Aufgaben:Exercise 1.3Z: Rayleigh Fading Revisited - Revision history

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Oezdemir at 20:42, 30 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 &bdquo;Stochastic Signal Theory".
+* 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 &bdquo;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".

@@ Line 30: / Line 30: @@
 ''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 &bdquo;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 &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 &bdquo;Stochastic Signal Theory".

← Older revision		Revision as of 12:38, 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^]]

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 </quiz>
-===Solutions===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>YES</u>:

← 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:EN_Mob_A_1_3Z.png\|right\|frame\|Two channels, characterized by complex factor  $z(t)$]]		[[File:EN_Mob_A_1_3Z.png\|right\|frame\|Two channels, characterized by complex factor  $z(t)$]]

@@ Line 100: / Line 100: @@
-'''(5)'''&nbsp; <u>Solution 3</u> is correct, as already explained in the sample solution for '''(4)'''.
+'''(5)'''&nbsp; <u>Solution 3</u> is correct, as already explained in the sample solution for&nbsp; '''(4)'''.
@@ Line 118: / Line 118: @@
-'''(7)'''&nbsp; For the channel &nbsp;$\rm R$&nbsp; and $\sigma = 0.5$, we have
+'''(7)'''&nbsp; For the channel &nbsp;$\rm R$&nbsp; and $\sigma = 0.5$ &nbsp; &rArr; &nbsp; $\sigma^2 = 0.25$, we have
 :$${\rm Pr}(|z(t)| > 1) = {\rm e}^{-2} \hspace{0.15cm} \underline{\approx 0.135}
   \hspace{0.05cm}.$$

@@ Line 4: / Line 4: @@
 The graph shows the multiplicative factor&nbsp; $z(t) = x(t) + {\rm j} \cdot y(t)$&nbsp; of two mobile radio channels&nbsp; (both without multipath propagation)&nbsp; in 2D&ndash;representation.&nbsp; The following is assumed:
 * The channel &nbsp;$\rm R$&nbsp; $($the designation results from the color "Red" of the point cloud$)$&nbsp; is Rayleigh distributed with&nbsp; $\sigma_{\rm R} = 0.5$.
-* The probability density functions (PDF) of the magnitude&nbsp; $a(t) = |z(t)|$&nbsp; or &nbsp; the squared magnitude $p(t) = |z(t)|^2$&nbsp; are $($with &nbsp;$\sigma = \sigma_{\rm R})$:
+* The probability density functions&nbsp $\rm (PDF)$&nbsp; of the magnitude&nbsp; $a(t) = |z(t)|$&nbsp; or &nbsp; the squared magnitude $p(t) = |z(t)|^2$&nbsp; are $($with &nbsp;$\sigma = \sigma_{\rm R})$:
 :$$f_a(a) =
 \left\{ \begin{array}{c} a/\sigma^2 \cdot {\rm e}^{ -{a^2}/(2\sigma^2)} \\
   \end{array} \right.\quad
-\begin{array}{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} a \ge 0
+\begin{array}{*{1}c} {\rm for}\hspace{0.15cm} a \ge 0
-\\  {\rm f\ddot{u}r}\hspace{0.15cm} a < 0 \\ \end{array}
+\\  {\rm for}\hspace{0.15cm} a < 0 \\ \end{array}
   \hspace{0.05cm},$$
 :$$f_p(p) =
 \left\{ \begin{array}{c} 1/(2\sigma^2) \cdot {\rm e}^{ -{p}/(2\sigma^2)} \\
   \end{array} \right.\quad
-\begin{array}{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} p \ge 0
+\begin{array}{*{1}c} {\rm for}\hspace{0.15cm} p \ge 0
-\\  {\rm f\ddot{u}r}\hspace{0.15cm} p < 0 \\ \end{array}
+\\  {\rm for}\hspace{0.15cm} p < 0 \\ \end{array}
 .$$
-* For channel &nbsp;$\rm B$&nbsp; ("Blue") only the point cloud is given. It must be estimated whether Rayleigh fading is also present here, and if YES, how large the parameter&nbsp; $\sigma = \sigma_{\rm B}$&nbsp; is for this channel.
+* For channel &nbsp;$\rm B$&nbsp; ("Blue") only the point cloud is given.&nbsp It must be estimated whether Rayleigh fading is also present here, and if YES, how large the parameter&nbsp; $\sigma = \sigma_{\rm B}$&nbsp; is for this channel.
-* Finally, subtask '''(3)''' also refers to the PDF&nbsp; $f_{\it \phi}(\phi)$&nbsp; of the phase function&nbsp; $\phi(t)$&nbsp;, which is defined as follows:
+* Finally, subtask&nbsp; '''(3)'''&nbsp; also refers to the PDF&nbsp; $f_{\it \phi}(\phi)$&nbsp; of the phase function&nbsp; $\phi(t)$&nbsp;, which is defined as follows:
 :$$\phi(t) = \arctan \hspace{0.15cm} \frac{y(t)}{x(t)}

@@ Line 35: / Line 35: @@
-===Questionnaire===
+===Questions===
 <quiz display=simple>
 {Can the channel &nbsp;$\rm B$&nbsp; also be modeled as Rayleigh fading?
@@ Line 75: / Line 75: @@
 </quiz>
-===Sample solution===
+===Solutions===
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>YES</u>:

@@ Line 78: / Line 78: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>YES</u>:
-*You can see the rotational symmetry here, too, if you consider that only $N = 10\hspace{0.05cm}000$ samples were displayed in the complex plane.
+*You can see the rotational symmetry here, too, if you consider that only&nbsp; $N = 10\hspace{0.05cm}000$&nbsp; samples were displayed in the complex plane.
-*Furthermore the following questions would not make sense if you answer NO.
+*Furthermore the following questions would not make sense if you answer "NO".
-'''(2)'''&nbsp; By measuring the two drawn circles you can see that for the &bdquo;blue&rdquo; channel the scattering of real&ndash; and imaginary part are larger by a factor of about 1.4 (exactly: $\sqrt{2}$) than for the &bdquo;red&rdquo; channel:
+'''(2)'''&nbsp; By measuring the two drawn circles you can see that for the "blue" channel the scattering of real&ndash; and imaginary part are larger by a factor of about&nbsp; $1.4$&nbsp; <br>$($exactly:&nbsp; $\sqrt{2})$&nbsp; than for the "red" channel:
 $$\sigma_{\rm B} = \sigma_{\rm R} \cdot \sqrt{2} = 0.5 \cdot \sqrt{2}= {1}/{\sqrt{2}}\hspace{0.15cm} \underline{\approx 0.707}
   \hspace{0.05cm}.$$
@@ Line 89: / Line 89: @@
 '''(3)'''&nbsp; <u>NO</u>:
-*In both cases $f_{\it \phi}(\phi)$ describes a uniform distribution between $-\pi$ and $+\pi$.
+*In both cases&nbsp; $f_{\it \phi}(\phi)$&nbsp; describes a uniform distribution between&nbsp; $-\pi$&nbsp; and&nbsp; $+\pi$.
-*The larger amplitudes of channel &nbsp;$\rm B$&nbsp; are not important for the phase function $\phi(t)$.
+*The larger amplitudes of channel&nbsp; &nbsp;$\rm B$&nbsp; are not important for the phase function&nbsp; $\phi(t)$.
 '''(4)'''&nbsp; <u>Solution 2</u> is correct:
-*With Rayleigh fading, the real part $x(t)$ and the imaginary part $y(t)$ are each Gaussian distributed.
+*With Rayleigh fading, the real part&nbsp; $x(t)$&nbsp; and the imaginary part&nbsp; $y(t)$&nbsp; are each Gaussian distributed.
-*The exponential distribution results for the square of the absolute value $p(t) = |z(t)|^2$.
+*The exponential distribution results for the square of the absolute value&nbsp; $p(t) = |z(t)|^2$.
@@ Line 104: / Line 104: @@
-'''(6)'''&nbsp; The magnitude $a(t)$ is Rayleigh distributed. Therefore, the following holds for the desired probability:
+'''(6)'''&nbsp; The magnitude&nbsp; $a(t)$ &nbsp;is Rayleigh distributed.&nbsp; Therefore, the following holds for the desired probability:
-$${\rm Pr}(a > A) = \int_{A}^{\infty}\frac{a}{\sigma^2} \cdot {\rm e}^{ -{a^2}/(2\sigma^2)} \hspace{0.15cm}{\rm d}a \hspace{0.05cm}.$$
+:$${\rm Pr}(a > A) = \int_{A}^{\infty}\frac{a}{\sigma^2} \cdot {\rm e}^{ -{a^2}/(2\sigma^2)} \hspace{0.15cm}{\rm d}a \hspace{0.05cm}.$$
 *In some formula collections you can find the solution for this integral, but not in all.
-*But it is also valid with the one-sided&ndash;exponentially distributed random variable $p = a^2$:
+*But it is also valid with the one-sided exponentially distributed random variable&nbsp; $p = a^2$:
-$${\rm Pr}(a > A) = {\rm Pr}(p > A^2) = \frac{1}{2\sigma^2} \cdot\int_{A^2}^{\infty} {\rm e}^{ -{p}/(2\sigma^2)} \hspace{0.15cm}{\rm d}p \hspace{0.05cm}.$$
+:$${\rm Pr}(a > A) = {\rm Pr}(p > A^2) = \frac{1}{2\sigma^2} \cdot\int_{A^2}^{\infty} {\rm e}^{ -{p}/(2\sigma^2)} \hspace{0.15cm}{\rm d}p \hspace{0.05cm}.$$
 *This integral is elementary and gives the result:
-$${\rm Pr}(a > A) = {\rm e}^{ -{A^2}/(2\sigma^2)} \hspace{0.05cm}.$$
+:$${\rm Pr}(a > A) = {\rm e}^{ -{A^2}/(2\sigma^2)} \hspace{0.05cm}.$$
 The correct answer is therefore <u>solution 2</u>.
@@ Line 119: / Line 119: @@
 '''(7)'''&nbsp; For the channel &nbsp;$\rm R$&nbsp; and $\sigma = 0.5$, we have
-$${\rm Pr}(|z(t)| > 1) = {\rm e}^{-2} \hspace{0.15cm} \underline{\approx 0.135}
+:$${\rm Pr}(|z(t)| > 1) = {\rm e}^{-2} \hspace{0.15cm} \underline{\approx 0.135}
   \hspace{0.05cm}.$$
-*In the upper graph this corresponds to the number of all points outside the circle drawn, based on the number $N = 10,000$ of all points.
+*In the upper graph this corresponds to the number of all points outside the circle drawn, based on the number&nbsp; $N = 10,000$&nbsp; of all points.
-*For the channel &nbsp;$\rm B$&nbsp; because of the double variance $\sigma^2 = 0.5$, we have ${\rm Pr}(|z(t)|>1) = {\rm e}^{-1}. \ \underline {\approx \ 0.368}$.
+*For the channel &nbsp;$\rm B$&nbsp; because of the double variance&nbsp; $\sigma^2 = 0.5$, we have&nbsp; ${\rm Pr}(|z(t)|>1) = {\rm e}^{-1} \ \underline {\approx \ 0.368}$.
-*The (not drawn) reference circle would also have radius 1 in the bottom graph.
+*The (not drawn) reference circle would also have radius&nbsp; $1$&nbsp; in the bottom graph.
-*The circle drawn in the bottom graph has a larger radius than $A = 1$, namely $A = \sqrt{2}\approx $1,414$.
+*The circle drawn in the bottom graph has a larger radius than&nbsp; $A = 1$,&nbsp; namely&nbsp; $A = \sqrt{2}\approx 1.414$.
 {{ML-Fuß}}
 [[Category:Exercises for Mobile Communications|^1.2 PDF of Rayleigh Fading^]]

@@ Line 46: / Line 46: @@
 $\sigma_{\rm B}\ = \ $ { 0.707 3% }
-{Are the probability density functions&nbsp; (PDFs)&nbsp; $f_{\it \phi}(\phi)$&nbsp; of the phases different for channel &nbsp;$\rm R$&nbsp; and &nbsp;$\rm B$&nbsp; and if YES,&nbsp; how?
+{Are the probability density functions&nbsp; (PDFs)&nbsp; $f_{\it \phi}(\phi)$&nbsp; of the phases different for channel &nbsp;$\rm R$&nbsp; and &nbsp;$\rm B$&nbsp; and if &nbsp; YES, &nbsp; how?
 |type="()"}
 - Yes.