Aufgaben:Exercise 3.1Z: Triangular PDF - Revision history

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@@ Line 68: / Line 68: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:StS_Z_3_1_bc_version2.png|right|frame|Height and area of triangular PDF]]
+[[File:EN_Sto_Z_3_1_bc.png|right|frame|Height and area of triangular PDF]]
 '''(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

← Older revision		Revision as of 14:44, 2 January 2022
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−	{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/~~Probability_Density_Function_(PDF)~~	+	{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Probability_Density_Function
	}}		}}

@@ Line 22: / Line 22: @@
 Hints:
-*The task belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Probability_Density_Function_(PDF)|Probability Density Function]].
+*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".

@@ Line 10: / Line 10: @@
 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)
-$$y(t)=\left\{\begin{array}{*{4}{c}} -2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} x(t)<-2\hspace{0.05cm} {\rm V} , \\ x(t) &  {\rm if}\hspace{0.1cm}-2\hspace{0.05cm} {\rm V} \le  x(t)\le +2\hspace{0.05cm} {\rm V},  \\ +2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} {\it x}({\it t})>+2\hspace{0.05cm} {\rm V},  \\\end{array}\right.$$
+:$$y(t)=\left\{\begin{array}{*{4}{c}} -2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} x(t)<-2\hspace{0.05cm} {\rm V} , \\ x(t) &  {\rm if}\hspace{0.1cm}-2\hspace{0.05cm} {\rm V} \le  x(t)\le +2\hspace{0.05cm} {\rm V},  \\ +2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} {\it x}({\it t})>+2\hspace{0.05cm} {\rm V},  \\\end{array}\right.$$
 so the signal&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 />
@@ Line 37: / Line 37: @@
 {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="{}"}
-${\rm Pr}(|\hspace{0.05cm}x\hspace{0.05cm}| < 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% }

@@ Line 69: / Line 69: @@
 {{ML-Kopf}}
 [[File:StS_Z_3_1_bc_version2.png|right|frame|Height and area of triangular PDF]]
-'''(1)'''&nbsp; The area under the PDF must always yield the value&nbsp; $1$&nbsp;. It follows that:
+'''(1)'''&nbsp; The area under the PDF must always yield the value&nbsp; $1$.&nbsp; It follows that:
 :$${A}/{ 2}\cdot {4\hspace{0.05cm}\rm V}=1\hspace{0.5cm}\Rightarrow\hspace{0.5cm} A
 \hspace{0.15cm}\underline{=\rm 0.5\;{1}/{V}}.$$
@@ Line 82: / Line 82: @@
-'''(3)'''&nbsp; With&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$ one obtains the PDF shown on the right.
+'''(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, which can be determined, for example, using the rectangle of equal area:
+*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, this probability is by definition zero &nbsp; &rArr; &nbsp; ${\rm Pr}(x =2\hspace{0.05cm} {\rm V}) \;\underline {= 0}$.
+'''(4)'''&nbsp; Since&nbsp; $x$&nbsp; represents a continuous random variable,&nbsp; this probability is by definition zero &nbsp; &rArr; &nbsp; ${\rm Pr}(x =2\hspace{0.05cm} {\rm V}) \;\underline {= 0}$.
@@ Line 97: / Line 97: @@
 *The PDF&nbsp; $f_y(y)$&nbsp; includes a continuous component (drawn in blue),
-*but also the (red) Dirac function at&nbsp; $y = +2\hspace{0.05cm} {\rm V}$&nbsp; with weight&nbsp; ${\rm Pr}(x >2\hspace{0.05cm} {\rm V})$.
+*but also the (red) Dirac delta function at&nbsp; $y = +2\hspace{0.05cm} {\rm V}$&nbsp; with weight&nbsp; ${\rm Pr}(x >2\hspace{0.05cm} {\rm V})$.
-'''(6)'''&nbsp; Opposite is the probability density of the random variable&nbsp; $y$&nbsp;.
-*From the right figure for the subtask&nbsp; '''(3)'''&nbsp; one can see the relation:
+'''(6)'''&nbsp; The graphic on the right shows the probability density function of the random variable&nbsp; $y$.
 :$${\rm Pr}( y=2\hspace{0.05cm} {\rm V}) = {\rm Pr}( x> 2\hspace{0.05cm} {\rm V}) =  \frac{1}{2}\cdot\frac{1}{6\hspace{0.05cm} {\rm V}}\cdot2{\hspace{0.05cm} {\rm V}} = {1}/{6}\hspace{0.15cm}\underline{=0.167}.$$
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID109__Sto_Z_3_1.png|right|frame|Triangular PDF and characteristic curve&nbsp; $y(x)$]]
+[[File:P_ID109__Sto_Z_3_1.png|right|frame|Triangular PDF and <br>characteristic curve&nbsp; $y(x)$]]
-We consider a continuous random variable&nbsp; $x$&nbsp; with the PDF outlined above.&nbsp;
+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, the maximum value&nbsp; $x_{\rm max}$&nbsp; is a free parameter, allowing values between&nbsp; $+2\hspace{0.05cm}\rm V$&nbsp; and&nbsp; $+4\hspace{0.05cm} \rm V$&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; $+4\hspace{0.05cm} \rm V$.
@@ Line 12: / Line 12: @@
 $$y(t)=\left\{\begin{array}{*{4}{c}} -2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} x(t)<-2\hspace{0.05cm} {\rm V} , \\ x(t) &  {\rm if}\hspace{0.1cm}-2\hspace{0.05cm} {\rm V} \le  x(t)\le +2\hspace{0.05cm} {\rm V},  \\ +2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} {\it x}({\it t})>+2\hspace{0.05cm} {\rm V},  \\\end{array}\right.$$
-so the signal&nbsp; $y(t)$&nbsp; or the new random variable&nbsp; $y$, which is considered in the last two subquestions&nbsp; '''(5)'''&nbsp; and&nbsp; '''(6)'''&nbsp; is obtained. <br />
+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, set&nbsp; $x_{\rm max} = 4\hspace{0.05cm} {\rm V} $&nbsp;.
+* For all other subtasks,&nbsp; set&nbsp; $x_{\rm max} = 4\hspace{0.05cm} {\rm V} $&nbsp;.
@@ Line 26: / Line 24: @@
 *The task belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Probability_Density_Function_(PDF)|Probability Density Function]].
-*The topic of this chapter is illustrated with examples in the (German) learning video&nbsp; [[Wahrscheinlichkeit_und_WDF_(Lernvideo)|Wahrscheinlichkeit und WDF]] $\Rightarrow$ Probability and PDF.
+*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".
@@ Line 32: / Line 30: @@
 <quiz display=simple>
-{Let&nbsp; $x_{\rm max} = +2\hspace{0.05cm} {\rm V}$.&nbsp; Calculate the parameter&nbsp; $A = f_x(0)$.
+{Let be&nbsp; $x_{\rm max} = +2\hspace{0.05cm} {\rm V}$.&nbsp; Calculate the parameter&nbsp; $A = f_x(0)$.
 |type="{}"}
 $A \ = \ $ { 0.5 3% } $\ \rm 1/V$.
-{Further, let&nbsp; $x_{\rm max} = +2\hspace{0.05cm} {\rm V}$.&nbsp; With what probability is&nbsp; $|x(t)|$&nbsp; less than&nbsp; $x_{\rm max}/2$?
+{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="{}"}
 ${\rm Pr}(|\hspace{0.05cm}x\hspace{0.05cm}| < 2\hspace{0.05cm} {\rm V}) \ = \ $ { 0.75 3% }
-{Now let&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; $+3\hspace{0.05cm} {\rm V}$&nbsp; ?
+{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="{}"}
 ${\rm Pr}(1\hspace{0.05cm} {\rm V} < x <3\hspace{0.05cm} {\rm V}) \ = \ $ { 0.333 3% }
-{ Let further&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.&nbsp; What is the probability that&nbsp; $x$&nbsp; is exactly equal to $+2\hspace{0.05cm} {\rm V}$&nbsp;?
+{ 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="{}"}
 ${\rm Pr}(x =2\hspace{0.05cm} {\rm V})\ = \ $ { 0. }
-{ Let further&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.&nbsp; Which of the following statements is true?
+{ Let further be&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$.&nbsp; Which of the following statements is true?
 |type="[]"}
 - $y$&nbsp; is a continuous random variable.
@@ Line 59: / Line 57: @@
-{What is the probability with&nbsp; $x_{\rm max} = +4\hspace{0.05cm} {\rm V}$ that&nbsp; $y$&nbsp; is exactly equal&nbsp; $+2\hspace{0.05cm} {\rm V}$&nbsp;?
+{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="{}"}
 ${\rm Pr}(y =2\hspace{0.05cm} {\rm V})\ = \ $ { 0.167 3% }

@@ Line 4: / Line 4: @@
 [[File:P_ID109__Sto_Z_3_1.png|right|frame|Dreieck-WDF und Kennlinie&nbsp; $y(x)$]]
-Wir betrachten eine kontinuierliche Zufallsgr&ouml;&szlig;e&nbsp; $x$&nbsp; mit der oben skizzierten WDF.&nbsp;
+We consider a continuous random variable&nbsp; $x$&nbsp; with the PDF outlined above.&nbsp;
-*Der Minimalwert des Signals ist&nbsp; $x_{\rm min} = -2\hspace{0.05cm} {\rm V}$.&nbsp;
+*The minimum value of the signal is&nbsp; $x_{\rm min} = -2\hspace{0.05cm} {\rm V}$.&nbsp;
-*Dagegen ist der maximale Wert&nbsp; $x_{\rm max}$&nbsp; ein freier Parameter, der Werte zwischen&nbsp; $+2\hspace{0.05cm}\rm V$&nbsp; und&nbsp; $+4\hspace{0.05cm} \rm V$&nbsp; annehmen kann.
+*On the other hand, the maximum value&nbsp; $x_{\rm max}$&nbsp; is a free parameter, allowing values between&nbsp; $+2\hspace{0.05cm}\rm V$&nbsp; and&nbsp; $+4\hspace{0.05cm} \rm V$&nbsp;.
-Die Zufallsgr&ouml;&szlig;e&nbsp; $x$&nbsp; soll hier als der Momentanwert eines Zufallssignals aufgefasst werden.&nbsp; Gibt man dieses Signal&nbsp; $x(t)$&nbsp; auf einen Amplitudenbegrenzer mit der Kennlinie&nbsp; (siehe untere Skizze)
+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)
-$$y(t)=\left\{\begin{array}{*{4}{c}} -2\hspace{0.05cm} {\rm V} & {\rm falls}\hspace{0.1cm} x(t)<-2\hspace{0.05cm} {\rm V} , \\ x(t) & {\rm falls}\hspace{0.1cm}-2\hspace{0.05cm} {\rm V} \le x(t)\le +2\hspace{0.05cm} {\rm V}, \\ +2\hspace{0.05cm} {\rm V} & {\rm falls}\hspace{0.1cm} {\it x}({\it t})>+2\hspace{0.05cm} {\rm V}, \\\end{array}\right.$$
+$$y(t)=\left\{\begin{array}{*{4}{c}} -2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} x(t)<-2\hspace{0.05cm} {\rm V} , \\ x(t) &  {\rm if}\hspace{0.1cm}-2\hspace{0.05cm} {\rm V} \le  x(t)\le +2\hspace{0.05cm} {\rm V},  \\ +2\hspace{0.05cm} {\rm V} & {\rm if}\hspace{0.1cm} {\it x}({\it t})>+2\hspace{0.05cm} {\rm V},  \\\end{array}\right.$$
-so entsteht das Signal&nbsp; $y(t)$&nbsp; bzw. die neue Zufallsgr&ouml;&szlig;e&nbsp; $y$, die in den beiden letzten Teilfragen&nbsp; '''(5)'''&nbsp; und&nbsp; '''(6)'''&nbsp; betrachtet wird. <br />
+so the signal&nbsp; $y(t)$&nbsp; or the new random size&ouml;&aerospace;e&nbsp; $y$, 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} $.