Aufgaben:Exercise 4.10Z: Correlation Duration - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.4 Auto-Correlation Function^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.4 Auto-Correlation Function^]]
	[[de:~~Zusatzaufgaben~~:4.~~10_Korrelationsdauer~~]]		[[de:Aufgaben:Aufgabe 4.10Z: Korrelationsdauer]]

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	*From this follows:		*From this follows:
	:$$\varphi_y(\tau) = 1.8 \hspace{0.1cm} {\rm mW} + 3.2 \hspace{0.1cm} {\rm mW} \cdot {\rm e}^{- \pi \hspace{0.03cm} \cdot \hspace{0.03cm}(\tau / {\rm \nabla} \tau_y)^2} \hspace{0.3cm }\Rightarrow \hspace{0.3cm }\varphi_y(\tau = 10\hspace{0.05 cm}{\rm µ s})		:$$\varphi_y(\tau) = 1.8 \hspace{0.1cm} {\rm mW} + 3.2 \hspace{0.1cm} {\rm mW} \cdot {\rm e}^{- \pi \hspace{0.03cm} \cdot \hspace{0.03cm}(\tau / {\rm \nabla} \tau_y)^2} \hspace{0.3cm }\Rightarrow \hspace{0.3cm }\varphi_y(\tau = 10\hspace{0.05 cm}{\rm µ s})\hspace{0.15 cm}\underline{=1.938\hspace{0.05 cm}\rm mW}.$$
	\hspace{0.15 cm}\underline{=1.938\hspace{0.05 cm}\rm mW}.$$

	*With positive mean  $m_y$  $($having the same magnitude$)$,  there would be no change in the ACF,  since  $m_y$  is squared in the ACF equation.		*With positive mean  $m_y$  $($having the same magnitude$)$,  there would be no change in the ACF,  since  $m_y$  is squared in the ACF equation.

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.4 Auto-Correlation Function^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.4 Auto-Correlation Function^]]
			[[de:Zusatzaufgaben:4.10_Korrelationsdauer]]

Guenter at 17:52, 20 March 2022

2022-03-20T17:52:52Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The ~~quadratic mean~~ results to  $m_{2x} = R \cdot P_x = 50 \hspace{0.05 cm}{\rm \Omega}\cdot 5 \hspace{0.05 cm}{\rm mW}= 0.25 \hspace{0.05 cm}{\rm V}^2.$		'''(1)'''  The second moment results to  $m_{2x} = R \cdot P_x = 50 \hspace{0.05 cm}{\rm \Omega}\cdot 5 \hspace{0.05 cm}{\rm mW}= 0.25 \hspace{0.05 cm}{\rm V}^2.$
	*From this follows the standard deviation  $\sigma_x\hspace{0.15 cm}\underline{= 0.5\hspace{0.05 cm}{\rm V}}$.		*From this follows the standard deviation  $\sigma_x\hspace{0.15 cm}\underline{= 0.5\hspace{0.05 cm}{\rm V}}$.



	'''(2)'''  Because  $P_x = \varphi_x (\tau = 0)$  holds for the ACF in general:		'''(2)'''  Because of  $P_x = \varphi_x (\tau = 0)$  holds for the ACF in general:
	:$$\varphi_x (\tau) = 5 \hspace{0.1cm} {\rm mW} \cdot {\rm e}^{- \pi \hspace{0.03cm} \cdot \hspace{0.03cm}(\tau / {\rm \nabla} \tau_x)^2}.$$		:$$\varphi_x (\tau) = 5 \hspace{0.1cm} {\rm mW} \cdot {\rm e}^{- \pi \hspace{0.03cm} \cdot \hspace{0.03cm}(\tau / {\rm \nabla} \tau_x)^2}.$$
	*From this we obtain:		*From this we obtain:
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	*From this follows  $T_{\rm K}\hspace{0.15 cm}\underline{= 2.35\hspace{0.05 cm}{\rm µ s}}$.		*From this follows  $T_{\rm K}\hspace{0.15 cm}\underline{= 2.35\hspace{0.05 cm}{\rm µ s}}$.
	*With ~~other~~ ACF form, a different ratio is obtained for  $T_{\rm K} / {\rm \nabla} \tau_x$.		*With another ACF form,  a different ratio is obtained for  $T_{\rm K} / {\rm \nabla} \tau_x$.




	'''(4)'''  Because  $P_x = P_y$  the ~~root mean square values~~ of  $x$  and  $y$  are equal~~, respectively~~  $0.25\hspace{0.05 cm}\rm V^2$.		'''(4)'''  Because of  $P_x = P_y$  the second order moments of  $x$  and  $y$  are equal   $0.25\hspace{0.05 cm}\rm V^2$.
	*Taking into account the mean value  $m_y = -0.3 \hspace{0.05 cm}\rm V$  holds:		*Taking into account the mean value  $m_y = -0.3 \hspace{0.05 cm}\rm V$  holds:
	:$$m_y^2 + \sigma_y^2 = \rm 0.25 \hspace{0.05 cm} V^2.$$		:$$m_y^2 + \sigma_y^2 = \rm 0.25 \hspace{0.05 cm} V^2.$$
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	'''(5)'''  In terms of unit resistance  $ R = 1 \hspace{0.05 cm}{\rm \Omega}$  the ~~ASF~~ of the process  $\{y_i(t)\}$ is:		'''(5)'''  In terms of unit resistance  $ R = 1 \hspace{0.05 cm}{\rm \Omega}$  the ACF of the process  $\{y_i(t)\}$ is:
	:$$\varphi_y (\tau) = m_y^2 + \sigma_y^2 \cdot {\rm e}^{- \pi \hspace{0.03cm} \cdot \hspace{0.03cm}(\tau / {\rm \nabla} \tau_y)^2}.$$		:$$\varphi_y (\tau) = m_y^2 + \sigma_y^2 \cdot {\rm e}^{- \pi \hspace{0.03cm} \cdot \hspace{0.03cm}(\tau / {\rm \nabla} \tau_y)^2}.$$

	*On the right you can see the ~~function progression~~.  Related to the resistor  $ R = 50 \hspace{0.05 cm}{\rm \Omega}$  results in the following ACF values:		*On the right you can see the ACF curve.  Related to the resistor  $ R = 50 \hspace{0.05 cm}{\rm \Omega}$  results in the following ACF values:
	:$$\varphi_y (\tau = 0) = 5 \hspace{0.1cm} {\rm mW} , \hspace{0.5cm} \varphi_y (\tau \rightarrow \infty) = 1.8\hspace{0.1cm} {\rm mW} .$$		:$$\varphi_y (\tau = 0) = 5 \hspace{0.1cm} {\rm mW} , \hspace{0.5cm} \varphi_y (\tau \rightarrow \infty) = 1.8\hspace{0.1cm} {\rm mW} .$$

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	\hspace{0.15 cm}\underline{=1.938\hspace{0.05 cm}\rm mW}.$$		\hspace{0.15 cm}\underline{=1.938\hspace{0.05 cm}\rm mW}.$$

	*With positive mean  $m_y$  (having the same ~~amplitude~~), there would be no change in the ~~ASF~~, since  $m_y$  is squared in the ~~ASF~~ equation.		*With positive mean  $m_y$  $($having the same magnitude$)$,  there would be no change in the ACF,  since  $m_y$  is squared in the ACF equation.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 17:40, 20 March 2022

2022-03-20T17:40:52Z

← Older revision		Revision as of 19:40, 20 March 2022
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	}}		}}

	[[File:P_ID393__Sto_Z_4_10.png\|right\|frame\|~~Sample functions~~ of ergodic processes]]		[[File:P_ID393__Sto_Z_4_10.png\|right\|frame\|Pattern signals of ergodic processes]]
	The graphic shows pattern signals of two random processes  $\{x_i(t)\}$  and  $\{y_i(t)\}$  with equal power		The graphic shows pattern signals of two random processes  $\{x_i(t)\}$  and  $\{y_i(t)\}$  with equal power
	:$$P_x = P_y = 5\hspace{0.05 cm} \rm mW.$$		:$$P_x = P_y = 5\hspace{0.05 cm} \rm mW.$$

Guenter at 17:40, 20 March 2022

2022-03-20T17:40:13Z

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	[[File:P_ID393__Sto_Z_4_10.png\|right\|frame\|Sample functions of ergodic processes]]		[[File:P_ID393__Sto_Z_4_10.png\|right\|frame\|Sample functions of ergodic processes]]
	The ~~adjacent image~~ shows pattern signals of two random processes  $\{x_i(t)\}$  and  $\{y_i(t)\}$  ~~each~~ with equal power  $P_x = P_y = 5\hspace{0.05 cm} \rm mW$~~. ~~ Assuming here the resistance  $R = 50\hspace{0.05 cm}\rm \Omega$.		The graphic shows pattern signals of two random processes  $\{x_i(t)\}$  and  $\{y_i(t)\}$  with equal power
			:$$P_x = P_y = 5\hspace{0.05 cm} \rm mW.$$
			Assuming here the resistance  $R = 50\hspace{0.05 cm}\rm \Omega$.


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			As can be seen from the diagram below,  the random process  $\{y_i(t)\}$  has much stronger internal statistical bindings than the random process  $\{x_i(t)\}$.
			Or,  to put it another way:
	As can be seen from the ~~picture~~ below, the random process  $\{y_i(t)\}$  has much stronger internal statistical bindings than the random process  $\{x_i(t)\}$.
	Or, to put it another way:
	*The random process  $\{y_i(t)\}$  is lower frequency than  $\{x_i(t)\}$.		*The random process  $\{y_i(t)\}$  is lower frequency than  $\{x_i(t)\}$.
	*The equivalent ACF duration is  $\nabla \tau_y = 10 \hspace{0.05 cm}\rm µ s $.		*The equivalent ACF duration is  $\nabla \tau_y = 10 \hspace{0.05 cm}\rm µ s $.
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	From the sketch it can also be seen that  $\{y_i(t)\}$  in contrast to  $\{x_i(t)\}$  is not ~~equal signal~~ free.  The ~~equal~~ signal component is rather  $m_y = -0.3 \hspace{0.05 cm}\rm V$.		From the sketch it can also be seen that  $\{y_i(t)\}$  in contrast to  $\{x_i(t)\}$  is not DC free.  The DC signal component is rather  $m_y = -0.3 \hspace{0.05 cm}\rm V$.





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	Hint:		'''Hint''':
	*~~Die Aufgabe gehört zum Kapitel~~  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function\|Auto-Correlation Function]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function\|Auto-Correlation Function]].
	*Reference is made in particular to the ~~page~~  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function#Interpretation_of_the_auto-correlation_function\|Interpretation of the auto-correlation function]].		*Reference is made in particular to the section  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function#Interpretation_of_the_auto-correlation_function\|Interpretation of the auto-correlation function]].


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	{What ACF~~–~~values result ~~für~~  $\tau = 2\hspace{0.05 cm}\rm µs$  resp.  $\tau = 5\hspace{0.05 cm}\rm µ s$?		{What ACF values result for  $\tau = 2\hspace{0.05 cm}\rm µs$  resp.  $\tau = 5\hspace{0.05 cm}\rm µ s$?
	\|type="{}"}		\|type="{}"}
	$\varphi_x(\tau = 2\hspace{0.05 cm}{\rm µ s}) \ = \ $ { 3.025 3% } $\ \rm mW$		$\varphi_x(\tau = 2\hspace{0.05 cm}{\rm µ s}) \ = \ $ { 3.025 3% } $\ \rm mW$
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	{What is the correlation time  $T_{\rm K}$, i.e. the time at which the ACF has dropped to half of the maximum?		{What is the correlation time  $T_{\rm K}$,  i.e. the time at which the ACF has dropped to half of the maximum?
	\|type="{}"}		\|type="{}"}
	$T_{\rm K} \ = \ $ { 2.35 3% } $\ \rm µ s$		$T_{\rm K} \ = \ $ { 2.35 3% } $\ \rm µ s$
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	{Calculate the ACF  $\varphi_x(\tau)$.  What is the ACF value at  $\tau = 10\hspace{0.05 cm}\rm µ s$?  What would be the ACF~~–~~curve with positive mean  $(m_y = +0.3 \hspace{0.05 cm}\rm V)$?		{Calculate the ACF  $\varphi_x(\tau)$.  What is the ACF value at  $\tau = 10\hspace{0.05 cm}\rm µ s$?  What would be the ACF curve with positive mean  $(m_y = +0.3 \hspace{0.05 cm}\rm V)$?
	\|type="{}"}		\|type="{}"}
	$\varphi_y(\tau = 10\hspace{0.05 cm}{\rm µ s}) \ = \ $ { 1.938 3% } $\ \rm mW$		$\varphi_y(\tau = 10\hspace{0.05 cm}{\rm µ s}) \ = \ $ { 1.938 3% } $\ \rm mW$

Noah at 23:08, 28 February 2022

2022-02-28T23:08:20Z

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Guenter at 08:57, 28 January 2022

2022-01-28T08:57:17Z

← Older revision		Revision as of 10:57, 28 January 2022
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	[[Category:Theory of Stochastic Signals: Exercises\|^4.4 ~~Autokorrelationsfunktion (AKF)~~^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.4 Auto-Correlation Function^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 4.10Z: Korrelationsdauer to Aufgaben:Exercise 4.10Z: Correlation Duration

2022-01-28T08:43:13Z

Guenter moved page Aufgaben:Aufgabe 4.10Z: Korrelationsdauer to Aufgaben:Exercise 4.10Z: Correlation Duration

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

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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~~\|^4.4 Autokorrelationsfunktion (AKF)^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.4 Autokorrelationsfunktion (AKF)^]]