Difference between revisions of "Aufgaben:Exercise 3.3: Moments for Cosine-square PDF"

From LNTwww
Line 3: Line 3:
 
}}
 
}}
  
[[File:P_ID119__Sto_A_3_3.png|right|frame|Cosine&ndash;Rectangular PDF <br>and a similar PDF]]
+
[[File:P_ID119__Sto_A_3_3.png|right|frame|Squared cosine PDF <br>and a similar PDF]]
 
As in&nbsp; [[Aufgaben:Exercise_3.1:_cos²-PDF_and_PDF_with_Dirac_Functions|Exercise 3.1]]&nbsp; and&nbsp; [[Aufgaben:Exercise_3.2:_cos²-CDF_and_CDF_with_Step_Functions|Exercise 3. 2]]&nbsp; we consider the random variable restricted to the range of values from&nbsp; $-2$&nbsp; to&nbsp; $+2$&nbsp; with the following PDF in this section:
 
As in&nbsp; [[Aufgaben:Exercise_3.1:_cos²-PDF_and_PDF_with_Dirac_Functions|Exercise 3.1]]&nbsp; and&nbsp; [[Aufgaben:Exercise_3.2:_cos²-CDF_and_CDF_with_Step_Functions|Exercise 3. 2]]&nbsp; we consider the random variable restricted to the range of values from&nbsp; $-2$&nbsp; to&nbsp; $+2$&nbsp; with the following PDF in this section:
 
:$$f_x(x)= {1}/{2}\cdot \cos^2({\pi}/{4}\cdot { x}).$$
 
:$$f_x(x)= {1}/{2}\cdot \cos^2({\pi}/{4}\cdot { x}).$$
Line 11: Line 11:
  
 
*Both density functions are shown in the graph.  
 
*Both density functions are shown in the graph.  
*Outside the ranges&nbsp; $-2 < x < +2$ &nbsp;resp.&nbsp; $0 < x < +2$&nbsp; respectively&nbsp; $f_x(x) = 0$ &nbsp;resp.&nbsp; $f_y(y) = 0$ holds.  
+
*Outside the ranges&nbsp; $-2 < x < +2$ &nbsp;resp.&nbsp; $0 < x < +2$&nbsp;, the followings holds: &nbsp; $f_x(x) = 0$ &nbsp;resp.&nbsp; $f_y(y) = 0$.
*Both random magnitudes&ouml;&ahead;can be taken as (normalized) instantaneous values of the associated random signals&nbsp; $x(t)$&nbsp; or&nbsp; $y(t)$&nbsp; respectively.
+
*Both random variables can be taken as (normalized) instantaneous values of the associated random signals&nbsp; $x(t)$&nbsp; or&nbsp; $y(t)$&nbsp; respectively.
  
  
Line 29: Line 29:
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welche der folgenden Aussagen treffen bei jeder beliebigen WDF&nbsp; $f_x(x)$&nbsp; zu?  
+
{Which of the following statements are true for any given PDF&nbsp; $f_x(x)$&nbsp;?  
<br>Verwendet sind folgende Größen: &nbsp; linearer Mittelwert&nbsp; $m_1$,&nbsp; quadratischer Mittelwert&nbsp; $m_2$,&nbsp; Varianz&nbsp; $\sigma^2$.
+
<br>Used are the following quantities: &nbsp; linear mean&nbsp; $m_1$,&nbsp; root mean square&nbsp; $m_2$,&nbsp; variance&nbsp; $\sigma^2$.
 
|type="[]"}
 
|type="[]"}
- $m_2 = 0$,&nbsp;&nbsp;&nbsp;falls &nbsp; $m_1 \ne 0$.
+
- $m_2 = 0$,&nbsp;&nbsp;if &nbsp; $m_1 \ne 0$.
- $m_2 = 0$,&nbsp;&nbsp;&nbsp;falls &nbsp; $m_1 = 0$.
+
- $m_2 = 0$,&nbsp;&nbsp;if &nbsp; $m_1 = 0$.
+ $m_1 = 0$,&nbsp;&nbsp;&nbsp;falls &nbsp; $m_2 = 0$.
+
+ $m_1 = 0$,&nbsp;&nbsp;if &nbsp; $m_2 = 0$.
+ $m_2 > \sigma^2$,&nbsp;&nbsp;&nbsp;falls &nbsp; $m_1 \ne 0$.
+
+ $m_2 > \sigma^2$,&nbsp;&nbsp;if &nbsp; $m_1 \ne 0$.
+ $m_1 = 0$,&nbsp;&nbsp;&nbsp;falls &nbsp; $f_x(-x) = f_x(x)$.
+
+ $m_1 = 0$,&nbsp;&nbsp;&nbsp;if &nbsp; $f_x(-x) = f_x(x)$.
- $f_x(-x) = f_x(x)$,&nbsp;&nbsp;&nbsp;falls &nbsp; $m_1 = 0$.
+
- $f_x(-x) = f_x(x)$,&nbsp;&nbsp;&nbsp;if &nbsp; $m_1 = 0$.
  
  
{Wie gro&szlig; ist der Gleichanteil (lineare Mittelwert) des Signals&nbsp; $x(t)$?
+
{How large is the DC component (linear mean) of the signal&nbsp; $x(t)$?
 
|type="{}"}
 
|type="{}"}
 
$m_x \ = \ $ { 0. }
 
$m_x \ = \ $ { 0. }
  
  
{Wie gro&szlig; ist der Effektivwert des Signals&nbsp; $x(t)$?
+
{What is the rms value (Effektivwert?) of the signal&nbsp; $x(t)$?
 
|type="{}"}
 
|type="{}"}
$\sigma_x \ = \ $ { 0.722 3% }
+
$\sigma_x \ = \ $ { 0.722 3% }
  
  
{Die Zufallsgr&ouml;&szlig;e&nbsp; $y$&nbsp; lässt sich aus&nbsp; $x$&nbsp; ableiten. Welche Zuordnung gilt?
+
{The random variable&nbsp; $y$&nbsp; can be derived from&nbsp; $x$&nbsp;. Which assignment is valid?
 
|type="()"}
 
|type="()"}
 
+ $y = 1+x/2.$
 
+ $y = 1+x/2.$
Line 60: Line 60:
  
  
{Wie gro&szlig; ist der Gleichanteil des Signals&nbsp; $y(t)$?
+
{How large is the DC component of the signal $y(t)$?
 
|type="{}"}
 
|type="{}"}
$m_y\ = \ $   { 1 3% }
+
$m_y\ = $ { 1 3% }
  
  
{Wie gro&szlig; ist der Effektivwert des Signals&nbsp; $y(t)$?
+
{What is the rms value (Effektivwert) of the signal&nbsp; $y(t)$?
 
|type="{}"}
 
|type="{}"}
$\sigma_y\ = \ $ { 0.361 3% }
+
$\sigma_y\ = \ $ { 0.361 3% }
 
 
  
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
 
'''(1)'''&nbsp; Unter allen Umst&auml;nden richtig sind <u>die Aussagen 3, 4 und 5</u>:
 
'''(1)'''&nbsp; Unter allen Umst&auml;nden richtig sind <u>die Aussagen 3, 4 und 5</u>:
Line 95: Line 94:
  
 
*Mit der Beziehung&nbsp; $\cos^2(\alpha) = 0.5 \cdot \big[1 + \cos(2\alpha)\big]$&nbsp; folgt daraus:
 
*Mit der Beziehung&nbsp; $\cos^2(\alpha) = 0.5 \cdot \big[1 + \cos(2\alpha)\big]$&nbsp; folgt daraus:
:$$\sigma_x^2=\int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{\rm 2} \hspace{0.1cm}{\rm d}x + \int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{2}\cdot \cos({\pi}/{\rm 2}\cdot\it x) \hspace{0.1cm} {\rm d}x.$$
+
:$$\sigma_x^2=\int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{\rm 2} \hspace{0.1cm}{\rm d}x + \int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{2}\cdot \cos({\pi}/{\rm 2}\cdot\it x) \hspace{0.1cm} {\rm d}x.$$
  
*Diese beiden Standardintegrale findet man in Tabellen. Man erh&auml;lt mit&nbsp; $a = \pi/2$:
+
*These two standard integrals can be found in tables. One obtains with&nbsp; $a = \pi/2$:
:$$\sigma_x^{\rm 2}=\left[\frac{x^{\rm 3}}{\rm 6} + \frac{x}{a^2}\cdot {\cos}(a x) + \left( \frac{x^{\rm2}}{{\rm2}a} - \frac{1}{a^3} \right) \cdot \sin(a \cdot x)\right]_{x=0}^{x=2} \hspace{0.5cm}  
+
:$$\sigma_x^{\rm 2}=\left[\frac{x^{\rm 3}}{\rm 6} + \frac{x}{a^2}\cdot {\cos}(a x) + \left( \frac{x^{\rm2}}{{\rm2}a} - \frac{1}{a^3} \right) \cdot \sin(a \cdot x)\right]_{x=0}^{x=2} \hspace{0.5cm}  
\Rightarrow \hspace{0.5cm} \sigma_{x}^{\rm 2}=\frac{\rm 4}{\rm 3}-\frac{\rm 8}{\rm \pi^2}\approx 0.524\hspace{0.5cm} \Rightarrow \hspace{0.5cm}\sigma_x \hspace{0.15cm}\underline{\approx 0.722}.$$
+
\rightarrow \hspace{0.5cm} \sigma_{x}^{\rm 2}=\frac{\rm 4}{\rm 3}-\frac{\rm 8}{\rm \pi^2}\approx 0.524\hspace{0.5cm} \Rightarrow \hspace{0.5cm}\sigma_x \hspace{0.15cm}\underline{\approx 0.722}.$$
  
  
  
'''(4)'''&nbsp; Richtig ist der <u>erstgenannte Vorschlag</u>:
+
'''(4)'''&nbsp; Correct is the <u>first mentioned suggestion</u>:
* Die Variante&nbsp; $y = 2x$&nbsp; w&uuml;rde eine zwischen&nbsp; $-4$&nbsp; und&nbsp; $+4$&nbsp; verteilte Zufallsgr&ouml;&szlig;e liefern.  
+
*The variant&nbsp; $y = 2x$&nbsp; would yield a random size distributed between&nbsp; $-4$&nbsp; and&nbsp; $+4$&nbsp; .  
*Beim letzten Vorschlag&nbsp; $y = x/2-1$&nbsp; w&auml;re der Mittelwert&nbsp; $m_y = -1$.
+
*In the last proposition&nbsp; $y = x/2-1$&nbsp; the mean&nbsp; $m_y = -1$ would w&auml;re.
  
  
  
'''(5)'''&nbsp; Aus der Grafik auf dem Angabenblatt ist bereits offensichtlich, dass&nbsp; $m_y \hspace{0.15cm}\underline{=+1}$&nbsp; gelten muss.
+
'''(5)'''&nbsp; From the graph on the specification sheet it is already obvious that&nbsp; $m_y \hspace{0.15cm}\underline{=+1}$&nbsp; must hold.
  
  
  
'''(6)'''&nbsp; Der Mittelwert &auml;ndert nichts an der Varianz und an der Streuung.  
+
'''(6)'''&nbsp; The mean value &auml;does not change the variance and dispersion.  
*Durch die Stauchung um den Faktor&nbsp; $2$&nbsp; wird die Streuung gegen&uuml;ber Teilaufgabe&nbsp; '''(3)'''&nbsp; ebenfalls um diesen Faktor kleiner:
+
*By compressing by the factor&nbsp; $2$&nbsp; the scatter becomes smaller against&uuml;ber Teilaufgabe&nbsp; '''(3)'''&nbsp; also by this factor:
 
:$$\sigma_y=\sigma_x/\rm 2\hspace{0.15cm}\underline{\approx 0.361}.$$
 
:$$\sigma_y=\sigma_x/\rm 2\hspace{0.15cm}\underline{\approx 0.361}.$$
 
{{ML-Fuß}}
 
{{ML-Fuß}}

Revision as of 19:11, 27 December 2021

Squared cosine PDF
and a similar PDF

As in  Exercise 3.1  and  Exercise 3. 2  we consider the random variable restricted to the range of values from  $-2$  to  $+2$  with the following PDF in this section:

$$f_x(x)= {1}/{2}\cdot \cos^2({\pi}/{4}\cdot { x}).$$

Next to this, we consider a second random variable  $y$ that returns only values between  $0$  and  $2$  with the following PDF:

$$f_y(y)=\sin^2({\pi}/{2}\cdot y).$$
  • Both density functions are shown in the graph.
  • Outside the ranges  $-2 < x < +2$  resp.  $0 < x < +2$ , the followings holds:   $f_x(x) = 0$  resp.  $f_y(y) = 0$.
  • Both random variables can be taken as (normalized) instantaneous values of the associated random signals  $x(t)$  or  $y(t)$  respectively.





Hints:

  • To solve this problem, you can use the following indefinite integral:
$$\int x^{2}\cdot {\cos}(ax)\,{\rm d}x=\frac{2 x}{ a^{ 2}}\cdot \cos(ax)+ \left [\frac{x^{\rm 2}}{\it a} - \frac{\rm 2}{\it a^{\rm 3}} \right ]\cdot \rm sin(\it ax \rm ) .$$


Questions

1

Which of the following statements are true for any given PDF  $f_x(x)$ ?
Used are the following quantities:   linear mean  $m_1$,  root mean square  $m_2$,  variance  $\sigma^2$.

$m_2 = 0$,  if   $m_1 \ne 0$.
$m_2 = 0$,  if   $m_1 = 0$.
$m_1 = 0$,  if   $m_2 = 0$.
$m_2 > \sigma^2$,  if   $m_1 \ne 0$.
$m_1 = 0$,   if   $f_x(-x) = f_x(x)$.
$f_x(-x) = f_x(x)$,   if   $m_1 = 0$.

2

How large is the DC component (linear mean) of the signal  $x(t)$?

$m_x \ = \ $

3

What is the rms value (Effektivwert?) of the signal  $x(t)$?

$\sigma_x \ = \ $

4

The random variable  $y$  can be derived from  $x$ . Which assignment is valid?

$y = 1+x/2.$
$y = 2x.$
$y = x/2-1.$

5

How large is the DC component of the signal $y(t)$?

$m_y\ = $

6

What is the rms value (Effektivwert) of the signal  $y(t)$?

$\sigma_y\ = \ $


Solution

(1)  Unter allen Umständen richtig sind die Aussagen 3, 4 und 5:

  • Die erste Aussage ist nie erfüllt, wie aus dem  Satz von Steiner  ersichtlich ist.
  • Die zweite Aussage gilt nur im (trivialen) Sonderfall  $x = 0$.


Es gibt aber auch mittelwertfreie Zufallsgrößen mit unsymmetrischer WDF.

  • Das bedeutet:  Die Aussage 6 trifft nicht immer zu.


(2)  Aufgrund der WDF-Symmetrie bezüglich  $x = 0$  ergibt sich für den linearen Mittelwert  $m_x \hspace{0.15cm}\underline{= 0}$.


(3)  Der Effektivwert des Signals  $x(t)$  ist gleich der Streuung  $\sigma_x$  bzw. gleich der Wurzel aus der Varianz  $\sigma_x^2$.

  • Da die Zufallsgröße  $x$  den Mittelwert  $m_x {= 0}$  aufweist, ist die Varianz nach dem Satz von Steiner gleich dem quadratischen Mittelwert.
  • Dieser wird in Zusammenhang mit Signalen auch als die Leistung  $($bezogen auf  $1 \ \rm \Omega)$  bezeichnet. Somit gilt:
$$\sigma_x^{\rm 2}=\int_{-\infty}^{+\infty}x^{\rm 2}\cdot f_x(x)\hspace{0.1cm}{\rm d}x=2 \cdot \int_{\rm 0}^{\rm 2} x^2/2 \cdot \cos^2({\pi}/4\cdot\it x)\hspace{0.1cm} {\rm d}x.$$
  • Mit der Beziehung  $\cos^2(\alpha) = 0.5 \cdot \big[1 + \cos(2\alpha)\big]$  folgt daraus:
$$\sigma_x^2=\int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{\rm 2} \hspace{0.1cm}{\rm d}x + \int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{2}\cdot \cos({\pi}/{\rm 2}\cdot\it x) \hspace{0.1cm} {\rm d}x.$$
  • These two standard integrals can be found in tables. One obtains with  $a = \pi/2$:
$$\sigma_x^{\rm 2}=\left[\frac{x^{\rm 3}}{\rm 6} + \frac{x}{a^2}\cdot {\cos}(a x) + \left( \frac{x^{\rm2}}{{\rm2}a} - \frac{1}{a^3} \right) \cdot \sin(a \cdot x)\right]_{x=0}^{x=2} \hspace{0.5cm} \rightarrow \hspace{0.5cm} \sigma_{x}^{\rm 2}=\frac{\rm 4}{\rm 3}-\frac{\rm 8}{\rm \pi^2}\approx 0.524\hspace{0.5cm} \Rightarrow \hspace{0.5cm}\sigma_x \hspace{0.15cm}\underline{\approx 0.722}.$$


(4)  Correct is the first mentioned suggestion:

  • The variant  $y = 2x$  would yield a random size distributed between  $-4$  and  $+4$  .
  • In the last proposition  $y = x/2-1$  the mean  $m_y = -1$ would wäre.


(5)  From the graph on the specification sheet it is already obvious that  $m_y \hspace{0.15cm}\underline{=+1}$  must hold.


(6)  The mean value ädoes not change the variance and dispersion.

  • By compressing by the factor  $2$  the scatter becomes smaller againstüber Teilaufgabe  (3)  also by this factor:
$$\sigma_y=\sigma_x/\rm 2\hspace{0.15cm}\underline{\approx 0.361}.$$