Aufgaben:Exercise 1.3: Rayleigh Fading - Versionsgeschichte

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

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← Nächstältere Version		Version vom 24. Februar 2026, 15:40 Uhr
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	* The magnitude  $a(t) = \|z(t)\|$  has a Rayleigh PDF, from which the name <i>Rayleigh Fading</i> is derived:		* The magnitude  $a(t) = \|z(t)\|$  has a Rayleigh PDF, from which the name <i>Rayleigh Fading</i> 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 f\ddot{u}r}\hspace{0.15cm} a \ge 0\\ {\rm f\ddot{u}r}\hspace{0.15cm} a < 0 \\ \end{array}\hspace{0.05cm}.$$
	\left\{ \begin{array}{c} a/\sigma^2 \cdot {\rm e}^ { -a^2/(2\sigma^2)} \\
	0 \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} a \ge 0
	\\ {\rm f\ddot{u}r}\hspace{0.15cm} a < 0 \\ \end{array}
	\hspace{0.05cm}.$$

	* 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 f\ddot{u}r}\hspace{0.15cm} p \ge 0\\ {\rm f\ddot{u}r}\hspace{0.15cm} p < 0 \\ \end{array}\hspace{0.05cm}.$$
	0 \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} p \ge 0
	\\ {\rm f\ddot{u}r}\hspace{0.15cm} p < 0 \\ \end{array}
	\hspace{0.05cm}.$$

	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.
Zeile 71:		Zeile 61:
	{{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 amount $a(t)$. Since $p(t)$ is a power quantity		*The logarithmic representation of the squared magnitude $p(t)$ is identical to the logarithmic representation of the amount $a(t)$. 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 $\underline{\approx 6\,\,{\rm dB}}$.
Zeile 89:		Zeile 76:
	'''(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
	$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{p}{2\sigma^2}]		$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{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:

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

	The graph shows		The graph shows
Zeile 109:		Zeile 92:
	'''(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}.$$
	\left [ - {\rm e}^{ -{p}/(2\sigma^2)}\hspace{0.15cm} \right ]_{0.5}^{2} ={\rm e}^{-0.25}- {\rm e}^{-1} \approx 0.779 - 0.368 \hspace{0.15cm} \underline{ = 0.411} \hspace{0.05cm}.$$

Guenter am 11. Mai 2020 um 13:06 Uhr

2020-05-11T13:06:51Z

← Nächstältere Version		Version vom 11. Mai 2020, 15:06 Uhr
Zeile 91:		Zeile 91:
	$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{p}{2\sigma^2}]		$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{p}{2\sigma^2}]
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	[[File:~~P_ID2112__Mob_A_1_3c~~.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:

Javier am 25. März 2020 um 16:52 Uhr

2020-03-25T16:52:01Z

← Nächstältere Version		Version vom 25. März 2020, 18:52 Uhr
Zeile 107:		Zeile 107:


	'''(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 $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$ to $p_2 = 2$.		'''(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 =

Javier am 25. März 2020 um 16:51 Uhr

2020-03-25T16:51:25Z

← Nächstältere Version		Version vom 25. März 2020, 18:51 Uhr
Zeile 107:		Zeile 107:


	'''(4)''  From $10 \cdot {\rm lg} \ p_1 = \ –3 \ \ \rm dB$ follows $p_1 = 0.5$ ~~and the~~ upper limit of the integration range results from the condition $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$ to $p_2 = 2$.		'''(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 $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$ to $p_2 = 2$.
	*This gives ~~you~~, according to the above ~~graphic~~:		*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}.$$		\left [ - {\rm e}^{ -{p}/(2\sigma^2)}\hspace{0.15cm} \right ]_{0.5}^{2} ={\rm e}^{-0.25}- {\rm e}^{-1} \approx 0.779 - 0.368 \hspace{0.15cm} \underline{ = 0.411} \hspace{0.05cm}.$$

Javier am 25. März 2020 um 16:50 Uhr

2020-03-25T16:50:31Z

← Nächstältere Version		Version vom 25. März 2020, 18:50 Uhr
Zeile 91:		Zeile 91:
	$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{p}{2\sigma^2}]		$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{p}{2\sigma^2}]
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	[[File:P_ID2112__Mob_A_1_3c.png\|right\|frame\|~~WDF~~ and probability regions ]]		[[File:P_ID2112__Mob_A_1_3c.png\|right\|frame\|PDF and probability regions ]]
	*It follows:		*It follows:

Javier am 25. März 2020 um 16:49 Uhr

2020-03-25T16:49:56Z

← Nächstältere Version		Version vom 25. März 2020, 18:49 Uhr
Zeile 70:		Zeile 70:
	===Sample solution===		===Sample solution===
	{{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 ~~square of the amount~~ $p(t)$ is identical to the logarithmic representation of the amount $a(t)$. Since $p(t)$ is a power quantity		*The logarithmic representation of the squared magnitude $p(t)$ is identical to the logarithmic representation of the amount $a(t)$. Since $p(t)$ is a power quantity
	$$~~10 \cdot~~ {\rm lg}\~~hspace{0.15cm}~~ p(t) = ~~10 \cdot~~ {\rm lg}\~~hspace{0.15cm}~~a(t)^2 ~~= 20 \cdot {~~\~~rm lg}~~\hspace{0.15cm} ~~a(t)~~		:$${\rm Max} \left [ p(t) \right ] = {\rm Max} \left [ a(t)^2 \right ] \hspace{0.15cm} \underline{= 4}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

Zeile 88:		Zeile 88:

	'''(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 $p ≥ 0$ ~~applies accordingly:~~		*The absolute square is known to be exponentially distributed, and for $p ≥ 0$ we have
	$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{p}{2\sigma^2}]		$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{p}{2\sigma^2}]
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
Zeile 100:		Zeile 100:
	\underline{\sigma = 1} \hspace{0.05cm}.$$		\underline{\sigma = 1} \hspace{0.05cm}.$$

	The ~~graphic~~ shows		The graph shows
	* left the probability  ${\rm Pr}(p(t) ≤ 1)$,		* on the left side the probability  ${\rm Pr}(p(t) ≤ 1)$,
	* right the probability  ${\rm Pr}(0.5 \le p(t) ≤ 2)$.		* on the right side the probability  ${\rm Pr}(0.5 \le p(t) ≤ 2)$.

Javier am 25. März 2020 um 16:46 Uhr

2020-03-25T16:46:51Z

← Nächstältere Version		Version vom 25. März 2020, 18:46 Uhr
Zeile 63:		Zeile 63:
	$\sigma \ = \ $ { 1 3% }		$\sigma \ = \ $ { 1 3% }

	{What is the probability that the logarithmic Rayleigh coefficient   ~~⇒  ~~ $10 \cdot {\rm lg} \ p(t)$  is ~~in the range~~ between   $\text{-3 dB}$  and  $\text{+3 dB}$?		{What is the probability that the logarithmic Rayleigh coefficient   $10 \cdot {\rm lg} \ p(t)$  is between   $\text{-3 dB}$  and  $\text{+3 dB}$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(\|10 \cdot {\rm lg} \ p(t)\| < 3 \ \rm dB) \ = \ $ { 0.411 3% }		${\rm Pr}(\|10 \cdot {\rm lg} \ p(t)\| < 3 \ \rm dB) \ = \ $ { 0.411 3% }

Javier am 25. März 2020 um 16:45 Uhr

2020-03-25T16:45:55Z

← Nächstältere Version		Version vom 25. März 2020, 18:45 Uhr
Zeile 63:		Zeile 63:
	$\sigma \ = \ $ { 1 3% }		$\sigma \ = \ $ { 1 3% }

	{What is the probability of the logarithmic Rayleigh~~–size~~   ⇒   $10 \cdot {\rm lg} \ p(t)$  in the range ~~between~~ between  $\text{-3 dB}$  and  $\text{+3 dB}$?		{What is the probability that the logarithmic Rayleigh coefficient   ⇒   $10 \cdot {\rm lg} \ p(t)$  is in the range between   $\text{-3 dB}$  and  $\text{+3 dB}$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(\|10 \cdot {\rm lg} \ p(t)\| < 3 \ \rm dB) \ = \ $ { 0.411 3% }		${\rm Pr}(\|10 \cdot {\rm lg} \ p(t)\| < 3 \ \rm dB) \ = \ $ { 0.411 3% }

Javier am 25. März 2020 um 16:44 Uhr

2020-03-25T16:44:54Z

← Nächstältere Version		Version vom 25. März 2020, 18:44 Uhr
Zeile 56:		Zeile 56:
	{What is the maximum value of  $p(t) = \|z(t)\|^2$  both in linear and logarithmic representation?		{What is the maximum value of  $p(t) = \|z(t)\|^2$  both in linear and logarithmic representation?
	\|type="{}"}		\|type="{}"}
	${\rm Max}\big[p(t)\big] \ = \ $ {~~\ $~~ 4 3% }		${\rm Max}\big[p(t)\big] \ = \ $ { 4 3% }
	${\rm Max}\big [10 \cdot {\rm lg} \ p(t)\big] \ = \ $ { 6 3% } $ \ \rm dB$		${\rm Max}\big[10 \cdot {\rm lg} \ p(t)\big] \ = \ $ { 6 3% } $ \ \rm dB$

	{Let   ${\rm Pr}\big[a(t) ≤ 1\big] = $0.394 Determine the Rayleigh~~–~~parameter  $\sigma$.		{Let   ${\rm Pr}\big[a(t) ≤ 1\big] = $0.394 Determine the Rayleigh parameter  $\sigma$.
	\|type="{}"}		\|type="{}"}
	$\sigma \ = \ $ { 1 3% }		$\sigma \ = \ $ { 1 3% }