Difference between revisions of "Aufgaben:Exercise 4.13: Gaussian ACF and PSD"

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{{quiz-Header|Buchseite=Stochastische Signaltheorie/Leistungsdichtespektrum (LDS)
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{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Power-Spectral_Density
 
}}
 
}}
  
[[File:P_ID411__Sto_A_4_13.png|right|frame|Zweimal gaußförmige AKF]]
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[[File:P_ID411__Sto_A_4_13.png|right|frame|Two Gaussian ACF]]
Der hier betrachtete Zufallsprozess  $\{x_i(t)\}$  sei durch die oben skizzierte Autokorrelationsfunktion (AKF) charakterisiert.  Dieser Zufallsprozess ist mittelwertfrei und die äquivalente AKF-Dauer beträgt  ${ {\rm \nabla} }\tau_x = 5 \hspace{0.08cm} \rm µ s$:
+
Let the random process considered here  $\{x_i(t)\}$  be characterized by the autocorrelation function (ACF) outlined above  This random process is mean-free and the equivalent ACF duration is  ${ {\rm \nabla} }\tau_x = 5 \hspace{0.08cm} \rm µ s$:
 
:$$\varphi_x(\tau)=\rm 0.25 V^2\cdot \rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\tau}{/ 5 \hspace{0.08cm}{\rm µ}s })^2} .$$
 
:$$\varphi_x(\tau)=\rm 0.25 V^2\cdot \rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\tau}{/ 5 \hspace{0.08cm}{\rm µ}s })^2} .$$
 
+
The bottom figure shows the ACF of the process  $\{y_i(t)\}$  This reads with the equivalent ACF duration  ${ {\rm \nabla} }\tau_y = 10 \hspace{0.08cm} \rm µ s$:
 
 
Im unteren Bild ist die AKF des Prozesses  $\{y_i(t)\}$  dargestellt.  Diese lautet mit der äquivalenten AKF-Dauer  ${ {\rm \nabla} }\tau_y = 10 \hspace{0.08cm} \rm µ s$:
 
 
:$$ \varphi_y(\tau)=\rm 0.16 V^2 + \rm 0.09 V^2\cdot\rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\tau}/{\nabla \it \tau_y})^2} .$$
 
:$$ \varphi_y(\tau)=\rm 0.16 V^2 + \rm 0.09 V^2\cdot\rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\tau}/{\nabla \it \tau_y})^2} .$$
  
In dieser Aufgabe werden die Leistungsdichtespektren der beiden Prozesse gesucht.
+
In this exercise, the Power spectral densities of the two processes are sought.
  
  
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''Hinweise:''
+
Hints:
*Die Aufgabe gehört zum  Kapitel  [[Theory_of_Stochastic_Signals/Leistungsdichtespektrum_(LDS)|Leistungsdichtespektrum]].
+
*This exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Power-Spectral_Density|Power-Spectral Density]].
*Bezug genommen wird auch auf das  Kapitel  [[Theory_of_Stochastic_Signals/Autokorrelationsfunktion_(AKF)|Autokorrelationsfunktion]].
+
*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function|Auto-correlation function]].
 
   
 
   
*Zur Lösung dieser Aufgabe können Sie die  folgende Fourierkorrespondenz benutzen:
+
*To solve this exercise you can use the following Fourier correspondence:
:$$\rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\it f}/{\rm \Delta\it f})^2}\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \ {\rm \Delta \it f} \cdot \rm e^{-\pi \hspace{0.05cm}\cdot \hspace{0.05cm} ({\rm \Delta\it f} \hspace{0.05cm}\cdot \hspace{0.05cm}\it t {\rm )}^{\rm 2}}.$$
+
:$$\rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\it f}/{\rm \Delta\it f})^2}\ \bullet\!\!-\!\!-\!\!-\!-\!\!\circ\, \ {\rm \Delta \it f} \cdot \rm e^{-\pi \hspace{0.05cm}\cdot \hspace{0.05cm} ({\rm \Delta\it f} \hspace{0.05cm}\cdot \hspace{0.05cm}\it t {\rm )}^{\rm 2}}.$$
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Wie gro&szlig; ist die &auml;quivalente LDS-Bandbreite des Prozesses&nbsp; $\{x_i(t)\}$?
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{What is the &auml;equivalent PSD bandwidth of the process&nbsp; $\{x_i(t)\}$?
 
|type="{}"}
 
|type="{}"}
$ {\rm \nabla} \hspace{-0.05cm} f_x \ = \ $ { 200 3% } $\   \rm kHz$
+
$ {\rm \nabla} \hspace{-0.05cm} f_x \ = \ $ { 200 3% } $\ \rm kHz$
  
  
{Wie lautet&nbsp; ${\it \Phi}_x(f)$?&nbsp; Geben Sie die LDS-Werte f&uuml;r&nbsp; $f= 0$&nbsp; und&nbsp; $f = 200 \hspace{0.08cm} \rm kHz$&nbsp; ein.
+
{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.
 
|type="{}"}
 
|type="{}"}
${\it \Phi}_x(f = 0)\ = \ $ { 1.25 3% }   $\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$
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${\it \Phi}_x(f = 0)\ = \ $ { 1.25 3% } $\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$
${\it \Phi}_x(f = 200 \hspace{0.08cm} \rm kHz)\ = \ $ { 0.054 3% }   $\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$
+
${\it \Phi}_x(f = 200 \hspace{0.08cm} \rm kHz)\ = \ $ { 0.054 3% } $\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$
  
  
{Welche Aussagen gelten, wenn der Zufallsprozess keine periodischen Anteile besitzt?&nbsp; Vorausgesetzt wird desweiteren eine konstante Leistung.
+
{Which statements are valid, if the random process has no periodic parts?&nbsp; Furthermore, a constant power is assumed.
 
|type="[]"}
 
|type="[]"}
+ Die Prozessleistung ist das Integral &uuml;ber das LDS.
+
+ The process power is the integral over the PSD.
+ Bei mittelwertfreiem Prozess ist das LDS stets kontinuierlich.
+
+ If the process is zero mean, the PSD is always continuous.
- Je breiter die AKF ist, um so breiter ist auch das LDS.
+
- The wider the ACF, the wider the PSD.
+ Eine breitere AKF bewirkt h&ouml;here LDS-Werte.
+
+ 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; $f = 200 \hspace{0.08cm} \rm kHz$?
 
 
{Berechnen Sie das Leistungsdichtespektrum&nbsp; ${\it \Phi}_y(f)$.&nbsp; Welche Werte ergeben sich f&uuml;r den kontinuierlichen LDS-Anteil bei&nbsp; $f= 0$&nbsp; und&nbsp; $f = 200 \hspace{0.08cm} \rm kHz$?
 
 
|type="{}"}
 
|type="{}"}
${\it \Phi}_y(f = 0)\ = \ $ { 0.9 3% } $\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$
+
${\it \Phi}_y(f = 0)\ = \ $ { 0.9 3% } $\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$
${\it \Phi}_y(f = 200 \hspace{0.08cm} \rm kHz)\ = \ $ { 0.0165 3% } $\ \cdot 10^{-6} \hspace{0.05cm} \rm V^2\hspace{-0.1cm}/Hz$
+
${\it \Phi}_y(f = 200 \hspace{0.08cm} \rm kHz)\ = \ $ { 0.0165 3% } $\ \cdot 10^{-6} \hspace{0.05cm} \rm V^2\hspace{-0.1cm}/Hz$
  
  
{Welche der folgenden Aussagen stimmen bez&uuml;glich des Prozesses&nbsp; $\{y_i(t)\}$?
+
{Which of the following statements are true regarding the process&nbsp; $\{y_i(t)\}$?
 
|type="[]"}
 
|type="[]"}
- Das LDS beinhaltet einen Dirac bei der Frequenz  $ f = {\rm \nabla} \hspace{-0.05cm} f_y$.
+
- The PSD involves a Dirac at frequency $ f = {\rm \nabla} \hspace{-0.05cm} f_y$.
+ Das LDS beinhaltet einen Dirac bei der Frequenz $f= 0$.
+
+ The PSD involves a Dirac at frequency $f= 0$.
- Diracgewicht und kontinuierliches LDS haben gleiche Einheit.
+
- Dirac weight and continuous PSD have the same unit.
 
 
  
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Die &auml;quivalente LDS-Bandbreite ist der Kehrwert der &auml;quivalenten AKF-Dauer:  
+
'''(1)'''&nbsp; The equivalent PSD bandwidth is the reciprocal of the equivalent ACF duration:  
 
:$$\nabla f_x = 1 / \nabla \tau_x \hspace{0.15cm}\underline{= {\rm 200\hspace{0.1cm}kHz}}.$$
 
:$$\nabla f_x = 1 / \nabla \tau_x \hspace{0.15cm}\underline{= {\rm 200\hspace{0.1cm}kHz}}.$$
  
  
'''(2)'''&nbsp; Die angegebene Fourierkorrespondenz kann man wie folgt an die Aufgabenstellung anpassen:
+
'''(2)'''&nbsp; One can adapt the given Fourier correspondence to the task as follows:
:$$K\cdot{\rm e}^{-\pi({\tau}/{\nabla\tau_x})^2}\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ \frac{\it K}{\nabla \it f_x}\cdot{\rm e}^{-\pi({f}/{\nabla f_x})^2}.$$
+
:$$K\cdot{\rm e}^{-\pi({\tau}/{\nabla\tau_x})^2}\ \circ\!\!-\!\!-\!\!-\!\!-\!-\!\bullet\,\ \frac{\it K}{\nabla \it f_x}\cdot{\rm e}^{-\pi({f}/{\nabla f_x})^2}.$$
  
*Mit&nbsp; $K = 0.25 \hspace{0.05cm}\rm V^2$&nbsp; und&nbsp; $ {\rm \nabla} \hspace{-0.05cm} f_x = 200\hspace{0.05cm} \rm kHz$&nbsp; erh&auml;lt man:
+
*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:
 
:$${\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}\hspace{0.3cm}
 
\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},
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'''(3)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1, 2 und 4</u>:
+
'''(3)'''&nbsp; Correct <u>solutions 1, 2, and 4</u>:
*Ein mittelwertfreier Prozess hat stets ein kontinuierliches LDS zur Folge.&nbsp; Dieses ist um so schmaler, je breiter die AKF ist (Reziprozit&auml;tsgesetz).
+
*A mean-free process always results in a continuous PSD.&nbsp; This is narrower the wider the ACF is (reciprocitylaw).
*Die Prozessleistung ist gleich dem Integral &uuml;ber das LDS.
+
*The process power is equal to the integral of the PSD.
*Deshalb muss bei konstanter Leistung eine breitere AKF&nbsp; (schmaleres LDS)&nbsp; durch h&ouml;here LDS-Werte ausgeglichen werden.  
+
*Therefore, at constant power, a wider ACF&nbsp; (narrower PSD)&nbsp; must be compensated by higher PSD values.  
*Ein Gleichanteil oder periodische Anteile führen stets zu Diracfunktionen im LDS;&nbsp; ansonsten ist das LDS stets wertkontinuierlich.  
+
*A DC component or periodic components always result in dirac functions in the PSD;&nbsp; otherwise, the PSD is always continuous in value.  
  
  
  
'''(4)'''&nbsp; Analog zu Teilaufgabe&nbsp; '''(2)'''&nbsp; gilt mit&nbsp; $ {\rm \nabla} \hspace{-0.05cm} f_y = 100\hspace{0.05cm} \rm kHz$:
+
'''(4)'''&nbsp; Analogous to subtask&nbsp; '''(2)'''&nbsp; holds with&nbsp; $ {\rm \nabla} \hspace{-0.05cm} f_y = 100\hspace{0.05cm} \rm kHz$:
 
:$${\it \Phi_y}(f)=\frac{\rm 0.09 V^2}{\nabla\it f_y}\cdot\rm e^{-\pi({\it f}/{\nabla\it f_y})^2}+\it m_y^{\rm 2}\cdot\delta(f).$$
 
:$${\it \Phi_y}(f)=\frac{\rm 0.09 V^2}{\nabla\it f_y}\cdot\rm e^{-\pi({\it f}/{\nabla\it f_y})^2}+\it m_y^{\rm 2}\cdot\delta(f).$$
  
*Aufgrund des Gleichanteils gibt es zus&auml;tzlich zum kontinuierlichen LDS-Anteil noch einen Dirac bei der Frequenz $f = 0$.  
+
*Because of the DC component, there is a Dirac at frequency $f = 0$ in addition to the continuous PSD component.  
*Der kontinuierliche LDS&ndash;Anteil bei $f= 0$ betr&auml;gt&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 continuous PSD&ndash;part at $f= 0$ is&nbsp; ${\it \Phi_y}(f = 0)=\hspace{0.15cm}\underline{\rm 0.9 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}.$  
*Der Anteil bei $f = 2 \cdot {\rm \nabla} \hspace{-0.05cm} f_y = 200 \hspace{0.05cm}\rm kHz$&nbsp; ist um den Faktor&nbsp; ${\rm e}^{-4} \approx 0.0183$&nbsp; geringer &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 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; ${\it \Phi_y}(f )=\hspace{0.15cm}\underline{\rm 0.0165 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}.$
  
  
'''(5)'''&nbsp; Richtig ist  <u>nur der zweite Lösungsvorschlag</u>:
+
'''(5)'''&nbsp; Correct is <u>only the second proposed solution</u>:
*Das LDS eines mittelwertbehafteten Prozesses beinhaltet allgemein eine Diracfunktion bei $f=0$&nbsp; mit Gewicht&nbsp; $m_y^2$.
+
*The PSD of a mean-valued process generally involves a Dirac function at $f=0$&nbsp; with weight&nbsp; $m_y^2$.
* Im vorliegenden Fall ist dieser Wert gleich&nbsp; $0.16 \ \rm V^2$.  
+
*In the present case, this value is equal to&nbsp; $0.16 \ \rm V^2$.  
*Da&nbsp; $\delta(f)$&nbsp; die Einheit&nbsp; $\rm 1/Hz = s$&nbsp; besitzt, unterscheiden sich die Einheiten des kontinuierlichen und des diskreten LDS-Anteils.  
+
*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}}
 
{{ML-Fuß}}

Revision as of 20:18, 7 March 2022

Two Gaussian ACF

Let the random process considered here  $\{x_i(t)\}$  be characterized by the autocorrelation function (ACF) outlined above  This random process is mean-free and the equivalent ACF duration is  ${ {\rm \nabla} }\tau_x = 5 \hspace{0.08cm} \rm µ s$:

$$\varphi_x(\tau)=\rm 0.25 V^2\cdot \rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\tau}{/ 5 \hspace{0.08cm}{\rm µ}s })^2} .$$

The bottom figure shows the ACF of the process  $\{y_i(t)\}$  This reads with the equivalent ACF duration  ${ {\rm \nabla} }\tau_y = 10 \hspace{0.08cm} \rm µ s$:

$$ \varphi_y(\tau)=\rm 0.16 V^2 + \rm 0.09 V^2\cdot\rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\tau}/{\nabla \it \tau_y})^2} .$$

In this exercise, the Power spectral densities of the two processes are sought.





Hints:

  • To solve this exercise you can use the following Fourier correspondence:
$$\rm e^{-\pi \hspace{0.03cm}\cdot \hspace{0.03cm} ({\it f}/{\rm \Delta\it f})^2}\ \bullet\!\!-\!\!-\!\!-\!-\!\!\circ\, \ {\rm \Delta \it f} \cdot \rm e^{-\pi \hspace{0.05cm}\cdot \hspace{0.05cm} ({\rm \Delta\it f} \hspace{0.05cm}\cdot \hspace{0.05cm}\it t {\rm )}^{\rm 2}}.$$


Questions

1

What is the äequivalent PSD bandwidth of the process  $\{x_i(t)\}$?

$ {\rm \nabla} \hspace{-0.05cm} f_x \ = \ $

$\ \rm kHz$

2

What is  ${\it \Phi}_x(f)$?  Give the PSD values for  $f= 0$  and  $f = 200 \hspace{0.08cm} \rm kHz$  on.

${\it \Phi}_x(f = 0)\ = \ $

$\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$
${\it \Phi}_x(f = 200 \hspace{0.08cm} \rm kHz)\ = \ $

$\ \cdot 10^{-6} \ \rm V^2\hspace{-0.1cm}/Hz$

3

Which statements are valid, if the random process has no periodic parts?  Furthermore, a constant power is assumed.

The process power is the integral over the PSD.
If the process is zero mean, the PSD is always continuous.
The wider the ACF, the wider the PSD.
A wider ACF results in higher PSD values.

4

Which of the following statements are true regarding the process  $\{y_i(t)\}$?

The PSD involves a Dirac at frequency $ f = {\rm \nabla} \hspace{-0.05cm} f_y$.
The PSD involves a Dirac at frequency $f= 0$.
Dirac weight and continuous PSD have the same unit.


Solution

(1)  The equivalent PSD bandwidth is the reciprocal of the equivalent ACF duration:

$$\nabla f_x = 1 / \nabla \tau_x \hspace{0.15cm}\underline{= {\rm 200\hspace{0.1cm}kHz}}.$$


(2)  One can adapt the given Fourier correspondence to the task as follows:

$$K\cdot{\rm e}^{-\pi({\tau}/{\nabla\tau_x})^2}\ \circ\!\!-\!\!-\!\!-\!\!-\!-\!\bullet\,\ \frac{\it K}{\nabla \it f_x}\cdot{\rm e}^{-\pi({f}/{\nabla f_x})^2}.$$
  • With  $K = 0.25 \hspace{0.05cm}\rm V^2$  and  $ {\rm \nabla} \hspace{-0.05cm} f_x = 200\hspace{0.05cm} \rm kHz$  obtains:
$${\it \Phi_x}(f)=1.25\cdot\rm 10^{-\rm 6}\hspace{0.1cm}\frac{V^2}{Hz}\cdot\rm e^{-\pi({\it f}/{\nabla\it f_x})^2}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\it \Phi_x}(f = 0)=\hspace{0.15cm}\underline{\rm 1.25 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}, \hspace{0.5cm}{\it \Phi_x}(f = 200 \hspace{0.05cm} \rm kHz)=\hspace{0.15cm}\underline{\rm 0.054 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}.$$


(3)  Correct solutions 1, 2, and 4:

  • A mean-free process always results in a continuous PSD.  This is narrower the wider the ACF is (reciprocitylaw).
  • The process power is equal to the integral of the PSD.
  • Therefore, at constant power, a wider ACF  (narrower PSD)  must be compensated by higher PSD values.
  • A DC component or periodic components always result in dirac functions in the PSD;  otherwise, the PSD is always continuous in value.


(4)  Analogous to subtask  (2)  holds with  $ {\rm \nabla} \hspace{-0.05cm} f_y = 100\hspace{0.05cm} \rm kHz$:

$${\it \Phi_y}(f)=\frac{\rm 0.09 V^2}{\nabla\it f_y}\cdot\rm e^{-\pi({\it f}/{\nabla\it f_y})^2}+\it m_y^{\rm 2}\cdot\delta(f).$$
  • Because of the DC component, there is a Dirac at frequency $f = 0$ in addition to the continuous PSD component.
  • The continuous PSD–part at $f= 0$ is  ${\it \Phi_y}(f = 0)=\hspace{0.15cm}\underline{\rm 0.9 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}.$
  • The fraction at $f = 2 \cdot {\rm \nabla} \hspace{-0.05cm} f_y = 200 \hspace{0.05cm}\rm kHz$  is increased by a factor  ${\rm e}^{-4} \approx 0.0183$  lower   ⇒   ${\it \Phi_y}(f )=\hspace{0.15cm}\underline{\rm 0.0165 \cdot 10^{-6} \hspace{0.1cm} V^2\hspace{-0.1cm}/Hz}.$


(5)  Correct is only the second proposed solution:

  • The PSD of a mean-valued process generally involves a Dirac function at $f=0$  with weight  $m_y^2$.
  • In the present case, this value is equal to  $0.16 \ \rm V^2$.
  • Since  $\delta(f)$  has unit  $\rm 1/Hz = s$ , the units of the continuous and discrete PSD components differ.