Aufgaben:Exercise 4.14Z: Echo Detection - Revision history

Guenter at 15:51, 10 April 2022

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Noah at 11:13, 14 March 2022

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Guenter at 09:55, 28 February 2022

2022-02-28T09:55:25Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.14Z: Auffinden von Echos to Aufgaben:Exercise 4.14Z: Echo Detection

2022-02-28T09:51:58Z

Guenter moved page Aufgabe 4.14Z: Auffinden von Echos to Exercise 4.14Z: Echo Detection

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2021-05-28T14:38:22Z

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2021-05-28T14:37:29Z

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@@ Line 95: / Line 95: @@
 *Outside of the range&nbsp; $|f| \le B_x$&nbsp; the power-spectral density&nbsp; ${\it \Phi}_{x}(f)$&nbsp; &ndash; and correspondingly the cross power-spectral density&nbsp; ${\it \Phi}_{xy}(f)$ &ndash; is identically zero.
 *In contrast,&nbsp; inside this interval holds&nbsp; ${\it \Phi}_{x}(f) = N_0/2$.&nbsp; It follows in this range:
-[[File:P_ID450__Sto_Z_4_14_d.png|right|frame|ACF and CCF in white noise]]
+[[File:P_ID450__Sto_Z_4_14_d.png|right|frame|ACF and CCF with white noise]]
 :$${\it \Phi}_{xy} (f) = {N_0}/{2} \left( \alpha_1 \cdot {\rm e}^{-{\rm j}2 \pi f t_1} \hspace{0.15cm}+ \hspace{0.15cm}\alpha_2 \cdot {\rm e}^{-{\rm j}2 \pi f t_2} \right). $$

@@ Line 48: / Line 48: @@
-{Which of the following statements are true if you use the approximation&nbsp; $\varphi_{xy}(\tau) \approx N_0/2 \cdot \delta(\tau)$&nbsp; instead of the ACF calculated in&nbsp; '''(1)'''?
+{Which of the following statements are true if you use the approximation&nbsp; $\varphi_{x}(\tau) \approx N_0/2 \cdot \delta(\tau)$&nbsp; instead of the ACF calculated in&nbsp; '''(1)'''?
 |type="[]"}
 + The noise is now&nbsp; "true white"&nbsp; &ndash; so it is not bandlimited.

@@ Line 66: / Line 66: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The two-sided power spectral density spectrum&nbsp; ${\it \Phi}_{x}(f)$&nbsp; is constantly equal&nbsp; $N_0/2$ in the range&nbsp; $\pm B_x$&nbsp; .
+'''(1)'''&nbsp; The two-sided power-spectral density&nbsp; ${\it \Phi}_{x}(f)$&nbsp; is constantly equal&nbsp; $N_0/2$&nbsp; in the range&nbsp; $\pm B_x$.
-*Whose Fourier transform is the ACF:
+*The Fourier transform is the ACF:
-:$$\varphi_x (\tau) = {N_0}/{2} \cdot 2 B_x \cdot {\rm si} (2 \pi B_x \tau) = 0.02 \hspace {0.08cm}{\rm W} \cdot {\rm si} (2 \pi B_x \tau).$$
+:$$\varphi_x (\tau) = {N_0}/{2} \cdot 2 B_x \cdot {\rm sinc} (2 B_x \tau) = 0.02 \hspace {0.08cm}{\rm W} \cdot {\rm sinc} (2 B_x \tau).$$
-*Converted from&nbsp; $R = 50 \hspace{0.08cm} \rm \Omega$&nbsp; to&nbsp; $R = 1 \hspace{0.08cm} \rm \Omega$&nbsp; thus one obtains&nbsp; $($multiplication by&nbsp; $R = 50 \hspace{0.08cm} \rm \Omega)$:
+*Converted from &nbsp; $R = 50 \hspace{0.08cm} \rm \Omega$ &nbsp; to &nbsp; $R = 1 \hspace{0.08cm} \rm \Omega$ &nbsp; one obtains&nbsp; $($multiplication by&nbsp; $R = 50 \hspace{0.08cm} \rm \Omega)$:
-:$$\varphi_x (\tau) = 0.02 \hspace {0.05cm}{\rm VA} \cdot 50 \hspace {0.05cm}{\rm V/A}\cdot {\rm si} (2 \pi B_x \tau)= 1 \hspace {0.05cm}{\rm V}^2 \cdot {\rm si} (2 \pi B_x \tau).$$
+:$$\varphi_x (\tau) = 0.02 \hspace {0.05cm}{\rm VA} \cdot 50 \hspace {0.05cm}{\rm V/A}\cdot {\rm sinc} (2 B_x \tau)= 1 \hspace {0.05cm}{\rm V}^2 \cdot {\rm sinc} (2 B_x \tau).$$
 *The rms value is the square root of the ACF value at&nbsp; $\tau = 0$: &nbsp;
 :$$\sigma_x \hspace{0.15cm}\underline{= 1 \hspace {0.08cm}{\rm V}}.$$
 '''(2)'''&nbsp; For the cross correlation function (CCF), in the present case:
 :$$\varphi_{xy} (\tau) = \overline {x(t) \hspace{0.05cm}\cdot \hspace{0.05cm}y(t+\tau)} = \overline {x(t) \hspace{0.05cm}\cdot \hspace{0.05cm}\big [ \alpha_1 \cdot x(t- t_1+ \tau)\hspace{0.1cm}+\hspace{0.1cm} \alpha_2 \cdot x(t- t_2+ \tau)\big] } . $$
-*After splitting the averaging on the two terms, we get:
+*After splitting the averaging on two terms,&nbsp; we get:
 :$$\varphi_{xy} (\tau) = \alpha_1 \cdot \overline {x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} x(t- t_1+ \tau)} \hspace{0.1cm}+\hspace{0.1cm} \alpha_2 \cdot \overline {x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} x(t- t_2+ \tau)} .$$
-*Using the ACF&nbsp; $\varphi_x(\tau)$&nbsp; can also be written herefor:
+*Using the ACF&nbsp; $\varphi_x(\tau)$&nbsp; can also be written:
-:$$\varphi_{xy} (\tau) = \alpha_1 \cdot {\varphi_{x}(\tau- t_1)} \hspace{0.1cm}+\hspace{0.1cm} \alpha_2\cdot {\varphi_{x}(\tau- t_2)} = 1 \hspace {0.08cm}{\rm V}^2 \cdot \big[ \alpha_1 \cdot {\rm si} (2 \pi B_x (\tau - t_1)) + \alpha_2 \cdot {\rm si} (2 \pi B_x (\tau - t_2))  \big].$$
+:$$\varphi_{xy} (\tau) = \alpha_1 \cdot {\varphi_{x}(\tau- t_1)} \hspace{0.1cm}+\hspace{0.1cm} \alpha_2\cdot {\varphi_{x}(\tau- t_2)} = 1 \hspace {0.08cm}{\rm V}^2 \cdot \big[ \alpha_1 \cdot {\rm sinc} (2 B_x (\tau - t_1)) + \alpha_2 \cdot {\rm sinc} (2  B_x (\tau - t_2))  \big].$$
-*The si function exhibits equidistant zero crossings at multiples of&nbsp; $1/(2B_x) = 25 \hspace{0.08cm} &micro; \rm s$&nbsp; respectively, related to the centers at&nbsp; $t_1 = 200 \hspace{0.08cm} {\rm ms}$&nbsp; and&nbsp; $t_2 = 250 \hspace{0.08cm} {\rm ms}$.
+*The sinc&ndash;function exhibits equidistant zero crossings at all multiples of&nbsp; $1/(2B_x) = 25 \hspace{0.08cm} &micro; \rm s$,&nbsp; related to the centers at&nbsp; $t_1 = 200 \hspace{0.08cm} {\rm ms}$&nbsp; and&nbsp; $t_2 = 250 \hspace{0.08cm} {\rm ms}$.&nbsp; This results in the CCF values:
 :$$\varphi_{xy} (\tau = 0) \hspace{0.15cm}\underline{= 0},\hspace{0.5cm}\varphi_{xy} (\tau = t_1)= \alpha_1 \cdot \varphi_{x} (\tau = 0) \hspace{0.15cm}\underline{= 0.5\,{\rm V}^2} ,\hspace{0.5cm} \varphi_{xy} (\tau = t_2)= \alpha_2 \cdot \varphi_{x} (\tau = 0) \hspace{0.15cm}\underline{= 0.1\,{\rm V}^2} .$$
-'''(3)'''&nbsp; The cross power spectral density is the Fourier transform of the CCF, just as the power spectral density spectrum (PSD) gives the Fourier transform of the ACF. For this holds:
+'''(3)'''&nbsp; The cross power-spectral density is the Fourier transform of the CCF,&nbsp; just as the power-spectral density&nbsp;$\rm  (PSD)$&nbsp; gives the Fourier transform of the ACF.&nbsp; It holds:
 :$${\it \Phi}_{xy} (f) = \alpha_1 \cdot {\it \Phi}_{x} (f) \cdot {\rm e}^{-{\rm j}2 \pi f t_1} \hspace{0.15cm}+ \hspace{0.15cm}\alpha_2 \cdot {\it \Phi}_{x} (f) \cdot {\rm e}^{-{\rm j}2 \pi f t_2}. $$
-*Outside the range&nbsp; $|f| \le B_x$&nbsp; the PSD&nbsp; ${\it \Phi}_{x}(f)$&nbsp; &ndash; and correspondingly the KPSD&nbsp; ${\it \Phi}_{xy}(f)$ &ndash; is identically zero.
+*Outside of the range&nbsp; $|f| \le B_x$&nbsp; the power-spectral density&nbsp; ${\it \Phi}_{x}(f)$&nbsp; &ndash; and correspondingly the cross power-spectral density&nbsp; ${\it \Phi}_{xy}(f)$ &ndash; is identically zero.
-*In contrast, inside this interval holds&nbsp; ${\it \Phi}_{x}(f) = N_0/2$. It follows in this range:
+*In contrast,&nbsp; inside this interval holds&nbsp; ${\it \Phi}_{x}(f) = N_0/2$.&nbsp; It follows in this range:
 [[File:P_ID450__Sto_Z_4_14_d.png|right|frame|ACF and CCF in white noise]]
 :$${\it \Phi}_{xy} (f) = {N_0}/{2} \left( \alpha_1 \cdot {\rm e}^{-{\rm j}2 \pi f t_1} \hspace{0.15cm}+ \hspace{0.15cm}\alpha_2 \cdot {\rm e}^{-{\rm j}2 \pi f t_2} \right). $$
@@ Line 102: / Line 103: @@
-'''(4)'''&nbsp; Correct are <u>the proposed solutions 1 and 3</u>:
+'''(4)'''&nbsp; Correct are&nbsp; <u>the proposed solutions 1 and 3</u>:
-*The Fourier transform of a dirac-shaped ACF leads to a constant PSD for all frequencies&nbsp; $f$&nbsp; that is, in fact, to "true white noise".
+*The Fourier transform of a Dirac-shaped ACF leads to a constant PSD for all frequencies&nbsp; $f$ &nbsp; &rArr; &nbsp;  "true white noise".
-*This has an infinitely large power, and for the CCF can then be written according to the above graph:
+*This has an infinitely large power,&nbsp; and for the CCF can then be written according to the above graph:
 :$$\varphi_{xy} (\tau) = \alpha_1 \cdot { N_0}/{2} \cdot {\rm \delta}( \tau - t_1) \hspace {0.1cm}+ \hspace {0.1cm}   \alpha_2 \cdot { N_0}/{2} \cdot { \rm \delta}( \tau - t_2) .$$
-*In the frequency domain, indeed, no difference is detectable for&nbsp; $|f| \le B_x$&nbsp; compared to the subtask&nbsp; '''(3)'''&nbsp; .
+*In the frequency domain,&nbsp; no difference is detectable for&nbsp; $|f| \le B_x$&nbsp; compared to subtask&nbsp; '''(3)'''&nbsp; .
-*But since true white noise is now present, but here the cross power spectral density is not limited to this range.
+*But since&nbsp; "true white noise"&nbsp; is now present,&nbsp; here the cross power-spectral density is not limited to this range.
 '''(5)'''&nbsp; For the ACF of the echoed signal, &nbsp; $\varphi_{y} (\tau) = \overline {y(t) \hspace{0.05cm}\cdot \hspace{0.05cm}y(t+\tau)}$.
-*This ACF $\varphi_{y} (\tau)$ can consequently be represented as the following sum:
+*This ACF&nbsp; $\varphi_{y} (\tau)$&nbsp; can consequently be represented as the following sum:
 :$$\alpha_1^2 \cdot \overline {x(t - t_1) \cdot x(t - t_1+ \tau)} \hspace{0.03cm} + \hspace{0.03cm} \alpha_1\hspace{0.02cm}\alpha_2 \cdot \overline {x(t - t_1) \cdot x(t - t_2+ \tau)} + \hspace{0.05cm} \alpha_2\hspace{0.02cm}\alpha_1 \cdot \overline {x(t - t_2) \cdot x(t - t_1+ \tau)}\hspace{0.03cm} + \hspace{0.03cm} \alpha_2^2 \cdot \overline {x(t - t_2) \cdot x(t - t_2+ \tau)}. $$
@@ Line 118: / Line 119: @@
 :$$\overline {x(t - t_1) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t - t_1+ \tau)} = \overline {x(t - t_2) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t - t_2+ \tau)} = \overline {x(t ) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t + \tau)} =\varphi_x(\tau).$$
-*In contrast, for the second and third mean values, we obtain with&nbsp; $\delta t = t_2 - t_1= 50 \ \rm ms$:
+*In contrast,&nbsp; for the second and third mean values,&nbsp; we obtain with&nbsp; $\Delta t = t_2 - t_1= 50 \ \rm ms$:
 :$$\overline {x(t - t_1) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t - t_2+ \tau)} = \overline {x(t ) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t + t_1- t_2+ \tau)} =\varphi_x(\tau - \Delta t),$$
 :$$\overline {x(t - t_2) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t - t_1+ \tau)} = \overline {x(t ) \hspace{0.05cm}\cdot \hspace{0.05cm}x(t + t_2- t_1+ \tau)} =\varphi_x(\tau + \Delta t).$$
-*Total, this again results in a symmetric ACF, as shown in the graph below:
+*Total,&nbsp; this again results in a symmetric ACF,&nbsp; as shown in the graph below:
 :$$\varphi_{y} (\tau) = {N_0}/{2} \cdot \left[ ( \alpha_1^2 \hspace{0.1cm} + \hspace{0.1cm} \alpha_2^2 ) \cdot {\rm \delta} (\tau) \hspace{0.1cm} + \hspace{0.1cm} \alpha_1 \cdot \alpha_2 \cdot {\rm \delta}(\tau - \delta t) \hspace{0.1cm} + \hspace{0.1cm} \alpha_1 \cdot \alpha_2 \cdot {\rm \delta}(\tau + \Delta t) \right].$$
 :$$\Rightarrow \hspace{0.3cm}\varphi_{y} (\tau = 0 ) \hspace{0.15cm}\underline{= 0.13 \cdot 10^{-6}\, {\rm W/Hz}},

@@ Line 19: / Line 19: @@
 Hints:
-*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Cross-Correlation_Function_and_Cross_Power_Density|Cross-Correlation Function and Cross Power Density]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Cross-Correlation_Function_and_Cross_Power_Density|Cross-Correlation Function and Cross Power-Spectral Density]].
 *Use for numerical calculations the parameter values.
@@ Line 31: / Line 31: @@
 <quiz display=simple>
-{Apply the ACF&nbsp; $\varphi_x(\tau)$&nbsp; at the transmitter.&nbsp; What is this converted to the resistor&nbsp; $R = 50 \hspace{0.08cm} \rm \Omega$&nbsp;?&nbsp; What is the rms value&nbsp; $\sigma_x$&nbsp;?
+{Apply the&nbsp; $\rm ACF$&nbsp; $\varphi_x(\tau)$&nbsp; at the transmitter.&nbsp; What means this converted to the resistor&nbsp; $R = 50 \hspace{0.08cm} \rm \Omega$&nbsp;?&nbsp; What is the rms value&nbsp; $\sigma_x$&nbsp;?
 |type="{}"}
 $\sigma_x \ = \ $ { 1 3% } $\ \rm V$
-{Calculate the cross-correlation function (CCF)&nbsp; $\varphi_{xy}(\tau)$&nbsp; between transmit&ndash; and receive signal. <br>What values result for&nbsp; $\tau = 0$,&nbsp; $\tau = t_1 = 200 \hspace{0.08cm} \rm ms$&nbsp; und&nbsp; $\tau = t_2 = 250 \hspace{0.08cm} \rm ms$&nbsp;?
+{Calculate the cross-correlation function&nbsp; $\rm (CCF)$ &nbsp; $\varphi_{xy}(\tau)$&nbsp; between transmitted and received signal. <br>What values result for&nbsp; $\tau = 0$, &nbsp; $\tau = t_1 = 200 \hspace{0.08cm} \rm ms$ &nbsp; and &nbsp; $\tau = t_2 = 250 \hspace{0.08cm} \rm ms$&nbsp;?
 |type="{}"}
 $\varphi_{xy}(\tau= 0) \ = \ $ { 0. } $\ \rm V^2$
@@ Line 43: / Line 43: @@
-{Calculate the cross power spectral density &nbsp; ${\it \Phi}_{xy}(f)$.&nbsp; What value is obtained at frequency $f = 0$?
+{Calculate the cross power-spectral density &nbsp; ${\it \Phi}_{xy}(f)$.&nbsp; What value is obtained at frequency $f = 0$?
 |type="{}"}
 ${\it \Phi}_{xy}(f =0)\ = \ $ { 15 3% } $\ \cdot 10^{-6}\ \rm V^2\hspace{-0.1cm}/Hz$
-{Which of the following statements are true if you use the approximation&nbsp; $\varphi_{xy}(\tau) \approx N_0/2 \cdot \delta(\tau)$&nbsp; instead of the ACF calculated in&nbsp; '''(1)''''?
+{Which of the following statements are true if you use the approximation&nbsp; $\varphi_{xy}(\tau) \approx N_0/2 \cdot \delta(\tau)$&nbsp; instead of the ACF calculated in&nbsp; '''(1)'''?
 |type="[]"}
-+ The noise is now "true" white &ndash; so it is not bandlimited.
++ The noise is now&nbsp; "true white"&nbsp; &ndash; so it is not bandlimited.
-- The noise power is reduced compared to the sub-task&nbsp; '''(1)'''&nbsp;.
+- The noise power is reduced compared to subtask&nbsp; '''(1)'''.
-+ The cross-correlation function is the sum of weighted and shifted diracs.
++ The cross-correlation function is the sum of weighted and shifted Dirac delta functions.
-- The cross power spectral density is calculated as in subtask&nbsp; '''(3)'''&nbsp;.
+- The cross power-spectral density is calculated as in subtask&nbsp; '''(3)'''.
-{Calculate using the approximation&nbsp; $\varphi_{xy}(\tau) \approx N_0/2 \cdot \delta(\tau)$&nbsp; the ACF $\varphi_y(\tau)$.  &nbsp; What weights result for&uuml;r&nbsp; $\tau = 0$ &nbsp;and&nbsp; $\tau = \delta t = t_2 - t_1$&nbsp;?
+{Calculate the&nbsp; $\rm ACF$&nbsp; $\varphi_y(\tau)$&nbsp; using the approximation&nbsp; $\varphi_{xy}(\tau) \approx N_0/2 \cdot \delta(\tau)$.   &nbsp; What weights result for &nbsp; $\tau = 0$ &nbsp; and &nbsp; $\tau = \Delta t = t_2 - t_1$?
 |type="{}"}
 $\varphi_{y}(\tau= 0) \ = \ $ { 0.13 3% } $\ \cdot 10^{-6}\ \rm W\hspace{-0.1cm}/Hz$

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Sto_Z_4_14.png|right|Echo&ndash;Measurement|frame]]
+[[File:EN_Sto_Z_4_14.png|right|Echo measuring arrangement|frame]]
-To measure acoustic echoes in rooms &ndash; for example caused by reflections at a wall &ndash; the adjacent setup can be used.
+To measure acoustic echoes in rooms&nbsp; &ndash; for example caused by reflections at a wall &ndash;&nbsp; the adjacent setup can be used:
-*The noise generator produces a "white noise in the relevant frequency range"&nbsp; $x(t)$&nbsp; with power density&nbsp; $N_0 = 10^{-6} \hspace{0.08cm} \rm W/Hz$.
+*The noise generator produces a&nbsp; "white noise in the relevant frequency range"&nbsp; $x(t)$&nbsp; with power density&nbsp; $N_0 = 10^{-6} \hspace{0.08cm} \rm W/Hz$.
 *This is bandlimited to&nbsp; $B_x = 20 \hspace{0.08cm} \rm kHz$&nbsp; and is given to a loudspeaker.
-*The entire measurement setup is designed for the resistance value&nbsp; $R = 50 \hspace{0.08cm} \rm \Omega$&nbsp;.
+*The entire measurement setup is designed for the resistance value&nbsp; $R = 50 \hspace{0.08cm} \rm \Omega$.
-In the most general case, the signal recorded by the microphone can be described as follows:
+In the most general case,&nbsp; the signal recorded by the microphone can be described as follows:
 :$$y(t) = \sum_{\mu = 1}^M \alpha_\mu \cdot x ( t - t_\mu ) .$$
-Here denote&nbsp; $\alpha_\mu$&nbsp; damping factors and&nbsp; $t_\mu$&nbsp; travel times.
+Here the&nbsp; $\alpha_\mu$&nbsp;  denote damping factors and&nbsp; $t_\mu$&nbsp; delay times.

← Older revision		Revision as of 09:55, 28 February 2022
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−	[[Category:Theory of Stochastic Signals: Exercises\|^4.6 ~~KKF und Kreuzleistungsdichte~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^4.6 CCF and Cross Power-Spectral Density^]]

← Older revision	Revision as of 09:51, 28 February 2022
(No difference)

@@ Line 5: / Line 5: @@
 [[File:EN_Sto_Z_4_14.png|right|Echo&ndash;Messung|frame]]
 Zur Messung akustischer Echos in R&auml;umen &ndash; zum Beispiel bedingt durch Reflexionen an einer Wand &ndash; kann die nebenstehende Anordnung verwendet werden.
-*Der Rauschgenerator erzeugt ein "im relevanten Frequenzbereich Wei&szlig;es Rauschen&rdquo;&nbsp; $x(t)$&nbsp; mit Leistungsdichte&nbsp; $N_0 = 10^{-6} \hspace{0.08cm} \rm W/Hz$.
+*Der Rauschgenerator erzeugt ein "im relevanten Frequenzbereich Wei&szlig;es Rauschen"&nbsp; $x(t)$&nbsp; mit Leistungsdichte&nbsp; $N_0 = 10^{-6} \hspace{0.08cm} \rm W/Hz$.
 *Dieses ist bandbegrenzt auf&nbsp; $B_x = 20 \hspace{0.08cm} \rm kHz$&nbsp; und wird auf einen Lautsprecher gegeben.
 *Die gesamte Messeinrichtung ist für den Widerstandswert&nbsp; $R = 50 \hspace{0.08cm} \rm \Omega$&nbsp; ausgelegt.
@@ Line 54: / Line 54: @@
 {Welche der folgenden Aussagen sind zutreffend, wenn Sie anstelle der in&nbsp; '''(1)'''&nbsp; berechneten AKF die N&auml;herung&nbsp; $\varphi_{xy}(\tau) \approx N_0/2 \cdot \delta(\tau)$&nbsp; verwenden?
 |type="[]"}
-+ Das Rauschen ist nun "echt&rdquo; wei&szlig; &ndash; also nicht bandbegrenzt.
++ Das Rauschen ist nun "echt" wei&szlig; &ndash; also nicht bandbegrenzt.
 - Die Rauschleistung wird gegen&uuml;ber der Teilaufgabe&nbsp; '''(1)'''&nbsp; vermindert.
 + Die Kreuzkorrelationsfunktion ist die Summe gewichteter und verschobener Diracs.
@@ Line 110: / Line 110: @@
 '''(4)'''&nbsp; Richtig sind <u>die Lösungsvorschläge 1 und 3</u>:
-*Die Fouriertransformierte einer diracf&ouml;rmigen AKF f&uuml;hrt zu einem f&uuml;r alle Frequenzen&nbsp; $f$&nbsp; konstanten LDS, das hei&szlig;t tats&auml;chlich zu "echt Wei&szlig;em Rauschen&rdquo;.
+*Die Fouriertransformierte einer diracf&ouml;rmigen AKF f&uuml;hrt zu einem f&uuml;r alle Frequenzen&nbsp; $f$&nbsp; konstanten LDS, das hei&szlig;t tats&auml;chlich zu "echt Wei&szlig;em Rauschen".
 *Dieses besitzt eine unendlich gro&szlig;e Leistung, und f&uuml;r die KKF kann dann gemäß der oberen Grafik geschrieben werden:
 :$$\varphi_{xy} (\tau) =  \alpha_1 \cdot { N_0}/{2} \cdot {\rm \delta}( \tau - t_1) \hspace {0.1cm}+ \hspace {0.1cm}   \alpha_2 \cdot { N_0}/{2} \cdot {\rm \delta}( \tau - t_2) .$$