Difference between revisions of "Aufgaben:Exercise 3.10: Rayleigh Fading"

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{{quiz-Header|Buchseite=Stochastische Signaltheorie/Weitere Verteilungen
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{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Further_Distributions
 
}}
 
}}
  
[[File:P_ID177__Sto_A_3_10.png|right|frame|Zur Rayleigh-WDF]]
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[[File:P_ID177__Sto_A_3_10.png|right|frame|Rayleigh distribution:  real part, imaginary part, magnitude ]]
H&auml;ufig beschreibt man ein Bandpass&uuml;bertragungssystem im sogenannten <i>&auml;quivalenten Tiefpassbereich</i>.&nbsp; Diese Darstellung f&uuml;hrt entsprechend dem Kapitel &bdquo;Bandpassartige Signale&rdquo; im Buch&nbsp; [[Signaldarstellung]]&nbsp; zu einem komplexen Signal:
+
One often describes a band-pass transmission system in the so-called&nbsp; "equivalent low-pass domain".&nbsp; This representation leads according to the chapter&nbsp; "Band-pass signals"&nbsp; in the book&nbsp; [[Signal Representation]]&nbsp; to a complex signal:
 
:$$z(t)=x(t)+ {\rm j} \cdot y(t).$$
 
:$$z(t)=x(t)+ {\rm j} \cdot y(t).$$
  
Der Realteil&nbsp; $x(t)$&nbsp; kennzeichnet hierbei die&nbsp; <i>Inphasekomponente</i>&nbsp; und der Imagin&auml;rteil&nbsp; $y(t)$&nbsp; die&nbsp; <i>Quadraturkomponente</i>.
+
The real part&nbsp; $x(t)$&nbsp; denotes the&nbsp; "inphase component"&nbsp; and the imaginary part&nbsp; $y(t)$&nbsp; the&nbsp; "quadrature component".
  
Bei einem Mobilfunksystem, bei dem zwischen dem mobilen Teilnehmer und der Basisstation keine Sichtverbindung besteht, gelangt somit das Funksignal ausschlie&szlig;lich auf indirekten Wegen (Brechung, Streuung, Reflexion usw.) zum Empf&auml;nger.&nbsp;  
+
In a mobile radio system where there is no line of sight between the mobile user and the base station,&nbsp; the radio signal thus reaches the receiver exclusively by indirect means&nbsp; (refraction,&nbsp; scattering,&nbsp; reflection, etc.);  
  
In diesem Fall ist folgendes Modell anwendbar:
+
In this case,&nbsp; the following model is applicable:
*Der Realteil&nbsp; $x(t)$&nbsp; und auch der&nbsp; Imagin&auml;rteil $y(t)$&nbsp; sind jeweils gau&szlig;verteilt und mittelwertfrei.
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*The real part&nbsp; $x(t)$&nbsp; and also the imaginary part&nbsp; $y(t)$&nbsp; are both Gaussian distributed and zero mean.
*$x(t)$&nbsp; und&nbsp; $y(t)$&nbsp; besitzen jeweils die gleiche Streuung&nbsp; $\sigma$&nbsp; und sind voneinander unabh&auml;ngig.
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*$x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; each have the same standard deviation&nbsp; $\sigma$&nbsp; and are independent of each other.
*Innere Bindungen der Signale&nbsp; $x(t)$&nbsp; und&nbsp; $y(t)$&nbsp; aufgrund des Dopplereffekts sollen hier nicht beachtet werden.
+
*Internal bindings of the signals&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; due to the Doppler effect shall not be considered here.
  
  
Das komplexe Signal&nbsp; $z(t)$&nbsp; kann man auch nach Betrag und Phase darstellen:
+
The complex signal&nbsp; $z(t)$&nbsp; can also be represented by magnitude and phase:
:$$ z(t)= a(t)\cdot {\rm e}^{{\rm j\phi(t)}.$$
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:$$ z(t)= a(t)\cdot {\rm e}^{\rm j \phi(t)}.$$
*Aufgrund der Symmetrie bez&uuml;glich&nbsp; $x(t)$&nbsp; und&nbsp; $y(t)$&nbsp; ist die Phase&nbsp; $\phi(t)$&nbsp; gleichverteilt.  
+
*Because of symmetry with respect to&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; the phase&nbsp; $\phi(t)$&nbsp; is uniformly distributed.  
*Dagegen ist der Betrag&nbsp; $a(t) = |z(t)|$&nbsp; rayleighverteilt, was zu der Namensgebung &bdquo;Rayleighfading&rdquo; gef&uuml;hrt hat.
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*In contrast,&nbsp; the magnitude&nbsp; $a(t) = |z(t)|$&nbsp; is Rayleigh distributed,&nbsp; which has led to the naming&nbsp; "Rayleigh fading".
*Als weitere Gr&ouml;&szlig;e definieren wir noch die Momentanleistung
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*As a further quantity we define the instantaneous power
 
:$$ p(t)=x^{\rm 2}(t)+y^{\rm 2}( t)=a^{\rm 2}(t).$$
 
:$$ p(t)=x^{\rm 2}(t)+y^{\rm 2}( t)=a^{\rm 2}(t).$$
  
Die Zufallsgröße&nbsp; $p$&nbsp; ist hier&nbsp; [[Stochastische_Signaltheorie/Exponentialverteilte_Zufallsgrößen#Einseitige_Exponentialverteilung|(einseitig) exponentialverteilt]].&nbsp; Deren WDF lautet f&uuml;r&nbsp; $p>0$:
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The random variable&nbsp; $p$&nbsp; is&nbsp; [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#One-sided_exponential_distribution|(one-sided) exponentially distributed]].&nbsp; Its PDF is for&nbsp; $p>0$:
 
:$$f_p(p)=\frac{1}{2\sigma^{\rm 2}}\cdot {\rm e}^{ -p/(\sigma^{\rm 2})}.$$
 
:$$f_p(p)=\frac{1}{2\sigma^{\rm 2}}\cdot {\rm e}^{ -p/(\sigma^{\rm 2})}.$$
  
F&uuml;r alle negativen $p$&ndash;Werte gilt natürlich&nbsp; $f_p(p)= 0$, da&nbsp; $p$&nbsp; eine Leistung kennzeichnet.
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For all negative&nbsp; $p$&ndash;values, of course&nbsp; $f_p(p)= 0$&nbsp; holds,&nbsp; since&nbsp; $p$&nbsp; denotes a power.
  
  
  
  
 +
Hints:
 +
*This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Further_Distributions|Further Distributions]].
 +
*In particular, reference is made to the section&nbsp; [[Theory_of_Stochastic_Signals/Further_Distributions#Rayleigh_PDF|Rayleigh PDF]].
 +
*The standard deviation of the two Gaussian random variables&nbsp; $x$&nbsp; and&nbsp; $y$ &nbsp;are each&nbsp; $\sigma = 1$.
 +
*All variables are therefore to be understood as normalized.
  
''Hinweise:''
 
*Die Aufgabe gehört zum  Kapitel&nbsp; [[Stochastische_Signaltheorie/Weitere_Verteilungen|Weitere Verteilungen]].
 
*Insbesondere wird Bezug genommen auf die Seite&nbsp; [[Stochastische_Signaltheorie/Weitere_Verteilungen#Rayleighverteilung|Rayleighverteilung]].
 
*Die Streuungen der beiden Gau&szlig;schen Zufallsgr&ouml;&szlig;en&nbsp; $x$&nbsp; und&nbsp; $y$ &nbsp;seien jeweils&nbsp; $\sigma = 1$.
 
*Alle Gr&ouml;&szlig;en sind deshalb als normiert zu verstehen.
 
 
   
 
   
  
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===Fragebogen===
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===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welche der folgenden Aussagen treffen stets zu?
+
{Which of the following statements are always true?
 
|type="[]"}
 
|type="[]"}
+ Kleine Momentanleistungen sind wahrscheinlicher als gro&szlig;e.
+
+ Small instantaneous powers are more likely than large ones.
- Die Phase&nbsp; $\phi(t) = \pi/2$&nbsp; bedeutet auch &bdquo;Imagin&auml;rteil $y(t) = 0$&rdquo;.
+
- The phase&nbsp; $\phi(t) = \pi/2$&nbsp; also means&nbsp; "imaginary part $y(t) = 0$".
+ Die Phase&nbsp; $\phi(t) = -\pi/2$&nbsp; bedeutet auch &bdquo;Realteil $x(t) = 0$&rdquo;.
+
+ The phase&nbsp; $\phi(t) = -\pi/2$&nbsp; also means&nbsp; "real part $x(t) = 0$".
  
  
{Mit welcher Wahrscheinlichkeit ist die (normierte) Momentanleistung&nbsp; $p(t)$&nbsp; größer als&nbsp; $4$?
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{With what probability is the (normalized) instantaneous power&nbsp; $p(t)$&nbsp; greater than&nbsp; $4$?
 
|type="{}"}
 
|type="{}"}
${\rm Pr}(p(t) > 4) \ = \ $ { 0.135 3% }
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${\rm Pr}(p(t) > 4) \ = \ $ { 13.5 3% } $\ \%$
  
  
{Wie gro&szlig; ist die Wahrscheinlichkeit, dass der Betrag&nbsp; $a(t)$&nbsp; größer als&nbsp; $2$&nbsp; ist?
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{What is the probability that the magnitude&nbsp; $a(t)$&nbsp; is greater than&nbsp; $2$?
 
|type="{}"}
 
|type="{}"}
${\rm Pr}(a(t) > 2) \ = \ $ { 0.135 3% }
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${\rm Pr}(a(t) > 2) \ = \ $ { 13.5 3% } $\ \%$
  
  
{Berechnen Sie ausgehend von&nbsp; $f_p(p)$&nbsp; die WDF der Zufallsgr&ouml;&szlig;e&nbsp; $a$.&nbsp; Welcher WDF-Wert ergibt sich f&uuml;r&nbsp; $a=1$?
+
{Calculate,&nbsp; starting from&nbsp; $f_p(p)$&nbsp; the PDF of the random variable&nbsp; $a$.&nbsp; What PDF value results for&nbsp; $a=1$?
 
|type="{}"}
 
|type="{}"}
 
$f_a(a = 1)\ = \ $ { 0.607 3% }
 
$f_a(a = 1)\ = \ $ { 0.607 3% }
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</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig sind <u>der erste und der dritte Lösungsvorschlag</u>:
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'''(1)'''&nbsp; Correct are&nbsp; <u>the first and the third proposed solutions</u>:
*Die erste Aussage trifft aufgrund der Exponentialverteilung für $p(t)$ zu.
+
*The first statement is true because of the exponential distribution for&nbsp; $p(t)$.
* Ein Phasenwinkel $\phi(t) = \pm \pi/2 (\pm 90^\circ)$ bedeutet, dass der Realteil $x(t) = 0$ ist.
+
* A phase angle&nbsp; $\phi(t) = \pm \pi/2 \ (\pm 90^\circ)$&nbsp; means that the real part&nbsp; $x(t) = 0$.
* Bei positivem Imagin&auml;rteil &nbsp;&#8658;&nbsp; $y(t) > 0$ ist der Phasenwinkel $\phi(t) = +90^\circ$, bei negativem Imagin&auml;rteil betr&auml;gt der Phasenwinkel $\phi(t) = -90^\circ$.  
+
* For positive imaginary part&nbsp;&#8658;&nbsp; $y(t) > 0$&nbsp; the phase angle&nbsp; $\phi(t) = +90^\circ$,&nbsp; for negative imaginary part the phase angle&nbsp; $\phi(t) = -90^\circ$.  
  
  
'''(2)'''&nbsp; Mit $\sigma = 1$ gilt f&uuml;r die WDF der Momentanleistung:  
+
 
 +
'''(2)'''&nbsp; With&nbsp; $\sigma = 1$&nbsp; holds for the PDF of the instantaneous power:  
 
:$$f_p(p)= {\rm 1}/{\rm 2}\cdot{\rm e}^{-p/\rm 2}.$$
 
:$$f_p(p)= {\rm 1}/{\rm 2}\cdot{\rm e}^{-p/\rm 2}.$$
  
Die gesuchte Wahrscheinlichkeit ist demnach:
+
*The probability we are looking for is thus:
:$${\rm Pr}(p(t) > 4) = \int_{\rm 4}^{\infty}{1}/{2}\cdot{\rm e}^{-p/\rm 2}\,{\rm d} p={\rm e}^{\rm -2} \rm \hspace{0.15cm}\underline{=0.135}.$$
+
:$${\rm Pr}(p(t) > 4) = \int_{\rm 4}^{\infty}{1}/{2}\cdot{\rm e}^{-p/\rm 2}\,{\rm d} p={\rm e}^{\rm -2} \rm \hspace{0.15cm}\underline{=13.5\%}.$$
  
  
'''(3)'''&nbsp; Da $p(t) =a^2(t)$ gilt und zudem &nbsp;$a(t) < 0$&nbsp; nicht möglich ist, ist das Ereignis &nbsp;$a(t) > 2$&nbsp; identisch mit dem Ereignis &nbsp;$p(t) > 4$.
 
  
Es ergibt sich die gleiche Wahrscheinlichkeit ${\rm Pr}(a(t) > 2) \hspace{0.15cm}\underline{=0.135}$ wie unter '''(2)''' berechnet.
+
'''(3)'''&nbsp; Since&nbsp; $p(t) =a^2(t)$&nbsp; holds and moreover &nbsp;$a(t) < 0$&nbsp; is not possible,&nbsp; the event &nbsp;$a(t) > 2$&nbsp; is identical to the event &nbsp;$p(t) > 4$:
  
 +
*It results in the same probability&nbsp; ${\rm Pr}(a(t) > 2)  \hspace{0.15cm}\underline{=13.5\%}$&nbsp; as calculated in&nbsp; '''(2)'''&nbsp;.
  
'''(4)'''&nbsp; Allgemein gilt:
+
 
 +
'''(4)'''&nbsp; In general:
 
:$$f_a(a)=\frac{f_p(p)}{|g'(p)|}\Bigg |_{\, p=h(a)}.$$
 
:$$f_a(a)=\frac{f_p(p)}{|g'(p)|}\Bigg |_{\, p=h(a)}.$$
  
Die Transformationskennlinie lautet:
+
*The transformation characteristic is:
:$$g\hspace{0.05cm}'(p)=\frac{ {\rm d} a}{ {\rm d} p}=\frac{1}{2 \cdot \sqrt{p}}.$$
+
:$$g\hspace{0.05cm}'(p)=\frac{ {\rm d} a}{ {\rm d} p}=\frac{1}{2 \cdot \sqrt{p}}.$$
  
Diese ist stets positiv. Daraus folgt für die Rayleigh&ndash;WDF:
+
*This is always positive.&nbsp; It follows for the Rayleigh PDF:
 
:$$f_a(a)=\sqrt {p}\cdot {\rm e}^{ -p/\rm 2}\Big|_{ p=a^{\rm 2}}= a\cdot {\rm e}^{\, -a^{\rm 2}/\rm 2}.$$
 
:$$f_a(a)=\sqrt {p}\cdot {\rm e}^{ -p/\rm 2}\Big|_{ p=a^{\rm 2}}= a\cdot {\rm e}^{\, -a^{\rm 2}/\rm 2}.$$
  
F&uuml;r $a = 1$ ergibt sich somit der Wert $f_a(a = 1)= {\rm e}^{-0.5}\hspace{0.15cm}\underline{=0.607} $.
+
*For&nbsp; $a = 1$&nbsp; this gives the value&nbsp; $f_a(a = 1)= {\rm e}^{-0.5}\hspace{0.15cm}\underline{=0.607} $.
 
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Stochastische Signaltheorie|^3.7 Weitere Verteilungen^]]
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[[Category:Theory of Stochastic Signals: Exercises|^3.7 Further Distributions^]]

Latest revision as of 12:01, 17 February 2022

Rayleigh distribution:  real part, imaginary part, magnitude

One often describes a band-pass transmission system in the so-called  "equivalent low-pass domain".  This representation leads according to the chapter  "Band-pass signals"  in the book  Signal Representation  to a complex signal:

$$z(t)=x(t)+ {\rm j} \cdot y(t).$$

The real part  $x(t)$  denotes the  "inphase component"  and the imaginary part  $y(t)$  the  "quadrature component".

In a mobile radio system where there is no line of sight between the mobile user and the base station,  the radio signal thus reaches the receiver exclusively by indirect means  (refraction,  scattering,  reflection, etc.);

In this case,  the following model is applicable:

  • The real part  $x(t)$  and also the imaginary part  $y(t)$  are both Gaussian distributed and zero mean.
  • $x(t)$  and  $y(t)$  each have the same standard deviation  $\sigma$  and are independent of each other.
  • Internal bindings of the signals  $x(t)$  and  $y(t)$  due to the Doppler effect shall not be considered here.


The complex signal  $z(t)$  can also be represented by magnitude and phase:

$$ z(t)= a(t)\cdot {\rm e}^{\rm j \phi(t)}.$$
  • Because of symmetry with respect to  $x(t)$  and  $y(t)$  the phase  $\phi(t)$  is uniformly distributed.
  • In contrast,  the magnitude  $a(t) = |z(t)|$  is Rayleigh distributed,  which has led to the naming  "Rayleigh fading".
  • As a further quantity we define the instantaneous power
$$ p(t)=x^{\rm 2}(t)+y^{\rm 2}( t)=a^{\rm 2}(t).$$

The random variable  $p$  is  (one-sided) exponentially distributed.  Its PDF is for  $p>0$:

$$f_p(p)=\frac{1}{2\sigma^{\rm 2}}\cdot {\rm e}^{ -p/(\sigma^{\rm 2})}.$$

For all negative  $p$–values, of course  $f_p(p)= 0$  holds,  since  $p$  denotes a power.



Hints:

  • This exercise belongs to the chapter  Further Distributions.
  • In particular, reference is made to the section  Rayleigh PDF.
  • The standard deviation of the two Gaussian random variables  $x$  and  $y$  are each  $\sigma = 1$.
  • All variables are therefore to be understood as normalized.




Questions

1

Which of the following statements are always true?

Small instantaneous powers are more likely than large ones.
The phase  $\phi(t) = \pi/2$  also means  "imaginary part $y(t) = 0$".
The phase  $\phi(t) = -\pi/2$  also means  "real part $x(t) = 0$".

2

With what probability is the (normalized) instantaneous power  $p(t)$  greater than  $4$?

${\rm Pr}(p(t) > 4) \ = \ $

$\ \%$

3

What is the probability that the magnitude  $a(t)$  is greater than  $2$?

${\rm Pr}(a(t) > 2) \ = \ $

$\ \%$

4

Calculate,  starting from  $f_p(p)$  the PDF of the random variable  $a$.  What PDF value results for  $a=1$?

$f_a(a = 1)\ = \ $


Solution

(1)  Correct are  the first and the third proposed solutions:

  • The first statement is true because of the exponential distribution for  $p(t)$.
  • A phase angle  $\phi(t) = \pm \pi/2 \ (\pm 90^\circ)$  means that the real part  $x(t) = 0$.
  • For positive imaginary part ⇒  $y(t) > 0$  the phase angle  $\phi(t) = +90^\circ$,  for negative imaginary part the phase angle  $\phi(t) = -90^\circ$.


(2)  With  $\sigma = 1$  holds for the PDF of the instantaneous power:

$$f_p(p)= {\rm 1}/{\rm 2}\cdot{\rm e}^{-p/\rm 2}.$$
  • The probability we are looking for is thus:
$${\rm Pr}(p(t) > 4) = \int_{\rm 4}^{\infty}{1}/{2}\cdot{\rm e}^{-p/\rm 2}\,{\rm d} p={\rm e}^{\rm -2} \rm \hspace{0.15cm}\underline{=13.5\%}.$$


(3)  Since  $p(t) =a^2(t)$  holds and moreover  $a(t) < 0$  is not possible,  the event  $a(t) > 2$  is identical to the event  $p(t) > 4$:

  • It results in the same probability  ${\rm Pr}(a(t) > 2) \hspace{0.15cm}\underline{=13.5\%}$  as calculated in  (2) .


(4)  In general:

$$f_a(a)=\frac{f_p(p)}{|g'(p)|}\Bigg |_{\, p=h(a)}.$$
  • The transformation characteristic is:
$$g\hspace{0.05cm}'(p)=\frac{ {\rm d} a}{ {\rm d} p}=\frac{1}{2 \cdot \sqrt{p}}.$$
  • This is always positive.  It follows for the Rayleigh PDF:
$$f_a(a)=\sqrt {p}\cdot {\rm e}^{ -p/\rm 2}\Big|_{ p=a^{\rm 2}}= a\cdot {\rm e}^{\, -a^{\rm 2}/\rm 2}.$$
  • For  $a = 1$  this gives the value  $f_a(a = 1)= {\rm e}^{-0.5}\hspace{0.15cm}\underline{=0.607} $.