Aufgaben:Exercise 3.10: Rayleigh Fading - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]
	[[de:Aufgaben:3.~~10_Rayleighfading~~]]		[[de:Aufgaben:Aufgabe 3.10: Rayleighfading]]

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	[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]
			[[de:Aufgaben:3.10_Rayleighfading]]

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	In this case,  the following model is applicable:		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.		*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 ~~root mean square~~  $\sigma$  and are independent of each other.		*$x(t)$  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.		*Internal bindings of the signals  $x(t)$  and  $y(t)$  due to the Doppler effect shall not be considered here.

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	*This exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Further_Distributions\|Further Distributions]].		*This exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Further_Distributions\|Further Distributions]].
	*In particular, reference is made to the section  [[Theory_of_Stochastic_Signals/Further_Distributions#Rayleigh_PDF\|Rayleigh PDF]].		*In particular, reference is made to the section  [[Theory_of_Stochastic_Signals/Further_Distributions#Rayleigh_PDF\|Rayleigh PDF]].
	*The ~~root mean square~~ of the two Gaussian random variables  $x$  and  $y$  are each  $\sigma = 1$.		*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.		*All variables are therefore to be understood as normalized.

Guenter at 10:17, 3 February 2022

2022-02-03T10:17:08Z

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	}}		}}

	[[File:P_ID177__Sto_A_3_10.png\|right\|frame\|Rayleigh ~~PDF~~]]		[[File:P_ID177__Sto_A_3_10.png\|right\|frame\|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:		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:
	:$$z(t)=x(t)+ {\rm j} \cdot y(t).$$		:$$z(t)=x(t)+ {\rm j} \cdot y(t).$$

Guenter at 10:00, 3 February 2022

2022-02-03T10:00:12Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Correct are <u>the first and the third proposed solutions</u>:		'''(1)'''  Correct are  <u>the first and the third proposed solutions</u>:
	*The first statement is true because of the exponential distribution for  $p(t)$~~ ~~.		*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  $x(t) = 0$~~ ~~.		* 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  $\phi(t) = -90^\circ$.		* For positive imaginary part ⇒  $y(t) > 0$  the phase angle  $\phi(t) = +90^\circ$,  for negative imaginary part the phase angle  $\phi(t) = -90^\circ$.


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	'''(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  $p(t) > 4$:		'''(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)''' .		*It results in the same probability  ${\rm Pr}(a(t) > 2) \hspace{0.15cm}\underline{=13.5\%}$  as calculated in  '''(2)''' .

Guenter at 09:54, 3 February 2022

2022-02-03T09:54:07Z

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	[[File:P_ID177__Sto_A_3_10.png\|right\|frame\|Rayleigh PDF]]		[[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>~~.  This representation leads according to the chapter "~~Bandpass~~-~~like~~ signals" in the book  [[Signal Representation]]  to a complex signal:		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:
	:$$z(t)=x(t)+ {\rm j} \cdot y(t).$$		:$$z(t)=x(t)+ {\rm j} \cdot y(t).$$

	The real part  $x(t)$  denotes the  ~~<i>~~inphase component~~</i>~~  and the imaginary part  $y(t)$  the  ~~<i>~~quadrature component~~</i>~~.		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 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:		In this case,  the following model is applicable:
	*The real part $x(t)$  and also the imaginary part $y(t)$  are both distributed and zero mean.		*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 ~~rms~~  $\sigma$  and are independent of each other.		*$x(t)$  and  $y(t)$  each have the same root mean square  $\sigma$  and are independent of each other.
	*Internal ~~binding~~ of the signals  $x(t)$  and  $y(t)$  due to the Doppler effect shall not be considered here.		*Internal bindings of the signals  $x(t)$  and  $y(t)$  due to the Doppler effect shall not be considered here.


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	:$$ z(t)= a(t)\cdot {\rm e}^{\rm j \phi(t)}.$$		:$$ z(t)= a(t)\cdot {\rm e}^{\rm j \phi(t)}.$$
	*Because of symmetry with respect to  $x(t)$  and  $y(t)$  the phase  $\phi(t)$  is uniformly distributed.		*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".		*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		*As a further quantity we define the instantaneous power
	:$$ p(t)=x^{\rm 2}(t)+y^{\rm 2}( t)=a^{\rm 2}(t).$$		:$$ p(t)=x^{\rm 2}(t)+y^{\rm 2}( t)=a^{\rm 2}(t).$$
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	:$$f_p(p)=\frac{1}{2\sigma^{\rm 2}}\cdot {\rm e}^{ -p/(\sigma^{\rm 2})}.$$		:$$f_p(p)=\frac{1}{2\sigma^{\rm 2}}\cdot {\rm e}^{ -p/(\sigma^{\rm 2})}.$$

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


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	Hints:		Hints:
	*This exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Further_Distributions\|Further Distributions]].		*This exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Further_Distributions\|Further Distributions]].
	*In particular, reference is made to the ~~page~~  [[Theory_of_Stochastic_Signals/Further_Distributions#Rayleigh_PDF\|Rayleigh PDF]].		*In particular, reference is made to the section  [[Theory_of_Stochastic_Signals/Further_Distributions#Rayleigh_PDF\|Rayleigh PDF]].
	*The ~~rms~~ of the two Gaussian random variables  $x$  and  $y$  are each  $\sigma = 1$.		*The root mean square of the two Gaussian random variables  $x$  and  $y$  are each  $\sigma = 1$.
	*All variables are therefore to be understood as normalized.		*All variables are therefore to be understood as normalized.

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


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	{What is the probability that the magnitude  $a(t)$  is greater than  $2$~~ ~~ ?		{What is the probability that the magnitude  $a(t)$  is greater than  $2$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(a(t) > 2) \ = \ $ { 13.5 3% } $\ \%$		${\rm Pr}(a(t) > 2) \ = \ $ { 13.5 3% } $\ \%$


	{Calculate, starting from  $f_p(p)$  the PDF of the random variable  $a$.  What PDF value results for  $a=1$?		{Calculate,  starting from  $f_p(p)$  the PDF of the random variable  $a$.  What PDF value results for  $a=1$?
	\|type="{}"}		\|type="{}"}
	$f_a(a = 1)\ = \ $ { 0.607 3% }		$f_a(a = 1)\ = \ $ { 0.607 3% }

Noah at 14:20, 12 January 2022

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	{Calculate, starting from  $f_p(p)$  the ~~WDF~~ of the random variable  $a$.  What PDF value results for  $a=1$?		{Calculate, starting from  $f_p(p)$  the PDF of the random variable  $a$.  What PDF value results for  $a=1$?
	\|type="{}"}		\|type="{}"}
	$f_a(a = 1)\ = \ $ { 0.607 3% }		$f_a(a = 1)\ = \ $ { 0.607 3% }
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	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Richtig sind~~ <u>~~der erste und der dritte Lösungsvorschlag~~</u>:		'''(1)'''  Correct are <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  $p(t)$ .
	* ~~Ein Phasenwinkel~~  $\phi(t) = \pm \pi/2 \ (\pm 90^\circ)$  ~~bedeutet, dass der Realteil~~  $x(t) = 0$  ~~ist~~.		* A phase angle  $\phi(t) = \pm \pi/2 \ (\pm 90^\circ)$  means that the real part  $x(t) = 0$ .
	* ~~Bei positivem Imaginärteil~~  ⇒  $y(t) > 0$  ~~ist der Phasenwinkel~~  $\phi(t) = +90^\circ$, ~~bei negativem Imaginärteil beträgt der Phasenwinkel~~  $\phi(t) = -90^\circ$.		* 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)'''  ~~Mit~~  $\sigma = 1$  ~~gilt für die WDF der Momentanleistung~~:		'''(2)'''  With  $\sigma = 1$  holds for the PDF of the instantaneous power:
	:$$f_p(p)= {\rm 1}/{\rm 2}\cdot{\rm e}^{-p/\rm 2}.$$		:$$f_p(p)= {\rm 1}/{\rm 2}\cdot{\rm e}^{-p/\rm 2}.$$

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



	'''(3)'''  Da  $p(t) =a^2(t)$  ~~gilt und zudem~~  $a(t) < 0$  ~~nicht möglich ist~~, ~~ist das Ereignis~~  $a(t) > 2$  ~~identisch mit dem Ereignis~~  $p(t) > 4$:		'''(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  $p(t) > 4$:

	*~~Es ergibt sich die gleiche Wahrscheinlichkeit~~  ${\rm Pr}(a(t) > 2) \hspace{0.15cm}\underline{=13.5\%}$  ~~wie unter~~  '''(2)'''  ~~berechnet~~.		*It results in the same probability  ${\rm Pr}(a(t) > 2) \hspace{0.15cm}\underline{=13.5\%}$  as calculated in  '''(2)''' .


	'''(4)'''  ~~Allgemein gilt~~:		'''(4)'''  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~~–WDF~~:		*This is always positive.  It follows for the Rayleigh PDF:
	:$$f_a(a)=\sqrt {p}\cdot {\rm e}^{ -p/\rm 2}\Big\|_{ p=a^{\rm 2}}= a\cdot {\rm e}^{\, -a^{\rm 2}/\rm 2}.$$		:$$f_a(a)=\sqrt {p}\cdot {\rm e}^{ -p/\rm 2}\Big\|_{ p=a^{\rm 2}}= a\cdot {\rm e}^{\, -a^{\rm 2}/\rm 2}.$$

	*~~Für~~  $a = 1$  ~~ergibt sich somit der Wert~~  $f_a(a = 1)= {\rm e}^{-0.5}\hspace{0.15cm}\underline{=0.607} $.		*For  $a = 1$  this gives the value  $f_a(a = 1)= {\rm e}^{-0.5}\hspace{0.15cm}\underline{=0.607} $.

	{{ML-Fuß}}		{{ML-Fuß}}

Noah at 13:53, 12 January 2022

2022-01-12T13:53:38Z

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	In this case, the following model is applicable:		In this case, the following model is applicable:
	*The real part $x(t)$  and also the imaginary part $y(t)$  are both distributed and zero mean.		*The real part $x(t)$  and also the imaginary part $y(t)$  are both distributed and zero mean.
	*$x(t)$  and  $y(t)$  each have the same ~~dispersion~~  $\sigma$  and are independent of each other.		*$x(t)$  and  $y(t)$  each have the same rms  $\sigma$  and are independent of each other.
	*Internal binding of the signals  $x(t)$  and  $y(t)$  due to the Doppler effect shall not be considered here.		*Internal binding of the signals  $x(t)$  and  $y(t)$  due to the Doppler effect shall not be considered here.

Noah at 13:53, 12 January 2022

2022-01-12T13:53:23Z

Noah at 13:48, 12 January 2022

2022-01-12T13:48:15Z

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	{{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]]
	~~Häufig beschreibt man ein Bandpassübertragungssystem im sogenannten~~ <i>~~äquivalenten Tiefpassbereich~~</i>.  ~~Diese Darstellung führt entsprechend dem Kapitel~~ "~~Bandpassartige Signale~~" ~~im Buch~~  [[~~Signaldarstellung~~]]  ~~zu einem komplexen Signal~~:		One often describes a bandpass transmission system in the so-called <i>equivalent lowpass domain</i>.  This representation leads according to the chapter "Bandpass-like signals" in the book  [[Signal Representation]]  to a complex signal:
	:$$z(t)=x(t)+ {\rm j} \cdot y(t).$$		:$$z(t)=x(t)+ {\rm j} \cdot y(t).$$

	~~Der Realteil~~  $x(t)$  ~~kennzeichnet hierbei die~~  <i>~~Inphasekomponente~~</i>  ~~und der Imaginärteil~~  $y(t)$  ~~die~~  <i>~~Quadraturkomponente~~</i>.		The real part  $x(t)$  denotes the  <i>inphase component</i>  and the imaginary part  $y(t)$  the  <i>quadrature component</i>.

	~~Bei einem Mobilfunksystem~~, ~~bei dem zwischen dem mobilen Teilnehmer und der Basisstation keine Sichtverbindung besteht~~, ~~gelangt somit das Funksignal ausschließlich auf indirekten Wegen (Brechung~~, ~~Streuung~~, ~~Reflexion usw~~.) ~~zum Empfänger.&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 ~~ $x(t)$  ~~und auch der  Imaginärteil~~ $y(t)$  ~~sind jeweils gaußverteilt und mittelwertfrei~~.		*The real part $x(t)$  and also the imaginary part $y(t)$  are both distributed and zero mean.
	*$x(t)$  ~~und~~  $y(t)$  ~~besitzen jeweils die gleiche Streuung~~  $\sigma$  ~~und sind voneinander unabhängig~~.		*$x(t)$  and  $y(t)$  each have the same dispersion  $\sigma$  and are independent of each other.
	*~~Innere Bindungen der Signale~~  $x(t)$  ~~und~~  $y(t)$  ~~aufgrund des Dopplereffekts sollen hier nicht beachtet werden~~.		*Internal binding of the signals  $x(t)$  and  $y(t)$  due to the Doppler effect shall not be considered here.


	~~Das komplexe Signal~~  $z(t)$  ~~kann man auch nach Betrag und Phase darstellen~~:		The complex signal  $z(t)$  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 bezüglich~~  $x(t)$  ~~und~~  $y(t)$  ~~ist die Phase~~  $\phi(t)$  ~~gleichverteilt~~.		*Because of symmetry with respect to  $x(t)$  and  $y(t)$  the phase  $\phi(t)$  is uniformly distributed.
	*~~Dagegen ist der Betrag~~  $a(t) = \|z(t)\|$  ~~rayleighverteilt~~, ~~was zu der Namensgebung~~ "~~Rayleighfading~~" ~~geführt hat~~.		*In contrast, the magnitude  $a(t) = \|z(t)\|$  is Rayleigh distributed, which has led to the naming "Rayleigh fading".
	*~~Als weitere Größe definieren wir noch die Momentanleistung~~		*As a further quantity we define the instantaneous power
	:$$ p(t)=x^{\rm 2}(t)+y^{\rm 2}( t)=a^{\rm 2}(t).$$		:$$ p(t)=x^{\rm 2}(t)+y^{\rm 2}( t)=a^{\rm 2}(t).$$

	~~Die Zufallsgröße~~  $p$  ~~ist hier~~  [[Theory_of_Stochastic_Signals/~~Exponentialverteilte_Zufallsgrößen~~#~~Einseitige_Exponentialverteilung~~\|(~~einseitig~~) ~~exponentialverteilt~~]].  ~~Deren WDF lautet für~~  $p>0$:		The random variable  $p$  here  [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#One-sided_exponential_distribution\|(one-sided) exponentially distributed]].  Its PDF is for  $p>0$:
	:$$f_p(p)=\frac{1}{2\sigma^{\rm 2}}\cdot {\rm e}^{ -p/(\sigma^{\rm 2})}.$$		:$$f_p(p)=\frac{1}{2\sigma^{\rm 2}}\cdot {\rm e}^{ -p/(\sigma^{\rm 2})}.$$

	~~Für alle negativen~~ $p$–~~Werte gilt natürlich~~  $f_p(p)= 0$, da  $p$  ~~eine Leistung kennzeichnet~~.		For all negative $p$–values, of course  $f_p(p)= 0$ holds, since  $p$  denotes a power.




			Hints:
			*This exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Further_Distributions\|Further Distributions]].
			*In particular, reference is made to the page  [[Theory_of_Stochastic_Signals/Further_Distributions#Rayleigh_PDF\|Rayleigh PDF]].
			*The rms of the two Gaussian random variables  $x$  and  $y$  are each  $\sigma = 1$.
			*All variables are therefore to be understood as normalized.

	~~''Hinweise:''~~
	*Die Aufgabe gehört zum Kapitel  [[Theory_of_Stochastic_Signals/Weitere_Verteilungen\|Weitere Verteilungen]].
	*Insbesondere wird Bezug genommen auf die Seite  [[Theory_of_Stochastic_Signals/Weitere_Verteilungen#Rayleighverteilung\|Rayleighverteilung]].
	*Die Streuungen der beiden Gaußschen Zufallsgrößen  $x$  und  $y$  seien jeweils  $\sigma = 1$.
	*Alle Größen sind deshalb als normiert zu verstehen.


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	===~~Fragebogen~~===		===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 große~~.		+ Small instantaneous powers are more likely than large ones.
	- ~~Die Phase~~  $\phi(t) = \pi/2$  ~~bedeutet auch~~ "Imaginärteil $y(t) = 0$".		- The phase  $\phi(t) = \pi/2$  also means "Imaginärteil $y(t) = 0$".
	+ ~~Die Phase~~  $\phi(t) = -\pi/2$  ~~bedeutet auch~~ "~~Realteil~~ $x(t) = 0$".		+ The phase  $\phi(t) = -\pi/2$  also means "real part $x(t) = 0$".


	{~~Mit welcher Wahrscheinlichkeit ist die~~ (~~normierte~~) ~~Momentanleistung~~  $p(t)$  ~~größer als~~  $4$?		{With what probability is the (normalized) instantaneous power  $p(t)$  greater than  $4$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(p(t) > 4) \ = \ $ { 13.5 3% } $\ \%$		${\rm Pr}(p(t) > 4) \ = \ $ { 13.5 3% } $\ \%$


	{~~Wie groß ist die Wahrscheinlichkeit, dass der Betrag~~  $a(t)$  ~~größer als~~  $2$  ~~ist~~?		{What is the probability that the magnitude  $a(t)$  is greater than  $2$  ?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(a(t) > 2) \ = \ $ { 13.5 3% } $\ \%$		${\rm Pr}(a(t) > 2) \ = \ $ { 13.5 3% } $\ \%$


	{~~Berechnen Sie ausgehend von~~  $f_p(p)$  ~~die~~ WDF ~~der Zufallsgröße~~  $a$.  ~~Welcher~~ WDF~~-Wert ergibt sich für~~  $a=1$?		{Calculate, starting from  $f_p(p)$  the WDF of the random variable  $a$.  What WDF value results for  $a=1$?
	\|type="{}"}		\|type="{}"}
	$f_a(a = 1)\ = \ $ { 0.607 3% }		$f_a(a = 1)\ = \ $ { 0.607 3% }