Aufgaben:Exercise 1.3: Rayleigh Fading - Revision history

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Guenter at 16:33, 4 December 2020

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Oezdemir at 20:41, 30 July 2020

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Guenter at 13:42, 26 July 2020

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

@@ Line 47: / Line 47: @@
 ''Notes:''
 * This task belongs to chapter&nbsp; [[Mobile_Communications/Probability_density_of_Rayleigh_fading|Probability density of Rayleigh fading]]&nbsp; of this book.
-* A similar topic is treated with a different approach in chapter&nbsp; [[Theory_of_Stochastic_Signals/Further_distributions|Further distributions]]&nbsp; of the book &bdquo;Stochastic Signal Theory&rdquo;.
+* A similar topic is treated with a different approach in chapter&nbsp; [[Theory_of_Stochastic_Signals/Further_distributions|Further distributions]]&nbsp; of the book "Stochastic Signal Theory&rdquo;.
-* To check your results, you can use the interactive applet&nbsp; [[Applets:PDF,_CDF_and_Moments_of_Special_Distributions|PDF, CDF and Moments of Special Distributions]]&nbsp; of the book &bdquo;Stochastic Signal Theory&rdquo;.
+* To check your results, you can use the interactive applet&nbsp; [[Applets:PDF,_CDF_and_Moments_of_Special_Distributions|PDF, CDF and Moments of Special Distributions]]&nbsp; of the book "Stochastic Signal Theory&rdquo;.

← Older revision		Revision as of 12:37, 23 March 2021
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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^]]

@@ Line 9: / Line 9: @@
 An example of such a Rayleigh channel occurs in urban mobile communications when narrow-band signals are used with ranges between&nbsp; $50$&nbsp; and&nbsp; $100$&nbsp; meters.
-Looking at the radio signals&nbsp; $s(t)$&nbsp; and&nbsp; $r(t)$&nbsp; in the equivalent low pass range&nbsp; $($that is, around the frequency &nbsp;$f = 0)$,&nbsp; the signal transmission is described completely by the equation
+Looking at the radio signals&nbsp; $s(t)$&nbsp; and&nbsp; $r(t)$&nbsp; in the equivalent low-pass range&nbsp; $($that is, around the frequency &nbsp;$f = 0)$,&nbsp; the signal transmission is described completely by the equation
 :$$r(t)= z(t) \cdot s(t)$$

@@ Line 73: / Line 73: @@
 </quiz>
-===Solutions===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; From&nbsp; ${\rm Max}[a(t)] = 2$&nbsp; follows directly:

← Older revision		Revision as of 13:56, 15 December 2020
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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]]

@@ Line 109: / Line 109: @@
 * on the left side the probability&nbsp; ${\rm Pr}(p(t) &#8804; 1)$,
 * on the right side the probability&nbsp; ${\rm Pr}(0.5 \le p(t) &#8804; 2)$.
+<br clear=all>
+'''(4)'''&nbsp; From&nbsp; $10 \cdot {\rm lg} \ p_1 = \ -3 \ \ \rm dB$&nbsp; follows&nbsp; $p_1 = 0.5$.&nbsp; The upper limit of the integration range results from the condition&nbsp; $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$,&nbsp; so&nbsp; $p_2 = 2$.
 *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 =

@@ Line 18: / Line 18: @@
 * The real part&nbsp; $x(t)$&nbsp; and the imaginary part&nbsp; $y(t)$&nbsp; are Gaussian mean-free random variables, both with equal variance&nbsp; $\sigma^2$.&nbsp; Within the components&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; there may be statistical dependence, but this is not relevant for the solution of the present task.&nbsp; We assume that&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; are uncorrelated.
-* The magnitude&nbsp; $a(t) = |z(t)|$&nbsp; has a Rayleigh PDF, from which the name <i>Rayleigh Fading</i> is derived:
+* The magnitude&nbsp; $a(t) = |z(t)|$&nbsp; has a Rayleigh PDF, from which the name "Rayleigh Fading" is derived:
 :$$f_a(a) =
 \left\{ \begin{array}{c} a/\sigma^2 \cdot {\rm e}^ { -a^2/(2\sigma^2)} \\
   \end{array} \right.\quad
-\begin{array}{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} a \ge 0
+\begin{array}{*{1}c} {\rm for}\hspace{0.15cm} a \ge 0
-\\  {\rm f\ddot{u}r}\hspace{0.15cm} a < 0 \\ \end{array}
+\\  {\rm for}\hspace{0.15cm} a < 0 \\ \end{array}
   \hspace{0.05cm}.$$
@@ Line 29: / Line 29: @@
 :$$f_p(p) = \left\{ \begin{array}{c} 1/(2\sigma^2) \cdot {\rm e}^ { -p/(2\sigma^2)} \\
   \end{array} \right.\quad
-\begin{array}{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} p \ge 0
+\begin{array}{*{1}c} {\rm for}\hspace{0.15cm} p \ge 0
-\\  {\rm f\ddot{u}r}\hspace{0.15cm} p < 0 \\ \end{array}
+\\  {\rm for}\hspace{0.15cm} p < 0 \\ \end{array}
   \hspace{0.05cm}.$$
@@ Line 37: / Line 37: @@
   \hspace{0.05cm}.$$
-In the lower graph, the value range between&nbsp; $\text{-3 dB}$&nbsp; and&nbsp; $\text{+3 dB}$&nbsp; of the logarithmic Rayleigh coefficient&nbsp; $20 \cdot {\rm lg} \ a(t)$ is highlighted in green.&nbsp; The subtask '''(4)''' refers to this.
+In the lower graph, the value range between&nbsp; $\text{-3 dB}$&nbsp; and&nbsp; $\text{+3 dB}$&nbsp; of the logarithmic Rayleigh coefficient&nbsp; $20 \cdot {\rm lg} \ a(t)$&nbsp; is highlighted in green.&nbsp; The subtask '''(4)'''&nbsp; refers to this.

@@ Line 52: / Line 52: @@
-===Questionnaire===
+===Questions===
 <quiz display=simple>
@@ Line 73: / Line 73: @@
 </quiz>
-===Sample solution===
+===Solutions===
 {{ML-Kopf}}
 '''(1)'''&nbsp; From&nbsp; ${\rm Max}[a(t)] = 2$&nbsp; follows directly:

@@ Line 38: / Line 38: @@
 In the lower graph, the value range between&nbsp; $\text{-3 dB}$&nbsp; and&nbsp; $\text{+3 dB}$&nbsp; of the logarithmic Rayleigh coefficient&nbsp; $20 \cdot {\rm lg} \ a(t)$ is highlighted in green.&nbsp; The subtask '''(4)''' refers to this.
@@ Line 73: / Line 75: @@
 ===Sample solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; From ${\rm Max}[a(t)] = 2$ follows directly:
+'''(1)'''&nbsp; From&nbsp; ${\rm Max}[a(t)] = 2$&nbsp; follows directly:
 :$${\rm Max} \left [ 20 \cdot {\rm lg}\hspace{0.15cm}a(t) \right ] = 20 \cdot {\rm lg}\hspace{0.15cm}(2) \hspace{0.15cm} \underline{\approx 6\,\,{\rm dB}}
    \hspace{0.05cm}.$$
-'''(2)'''&nbsp; The maximum value of the square $p(t) = a(t)^2$ is
+'''(2)'''&nbsp; The maximum value of the square&nbsp; $p(t) = a(t)^2$&nbsp; 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}.$$
-*The logarithmic representation of the squared magnitude $p(t)$ is identical to the logarithmic representation of the amount $a(t)$. Since $p(t)$ is a power quantity
+*The logarithmic representation of the squared magnitude&nbsp; $p(t)$&nbsp; is identical to the logarithmic representation of the magnitude&nbsp; $a(t)$.&nbsp;
 :$${\rm Max} \left [  p(t) \right ] = {\rm Max} \left [  a(t)^2 \right ]  \hspace{0.15cm} \underline{= 4}
    \hspace{0.05cm}.$$
-*The maximum value is thus also $\underline{\approx 6\,\,{\rm dB}}$.
+*The maximum value is thus also&nbsp; $\underline{\approx 6\,\,{\rm dB}}$.
-'''(3)'''&nbsp; The condition $a(t) &#8804; 1$ is equivalent to the requirement $p(t) = a(t)^2 &#8804; 1$.
+'''(3)'''&nbsp; The condition&nbsp; $a(t) &#8804; 1$&nbsp; is equivalent to the requirement&nbsp; $p(t) = a(t)^2 &#8804; 1$.
-*The absolute square is known to be exponentially distributed, and for $p &#8805; 0$ we have
+*The absolute square is known to be exponentially distributed, and for&nbsp; $p &#8805; 0$&nbsp; we have
-$$f_p(p) = \frac{1}{2\sigma^2} \cdot {\rm exp} [ -\frac{p}{2\sigma^2}]
+:$$f_p(p) =  \frac{1}{2\sigma^2} \cdot {\rm e}^{ -p/(2\sigma^2)}
   \hspace{0.05cm}.$$
 [[File:EN_Mob_A_1_3_c.png|right|frame|PDF and probability regions ]]
 *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 e}^{ -p/(2\sigma^2)} \hspace{0.15cm}{\rm d}p =
-- {\rm exp} [ -\frac{1}{2\sigma^2}] = 0.394$$
+- {\rm e}^{ -1/(2\sigma^2)} = 0.394$$
-$$\Rightarrow \hspace{0.3cm} {\rm exp} [ -\frac{1}{2\sigma^2}] = 0.606 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
+:$$\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}
+  \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}.$$
@@ Line 109: / Line 112: @@
+'''(4)'''&nbsp; From&nbsp; $10 \cdot {\rm lg} \ p_1 = \ -3 \ \ \rm dB$ follows $p_1 = 0.5$.&nbsp; The upper limit of the integration range results from the condition&nbsp; $10 \cdot {\rm lg} \ p_2 = +3 \ \ \rm dB$, so&nbsp; $p_2 = 2$.
 *This gives, according to the above graph:
 $${\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 =