Aufgaben:Exercise 1.3Z: Rayleigh Fading Revisited - Versionsgeschichte

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Javier verschob die Seite Exercises:Exercise 1.3Z: Rayleigh Fading Revisited nach Exercise 1.3Z: Rayleigh Fading Revisited

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@@ Zeile 1: / Zeile 1: @@
 {{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; $z(t)$]]
+[[Datei:EN_Mob_A_1_3Z.png|right|frame|Two channels, characterized by complex factor&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; $\sigma_{\rm R} = 0.5$.

@@ Zeile 118: / Zeile 118: @@
-'''(7)'''&nbsp; For the channel &nbsp;$\rm R$&nbsp; applies with $\sigma = 0.5$:
+'''(7)'''&nbsp; For the channel &nbsp;$\rm R$&nbsp; and $\sigma = 0.5$, we have
 $${\rm Pr}(|z(t)| > 1) = {\rm e}^{-2} \hspace{0.15cm} \underline{\approx 0.135}
   \hspace{0.05cm}.$$
@@ Zeile 124: / Zeile 124: @@
 *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 $\sigma^2 = 0.5$, ${\rm Pr}(|z(t)|>1) = {\rm e}^{\rm &ndash;1} applies. \ \underline {\approx \ 0.368}$.
+*For the channel &nbsp;$\rm B$&nbsp; because of the double variance $\sigma^2 = 0.5$, we have ${\rm Pr}(|z(t)|>1) = {\rm e}^{-1}. \ \underline {\approx \ 0.368}$.
-*The (not drawn) reference circle would also have the radius 1 in the lower graph.
+*The (not drawn) reference circle would also have radius 1 in the bottom graph.
-*The circle drawn in the lower graphic 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 $A = 1$, namely $A = \sqrt{2}\approx $1,414$.
 {{ML-Fuß}}
 [[Category:Exercises for Mobile Communications|^1.2 PDF of Rayleigh Fading^]]

@@ Zeile 122: / Zeile 122: @@
   \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 $N = 10,000$ of all points.
 *For the channel &nbsp;$\rm B$&nbsp; because of the double variance $\sigma^2 = 0.5$, ${\rm Pr}(|z(t)|>1) = {\rm e}^{\rm &ndash;1} applies. \ \underline {\approx \ 0.368}$.

@@ Zeile 127: / Zeile 127: @@
 *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.
+*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^]]

@@ Zeile 88: / Zeile 88: @@
-'''(3)'''&nbsp; <u>correct is NO</u>:
+'''(3)'''&nbsp; <u>NO</u>:
-*In both cases $f_{\it \phi}(\phi)$ describes an equal distribution between $-\pi$ and $+\pi$.
+*In both cases $f_{\it \phi}(\phi)$ describes a uniform 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>:
+'''(4)'''&nbsp; <u>Solution 2</u> is correct:
-*With Rayleigh&ndash;Fading, the real part $x(t)$ and the imaginary part $y(t)$ are each Gaussian distributed.
+*With Rayleigh 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)'''.
+'''(5)'''&nbsp; <u>Solution 3</u> is correct, 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:
+'''(6)'''&nbsp; The magnitude $a(t)$ is Rayleigh distributed. 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}.$$
 *In some formula collections you can find the solution for this integral, but not in all.
@@ Zeile 114: / Zeile 114: @@
 $${\rm Pr}(a > A) = {\rm e}^{ -{A^2}/(2\sigma^2)} \hspace{0.05cm}.$$
-Correct is therefore the <u>solution 2</u>.
+The correct answer is therefore <u>solution 2</u>.

@@ Zeile 17: / Zeile 17: @@
 \\  {\rm f\ddot{u}r}\hspace{0.15cm} p < 0 \\ \end{array}
 .$$
-* From channel &nbsp;$\rm B$&nbsp; ("Blue") only the point cloud is given. It must be estimated whether <i>Rayleigh&ndash;Fading</i>&nbsp; is also present here, and if YES, how large the parameter&nbsp; $\sigma = \sigma_{\rm B}$&nbsp; is for this channel.
+* For channel &nbsp;$\rm B$&nbsp; ("Blue") only the point cloud is given. It must be estimated whether Rayleigh fading is also present here, and if YES, how large the parameter&nbsp; $\sigma = \sigma_{\rm B}$&nbsp; is for this channel.
-* Finally, subtask '''(3)''' also refers to PDF&nbsp; $f_{\it \phi}(\phi)$&nbsp; the phase function&nbsp; $\phi(t)$&nbsp;. This is defined as follows:
+* Finally, subtask '''(3)''' also refers to PDF&nbsp; $f_{\it \phi}(\phi)$&nbsp; of the phase function&nbsp; $\phi(t)$&nbsp;, which is defined as follows:
 $$\phi(t) = \arctan \hspace{0.15cm} \frac{y(t)}{x(t)}
@@ Zeile 29: / Zeile 29: @@
 ''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.
+* This task belongs to chapter&nbsp; [[Mobile_Kommunikation/Wahrscheinlichkeitsdichte_des_Rayleigh%E2%80%93Fadings|Wahrscheinlichkeitsdichte des 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;.
+* A similar topic is treated with a different approach in chapter&nbsp; [[Stochastische_Signaltheorie/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;.
@@ Zeile 37: / Zeile 37: @@
 ===Questionnaire===
 <quiz display=simple>
-Can the channel &nbsp;$\rm B$&nbsp; also be modeled by &bdquo;<i>Rayleigh</i>&rdquo;?
+{Can the channel &nbsp;$\rm B$&nbsp; also be modeled as Rayleigh fading?
 |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; $\sigma_{\rm R} = 0.5$.
+{Estimate the Rayleigh parameter of channel $\rm B$. 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?
+{Are the probability density functions (PDFs) $f_{\it \phi}(\phi)$ of the phases different for channel &nbsp;$\rm R$&nbsp; and &nbsp;$\rm B$&nbsp; and if YES, how?
 |type="()"}
-- Yeah.
+- Yes.
 + No.
-{What is the course of WDF&nbsp; $f_a(a)$&nbsp; with&nbsp; $a(t) = |z(t)|$ in both cases?
+{What is the PDF&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 Gaussian distributed.
-+ The amount&nbsp; $a(t)$&nbsp; is rayleigh distributed.
++ The amount&nbsp; $a(t)$&nbsp; is Rayleigh distributed.
-- The amount&nbsp; $a(t)$&nbsp; is positive&ndash;exponentially distributed.
+- The amount&nbsp; $a(t)$&nbsp; is positive and exponentially distributed.
-{What is the course of the WDF&nbsp; $f_p(p)$&nbsp; in both cases with &nbsp;$p(t) = |z(t)|^2$?
+{What is the PDF&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 Gaussian distributed.
-- The amount&nbsp; $p(t)$&nbsp; is rayleigh distributed.
+- The amount&nbsp; $p(t)$&nbsp; is Rayleigh distributed.
-+ The amount&nbsp; $p(t)$&nbsp; is positive&ndash;exponentially distributed.
++ The amount&nbsp; $p(t)$&nbsp; is positive and exponentially distributed.
-{What is the probability that the amount&nbsp; $a(t) = |z(t)|$&nbsp; is greater than a given value&nbsp; $A$?
+{What is the probability that the magnitude&nbsp; $a(t) = |z(t)|$&nbsp; is greater than a given value&nbsp; $A$?
 |type="()"}
-- It applies&nbsp; ${\rm Pr}(|z(t)|) > A) = {\rm ln}(A/\sigma).$
+- ${\rm Pr}(|z(t)|) > A) = {\rm ln}(A/\sigma).$
-+ The following applies&nbsp; ${\rm Pr}(|z(t)|) > A) = {\rm e}^{-A^2/2\sigma^2}.$
++ ${\rm Pr}(|z(t)|) > A) = {\rm e}^{-A^2/2\sigma^2}.$
-- It applies&nbsp; ${\rm Pr}(|z(t)|) > A) = {\rm e}^{-A/\sigma}.$
+- ${\rm Pr}(|z(t)|) > A) = {\rm e}^{-A/\sigma}.$
 {Calculate the probability&nbsp; ${\rm Pr}(|z(t)| > 1)$ for both channels.
@@ Zeile 73: / Zeile 73: @@
 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
+</quiz>
 ===Sample solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>correct is YES</u>:
+'''(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.
 *Furthermore the following questions would not make sense if you answer NO.

@@ Zeile 6: / Zeile 6: @@
 * 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;$\sigma = \sigma_{\rm R})$:
 :$$f_a(a) =
-\left\{ \begin{array}{c} a/\sigma^2 \cdot {\rm e}^{ -{a^2}/(2\sigma^2)} \\\
+\left\{ \begin{array}{c} a/\sigma^2 \cdot {\rm e}^{ -{a^2}/(2\sigma^2)} \\
 \end{array} \right.\quad
-\begin{array}{*{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} a \ge 0
+\begin{array}{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} a \ge 0
-\\ {\rm f\ddot{u}r}\hspace{0.15cm} a < 0 \\\ \\ \end{array}
+\\  {\rm f\ddot{u}r}\hspace{0.15cm} a < 0 \\ \end{array}
   \hspace{0.05cm},$$
 :$$f_p(p) =
-\left\{ \begin{array}{c} 1/(2\sigma^2) \cdot {\rm e}^{ -{p}/(2\sigma^2)} \\\
+\left\{ \begin{array}{c} 1/(2\sigma^2) \cdot {\rm e}^{ -{p}/(2\sigma^2)} \\
 \end{array} \right.\quad
-\begin{array}{*{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} p \ge 0
+\begin{array}{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} p \ge 0
-\\ {\rm f\ddot{u}r}\hspace{0.15cm} p < 0 \\\ \\ \end{array}
+\\  {\rm f\ddot{u}r}\hspace{0.15cm} p < 0 \\ \end{array}
 .$$
 * 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.

@@ Zeile 2: / Zeile 2: @@
 [[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; two mobile radio channels (both without multipath propagation) in 2D&ndash;representation. The default is saved:
+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; the point cloud) is rayleigh distributed with&nbsp; $\sigma_{\rm R} = 0.5$.
 * For the probability density function (WDF) of amount&nbsp; $a(t) = |z(t)|$&nbsp; or &nbsp; amount square $p(t) = |z(t)|^2$&nbsp; the following equations $($with &nbsp;$\sigma = \sigma_{\rm R})$ thus apply: