Aufgaben:Exercise 3.4: Characteristic Function - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^3.3 Expected Values and Moments^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.3 Expected Values and Moments^]]
	[[de:Aufgaben:3.~~4_Charakteristische_Funktion~~]]		[[de:Aufgaben:Aufgabe 3.4: Charakteristische Funktion]]

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	*For  ${\it \Omega} = \pi/2$  we thus obtain a purely real numerical value:		*For  ${\it \Omega} = \pi/2$  we thus obtain a purely real numerical value:
	:$${\rm Re}[C_y ({\it \Omega} = {\rm{\pi }}/2 )] = \frac{{\sin( {{\rm{\pi }}/2})}}{{{\rm{\pi }}/2}} \cdot {\rm{e}}^{{\rm{j\pi }}} = - \frac{2}{{\rm{\pi }}}		:$${\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} .$$
	\hspace{0.15cm}\underline{\approx -0.637}, \hspace{0.5cm}
	{\rm Im}[C_y ({\it \Omega} = {\rm{\pi }}/2 )] \hspace{0.15cm}\underline{= 0} .$$


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	*Thus,  the resulting PDF  $f_z(z)$  is the convolution of these two rectangular functions.		*Thus,  the resulting PDF  $f_z(z)$  is the convolution of these two rectangular functions.
	*The three PDF parameters are thus:		*The three PDF parameters are thus:
	:$$\hspace{0.15cm}\underline{a = 1},\quad \hspace{0.15cm}\underline{b = 5},		:$$\hspace{0.15cm}\underline{a = 1},\quad \hspace{0.15cm}\underline{b = 5},\quad c = 1/6 \hspace{0.15cm}\underline{= 0.167}.$$
	\quad c = 1/6 \hspace{0.15cm}\underline{= 0.167}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Theory of Stochastic Signals: Exercises\|^3.3 Expected Values and Moments^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.3 Expected Values and Moments^]]
			[[de:Aufgaben:3.4_Charakteristische_Funktion]]

Guenter at 15:48, 6 January 2022

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	'''(1)'''  Correct are  <u>the proposed solutions 2, 3 and 4</u>:		'''(1)'''  Correct are  <u>the proposed solutions 2, 3 and 4</u>:
	* $C_x( {\it \Omega} )$  is not the Fourier transform to  $f_x(x)$,  but the inverse Fourier transform:		* $C_x( {\it \Omega} )$  is not the Fourier transform to  $f_x(x)$,  but the inverse Fourier transform:
	:$$C_x( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_x }( x )\cdot {\rm{e}}^{\hspace{0.03cm}{\rm{j}}\hspace{0.~~03cm~~}{\it \Omega x}} \hspace{0.1cm}{\rm{d}}x .$$		:$$C_x( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_x }( x )\cdot {\rm{e}}^{\hspace{0.03cm}{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega\hspace{0.05cm}\cdot \hspace{0.05cm} x}} \hspace{0.1cm}{\rm{d}}x .$$
	*Also for this,  the real part is always even and the imaginary part odd.  For  ${\it \Omega} = 0$  holds:		*Also for this,  the real part is always even and the imaginary part odd.  For  ${\it \Omega} = 0$  holds:
	:$$C_x( {\it \Omega} = 0 ) = \int_{ - \infty }^{ + \infty } {f_x }( x ) \hspace{0.1cm}{\rm{d}}x = 1.$$		:$$C_x( {\it \Omega} = 0 ) = \int_{ - \infty }^{ + \infty } {f_x }( x ) \hspace{0.1cm}{\rm{d}}x = 1.$$

Guenter at 14:53, 6 January 2022

2022-01-06T14:53:15Z

Guenter at 16:43, 5 January 2022

2022-01-05T16:43:28Z

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	===Solutions===		===Solutions===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Correct are <u>the proposed solutions 2, 3 and 4</u>:		'''(1)'''  Correct are  <u>the proposed solutions 2, 3 and 4</u>:
	* $C_x( {\it \Omega} )$  is not the Fourier transform to  $f_x(x)$,  but the Fourier ~~retransform~~:		* $C_x( {\it \Omega} )$  is not the Fourier transform to  $f_x(x)$,  but the inverse Fourier transform:
	:$$C_x( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_x }( x )\cdot {\rm{e}}^{\hspace{0.03cm}{\rm{j}}\hspace{0.03cm}{\it \Omega x}} \hspace{0.1cm}{\rm{d}}x .$$		:$$C_x( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_x }( x )\cdot {\rm{e}}^{\hspace{0.03cm}{\rm{j}}\hspace{0.03cm}{\it \Omega x}} \hspace{0.1cm}{\rm{d}}x .$$
	*Also for this, the real part is always even and the imaginary part odd.  For  ${\it \Omega} = 0$  holds:		*Also for this,  the real part is always even and the imaginary part odd.  For  ${\it \Omega} = 0$  holds:
	:$$C_x( {\it \Omega} = 0 ) = \int_{ - \infty }^{ + \infty } {f_x }( x ) \hspace{0.1cm}{\rm{d}}x = 1.$$		:$$C_x( {\it \Omega} = 0 ) = \int_{ - \infty }^{ + \infty } {f_x }( x ) \hspace{0.1cm}{\rm{d}}x = 1.$$
	*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 still ~~has~~ a complex characteristic function.		*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:		'''(2)'''  According to the general definition:
	:$$C_y( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_y }( y )\cdot {\rm{e}}^{{\rm{j}}{\it \Omega y}} \hspace{0.1cm}{\rm{d}}y = 0.5\int_1^3 {{\rm{e}}^{{\rm{j}}\Omega y} \hspace{0.1cm}{\rm{d}}y.} $$		:$$C_y( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_y }( y )\cdot {\rm{e}}^{{\rm{j}}\hspace{0.01cm}{\it \Omega\hspace{0.03cm} y}} \hspace{0.1cm}{\rm{d}}y = 0.5\int_1^3 {{\rm{e}}^{{\rm{j}}\hspace{0.01cm}\Omega\hspace{0.03cm} y} \hspace{0.1cm}{\rm{d}}y.} $$

	*After solving this integral, we get:		*After solving this integral, we get:

Guenter at 16:35, 5 January 2022

2022-01-05T16:35:54Z

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	}}		}}

	[[File:P_ID619__Sto_A_3_4.png\|right\|frame\|Rectangular ~~PDF~~ and trapezoidal PDF]]		[[File:P_ID619__Sto_A_3_4.png\|right\|frame\|Rectangular and trapezoidal PDF]]
	Given here are ~~the~~ three random variables  $x$,  $y$  and  $z$ , mostly by their respective probability density functions:		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 ~~and~~ a continuous random variable, and can have any PDF  $f_x(x)$  The mean is generally equal  $m_x$.		*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 ~~only~~ take values in the range between  $1$  to  $3$  with equal probability.  ~~The mean is~~  $$m_y = 2.$$		*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:		*The random variable  $z$  has the following characteristic function:
	:$$C_z ({\it \Omega} ) = {\mathop{\rm si}\nolimits}( {3{\it \Omega}} ) \cdot {\mathop{\rm si}\nolimits} ( {2{\it \Omega} } ).$$		:$$C_z ({\it \Omega} ) = {\mathop{\rm si}\nolimits}( {3{\it \Omega}} ) \cdot {\mathop{\rm si}\nolimits} ( {2{\it \Omega} } ).$$
	: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  $a$,  $b$~~  and~~  $c$  of this PDF.		: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.




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	Hints:		Hints:
	*This exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments\|~~expected~~ values and moments]].		*This exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments\|Expected values and moments]].
	*Reference is made to the ~~page~~  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments#Characteristic_function\|~~charakteristic~~ funcion]] .		*Reference is made to the section  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments#Characteristic_function\|Charakteristic funcion]] .

	*The characteristic function of a between  $\pm a$  uniformly distributed random variable  $z$  is:		*The characteristic function of a between  $\pm a$  uniformly distributed random variable  $z$  is:
	:$$~~C_z~~ ( {\it \Omega} ) = {\mathop{\rm si}\nolimits} ( {a {\it \Omega} } )\quad {\rm{with}}\quad {\mathop{\rm si}\nolimits}( x ) = \sin ( x )/x.$$		:$$C ( {\it \Omega} ) = {\mathop{\rm si}\nolimits} ( {a {\it \Omega} } )\quad {\rm{with}}\quad {\mathop{\rm si}\nolimits}( x ) = \sin ( x )/x.$$


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	<quiz display=simple>		<quiz display=simple>
	{Which statements are valid with respect to the characteristic function  $C_x ( {\it \Omega} )$  always – that is, at any PDF &~~ndash~~;?		{Which statements are valid with respect to the characteristic function  $C_x ( {\it \Omega} )$  always – that is, at any PDF ?
	\|type="[]"}		\|type="[]"}
	- $C_x ( {\it \Omega} )$  is the Fourier transform of  $f_x(x)$.		- $C_x ( {\it \Omega} )$  is the Fourier transform of  $f_x(x)$.
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	+ The imaginary part of  $C_x ( {\it \Omega} )$  is an odd 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  $C_x ( {\it \Omega} ) = 1$.		+ The value at location  ${\it \Omega} = 0$  is always  $C_x ( {\it \Omega} ) = 1$.
	- For zero mean random variable  $(m_x = 0)$  is  $C_x ( {\it \Omega} )$  always real.		- For a zero mean random variable  $(m_x = 0)$  ⇒   $C_x ( {\it \Omega} )$  is always real.


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	{Determine the ~~characteristics~~  $a$,  $b$  and  $c$  of the PDF  $f_z(z)$.		{Determine the characteristic parameters  $a$,  $b$  and  $c$  of the PDF  $f_z(z)$.
	\|type="{}"}		\|type="{}"}
	$a \ = \ $ { 1 3% }		$a \ = \ $ { 1 3% }

Noah at 22:08, 28 December 2021

2021-12-28T22:08:27Z

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Noah at 21:29, 26 December 2021

2021-12-26T21:29:55Z

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	{{quiz-Header\|Buchseite=~~Stochastische Signaltheorie~~/~~Kapitel~~		{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Expected_Values_and_Moments
	}}		}}

Guenter at 10:07, 20 December 2021

2021-12-20T10:07:10Z

← Older revision		Revision as of 12:07, 20 December 2021
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	[[Category:Theory of Stochastic Signals: Exercises\|^3.3 ~~Erwartungswerte und Momente~~^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.3 Expected Values and Moments^]]

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	'''(2)'''  According to the general definition:		'''(2)'''  According to the general definition:
	:$$C_y( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_y }( y )\cdot {\rm{e}}^{{\rm{j}}\hspace{0.~~01cm~~}{\it \Omega\hspace{0.~~03cm~~} y}} \hspace{0.1cm}{\rm{d}}y = 0.5\int_1^3 {{\rm{e}}^{{\rm{j}}\hspace{0.~~01cm~~}\Omega\hspace{0.~~03cm~~} y} \hspace{0.1cm}{\rm{d}}y.} $$		:$$C_y( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_y }( y )\cdot {\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega\hspace{0.01cm}\hspace{0.05cm}\cdot \hspace{0.05cm} y}} \hspace{0.1cm}{\rm{d}}y = 0.5\int_1^3 {{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}\Omega\hspace{0.05cm}\cdot \hspace{0.05cm} y} \hspace{0.1cm}{\rm{d}}y.} $$

	*After solving this integral, we get:		*After solving this integral, we get:
	:$$C_y ( {\it \Omega } ) = \frac{{{\rm{e}}^{{\rm{j}}3{\it \Omega } } - {\rm{e}}^{{\rm{j}}{\it \Omega } } }}{{2{\rm{j}}{\it \Omega } }} = \frac{{{\rm{e}}^{{\rm{j}}{\it \Omega } } - {\rm{e}}^{{\rm{ - j}}{\it \Omega }} }}{{2{\rm{j}}{\it \Omega } }} \cdot {\rm{e}}^{{\rm{j2}}{\it \Omega } } .$$		:$$C_y ( {\it \Omega } ) = \frac{{{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}3{\it \Omega } } - {\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } } }}{{2{\rm{j}}{\it \Omega } }} = = \frac{{{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } } - {\rm{e}}^{{\rm{ - j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega }} }}{{2{\rm{j}}{\it \Omega } }} \cdot {\rm{e}}^{{\rm{j\hspace{0.05cm}\cdot \hspace{0.05cm}2}}{\it \Omega } } .$$

	*Using Euler's theorem, this can also be written:		*Using Euler's theorem, this can also be written:
	:$$C_y ( {\it \Omega } ) = \frac{{\sin ( {\it \Omega } )}}{{\it \Omega } } \cdot {\rm{e}}^{{\rm{j2}}{\it \Omega } } .$$		:$$C_y ( {\it \Omega } ) = \frac{{\sin ( {\it \Omega } )}}{{\it \Omega } } \cdot {\rm{e}}^{{\rm{j2}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } } = {\rm si} ( {\it \Omega } ) \cdot {\rm{e}}^{{\rm{j2}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } }.$$

	*For  ${\it \Omega} = \pi/2$  we thus obtain a purely real numerical value:		*For  ${\it \Omega} = \pi/2$  we thus obtain a purely real numerical value:
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	'''(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  $\pm 2$~~ ~~ .		'''(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  $\pm 2$.
			[[File:P_ID620__Sto_A_3_4_c_neu.png\|right\|frame\|Construction of the trapezoidal PDF]]
	*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:		*In the characteristic function, these two proportions are multiplicatively linked.
	~~[[File:P_ID620__Sto_A_3_4_c_neu~~.~~png\|center\|frame\|Construction of trapezoidal PDF]]~~		*Thus,  the resulting PDF  $f_z(z)$  is the convolution of these two rectangular functions.
	*The three PDF parameters are thus:		*The three PDF parameters are thus:
	:$$\hspace{0.15cm}\underline{a = 1},\quad \hspace{0.15cm}\underline{b = 5},		:$$\hspace{0.15cm}\underline{a = 1},\quad \hspace{0.15cm}\underline{b = 5},