Aufgaben:Exercise 3.5: Triangular and Trapezoidal Signal - Revision history

Bene: Text replacement - "rms value" to "standard deviation"

2022-02-17T11:11:50Z

Text replacement - "rms value" to "standard deviation"

Guenter at 12:02, 20 January 2022

2022-01-20T12:02:57Z

Guenter at 14:45, 12 January 2022

2022-01-12T14:45:13Z

Guenter at 14:44, 12 January 2022

2022-01-12T14:44:07Z

Guenter at 14:27, 12 January 2022

2022-01-12T14:27:33Z

Noah at 20:57, 29 December 2021

2021-12-29T20:57:43Z

Noah at 23:34, 28 December 2021

2021-12-28T23:34:12Z

Noah at 23:33, 28 December 2021

2021-12-28T23:33:20Z

Guenter at 10:10, 20 December 2021

2021-12-20T10:10:58Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.5: Dreieck- und Trapezsignal to Aufgaben:Exercise 3.5: Triangular and Trapezoidal Signal

2021-12-20T10:08:54Z

Guenter moved page Aufgabe 3.5: Dreieck- und Trapezsignal to Exercise 3.5: Triangular and Trapezoidal Signal

@@ Line 7: / Line 7: @@
 *The amplitude values are&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; and&nbsp; $2\hspace{0.05cm} \rm V$.
 *Let the duration of a rectangle and the distance between two successive rectangular pulses be equal to each&nbsp; $T$.
-*The random variable&nbsp; $x$&nbsp; &ndash; the instantaneous value of the rectangular signal&nbsp; $x(t)$&nbsp; &ndash; thus has the characteristic values&nbsp; (mean,&nbsp; rms value):
+*The random variable&nbsp; $x$&nbsp; &ndash; the instantaneous value of the rectangular signal&nbsp; $x(t)$&nbsp; &ndash; thus has the characteristic values&nbsp; (mean,&nbsp; standard deviation):
 :$$m_x = \sigma_x = 1\hspace{0.05cm} \rm V.$$
@@ Line 58: / Line 58: @@
-{What is the rms value of the signal&nbsp; $y_1(t)$?
+{What is the standard deviation of the signal&nbsp; $y_1(t)$?
 |type="{}"}
 $\sigma_{y1} \ = \ $ { 0.577 3% } $\ \rm V$
@@ Line 78: / Line 78: @@
-{What is the rms value of the signal&nbsp; $y_2(t)$?
+{What is the standard deviation of the signal&nbsp; $y_2(t)$?
 |type="{}"}
 $\sigma_{y2} \ = \ $ { 0.409 3% } $\ \rm V$
@@ Line 120: / Line 120: @@
 '''(4)'''&nbsp; The variance can be determined using Steiner's theorem:
 :$$4/3\hspace{0.05cm} \rm V^2 - 1\hspace{0.05cm} \rm V^2 = 1/3\hspace{0.05cm} \rm V^2.$$
-*The root of this is the standard deviation&nbsp;  (rms value)&nbsp; we are looking for: &nbsp; &nbsp;
+*The root of this is the standard deviation&nbsp;  (standard deviation)&nbsp; we are looking for: &nbsp; &nbsp;
 :$$\sigma_{y_{\rm 1}}\hspace{0.15cm}\underline{=0.577 \, \rm V}.$$
@@ Line 143: / Line 143: @@
 *The first part goes back to the continuous PDF,&nbsp; the second part to the PDF Dirac function at&nbsp; $1\hspace{0.05cm} \rm V$.
-*The Dirac function at&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; does not contribute to the power.&nbsp; It follows for the rms value:
+*The Dirac function at&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; does not contribute to the power.&nbsp; It follows for the standard deviation:
 :$$\sigma_{y_{\rm 2}} = \sqrt{{\rm 5}/{\rm 12}\rm V^2 -{1}/{4}\rm V^2}=
 \sqrt{{\rm 1}/{\rm 6}\rm V^2}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID125__Sto_A_3_5.png|right|frame|rectangle signal,&nbsp; triangle signal <br>and trapezoid signal ]]
+[[File:P_ID125__Sto_A_3_5.png|right|frame|Rectangle signal,&nbsp; triangle signal <br>and trapezoid signal ]]
 We start from the rectangular signal&nbsp; $x(t)$&nbsp; according to the upper graph.
 *The amplitude values are&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; and&nbsp; $2\hspace{0.05cm} \rm V$.

@@ Line 104: / Line 104: @@
 '''(2)'''&nbsp; The linear mean&nbsp; $m_x = 1\hspace{0.05cm} \rm V$&nbsp; can be read directly from the data sketch,&nbsp; but could also be formally calculated using the equation for the uniform distribution $($between&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; and&nbsp; $2\hspace{0.05cm} \rm V)$.&nbsp; Another solution is provided by the relation:
 :$$m_{y_{\rm 1}}=m_x\cdot H_{\rm 1}( f= 0) = 1\hspace{0.05cm} \rm V \cdot 1 \hspace{0.15cm}\underline{ =\rm 1\hspace{0.05cm} \rm V}.$$
@@ Line 117: / Line 116: @@
 =\rm\frac{1}{2V}\cdot \int_0^{2V} \hspace{-0.3cm}\it y_{\rm 1}^{\rm 2}\hspace{0.1cm}{\rm d}y_{\rm 1} =\rm \frac{8\,{\rm V^3}}{3 \cdot 2\,{\rm V}}
 \hspace{0.15cm}\underline{= \rm 1.333\, V^2}.$$
@@ Line 124: / Line 122: @@
 *The root of this is the standard deviation&nbsp;  (rms value)&nbsp; we are looking for: &nbsp; &nbsp;
 :$$\sigma_{y_{\rm 1}}\hspace{0.15cm}\underline{=0.577 \, \rm V}.$$
 '''(5)'''&nbsp; The probability we are looking for is the integral over the PDF of&nbsp; $0.75\hspace{0.05cm} \rm V$&nbsp; to $2\hspace{0.05cm} \rm V$, i.e.
 :$${\rm Pr}(y_1 > 0.75\hspace{0.05cm} \rm V) \hspace{0.15cm}\underline{ = 62.5\%}.$$
@@ Line 135: / Line 131: @@
 :$$f_{y2}(y_2= 0.5\hspace{0.05cm} \rm V) \hspace{0.15cm}\underline{=0.5 \cdot\rm 1/V}.$$
 *At&nbsp; $y_2 = 0.5 \hspace{0.05cm} \rm V$&nbsp; there is therefore only the continuous part.
 '''(7)'''&nbsp; The mean value&nbsp; $m_{y_{\rm 2}}\hspace{0.15cm}\underline{ =\rm 0.5\hspace{0.05cm} \rm V}$&nbsp; can be read directly from the above PDF sketch or calculated as in subtask&nbsp; '''(2)'''&nbsp; as follows:
 :$$m_{y_{\rm 2}} = m_x\cdot H_{\rm 2}( f = 0) = 1\hspace{0.05cm} \rm V \cdot 0.5 {\hspace{0.1cm} = \rm 0.5\hspace{0.05cm} \rm V}.$$
@@ Line 153: / Line 147: @@
 \sqrt{{\rm 1}/{\rm 6}\rm V^2}
 \hspace{0.15cm}\underline{=0.409\, \rm V}.$$

@@ Line 93: / Line 93: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID121__Sto_A_3_5_a_neu.png|right|frame|Amplitude limit, readable in the PDF]]
+[[File:P_ID121__Sto_A_3_5_a_neu.png|right|frame|Amplitude limit,&nbsp; <br>readable in the PDF]]
-'''(1)'''&nbsp; Correct are <u>the proposed solutions 1, 3 and 4</u>:
+'''(1)'''&nbsp; Correct are&nbsp; <u>the proposed solutions 1, 3 and 4</u>:
-*The random variable&nbsp; $y_1$&nbsp; is uniformly distributed and thus just like&nbsp; $x$&nbsp; also a continuous random variable.
+*The random variable&nbsp; $y_1$&nbsp; is uniformly distributed and thus just like&nbsp; $x$&nbsp; also a continuous valued random variable.
 *The PDF of&nbsp; $y_2$&nbsp; exhibits discrete proportions at&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; and&nbsp; $2\hspace{0.05cm} \rm V$&nbsp; on.
-*There are, of course, continuous components between these two boundaries.&nbsp;
+*There are,&nbsp; of course,&nbsp; continuous valued components between these two boundaries.&nbsp;
 *In this domain holds&nbsp; $f_{y2}(y_2) = 1/2$.
-'''(2)'''&nbsp; The linear mean&nbsp; $m_x = 1\hspace{0.05cm} \rm V$&nbsp; can be read directly from the data sketch, but could also be formally calculated using the equation for the uniform distribution $($between&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; and&nbsp; $2\hspace{0.05cm} \rm V)$&nbsp; can be calculated.&nbsp; Another solution is provided by the relation:
+'''(2)'''&nbsp; The linear mean&nbsp; $m_x = 1\hspace{0.05cm} \rm V$&nbsp; can be read directly from the data sketch,&nbsp; but could also be formally calculated using the equation for the uniform distribution $($between&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; and&nbsp; $2\hspace{0.05cm} \rm V)$.&nbsp; Another solution is provided by the relation:
 :$$m_{y_{\rm 1}}=m_x\cdot H_{\rm 1}( f= 0) = 1\hspace{0.05cm} \rm V \cdot 1 \hspace{0.15cm}\underline{ =\rm 1\hspace{0.05cm} \rm V}.$$
-'''(3)'''&nbsp; Actually, the averaging should be done over the whole time domain (both sides to infinity).
+'''(3)'''&nbsp; Actually,&nbsp; the averaging should be done over the whole time domain (both sides to infinity).
-* However, for reasons of symmetry, the averaging over the time interval $0 \le t \le T$ is sufficient:
+* However,&nbsp; for reasons of symmetry,&nbsp; the averaging over the time interval&nbsp; $0 \le t \le T$&nbsp; is sufficient:
 :$$P_{y_{\rm 1}}=\rm\frac{1}{\it T}\cdot \int_{\rm 0}^{\it T} \hspace{-0.15cm}\it y_{\rm 1}{\rm (}\it t{\rm {\rm )}}^{\rm 2}\it \hspace{0.05cm}\hspace{0.1cm}{\rm d}t=\rm\frac{1}{\it T}\cdot \int_{\rm 0}^{\it T} \hspace{-0.15cm}(\rm 2V \cdot \it\frac{t}{T}{\rm )}^{\rm 2} \hspace{0.1cm}{\rm d} t
 = \rm {4}/{3}\, V^2 \hspace{0.15cm}\underline{= \rm 1.333\, V^2}.$$
-*Sharp averaging gives the same result. Using the WDF&nbsp; $f_{y1}(y_1) = 1/(2\hspace{0.05cm} \rm V)$&nbsp; namely:
+*Sharp averaging gives the same result.&nbsp; Using the PDF&nbsp; $f_{y1}(y_1) = 1/(2\hspace{0.05cm} \rm V)$&nbsp; namely:
 :$$P_{y_{\rm 1}}=
 \int_0^{2V} \hspace{-0.3cm}\it y_{\rm 1}^{\rm 2}\cdot f_{\it y_{\rm 1}}{\rm (}\it y_{\rm 1}{\rm )}\it \hspace{0.1cm}{\rm d}y_{\rm 1}
@@ Line 120: / Line 120: @@
-'''(4)'''&nbsp; The variance can be determined using Steiner's theorem and gives&nbsp;
+'''(4)'''&nbsp; The variance can be determined using Steiner's theorem:
 :$$4/3\hspace{0.05cm} \rm V^2 - 1\hspace{0.05cm} \rm V^2 = 1/3\hspace{0.05cm} \rm V^2.$$
-*The root of this is the scattering we are looking for (the rms value): &nbsp; &nbsp;
+*The root of this is the standard deviation&nbsp;  (rms value)&nbsp; we are looking for: &nbsp; &nbsp;
 :$$\sigma_{y_{\rm 1}}\hspace{0.15cm}\underline{=0.577 \, \rm V}.$$
@@ Line 132: / Line 132: @@
-'''(6)'''&nbsp; The WDF consists of two Dirac functions at&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; and&nbsp; $1\hspace{0.05cm} \rm V$&nbsp; $($each with weight&nbsp; $1/4)$&nbsp; and a constant continuous component of
+'''(6)'''&nbsp; The PDF consists of two Dirac delta functions at&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; and&nbsp; $1\hspace{0.05cm} \rm V$&nbsp; $($each with weight&nbsp; $1/4)$&nbsp; and a constant continuous component of
 :$$f_{y2}(y_2= 0.5\hspace{0.05cm} \rm V) \hspace{0.15cm}\underline{=0.5 \cdot\rm 1/V}.$$
 *At&nbsp; $y_2 = 0.5 \hspace{0.05cm} \rm V$&nbsp; there is therefore only the continuous part.
@@ Line 143: / Line 143: @@
-'''(8)'''&nbsp; With above PDF, for given power:
+'''(8)'''&nbsp; With the above PDF,&nbsp; for given power:
 :$$P_{y_{\rm 2}}=\int_{-\infty}^{+\infty}\hspace{-0.3cm}y_{\rm 2}^{\rm 2}\cdot f_{\it y_{\rm 2}}{\rm (}\it y_{\rm 2})\hspace{0.1cm}{\rm d}y_{\rm 2}=\rm \frac{1}{2}\cdot\frac{1}{3}\cdot 1\,V^2+\rm \frac{1}{4}\cdot 1\,V^2
 = 5/12 \,V^2
 \hspace{0.15cm}{ =\rm 0.417\,V^2}.$$
-*The first part goes back to the continuous PDF, the second to the PDF Dirac function at&nbsp; $1\hspace{0.05cm} \rm V$.
+*The first part goes back to the continuous PDF,&nbsp; the second part to the PDF Dirac function at&nbsp; $1\hspace{0.05cm} \rm V$.
-*The Dirac function at&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; does not contribute to the power.&nbsp; It follows for the rms:
+*The Dirac function at&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; does not contribute to the power.&nbsp; It follows for the rms value:
 :$$\sigma_{y_{\rm 2}} = \sqrt{{\rm 5}/{\rm 12}\rm V^2 -{1}/{4}\rm V^2}=
 \sqrt{{\rm 1}/{\rm 6}\rm V^2}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID125__Sto_A_3_5.png|right|frame|rectangle&ndash;, triangle&ndash; and trapezoid signal ]]
+[[File:P_ID125__Sto_A_3_5.png|right|frame|rectangle signal,&nbsp; triangle signal <br>and trapezoid signal ]]
 We start from the rectangular signal&nbsp; $x(t)$&nbsp; according to the upper graph.
 *The amplitude values are&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; and&nbsp; $2\hspace{0.05cm} \rm V$.
 *Let the duration of a rectangle and the distance between two successive rectangular pulses be equal to each&nbsp; $T$.
-*The random variable&nbsp; $x$&nbsp; &ndash; the instantaneous value of the rectangular signal&nbsp; $x(t)$&nbsp; &ndash; thus has the following characteristic values:
+*The random variable&nbsp; $x$&nbsp; &ndash; the instantaneous value of the rectangular signal&nbsp; $x(t)$&nbsp; &ndash; thus has the characteristic values&nbsp; (mean,&nbsp; rms value):
 :$$m_x = \sigma_x = 1\hspace{0.05cm} \rm V.$$
@@ Line 15: / Line 15: @@
 then the triangular signal&nbsp; $y_1(t) = x(t) \star h_1(t)$&nbsp; is obtained at its output according to the convolution with
-*the minimum values&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; $($at $t = 0, 2T, 4T,$ ...$)$,
+*the minimum values&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; $($at&nbsp; $t = 0,&nbsp; 2T,&nbsp; 4T,&nbsp;$ ...$)$,
-*the maximum values&nbsp; $2\hspace{0.05cm} \rm V$&nbsp; $($bei $t = T, 3T, 5T,$ ...$)$.
+*the maximum values&nbsp; $2\hspace{0.05cm} \rm V$&nbsp; $($at&nbsp; $t = T,&nbsp; 3T,&nbsp; 5T,&nbsp;$ ...$)$.
 Thus, this low-pass filter is an integrator over the time duration&nbsp; $T$.<br>
-If, on the other hand, we apply the signal &nbsp; $x(t)$&nbsp; to the input of a filter with the impulse response
+If,&nbsp; on the other hand,&nbsp; we apply the signal &nbsp; $x(t)$&nbsp; to the input of a filter with the impulse response
 :$$h_{\rm 2}(t)=\left \{ \begin{array}{*{4}{c}} 1/T & {\; \rm for}\hspace{0.2cm}{ 0\le t \le T/2} \\\ 0 & {\rm else} \end{array} \right. , $$
@@ Line 34: / Line 34: @@
 *This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Uniformly_Distributed_Random_Variables|Uniformly distributed random variables]].
-*For the corresponding frequency bands $H_1(f=0)= 1$&nbsp; or $H_2(f=0)= 0.5$.
+*For the corresponding frequency responses&nbsp; $H_1(f=0)= 1$&nbsp; or&nbsp; $H_2(f=0)= 0.5$.
@@ Line 42: / Line 42: @@
 {Which of the following statements are true?
 |type="[]"}
-+ $y_1(t)$&nbsp; is a continuous random variable.
++ $y_1(t)$&nbsp; is a continuous valued random variable.
 - $y_1(t)$&nbsp; has a triangular PDF.
 + $y_1(t)$&nbsp; is uniformly distributed.
-+ $y_2(t)$&nbsp; has continuous and discrete components.
++ $y_2(t)$&nbsp; has continuous and discrete  valued  components.
-{How large; is the uniform part of the signal&nbsp; $y_1(t)$?&nbsp; Check this value&nbsp; $m_{y1}$&nbsp; also by using the variables&nbsp; $m_x$&nbsp; and&nbsp; $H_1(f=0)$.
+{How large is the uniform part of the signal&nbsp; $y_1(t)$?&nbsp; Check this value&nbsp; $m_{y1}$&nbsp; also by using the variables&nbsp; $m_x$&nbsp; and&nbsp; $H_1(f=0)$.
 |type="{}"}
 $m_{y1} \ = \ $ { 1 3% } $\ \rm V$
-{Determine the power of the signal&nbsp; $y_1(t)$ by both time averaging and coulter averaging.
+{Determine the power of the signal&nbsp; $y_1(t)$&nbsp; by both time averaging and coulter averaging.
 |type="{}"}
 $P_{y1} \ = \ $ { 1.333 3% } $\ \rm V^2$
-{What is the rms of the signal $y_1(t)$?
+{What is the rms value of the signal&nbsp; $y_1(t)$?
 |type="{}"}
 $\sigma_{y1} \ = \ $ { 0.577 3% } $\ \rm V$
@@ Line 68: / Line 68: @@
-{Determine the PDF of the signal&nbsp; $y_2(t)$&nbsp; and sketch it.&nbsp; As a check, enter the PDF value at the point&nbsp; $y_2 = 0.5\hspace{0.05cm} \rm V$&nbsp; Enter.
+{Determine the PDF of the signal&nbsp; $y_2(t)$&nbsp; and sketch it.&nbsp; As a check, enter the PDF value at the point&nbsp; $y_2 = 0.5\hspace{0.05cm} \rm V$.
 |type="{}"}
 $f_{y2}(y_2= 0.5\hspace{0.05cm} \rm V) \ = \ $ { 0.5 3% } $\ \rm 1/V$
-{What is the DC component of the signal&nbsp; $y_2(t)$?&nbsp; &Check this value&nbsp; $m_{y2}$&nbsp; also using the quantities&nbsp; $m_x$&nbsp; and&nbsp; $H_2(f=0)$.
+{What is the DC component of the signal&nbsp; $y_2(t)$?&nbsp; Check this value&nbsp; $m_{y2}$&nbsp; also using the quantities&nbsp; $m_x$&nbsp; and&nbsp; $H_2(f=0)$.
 |type="{}"}
 $m_{y2} \ = \ ${ 0.5 3% } $\ \rm V$
-{What is the rms value of the signal $y_2(t)$?
+{What is the rms value of the signal&nbsp; $y_2(t)$?
 |type="{}"}
 $\sigma_{y2} \ = \ $ { 0.409 3% } $\ \rm V$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Gleichverteilte Zufallsgröße
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Uniformly_Distributed_Random_Variables
 }}
-[[File:P_ID125__Sto_A_3_5.png|right|frame|Rechteck&ndash;, Dreieck&ndash; und Trapezsignal ]]
+[[File:P_ID125__Sto_A_3_5.png|right|frame|rectangle&ndash;, triangle&ndash; and trapezoid signal ]]
-Wir gehen vom Rechtecksignal&nbsp; $x(t)$&nbsp; gem&auml;&szlig; der oberen Grafik aus.
+We start from the rectangular signal&nbsp; $x(t)$&nbsp; according to the upper graph.
-*Die Amplitudenwerte seien&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; und&nbsp; $2\hspace{0.05cm} \rm V$.
+*The amplitude values are&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; and&nbsp; $2\hspace{0.05cm} \rm V$.
-*Die Dauer eines Rechtecks sowie der Abstand zweier aufeinander folgender Rechteckimpulse seien jeweils gleich&nbsp; $T$.
+*Let the duration of a rectangle and the distance between two successive rectangular pulses be equal to each&nbsp; $T$.
-*Die Zufallsgr&ouml;&szlig;e&nbsp; $x$&nbsp; &ndash; der Momentanwert des Rechtecksignals&nbsp; $x(t)$&nbsp; &ndash; hat somit folgende Kennwerte:
+*The random variable&nbsp; $x$&nbsp; &ndash; the instantaneous value of the rectangular signal&nbsp; $x(t)$&nbsp; &ndash; thus has the following characteristic values:
 :$$m_x = \sigma_x = 1\hspace{0.05cm} \rm V.$$
-Gibt man nun dieses Signal auf ein lineares Filter mit der Impulsantwort
+If we now apply this signal to a linear filter with the impulse response
-:$$h_{\rm 1}(t)=\left \{ \begin{array}{*{4}{c}} 1/T & {\; \rm f\ddot{u}r}\hspace{0.2cm}{ 0\le  t \le  T} \\\ 0 & {\rm sonst} \end{array} \right. ,  $$
+:$$h_{\rm 1}(t)=\left \{ \begin{array}{*{4}{c}} 1/T & {\; \rm for}\hspace{0.2cm}{ 0\le  t \le  T} \\\ 0 & {\rm else} \end{array} \right. ,  $$
-so ergibt sich an dessem Ausgang entsprechend der Faltung das Dreiecksignal&nbsp; $y_1(t) = x(t) \star  h_1(t)$&nbsp; mit
+then the triangular signal&nbsp; $y_1(t) = x(t) \star h_1(t)$&nbsp; is obtained at its output according to the convolution with
-*den Minimalwerten&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; $($bei $t = 0, 2T, 4T,$ ...$)$,
+*the minimum values&nbsp; $0\hspace{0.05cm} \rm V$&nbsp; $($at $t = 0, 2T, 4T,$ ...$)$,
-*den Maximalwerten&nbsp; $2\hspace{0.05cm} \rm V$&nbsp; $($bei $t = T, 3T, 5T,$ ...$)$.
+*the maximum values&nbsp; $2\hspace{0.05cm} \rm V$&nbsp; $($bei $t = T, 3T, 5T,$ ...$)$.
-Bei diesem Tiefpassfilter handelt es sich also um einen Integrator &uuml;ber die Zeitdauer&nbsp; $T$.<br>
+Thus, this low-pass filter is an integrator over the time duration&nbsp; $T$.<br>
-Legt man dagegen das Signal&nbsp; $x(t)$&nbsp; an den Eingang eines Filters mit der Impulsantwort
+If, on the other hand, we apply the signal &nbsp; $x(t)$&nbsp; to the input of a filter with the impulse response
-:$$h_{\rm 2}(t)=\left \{ \begin{array}{*{4}{c}} 1/T & {\; \rm f\ddot{u}r}\hspace{0.2cm}{ 0\le  t \le  T/2} \\\ 0 & {\rm sonst} \end{array} \right. ,  $$
+:$$h_{\rm 2}(t)=\left \{ \begin{array}{*{4}{c}} 1/T & {\; \rm for}\hspace{0.2cm}{ 0\le t \le T/2} \\\ 0 & {\rm else} \end{array} \right. , $$
-so ergibt sich das trapezf&ouml;rmige Signal&nbsp; $y_2(t) = x(t) \star  h_2(t)$.&nbsp; Dieses zweite Filter wirkt somit als ein Integrator &uuml;ber die Zeitdauer&nbsp; $T/2$.
+so the trapezoidal signal&nbsp; $y_2(t) = x(t) \star h_2(t)$.&nbsp; This second filter thus acts as an integrator over the time duration&nbsp; $T/2$.
 <br>
@@ Line 31: / Line 31: @@
-''Hinweise:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Gleichverteilte_Zufallsgröße|Gleichverteilte Zufallsgröße]].
+*This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Uniformly_Distributed_Random_Variables|Uniformly distributed random variables]].
-*F&uuml;r die zugeh&ouml;rigen Frequenzg&auml;nge gilt&nbsp; $H_1(f=0)= 1$&nbsp; bzw.&nbsp; $H_2(f=0)= 0.5$.
+*For the corresponding frequency bands $H_1(f=0)= 1$&nbsp; or $H_2(f=0)= 0.5$.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welche der folgenden Aussagen sind zutreffend?
+{Which of the following statements are true?
 |type="[]"}
-+ $y_1(t)$&nbsp; ist eine kontinuierliche Zufallsgr&ouml;&szlig;e.
++ $y_1(t)$&nbsp; is a continuous random variable.
-- $y_1(t)$&nbsp; besitzt eine dreieckf&ouml;rmige WDF.
+- $y_1(t)$&nbsp; has a triangular PDF.
-+ $y_1(t)$&nbsp; ist gleichverteilt.
++ $y_1(t)$&nbsp; is uniformly distributed.
-+ $y_2(t)$&nbsp; hat kontinuierliche und diskrete Anteile.
++ $y_2(t)$&nbsp; has continuous and discrete components.
-{Wie gro&szlig; ist der Gleichanteil des Signals&nbsp; $y_1(t)$?&nbsp; &Uuml;berpr&uuml;fen Sie diesen Wert&nbsp; $m_{y1}$&nbsp; auch anhand der Gr&ouml;&szlig;en&nbsp; $m_x$&nbsp; und&nbsp; $H_1(f=0)$.
+{How large; is the uniform part of the signal&nbsp; $y_1(t)$?&nbsp; &Check this value&nbsp; $m_{y1}$&nbsp; also by using the variables&nbsp; $m_x$&nbsp; and&nbsp; $H_1(f=0)$.
 |type="{}"}
-$m_{y1} \ = \ $  { 1 3% } $\ \rm V$
+$m_{y1} \ = \ $ { 1 3% } $\ \rm V$
-{Bestimmen Sie die Leistung des Signals&nbsp; $y_1(t)$ sowohl durch Zeitmittelung als auch durch Scharmittelung.
+{Determine the power of the signal&nbsp; $y_1(t)$ by both time averaging and coulter averaging.
 |type="{}"}
 $P_{y1} \ = \ $ { 1.333 3% } $\ \rm V^2$
-{Wie gro&szlig; ist der Effektivwert des Signals&nbsp; $y_1(t)$?
+{What is the rms of the signal $y_1(t)$?
 |type="{}"}
-$\sigma_{y1} \ = \ $  { 0.577 3% }  $\ \rm V$
+$\sigma_{y1} \ = \ $ { 0.577 3% } $\ \rm V$
-{Wie gro&szlig; ist die Wahrscheinlichkeit, dass&nbsp; $y_1(t)$&nbsp; gr&ouml;&szlig;er ist als&nbsp; $0.75\hspace{0.05cm} \rm V$?
+{What is the probability that&nbsp; $y_1(t)$&nbsp; is larger than&nbsp; $0.75\hspace{0.05cm} \rm V$?
 |type="{}"}
 ${\rm Pr}(y_1 > 0.75\hspace{0.05cm} \rm V) \ = \ $ { 62.5 3% } $ \ \%$
-{Ermitteln Sie die WDF des Signals&nbsp; $y_2(t)$&nbsp; und skizzieren Sie diese.&nbsp; Geben Sie zur Kontrolle den WDF-Wert an der Stelle&nbsp; $y_2 = 0.5\hspace{0.05cm} \rm V$&nbsp; ein.
+{Determine the PDF of the signal&nbsp; $y_2(t)$&nbsp; and sketch it.&nbsp; As a check, enter the PDF value at the point&nbsp; $y_2 = 0.5\hspace{0.05cm} \rm V$&nbsp; Enter.
 |type="{}"}
-$f_{y2}(y_2= 0.5\hspace{0.05cm} \rm V) \ = \ $  { 0.5 3% } $\ \rm 1/V$
+$f_{y2}(y_2= 0.5\hspace{0.05cm} \rm V) \ = \ $ { 0.5 3% } $\ \rm 1/V$
-{Wie gro&szlig; ist der Gleichanteil des Signals&nbsp; $y_2(t)$?&nbsp; &Uuml;berpr&uuml;fen Sie diesen Wert&nbsp; $m_{y2}$&nbsp; auch anhand der Gr&ouml;&szlig;en&nbsp; $m_x$&nbsp; und&nbsp; $H_2(f=0)$.
+{What is the DC component of the signal&nbsp; $y_2(t)$?&nbsp; &Check this value&nbsp; $m_{y2}$&nbsp; also using the quantities&nbsp; $m_x$&nbsp; and&nbsp; $H_2(f=0)$.
 |type="{}"}
 $m_{y2} \ = \ ${ 0.5 3% } $\ \rm V$
-{Wie gro&szlig; ist der Effektivwert des Signals&nbsp; $y_2(t)$?
+{What is the rms value of the signal $y_2(t)$?
 |type="{}"}
 $\sigma_{y2} \ = \ $ { 0.409 3% } $\ \rm V$
-{Wie gro&szlig; ist die Wahrscheinlichkeit, dass&nbsp; $y_2(t)$&nbsp; gr&ouml;&szlig;er ist als&nbsp; $0.75\hspace{0.05cm} \rm V$?
+{What is the probability that&nbsp; $y_2(t)$&nbsp; is larger than&nbsp; $0.75\hspace{0.05cm} \rm V$?
 |type="{}"}
 ${\rm Pr}(y_2 > 0.75\hspace{0.05cm} \rm V) \ = \ $ { 37.5 3% } $ \ \%$
@@ Line 91: / Line 91: @@
 </quiz>
-===Musterlösung===
+===Solutions===
 {{ML-Kopf}}
 [[File:P_ID121__Sto_A_3_5_a_neu.png|right|frame|Amplitudenbegrenzung, ablesbar in der WDF]]

← Older revision		Revision as of 10:10, 20 December 2021
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−	[[Category:Theory of Stochastic Signals: Exercises\|^3.4 ~~Gleichverteilte Zufallsgröße~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^3.4 Uniformly Distributed Random Variable^]]

← Older revision	Revision as of 10:08, 20 December 2021
(No difference)