Aufgaben:Exercise 5.1: Gaussian ACF and Gaussian Low-Pass - 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.~~1_Gaußsche_AKF_und_Gaußtiefpass~~]]		[[de:Aufgaben:Aufgabe 5.1: Gaußsche AKF und Gaußtiefpass]]

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	At the input of a low-pass filter with frequency response  $H(f)$,  there is a Gaussian distributed mean-free noise signal  $x(t)$  with the following auto-correlation function  $\rm (ACF)$:		At the input of a low-pass filter with frequency response  $H(f)$,  there is a Gaussian distributed mean-free noise signal  $x(t)$  with the following auto-correlation function  $\rm (ACF)$:

	:$${\it \varphi}_{x}(\tau) = \sigma_x^2 \cdot {\rm e}^{- \pi (\tau		:$${\it \varphi}_{x}(\tau) = \sigma_x^2 \cdot {\rm e}^{- \pi (\tau/{\rm \nabla} \tau_x)^2}.$$
	/{\rm \nabla} \tau_x)^2}.$$

	This ACF is shown in the accompanying diagram above.		This ACF is shown in the accompanying diagram above.
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	*Consider the following Fourier correspondence:		*Consider the following Fourier correspondence:
	:$${\rm e}^{- \pi (f/{\rm \Delta} f)^2} \hspace{0.15cm}		:$${\rm e}^{- \pi (f/{\rm \Delta} f)^2} \hspace{0.15cm}\bullet\!\!-\!\!\!-\!\!\!\hspace{0.03cm}\circ \hspace{0.15cm}{\rm \Delta} f \cdot{\rm e}^{- \pi ({\rm \Delta} f \hspace{0.03cm} \cdot \hspace{0.03cm} t)^2}.$$
	\bullet\!\!-\!\!\!-\!\!\!\hspace{0.03cm}\circ \hspace{0.15cm}{\rm \Delta} f \cdot
	{\rm e}^{- \pi ({\rm \Delta} f \hspace{0.03cm} \cdot \hspace{0.03cm} t)^2}.$$


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	'''(3)'''  The PSD is the Fourier transform of the ACF.		'''(3)'''  The PSD is the Fourier transform of the ACF.
	*With the given Fourier correspondence holds:		*With the given Fourier correspondence holds:
	:$${\it \Phi}_{x}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot		:$${\it \Phi}_{x}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot{\rm e}^{- \pi ({\rm \nabla} \tau_x \hspace{0.03cm}\cdot \hspace{0.03cm}f)^2} .$$
	{\rm e}^{- \pi ({\rm \nabla} \tau_x \hspace{0.03cm}\cdot \hspace{0.03cm}f)^2} .$$

	*At frequency  $f = 0$,  we obtain:		*At frequency  $f = 0$,  we obtain:
	:$${\it \Phi}_{x}(f = 0) = \sigma_x^2 \cdot {\rm \nabla} \tau_x =		:$${\it \Phi}_{x}(f = 0) = \sigma_x^2 \cdot {\rm \nabla} \tau_x =\rm 0.04 \hspace{0.1cm} V^2 \cdot 10^{-6} \hspace{0.1cm} s \hspace{0.15cm} \underline{= 40\cdot 10^{-9} \hspace{0.1cm} V^2 / Hz}.$$
	\rm 0.04 \hspace{0.1cm} V^2 \cdot 10^{-6} \hspace{0.1cm} s \hspace{0.15cm} \underline{= 40
	\cdot 10^{-9} \hspace{0.1cm} V^2 / Hz}.$$


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	'''(4)'''  <u>Solutions 1 and 3</u>  are correct:		'''(4)'''  <u>Solutions 1 and 3</u>  are correct:
	*In general,  ${\it \Phi}_{y}(f) = {\it \Phi}_{x}(f) \cdot \|H(f)\|^2$.  It follows:		*In general,  ${\it \Phi}_{y}(f) = {\it \Phi}_{x}(f) \cdot \|H(f)\|^2$.  It follows:
	:$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot		:$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot{\rm e}^{- \pi ({\rm \nabla} \tau_x \cdot f)^2}\cdot H_{\rm 0}^2\cdot{\rm e}^{- 2 \pi (f/ {\rm \Delta} f)^2} .$$
	{\rm e}^{- \pi ({\rm \nabla} \tau_x \cdot f)^2}\cdot H_{\rm 0}^2
	\cdot{\rm e}^{- 2 \pi (f/ {\rm \Delta} f)^2} .$$
	*By combining the two exponential functions,  we obtain:		*By combining the two exponential functions,  we obtain:
	:$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot H_0^2 \cdot		:$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot H_0^2 \cdot{\rm e}^{- \pi\cdot ({\rm \nabla} \tau_x^2 + 2/\Delta f^2 ) \hspace{0.1cm}\cdot f^2}.$$
	{\rm e}^{- \pi\cdot ({\rm \nabla} \tau_x^2 + 2/\Delta f^2 ) \hspace{0.1cm}\cdot f^2}.$$
	*Also  ${\it \Phi}_{y}(f)$  is Gaussian and never wider than  ${\it \Phi}_{x}(f)$.  For $f \to \infty$,  the approximation  ${\it \Phi}_{y}(f) \approx {\it \Phi}_{x}(f)$  holds.		*Also  ${\it \Phi}_{y}(f)$  is Gaussian and never wider than  ${\it \Phi}_{x}(f)$.  For $f \to \infty$,  the approximation  ${\it \Phi}_{y}(f) \approx {\it \Phi}_{x}(f)$  holds.
	*As  $\Delta f$  gets smaller,  ${\it \Phi}_{y}(f)$  gets narrower  (so the second statement is false).		*As  $\Delta f$  gets smaller,  ${\it \Phi}_{y}(f)$  gets narrower  (so the second statement is false).
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	'''(5)'''  Analogous to task  '''(1)''',  it can be written for the PSD of the output signal  $y(t)$:		'''(5)'''  Analogous to task  '''(1)''',  it can be written for the PSD of the output signal  $y(t)$:
	:$${\it \Phi}_{y}(f) = \sigma_y^2 \cdot {\rm \nabla} \tau_y \cdot		:$${\it \Phi}_{y}(f) = \sigma_y^2 \cdot {\rm \nabla} \tau_y \cdot{\rm e}^{- \pi \cdot {\rm \nabla} \tau_y^2 \cdot f^2 }.$$
	{\rm e}^{- \pi \cdot {\rm \nabla} \tau_y^2 \cdot f^2 }.$$

	*By comparing with the result from  '''(4)'''  we get:		*By comparing with the result from  '''(4)'''  we get:
	:$${{\rm \nabla} \tau_y^2} = {{\rm \nabla} \tau_x^2} + \frac {2}{{\rm		:$${{\rm \nabla} \tau_y^2} = {{\rm \nabla} \tau_x^2} + \frac {2}{{\rm\Delta} f^2}.$$
	\Delta} f^2}.$$
	*Solving the equation for  $\Delta f$  and considering the values  $\nabla \tau_x {= 1 \ \rm µ s}$  as well as  $\nabla \tau_y {= 3 \ \rm µ s}$,  it follows:		*Solving the equation for  $\Delta f$  and considering the values  $\nabla \tau_x {= 1 \ \rm µ s}$  as well as  $\nabla \tau_y {= 3 \ \rm µ s}$,  it follows:
	:$${\rm \Delta} f = \sqrt{\frac{2}{{\rm \nabla} \tau_y^2 - {\rm		:$${\rm \Delta} f = \sqrt{\frac{2}{{\rm \nabla} \tau_y^2 - {\rm\nabla} \tau_x^2}} = \sqrt{\frac{2}{9 - 1}} \hspace{0.1cm}\rm MHz\hspace{0.15cm} \underline{= 0.5\hspace{0.1cm} MHz} .$$
	\nabla} \tau_x^2}} = \sqrt{\frac{2}{9 - 1}} \hspace{0.1cm}\rm MHz
	\hspace{0.15cm} \underline{= 0.5\hspace{0.1cm} MHz} .$$


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	*Moreover,  since  $\nabla \tau_y = 3 \cdot \nabla \tau_x$  is given,  therefore  ${\it \Phi}_{y}(f= 0) = 3 \cdot {\it \Phi}_{x}(f= 0)$  must also hold.		*Moreover,  since  $\nabla \tau_y = 3 \cdot \nabla \tau_x$  is given,  therefore  ${\it \Phi}_{y}(f= 0) = 3 \cdot {\it \Phi}_{x}(f= 0)$  must also hold.
	*From this we obtain:		*From this we obtain:
	:$$H_{\rm 0} = \sqrt{\frac{\it \Phi_y (f \rm = 0)}{\it \Phi_x (f = \rm 0)}} = \sqrt		:$$H_{\rm 0} = \sqrt{\frac{\it \Phi_y (f \rm = 0)}{\it \Phi_x (f = \rm 0)}} = \sqrt{3}\hspace{0.15cm} \underline{=1.732}.$$
	{3}\hspace{0.15cm} \underline{=1.732}.$$

	{{ML-Fuß}}		{{ML-Fuß}}

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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:5.1_Gaußsche_AKF_und_Gaußtiefpass]]

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	<quiz display=simple>		<quiz display=simple>
	{What is the ~~rms value~~ of the filter input signal?		{What is the standard deviation of the filter input signal?
	\|type="{}"}		\|type="{}"}
	$\sigma_x \ = \ $ { 0.2 3% } $\ \rm V$		$\sigma_x \ = \ $ { 0.2 3% } $\ \rm V$

Guenter at 16:43, 9 February 2022

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Guenter at 16:34, 9 February 2022

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	}}		}}

	[[File:P_ID487__Sto_A_5_1.png\|right\|frame\|Gaussian ACF at the ~~<br>~~input and output]]		[[File:P_ID487__Sto_A_5_1.png\|right\|frame\|Gaussian ACF at the filter input and output]]
	At the input of a low-pass filter with frequency response  $H(f)$,  there is a Gaussian-distributed mean-free noise signal  $x(t)$  with the following auto-correlation function (ACF):		At the input of a low-pass filter with frequency response  $H(f)$,  there is a Gaussian distributed mean-free noise signal  $x(t)$  with the following auto-correlation function  $\rm (ACF)$:

	:$${\it \varphi}_{x}(\tau) = \sigma_x^2 \cdot {\rm e}^{- \pi (\tau		:$${\it \varphi}_{x}(\tau) = \sigma_x^2 \cdot {\rm e}^{- \pi (\tau
	/{\rm \nabla} \tau_x)^2}.$$		/{\rm \nabla} \tau_x)^2}.$$
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	This ACF is shown in the accompanying diagram above.		This ACF is shown in the accompanying diagram above.

	Let the filter be Gaussian with the DC gain  $H_0$  and the equivalent bandwidth  $\Delta f$.  Thus, for the frequency response, it can be written:		Let the filter be Gaussian with the DC gain  $H_0$  and the equivalent bandwidth  $\Delta f$.  Thus,  for the frequency response,  it can be written:
	:$$H(f) = H_{\rm 0} \cdot{\rm e}^{- \pi (f/ {\rm \Delta} f)^2}.$$		:$$H(f) = H_{\rm 0} \cdot{\rm e}^{- \pi (f/ {\rm \Delta} f)^2}.$$

	In the course of this task, the two filter parameters  $H_0$  and  $\Delta f$  are to be dimensioned so that the output signal  $y(t)$  has an ACF corresponding to the diagram below.		In the course of this task,  the two filter parameters  $H_0$  and  $\Delta f$  are to be dimensioned so that the output signal  $y(t)$  has an ACF corresponding to the diagram below.






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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/Auto-Correlation_Function_(ACF)\|Auto-Correlation Function]].		*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function_(ACF)\|Auto-Correlation Function]].
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	{From the sketched ACF, also determine the equivalent ACF duration  $\nabla\tau_x$  of the signal~~  $x(t)$~~.  How can this be determined in general?		{From the sketched ACF,  also determine the equivalent ACF duration  $\nabla\tau_x$  of the input signal.  How can this be determined in general?
	\|type="{}"}		\|type="{}"}
	$\nabla\tau_x \ = \ $ { 1 3% } $\ µ s$		$\nabla\tau_x \ = \ $ { 1 3% } $\ \rm µ s$


	{What is the power density ~~spectrum~~  ${\it Φ}_x(f)$ of the input signal?  What is the ~~PDS~~ value at $f= 0$?		{What is the power-spectral density  ${\it Φ}_x(f)$  of the input signal?  What is the PSD value at  $f= 0$?
	\|type="{}"}		\|type="{}"}
	${\it Φ}_x(f=0) \ = \ $ { 40 3% } $\ \cdot 10^{-9}\ \rm V^2/Hz$		${\it Φ}_x(f=0) \ = \ $ { 40 3% } $\ \cdot 10^{-9}\ \rm V^2/Hz$


	{Calculate the ~~PDS~~  ${\it Φ}_y(f)$  at the filter output in general as a function of  $\sigma_x$,  $\nabla \tau_x$,  $H_0$  and  $\Delta f$.  Which statements are true?		{Calculate the PSD  ${\it Φ}_y(f)$  at the filter output in general as a function of  $\sigma_x$,  $\nabla \tau_x$,  $H_0$  and  $\Delta f$.  Which statements are true?
	\|type="[]"}		\|type="[]"}
	+ The ~~PDS~~  ${\it Φ}_y(f)$  is also Gaussian.		+ The PSD  ${\it Φ}_y(f)$  is also Gaussian.
	- The smaller  $\Delta f$  is, the wider  ${\it Φ}_y(f)$.		- The smaller  $\Delta f$  is,  the wider  ${\it Φ}_y(f)$.
	+ $H_0$  only affects the height, but not the width  ${\it Φ}_y(f)$.		+ $H_0$  only affects the height,  but not the width of  ${\it Φ}_y(f)$.

Hwang at 14:47, 13 January 2022

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	[[Category:Theory of Stochastic Signals: Exercises\|^5.1 ~~Stochastische Systemtheorie~~^]]		[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastic Systems Theory^]]

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Guenter moved page Aufgaben:Aufgabe 5.1: Gaußsche AKF und Gaußtiefpass to Aufgaben:Exercise 5.1: Gaussian ACF and Gaussian Low-Pass

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Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

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	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^5.1 Stochastische Systemtheorie^]]		[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastische Systemtheorie^]]

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The variance is equal to the ACF value at  $\tau = 0$, so  $\sigma_x^2 = 0.04 \ \rm V^2$.		'''(1)'''  The variance is equal to the ACF value at  $\tau = 0$,  so  $\sigma_x^2 = 0.04 \ \rm V^2$.
	*From this follows  $\sigma_x\hspace{0.15cm}\underline {= 0.2 \ \rm V}$~~ ~~.		*From this follows  $\sigma_x\hspace{0.15cm}\underline {= 0.2 \ \rm V}$.



	'''(2)'''  The equivalent ACF duration can be determined via the rectangle of equal area.		'''(2)'''  The equivalent ACF duration can be determined via the rectangle of equal area.
	*According to the sketch, we obtain  $\nabla \tau_x\hspace{0.15cm}\underline {= 1 \ \rm µ s}$.		*According to the sketch,  we obtain  $\nabla \tau_x\hspace{0.15cm}\underline {= 1 \ \rm µ s}$.



	'''(3)'''  The ~~PDS~~ is the Fourier transform of the ACF.		'''(3)'''  The PSD is the Fourier transform of the ACF.
	*With the given Fourier correspondence holds:		*With the given Fourier correspondence holds:
	:$${\it \Phi}_{x}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot		:$${\it \Phi}_{x}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot
	{\rm e}^{- \pi ({\rm \nabla} \tau_x \hspace{0.03cm}\cdot \hspace{0.03cm}f)^2} .$$		{\rm e}^{- \pi ({\rm \nabla} \tau_x \hspace{0.03cm}\cdot \hspace{0.03cm}f)^2} .$$

	*At frequency $f = 0$,  we obtain:		*At frequency  $f = 0$,  we obtain:
	:$${\it \Phi}_{x}(f = 0) = \sigma_x^2 \cdot {\rm \nabla} \tau_x =		:$${\it \Phi}_{x}(f = 0) = \sigma_x^2 \cdot {\rm \nabla} \tau_x =
	\rm 0.04 \hspace{0.1cm} V^2 \cdot 10^{-6} \hspace{0.1cm} s \hspace{0.15cm} \underline{= 40		\rm 0.04 \hspace{0.1cm} V^2 \cdot 10^{-6} \hspace{0.1cm} s \hspace{0.15cm} \underline{= 40
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	'''(4)'''  <u>Solutions 1 and 3</u> are correct:		'''(4)'''  <u>Solutions 1 and 3</u>  are correct:
	*In general,  ${\it \Phi}_{y}(f) = {\it \Phi}_{x}(f) \cdot \|H(f)\|^2$.  It follows:		*In general,  ${\it \Phi}_{y}(f) = {\it \Phi}_{x}(f) \cdot \|H(f)\|^2$.  It follows:
	:$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot		:$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot
	{\rm e}^{- \pi ({\rm \nabla} \tau_x \cdot f)^2}\cdot H_{\rm 0}^2		{\rm e}^{- \pi ({\rm \nabla} \tau_x \cdot f)^2}\cdot H_{\rm 0}^2
	\cdot{\rm e}^{- 2 \pi (f/ {\rm \Delta} f)^2} .$$		\cdot{\rm e}^{- 2 \pi (f/ {\rm \Delta} f)^2} .$$
	*By combining the two exponential functions, we obtain:		*By combining the two exponential functions,  we obtain:
	:$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot H_0^2 \cdot		:$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot H_0^2 \cdot
	{\rm e}^{- \pi\cdot ({\rm \nabla} \tau_x^2 + 2/\Delta f^2 ) \hspace{0.1cm}\cdot f^2}.$$		{\rm e}^{- \pi\cdot ({\rm \nabla} \tau_x^2 + 2/\Delta f^2 ) \hspace{0.1cm}\cdot f^2}.$$
	*Also ${\it \Phi}_{y}(f)$  is Gaussian and never wider than  ${\it \Phi}_{x}(f)$.  For $f \to \infty$,  the approximation  ${\it \Phi}_{y}(f) \approx {\it \Phi}_{x}(f)$ holds.		*Also  ${\it \Phi}_{y}(f)$  is Gaussian and never wider than  ${\it \Phi}_{x}(f)$.  For $f \to \infty$,  the approximation  ${\it \Phi}_{y}(f) \approx {\it \Phi}_{x}(f)$  holds.
	*As  $\Delta f$  gets smaller,  ${\it \Phi}_{y}(f)$  gets narrower  (so the second statement is false).		*As  $\Delta f$  gets smaller,  ${\it \Phi}_{y}(f)$  gets narrower  (so the second statement is false).
	*$H_0$  actually affects only the ~~PDS~~ height, but not the width of the ~~PDS~~.		*$H_0$  actually affects only the PSD height,  but not the width of the PSD.



	'''(5)'''  Analogous to task  '''(1)''',  it can be written for the ~~PDS~~ of the output signal  $y(t)$~~ ~~:		'''(5)'''  Analogous to task  '''(1)''',  it can be written for the PSD of the output signal  $y(t)$:
	:$${\it \Phi}_{y}(f) = \sigma_y^2 \cdot {\rm \nabla} \tau_y \cdot		:$${\it \Phi}_{y}(f) = \sigma_y^2 \cdot {\rm \nabla} \tau_y \cdot
	{\rm e}^{- \pi \cdot {\rm \nabla} \tau_y^2 \cdot f^2 }.$$		{\rm e}^{- \pi \cdot {\rm \nabla} \tau_y^2 \cdot f^2 }.$$
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	:$${{\rm \nabla} \tau_y^2} = {{\rm \nabla} \tau_x^2} + \frac {2}{{\rm		:$${{\rm \nabla} \tau_y^2} = {{\rm \nabla} \tau_x^2} + \frac {2}{{\rm
	\Delta} f^2}.$$		\Delta} f^2}.$$
	*Solving the equation for  $\Delta f$  and considering the values  $\nabla \tau_x {= 1 \ \rm µ s}$  as well as  $\nabla \tau_y {= 3 \ \rm µ s}$, it follows:		*Solving the equation for  $\Delta f$  and considering the values  $\nabla \tau_x {= 1 \ \rm µ s}$  as well as  $\nabla \tau_y {= 3 \ \rm µ s}$,  it follows:
	:$${\rm \Delta} f = \sqrt{\frac{2}{{\rm \nabla} \tau_y^2 - {\rm		:$${\rm \Delta} f = \sqrt{\frac{2}{{\rm \nabla} \tau_y^2 - {\rm
	\nabla} \tau_x^2}} = \sqrt{\frac{2}{9 - 1}} \hspace{0.1cm}\rm MHz		\nabla} \tau_x^2}} = \sqrt{\frac{2}{9 - 1}} \hspace{0.1cm}\rm MHz
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	'''(6)'''  The condition  $\sigma_y = \sigma_x$  is equivalent to  $\varphi_y(\tau = 0)= \varphi_x(\tau = 0)$.		'''(6)'''  The condition  $\sigma_y = \sigma_x$  is equivalent to  $\varphi_y(\tau = 0)= \varphi_x(\tau = 0)$.
	*Moreover, since  $\nabla \tau_y = 3 \cdot \nabla \tau_x$  is given, therefore  ${\it \Phi}_{y}(f= 0) = 3 \cdot {\it \Phi}_{x}(f= 0)$  must also hold.		*Moreover,  since  $\nabla \tau_y = 3 \cdot \nabla \tau_x$  is given,  therefore  ${\it \Phi}_{y}(f= 0) = 3 \cdot {\it \Phi}_{x}(f= 0)$  must also hold.
	*From this we obtain:		*From this we obtain:
	:$$H_{\rm 0} = \sqrt{\frac{\it \Phi_y (f \rm = 0)}{\it \Phi_x (f = \rm 0)}} = \sqrt		:$$H_{\rm 0} = \sqrt{\frac{\it \Phi_y (f \rm = 0)}{\it \Phi_x (f = \rm 0)}} = \sqrt