Aufgaben:Exercise 5.2: Determination of the Frequency Response - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastic Systems Theory^]]		[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastic Systems Theory^]]
	[[de:Aufgaben:5.~~2_Bestimmung_des_Frequenzgangs~~]]		[[de:Aufgaben:Aufgabe 5.2: Bestimmung des Frequenzgangs]]

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	:$$\varphi _x ( \tau ) = {N_0 }/{2} \cdot \delta ( \tau ).$$		:$$\varphi _x ( \tau ) = {N_0 }/{2} \cdot \delta ( \tau ).$$
	*The measured cross-correlation function  $\rm (CCF)$  between the signals  $x(t)$  and  $y(t)$  can be approximated as follows <br>$($valid only for positive times  $t)$:		*The measured cross-correlation function  $\rm (CCF)$  between the signals  $x(t)$  and  $y(t)$  can be approximated as follows <br>$($valid only for positive times  $t)$:
	:$$\varphi _{xy} \left( \tau \right) = K \cdot {\rm{e}}^{ - \tau /T_0 },\hspace{0.5cm}\text{with } \ K = 0.628 \cdot 10^{-12} \hspace{0.05cm} {\rm W}, \ T_0 = 1 \hspace{0.05cm} \rm ms.$$		:$$\varphi _{xy} \left( \tau \right) = K \cdot {\rm{e}}^{- \tau /T_0},\hspace{0.5cm}\text{with } \ K = 0.628 \cdot 10^{-12} \hspace{0.05cm} {\rm W}, \ T_0 = 1 \hspace{0.05cm} \rm ms.$$
	*The ACF  $\varphi_y(\tau)$  of the output signal  $y(t)$ is also measured.		*The ACF  $\varphi_y(\tau)$  of the output signal  $y(t)$ is also measured.

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	'''(2)'''  For Dirac-shaped input ACF  $\varphi_x(\tau)$,  the impulse response  $h(t)$  is equal in shape to the CCF:		'''(2)'''  For Dirac-shaped input ACF  $\varphi_x(\tau)$,  the impulse response  $h(t)$  is equal in shape to the CCF:
	:$$h(t) = \frac{{K \cdot {\rm{e}}^{ - t/T_0 } }}{N_0 /2} = 1.256 \cdot 10^{ - 2} \frac{1}{{\rm{s}}} \cdot {\rm{e}}^{ - t/T_0 } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}h(t = T_0)\hspace{0.15cm}\underline{ = 4.62 \cdot 10^{-3}\ \rm 1/s}.$$		:$$h(t) = \frac{{K \cdot {\rm{e}}^{- t/T_0} }}{N_0 /2} = 1.256 \cdot 10^{ - 2} \frac{1}{{\rm{s}}} \cdot {\rm{e}}^{- t/T_0} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}h(t = T_0)\hspace{0.15cm}\underline{ = 4.62 \cdot 10^{-3}\ \rm 1/s}.$$



	'''(3)'''  The Fourier correspondence given is with  $T_0 = 1/\omega_0$  and the constant  $C= N_0/2 \cdot T_0/K$:		'''(3)'''  The Fourier correspondence given is with  $T_0 = 1/\omega_0$  and the constant  $C= N_0/2 \cdot T_0/K$:
	:$$h(t) = \frac{C}{T_0 } \cdot {\rm{e}}^{ - t/T_0 }\hspace{0.15cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\hspace{0.15cm} H( \omega ) = \frac{C}{{1 + {\rm{j}}\cdot \omega T_0 }}.$$		:$$h(t) = \frac{C}{T_0 } \cdot {\rm{e}}^{- t/T_0}\hspace{0.15cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\hspace{0.15cm} H( \omega ) = \frac{C}{{1 + {\rm{j}}\cdot \omega T_0 }}.$$

	*The constant results in  $C = 0.08$.  With  $H(f) = 2 \pi \cdot H(\omega)$  it follows:		*The constant results in  $C = 0.08$.  With  $H(f) = 2 \pi \cdot H(\omega)$  it follows:

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	[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastic Systems Theory^]]		[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastic Systems Theory^]]
			[[de:Aufgaben:5.2_Bestimmung_des_Frequenzgangs]]

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	[[File:~~EN_Sto_A_5_2_XXX~~.png\|right\|frame\|Measuring the frequency response  $H(f)$]]		[[File:EN_Sto_A_5_2_neu2.png\|right\|frame\|Measuring the frequency response  $H(f)$]]
	We consider the illustrated measurement setup for the determination of the frequency response  $H(f)$  highlighted in blue.		We consider the illustrated measurement setup for the determination of the frequency response  $H(f)$  highlighted in blue.
	*The input signal  $x(t)$  is white Gaussian noise with the noise power density  $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.		*The input signal  $x(t)$  is white Gaussian noise with the noise power density  $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.

Guenter at 15:19, 10 February 2022

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Guenter at 15:12, 10 February 2022

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	''Notes:''		''Notes:''
	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Stochastic_System_Theory\|Stochastic System Theory]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Stochastic_System_Theory\|Stochastic System Theory]].
	*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/~~Leistungsdichtespektrum_(LDS)~~\|Power Density ~~Spectrum~~]].		*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/Power-Spectral_Density\|Power-Spectral Density]].

	*Please also note the following Fourier transform  $($in  $\omega)$:		*Please also note the following Fourier transform  $($in  $\omega)$:
	:$$H( \omega ) = \frac{1}{{1 + {\rm{j}}\cdot \omega /\omega _0 }}\hspace{0.15cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\,\hspace{0.15cm}h(t) = \omega _0 \cdot {\rm{e}}^{ - \omega _0 t} \hspace{0.3cm}(t \ge 0).$$		:$$H( \omega ) = \frac{1}{{1 + {\rm{j}}\cdot \omega /\omega _0 }}\hspace{0.15cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\,\hspace{0.15cm}h(t) = \omega _0 \cdot {\rm{e}}^{ - \omega _0 t} \hspace{0.3cm}(t \ge 0).$$
	:For negative  $t$ values, on the other hand,  $h(t) =0$ at all times.		:For negative  $t$–values,  on the other hand,  $h(t) =0$  at all times.


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	{Calculate the power density ~~spectrum~~ of the output signal  $y(t)$.  What value results for frequency  $f = 1/(2\pi T_0)$?		{Calculate the power-spectral density of the output signal  $y(t)$.  What value results for frequency  $f = 1/(2\pi T_0)$?
	\|type="{}"}		\|type="{}"}
	${\it \Phi}_y(f = 1/(2\pi T_0)) \ = \ $ { 6.25 3% } $\ \cdot 10^{-12}\ \rm W/Hz$		${\it \Phi}_y(f = 1/(2\pi T_0)) \ = \ $ { 6.25 3% } $\ \cdot 10^{-12}\ \rm W/Hz$

Guenter at 15:08, 10 February 2022

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	[[File:EN_Sto_A_5_2_XXX.png\|right\|frame\|~~Determining~~ the frequency response  $H(f)$]]		[[File:EN_Sto_A_5_2_XXX.png\|right\|frame\|Measuring the frequency response  $H(f)$]]
	We consider the illustrated measurement setup for the determination of the frequency response  $H(f)$ highlighted in blue.		We consider the illustrated measurement setup for the determination of the frequency response  $H(f)$  highlighted in blue.
	*The input signal  $x(t)$  is white Gaussian noise with the noise power density  $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.		*The input signal  $x(t)$  is white Gaussian noise with the noise power density  $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.
	*Thus, the auto-correlation function (ACF) is:		*Thus,  the auto-correlation function  $\rm (ACF)$  is:
	:$$\varphi _x ( \tau ) = {N_0 }/{2} \cdot \delta ( \tau ).$$		:$$\varphi _x ( \tau ) = {N_0 }/{2} \cdot \delta ( \tau ).$$
	*The measured cross-correlation function (CCF) between the signals  $x(t)$  and  $y(t)$  can be approximated as follows <br>$($valid only for positive times  $t)$:		*The measured cross-correlation function  $\rm (CCF)$  between the signals  $x(t)$  and  $y(t)$  can be approximated as follows <br>$($valid only for positive times  $t)$:
	:$$\varphi _{xy} \left( \tau \right) = K \cdot {\rm{e}}^{ - \tau /T_0 },\hspace{0.5cm}\text{~~mit~~ } \ K = 0.628 \cdot 10^{-12} \hspace{0.05cm} {\rm W}, \ T_0 = 1 \hspace{0.05cm} \rm ms.$$		:$$\varphi _{xy} \left( \tau \right) = K \cdot {\rm{e}}^{ - \tau /T_0 },\hspace{0.5cm}\text{with } \ K = 0.628 \cdot 10^{-12} \hspace{0.05cm} {\rm W}, \ T_0 = 1 \hspace{0.05cm} \rm ms.$$
	*The ACF  $\varphi_y(\tau)$  of the output signal  $y(t)$ is also measured.		*The ACF  $\varphi_y(\tau)$  of the output signal  $y(t)$ is also measured.

Guenter at 16:46, 8 February 2022

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	[[File:~~EN_Sto_A_5_2_neu~~.png\|right\|frame\|Determining the frequency response  $H(f)$]]		[[File:EN_Sto_A_5_2_XXX.png\|right\|frame\|Determining the frequency response  $H(f)$]]
	We consider the illustrated measurement setup for the determination of the frequency response  $H(f)$ highlighted in blue.		We consider the illustrated measurement setup for the determination of the frequency response  $H(f)$ highlighted in blue.
	*The input signal  $x(t)$  is white Gaussian noise with the noise power density  $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.		*The input signal  $x(t)$  is white Gaussian noise with the noise power density  $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.

Guenter at 16:07, 8 February 2022

2022-02-08T16:07:03Z

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	[[File:~~EN_Sto_A_5_2~~.png\|right\|frame\|Determining the frequency response  $H(f)$]]		[[File:EN_Sto_A_5_2_neu.png\|right\|frame\|Determining the frequency response  $H(f)$]]
	We consider the illustrated measurement setup for the determination of the frequency response  $H(f)$ highlighted in blue.		We consider the illustrated measurement setup for the determination of the frequency response  $H(f)$ highlighted in blue.
	*The input signal  $x(t)$  is white Gaussian noise with the noise power density  $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.		*The input signal  $x(t)$  is white Gaussian noise with the noise power density  $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.

Hwang at 16:11, 13 January 2022

2022-01-13T16:11:52Z

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			Notes:



	''Notes:''
	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Stochastic_System_Theory\|Stochastic System Theory]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Stochastic_System_Theory\|Stochastic System Theory]].
	*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/Power-Spectral_Density\|Power-Spectral Density]].		*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/Power-Spectral_Density\|Power-Spectral Density]].
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	:$$\varphi _{xy} ( \tau ) = h( \tau ) * \varphi _x ( \tau )\quad \Rightarrow \quad H( f ) = \frac{{{\it \Phi} _{xy} ( f )}}{{{\it \Phi} _x ( f )}},$$		:$$\varphi _{xy} ( \tau ) = h( \tau ) * \varphi _x ( \tau )\quad \Rightarrow \quad H( f ) = \frac{{{\it \Phi} _{xy} ( f )}}{{{\it \Phi} _x ( f )}},$$
	:$$\varphi _y ( \tau) = \varphi _{xy} ( \tau) * h(- \tau)\quad \Rightarrow \quad H^{\star}( f ) = \frac{{{\it \Phi} _y ( f )}}{{{\it \Phi} _{xy} ( f )}}.$$		:$$\varphi _y ( \tau) = \varphi _{xy} ( \tau) * h(- \tau)\quad \Rightarrow \quad H^{\star}( f ) = \frac{{{\it \Phi} _y ( f )}}{{{\it \Phi} _{xy} ( f )}}.$$
	*In contrast, the first statement is false:   ~~the~~ phase relations are lost in the ACF calculation.		*In contrast,  the first statement is false:   The phase relations are lost in the ACF calculation.
	*The associated spectral functions to  $\varphi_x(\tau)$  and  $\varphi_x(\tau)$  – namely  ${\it \Phi}_x(f)$  and  ${\it \Phi}_y(f)$  – are purely real, so only the magnitude  $\|H(f)\|$  can be given.		*The associated spectral functions to  $\varphi_x(\tau)$  and  $\varphi_x(\tau)$  – namely  ${\it \Phi}_x(f)$  and  ${\it \Phi}_y(f)$  – are purely real,  so only the magnitude  $\|H(f)\|$  can be given.



	'''(2)'''  For ~~dirac~~-shaped input ACF  $\varphi_x(\tau)$,  the impulse response  $h(t)$  is equal in shape to the CCF:		'''(2)'''  For Dirac-shaped input ACF  $\varphi_x(\tau)$,  the impulse response  $h(t)$  is equal in shape to the CCF:
	:$$h(t) = \frac{{K \cdot {\rm{e}}^{ - t/T_0 } }}{N_0 /2} = 1.256 \cdot 10^{ - 2} \frac{1}{{\rm{s}}} \cdot {\rm{e}}^{ - t/T_0 } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}h(t = T_0)\hspace{0.15cm}\underline{ = 4.62 \cdot 10^{-3}\ \rm 1/s}.$$		:$$h(t) = \frac{{K \cdot {\rm{e}}^{ - t/T_0 } }}{N_0 /2} = 1.256 \cdot 10^{ - 2} \frac{1}{{\rm{s}}} \cdot {\rm{e}}^{ - t/T_0 } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}h(t = T_0)\hspace{0.15cm}\underline{ = 4.62 \cdot 10^{-3}\ \rm 1/s}.$$

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	*The constant results in  $C = 0.08$.  With  $H(f) = 2 \pi \cdot H(\omega)$  it follows:		*The constant results in  $C = 0.08$.  With  $H(f) = 2 \pi \cdot H(\omega)$  it follows:
	:$$H(f) = \frac{0.5}{1 + {\rm{j\cdot 2\pi }}\cdot ~~fT_0~~ } \hspace{0.3cm}\Rightarrow \hspace{0.3cm} H(f= 0) \hspace{0.15cm}\underline{=0.5}.$$		:$$H(f) = \frac{0.5}{1 + {\rm{j\cdot 2\pi }}\cdot f \cdot T_0 } \hspace{0.3cm}\Rightarrow \hspace{0.3cm} H(f= 0) \hspace{0.15cm}\underline{=0.5}.$$



	'''(4)'''  For the output ~~PDS~~, in general, or specifically here:		'''(4)'''  For the output PSD,  in general or specifically here:
	:$${\it \Phi}_y (f) = {\it \Phi} _x (f) \cdot \left\| {H(f)} \right\|^2 = \frac{N_0 }{2} \cdot \frac{0.5^2 }{{\left( {1 + {\rm{j\cdot 2\pi }}\cdot ~~fT_0~~ } \right)\left( {1 - {\rm{j\cdot 2\pi }}\cdot ~~fT_0~~ } \right)}} = {N_0 }/{8} \cdot \frac{1}{1 + \left( {{\rm{2\pi }}\cdot ~~fT_0~~ } \right)^2 }.$$		:$${\it \Phi}_y (f) = {\it \Phi} _x (f) \cdot \left\| {H(f)} \right\|^2 = \frac{N_0 }{2} \cdot \frac{0.5^2 }{{\left( {1 + {\rm{j\cdot 2\pi }}\cdot f \cdot T_0 } \right)\left( {1 - {\rm{j\cdot 2\pi }}\cdot f \cdot T_0 } \right)}} = {N_0 }/{8} \cdot \frac{1}{1 + \left( {{\rm{2\pi }}\cdot f\cdot T_0 } \right)^2 }.$$

	*At the given frequency  $f = 1/(2\pi T_0)$,    ${\it \Phi}_y (f)$ has dropped by half compared to its maximum at~~ ~~ $f=0$ :		*At the given frequency  $f = 1/(2\pi T_0)$,    ${\it \Phi}_y (f)$  has dropped by half compared to its maximum at $f=0$ :
	:$${\it \Phi}_y (f = 1/(2 \pi T_0)) ={N_0 }/{16}\hspace{0.15cm} \underline{ = 6.25 \cdot 10^{ - 12} \;{\rm{W/Hz}}}.$$		:$${\it \Phi}_y (f = 1/(2 \pi T_0)) ={N_0 }/{16}\hspace{0.15cm} \underline{ = 6.25 \cdot 10^{ - 12} \;{\rm{W/Hz}}}.$$