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

Guenter at 15:54, 22 February 2022

2022-02-22T15:54:02Z

Guenter at 15:19, 10 February 2022

2022-02-10T15:19:30Z

Guenter at 15:12, 10 February 2022

2022-02-10T15:12:19Z

Guenter at 15:08, 10 February 2022

2022-02-10T15:08:11Z

Guenter at 16:46, 8 February 2022

2022-02-08T16:46:49Z

Guenter at 16:07, 8 February 2022

2022-02-08T16:07:03Z

Hwang at 16:11, 13 January 2022

2022-01-13T16:11:52Z

Guenter at 10:29, 12 January 2022

2022-01-12T10:29:13Z

Guenter: Guenter moved page Aufgaben:Aufgabe 5.2: Bestimmung des Frequenzgangs to Aufgaben:Exercise 5.2: Determination of the Frequency Response

2022-01-12T08:55:32Z

Guenter moved page Aufgabe 5.2: Bestimmung des Frequenzgangs to Exercise 5.2: Determination of the Frequency Response

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

2021-03-23T12:43:31Z

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

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Sto_A_5_2_XXX.png|right|frame|Measuring the frequency response&nbsp; $H(f)$]]
+[[File:EN_Sto_A_5_2_neu2.png|right|frame|Measuring the frequency response&nbsp; $H(f)$]]
 We consider the illustrated measurement setup for the determination of the frequency response&nbsp; $H(f)$&nbsp; highlighted in blue.
 *The input signal&nbsp; $x(t)$&nbsp; is white Gaussian noise with the noise power density&nbsp; $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.

@@ Line 15: / Line 15: @@
 Notes:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Stochastic_System_Theory|Stochastic System Theory]].
 *Reference is also made to the chapter&nbsp;  [[Theory_of_Stochastic_Signals/Power-Spectral_Density|Power-Spectral Density]].
@@ Line 62: / Line 58: @@
 :$$\varphi _{xy} ( \tau ) = h( \tau ) * \varphi _x ( \tau )\quad  \Rightarrow \quad H( f ) = \frac{{{\it \Phi} _{xy} ( f )}}{{{\it \Phi} _x ( f )}},$$
 :$$\varphi _y ( \tau) = \varphi _{xy} ( \tau) * h(- \tau)\quad  \Rightarrow \quad H^{\star}( f ) = \frac{{{\it \Phi} _y ( f )}}{{{\it \Phi} _{xy} ( f )}}.$$
-*In contrast, the first statement is false: &nbsp; the phase relations are lost in the ACF calculation.
+*In contrast,&nbsp; the first statement is false: &nbsp; The phase relations are lost in the ACF calculation.
-*The associated spectral functions to&nbsp; $\varphi_x(\tau)$&nbsp; and&nbsp; $\varphi_x(\tau)$&nbsp; &ndash; namely&nbsp; ${\it \Phi}_x(f)$&nbsp; and&nbsp;  ${\it \Phi}_y(f)$&nbsp; &ndash; are purely real, so only the magnitude&nbsp; $|H(f)|$&nbsp; can be given.
+*The associated spectral functions to&nbsp; $\varphi_x(\tau)$&nbsp; and&nbsp; $\varphi_x(\tau)$&nbsp; &ndash; namely&nbsp; ${\it \Phi}_x(f)$&nbsp; and&nbsp;  ${\it \Phi}_y(f)$&nbsp; &ndash; are purely real,&nbsp; so only the magnitude&nbsp; $|H(f)|$&nbsp; can be given.
-'''(2)'''&nbsp; For dirac-shaped input ACF&nbsp; $\varphi_x(\tau)$,&nbsp; the impulse response&nbsp; $h(t)$&nbsp; is equal in shape to the CCF:
+'''(2)'''&nbsp; For Dirac-shaped input ACF&nbsp; $\varphi_x(\tau)$,&nbsp; the impulse response&nbsp; $h(t)$&nbsp; is equal in shape to the CCF:
 :$$h(t) = \frac{{K \cdot {\rm{e}}^{ - t/T_0 } }}{N_0 /2} = 1.256 \cdot 10^{ - 2} \frac{1}{{\rm{s}}} \cdot {\rm{e}}^{ - t/T_0 } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}h(t = T_0)\hspace{0.15cm}\underline{ = 4.62 \cdot 10^{-3}\ \rm 1/s}.$$
@@ Line 76: / Line 72: @@
 *The constant results in&nbsp; $C = 0.08$.&nbsp; With&nbsp; $H(f) = 2 \pi \cdot  H(\omega)$&nbsp; 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)'''&nbsp; For the output PDS, in general, or specifically here:
+'''(4)'''&nbsp; For the output PSD,&nbsp; 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&nbsp; $f = 1/(2\pi T_0)$,&nbsp; &nbsp; ${\it \Phi}_y (f)$&nbsp;has dropped by half compared to its maximum at&nbsp; $f=0$&nbsp;:
+*At the given frequency&nbsp; $f = 1/(2\pi T_0)$,&nbsp; &nbsp; ${\it \Phi}_y (f)$&nbsp; has dropped by half compared to its maximum at $f=0$&nbsp;:
 :$${\it \Phi}_y (f = 1/(2 \pi T_0)) ={N_0 }/{16}\hspace{0.15cm} \underline{ = 6.25 \cdot 10^{ - 12} \;{\rm{W/Hz}}}.$$

@@ Line 21: / Line 21: @@
 ''Notes:''
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Stochastic_System_Theory|Stochastic System Theory]].
-*Reference is also made to the chapter&nbsp;  [[Theory_of_Stochastic_Signals/Leistungsdichtespektrum_(LDS)|Power Density Spectrum]].
+*Reference is also made to the chapter&nbsp;  [[Theory_of_Stochastic_Signals/Power-Spectral_Density|Power-Spectral Density]].
 *Please also note the following Fourier transform&nbsp; $($in&nbsp; $\omega)$:
 :$$H( \omega  ) = \frac{1}{{1 + {\rm{j}}\cdot \omega /\omega _0 }}\hspace{0.15cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\,\hspace{0.15cm}h(t) = \omega _0  \cdot {\rm{e}}^{ - \omega _0 t} \hspace{0.3cm}(t \ge 0).$$
-:For negative&nbsp; $t$ values, on the other hand,&nbsp;  $h(t) =0$ at all times.
+:For negative&nbsp; $t$&ndash;values,&nbsp; on the other hand,&nbsp;  $h(t) =0$&nbsp; at all times.
@@ Line 49: / Line 48: @@
-{Calculate the power density spectrum of the output signal&nbsp; $y(t)$.&nbsp; What value results for frequency&nbsp; $f = 1/(2\pi T_0)$?
+{Calculate the power-spectral density of the output signal&nbsp; $y(t)$.&nbsp; What value results for frequency&nbsp; $f = 1/(2\pi T_0)$?
 |type="{}"}
 ${\it \Phi}_y(f = 1/(2\pi T_0)) \ =  \ $ { 6.25 3% } $\ \cdot 10^{-12}\  \rm W/Hz$

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Sto_A_5_2_XXX.png|right|frame|Determining the frequency response&nbsp; $H(f)$]]
+[[File:EN_Sto_A_5_2_XXX.png|right|frame|Measuring the frequency response&nbsp; $H(f)$]]
-We consider the illustrated measurement setup for the determination of the frequency response&nbsp; $H(f)$ highlighted in blue.
+We consider the illustrated measurement setup for the determination of the frequency response&nbsp; $H(f)$&nbsp; highlighted in blue.
 *The input signal&nbsp; $x(t)$&nbsp; is white Gaussian noise with the noise power density&nbsp; $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.
-*Thus, the auto-correlation function (ACF) is:
+*Thus,&nbsp; the auto-correlation function&nbsp; $\rm (ACF)$&nbsp; is:
 :$$\varphi _x ( \tau ) = {N_0 }/{2} \cdot \delta ( \tau  ).$$
-*The measured cross-correlation function (CCF) between the signals&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; can be approximated as follows <br>$($valid only for positive times&nbsp; $t)$:
+*The measured cross-correlation function&nbsp; $\rm (CCF)$&nbsp; between the signals&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; can be approximated as follows <br>$($valid only for positive times&nbsp; $t)$:
-:$$\varphi _{xy} \left( \tau  \right) = K \cdot {\rm{e}}^{ - \tau /T_0 },\hspace{0.5cm}\text{mit } \ K = 0.628 \cdot 10^{-12} \hspace{0.05cm} {\rm W}, \  T_0 = 1 \hspace{0.05cm} \rm ms.$$
+:$$\varphi _{xy} \left( \tau  \right) = K \cdot {\rm{e}}^{ - \tau /T_0 },\hspace{0.5cm}\text{with } \ K = 0.628 \cdot 10^{-12} \hspace{0.05cm} {\rm W}, \  T_0 = 1 \hspace{0.05cm} \rm ms.$$
 *The ACF&nbsp; $\varphi_y(\tau)$&nbsp; of the output signal&nbsp; $y(t)$ is also measured.

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Sto_A_5_2_neu.png|right|frame|Determining the frequency response&nbsp; $H(f)$]]
+[[File:EN_Sto_A_5_2_XXX.png|right|frame|Determining the frequency response&nbsp; $H(f)$]]
 We consider the illustrated measurement setup for the determination of the frequency response&nbsp; $H(f)$ highlighted in blue.
 *The input signal&nbsp; $x(t)$&nbsp; is white Gaussian noise with the noise power density&nbsp; $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Stochastische Systemtheorie
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Stochastic_System_Theory
 }}
-[[File:EN_Sto_A_5_2.png|right|frame|Zur Bestimmung des Frequenzgangs&nbsp; $H(f)$]]
+[[File:EN_Sto_A_5_2.png|right|frame|Determining the frequency response&nbsp; $H(f)$]]
-Wir betrachten die abgebildete Messanordnung zur Bestimmung des blau hervorgehobenen Frequenzgangs&nbsp; $H(f)$.
+We consider the illustrated measurement setup for the determination of the frequency response&nbsp; $H(f)$ highlighted in blue.
-*Das Eingangssignal&nbsp; $x(t)$&nbsp;  ist weißes Gaußsches Rauschen mit der Rauschleistungsdichte&nbsp; $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.
+*The input signal&nbsp; $x(t)$&nbsp; is white Gaussian noise with the noise power density&nbsp; $N_0 = 10^{-10} \hspace{0.05cm} \rm W/Hz$.
-*Somit gilt für die Autokorrelationsfunktion (AKF):
+*Thus, the auto-correlation function (ACF) is:
 :$$\varphi _x ( \tau ) = {N_0 }/{2} \cdot \delta ( \tau  ).$$
-*Die gemessene Kreuzkorrelationsfunktion (KKF) zwischen den Signalen&nbsp; $x(t)$&nbsp; und&nbsp; $y(t)$&nbsp; kann wie folgt angenähert werden <br>$($nur gültig für positive Zeiten&nbsp; $t)$:
+*The measured cross-correlation function (CCF) between the signals&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; can be approximated as follows <br>$($valid only for positive times&nbsp; $t)$:
 :$$\varphi _{xy} \left( \tau  \right) = K \cdot {\rm{e}}^{ - \tau /T_0 },\hspace{0.5cm}\text{mit } \ K = 0.628 \cdot 10^{-12} \hspace{0.05cm} {\rm W}, \  T_0 = 1 \hspace{0.05cm} \rm ms.$$
-*Gemessen wird außerdem die AKF&nbsp; $\varphi_y(\tau)$&nbsp; des Ausgangssignals&nbsp; $y(t)$.
+*The ACF&nbsp; $\varphi_y(\tau)$&nbsp; of the output signal&nbsp; $y(t)$ is also measured.
@@ Line 19: / Line 19: @@
-''Hinweise:''
+''Notes:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Stochastische_Systemtheorie|Stochastische Systemtheorie]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Stochastic_System_Theory|Stochastic System Theory]].
-*Bezug genommen wird auch auf das  Kapitel&nbsp;  [[Theory_of_Stochastic_Signals/Leistungsdichtespektrum_(LDS)|Leistungsdichtespektrum]].
+*Reference is also made to the chapter&nbsp;  [[Theory_of_Stochastic_Signals/Leistungsdichtespektrum_(LDS)|Power Density Spectrum]].
-*Beachten Sie bitte auch die folgende Fouriertransformation&nbsp; $($in&nbsp; $\omega)$:
+*Please also note the following Fourier transform&nbsp; $($in&nbsp; $\omega)$:
 :$$H( \omega  ) = \frac{1}{{1 + {\rm{j}}\cdot \omega /\omega _0 }}\hspace{0.15cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\,\hspace{0.15cm}h(t) = \omega _0  \cdot {\rm{e}}^{ - \omega _0 t} \hspace{0.3cm}(t \ge 0).$$
-:Für negative&nbsp; $t$&ndash;Werte ist dagegen stets&nbsp;  $h(t) =0$.
+:For negative&nbsp; $t$ values, on the other hand,&nbsp;  $h(t) =0$ at all times.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welche der folgenden Aussagen treffen zu?&nbsp; Man kann den Frequenzgang&nbsp; $H(f)$&nbsp; nach Betrag und Phase vollständig bestimmen, wenn
+{Which of the following statements is true?&nbsp; One can completely determine the frequency response&nbsp; $H(f)$&nbsp; by magnitude and phase if
 |type="[]"}
-- die Funktionen&nbsp; $\varphi_x(\tau)$&nbsp; und&nbsp; $\varphi_y(\tau)$&nbsp; bekannt sind,
+- the functions&nbsp; $\varphi_x(\tau)$&nbsp; and&nbsp; $\varphi_y(\tau)$&nbsp; are known,
-+ die Funktionen&nbsp; $\varphi_x(\tau)$&nbsp; und&nbsp; $\varphi_{xy}(\tau)$&nbsp; bekannt sind,
++ the functions&nbsp; $\varphi_x(\tau)$&nbsp; and&nbsp; $\varphi_{xy}(\tau)$&nbsp; are known,
-+ die Funktionen&nbsp; $\varphi_{xy}(\tau)$&nbsp; und&nbsp; $\varphi_y(\tau)$&nbsp; bekannt sind.
++ the functions&nbsp; $\varphi_{xy}(\tau)$&nbsp; and&nbsp; $\varphi_y(\tau)$&nbsp; are known.
-{Berechnen Sie die Impulsantwort&nbsp; $h(t)$.&nbsp; Welcher Wert ergibt sich für&nbsp; $t=T_0$?
+{Calculate the impulse response&nbsp; $h(t)$.&nbsp; What is the value for&nbsp; $t=T_0$?
 |type="{}"}
 $h(t = T_0) \ = \ $  { 4.62 3% } $\ \cdot 10^{-3} \ \rm 1/s$
-{Wie lautet der Frequenzgang&nbsp; $H(f)$?&nbsp; Welcher Wert ergibt sich für $f= 0$?
+{What is the frequency response&nbsp; $H(f)$?&nbsp; What value results for $f= 0$?
 |type="{}"}
 $H(f = 0) \ =  \ $ { 0.5 3% }
-{Berechnen Sie das Leistungsdichtespektrum des Ausgangssignals&nbsp; $y(t)$.&nbsp; Welcher Wert ergibt sich bei der Frequenz&nbsp; $f = 1/(2\pi T_0)$?
+{Calculate the power density spectrum of the output signal&nbsp; $y(t)$.&nbsp; What value results for frequency&nbsp; $f = 1/(2\pi T_0)$?
 |type="{}"}
 ${\it \Phi}_y(f = 1/(2\pi T_0)) \ =  \ $ { 6.25 3% } $\ \cdot 10^{-12}\  \rm W/Hz$
@@ Line 57: / Line 57: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die <u>Aussagen 2 und 3</u> sind zutreffend.
+'''(1)'''&nbsp; <u>Statements 2 and 3</u> are true.
-*Es gelten folgende Gleichungen:
+*The following equations are valid:
 :$$\varphi _{xy} ( \tau ) = h( \tau ) * \varphi _x ( \tau )\quad  \Rightarrow \quad H( f ) = \frac{{{\it \Phi} _{xy} ( f )}}{{{\it \Phi} _x ( f )}},$$
 :$$\varphi _y ( \tau) = \varphi _{xy} ( \tau) * h(- \tau)\quad  \Rightarrow \quad H^{\star}( f ) = \frac{{{\it \Phi} _y ( f )}}{{{\it \Phi} _{xy} ( f )}}.$$
-*Dagegen ist die erste Aussage falsch: &nbsp; Bei der AKF-Berechnung gehen die Phasenbeziehungen verloren.
+*In contrast, the first statement is false: &nbsp; the phase relations are lost in the ACF calculation.
-*Die zugehörigen Spektralfunktionen zu&nbsp; $\varphi_x(\tau)$&nbsp; und&nbsp; $\varphi_x(\tau)$&nbsp; &ndash; nämlich&nbsp; ${\it \Phi}_x(f)$&nbsp; und&nbsp;  ${\it \Phi}_y(f)$&nbsp; &ndash; sind rein reell, so dass nur der Betrag&nbsp; $|H(f)|$&nbsp; angegeben werden kann.
+*The associated spectral functions to&nbsp; $\varphi_x(\tau)$&nbsp; and&nbsp; $\varphi_x(\tau)$&nbsp; &ndash; namely&nbsp; ${\it \Phi}_x(f)$&nbsp; and&nbsp;  ${\it \Phi}_y(f)$&nbsp; &ndash; are purely real, so only the magnitude&nbsp; $|H(f)|$&nbsp; can be given.
-'''(2)'''&nbsp; Bei diracförmiger Eingangs-AKF&nbsp; $\varphi_x(\tau)$&nbsp; ist die Impulsantwort&nbsp; $h(t)$&nbsp; formgleich mit der KKF:
+'''(2)'''&nbsp; For dirac-shaped input ACF&nbsp; $\varphi_x(\tau)$,&nbsp; the impulse response&nbsp; $h(t)$&nbsp; is equal in shape to the CCF:
 :$$h(t) = \frac{{K \cdot {\rm{e}}^{ - t/T_0 } }}{N_0 /2} = 1.256 \cdot 10^{ - 2} \frac{1}{{\rm{s}}} \cdot {\rm{e}}^{ - t/T_0 } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}h(t = T_0)\hspace{0.15cm}\underline{ = 4.62 \cdot 10^{-3}\ \rm 1/s}.$$
-'''(3)'''&nbsp; Die angegebene Fourierkorrespondenz lautet mit&nbsp; $T_0 = 1/\omega_0$&nbsp; und der Konstanten&nbsp; $C= N_0/2 \cdot T_0/K$:
+'''(3)'''&nbsp; The Fourier correspondence given is with&nbsp; $T_0 = 1/\omega_0$&nbsp; and the constant&nbsp; $C= N_0/2 \cdot T_0/K$:
 :$$h(t) = \frac{C}{T_0 } \cdot {\rm{e}}^{ - t/T_0 }\hspace{0.15cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\hspace{0.15cm} H( \omega  ) = \frac{C}{{1 + {\rm{j}}\cdot \omega T_0 }}.$$
-*Die Konstante ergibt sich zu&nbsp; $C = 0.08$.&nbsp; Mit&nbsp; $H(f) = 2 \pi \cdot  H(\omega)$&nbsp; folgt daraus:
+*The constant results in&nbsp; $C = 0.08$.&nbsp; With&nbsp; $H(f) = 2 \pi \cdot  H(\omega)$&nbsp; it follows:
 :$$H(f) = \frac{0.5}{1 + {\rm{j\cdot 2\pi }}\cdot fT_0 } \hspace{0.3cm}\Rightarrow \hspace{0.3cm} H(f= 0) \hspace{0.15cm}\underline{=0.5}.$$
-'''(4)'''&nbsp; Für das Ausgangs-LDS gilt im Allgemeinen bzw. speziell hier:
+'''(4)'''&nbsp; For the output PDS, in general, or specifically here:
 :$${\it \Phi}_y (f) = {\it \Phi} _x (f) \cdot \left| {H(f)} \right|^2  = \frac{N_0 }{2} \cdot \frac{0.5^2 }{{\left( {1 + {\rm{j\cdot  2\pi }}\cdot fT_0 } \right)\left( {1 - {\rm{j\cdot 2\pi }}\cdot fT_0 } \right)}} = {N_0 }/{8} \cdot \frac{1}{1 + \left( {{\rm{2\pi }}\cdot fT_0 } \right)^2 }.$$
-*Bei der angegebenen Frequenz&nbsp; $f = 1/(2\pi T_0)$&nbsp; ist&nbsp; ${\it \Phi}_y (f)$&nbsp;  gegenüber seinem Maximum bei&nbsp; $f=0$&nbsp; um die Hälfte abgefallen:
+*At the given frequency&nbsp; $f = 1/(2\pi T_0)$,&nbsp; &nbsp; ${\it \Phi}_y (f)$&nbsp;has dropped by half compared to its maximum at&nbsp; $f=0$&nbsp;:
 :$${\it \Phi}_y (f = 1/(2 \pi T_0)) ={N_0 }/{16}\hspace{0.15cm} \underline{ = 6.25 \cdot 10^{ - 12} \;{\rm{W/Hz}}}.$$

← Older revision		Revision as of 10:29, 12 January 2022
Line 91:		Line 91:


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

← Older revision	Revision as of 08:55, 12 January 2022
(No difference)

← Older revision		Revision as of 12:43, 23 March 2021
Line 91:		Line 91:


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