Difference between revisions of "Aufgaben:Exercise 3.1Z: Triangular PDF"

From LNTwww
 
(19 intermediate revisions by 4 users not shown)
Line 1: Line 1:
  
{{quiz-Header|Buchseite=Stochastische Signaltheorie/Wahrscheinlichkeitsdichtefunktion (WDF)
+
{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Probability_Density_Function
 
}}
 
}}
  
[[File:P_ID109__Sto_Z_3_1.png|right|Dreieck-WDF und Kennlinie]]
+
[[File:P_ID109__Sto_Z_3_1.png|right|frame|Triangular PDF and <br>characteristic curve&nbsp; $y(x)$]]
Wir betrachten eine kontinuierliche Zufallsgr&ouml;&szlig;e $x$ mit der oben skizzierten WDF. Der Minimalwert des Signals ist $x_{min} = -2\hspace{0.05cm} {\rm V}$. Dagegen ist der maximale Wert $x_{max}$ ein freier Parameter, der Werte zwischen $2\hspace{0.05cm}\rm V$ und $4\hspace{0.05cm} \rm V$ annehmen kann.
+
We consider a continuous random variable&nbsp; $x$&nbsp; with the PDF outlined on the right.&nbsp;
 +
*The minimum value of the signal is&nbsp; $x_{\rm min} = -2\hspace{0.05cm} {\rm V}$.&nbsp;
 +
*On the other hand,&nbsp; the maximum value&nbsp; $x_{\rm max}$&nbsp; is a free parameter,&nbsp; allowing values between&nbsp; $+2\hspace{0.05cm}\rm V$&nbsp; and&nbsp; $+4\hspace{0.05cm} \rm V$.
  
Die Zufallsgr&ouml;&szlig;e $x$ soll hier als der Momentanwert eines Zufallssignals aufgefasst werden. Gibt man dieses Signal $x(t)$ auf einen Amplitudenbegrenzer mit der Kennlinie (siehe untere Skizze)
 
$$y(t)=\left\{\begin{array}{*{4}{c}} -2\hspace{0.05cm} {\rm V} & {\rm falls}\hspace{0.1cm} x(t)<-2\hspace{0.05cm} {\rm V} , \\ x(t) &  {\rm falls}\hspace{0.1cm}-2\hspace{0.05cm} {\rm V} \le  x(t)\le +2\hspace{0.05cm} {\rm V},  \\ +2\hspace{0.05cm} {\rm V} & {\rm falls}\hspace{0.1cm} {\it x}({\it t})>+2\hspace{0.05cm} {\rm V},  \\\end{array}\right.$$
 
  
so entsteht das Signal $y(t)$ bzw. die neue Zufallsgr&ouml;&szlig;e $y$, die in den beiden letzten Teilfragen (5) und (6) betrachtet wird. <br />
+
The random variable&nbsp; $x$&nbsp; is to be understood here as the instantaneous value of a random signal.&nbsp; If this signal&nbsp; $x(t)$&nbsp; is applied to an amplitude limiter with the characteristic curve&nbsp; (see sketch below)
 +
:$$y(t)=\left\{\begin{array}{*{4}{c}} -2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} x(t)<-2\hspace{0.05cm} {\rm V} , \\ x(t) &  {\rm if}\hspace{0.1cm}-2\hspace{0.05cm} {\rm V} \le  x(t)\le +2\hspace{0.05cm} {\rm V},  \\ +2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} {\it x}({\it t})>+2\hspace{0.05cm} {\rm V},  \\\end{array}\right.$$
  
F&uuml;r die Teilaufgaben (1) und (2) gelte $x_{\rm max} = 2\hspace{0.05cm} {\rm V} $; f&uuml;r alle weiteren Teilaufgaben ist $x_{\rm max} = 4\hspace{0.05cm} {\rm V} $ zu setzen.
+
so the signal&nbsp; $y(t)$&nbsp; or the new random variable&nbsp; $y$,&nbsp; which is considered in the last two subquestions&nbsp; '''(5)'''&nbsp; and&nbsp; '''(6)'''&nbsp; is obtained. <br />
  
 +
*For the subtasks&nbsp; '''(1)'''&nbsp; and&nbsp; '''(2)'''&nbsp; apply&nbsp; $x_{\rm max} = 2\hspace{0.05cm} {\rm V} $.
 +
* For all other subtasks,&nbsp; set&nbsp; $x_{\rm max} = 4\hspace{0.05cm} {\rm V} $&nbsp;.
  
''Hinweise:''
 
*Die Aufgabe gehört zum  Kapitel [[Stochastische_Signaltheorie/Wahrscheinlichkeitsdichtefunktion_(WDF)|Wahrscheinlichkeitsdichtefunktion]].
 
*Sollte die Eingabe des Zahlenwertes &bdquo;0&rdquo; erforderlich sein, so geben Sie bitte &bdquo;0.&rdquo; ein.
 
*Eine Zusammenfassung der hier behandelten Thematik bietet das folgende Lernvideo:<br />
 
:[[Wahrscheinlichkeit und Wahrscheinlichkeitsdichtefunktion]]
 
  
  
===Fragebogen===
+
 
 +
 
 +
Hints:
 +
*The task belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Probability_Density_Function|Probability Density Function]].
 +
 +
*The topic of this chapter is illustrated with examples in the&nbsp; (German language)&nbsp; learning video&nbsp; <br>&nbsp; &nbsp;[[Wahrscheinlichkeit_und_WDF_(Lernvideo)|"Wahrscheinlichkeit und WDF"]] &nbsp; $\Rightarrow$ &nbsp; "Probability and PDF".
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Es sei $x_{\rm max} = 2\hspace{0.05cm} {\rm V}$. Berechnen Sie den Parameter $A = f_x(0)$.
+
{Let be&nbsp; $x_{\rm max} = +2\hspace{0.05cm} {\rm V}$.&nbsp; Calculate the parameter&nbsp; $A = f_x(0)$.
 
|type="{}"}
 
|type="{}"}
$x_{\rm max} = 2\hspace{0.05cm} {\rm V}$: &nbsp; $A \ =$ { 0.5 3% } $\ \rm 1/V$
+
$A \ = \ $ { 0.5 3% } $\ \rm 1/V$.
  
  
{Mit welcher Wahrscheinlichkeit ist $|x(t)|$ kleiner als $x_{\rm max}/2$?
+
{Let&nbsp; $x_{\rm max} = +2\hspace{0.05cm} {\rm V}$.&nbsp; What is the probability that&nbsp; $|x(t)|$&nbsp; less than&nbsp; $x_{\rm max}/2$?
 
|type="{}"}
 
|type="{}"}
$x_{\rm max} = 2\hspace{0.05cm} {\rm V}$: &nbsp; ${\rm Pr}(|x| < 2\hspace{0.05cm} {\rm V}) \ = $ { 0.75 3% }
+
${\rm Pr}(|\hspace{0.05cm}x\hspace{0.05cm}| < 1\hspace{0.05cm} {\rm V}) \ = \ $ { 0.75 3% }
  
  
{Nun gelte $x_{\rm max} = 4\hspace{0.05cm} {\rm V}$. Wie gro&szlig; ist die Wahrscheinlichkeit, dass $x$ zwischen $1\hspace{0.05cm} {\rm V}$ und $3\hspace{0.05cm} {\rm V}$ liegt?
+
{Now let be&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.&nbsp; What is the probability that&nbsp; $x$&nbsp; lies between&nbsp; $+1\hspace{0.05cm} {\rm V}$&nbsp; and&nbsp; $+3\hspace{0.05cm} {\rm V}$?
 
|type="{}"}
 
|type="{}"}
$x_{\rm max} = 4\hspace{0.05cm} {\rm V}$: &nbsp; ${\rm Pr}(1\hspace{0.05cm} {\rm V} < x <3\hspace{0.05cm} {\rm V}) \ =$ { 0.333 3% }
+
${\rm Pr}(1\hspace{0.05cm} {\rm V} < x <3\hspace{0.05cm} {\rm V}) \ = \ $ { 0.333 3% }
  
  
{Wie gro&szlig; ist die Wahrscheinlichkeit, dass $x$ genau gleich $2\hspace{0.05cm} {\rm V}$ ist?
+
{ Let further be&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.&nbsp; What is the probability that&nbsp; $x$&nbsp; is exactly equal to $+2\hspace{0.05cm} {\rm V}$?
 
|type="{}"}
 
|type="{}"}
$x_{\rm max} = 4\hspace{0.05cm} {\rm V}$: &nbsp; ${\rm Pr}(x =2\hspace{0.05cm} {\rm V})\ =$ { 0. }
+
${\rm Pr}(x =2\hspace{0.05cm} {\rm V})\ = \ $ { 0. }
  
  
{Es sei weiterhin $x_{\rm max} = 4\hspace{0.05cm} {\rm V}$. Welche der nachfolgenden Aussagen sind zutreffend?
+
{ Let further be&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.&nbsp; Which of the following statements is true?
 
|type="[]"}
 
|type="[]"}
- $y$ ist eine kontinuierliche Zufallsgr&ouml;&szlig;e.
+
- $y$&nbsp; is a continuous random variable.
- $y$ ist eine diskrete Zufallsgr&ouml;&szlig;e.
+
- $y$&nbsp; is a discrete random variable.
+ $y$ ist eine gemischt kontinuierlich-diskrete Zufallsgr&ouml;&szlig;e.
+
+ $y$&nbsp; is a mixed continuous-discrete random variable.
  
  
{Wie gro&szlig; ist die Wahrscheinlichkeit, dass $y$ genau gleich $2\hspace{0.05cm} {\rm V}$ ist?
+
{Let further be&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.&nbsp; What is the probability  that&nbsp; $y$&nbsp; is exactly equal&nbsp; $+2\hspace{0.05cm} {\rm V}$?
 
|type="{}"}
 
|type="{}"}
$x_{\rm max} = 4\hspace{0.05cm} {\rm V}$: &nbsp; ${\rm Pr}(y =2\hspace{0.05cm} {\rm V})\ =$ { 0.167 3% }
+
${\rm Pr}(y =2\hspace{0.05cm} {\rm V})\ = \ $ { 0.167 3% }
 +
 
  
  
Line 59: Line 66:
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Die Fl&auml;che unter der WDF muss stets den Wert $1$ ergeben. Daraus folgt:
+
[[File:EN_Sto_Z_3_1_bc.png|right|frame|Height and area of triangular PDF]]
$${A}/{ 2}\cdot {4\hspace{0.5cm}V}=1\hspace{0.5cm}\Rightarrow\hspace{0.5cm} A
+
'''(1)'''&nbsp; The area under the PDF must always yield the value&nbsp; $1$.&nbsp; It follows that:
 +
:$${A}/{ 2}\cdot {4\hspace{0.05cm}\rm V}=1\hspace{0.5cm}\Rightarrow\hspace{0.5cm} A
 
\hspace{0.15cm}\underline{=\rm 0.5\;{1}/{V}}.$$
 
\hspace{0.15cm}\underline{=\rm 0.5\;{1}/{V}}.$$
  
[[File:StS_Z_3_1_bc_version2.png|right|Höhe und Fläche der Dreieck-WDF]]
 
'''(2)'''&nbsp; Mit $x_{\rm max} = 2\hspace{0.05cm} {\rm V}$ ergibt sich die WDF nach der linken Grafik. Die Schraffierung markiert die gesuchte Wahrscheinlichkeit und man erhält durch einfache geometrische &Uuml;berlegungen:
 
$${\rm Pr}(|x|<\rm 1\hspace{0.05cm} V)\hspace{0.15cm}\underline{=\rm 0.75}.$$
 
  
'''(3)'''&nbsp; Mit $x_{\rm max} = 4\hspace{0.05cm} {\rm V}$ erh&auml;lt man die rechts dargestellte WDF und den Maximalwert $A = 1/(3\hspace{0.05cm} {\rm V})$. Die schraffierte Fl&auml;che gibt wieder die gesuchte Wahrscheinlichkeit an, die man zum Beispiel &uuml;ber das fl&auml;chengleiche Rechteck bestimmen kann:
 
$${\rm Pr}(1\hspace{0.05cm} {\rm V}< x<3\hspace{0.05cm} {\rm V})=\rm \frac{1}{6\hspace{0.05cm} {\rm V}}\cdot 2\hspace{0.05cm} {\rm V}=\hspace{0.15cm}\underline{0.333}.$$
 
  
'''(4)'''&nbsp; Diese Wahrscheinlichkeit ist definitionsgem&auml;&szlig; gleich null ${\rm Pr}(x =2\hspace{0.05cm} {\rm V}) \;\underline {= 0}$, da $x$ eine kontinuierliche Zufallsgr&ouml;&szlig;e darstellt.
+
'''(2)'''&nbsp; With&nbsp; $x_{\rm max} = +2\hspace{0.05cm} {\rm V}$&nbsp; the PDF is obtained according to the left graph.
 +
*The shading marks the probability we are looking for.
 +
*One obtains by simple geometric considerations:
 +
:$${\rm Pr}(|x|<\rm 1\hspace{0.05cm} V)\hspace{0.15cm}\underline{=\rm 0.75}.$$
 +
 
 +
 
 +
 
 +
'''(3)'''&nbsp; With&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$&nbsp; one obtains the PDF shown on the right.
 +
*The maximum value is now &nbsp; $A = 1/(3\hspace{0.05cm} {\rm V})$.
 +
*The shaded area again indicates the probability we are looking for,&nbsp; which can be determined using the rectangle of equal area:
 +
:$${\rm Pr}(1\hspace{0.05cm} {\rm V}< x<3\hspace{0.05cm} {\rm V})=\rm \frac{1}{6\hspace{0.05cm} {\rm V}}\cdot 2\hspace{0.05cm} {\rm V}=\hspace{0.15cm}\underline{0.333}.$$
 +
 
 +
 
 +
 
 +
'''(4)'''&nbsp; Since&nbsp; $x$&nbsp; represents a continuous random variable,&nbsp; this probability is by definition zero &nbsp; &rArr; &nbsp; ${\rm Pr}(x =2\hspace{0.05cm} {\rm V}) \;\underline {= 0}$.
 +
 
 +
 
 +
 
 +
[[File:P_ID113__Sto_Z_3_1_f.png|right|frame|Mixed continuous/discrete PDF]]
 +
'''(5)'''&nbsp; <u>Only the last statement</u> of the given answers is true:
 +
 
 +
*The PDF&nbsp; $f_y(y)$&nbsp; includes a continuous component (drawn in blue),  
 +
*but also the (red) Dirac delta function at&nbsp; $y = +2\hspace{0.05cm} {\rm V}$&nbsp; with weight&nbsp; ${\rm Pr}(x >2\hspace{0.05cm} {\rm V})$.
  
[[File:P_ID113__Sto_Z_3_1_f.png|right|Gemischt kontinuierlich/diskrete WDF]]
 
'''(5)'''&nbsp; <u>Nur die letzte Aussage</u> der vorgegebenen Antworten ist zutreffend:
 
  
Die WDF $f_y(y)$ beinhaltet einen kontinuierlichen Anteil, aber auch eine Diracfunktion an der Stelle $y = 2\hspace{0.05cm} {\rm V}$ mit dem Gewicht ${\rm Pr}(x >2\hspace{0.05cm} {\rm V})$ .
 
  
'''(6)'''&nbsp; Nebenstehend ist die Wahrscheinlichkeitsdichte der Zufallsgr&ouml;&szlig;e $y$ dargestellt. Aus der oberen rechten Abbildung zur Teilaufgabe (3) erkennt man den Zusammenhang:
+
'''(6)'''&nbsp; The graphic on the right shows the probability density function of the random variable&nbsp; $y$.  
$${\rm Pr}( y=2\hspace{0.05cm} {\rm V})) = {\rm Pr}( x> 2\hspace{0.05cm} {\rm V})) =  \frac{1}{2}\cdot\frac{1}{6\hspace{0.05cm} {\rm V}}\cdot2{\hspace{0.05cm} {\rm V}} = \hspace{-0.15cm}{1}/{6}\hspace{0.15cm}\underline{=0.167}.$$
+
*From the right figure for subtask&nbsp; '''(3)'''&nbsp; one can see the relation:
 +
:$${\rm Pr}( y=2\hspace{0.05cm} {\rm V}) = {\rm Pr}( x> 2\hspace{0.05cm} {\rm V}) =  \frac{1}{2}\cdot\frac{1}{6\hspace{0.05cm} {\rm V}}\cdot2{\hspace{0.05cm} {\rm V}} = {1}/{6}\hspace{0.15cm}\underline{=0.167}.$$
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Stochastische Signaltheorie|^3.1 Wahrscheinlichkeitsdichtefunktion^]]
+
[[Category:Theory of Stochastic Signals: Exercises|^3.1 Probability Density Function^]]

Latest revision as of 16:55, 8 February 2022

Triangular PDF and
characteristic curve  $y(x)$

We consider a continuous random variable  $x$  with the PDF outlined on the right. 

  • The minimum value of the signal is  $x_{\rm min} = -2\hspace{0.05cm} {\rm V}$. 
  • On the other hand,  the maximum value  $x_{\rm max}$  is a free parameter,  allowing values between  $+2\hspace{0.05cm}\rm V$  and  $+4\hspace{0.05cm} \rm V$.


The random variable  $x$  is to be understood here as the instantaneous value of a random signal.  If this signal  $x(t)$  is applied to an amplitude limiter with the characteristic curve  (see sketch below)

$$y(t)=\left\{\begin{array}{*{4}{c}} -2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} x(t)<-2\hspace{0.05cm} {\rm V} , \\ x(t) & {\rm if}\hspace{0.1cm}-2\hspace{0.05cm} {\rm V} \le x(t)\le +2\hspace{0.05cm} {\rm V}, \\ +2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} {\it x}({\it t})>+2\hspace{0.05cm} {\rm V}, \\\end{array}\right.$$

so the signal  $y(t)$  or the new random variable  $y$,  which is considered in the last two subquestions  (5)  and  (6)  is obtained.

  • For the subtasks  (1)  and  (2)  apply  $x_{\rm max} = 2\hspace{0.05cm} {\rm V} $.
  • For all other subtasks,  set  $x_{\rm max} = 4\hspace{0.05cm} {\rm V} $ .



Hints:

  • The topic of this chapter is illustrated with examples in the  (German language)  learning video 
       "Wahrscheinlichkeit und WDF"   $\Rightarrow$   "Probability and PDF".


Questions

1

Let be  $x_{\rm max} = +2\hspace{0.05cm} {\rm V}$.  Calculate the parameter  $A = f_x(0)$.

$A \ = \ $

$\ \rm 1/V$.

2

Let  $x_{\rm max} = +2\hspace{0.05cm} {\rm V}$.  What is the probability that  $|x(t)|$  less than  $x_{\rm max}/2$?

${\rm Pr}(|\hspace{0.05cm}x\hspace{0.05cm}| < 1\hspace{0.05cm} {\rm V}) \ = \ $

3

Now let be  $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.  What is the probability that  $x$  lies between  $+1\hspace{0.05cm} {\rm V}$  and  $+3\hspace{0.05cm} {\rm V}$?

${\rm Pr}(1\hspace{0.05cm} {\rm V} < x <3\hspace{0.05cm} {\rm V}) \ = \ $

4

Let further be  $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.  What is the probability that  $x$  is exactly equal to $+2\hspace{0.05cm} {\rm V}$?

${\rm Pr}(x =2\hspace{0.05cm} {\rm V})\ = \ $

5

Let further be  $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.  Which of the following statements is true?

$y$  is a continuous random variable.
$y$  is a discrete random variable.
$y$  is a mixed continuous-discrete random variable.

6

Let further be  $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.  What is the probability that  $y$  is exactly equal  $+2\hspace{0.05cm} {\rm V}$?

${\rm Pr}(y =2\hspace{0.05cm} {\rm V})\ = \ $


Solution

Height and area of triangular PDF

(1)  The area under the PDF must always yield the value  $1$.  It follows that:

$${A}/{ 2}\cdot {4\hspace{0.05cm}\rm V}=1\hspace{0.5cm}\Rightarrow\hspace{0.5cm} A \hspace{0.15cm}\underline{=\rm 0.5\;{1}/{V}}.$$


(2)  With  $x_{\rm max} = +2\hspace{0.05cm} {\rm V}$  the PDF is obtained according to the left graph.

  • The shading marks the probability we are looking for.
  • One obtains by simple geometric considerations:
$${\rm Pr}(|x|<\rm 1\hspace{0.05cm} V)\hspace{0.15cm}\underline{=\rm 0.75}.$$


(3)  With  $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$  one obtains the PDF shown on the right.

  • The maximum value is now   $A = 1/(3\hspace{0.05cm} {\rm V})$.
  • The shaded area again indicates the probability we are looking for,  which can be determined using the rectangle of equal area:
$${\rm Pr}(1\hspace{0.05cm} {\rm V}< x<3\hspace{0.05cm} {\rm V})=\rm \frac{1}{6\hspace{0.05cm} {\rm V}}\cdot 2\hspace{0.05cm} {\rm V}=\hspace{0.15cm}\underline{0.333}.$$


(4)  Since  $x$  represents a continuous random variable,  this probability is by definition zero   ⇒   ${\rm Pr}(x =2\hspace{0.05cm} {\rm V}) \;\underline {= 0}$.


Mixed continuous/discrete PDF

(5)  Only the last statement of the given answers is true:

  • The PDF  $f_y(y)$  includes a continuous component (drawn in blue),
  • but also the (red) Dirac delta function at  $y = +2\hspace{0.05cm} {\rm V}$  with weight  ${\rm Pr}(x >2\hspace{0.05cm} {\rm V})$.


(6)  The graphic on the right shows the probability density function of the random variable  $y$.

  • From the right figure for subtask  (3)  one can see the relation:
$${\rm Pr}( y=2\hspace{0.05cm} {\rm V}) = {\rm Pr}( x> 2\hspace{0.05cm} {\rm V}) = \frac{1}{2}\cdot\frac{1}{6\hspace{0.05cm} {\rm V}}\cdot2{\hspace{0.05cm} {\rm V}} = {1}/{6}\hspace{0.15cm}\underline{=0.167}.$$