Aufgaben:Exercise 3.10: Rayleigh Fading - Revision history

Bene: Text replacement - "root mean square" to "standard deviation"

2022-02-17T11:01:02Z

Text replacement - "root mean square" to "standard deviation"

Guenter at 10:17, 3 February 2022

2022-02-03T10:17:08Z

Guenter at 10:00, 3 February 2022

2022-02-03T10:00:12Z

Guenter at 09:54, 3 February 2022

2022-02-03T09:54:07Z

Noah at 14:20, 12 January 2022

2022-01-12T14:20:19Z

Noah at 13:53, 12 January 2022

2022-01-12T13:53:38Z

Noah at 13:53, 12 January 2022

2022-01-12T13:53:23Z

Noah at 13:48, 12 January 2022

2022-01-12T13:48:15Z

Guenter at 10:31, 20 December 2021

2021-12-20T10:31:44Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.10: Rayleighfading to Aufgaben:Exercise 3.10: Rayleigh Fading

2021-12-20T10:28:32Z

Guenter moved page Aufgabe 3.10: Rayleighfading to Exercise 3.10: Rayleigh Fading

@@ Line 13: / Line 13: @@
 In this case,&nbsp; the following model is applicable:
 *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; and&nbsp; $y(t)$&nbsp; each have the same root mean square&nbsp; $\sigma$&nbsp; and are independent of each other.
+*$x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; each have the same standard deviation&nbsp; $\sigma$&nbsp; and are independent of each other.
 *Internal bindings of the signals&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; due to the Doppler effect shall not be considered here.
@@ Line 35: / Line 35: @@
 *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 root mean square of the two Gaussian random variables&nbsp; $x$&nbsp; and&nbsp; $y$ &nbsp;are each&nbsp; $\sigma = 1$.
+*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.

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID177__Sto_A_3_10.png|right|frame|Rayleigh PDF]]
+[[File:P_ID177__Sto_A_3_10.png|right|frame|Rayleigh distribution:&nbsp; real part, imaginary part, magnitude ]]
 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:
 :$$z(t)=x(t)+ {\rm j} \cdot y(t).$$

@@ Line 72: / Line 72: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Correct are <u>the first and the third proposed solutions</u>:
+'''(1)'''&nbsp; Correct are&nbsp; <u>the first and the third proposed solutions</u>:
-*The first statement is true because of the exponential distribution for&nbsp; $p(t)$&nbsp;.
+*The first statement is true because of the exponential distribution for&nbsp; $p(t)$.
-* A phase angle&nbsp; $\phi(t) = \pm \pi/2 \ (\pm 90^\circ)$&nbsp; means that the real part&nbsp; $x(t) = 0$&nbsp;.
+* A phase angle&nbsp; $\phi(t) = \pm \pi/2 \ (\pm 90^\circ)$&nbsp; means that the real part&nbsp; $x(t) = 0$.
-* For positive imaginary part&nbsp;&#8658;&nbsp; $y(t) > 0$&nbsp; the phase angle&nbsp; $\phi(t) = +90^\circ$, for negative imaginary part the phase angle&nbsp; $\phi(t) = -90^\circ$.
+* 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; $\phi(t) = -90^\circ$.
@@ Line 87: / Line 87: @@
-'''(3)'''&nbsp; Since&nbsp; $p(t) =a^2(t)$&nbsp; holds and moreover &nbsp;$a(t) < 0$&nbsp; is not possible, the event &nbsp;$a(t) > 2$&nbsp; is identical to the event &nbsp;$p(t) > 4$:
+'''(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;.

@@ Line 4: / Line 4: @@
 [[File:P_ID177__Sto_A_3_10.png|right|frame|Rayleigh PDF]]
-One often describes a bandpass transmission system in the so-called <i>equivalent lowpass domain</i>.&nbsp; This representation leads according to the chapter "Bandpass-like signals" in the book&nbsp; [[Signal Representation]]&nbsp; to a complex 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:
 :$$z(t)=x(t)+ {\rm j} \cdot y(t).$$
-The real part&nbsp; $x(t)$&nbsp; denotes the&nbsp; <i>inphase component</i>&nbsp; and the imaginary part&nbsp; $y(t)$&nbsp; the&nbsp; <i>quadrature component</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".
-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 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 this case, the following model is applicable:
+In this case,&nbsp; the following model is applicable:
-*The real part $x(t)$&nbsp; and also the imaginary part $y(t)$&nbsp; are both distributed and zero mean.
+*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; and&nbsp; $y(t)$&nbsp; each have the same rms&nbsp; $\sigma$&nbsp; and are independent of each other.
+*$x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; each have the same root mean square&nbsp; $\sigma$&nbsp; and are independent of each other.
-*Internal binding of the signals&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; due to the Doppler effect shall not be considered here.
+*Internal bindings of the signals&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; due to the Doppler effect shall not be considered here.
@@ Line 20: / Line 20: @@
 :$$ z(t)= a(t)\cdot {\rm e}^{\rm j \phi(t)}.$$
 *Because of symmetry with respect to&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; the phase&nbsp; $\phi(t)$&nbsp; is uniformly distributed.
-*In contrast, the magnitude&nbsp; $a(t) = |z(t)|$&nbsp; is Rayleigh distributed, which has led to the naming "Rayleigh fading".
+*In contrast,&nbsp; the magnitude&nbsp; $a(t) = |z(t)|$&nbsp; is Rayleigh distributed,&nbsp; which has led to the naming&nbsp; "Rayleigh fading".
 *As a further quantity we define the instantaneous power
 :$$ p(t)=x^{\rm 2}(t)+y^{\rm 2}( t)=a^{\rm 2}(t).$$
@@ Line 27: / Line 27: @@
 :$$f_p(p)=\frac{1}{2\sigma^{\rm 2}}\cdot {\rm e}^{ -p/(\sigma^{\rm 2})}.$$
-For all negative $p$&ndash;values, of course&nbsp; $f_p(p)= 0$ holds, since&nbsp; $p$&nbsp; denotes a power.
+For all negative&nbsp; $p$&ndash;values, of course&nbsp; $f_p(p)= 0$&nbsp; holds,&nbsp; since&nbsp; $p$&nbsp; denotes a power.
@@ Line 34: / Line 34: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Further_Distributions|Further Distributions]].
-*In particular, reference is made to the page&nbsp; [[Theory_of_Stochastic_Signals/Further_Distributions#Rayleigh_PDF|Rayleigh PDF]].
+*In particular, reference is made to the section&nbsp; [[Theory_of_Stochastic_Signals/Further_Distributions#Rayleigh_PDF|Rayleigh PDF]].
-*The rms of the two Gaussian random variables&nbsp; $x$&nbsp; and&nbsp; $y$ &nbsp;are each&nbsp; $\sigma = 1$.
+*The root mean square 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.
@@ Line 49: / Line 49: @@
 |type="[]"}
 + Small instantaneous powers are more likely than large ones.
-- The phase&nbsp; $\phi(t) = \pi/2$&nbsp; also means "imaginary part $y(t) = 0$".
+- The phase&nbsp; $\phi(t) = \pi/2$&nbsp; also means&nbsp; "imaginary part $y(t) = 0$".
-+ The phase&nbsp; $\phi(t) = -\pi/2$&nbsp; also means "real part $x(t) = 0$".
++ The phase&nbsp; $\phi(t) = -\pi/2$&nbsp; also means&nbsp; "real part $x(t) = 0$".
@@ Line 58: / Line 58: @@
-{What is the probability that the magnitude&nbsp; $a(t)$&nbsp; is greater than&nbsp; $2$&nbsp; ?
+{What is the probability that the magnitude&nbsp; $a(t)$&nbsp; is greater than&nbsp; $2$?
 |type="{}"}
 ${\rm Pr}(a(t) > 2) \ = \ $ { 13.5 3% } $\ \%$
-{Calculate, starting from&nbsp; $f_p(p)$&nbsp; the PDF of the random variable&nbsp; $a$.&nbsp; What PDF value results for&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="{}"}
 $f_a(a = 1)\ = \ $ { 0.607 3% }

@@ Line 63: / Line 63: @@
-{Calculate, starting from&nbsp; $f_p(p)$&nbsp; the WDF of the random variable&nbsp; $a$.&nbsp; What PDF value results for&nbsp; $a=1$?
+{Calculate, 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="{}"}
 $f_a(a = 1)\ = \ $ { 0.607 3% }
@@ Line 70: / Line 70: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Richtig sind <u>der erste und der dritte Lösungsvorschlag</u>:
+'''(1)'''&nbsp; Correct are <u>the first and the third proposed solutions</u>:
-*Die erste Aussage trifft aufgrund der Exponentialverteilung für&nbsp; $p(t)$&nbsp; zu.
+*The first statement is true because of the exponential distribution for&nbsp; $p(t)$&nbsp;.
-* Ein Phasenwinkel&nbsp; $\phi(t) = \pm \pi/2 \ (\pm 90^\circ)$&nbsp; bedeutet, dass der Realteil&nbsp; $x(t) = 0$&nbsp; ist.
+* A phase angle&nbsp; $\phi(t) = \pm \pi/2 \ (\pm 90^\circ)$&nbsp; means that the real part&nbsp; $x(t) = 0$&nbsp;.
-* Bei positivem Imagin&auml;rteil &nbsp;&#8658;&nbsp; $y(t) > 0$&nbsp; ist der Phasenwinkel&nbsp; $\phi(t) = +90^\circ$, bei negativem Imagin&auml;rteil betr&auml;gt der Phasenwinkel&nbsp; $\phi(t) = -90^\circ$.
+* For positive imaginary part&nbsp;&#8658;&nbsp; $y(t) > 0$&nbsp; the phase angle&nbsp; $\phi(t) = +90^\circ$, for negative imaginary part the phase angle&nbsp; $\phi(t) = -90^\circ$.
-'''(2)'''&nbsp; Mit&nbsp; $\sigma = 1$&nbsp; gilt f&uuml;r die WDF der Momentanleistung:
+'''(2)'''&nbsp; With&nbsp; $\sigma = 1$&nbsp; holds for the PDF of the instantaneous power:
 :$$f_p(p)= {\rm 1}/{\rm 2}\cdot{\rm e}^{-p/\rm 2}.$$
-*Die gesuchte Wahrscheinlichkeit ist demnach:
+*The probability we are looking for is thus:
 :$${\rm Pr}(p(t) > 4) = \int_{\rm 4}^{\infty}{1}/{2}\cdot{\rm e}^{-p/\rm 2}\,{\rm d} p={\rm e}^{\rm -2} \rm \hspace{0.15cm}\underline{=13.5\%}.$$
-'''(3)'''&nbsp; Da&nbsp; $p(t) =a^2(t)$&nbsp; gilt und zudem &nbsp;$a(t) < 0$&nbsp; nicht möglich ist, ist das Ereignis &nbsp;$a(t) > 2$&nbsp; identisch mit dem Ereignis &nbsp;$p(t) > 4$:
+'''(3)'''&nbsp; Since&nbsp; $p(t) =a^2(t)$&nbsp; holds and moreover &nbsp;$a(t) < 0$&nbsp; is not possible, the event &nbsp;$a(t) > 2$&nbsp; is identical to the event &nbsp;$p(t) > 4$:
-*Es ergibt sich die gleiche Wahrscheinlichkeit&nbsp; ${\rm Pr}(a(t) > 2)  \hspace{0.15cm}\underline{=13.5\%}$&nbsp; wie unter&nbsp; '''(2)'''&nbsp; berechnet.
+*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:
 :$$f_a(a)=\frac{f_p(p)}{|g'(p)|}\Bigg |_{\, p=h(a)}.$$
-*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.&nbsp; 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&uuml;r&nbsp; $a = 1$&nbsp; ergibt sich somit der Wert&nbsp; $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}}

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Weitere Verteilungen
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Further_Distributions
 }}
-[[File:P_ID177__Sto_A_3_10.png|right|frame|Zur Rayleigh-WDF]]
+[[File:P_ID177__Sto_A_3_10.png|right|frame|Rayleigh WDF]]
-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 "Bandpassartige Signale" im Buch&nbsp; [[Signaldarstellung]]&nbsp; zu einem komplexen Signal:
+One often describes a bandpass transmission system in the so-called <i>equivalent lowpass domain</i>.&nbsp; This representation leads according to the chapter "Bandpass-like signals" in the book&nbsp; [[Signal Representation]]&nbsp; to a complex signal:
 :$$z(t)=x(t)+ {\rm j} \cdot y(t).$$
-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; <i>inphase component</i>&nbsp; and the imaginary part&nbsp; $y(t)$&nbsp; the&nbsp; <i>quadrature component</i>.
-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, the radio signal thus reaches the receiver exclusively by indirect means (refraction, scattering, reflection, etc.);
-In diesem Fall ist folgendes Modell anwendbar:
+In this case, 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.
+*The real part $x(t)$&nbsp; and also the imaginary part $y(t)$&nbsp; are both 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.
+*$x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; each have the same dispersion&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 binding 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:
-:$$ z(t)= a(t)\cdot {\rm e}^{{\rm j}  \phi(t)}.$$
+:$$ z(t)= a(t)\cdot {\rm e}^{\rm j}  \phi(t)}.$$
-*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 "Rayleighfading" gef&uuml;hrt hat.
+*In contrast, the magnitude&nbsp; $a(t) = |z(t)|$&nbsp; is Rayleigh distributed, which has led to the naming "Rayleigh fading".
-*Als weitere Gr&ouml;&szlig;e definieren wir noch die Momentanleistung
+*As a further quantity we define the instantaneous power
 :$$ p(t)=x^{\rm 2}(t)+y^{\rm 2}( t)=a^{\rm 2}(t).$$
-Die Zufallsgröße&nbsp; $p$&nbsp; ist hier&nbsp; [[Theory_of_Stochastic_Signals/Exponentialverteilte_Zufallsgrößen#Einseitige_Exponentialverteilung|(einseitig) exponentialverteilt]].&nbsp; Deren WDF lautet f&uuml;r&nbsp; $p>0$:
+The random variable&nbsp; $p$&nbsp; here&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&uuml;r alle negativen $p$&ndash;Werte gilt natürlich&nbsp; $f_p(p)= 0$, da&nbsp; $p$&nbsp; eine Leistung kennzeichnet.
+For all negative $p$&ndash;values, of course&nbsp; $f_p(p)= 0$ holds, since&nbsp; $p$&nbsp; denotes a power.
@@ Line 43: / Line 43: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welche der folgenden Aussagen treffen stets zu?
+{Which of the following statements are always true?
 |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 "Imagin&auml;rteil $y(t) = 0$".
+- The phase&nbsp; $\phi(t) = \pi/2$&nbsp; also means "Imagin&auml;rteil $y(t) = 0$".
-+ Die Phase&nbsp; $\phi(t) = -\pi/2$&nbsp; bedeutet auch "Realteil $x(t) = 0$".
++ The phase&nbsp; $\phi(t) = -\pi/2$&nbsp; also means "real part $x(t) = 0$".
-{Mit welcher Wahrscheinlichkeit ist die (normierte) Momentanleistung&nbsp; $p(t)$&nbsp; größer als&nbsp; $4$?
+{With what probability is the (normalized) instantaneous power&nbsp; $p(t)$&nbsp; greater than&nbsp; $4$?
 |type="{}"}
 ${\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?
+{What is the probability that the magnitude&nbsp; $a(t)$&nbsp; is greater than&nbsp; $2$&nbsp; ?
 |type="{}"}
 ${\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, starting from&nbsp; $f_p(p)$&nbsp; the WDF of the random variable&nbsp; $a$.&nbsp; What WDF value results for&nbsp; $a=1$?
 |type="{}"}
 $f_a(a = 1)\ = \ $ { 0.607 3% }

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−	[[Category:Theory of Stochastic Signals: Exercises\|^3.7 ~~Weitere Verteilungen~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]