Aufgaben:Exercise 4.13: Gaussian ACF and PSD - Revision history

Guenter at 15:19, 25 March 2022

2022-03-25T15:19:35Z

Guenter at 14:43, 25 March 2022

2022-03-25T14:43:28Z

Noah at 19:35, 7 March 2022

2022-03-07T19:35:18Z

Noah at 19:18, 7 March 2022

2022-03-07T19:18:56Z

Show changes

Guenter at 09:45, 28 February 2022

2022-02-28T09:45:18Z

Guenter at 09:42, 28 February 2022

2022-02-28T09:42:55Z

Guenter: Guenter moved page Aufgaben:Exercise 4.13: Gaussian ACF to Aufgaben:Exercise 4.13: Gaussian ACF and PSD

2022-02-28T09:39:55Z

Guenter moved page Exercise 4.13: Gaussian ACF to Exercise 4.13: Gaussian ACF and PSD

Guenter: Guenter moved page Aufgaben:Aufgabe 4.13: Gaußförmige AKF to Aufgaben:Exercise 4.13: Gaussian ACF

2022-02-28T09:34:30Z

Guenter moved page Aufgabe 4.13: Gaußförmige AKF to Exercise 4.13: Gaussian ACF

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

2021-03-23T12:44:04Z

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

Javier: Text replacement - "[[Stochastische_Signaltheorie/" to "[[Theory_of_Stochastic_Signals/"

2020-07-09T09:47:37Z

Text replacement - "[[Stochastische_Signaltheorie/" to "[[Theory_of_Stochastic_Signals/"

@@ Line 64: / Line 64: @@
 '''(2)'''&nbsp; One can adapt the given Fourier correspondence to the task as follows:
-:$$K\cdot{\rm e}^{-\pi({\tau}/{\nabla\tau_x})^2}\ \circ\!\!-\!\!-\!\!-\!\!-\!-\!\bullet\,\ \frac{\it K}{\nabla \it f_x}\cdot{\rm e}^{-\pi({f}/{\nabla f_x})^2}.$$
+:$$K\cdot{\rm e}^{-\pi({\tau}/{\nabla\tau_x})^2}\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ \frac{\it K}{\nabla \it f_x}\cdot{\rm e}^{-\pi({f}/{\nabla f_x})^2}.$$
-*With&nbsp; $K = 0.25 \hspace{0.05cm}\rm V^2$&nbsp; and&nbsp; $ {\rm \nabla} \hspace{-0.05cm} f_x = 200\hspace{0.05cm} \rm kHz$&nbsp; obtains:
+*With&nbsp; $K = 0.25 \hspace{0.05cm}\rm V^2$&nbsp; and&nbsp; $ {\rm \nabla} \hspace{-0.05cm} f_x = 200\hspace{0.05cm} \rm kHz$&nbsp; one obtains:
-:$${\it \Phi_x}(f)=1.25\cdot\rm 10^{-\rm 6}\hspace{0.1cm}\frac{V^2}{Hz}\cdot\rm e^{-\pi({\it f}/{\nabla\it f_x})^2}\hspace{0.3cm}
+:$${\it \Phi_x}(f)=1.25\cdot\rm 10^{-\rm 6}\hspace{0.1cm}\frac{V^2}{Hz}\cdot\rm e^{-\pi({\it f}/{\nabla\it f_x})^2}$$
-\Rightarrow \hspace{0.3cm}{\it \Phi_x}(f = 0)=\hspace{0.15cm}\underline{\rm 1.25 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz},
+:$$\Rightarrow \hspace{0.3cm}{\it \Phi_x}(f = 0)=\hspace{0.15cm}\underline{\rm 1.25 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz},
   \hspace{0.5cm}{\it \Phi_x}(f = 200 \hspace{0.05cm} \rm kHz)=\hspace{0.15cm}\underline{\rm 0.054 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}.$$
-'''(3)'''&nbsp; Correct <u>solutions 1, 2, and 4</u>:
+'''(3)'''&nbsp; Correct are the&nbsp; <u>solutions 1, 2, and 4</u>:
-*A mean-free process always results in a continuous PSD.&nbsp; This is narrower the wider the ACF is (reciprocitylaw).
+*A mean-free process always results in a continuous PSD.&nbsp; This is narrower the wider the ACF is&nbsp; ("Reciprocity Law").
 *The process power is equal to the integral of the PSD.
-*Therefore, at constant power, a wider ACF&nbsp; (narrower PSD)&nbsp; must be compensated by higher PSD values.
+*Therefore,&nbsp; at constant power,&nbsp; a wider ACF&nbsp; (narrower PSD)&nbsp; must be compensated by higher PSD values.
-*A DC component or periodic components always result in dirac functions in the PSD;&nbsp; otherwise, the PSD is always continuous in value.
+*A DC component or periodic components always result in Dirac delta functions in the PSD;&nbsp; otherwise, the PSD is always continuous in value.
@@ Line 83: / Line 83: @@
 :$${\it \Phi_y}(f)=\frac{\rm 0.09 V^2}{\nabla\it f_y}\cdot\rm e^{-\pi({\it f}/{\nabla\it f_y})^2}+\it m_y^{\rm 2}\cdot\delta(f).$$
-*Because of the DC component, there is a Dirac at frequency $f = 0$ in addition to the continuous PSD component.
+*Because of the DC component,&nbsp; there is a Dirac delta function at frequency&nbsp; $f = 0$&nbsp; in addition to the continuous PSD component.
 *The continuous PSD&ndash;part at $f= 0$ is&nbsp; ${\it \Phi_y}(f = 0)=\hspace{0.15cm}\underline{\rm 0.9 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}.$
-*The fraction at $f = 2 \cdot {\rm \nabla} \hspace{-0.05cm} f_y = 200 \hspace{0.05cm}\rm kHz$&nbsp; is increased by a factor&nbsp; ${\rm e}^{-4} \approx 0.0183$&nbsp; lower &nbsp; &rArr; &nbsp; ${\it \Phi_y}(f )=\hspace{0.15cm}\underline{\rm 0.0165 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}.$
+*The continuous PSD&ndash;part at&nbsp; $f = 2 \cdot {\rm \nabla} \hspace{-0.05cm} f_y = 200 \hspace{0.05cm}\rm kHz$&nbsp; is lower by a factor&nbsp; ${\rm e}^{-4} \approx 0.0183$ &nbsp; &rArr; &nbsp; ${\it \Phi_y}(f )=\hspace{0.15cm}\underline{\rm 0.0165 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}.$
-'''(5)'''&nbsp; Correct is <u>only the second proposed solution</u>:
+'''(5)'''&nbsp; Correct is&nbsp; <u>only the second proposed solution</u>:
-*The PSD of a mean-valued process generally involves a Dirac function at $f=0$&nbsp; with weight&nbsp; $m_y^2$.
+*The PSD of a mean-valued process generally involves a Dirac delta function at $f=0$&nbsp; with weight&nbsp; $m_y^2$.
-*In the present case, this value is equal to&nbsp; $0.16 \ \rm V^2$.
+*In the present case,&nbsp; this value is equal to&nbsp; $0.16 \ \rm V^2$.
-*Since&nbsp; $\delta(f)$&nbsp; has unit&nbsp; $\rm 1/Hz = s$&nbsp;, the units of the continuous and discrete PSD components differ.
+*Since&nbsp; $\delta(f)$&nbsp; has unit&nbsp; $\rm 1/Hz = s$,&nbsp; the units of the continuous and discrete PSD components differ.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID411__Sto_A_4_13.png|right|frame|Two Gaussian ACF]]
+[[File:P_ID411__Sto_A_4_13.png|right|frame|Two Gaussian ACFs]]
-Let the random process considered here&nbsp; $\{x_i(t)\}$&nbsp; be characterized by the autocorrelation function (ACF) outlined above&nbsp; This random process is mean-free and the equivalent ACF duration is&nbsp; ${ {\rm \nabla}  }\tau_x = 5 \hspace{0.08cm} \rm &micro; s$:
+Let the random process considered here&nbsp; $\{x_i(t)\}$&nbsp; be characterized by the auto-correlation function&nbsp; $\rm (ACF)$&nbsp; outlined above&nbsp; This random process is mean-free and the equivalent ACF duration is&nbsp; ${ {\rm \nabla}  }\tau_x = 5 \hspace{0.08cm} \rm &micro; s$:
 :$$\varphi_x(\tau)=\rm 0.25 V^2\cdot \rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\tau}{/ 5 \hspace{0.08cm}{\rm &micro;}s })^2} .$$
-The bottom figure shows the ACF of the process&nbsp; $\{y_i(t)\}$&nbsp; This reads with the equivalent ACF duration&nbsp; ${ {\rm \nabla}  }\tau_y = 10 \hspace{0.08cm} \rm &micro; s$:
+The bottom graph shows the ACF of the process&nbsp; $\{y_i(t)\}$.&nbsp; This reads with the equivalent ACF duration&nbsp; ${ {\rm \nabla}  }\tau_y = 10 \hspace{0.08cm} \rm &micro; s$:
 :$$ \varphi_y(\tau)=\rm 0.16 V^2 + \rm 0.09 V^2\cdot\rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\tau}/{\nabla \it \tau_y})^2} .$$
-In this exercise, the Power spectral densities of the two processes are sought.
+In this exercise,&nbsp; the power-spectral densities of the two processes are sought.
@@ Line 20: / Line 15: @@
 Hints:
-*This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Power-Spectral_Density|Power-Spectral Density]].
+*This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Power-Spectral_Density|Power-spectral density]]&nbsp; $\rm (PSD)$.
-*Reference is also made to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Auto-Correlation_Function|Auto-correlation function]].
+*Reference is also made to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Auto-Correlation_Function|Auto-correlation function]]&nbsp; $\rm (ACF)$.
 *To solve this exercise you can use the following Fourier correspondence:
-:$$\rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\it f}/{\rm \Delta\it f})^2}\ \bullet\!\!-\!\!-\!\!-\!-\!\!\circ\, \ {\rm \Delta \it f} \cdot \rm e^{-\pi \hspace{0.05cm}\cdot \hspace{0.05cm} ({\rm \Delta\it f} \hspace{0.05cm}\cdot \hspace{0.05cm}\it t {\rm )}^{\rm 2}}.$$
+:$$\rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\it f}/{\rm \Delta\it f})^2}\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ {\rm \Delta \it f} \cdot \rm e^{-\pi \hspace{0.05cm}\cdot \hspace{0.05cm} ({\rm \Delta\it f} \hspace{0.05cm}\cdot \hspace{0.05cm}\it t {\rm )}^{\rm 2}}.$$
@@ Line 35: / Line 29: @@
-{What is&nbsp; ${\it \Phi}_x(f)$?&nbsp; Give the PSD values for&nbsp; $f= 0$&nbsp; and&nbsp; $f = 200 \hspace{0.08cm} \rm kHz$&nbsp; on.
+{What is the power-spectral density&nbsp; ${\it \Phi}_x(f)$?&nbsp; Give the PSD values for&nbsp; $f= 0$&nbsp; and&nbsp; $f = 200 \hspace{0.08cm} \rm kHz$.
 |type="{}"}
 ${\it \Phi}_x(f = 0)\ = \ $ { 1.25 3% } $\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$
@@ Line 41: / Line 35: @@
-{Which statements are valid, if the random process has no periodic parts?&nbsp; Furthermore, a constant power is assumed.
+{Which statements are valid,&nbsp; if the random process has no periodic parts?&nbsp; Furthermore,&nbsp; a constant power is assumed.
 |type="[]"}
 + The process power is the integral over the PSD.
-+ If the process is zero mean, the PSD is always continuous.
++ If the process is zero mean,&nbsp; the PSD is always continuous.
-- The wider the ACF, the wider the PSD.
+- The wider the ACF,&nbsp; the wider the PSD.
 + A wider ACF results in higher PSD values.
-{Calculate the Power spectral density spectrum&nbsp; ${\it \Phi}_y(f)$.&nbsp; What are the values for the continuous PSD component at&nbsp; $f= 0$&nbsp; and&nbsp; $f = 200 \hspace{0.08cm}  \rm kHz$?
 |type="{}"}
 ${\it \Phi}_y(f = 0)\ = \ $ { 0.9 3% } $\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$
@@ Line 55: / Line 50: @@
 {Which of the following statements are true regarding the process&nbsp; $\{y_i(t)\}$?
 |type="[]"}
-- The PSD involves a Dirac at frequency $ f = {\rm \nabla} \hspace{-0.05cm} f_y$.
+- The PSD involves a Dirac delta function at frequency $ f = {\rm \nabla} \hspace{-0.05cm} f_y$.
-+ The PSD involves a Dirac at frequency $f= 0$.
++ The PSD involves a Dirac delta function at frequency $f= 0$.
-- Dirac weight and continuous PSD have the same unit.
+- The Dirac weight and the continuous PSD have the same unit.

@@ Line 30: / Line 30: @@
 <quiz display=simple>
-{What is the &auml;equivalent PSD bandwidth of the process&nbsp; $\{x_i(t)\}$?
+{What is the equivalent PSD bandwidth of the process&nbsp; $\{x_i(t)\}$?
 |type="{}"}
 $ {\rm \nabla} \hspace{-0.05cm} f_x \ = \ $ { 200 3% } $\ \rm kHz$

← Older revision		Revision as of 09:45, 28 February 2022
Line 107:		Line 107:


−	[[Category:Theory of Stochastic Signals: Exercises\|^4.5 Power-~~spectral~~ Density^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^4.5 Power-Spectral Density^]]

← Older revision		Revision as of 09:42, 28 February 2022
Line 107:		Line 107:


−	[[Category:Theory of Stochastic Signals: Exercises\|^4.5 ~~Leistungsdichtespektrum (LDS)~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^4.5 Power-spectral Density^]]

← Older revision	Revision as of 09:39, 28 February 2022
(No difference)

← Older revision		Revision as of 12:44, 23 March 2021
Line 107:		Line 107:


−	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^4.5 Leistungsdichtespektrum (LDS)^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^4.5 Leistungsdichtespektrum (LDS)^]]

@@ Line 22: / Line 22: @@
 ''Hinweise:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Stochastische_Signaltheorie/Leistungsdichtespektrum_(LDS)|Leistungsdichtespektrum]].
+*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Leistungsdichtespektrum_(LDS)|Leistungsdichtespektrum]].
-*Bezug genommen wird auch auf das  Kapitel&nbsp; [[Stochastische_Signaltheorie/Autokorrelationsfunktion_(AKF)|Autokorrelationsfunktion]].
+*Bezug genommen wird auch auf das  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Autokorrelationsfunktion_(AKF)|Autokorrelationsfunktion]].
 *Zur L&ouml;sung dieser Aufgabe k&ouml;nnen Sie die  folgende Fourierkorrespondenz benutzen: