Aufgaben:Exercise 3.4: Characteristic Function: Difference between revisions

Latest revision as of 17:55, 16 March 2026

Given here are three random variables $x$, $y$ and $z$, mostly by their respective probability density functions:

Nothing else is known about the random variable $x$: This can be both a discrete or a continuous random variable, and can have any PDF $f_x(x)$ The mean is generally equal $m_x$.
The continuous random variable $y$ can take values in the range between $1$ to $3$ with equal probability. Mean: $m_y = 2.$
The random variable $z$ has the following characteristic function:

$$C_z ({\it \Omega} ) = {\mathop{\rm si}\nolimits}( {3{\it \Omega}} ) \cdot {\mathop{\rm si}\nolimits} ( {2{\it \Omega} } ).$$

Besides, the qualitative course of the WDF $f_z(z)$ according to the blue sketch is assumed to be known. To be determined are the PDF parameters $a$, $b$, $c$ of this PDF.

Hints:

This exercise belongs to the chapter Expected values and moments.
Reference is made to the section Charakteristic funcion .

The characteristic function of a between $\pm a$ uniformly distributed random variable $z$ is:

$$C ( {\it \Omega} ) = {\mathop{\rm si}\nolimits} ( {a {\it \Omega} } )\quad {\rm{with}}\quad {\mathop{\rm si}\nolimits}( x ) = \sin ( x )/x.$$

Questions

	$C_x ( {\it \Omega} )$ is the Fourier transform of $f_x(x)$.
	The real part of $C_x ( {\it \Omega} )$ is an even function in ${\it \Omega}$.
	The imaginary part of $C_x ( {\it \Omega} )$ is an odd function in ${\it \Omega}$.
	The value at location ${\it \Omega} = 0$ is always $C_x ( {\it \Omega} ) = 1$.
	For a zero mean random variable $(m_x = 0)$ ⇒ $C_x ( {\it \Omega} )$ is always real.

${\rm Re}\big[C_y(\Omega\ =\ \pi/2)\big] \ = \ $

${\rm Im}\big[C_y(\Omega\ =\ \pi/2)\big] \ = \ $

$a \ = \ $

$b \ = \ $

$c \ = \ $

Solutions

Solution

(1) Correct are the proposed solutions 2, 3 and 4:

$C_x( {\it \Omega} )$ is not the Fourier transform to $f_x(x)$, but the inverse Fourier transform:

$$C_x( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_x }( x )\cdot {\rm{e}}^{\hspace{0.03cm}{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega\hspace{0.05cm}\cdot \hspace{0.05cm} x}} \hspace{0.1cm}{\rm{d}}x .$$

Also for this, the real part is always even and the imaginary part odd. For ${\it \Omega} = 0$ holds:

$$C_x( {\it \Omega} = 0 ) = \int_{ - \infty }^{ + \infty } {f_x }( x ) \hspace{0.1cm}{\rm{d}}x = 1.$$

The last alternative does not always hold: A two-point distributed random variable $x \in \{-1, +3\}$ with probabilities $0.75$ and $0.25$ is zero mean $(m_x = 0)$, but has still a complex characteristic function.

(2) According to the general definition:

$$C_y( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_y }( y )\cdot {\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega\hspace{0.01cm}\hspace{0.05cm}\cdot \hspace{0.05cm} y}} \hspace{0.1cm}{\rm{d}}y = 0.5\int_1^3 {{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}\Omega\hspace{0.05cm}\cdot \hspace{0.05cm} y} \hspace{0.1cm}{\rm{d}}y.} $$

After solving this integral, we get:

$$C_y ( {\it \Omega } ) = \frac{{{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}3{\it \Omega } } - {\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } } }}{{2{\rm{j}}{\it \Omega } }} = = \frac{{{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } } - {\rm{e}}^{{\rm{ - j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega }} }}{{2{\rm{j}}{\it \Omega } }} \cdot {\rm{e}}^{{\rm{j\hspace{0.05cm}\cdot \hspace{0.05cm}2}}{\it \Omega } } .$$

Using Euler's theorem, this can also be written:

$$C_y ( {\it \Omega } ) = \frac{{\sin ( {\it \Omega } )}}{{\it \Omega } } \cdot {\rm{e}}^{{\rm{j2}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } } = {\rm si} ( {\it \Omega } ) \cdot {\rm{e}}^{{\rm{j2}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } }.$$

For ${\it \Omega} = \pi/2$ we thus obtain a purely real numerical value:

$${\rm Re}[C_y ({\it \Omega} = {\rm{\pi }}/2 )] = \frac{{\sin( {{\rm{\pi }}/2})}}{{{\rm{\pi }}/2}} \cdot {\rm{e}}^{{\rm{j\pi }}} = - \frac{2}{{\rm{\pi }}}\hspace{0.15cm}\underline{\approx -0.637}, \hspace{0.5cm}{\rm Im}[C_y ({\it \Omega} = {\rm{\pi }}/2 )] \hspace{0.15cm}\underline{= 0} .$$

(3) From the given correspondence it can be read that ${\rm si}(3 {\it \Omega} )$ is due to an between $\pm 3$ equally distributed random variable and ${\rm si}(2 {\it \Omega} )$ gives the transform of a uniform distribution between $\pm 2$.

In the characteristic function, these two proportions are multiplicatively linked.
Thus, the resulting PDF $f_z(z)$ is the convolution of these two rectangular functions.
The three PDF parameters are thus:

$$\hspace{0.15cm}\underline{a = 1},\quad \hspace{0.15cm}\underline{b = 5},\quad c = 1/6 \hspace{0.15cm}\underline{= 0.167}.$$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/*Kapitel*
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Expected_Values_and_Moments
 }}
-[[File:P_ID619__Sto_A_3_4.png|right|frame|Rechteck&ndash; und Trapez&ndash;WDF]]
+[[File:P_ID619__Sto_A_3_4.png|right|frame|Rectangular and trapezoidal PDF]]
-Gegeben seien hier die drei Zufallsgrößen $x$, $y$ und $z$ durch ihre jeweiligen Wahrscheinlichkeitsdichtefunktionen:
+Given here are three random variables&nbsp; $x$,&nbsp; $y$&nbsp; and&nbsp; $z$,&nbsp; mostly by their respective probability density functions:
-*Über die Zufallsgröße $x$ ist nichts weiter bekannt: &nbsp; Diese kann sowohl eine diskrete als auch eine kontinuierliche Zufallsgröße sein und eine beliebige WDF $f_x(x)$ besitzen. Der Mittelwert ist allgemein gleich $m_x$.
+*Nothing else is known about the random variable&nbsp; $x$:&nbsp; This can be both a discrete or a continuous random variable,&nbsp; and can have any PDF&nbsp; $f_x(x)$&nbsp; The mean is generally equal&nbsp; $m_x$.
-*Die Zufallsgröße $y$  kann nur Werte im Bereich zwischen $1$ bis $3$  mit  gleicher Wahrscheinlichkeit annehmen. Der Mittelwert ist :$$m_y = 2.$$
+*The continuous random variable&nbsp; $y$&nbsp; can take values in the range between&nbsp; $1$&nbsp; to&nbsp; $3$&nbsp; with equal probability.&nbsp; Mean:&nbsp; $m_y = 2.$
-*Die Zufallsgröße $z$ besitzt die folgende charakteristische Funktion:
+*The random variable&nbsp; $z$&nbsp; has the following characteristic function:
-:$$C_z ({\it \Omega}  ) = {\mathop{\rm si}\nolimits}( {3{\it \Omega}} ) \cdot {\mathop{\rm si}\nolimits} ( {2{\it \Omega} } ).$$
+:$$C_z ({\it \Omega} ) = {\mathop{\rm si}\nolimits}( {3{\it \Omega}} ) \cdot {\mathop{\rm si}\nolimits} ( {2{\it \Omega} } ).$$
-:Daneben wird noch der qualitative Verlauf der WDF $f_z(z)$ entsprechend der blauen Skizze als bekannt vorausgesetzt. Zu bestimmen sind die WDF-Parameter $a$, $b$ und $c$ dieser WDF.
+:Besides, the qualitative course of the WDF&nbsp; $f_z(z)$&nbsp; according to the blue sketch is assumed to be known.&nbsp; To be determined are the PDF parameters&nbsp; $a$,&nbsp; $b$,&nbsp; $c$&nbsp; of this PDF.
+Hints:
+*This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected values and moments]].
-''Hinweise:''
+*Reference is made to the section&nbsp;  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments#Characteristic_function|Charakteristic funcion]]&nbsp;.
-*Die Aufgabe gehört zum  Kapitel [[Stochastische_Signaltheorie/Erwartungswerte_und_Momente|Erwartungswerte und Momente]].
-*Insbesondere wird auf die Seite  [[Stochastische_Signaltheorie/Erwartungswerte_und_Momente#Charakteristische_Funktion|Charakteristische Funktion]] Bezug genommen.
-*Die charakteristische Funktion einer zwischen $\pm a$ gleichverteilten Zufallsgröße $z$ lautet:
+*The characteristic function of a between&nbsp; $\pm a$&nbsp; uniformly distributed random variable&nbsp; $z$&nbsp; is:
-:$$C_z ( {\it \Omega}  ) = {\mathop{\rm si}\nolimits} ( {a  {\it \Omega} } )\quad {\rm{mit}}\quad {\mathop{\rm si}\nolimits}( x ) = \sin ( x )/x.$$
+:$$C ( {\it \Omega} ) = {\mathop{\rm si}\nolimits} ( {a {\it \Omega} } )\quad {\rm{with}}\quad {\mathop{\rm si}\nolimits}( x ) = \sin ( x )/x.$$
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welche Aussagen sind bezüglich der charakteristischen Funktion $C_x ( {\it \Omega} )$ stets &ndash; also bei beliebiger WDF &ndash; gültig?
+{Which statements are valid with respect to the characteristic function&nbsp; $C_x ( {\it \Omega} )$&nbsp; always &ndash; that is, at any PDF&nbsp;?
 |type="[]"}
-- $C_x ( {\it \Omega} )$ ist die Fouriertransformierte von $f_x(x)$.
+- $C_x ( {\it \Omega} )$&nbsp; is the Fourier transform of&nbsp; $f_x(x)$.
-+ Der Realteil von $C_x ( {\it \Omega} )$ ist eine gerade Funktion in ${\it \Omega}$.
++ The real part of&nbsp; $C_x ( {\it \Omega} )$&nbsp; is an even function in&nbsp; ${\it \Omega}$.
-+ Der Imaginärteil von $C_x ( {\it \Omega} )$ ist eine ungerade Funktion in  ${\it \Omega}$.
++ The imaginary part of&nbsp; $C_x ( {\it \Omega} )$&nbsp; is an odd function in&nbsp; ${\it \Omega}$.
-+ Der Wert an der Stelle ${\it \Omega} = 0$ ist stets $C_x ( {\it \Omega} ) = 1$.
++ The value at location&nbsp; ${\it \Omega} = 0$&nbsp; is always&nbsp; $C_x ( {\it \Omega} ) = 1$.
-- Bei mittelwertfreier Zufallsgröße $(m_x = 0)$ ist $C_x ( {\it \Omega} )$ stets reell.
+- For a zero mean random variable&nbsp; $(m_x = 0)$&nbsp; &rArr; &nbsp; $C_x ( {\it \Omega} )$&nbsp; is always real.
-{Berechnen Sie die charakteristische Funktion $C_y( {\it \Omega} )$. Wie groß sind Real- und Imaginärteil bei ${\it \Omega}  = \pi/2$?
+{Calculate the characteristic function&nbsp; $C_y( {\it \Omega} )$.&nbsp; What are the real and imaginary parts at&nbsp; ${\it \Omega} = \pi/2$?
 |type="{}"}
-${\rm Re}\big[C_y(\Omega\ =\ \pi/2)\big] \ = \ $   { -0.657--0.617 }
+${\rm Re}\big[C_y(\Omega\ =\ \pi/2)\big] \ = \ $ { -0.657--0.617 }
-${\rm Im}\big[C_y(\Omega\ =\ \pi/2)\big] \ =  \ $  { 0. }
+${\rm Im}\big[C_y(\Omega\ =\ \pi/2)\big] \ = \ $ { 0. }
-{Bestimmen Sie die Kenngrößen $a$, $b$ und $c$ der WDF $f_z(z)$.
+{Determine the characteristic parameters&nbsp; $a$,&nbsp; $b$&nbsp; and&nbsp; $c$&nbsp; of the PDF&nbsp; $f_z(z)$.
 |type="{}"}
-$a \ =  \ $ { 1 3% }
+$a \ = \ $ { 1 3% }
-$b \ =  \ $ { 5 3% }
+$b \ = \ $ { 5 3% }
-$c \ =  \ $ { 0.167 3% }
+$c \ = \ $ { 0.167 3% }
 </quiz>
-===Musterlösung===
+===Solutions===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Richtig sind <u>die Lösungsvorschläge 2, 3 und 4</u>:
+'''(1)'''&nbsp; Correct are&nbsp; <u>the proposed solutions 2, 3 and 4</u>:
-* $C_x ( {\it \Omega} )$ ist nicht die Fouriertransformierte zu $f_x(x)$, sondern die Fourierrücktransformierte:
+* $C_x( {\it \Omega} )$&nbsp; is not the Fourier transform to&nbsp; $f_x(x)$,&nbsp; but the inverse Fourier transform:
-:$$C_x( {\it \Omega  } ) = \int_{ - \infty }^{ + \infty } {f_x }( x )\cdot {\rm{e}}^{\hspace{0.03cm}{\rm{j}}\hspace{0.03cm}{\it \Omega x}} \hspace{0.1cm}{\rm{d}}x .$$
+:$$C_x( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_x }( x )\cdot {\rm{e}}^{\hspace{0.03cm}{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega\hspace{0.05cm}\cdot \hspace{0.05cm} x}} \hspace{0.1cm}{\rm{d}}x .$$
-*Auch bei dieser ist der Realteil stets gerade und der Imaginärteil ungerade. Für ${\it \Omega}  = 0$  gilt:
+*Also for this,&nbsp; the real part is always even and the imaginary part odd.&nbsp; For&nbsp; ${\it \Omega} = 0$&nbsp; holds:
-:$$C_x( {\it \Omega}  = 0 ) = \int_{ - \infty }^{ + \infty } {f_x }( x ) \hspace{0.1cm}{\rm{d}}x = 1.$$
+:$$C_x( {\it \Omega} = 0 ) = \int_{ - \infty }^{ + \infty } {f_x }( x ) \hspace{0.1cm}{\rm{d}}x = 1.$$
-*Die letzte Alternative trifft nicht immer zu: &nbsp; Eine zweipunktverteilte Zufallsgröße $x \in  \{-1, +3\}$ mit den Wahrscheinlichkeiten $0.75$ und $0.25$ ist zwar mittelwertfrei $(m_x = 0)$, besitzt aber trotzdem eine komplexe charakteristische Funktion.
+*The last alternative does not always hold: &nbsp; A two-point distributed random variable&nbsp; $x \in \{-1, +3\}$&nbsp; with probabilities&nbsp; $0.75$&nbsp; and&nbsp; $0.25$&nbsp; is zero mean&nbsp; $(m_x = 0)$,&nbsp; but has still a complex characteristic function.
-'''(2)'''&nbsp; Entsprechend der allgemeinen Definition gilt:
-:$$C_y( {\it \Omega  } ) = \int_{ - \infty }^{ + \infty } {f_y }( y )\cdot {\rm{e}}^{{\rm{j}}{\it \Omega y}} \hspace{0.1cm}{\rm{d}}y   = 0.5\int_1^3 {{\rm{e}}^{{\rm{j}}\Omega y} \hspace{0.1cm}{\rm{d}}y.} $$
-Nach Lösen dieses Integrals ergibt sich:
+'''(2)'''&nbsp; According to the general definition:
-:$$C_y ( {\it \Omega } ) = \frac{{{\rm{e}}^{{\rm{j}}3{\it \Omega } }  - {\rm{e}}^{{\rm{j}}{\it \Omega } } }}{{2{\rm{j}}{\it \Omega } }} = \frac{{{\rm{e}}^{{\rm{j}}{\it \Omega } }  - {\rm{e}}^{{\rm{ - j}}{\it \Omega }} }}{{2{\rm{j}}{\it \Omega } }} \cdot {\rm{e}}^{{\rm{j2}}{\it \Omega } } .$$
+:$$C_y( {\it \Omega  } ) = \int_{ - \infty }^{ + \infty } {f_y }( y )\cdot {\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega\hspace{0.01cm}\hspace{0.05cm}\cdot \hspace{0.05cm} y}} \hspace{0.1cm}{\rm{d}}y   = 0.5\int_1^3 {{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}\Omega\hspace{0.05cm}\cdot \hspace{0.05cm} y} \hspace{0.1cm}{\rm{d}}y.} $$
-Mit dem Satz von Euler kann hierfür auch geschrieben werden:
+*After solving this integral, we get:
-:$$C_y ( {\it \Omega }  ) = \frac{{\sin ( {\it \Omega }  )}}{{\it \Omega } } \cdot {\rm{e}}^{{\rm{j2}}{\it \Omega } } .$$
+:$$C_y ( {\it \Omega } ) = \frac{{{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}3{\it \Omega } }  - {\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } } }}{{2{\rm{j}}{\it \Omega } }} =  = \frac{{{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } }  - {\rm{e}}^{{\rm{ - j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega }} }}{{2{\rm{j}}{\it \Omega } }} \cdot {\rm{e}}^{{\rm{j\hspace{0.05cm}\cdot \hspace{0.05cm}2}}{\it \Omega } } .$$
-Für ${\it \Omega}  = \pi/2$ erhält man somit einen rein reellen Zahlenwert:
+*Using Euler's theorem, this can also be written:
-:$${\rm Re}[C_y ({\it \Omega}  = {\rm{\pi }}/2 )] = \frac{{\sin( {{\rm{\pi }}/2})}}{{{\rm{\pi }}/2}} \cdot {\rm{e}}^{{\rm{j\pi }}}  =  - \frac{2}{{\rm{\pi }}}
+:$$C_y ( {\it \Omega }  ) = \frac{{\sin ( {\it \Omega }  )}}{{\it \Omega } } \cdot {\rm{e}}^{{\rm{j2}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } } = {\rm si} ( {\it \Omega }  ) \cdot {\rm{e}}^{{\rm{j2}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } }.$$
-\hspace{0.15cm}\underline{\approx -0.637}, \hspace{0.5cm}
-{\rm Im}[C_y ({\it \Omega}  = {\rm{\pi }}/2 )]  \hspace{0.15cm}\underline{= 0}  .$$
+*For&nbsp; ${\it \Omega} = \pi/2$&nbsp; we thus obtain a purely real numerical value:
+:$${\rm Re}[C_y ({\it \Omega}  = {\rm{\pi }}/2 )] = \frac{{\sin( {{\rm{\pi }}/2})}}{{{\rm{\pi }}/2}} \cdot {\rm{e}}^{{\rm{j\pi }}}  =  - \frac{2}{{\rm{\pi }}}\hspace{0.15cm}\underline{\approx -0.637}, \hspace{0.5cm}{\rm Im}[C_y ({\it \Omega}  = {\rm{\pi }}/2 )]  \hspace{0.15cm}\underline{= 0}  .$$
-'''(3)'''&nbsp; Aus der angegebenen Korrespondenz kann abgelesen werden, dass ${\rm si}(3 {\it \Omega} )$ auf eine zwischen $\pm 3$ gleichverteilte Zufallsgröße zurückgeht und ${\rm si}(2 {\it \Omega} )$ die Transformierte einer Gleichverteilung zwischen $\pm 2$ angibt.
-In der charakteristischen Funktion sind diese beiden Anteile multiplikativ verknüpft. Damit ist die resultierende WDF $f_z(z)$ die Faltung dieser beiden Rechteckfunktionen:
+'''(3)'''&nbsp; From the given correspondence it can be read that&nbsp; ${\rm si}(3 {\it \Omega} )$&nbsp; is due to an between&nbsp; $\pm 3$&nbsp; equally distributed random variable and&nbsp; ${\rm si}(2 {\it \Omega} )$&nbsp; gives the transform of a uniform distribution between&nbsp; $\pm 2$.
-[[File:P_ID620__Sto_A_3_4_c_neu.png|center|frame|Konstruktion der Trapez-WDF]]
+[[File:P_ID620__Sto_A_3_4_c_neu.png|right|frame|Construction of the trapezoidal PDF]]
-:Die drei WDF-Parameter lauten somit:
+*In the characteristic function, these two proportions are multiplicatively linked.
-:$$\hspace{0.15cm}\underline{a = 1},\quad \hspace{0.15cm}\underline{b = 5},
+*Thus,&nbsp; the resulting PDF&nbsp; $f_z(z)$&nbsp; is the convolution of these two rectangular functions.
-\quad c = 1/6 \hspace{0.15cm}\underline{= 0.167}.$$
+*The three PDF parameters are thus:
+:$$\hspace{0.15cm}\underline{a = 1},\quad \hspace{0.15cm}\underline{b = 5},\quad c = 1/6 \hspace{0.15cm}\underline{= 0.167}.$$
 {{ML-Fuß}}
-[[Category:Aufgaben zu Stochastische Signaltheorie|^3.3 Erwartungswerte und Momente^]]
+[[Category:Theory of Stochastic Signals: Exercises|^3.3 Expected Values and Moments^]]
+[[de:Aufgaben:Aufgabe 3.4: Charakteristische Funktion]]