Quelltext der Seite Aufgaben:Exercise 1.3Z: Rayleigh Fading Revisited

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

[[Datei:P_ID2107__Mob_Z_1_3.png|right|frame|Zwei Kanäle, gekennzeichnet durch komplexen Faktor&nbsp; $z(t)$]]
The graph shows the multiplicative factor&nbsp; $z(t) = x(t) + {\rm j} \cdot y(t)$&nbsp; of two mobile radio channels (both without multipath propagation) in 2D&ndash;representation. The following is assumed:
* The channel &nbsp;$\rm R$&nbsp; (the designation results from the color &bdquo;Red&rdquo; of the point cloud) 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;$\sigma = \sigma_{\rm R})$:
:$$f_a(a) =
\left\{ \begin{array}{c} a/\sigma^2 \cdot {\rm e}^{ -{a^2}/(2\sigma^2)} \\
0  \end{array} \right.\quad
\begin{array}{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} a \ge 0
\\  {\rm f\ddot{u}r}\hspace{0.15cm} a < 0 \\ \end{array}
 \hspace{0.05cm},$$
:$$f_p(p) =
\left\{ \begin{array}{c} 1/(2\sigma^2) \cdot {\rm e}^{ -{p}/(2\sigma^2)} \\
0  \end{array} \right.\quad
\begin{array}{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} p \ge 0
\\  {\rm f\ddot{u}r}\hspace{0.15cm} p < 0 \\ \end{array}
.$$
* From channel &nbsp;$\rm B$&nbsp; ("Blue") only the point cloud is given. It must be estimated whether <i>Rayleigh&ndash;Fading</i>&nbsp; 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 PDF&nbsp; $f_{\it \phi}(\phi)$&nbsp; the phase function&nbsp; $\phi(t)$&nbsp;. This is defined as follows:

$$\phi(t) = \arctan \hspace{0.15cm} \frac{y(t)}{x(t)}
 \hspace{0.05cm}.$$






''Notes:'' 
* The task belongs to chapter&nbsp; [[Mobile_Communications/Probability Density_of_Rayleigh%E2%80%93Fadings|Probability Density of Rayleigh&ndash;Fadings]] of this book. 
* A similar topic is treated with a different approach in chapter&nbsp; [[Stochastic_Signal Theory/Weitere_Verteilungen|Weitere Verteilungen]]&nbsp; of the book &bdquo;Stochastic Signal Theory&rdquo;.
* To check your results you can use the interactive applet&nbsp; [[Applets:WDF_VTF|WDF, VTF and Moments]]&nbsp; of the book &bdquo;Stochastic Signal Theory&rdquo;.
 


===Questionnaire===
<quiz display=simple>
Can the channel &nbsp;$\rm B$&nbsp; also be modeled by &bdquo;<i>Rayleigh</i>&rdquo;?
|type="()"}
+ Yes.
- No.

Estimate the Rayleigh&ndash parameter of channel $\rm B$&nbsp. Reminder: &nbsp; For channel &nbsp;$\rm R$&nbsp; this parameter has the value&nbsp; $\sigma_{\rm R} = 0.5$.
|type="{}"}
$\sigma_{\rm B}\ = \ $ { 0.707 3% } 

{Do the phases&ndash;probability density functions&nbsp; $f_{\it \phi}(\phi)$ differ from channel &nbsp;$\rm R$&nbsp; and &nbsp;$\rm B$&nbsp; and if YES, how?
|type="()"}
- Yeah.
+ No.

{What is the course of WDF&nbsp; $f_a(a)$&nbsp; with&nbsp; $a(t) = |z(t)|$ in both cases?
|type="()"}
- The amount&nbsp; $a(t)$&nbsp; is gaussian distributed.
+ The amount&nbsp; $a(t)$&nbsp; is rayleigh distributed.
- The amount&nbsp; $a(t)$&nbsp; is positive&ndash;exponentially distributed.

{What is the course of the WDF&nbsp; $f_p(p)$&nbsp; in both cases with &nbsp;$p(t) = |z(t)|^2$?
|type="()"}
- The amount&nbsp; $p(t)$&nbsp; is gaussian distributed.
- The amount&nbsp; $p(t)$&nbsp; is rayleigh distributed.
+ The amount&nbsp; $p(t)$&nbsp; is positive&ndash;exponentially distributed.

{What is the probability that the amount&nbsp; $a(t) = |z(t)|$&nbsp; is greater than a given value&nbsp; $A$?
|type="()"}
- It applies&nbsp; ${\rm Pr}(|z(t)|) > A) = {\rm ln}(A/\sigma).$
+ The following applies&nbsp; ${\rm Pr}(|z(t)|) > A) = {\rm e}^{-A^2/2\sigma^2}.$
- It applies&nbsp; ${\rm Pr}(|z(t)|) > A) = {\rm e}^{-A/\sigma}.$

{Calculate the probability&nbsp; ${\rm Pr}(|z(t)| > 1)$ for both channels.
|type="{}"}
Channel &nbsp;$\rm R$:$\hspace{0.4cm} {\rm Pr}(|z(t)| > 1)\ = \ $ { 0.135 3% }
Channel &nbsp;$\rm B$:$\hspace{0.4cm} {\rm Pr}(|z(t)| > 1)\ = \ $ { 0.368 3% }
</quiz

===Sample solution===
{{ML-Kopf}}
'''(1)'''&nbsp; <u>correct is 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. 
*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:
$$\sigma_{\rm B} = \sigma_{\rm R} \cdot \sqrt{2} = 0.5 \cdot \sqrt{2}= {1}/{\sqrt{2}}\hspace{0.15cm} \underline{\approx 0.707}
 \hspace{0.05cm}.$$


'''(3)'''&nbsp; <u>correct is NO</u>: 
*In both cases $f_{\it \phi}(\phi)$ describes an equal distribution between $-\pi$ and $+\pi$. 
*The larger amplitudes of channel &nbsp;$\rm B$&nbsp; are not important for the phase function $\phi(t)$.



'''(4)'''&nbsp; Correct is the <u>solution 2</u>: 
*With Rayleigh&ndash;Fading, the real part $x(t)$ and the imaginary part $y(t)$ are each Gaussian distributed. 
*The exponential distribution results for the square of the absolute value $p(t) = |z(t)|^2$.



'''(5)'''&nbsp; Correct here is the <u>solution 3</u>, as already explained in the sample solution for '''(4)'''.



'''(6)'''&nbsp; The amount $a(t)$ is rayleigh distributed. Therefore, the following applies for the searched 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}.$

*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 $p = a^2$:
$${\rm Pr}(a > A) = {\rm Pr}(p > A^2) = \frac{1}{2\sigma^2} \cdot\int_{A^2}^{\infty} {\rm e}^{ -{p}/(2\sigma^2)} \hspace{0.15cm}{\rm d}p \hspace{0.05cm}.$$

*This integral is elementary and gives the result:
$${\rm Pr}(a > A) = {\rm e}^{ -{A^2}/(2\sigma^2)} \hspace{0.05cm}.$$

Correct is therefore the <u>solution 2</u>.



'''(7)'''&nbsp; For the channel &nbsp;$\rm R$&nbsp; applies with $\sigma = 0.5$:
$${\rm Pr}(|z(t)| > 1) = {\rm e}^{-2} \hspace{0.15cm} \underline{\approx 0.135}
 \hspace{0.05cm}.$$

*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.

*For the channel &nbsp;$\rm B$&nbsp; because of the double variance $\sigma^2 = 0.5$, ${\rm Pr}(|z(t)|>1) = {\rm e}^{\rm &ndash;1} applies. \ \underline {\approx \ 0.368}$. 

*The (not drawn) reference circle would also have the radius 1 in the lower graph. 
*The circle drawn in the lower graphic has a larger radius than $A = $1, namely $A = \sqrt{2}\approx $1,414.
{{ML-Fuß}}


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