Aufgaben:Exercise 3.2: From the Spectrum to the Signal - Revision history

Guenter at 14:25, 21 April 2021

2021-04-21T14:25:32Z

Guenter at 14:19, 21 April 2021

2021-04-21T14:19:54Z

Guenter at 14:18, 21 April 2021

2021-04-21T14:18:22Z

Noah at 13:41, 26 March 2021

2021-03-26T13:41:58Z

Noah at 13:36, 26 March 2021

2021-03-26T13:36:16Z

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

2021-03-23T12:39:51Z

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

Noah at 21:47, 17 January 2021

2021-01-17T21:47:49Z

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 3.2: Vom Spektrum zum Signal to Aufgaben:Exercise 3.2: From the Spectrum to the Signal

2020-11-02T14:28:25Z

Oezdemir moved page Aufgabe 3.2: Vom Spektrum zum Signal to Exercise 3.2: From the Spectrum to the Signal

Oezdemir at 14:24, 2 November 2020

2020-11-02T14:24:15Z

Javier: Text replacement - "Signaldarstellung/Fouriertransformation und -rücktransformation" to "Signal_Representation/Fourier_Transform_and_Its_Inverse"

2020-09-01T09:05:38Z

Text replacement - "Signaldarstellung/Fouriertransformation und -rücktransformation" to "Signal_Representation/Fourier_Transform_and_Its_Inverse"

@@ Line 57: / Line 57: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; Correct is the <u>proposed solution 2</u> &nbsp; &rArr; &nbsp;$x(t)$&nbsp; is <u>purely real</u>:
-*For the imaginary signal component&nbsp; &rArr; &nbsp;  $x_{\rm I}(t)$&nbsp; the integrand is an odd function (even numerator, odd denominator)..
+*For the imaginary signal component&nbsp; &rArr; &nbsp;  $x_{\rm I}(t)$&nbsp; the integrand is an odd function&nbsp; (even numerator, odd denominator).
 *Thus the integral from&nbsp; $-\infty$&nbsp; bis&nbsp; $+\infty$&nbsp; is zero.
-*In contrast, for the real component&nbsp; $x_{\rm R}(t)$ &nbsp; &rArr; &nbsp;  even integrand (odd numerator, odd denominator) yields a non-zero value.
+*In contrast, for the real component&nbsp; $x_{\rm R}(t)$ &nbsp; &rArr; &nbsp;  even integrand&nbsp; (odd numerator, odd denominator)&nbsp; yields a non-zero value.
 '''(2)'''&nbsp; With the abbreviation&nbsp; $a = 2\pi t$&nbsp; can be written for the time signal:
 :$$x(t) = x_{\rm R} \left( t \right) = \frac{{4\,{\rm V}}}{\pi }\int_0^\infty  {\frac{{\sin( {af} )}}{f}}\hspace{0.1cm} {\rm d}f.$$
@@ Line 78: / Line 77: @@
-'''(3)'''&nbsp; At&nbsp; $t = 0$&nbsp; &nbsp; $x(t)$&nbsp; has a jump point. The right-hand limit value for&nbsp; $t \rightarrow 0$&nbsp; lautet&nbsp; $x_+ = +2\,\text{V}$.
+'''(3)'''&nbsp;  $x(t)$&nbsp; has a jumping point at&nbsp; $t = 0$.&nbsp; The right-hand limit value for&nbsp; $t \rightarrow 0$&nbsp; is&nbsp; $x_+ = +2\,\text{V}$.
-*If one approaches the jump point of negative times as close as desired, one obtains&nbsp; $x_– = -\hspace{-0.05cm}2\,\text{V}$. The following then applies to the actual signal value at&nbsp; $t = 0$&nbsp;:
+*If one approaches the jumping point of negative times as close as desired, one obtains&nbsp; $x_– = -\hspace{-0.05cm}2\,\text{V}$.
 :$$x( {t = 0} ) = {1}/{2}\cdot ( x_{+} +    x_{-} ) \hspace{0.15 cm}\underline{= 0}.$$
@@ Line 93: / Line 93: @@
 Here is a second solution:
-*The right-hand limit for&nbsp; $f → 0$&nbsp; is&nbsp; $X_+ = -\text{j} \cdot \infty$, the left-hand limit&nbsp; $X_- = \text{j} \cdot \infty$.
+*The right&ndash;hand limit for&nbsp; $f → 0$&nbsp; is&nbsp; $X_+ = -\text{j} \cdot \infty$,&ndash; and the left&ndash;hand limit&nbsp; $X_- = \text{j} \cdot \infty$.
-*So the relationship also applies with regard to the spectral value at &nbsp; $f = 0$&nbsp;:
+*So the relationship also applies with regard to the spectral value at &nbsp; $f = 0$:
 :$$X( {f = 0}) = {1}/{2}\cdot \left( {X_{ +}   + X_{-}  } \right) = 0.$$

@@ Line 9: / Line 9: @@
 :$$X(f) = \frac{{2\,{\rm V}}}{ { {\rm j}\pi f}}.$$
-The associated time function&nbsp; $x(t)$&nbsp; can be determined with the help of&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_Second_Fourier_Integral|The Second Fourier Integral]]&nbsp;:
+The associated time function&nbsp; $x(t)$&nbsp; can be determined with the help of&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_second_Fourier_integral|The second Fourier integral]]&nbsp;:
 :$$x(t)  = \int_{ - \infty }^{ + \infty } {X(f)}  \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d} f =  x_{\rm R} (t) + {\rm j} \cdot x_{\rm I} (t),$$

@@ Line 17: / Line 17: @@
 :$$x_{\rm R} (t) = 2\,{\rm V} \cdot \int_{ - \infty }^{ + \infty } {\frac{{\sin ( {2\pi ft} )}}{ {\pi f}}}\hspace{0.1cm} {\rm d}f, $$
 :$$x_{\rm I} (t) = -2\, {\rm V} \cdot \int_{ - \infty }^{ + \infty } {\frac{ {\cos ( {2\pi ft} )}}{ {\pi f}}} \hspace{0.1cm}{\rm d}f.$$
@@ Line 26: / Line 23: @@
 ''Hints:''
-*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse|Fourier Transform and Its Inverse]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse|Fourier Transform and its Inverse]].
 *If necessary, use the following information for the solution:
@@ Line 37: / Line 34: @@
 <quiz display=simple>
 {Which of the following statements are true for the time signal&nbsp; $x(t)$&nbsp;?
-|type="[]"}
+|type="()"}
 - $x(t)$&nbsp; is a complex function.
 + $x(t)$&nbsp; is purely real.
 - $x(t)$&nbsp; is purely imaginary.
-{Calculate the signal curve&nbsp; $x(t)$&nbsp; in the entire definition area. Which values occur at the times&nbsp; $t = 1\, \text{ms}$&nbsp; and&nbsp; $t = -\hspace{-0.05cm}1\, \text{ ms}$&nbsp; auf?
+{Calculate the signal curve&nbsp; $x(t)$&nbsp; in the entire definition area.&nbsp; Which values occur at the times&nbsp; $t = 1\, \text{ms}$&nbsp; and&nbsp; $t = -\hspace{-0.05cm}1\, \text{ ms}$?
 |type="{}"}
 $x(t=+1\, \text{ms}) \ = \ $ { 2 3% } $\ \text{V}$

@@ Line 36: / Line 36: @@
 <quiz display=simple>
-{Which of the following statements are true for the time signa&nbsp; $x(t)$&nbsp;?
+{Which of the following statements are true for the time signal&nbsp; $x(t)$&nbsp;?
 |type="[]"}
 - $x(t)$&nbsp; is a complex function.

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID495__Sig_A_3_2.png|right|frame|Spektraldarstellung der Sprungfunktion]]
+[[File:P_ID495__Sig_A_3_2.png|right|frame|Spectral representation of the unit step function]]
 Given the spectral function
@@ Line 59: / Line 59: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Correct is the <u>proposed solution 2</u> &nbsp; &rArr; &nbsp;$x(t)$&nbsp; is <u>purely real:</u>:
+'''(1)'''&nbsp; Correct is the <u>proposed solution 2</u> &nbsp; &rArr; &nbsp;$x(t)$&nbsp; is <u>purely real</u>:
-*For the imaginary signal component&nbsp; &rArr; &nbsp;  $x_{\rm I}(t)$&nbsp; the integrand is an odd function (gerader Zähler, ungerader Nenner).
+*For the imaginary signal component&nbsp; &rArr; &nbsp;  $x_{\rm I}(t)$&nbsp; the integrand is an odd function (even numerator, odd denominator)..
 *Thus the integral from&nbsp; $-\infty$&nbsp; bis&nbsp; $+\infty$&nbsp; is zero.
 *In contrast, for the real component&nbsp; $x_{\rm R}(t)$ &nbsp; &rArr; &nbsp;  even integrand (odd numerator, odd denominator) yields a non-zero value.

← Older revision		Revision as of 12:39, 23 March 2021
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	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:~~Exercises for~~ Signal Representation\|^3.1 Fourier Transform and Its Inverse^]]	+	[[Category:Signal Representation: Exercises\|^3.1 Fourier Transform and Its Inverse^]]

@@ Line 5: / Line 5: @@
 [[File:P_ID495__Sig_A_3_2.png|right|frame|Spektraldarstellung der Sprungfunktion]]
-Gegeben sei die Spektralfunktion
+Given the spectral function
 :$$X(f) = \frac{{2\,{\rm V}}}{ { {\rm j}\pi f}}.$$
-Die zugehörige Zeitfunktion&nbsp; $x(t)$&nbsp; kann mit Hilfe des&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Das_zweite_Fourierintegral|zweiten Fourierintegrals]]&nbsp; ermittelt werden:
+The associated time function&nbsp; $x(t)$&nbsp; can be determined with the help of&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_Second_Fourier_Integral|The Second Fourier Integral]]&nbsp;:
 :$$x(t)  = \int_{ - \infty }^{ + \infty } {X(f)}  \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d} f =  x_{\rm R} (t) + {\rm j} \cdot x_{\rm I} (t),$$
-wobei für den Realteil bzw. den Imaginärteil gilt:
+where holds for the real part and the imaginary part, respectively:
 :$$x_{\rm R} (t) = 2\,{\rm V} \cdot \int_{ - \infty }^{ + \infty } {\frac{{\sin ( {2\pi ft} )}}{ {\pi f}}}\hspace{0.1cm} {\rm d}f, $$
@@ Line 25: / Line 25: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum Kapitel&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse|Fouriertransformation und &ndash;rücktransformation]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse|Fourier Transform and Its Inverse]].
-*Benutzen Sie zur Lösung eventuell die nachfolgenden Angaben:
+*If necessary, use the following information for the solution:
 :$$x( {t = 0}) = \int_{ - \infty }^{ + \infty } {X( f )}\hspace{0.1cm} {\rm d}f,\hspace{0.5cm} X( {f = 0} ) = \int_{ - \infty }^{ + \infty } {x( t)}\hspace{0.1cm} {\rm d}t ,\hspace{0.5cm}\int_0^\infty  {\frac{{\sin ( {ax} )}}{x}}\hspace{0.1cm} {\rm d}x = {\rm sign} ( a ) \cdot{\pi }/{2}. $$
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welche der folgenden Aussagen treffen für das Zeitsignal&nbsp; $x(t)$&nbsp; zu?
+{Which of the following statements are true for the time signa&nbsp; $x(t)$&nbsp;?
 |type="[]"}
-- $x(t)$&nbsp; ist eine komplexe Funktion.
+- $x(t)$&nbsp; is a complex function.
-+ $x(t)$&nbsp; ist rein reell.
++ $x(t)$&nbsp; is purely real.
-- $x(t)$&nbsp; ist rein imaginär.
+- $x(t)$&nbsp; is purely imaginary.
-{Berechnen Sie den Signalverlauf&nbsp; $x(t)$&nbsp; im gesamten Definitionsgebiet. Welche Werte treten zu den Zeiten&nbsp; $t = 1\, \text{ms}$&nbsp; und&nbsp; $t = -\hspace{-0.05cm}1\, \text{ ms}$&nbsp; auf?
+{Calculate the signal curve&nbsp; $x(t)$&nbsp; in the entire definition area. Which values occur at the times&nbsp; $t = 1\, \text{ms}$&nbsp; and&nbsp; $t = -\hspace{-0.05cm}1\, \text{ ms}$&nbsp; auf?
 |type="{}"}
 $x(t=+1\, \text{ms}) \ = \ $ { 2 3% } $\ \text{V}$
 $x(t=-1 \text{ms})\hspace{0.2cm} = \ $  { -2.1--1.9 } $\ \text{V}$
-{Wie lautet der Signalwert zum Zeitpunkt&nbsp; $t = 0$?
+{What is the signal value at time&nbsp; $t = 0$?
 |type="{}"}
 $x(t=0) \ = \ $ { 0. } $\ \text{V}$
-{Wie groß ist der Spektralwert bei der Frequenz&nbsp; $f = 0$?
+{What is the spectral value at the frequency&nbsp; $f = 0$?
 |type="{}"}
 $X(f=0) \ = \ ${ 0. } $\ \text{V/Hz}$
@@ Line 57: / Line 57: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 2</u> &nbsp; &rArr; &nbsp;$x(t)$&nbsp; ist <u>rein reell</u>:
+'''(1)'''&nbsp; Correct is the <u>proposed solution 2</u> &nbsp; &rArr; &nbsp;$x(t)$&nbsp; is <u>purely real:</u>:
-*Beim imaginären Signalanteil &nbsp; &rArr; &nbsp;  $x_{\rm I}(t)$&nbsp; ist der Integrand eine ungerade Funktion (gerader Zähler, ungerader Nenner).
+*For the imaginary signal component&nbsp; &rArr; &nbsp;  $x_{\rm I}(t)$&nbsp; the integrand is an odd function (gerader Zähler, ungerader Nenner).
-*Somit ist das Integral von&nbsp; $-\infty$&nbsp; bis&nbsp; $+\infty$&nbsp; gleich Null.
+*Thus the integral from&nbsp; $-\infty$&nbsp; bis&nbsp; $+\infty$&nbsp; is zero.
-*Demgegenüber liefert beim reellen Anteil&nbsp; $x_{\rm R}(t)$ &nbsp; &rArr; &nbsp; gerader Integrand (ungerader Zähler, ungerader Nenner) einen von Null verschiedenen Wert.
+*In contrast, for the real component&nbsp; $x_{\rm R}(t)$ &nbsp; &rArr; &nbsp;  even integrand (odd numerator, odd denominator) yields a non-zero value.
-'''(2)'''&nbsp; Mit der Abkürzung&nbsp; $a = 2\pi t$&nbsp; kann für das Zeitsignal geschrieben werden:
+'''(2)'''&nbsp; With the abbreviation&nbsp; $a = 2\pi t$&nbsp; can be written for the time signal:
 :$$x(t) = x_{\rm R} \left( t \right) = \frac{{4\,{\rm V}}}{\pi }\int_0^\infty  {\frac{{\sin( {af} )}}{f}}\hspace{0.1cm} {\rm d}f.$$
-Dies führt unter Verwendung des angegebenen bestimmten Integrals zum Ergebnis:
+This leads to the result using the given definite integral:
 :$$x(t) = \frac{{4\,{\rm V}}}{\pi } \cdot \frac{\pi }{2} \cdot {\mathop{\rm sign}\nolimits} ( t ) = 2\;{\rm V} \cdot {\mathop{\rm sign}\nolimits} ( t ).$$
-*Für&nbsp; $t > 0$&nbsp; ist&nbsp; $x(t) = +2\,\text{V}$ .
+*For&nbsp; $t > 0$&nbsp; &nbsp; $x(t) = +2\,\text{V}$ .
-*Entsprechend gilt&nbsp; $x(t) = -\hspace{-0.1cm}2\,\text{V}$&nbsp; für&nbsp; $t < 0$.
+*Correspondingly,&nbsp; $x(t) = -\hspace{-0.1cm}2\,\text{V}$&nbsp; applies for&nbsp; $t < 0$.
-*Das Signal&nbsp; $x(t)$&nbsp; beschreibt also eine Sprungfunktion von&nbsp; $-\hspace{-0.05cm}2\,\text{V}$ auf $+2\,\text{V}$.
+*The signal&nbsp; $x(t)$&nbsp; thus describes a step function from&nbsp; $-\hspace{-0.05cm}2\,\text{V}$ auf $+2\,\text{V}$.
-'''(3)'''&nbsp; Bei&nbsp; $t = 0$&nbsp; besitzt&nbsp; $x(t)$&nbsp; eine Sprungstelle. Der rechtsseitige Grenzwert für&nbsp; $t \rightarrow 0$&nbsp; lautet&nbsp; $x_+ = +2\,\text{V}$.
+'''(3)'''&nbsp; At&nbsp; $t = 0$&nbsp; &nbsp; $x(t)$&nbsp; has a jump point. The right-hand limit value for&nbsp; $t \rightarrow 0$&nbsp; lautet&nbsp; $x_+ = +2\,\text{V}$.
-*Nähert man sich der Sprungstelle von negativen Zeiten beliebig nahe, so erhält man&nbsp; $x_– = -\hspace{-0.05cm}2\,\text{V}$. Für den tatsächlichen Signalwert bei&nbsp; $t = 0$&nbsp; gilt dann:
+*If one approaches the jump point of negative times as close as desired, one obtains&nbsp; $x_– = -\hspace{-0.05cm}2\,\text{V}$. The following then applies to the actual signal value at&nbsp; $t = 0$&nbsp;:
 :$$x( {t = 0} ) = {1}/{2}\cdot ( x_{+} +    x_{-} ) \hspace{0.15 cm}\underline{= 0}.$$
-*Zum gleichen Ergebnis kommt man bei Berücksichtigung der Beziehung
+*The same result is obtained by considering the relation
 :$$x( t = 0) = \int_{ - \infty }^{ + \infty } {X( f)}\hspace{0.1cm} {\rm d}f = 0.$$
-'''(4)'''&nbsp; Der Spektralwert bei&nbsp; $f = 0$&nbsp; ist gleich dem Integral von&nbsp; $-\infty$&nbsp; bis&nbsp; $+\infty$&nbsp; über die Zeitfunktion&nbsp; $x(t)$:
+'''(4)'''&nbsp; The spectral value at&nbsp; $f = 0$&nbsp; is equal to the integral from&nbsp; $-\infty$&nbsp; to&nbsp; $+\infty$&nbsp; over the time function&nbsp; $x(t)$:
 :$$X( f = 0) = \int_{ - \infty }^{ + \infty } {x( t)}\hspace{0.1cm} {\rm d}t \hspace{0.15 cm}\underline{= 0}.$$
-Hier noch ein zweiter Lösungsweg:
+Here is a second solution:
-*Der rechtsseitige Grenzwert für&nbsp; $f → 0$&nbsp; ist&nbsp; $X_+ = -\text{j} \cdot \infty$, der linksseitige Grenzwert&nbsp; $X_- = \text{j} \cdot \infty$.
+*The right-hand limit for&nbsp; $f → 0$&nbsp; is&nbsp; $X_+ = -\text{j} \cdot \infty$, the left-hand limit&nbsp; $X_- = \text{j} \cdot \infty$.
-*Auch bezüglich des Spektralwertes bei&nbsp; $f = 0$&nbsp; gilt also der Zusammenhang:
+*So the relationship also applies with regard to the spectral value at &nbsp; $f = 0$&nbsp;:
 :$$X( {f = 0}) = {1}/{2}\cdot \left( {X_{ +}   + X_{-}  } \right) = 0.$$

← Older revision		Revision as of 14:24, 2 November 2020
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−	[[Category:Exercises for Signal Representation\|^3. ~~Aperiodische Signale - Impulse~~^]]	+	[[Category:Exercises for Signal Representation\|^3.1 Fourier Transform and Its Inverse^]]

← Older revision		Revision as of 09:05, 1 September 2020
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−	{{quiz-Header\|Buchseite=~~Signaldarstellung~~/~~Fouriertransformation und -rücktransformation~~	+	{{quiz-Header\|Buchseite=Signal_Representation/Fourier_Transform_and_Its_Inverse
	}}		}}