5$.  The upper limit of the integration range results from the condition  $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$,  so  $p_2 = 2$.

	'''(4)'''  From  $10 \cdot {\rm lg} \ p_1 = \ -3 \ \ \rm dB$ follows $p_1 = 0.5$.  The upper limit of the integration range results from the condition  $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$, so  $p_2 = 2$.
	*This gives, according to the above graph:		*This gives, according to the above graph:
	$${\rm Pr}(-3\,\,{\rm dB}\le 10 \cdot {\rm lg}\hspace{0.15cm}p(t) \le +3\,\,{\rm dB}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \int_{0.5}^{2}f_p(p)\hspace{0.15cm}{\rm d}p =		$${\rm Pr}(-3\,\,{\rm dB}\le 10 \cdot {\rm lg}\hspace{0.15cm}p(t) \le +3\,\,{\rm dB}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \int_{0.5}^{2}f_p(p)\hspace{0.15cm}{\rm d}p =