Difference between revisions of "Aufgaben:Exercise 5.1: Gaussian ACF and Gaussian Low-Pass"

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{{quiz-Header|Buchseite=Stochastische Signaltheorie/Stochastische Systemtheorie
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{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Stochastic_System_Theory
 
}}
 
}}
  
[[File:P_ID487__Sto_A_5_1.png|right|frame|Gaußsche AKF am Eingang und Ausgang]]
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[[File:P_ID487__Sto_A_5_1.png|right|frame|Gaussian ACF at the filter input and output]]
Am Eingang eines Tiefpassfilters mit dem Frequenzgang $H(f)$ liegt ein gaußverteiltes mittelwertfreies Rauschsignal $x(t)$ mit folgender Autokorrelationsfunktion (AKF) an:
+
At the input of a low-pass filter with frequency response  $H(f)$,  there is a Gaussian distributed mean-free noise signal  $x(t)$  with the following auto-correlation function  $\rm (ACF)$:
:$${\it \varphi_{x}(\tau)} = \sigma_x^2 \cdot {\rm e}^{- \pi (\tau
+
 
 +
:$${\it \varphi}_{x}(\tau) = \sigma_x^2 \cdot {\rm e}^{- \pi (\tau
 
/{\rm \nabla} \tau_x)^2}.$$
 
/{\rm \nabla} \tau_x)^2}.$$
  
Diese AKF ist in der nebenstehenden Grafik oben dargestellt.
+
This ACF is shown in the accompanying diagram above.
  
Das Filter sei gaußförmig mit der Gleichsignalverstärkung $H_0$ und der äquivalenten Bandbreite $\Delta f$. Für den Frequenzgang kann somit geschrieben werden:
+
Let the filter be Gaussian with the DC gain  $H_0$  and the equivalent bandwidth  $\Delta f$.  Thus,  for the frequency response,  it can be written:
 
:$$H(f) = H_{\rm 0} \cdot{\rm e}^{-  \pi (f/ {\rm \Delta} f)^2}.$$
 
:$$H(f) = H_{\rm 0} \cdot{\rm e}^{-  \pi (f/ {\rm \Delta} f)^2}.$$
  
Im Verlaufe dieser Aufgabe sollen die beiden Filterparameter $H_0$ und $\Delta f$ so dimensioniert werden, dass das Ausgangssignal $y(t)$ eine AKF entsprechend der unteren Skizze aufweist.  
+
In the course of this task,  the two filter parameters  $H_0$  and  $\Delta f$  are to be dimensioned so that the output signal  $y(t)$  has an ACF corresponding to the diagram below.
  
  
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+
Notes:  
''Hinweise:''
+
*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Stochastic_System_Theory|Stochastic System Theory]].
*Die Aufgabe gehört zum  Kapitel [[Stochastische_Signaltheorie/Stochastische_Systemtheorie|Stochastische Systemtheorie]].
+
*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function_(ACF)|Auto-Correlation Function]].
*Bezug genommen wird auch auf das  Kapitel [[Stochastische_Signaltheorie/Autokorrelationsfunktion_(AKF)|Autokorrelationsfunktion]].
 
 
   
 
   
*Berücksichtigen Sie die folgende Fourierkorrespondenz:
+
*Consider the following Fourier correspondence:
 
:$${\rm e}^{-  \pi (f/{\rm \Delta} f)^2} \hspace{0.15cm}
 
:$${\rm e}^{-  \pi (f/{\rm \Delta} f)^2} \hspace{0.15cm}
 
\bullet\!\!-\!\!\!-\!\!\!\hspace{0.03cm}\circ \hspace{0.15cm}{\rm \Delta} f \cdot
 
\bullet\!\!-\!\!\!-\!\!\!\hspace{0.03cm}\circ \hspace{0.15cm}{\rm \Delta} f \cdot
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===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Wie groß ist der Effektivwert des Filtereingangssignals?
+
{What is the standard deviation of the filter input signal?
 
|type="{}"}
 
|type="{}"}
 
$\sigma_x \ = \ $ { 0.2 3% } $\ \rm V$
 
$\sigma_x \ = \ $ { 0.2 3% } $\ \rm V$
  
  
{Bestimmen Sie aus der skizzierten AKF auch die äquivalente AKF-Dauer des Signals $x(t)$. <br>Wie kann diese allgemein ermittelt werden?
+
{From the sketched ACF,&nbsp; also determine the equivalent ACF duration&nbsp; $\nabla\tau_x$&nbsp; of the input signal.&nbsp; How can this be determined in general?
 
|type="{}"}
 
|type="{}"}
$\nabla\tau_x \ =  \ $ { 1 3% } $\ &micro; s$
+
$\nabla\tau_x \ =  \ $ { 1 3% } $\ \rm &micro; s$
  
  
{Wie lautet das Leistungsdichtespektrum ${\it Φ}_x(f)$ des Eingangsignals? <br>Wie groß ist der LDS-Wert bei $f= 0$?
+
{What is the power-spectral density&nbsp; ${\it Φ}_x(f)$&nbsp; of the input signal?&nbsp; What is the PSD value at&nbsp; $f= 0$?
 
|type="{}"}
 
|type="{}"}
 
${\it Φ}_x(f=0) \ =  \ $ { 40 3% } $\ \cdot 10^{-9}\ \rm V^2/Hz$
 
${\it Φ}_x(f=0) \ =  \ $ { 40 3% } $\ \cdot 10^{-9}\ \rm V^2/Hz$
  
  
{Berechnen Sie das LDS ${\it Φ}_y(f)$ am Filterausgang allgemein als Funktion von $\sigma_x$, $\nabla \tau_x$, $H_0$ und $\Delta f$. <br>Welche Aussagen treffen zu?
+
{Calculate the PSD&nbsp; ${\it Φ}_y(f)$&nbsp; at the filter output in general as a function of&nbsp; $\sigma_x$,&nbsp; $\nabla \tau_x$,&nbsp; $H_0$&nbsp; and&nbsp; $\Delta f$.&nbsp; Which statements are true?
 
|type="[]"}
 
|type="[]"}
+ Das LDS ${\it Φ}_y(f)$ ist ebenfalls gaußförmig.
+
+ The PSD&nbsp; ${\it Φ}_y(f)$&nbsp; is also Gaussian.
- Je kleiner $\Delta f$ ist, um so breiter ist ${\it Φ}_y(f)$.
+
- The smaller&nbsp; $\Delta f$&nbsp; is,&nbsp; the wider&nbsp; ${\it Φ}_y(f)$.
+ $H_0$ beeinflusst nur die Höhe, aber nicht die Breite von ${\it Φ}_y(f)$.
+
+ $H_0$&nbsp; only affects the height,&nbsp; but not the width of&nbsp; ${\it Φ}_y(f)$.
  
  
{Wie groß muss die äquivalente Filterbandbreite $\Delta f$ gewählt werden, damit für die äquivalente AKF-Dauer $\nabla \tau_y = 3 \ \rm  &micro; s$ gilt?
+
{How large must the equivalent filter bandwidth&nbsp; $\Delta f$&nbsp; be chosen so that&nbsp; $\nabla \tau_y = 3 \ \rm  &micro; s$&nbsp; holds for the equivalent ACF duration?
 
|type="{}"}
 
|type="{}"}
 
$\Delta f \ =  \ $ { 0.5 3% } $\ \rm MHz$
 
$\Delta f \ =  \ $ { 0.5 3% } $\ \rm MHz$
  
  
{Wie groß muss man den Gleichsignalübertragungsfaktor $H_0$ wählen, damit die Bedingung $\sigma_y = \sigma_x$ erfüllt wird?
+
{How large must one select the DC signal transfer factor&nbsp; $H_0$&nbsp; so that the condition&nbsp; $\sigma_y = \sigma_x$&nbsp; is fulfilled?
 
|type="{}"}
 
|type="{}"}
 
$H_0 \ =  \ $ { 1.732 3% }
 
$H_0 \ =  \ $ { 1.732 3% }
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</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Die Varianz ist gleich dem AKF-Wert bei $\tau = 0$, also &nbsp;$\sigma_x^2 = 0.04 \ \rm V^2$. Daraus folgt &nbsp;$\sigma_x\hspace{0.15cm}\underline {= 0.2 \ \rm V}$&nbsp;.
+
'''(1)'''&nbsp; The variance is equal to the ACF value at&nbsp; $\tau = 0$,&nbsp;  so &nbsp;$\sigma_x^2 = 0.04 \ \rm V^2$.
 +
*From this follows &nbsp;$\sigma_x\hspace{0.15cm}\underline {= 0.2 \ \rm V}$.
 +
 
 +
 
  
 +
'''(2)'''&nbsp; The equivalent ACF duration can be determined via the rectangle of equal area.
 +
*According to the sketch,&nbsp; we obtain &nbsp;$\nabla \tau_x\hspace{0.15cm}\underline {= 1 \ \rm &micro; s}$.
  
'''(2)'''&nbsp; Die äquivalente AKF-Dauer kann man über das flächengleiche Rechteck ermitteln. Gemäß der Skizze erhält man &nbsp;$\nabla \tau_x\hspace{0.15cm}\underline {= 1 \ \rm \mu s}$.
 
  
  
'''(3)'''&nbsp; Das LDS ist die Fouriertransformierte der AKF.  
+
'''(3)'''&nbsp; The PSD is the Fourier transform of the ACF.  
*Mit der gegebenen Fourierkorrespondenz gilt:
+
*With the given Fourier correspondence holds:
 
:$${\it \Phi}_{x}(f) = \sigma_x^2 \cdot  {\rm \nabla} \tau_x \cdot
 
:$${\it \Phi}_{x}(f) = \sigma_x^2 \cdot  {\rm \nabla} \tau_x \cdot
 
{\rm e}^{- \pi ({\rm \nabla} \tau_x \hspace{0.03cm}\cdot \hspace{0.03cm}f)^2} .$$
 
{\rm e}^{- \pi ({\rm \nabla} \tau_x \hspace{0.03cm}\cdot \hspace{0.03cm}f)^2} .$$
  
*Bei der Frequenz $f = 0$ erhält man:
+
*At frequency&nbsp; $f = 0$,&nbsp; we obtain:
 
:$${\it \Phi}_{x}(f  = 0) = \sigma_x^2 \cdot  {\rm \nabla} \tau_x =
 
:$${\it \Phi}_{x}(f  = 0) = \sigma_x^2 \cdot  {\rm \nabla} \tau_x =
 
\rm 0.04 \hspace{0.1cm} V^2 \cdot 10^{-6} \hspace{0.1cm} s \hspace{0.15cm} \underline{= 40
 
\rm 0.04 \hspace{0.1cm} V^2 \cdot 10^{-6} \hspace{0.1cm} s \hspace{0.15cm} \underline{= 40
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'''(4)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 3</u>:
+
 
*Allgemein gilt ${\it \Phi}_{y}(f) = {\it \Phi}_{x}(f) \cdot |H(f)|^2$. Daraus folgt:
+
'''(4)'''&nbsp; <u>Solutions 1 and 3</u>&nbsp; are correct:
 +
*In general,&nbsp; ${\it \Phi}_{y}(f) = {\it \Phi}_{x}(f) \cdot |H(f)|^2$.&nbsp; It follows:
 
:$${\it \Phi}_{y}(f) =  \sigma_x^2 \cdot  {\rm \nabla} \tau_x \cdot
 
:$${\it \Phi}_{y}(f) =  \sigma_x^2 \cdot  {\rm \nabla} \tau_x \cdot
 
{\rm e}^{- \pi ({\rm \nabla} \tau_x \cdot f)^2}\cdot H_{\rm 0}^2
 
{\rm e}^{- \pi ({\rm \nabla} \tau_x \cdot f)^2}\cdot H_{\rm 0}^2
 
\cdot{\rm e}^{- 2 \pi (f/ {\rm \Delta} f)^2} .$$
 
\cdot{\rm e}^{- 2 \pi (f/ {\rm \Delta} f)^2} .$$
*Durch Zusammenfassen der beiden Exponentialfunktionen erhält man:
+
*By combining the two exponential functions,&nbsp; we obtain:
 
:$${\it \Phi}_{y}(f) =  \sigma_x^2 \cdot  {\rm \nabla} \tau_x \cdot H_0^2 \cdot
 
:$${\it \Phi}_{y}(f) =  \sigma_x^2 \cdot  {\rm \nabla} \tau_x \cdot H_0^2 \cdot
{\rm e}^{- \pi\cdot  ({\rm \nabla} \tau_x^2 + 2/(\Delta f^2) ) \hspace{0.1cm}\cdot f^2}.$$
+
{\rm e}^{- \pi\cdot  ({\rm \nabla} \tau_x^2 + 2/\Delta f^2  ) \hspace{0.1cm}\cdot f^2}.$$
*Auch ${\it \Phi}_{y}(f)$ ist gaußförmig und nie breiter als ${\it \Phi}_{x}(f)$. Für $f \to \infty$ gilt die Näherung ${\it \Phi}_{y}(f) \approx {\it \Phi}_{x}(f)$.  
+
*Also&nbsp; ${\it \Phi}_{y}(f)$&nbsp; is Gaussian and never wider than&nbsp; ${\it \Phi}_{x}(f)$.&nbsp; For $f \to \infty$,&nbsp; the approximation&nbsp; ${\it \Phi}_{y}(f) \approx {\it \Phi}_{x}(f)$&nbsp; holds.  
*Mit kleiner werdendem $\Delta f$ wird ${\it \Phi}_{y}(f)$ immer schmäler (also ist die zweite Aussage falsch).  
+
*As&nbsp; $\Delta f$&nbsp; gets smaller,&nbsp; ${\it \Phi}_{y}(f)$&nbsp; gets narrower&nbsp; (so the second statement is false).  
*$H_0$ beeinflusst tatsächlich nur die LDS-Höhe, aber nicht die Breite des LDS.
+
*$H_0$&nbsp; actually affects only the PSD height,&nbsp; but not the width of the PSD.
 
   
 
   
  
  
'''(5)'''&nbsp; Analog zum Aufgabenteil '''(1)''' kann für das LDS des Ausgangssignals $y(t)$ geschrieben werden:
+
'''(5)'''&nbsp; Analogous to task&nbsp; '''(1)''',&nbsp; it can be written for the PSD of the output signal&nbsp; $y(t)$:
 
:$${\it \Phi}_{y}(f) =  \sigma_y^2 \cdot  {\rm \nabla} \tau_y \cdot
 
:$${\it \Phi}_{y}(f) =  \sigma_y^2 \cdot  {\rm \nabla} \tau_y \cdot
 
{\rm e}^{- \pi  \cdot {\rm \nabla} \tau_y^2 \cdot f^2 }.$$
 
{\rm e}^{- \pi  \cdot {\rm \nabla} \tau_y^2 \cdot f^2 }.$$
  
*Durch Vergleich mit dem Ergebnis aus '''(4)''' ergibt sich:
+
*By comparing with the result from&nbsp; '''(4)'''&nbsp; we get:
 
:$${{\rm \nabla} \tau_y^2} = {{\rm \nabla} \tau_x^2} + \frac {2}{{\rm
 
:$${{\rm \nabla} \tau_y^2} = {{\rm \nabla} \tau_x^2} + \frac {2}{{\rm
 
\Delta} f^2}.$$
 
\Delta} f^2}.$$
*Löst man die Gleichung nach $\Delta f$ auf und berücksichtigt die Werte $\nabla \tau_x {= 1 \ \rm &micro; s}$ und   $\nabla \tau_y {= 3 \ \rm &micro; s}$,  so folgt:
+
*Solving the equation for&nbsp; $\Delta f$&nbsp; and considering the values&nbsp; $\nabla \tau_x {= 1 \ \rm &micro; s}$&nbsp; as well as&nbsp;   $\nabla \tau_y {= 3 \ \rm &micro; s}$,&nbsp; it follows:
 
:$${\rm \Delta} f = \sqrt{\frac{2}{{\rm \nabla} \tau_y^2 - {\rm
 
:$${\rm \Delta} f = \sqrt{\frac{2}{{\rm \nabla} \tau_y^2 - {\rm
 
\nabla} \tau_x^2}} = \sqrt{\frac{2}{9 - 1}} \hspace{0.1cm}\rm MHz
 
\nabla} \tau_x^2}} = \sqrt{\frac{2}{9 - 1}} \hspace{0.1cm}\rm MHz
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'''(6)'''&nbsp; Die Bedingung $\sigma_y = \sigma_x$ ist gleichbedeutend mit $\varphi_y(\tau = 0)= \varphi_x(\tau = 0)$.  
+
 
*Da zudem $\nabla \tau_y = 3 \cdot \nabla \tau_x$ vorgegeben ist, muss deshalb auch ${\it \Phi}_{y}(f= 0) =  3 \cdot {\it \Phi}_{x}(f= 0)$ gelten.  
+
'''(6)'''&nbsp; The condition&nbsp; $\sigma_y = \sigma_x$&nbsp; is equivalent to&nbsp; $\varphi_y(\tau = 0)= \varphi_x(\tau = 0)$.  
*Daraus erhält man:
+
*Moreover,&nbsp; since&nbsp; $\nabla \tau_y = 3 \cdot \nabla \tau_x$&nbsp; is given,&nbsp; therefore&nbsp; ${\it \Phi}_{y}(f= 0) =  3 \cdot {\it \Phi}_{x}(f= 0)$&nbsp; must also hold.
 +
*From this we obtain:
 
:$$H_{\rm 0} = \sqrt{\frac{\it \Phi_y (f \rm = 0)}{\it \Phi_x (f = \rm 0)}} = \sqrt
 
:$$H_{\rm 0} = \sqrt{\frac{\it \Phi_y (f \rm = 0)}{\it \Phi_x (f = \rm 0)}} = \sqrt
 
{3}\hspace{0.15cm} \underline{=1.732}.$$
 
{3}\hspace{0.15cm} \underline{=1.732}.$$
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[[Category:Aufgaben zu Stochastische Signaltheorie|^5.1 Stochastische Systemtheorie^]]
+
[[Category:Theory of Stochastic Signals: Exercises|^5.1 Stochastic Systems Theory^]]

Latest revision as of 12:11, 17 February 2022

Gaussian ACF at the filter input and output

At the input of a low-pass filter with frequency response  $H(f)$,  there is a Gaussian distributed mean-free noise signal  $x(t)$  with the following auto-correlation function  $\rm (ACF)$:

$${\it \varphi}_{x}(\tau) = \sigma_x^2 \cdot {\rm e}^{- \pi (\tau /{\rm \nabla} \tau_x)^2}.$$

This ACF is shown in the accompanying diagram above.

Let the filter be Gaussian with the DC gain  $H_0$  and the equivalent bandwidth  $\Delta f$.  Thus,  for the frequency response,  it can be written:

$$H(f) = H_{\rm 0} \cdot{\rm e}^{- \pi (f/ {\rm \Delta} f)^2}.$$

In the course of this task,  the two filter parameters  $H_0$  and  $\Delta f$  are to be dimensioned so that the output signal  $y(t)$  has an ACF corresponding to the diagram below.



Notes:

  • Consider the following Fourier correspondence:
$${\rm e}^{- \pi (f/{\rm \Delta} f)^2} \hspace{0.15cm} \bullet\!\!-\!\!\!-\!\!\!\hspace{0.03cm}\circ \hspace{0.15cm}{\rm \Delta} f \cdot {\rm e}^{- \pi ({\rm \Delta} f \hspace{0.03cm} \cdot \hspace{0.03cm} t)^2}.$$


Questions

1

What is the standard deviation of the filter input signal?

$\sigma_x \ = \ $

$\ \rm V$

2

From the sketched ACF,  also determine the equivalent ACF duration  $\nabla\tau_x$  of the input signal.  How can this be determined in general?

$\nabla\tau_x \ = \ $

$\ \rm µ s$

3

What is the power-spectral density  ${\it Φ}_x(f)$  of the input signal?  What is the PSD value at  $f= 0$?

${\it Φ}_x(f=0) \ = \ $

$\ \cdot 10^{-9}\ \rm V^2/Hz$

4

Calculate the PSD  ${\it Φ}_y(f)$  at the filter output in general as a function of  $\sigma_x$,  $\nabla \tau_x$,  $H_0$  and  $\Delta f$.  Which statements are true?

The PSD  ${\it Φ}_y(f)$  is also Gaussian.
The smaller  $\Delta f$  is,  the wider  ${\it Φ}_y(f)$.
$H_0$  only affects the height,  but not the width of  ${\it Φ}_y(f)$.

5

How large must the equivalent filter bandwidth  $\Delta f$  be chosen so that  $\nabla \tau_y = 3 \ \rm µ s$  holds for the equivalent ACF duration?

$\Delta f \ = \ $

$\ \rm MHz$

6

How large must one select the DC signal transfer factor  $H_0$  so that the condition  $\sigma_y = \sigma_x$  is fulfilled?

$H_0 \ = \ $


Solution

(1)  The variance is equal to the ACF value at  $\tau = 0$,  so  $\sigma_x^2 = 0.04 \ \rm V^2$.

  • From this follows  $\sigma_x\hspace{0.15cm}\underline {= 0.2 \ \rm V}$.


(2)  The equivalent ACF duration can be determined via the rectangle of equal area.

  • According to the sketch,  we obtain  $\nabla \tau_x\hspace{0.15cm}\underline {= 1 \ \rm µ s}$.


(3)  The PSD is the Fourier transform of the ACF.

  • With the given Fourier correspondence holds:
$${\it \Phi}_{x}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot {\rm e}^{- \pi ({\rm \nabla} \tau_x \hspace{0.03cm}\cdot \hspace{0.03cm}f)^2} .$$
  • At frequency  $f = 0$,  we obtain:
$${\it \Phi}_{x}(f = 0) = \sigma_x^2 \cdot {\rm \nabla} \tau_x = \rm 0.04 \hspace{0.1cm} V^2 \cdot 10^{-6} \hspace{0.1cm} s \hspace{0.15cm} \underline{= 40 \cdot 10^{-9} \hspace{0.1cm} V^2 / Hz}.$$


(4)  Solutions 1 and 3  are correct:

  • In general,  ${\it \Phi}_{y}(f) = {\it \Phi}_{x}(f) \cdot |H(f)|^2$.  It follows:
$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot {\rm e}^{- \pi ({\rm \nabla} \tau_x \cdot f)^2}\cdot H_{\rm 0}^2 \cdot{\rm e}^{- 2 \pi (f/ {\rm \Delta} f)^2} .$$
  • By combining the two exponential functions,  we obtain:
$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot H_0^2 \cdot {\rm e}^{- \pi\cdot ({\rm \nabla} \tau_x^2 + 2/\Delta f^2 ) \hspace{0.1cm}\cdot f^2}.$$
  • Also  ${\it \Phi}_{y}(f)$  is Gaussian and never wider than  ${\it \Phi}_{x}(f)$.  For $f \to \infty$,  the approximation  ${\it \Phi}_{y}(f) \approx {\it \Phi}_{x}(f)$  holds.
  • As  $\Delta f$  gets smaller,  ${\it \Phi}_{y}(f)$  gets narrower  (so the second statement is false).
  • $H_0$  actually affects only the PSD height,  but not the width of the PSD.


(5)  Analogous to task  (1),  it can be written for the PSD of the output signal  $y(t)$:

$${\it \Phi}_{y}(f) = \sigma_y^2 \cdot {\rm \nabla} \tau_y \cdot {\rm e}^{- \pi \cdot {\rm \nabla} \tau_y^2 \cdot f^2 }.$$
  • By comparing with the result from  (4)  we get:
$${{\rm \nabla} \tau_y^2} = {{\rm \nabla} \tau_x^2} + \frac {2}{{\rm \Delta} f^2}.$$
  • Solving the equation for  $\Delta f$  and considering the values  $\nabla \tau_x {= 1 \ \rm µ s}$  as well as  $\nabla \tau_y {= 3 \ \rm µ s}$,  it follows:
$${\rm \Delta} f = \sqrt{\frac{2}{{\rm \nabla} \tau_y^2 - {\rm \nabla} \tau_x^2}} = \sqrt{\frac{2}{9 - 1}} \hspace{0.1cm}\rm MHz \hspace{0.15cm} \underline{= 0.5\hspace{0.1cm} MHz} .$$


(6)  The condition  $\sigma_y = \sigma_x$  is equivalent to  $\varphi_y(\tau = 0)= \varphi_x(\tau = 0)$.

  • Moreover,  since  $\nabla \tau_y = 3 \cdot \nabla \tau_x$  is given,  therefore  ${\it \Phi}_{y}(f= 0) = 3 \cdot {\it \Phi}_{x}(f= 0)$  must also hold.
  • From this we obtain:
$$H_{\rm 0} = \sqrt{\frac{\it \Phi_y (f \rm = 0)}{\it \Phi_x (f = \rm 0)}} = \sqrt {3}\hspace{0.15cm} \underline{=1.732}.$$