Aufgaben:Exercise 3.8Z: Convolution of Two Rectangles - Revision history

Guenter: Guenter moved page Aufgaben:Exercise 3.8Z:Convolution of Two Rectangles to Aufgaben:Exercise 3.8Z: Convolution of Two Rectangles

2021-05-24T12:56:52Z

Guenter moved page Exercise 3.8Z:Convolution of Two Rectangles to Exercise 3.8Z: Convolution of Two Rectangles

Guenter at 12:57, 29 April 2021

2021-04-29T12:57:17Z

Guenter at 12:46, 29 April 2021

2021-04-29T12:46:45Z

Guenter at 12:39, 29 April 2021

2021-04-29T12:39:54Z

Guenter at 12:38, 29 April 2021

2021-04-29T12:38:34Z

Javier: Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

2021-03-23T12:38:46Z

Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

Noah at 21:45, 15 February 2021

2021-02-15T21:45:00Z

Noah at 21:01, 28 January 2021

2021-01-28T21:01:19Z

Noah at 21:00, 28 January 2021

2021-01-28T21:00:26Z

Noah at 20:59, 28 January 2021

2021-01-28T20:59:55Z

@@ Line 56: / Line 56: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID536__Sig_Z_3_8_a_neu.png|right|frame|To illustrate the convolution&nbsp; $x(t) \star h(t)$;<br>the abscissas  have been renamed to&nbsp; $"\tau"$]]
+[[File:P_ID536__Sig_Z_3_8_a_neu.png|right|frame|To illustrate the convolution&nbsp; $x(t) \star h(t)$;<br>the abscissas  have been renamed:&nbsp; $\tau$]]
 '''(1)'''&nbsp;  In general, the following applies to the convolution integral:
 :$$y(t) = \int_{ - \infty }^{ + \infty } {x( \tau  ) \cdot h( {t - \tau } )}\hspace{0.1cm} {\rm d}\tau.$$
-The signal value at time&nbsp; $t = 1 \,\text{ms}$&nbsp; ms can be calculated as follows:
+The signal value at time&nbsp; $t = 1 \,\text{ms}$&nbsp; can be calculated as follows:
-*Reflection of the impulse response&nbsp; ${h(\tau)}$,
+*Mirroring of the impulse response&nbsp; ${h(\tau)}$,
-*shift by&nbsp; $t = 1 \text{ ms}$&nbsp; to the right (violet curve in the sketch),
+*shifting by&nbsp; $t = 1 \text{ ms}$&nbsp; to the right (violet curve in the sketch),
 *multiplication of the two functions and integration.
-The product is also rectangular with the height&nbsp; $2 \text{ V} \cdot 300 \; \text{1/s}$&nbsp; and width&nbsp; $1 \,\text{ms}$. This results in for the area:
+The product is also rectangular with the height&nbsp; $2 \text{ V} \cdot 300 \; \text{1/s}$&nbsp; and width&nbsp; $1 \,\text{ms}$. This results for the area:
 :$$y( {t = 1\;{\rm{ms}}} ) \hspace{0.15 cm}\underline{= 0.6\;{\rm{V}}}{\rm{.}}$$
-The green rectangle illustrates the calculation of the second signal value. Now the resulting rectangle is twice as wide after the multiplication and we get:
+The green rectangle illustrates the calculation of the second signal value.&nbsp; Now the resulting rectangle is twice as wide after the multiplication and we get:
 :$$y( {t = 2\;{\rm{ms}}} ) = 2\;{\rm{V}} \cdot {\rm{300}}\;{1}/{{\rm{s}}} \cdot 2\;{\rm{ms}}\hspace{0.15 cm}\underline{={\rm{1.2}}\;{\rm{V}}}{\rm{.}}$$
-'''(2)'''&nbsp; Because of the symmetry of ${y(t)}$ with respect to the time&nbsp; $t = 2.5\, \text {ms}$&nbsp; holds:
+'''(2)'''&nbsp; Because of the symmetry of&nbsp; ${y(t)}$&nbsp; with respect to the time&nbsp; $t = 2.5\, \text {ms}$&nbsp; holds:
 :$$y( {t = 3\;{\rm{ms}}} ) = y( {t = 2\;{\rm{ms}}} ) \hspace{0.15 cm}\underline{= {\rm{1}}{\rm{.2}}\;{\rm{V}}}{\rm{,}}$$
 :$$y( {t = 4\;{\rm{ms}}} ) = y( {t = 1\;{\rm{ms}}} )\hspace{0.15 cm}\underline{ = 0.6\;{\rm{V}}}{\rm{.}}$$
-[[File:P_ID537__Sig_Z_3_8_c.png|right|frame|Convolution result&nbsp; $y(t)$]]
+[[File:P_ID537__Sig_Z_3_8_c.png|right|frame|Overall result&nbsp; $y(t)$]]
-'''(3)'''&nbsp;  In subtasks&nbsp; '''(1)'''&nbsp; and&nbsp; '''(2)'''&nbsp; the signal values were calculated at discrete time points.
+'''(3)'''&nbsp;  Proposed solutions <u>1 and 3</u> are correct:
 *All points are to be connected by straight line segments, since the integration over rectangular functions of increasing width results in a linear course.
 *This means:&nbsp; The output signal&nbsp; ${y(t)}$&nbsp; is trapezoidal.
@@ Line 84: / Line 85: @@
 *The associated spectrum is complex and reads:
 :$$Y(f) = 6 \cdot 10^{ - 3} \;{{\rm{V}}}/{{{\rm{Hz}}}} \cdot {\mathop{\rm si}\nolimits} ( {2\;{\rm{ms}}\cdot{\rm{\pi }}f} ) \cdot {\mathop{\rm si}\nolimits} ( {3\;{\rm{ms}}\cdot{\rm{\pi }}f}) \cdot {\rm{e}}^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \hspace{0.05cm}\cdot \hspace{0.05cm}2.5\;{\rm{ms}}\hspace{0.05cm}\cdot \hspace{0.05cm} \pi }}f} .$$
-*If the input pulse&nbsp; ${x(t)}$&nbsp; had the duration&nbsp; $T = 2\, \text {ms}$, had the duration&nbsp; ${y(t)}$&nbsp; would show a triangular waveform between&nbsp; ${t = 0}$&nbsp; and&nbsp; $t = 4  \text { ms}$&nbsp;.
+*If the input pulse&nbsp; ${x(t)}$&nbsp; would have the duration&nbsp; $T = 2\, \text {ms}$, the duration&nbsp; ${y(t)}$&nbsp; would show a triangular waveform between&nbsp; ${t = 0}$&nbsp; and&nbsp; $t = 4  \text { ms}$.&nbsp; The maximum&nbsp; $1.2 \, \text {V}$&nbsp; would then only result at the time&nbsp; $t = 2 \, \text {ms}$.
-*The maximum&nbsp; $1.2 \, \text {V}$&nbsp; would then only result at the time&nbsp; $t = 2 \, \text {ms}$.
 {{ML-Fuß}}

@@ Line 48: / Line 48: @@
 + The output signal&nbsp; ${y(t)}$&nbsp; has a trapezoidal shape.
 - The spectrum is &nbsp; ${Y(f)} = Y_0 \cdot \text{si}^{2}(\pi f T)$.
-+ With&nbsp; $T = 2 \,\text{ms}$&nbsp;, a triangular shape would result.
++ With&nbsp; $T = 2 \,\text{ms}$,&nbsp; a triangular shape would result.
@@ Line 56: / Line 56: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID536__Sig_Z_3_8_a_neu.png|right|frame|To illustrate the graphical convolution $x(t) \star h(t)$]]
+[[File:P_ID536__Sig_Z_3_8_a_neu.png|right|frame|To illustrate the convolution&nbsp; $x(t) \star h(t)$;<br>the abscissas  have been renamed to&nbsp; $"\tau"$]]
 '''(1)'''&nbsp;  In general, the following applies to the convolution integral:
 :$$y(t) = \int_{ - \infty }^{ + \infty } {x( \tau  ) \cdot h( {t - \tau } )}\hspace{0.1cm} {\rm d}\tau.$$
 The signal value at time&nbsp; $t = 1 \,\text{ms}$&nbsp; ms can be calculated as follows:

@@ Line 23: / Line 23: @@
 ''Hints:''
 *This exercise belongs to the chapter&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|"The Convolution Theorem and Operation"]].
-*It mainly refers to the page&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Graphical_Convolution|"Graphical Convolution"]].
+*It mainly refers to the page&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Graphical_convolution|"Graphical convolution"]].
 *The topic of this section is also illustrated in the interactive applet&nbsp; [[Applets:Graphical_Convolution| "Graphical Convolution"]].

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID535__Sig_Z_3_8.png|right|frame|For the convolution of two rectangles
+[[File:P_ID535__Sig_Z_3_8.png|right|frame|The convolution of two rectangles&nbsp; $x(t)$&nbsp; and&nbsp; $h(t)$
 ]]
-At the input of a causal LTI system (i.e. linear and time-invariant) with a rectangular impulse response&nbsp; ${h(t)}$&nbsp; of duration&nbsp; $2 \,\text{ms}$&nbsp;, a rectangular impulse&nbsp; ${x(t)}$&nbsp; of duration&nbsp; $T = 3 \,\text{ms}$&nbsp; and amplitude&nbsp; $A = 2\,\text{ V}$&nbsp; is applied. The square-wave functions each start at the time&nbsp; $t = 0$.
+At the input of a causal LTI system (i.e. linear and time-invariant)
 ===Questions===
@@ Line 35: / Line 38: @@
-{Determine the signal values for the time points&nbsp; $t = 3 \,\text{ms}$&nbsp; and&nbsp; $t = 4 \,\text{ms}$&nbsp; by exploiting the symmetry properties.
+{Determine the signal values for the times&nbsp; $t = 3 \,\text{ms}$&nbsp; and&nbsp; $t = 4 \,\text{ms}$&nbsp; by exploiting the symmetry properties.
 |type="{}"}
 $y(t = 3 \,\text{ms})\ = \ $ { 1.2 3% } &nbsp;$\text{V}$
@@ Line 44: / Line 47: @@
 |type="[]"}
 + The output signal&nbsp; ${y(t)}$&nbsp; has a trapezoidal shape.
-- The spectrum is: &nbsp; ${Y(f)} = Y_0 \cdot \text{si}^{2}(\pi f T)$.
+- The spectrum is &nbsp; ${Y(f)} = Y_0 \cdot \text{si}^{2}(\pi f T)$.
 + With&nbsp; $T = 2 \,\text{ms}$&nbsp;, a triangular shape would result.

← Older revision		Revision as of 12:38, 23 March 2021
Line 91:		Line 91:

	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:~~Exercises for~~ Signal Representation\|^3.4 The Convolution Theorem^]]	+	[[Category:Signal Representation: Exercises\|^3.4 The Convolution Theorem^]]

← Older revision	Revision as of 12:56, 24 May 2021
(No difference)

@@ Line 21: / Line 21: @@
 *This exercise belongs to the chapter&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|The Convolution Theorem and Operation]].
 *It mainly refers to the page&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Graphical_Convolution|Graphical Convolution]]
-*The topic of this section is also illustrated in the interactive applet&nbsp; [[Applets:Zur_Verdeutlichung_der_grafischen_Faltung|To illustrate the graphical convolution]] veranschaulicht.
+*The topic of this section is also illustrated in the interactive applet&nbsp; [[Applets:Zur_Verdeutlichung_der_grafischen_Faltung|To illustrate the graphical convolution]].

← Older revision		Revision as of 21:00, 28 January 2021
Line 86:		Line 86:


−	.Proposed solutions <u>1 and 3</u> are therefore correct.	+	Proposed solutions <u>1 and 3</u> are therefore correct.
	{{ML-Fuß}}		{{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID535__Sig_Z_3_8.png|right|frame|Zur Faltung zweier Rechtecke]]
+[[File:P_ID535__Sig_Z_3_8.png|right|frame|For the convolution of two rectangles
 At the input of a causal LTI system (i.e. linear and time-invariant) with a rectangular impulse response&nbsp; ${h(t)}$&nbsp; of duration&nbsp; $2 \,\text{ms}$&nbsp;, a rectangular impulse&nbsp; ${x(t)}$&nbsp; of duration&nbsp; $T = 3 \,\text{ms}$&nbsp; and amplitude&nbsp; $A = 2\,\text{ V}$&nbsp; is applied. The square-wave functions each start at the time&nbsp; $t = 0$.
@@ Line 18: / Line 19: @@
 ''Hinweise:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|Faltungssatz und Faltungsoperation]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|The Convolution Theorem and Operation]].
-*Sie bezieht sich vorwiegend auf die Seite&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Grafische_Faltung|Grafische Faltung]]
+*It mainly refers to the page&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Graphical_Convolution|Graphical Convolution]]
-*Die Thematik dieses Abschnitts wird auch im interaktiven Applet&nbsp; [[Applets:Zur_Verdeutlichung_der_grafischen_Faltung|Zur Verdeutlichung der grafischen Faltung]] veranschaulicht.
+*The topic of this section is also illustrated in the interactive applet&nbsp; [[Applets:Zur_Verdeutlichung_der_grafischen_Faltung|To illustrate the graphical convolution]] veranschaulicht.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Berechnen Sie die Signalwerte zu den Zeitpunkten&nbsp; $t = 1 \,\text{ms}$&nbsp; und&nbsp; $t = 2 \,\text{ms}$.
+{Calculate the signal values at the times &nbsp; $t = 1 \,\text{ms}$&nbsp; and&nbsp; $t = 2 \,\text{ms}$.
 |type="{}"}
 $y(t = 1 \,\text{ms})\ = \ $ { 0.6 3% } &nbsp;$\text{V}$
@@ Line 34: / Line 35: @@
-{Bestimmen Sie die Signalwerte für die Zeitpunkte&nbsp; $t = 3 \,\text{ms}$&nbsp; und&nbsp; $t = 4 \,\text{ms}$&nbsp; durch Ausnutzung der Symmetrieeigenschaften.
+{Determine the signal values for the time points&nbsp; $t = 3 \,\text{ms}$&nbsp; and&nbsp; $t = 4 \,\text{ms}$&nbsp; by exploiting the symmetry properties.
 |type="{}"}
 $y(t = 3 \,\text{ms})\ = \ $ { 1.2 3% } &nbsp;$\text{V}$
@@ Line 40: / Line 41: @@
-{Welche der folgenden Aussagen sind zutreffend?
+{Which of the following statements are true?
 |type="[]"}
-+ Das Ausgangssignal&nbsp; ${y(t)}$&nbsp; hat einen trapezförmigen Verlauf.
++ The output signal&nbsp; ${y(t)}$&nbsp; has a trapezoidal shape.
-- Das Spektrum lautet: &nbsp; ${Y(f)} = Y_0 \cdot \text{si}^{2}(\pi f T)$.
+- The spectrum is: &nbsp; ${Y(f)} = Y_0 \cdot \text{si}^{2}(\pi f T)$.
-+ Mit&nbsp; $T = 2 \,\text{ms}$&nbsp; würde sich eine Dreiecksform ergeben.
++ With&nbsp; $T = 2 \,\text{ms}$&nbsp;, a triangular shape would result.
@@ Line 50: / Line 51: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 [[File:P_ID536__Sig_Z_3_8_a_neu.png|right|frame|Zur Verdeutlichung der grafischen Faltung $x(t) \star h(t)$]]
-'''(1)'''&nbsp;  Allgemein gilt für das Faltungsintegral:
+'''(1)'''&nbsp;  In general, the following applies to the convolution integral:
 :$$y(t) = \int_{ - \infty }^{ + \infty } {x( \tau  ) \cdot h( {t - \tau } )}\hspace{0.1cm} {\rm d}\tau.$$
-''Hinweis:''&nbsp; Die Abszissen in nebenstehender Grafik wurden zu&nbsp; $\tau$&nbsp; umbenannt.
+''Hints:''&nbsp; The abscissas in the graph opposite have been renamed&nbsp; $\tau$&nbsp;.
-Der Signalwert zum Zeitpunkt&nbsp; $t = 1 \,\text{ms}$&nbsp; kann wie folgt berechnet werden:
+The signal value at time&nbsp; $t = 1 \,\text{ms}$&nbsp; ms can be calculated as follows:
-*Spiegelung der Impulsantwort&nbsp; ${h(\tau)}$,
+*Reflection of the impulse response&nbsp; ${h(\tau)}$,
-*Verschiebung um&nbsp; $t = 1 \text{ ms}$&nbsp; nach rechts (violette Kurve in der Skizze),
+*shift by&nbsp; $t = 1 \text{ ms}$&nbsp; to the right (violet curve in the sketch),
-*Multiplikation der beiden Funktionen sowie Integration.
+*multiplication of the two functions and integration.
-Das Produkt ist ebenfalls rechteckförmig mit der Höhe&nbsp; $2 \text{ V} \cdot 300 \; \text{1/s}$&nbsp; und der Breite&nbsp; $1 \,\text{ms}$. Daraus ergibt sich für die Fläche:
+The product is also rectangular with the height&nbsp; $2 \text{ V} \cdot 300 \; \text{1/s}$&nbsp; and width&nbsp; $1 \,\text{ms}$. This results in for the area:
 :$$y( {t = 1\;{\rm{ms}}} ) \hspace{0.15 cm}\underline{= 0.6\;{\rm{V}}}{\rm{.}}$$
-Das grüne Rechteck verdeutlicht die Berechnung des zweiten Signalwertes. Nun ist das resultierende Rechteck nach der Multiplikation doppelt so breit und man erhält:
+The green rectangle illustrates the calculation of the second signal value. Now the resulting rectangle is twice as wide after the multiplication and we get:
 :$$y( {t = 2\;{\rm{ms}}} ) = 2\;{\rm{V}} \cdot {\rm{300}}\;{1}/{{\rm{s}}} \cdot 2\;{\rm{ms}}\hspace{0.15 cm}\underline{={\rm{1.2}}\;{\rm{V}}}{\rm{.}}$$
-'''(2)'''&nbsp; Wegen der Symmetrie von ${y(t)}$ bezüglich des Zeitpunktes&nbsp; $t = 2.5\, \text {ms}$&nbsp; gilt:
+'''(2)'''&nbsp; Because of the symmetry of ${y(t)}$ with respect to the time&nbsp; $t = 2.5\, \text {ms}$&nbsp; holds:
 :$$y( {t = 3\;{\rm{ms}}} ) = y( {t = 2\;{\rm{ms}}} ) \hspace{0.15 cm}\underline{= {\rm{1}}{\rm{.2}}\;{\rm{V}}}{\rm{,}}$$
 :$$y( {t = 4\;{\rm{ms}}} ) = y( {t = 1\;{\rm{ms}}} )\hspace{0.15 cm}\underline{ = 0.6\;{\rm{V}}}{\rm{.}}$$
-[[File:P_ID537__Sig_Z_3_8_c.png|right|frame|Faltungsergebnis&nbsp; $y(t)$]]
+[[File:P_ID537__Sig_Z_3_8_c.png|right|frame|Convolution result&nbsp; $y(t)$]]
-'''(3)'''&nbsp;  In den Teilaufgaben&nbsp; '''(1)'''&nbsp; und&nbsp; '''(2)'''&nbsp; wurden die Signalwerte zu diskreten Zeitpunkten berechnet.
+'''(3)'''&nbsp;  In subtasks&nbsp; '''(1)'''&nbsp; and&nbsp; '''(2)'''&nbsp; the signal values were calculated at discrete time points.
-*Alle Punkte sind durch Geradenstücke zu verbinden, da die Integration über Rechteckfunktionen wachsender Breite einen linearen Verlauf ergibt.
+*All points are to be connected by straight line segments, since the integration over rectangular functions of increasing width results in a linear course.
-*Das heißt:&nbsp; Das Ausgangssignal&nbsp; ${y(t)}$&nbsp; ist trapezförmig.
+*This means:&nbsp; The output signal&nbsp; ${y(t)}$&nbsp; is trapezoidal.
-*Das dazugehörige Spektrum ist komplex und lautet:
+*The associated spectrum is complex and reads:
 :$$Y(f) = 6 \cdot 10^{ - 3} \;{{\rm{V}}}/{{{\rm{Hz}}}} \cdot {\mathop{\rm si}\nolimits} ( {2\;{\rm{ms}}\cdot{\rm{\pi }}f} ) \cdot {\mathop{\rm si}\nolimits} ( {3\;{\rm{ms}}\cdot{\rm{\pi }}f}) \cdot {\rm{e}}^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \hspace{0.05cm}\cdot \hspace{0.05cm}2.5\;{\rm{ms}}\hspace{0.05cm}\cdot \hspace{0.05cm} \pi }}f} .$$
-*Hätte der Eingangsimpuls&nbsp; ${x(t)}$&nbsp; die Dauer&nbsp; $T = 2\, \text {ms}$, so würde&nbsp; ${y(t)}$&nbsp; einen dreieckförmigen Signalverlauf zwischen&nbsp; ${t = 0}$&nbsp; und&nbsp; $t = 4  \text { ms}$&nbsp; zeigen.
+*If the input pulse&nbsp; ${x(t)}$&nbsp; had the duration&nbsp; $T = 2\, \text {ms}$, had the duration&nbsp; ${y(t)}$&nbsp; would show a triangular waveform between&nbsp; ${t = 0}$&nbsp; and&nbsp; $t = 4  \text { ms}$&nbsp;.
-*Das Maximum&nbsp; $1.2 \, \text {V}$&nbsp; ergäbe sich dann nur zum Zeitpunkt&nbsp; $t = 2 \, \text {ms}$.
+*The maximum&nbsp; $1.2 \, \text {V}$&nbsp; would then only result at the time&nbsp; $t = 2 \, \text {ms}$.
-Richtig sind somit die Lösungsvorschläge <u>1 und 3</u>.
+.Proposed solutions <u>1 and 3</u> are therefore correct.
 {{ML-Fuß}}