Aufgaben:Exercise 3.3: From the Signal to the Spectrum - Revision history

Guenter at 09:36, 23 April 2021

2021-04-23T09:36:05Z

Guenter at 09:08, 23 April 2021

2021-04-23T09:08:12Z

Guenter at 09:07, 23 April 2021

2021-04-23T09:07:21Z

Guenter: Guenter moved page Aufgaben:Exercise 3.3: From The Signal to the Spectrum to Aufgaben:Exercise 3.3: From the Signal to the Spectrum

2021-04-23T08:59:18Z

Guenter moved page Exercise 3.3: From The Signal to the Spectrum to Exercise 3.3: From the Signal to the Spectrum

Guenter at 08:57, 23 April 2021

2021-04-23T08:57:49Z

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

2021-03-23T12:39:08Z

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

Noah at 21:14, 20 January 2021

2021-01-20T21:14:35Z

Noah at 20:45, 20 January 2021

2021-01-20T20:45:32Z

Noah at 20:30, 20 January 2021

2021-01-20T20:30:41Z

Noah at 20:18, 20 January 2021

2021-01-20T20:18:44Z

@@ Line 21: / Line 21: @@
 ''Hints:''
 *This task belongs to the chapter&nbsp;  [[Signal_Representation/Special_Cases_of_Pulses|Special Cases of Pulses]].
+*Use one of the functions&nbsp; $\text{si}(x) = \sin(x)/x$&nbsp; or&nbsp; $\text{sinc}(x) = \sin(\pi x)/(\pi x)$.
 *The following trigonometric transformations are given:
@@ Line 66: / Line 66: @@
 :$${\mathop{\rm Im}\nolimits} [ {X( f)} ] = \frac{A}{\omega } \cdot \left( {\cos ( {\omega T}) - 1} \right) =  - \frac{A}{\omega } \cdot\left( {1 - \cos ( {\omega T} )} \right).$$
-*For the frequency&nbsp; $f = 1/(2T) = 10\, \text{kHz}$  &nbsp; &rArr; &nbsp;  $\omega \cdot T = \pi$&nbsp;  we get::
+*For the frequency&nbsp; $f = 1/(2T) = 10\, \text{kHz}$  &nbsp; &rArr; &nbsp;  $\omega \cdot T = \pi$&nbsp;  we get:
 :$${\mathop{\rm Re}\nolimits}[{X( {f = 10 \;{\rm{kHz}}} )}] = \frac{A}{ {2{\rm{\pi }}f}} \cdot \sin ( {\rm{\pi }} ) \hspace{0.15 cm}\underline{= 0},$$
@@ Line 85: / Line 85: @@
 :$$\left| {X( f )} \right| =  \left| \frac{A}{ {\rm\pi }f} \cdot \sin ( {\rm \pi }fT ) \right| = A \cdot T \cdot \left| \frac{\sin ( {\rm\pi }fT )}{ {\rm \pi}fT} \right|.$$
-*With the shortcut&nbsp; $\text{si}(x) = \sin(x)/x$&nbsp; the result is:
+*With the shortcuts&nbsp; $\text{si}(x) = \sin(x)/x$&nbsp; or&nbsp; $\text{sinc}(x) = \sin(\pi x)/(\pi x)$&nbsp; the results are:
-:$$\left| {X( f)} \right| = A \cdot T \cdot\left|{\rm si} ( { {\rm{\pi }}fT} ) \right|.$$
+:$$\left| {X( f)} \right| = A \cdot T \cdot\left|{\rm si} ( { {\rm{\pi }}fT} ) \right|,$$
 *The spectral value at the frequency&nbsp; $f = 1/T = \text{20 kHz}$&nbsp; is given by
@@ Line 93: / Line 94: @@
 :$$\left| {X( {f = 20\;{\rm{kHz}}} )} \right| = \frac{ {A \cdot T}}{ {\rm{\pi }}} \cdot \sin ( {\rm{\pi }} ) \hspace{0.15 cm}\underline{= 0}.$$
-*When calculating the value for&nbsp; $f = 0$&nbsp; the quotient&nbsp; $\text{0 divided by 0}$. By applying&nbsp; [https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule L'Hospitals rule]&nbsp; the limiting value can be calculated:
+*When calculating the value for&nbsp; $f = 0$&nbsp; the quotient is&nbsp; $\text{"0 divided by 0"}$.&nbsp; By applying&nbsp; [https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule L'Hospitals rule]&nbsp; the limiting value can be calculated:
 :$$\mathop {\lim }\limits_{x \to 0 } \frac{ {\sin \left( x \right)}}{x} = 1.$$
@@ Line 105: / Line 106: @@
-[[File:P_ID563__Sig_A_3_3_c.png|right|frame|Betragsspektrum des Rechteckimpulses]]
+[[File:P_ID563__Sig_A_3_3_c.png|right|frame|Magnitude spectrum of the rectangular pulse]]
-'''(3)'''&nbsp; Proposed<u> solutions 1 and 3</u> are correct::
+'''(3)'''&nbsp; The <u>proposed  solutions 1 and 3</u> are correct:
 *According to the result of subtask&nbsp; '''(2)'''&nbsp; the zeros occur at the distance&nbsp; $f_0 = 1/T$&nbsp;.
 *With&nbsp; $f_0 = 1/(2T) = f = 10 \;{\rm{kHz}}$&nbsp;  the real part is&nbsp; $0$, but not the imaginary part.
-*With the arguments&nbsp; $f \cdot T = 0.5, 1.5, 2.5,\hspace{0.05cm}\text{ ... }$&nbsp; the sine function is in each case equal in magnitude to&nbsp; $1$, and it holds:
+*With the arguments&nbsp; $f \cdot T = 0.5,\ 1.5,\ 2.5,\hspace{0.05cm}\text{ ... }$&nbsp; the sine function is in each case equal in magnitude to&nbsp; $1$, and it holds:
 :$$\left| {X( f )} \right| = \frac{A}{ {{\rm{\pi }}\left| f \right|}} = X_{\rm S} ( f ).$$
-*At other frequencies,&nbsp; $X_{\rm S}(f)$&nbsp; als obere Schranke, serves as an upper bound, i.e.&nbsp; $|Xf)| \leq X_{\rm S}(f)$ always applies.
+*At other frequencies,&nbsp; $X_{\rm S}(f)$&nbsp; serves as an upper bound, i.e.&nbsp; $|Xf)| \leq X_{\rm S}(f)$&nbsp; always applies.
-*In the sketch, this bound is drawn as a violet curve in addition to&nbsp; $|X(f)|$&nbsp;.
+*In the sketch, this bound is drawn as a violet curve in addition to&nbsp; $|X(f)|$.
@@ Line 126: / Line 127: @@
 :$$\varphi ( f ) = \arctan \left( {\frac{ {1 - \cos ( {\omega T} )}}{ {\sin ( {\omega T} )}}} \right).$$
-*The argument of this function is equal to&nbsp; $\tan(\omega T/2) = \tan(\pi fT)$ according to the specification. From this follows a linearly increasing course with the frequency:
+*The argument of this function is equal to&nbsp; $\tan(\omega T/2) = \tan(\pi fT)$&nbsp; according to the specification.&nbsp; From this follows a linearly increasing course with frequency:
 :$$\varphi ( f ) = \arctan \left( {\tan ( { {\rm{\pi }}fT} )} \right) = {\rm{\pi }}fT.$$

@@ Line 20: / Line 20: @@
 ''Hints:''
-*This task belongs to the chapter&nbsp;  [[Signal_Representation/Special_Cases_of_Impulse_Signals|Special Cases of Impulse Signals]].
+*This task belongs to the chapter&nbsp;  [[Signal_Representation/Special_Cases_of_Pulses|Special Cases of Pulses]].
 *The following trigonometric transformations are given:

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Sig_A_3_3.png|250px|right|frame|Rechteckimpuls und zugehöriges Spektrum]]
+[[File:EN_Sig_A_3_3.png|250px|right|frame|Rectangular pulse and its spectrum]]
-A rectangular pulse&nbsp; $x(t)$&nbsp; with a duration of&nbsp; $T = 50\,\text{µs}$&nbsp; and  the height of&nbsp; $A = 2\,\text{V}$ is considered. At the jump points at&nbsp; $t = 0$&nbsp; and&nbsp; $t = T$&nbsp; the signal value is&nbsp; $A/2$, in each case, but this has no influence on the solution of the task.
+A rectangular pulse&nbsp; $x(t)$&nbsp; with duration&nbsp; $T = 50\,\text{µs}$&nbsp; and  height&nbsp; $A = 2\,\text{V}$ is considered.&nbsp; At the jumping points at&nbsp; $t = 0$&nbsp; and&nbsp; $t = T$&nbsp; the signal value is&nbsp; $A/2$&nbsp; in each case, but this has no influence on the solution of the task.
-In the graphic below, the corresponding spectral function is sketched qualitatively according to magnitude and phase. It is valid:
+In the lower graph, the corresponding spectral function is sketched qualitatively according to magnitude and phase.&nbsp; It is valid:
 :$$X( f ) = \left| {X( f )} \right| \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \varphi ( f )} .$$
@@ Line 20: / Line 20: @@
 ''Hints:''
-*This exercise belongs to the chapter&nbsp;  [[Signal_Representation/Special_Cases_of_Impulse_Signals|Special Cases of Impulse Signals]].
+*This task belongs to the chapter&nbsp;  [[Signal_Representation/Special_Cases_of_Impulse_Signals|Special Cases of Impulse Signals]].
-*The following trigonometric transformations are also given:
+*The following trigonometric transformations are given:
 :$$\sin ^2( \alpha ) = {1}/{2} \cdot \big( {1 - \cos ( {2\alpha } )} \big),\hspace{0.5cm} \tan( {\alpha /2} ) = \frac{ {1 - \cos ( \alpha  )}}{ {\sin ( \alpha  )}}.$$
@@ Line 31: / Line 31: @@
 <quiz display=simple>
-{Calculate the spectral function&nbsp; $X(f)$. in general. What value results at the frequency&nbsp; $f=10\, \text{kHz}$?
+{Calculate the spectral function&nbsp; $X(f)$&nbsp; in general.&nbsp; What value results at the frequency&nbsp; $f=10\, \text{kHz}$?
 |type="{}"}
 ${\rm Re}\big[X(f=10 \,\text{kHz})\big] \ = \ $ { 0. } &nbsp; $\text{mV/Hz}$
 ${\rm Im}\big[X(f=10\, \text{kHz})\big]\ = \ $ { -0.064--0.06 } &nbsp; $\text{mV/Hz}$
-{Calculate the magnitude function&nbsp; $|X(f)|$&nbsp; in general. What values result for the frequencies&nbsp; $f = 0$&nbsp; and&nbsp; $f=20 \,\text{kHz}$?
+{Calculate the magnitude function&nbsp; $|X(f)|$&nbsp; in general.&nbsp; What values result for the frequencies&nbsp; $f = 0$&nbsp; and&nbsp; $f=20 \,\text{kHz}$?
 |type="{}"}
 $|X(f=0)|\ = \ $ { 0.1 3% } &nbsp; $\text{mV/Hz}$
 $|X(f=20\, \text{kHz})|\ = \ $ { 0. } &nbsp; $\text{mV/Hz}$
-{Which of the following statements are true regarding&nbsp; $|X(f)|$&nbsp;?
+{Which of the following statements are true regarding&nbsp; $|X(f)|$?
 |type="[]"}
 + $|X(f)|$&nbsp; has zeros at multiples of&nbsp; $f_0 = 1/T$.
@@ Line 47: / Line 47: @@
 + In the middle between two zeros&nbsp; $|X(f)| = |A/(\pi f)|$ is true.
-{Calculate the phase function&nbsp; $\varphi (f)$. What phase angle (in degrees) results at the frequency&nbsp; $f=10\, \text{kHz}$?
+{Calculate the phase function&nbsp; $\varphi (f)$.&nbsp; What phase angle (in degrees) results at the frequency&nbsp; $f=10\, \text{kHz}$?
 |type="{}"}
 $\varphi (f=10\, \text{kHz})\ = \ $ { 90 1% } &nbsp; $\text{deg}$

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← Older revision		Revision as of 12:39, 23 March 2021
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@@ Line 93: / Line 93: @@
 :$$\left| {X( {f = 20\;{\rm{kHz}}} )} \right| = \frac{ {A \cdot T}}{ {\rm{\pi }}} \cdot \sin ( {\rm{\pi }} ) \hspace{0.15 cm}\underline{= 0}.$$
-*When calculating the value for&nbsp; $f = 0$&nbsp; the quotient&nbsp; $\text{0 divided by 0}$. By applying&nbsp; [https://de.wikipedia.org/wiki/Regel_von_de_l%E2%80%99Hospital l'Hospitalschen Regel]&nbsp; the limiting value can be calculated:
+*When calculating the value for&nbsp; $f = 0$&nbsp; the quotient&nbsp; $\text{0 divided by 0}$. By applying&nbsp; [https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule L'Hospitals rule]&nbsp; the limiting value can be calculated:
 :$$\mathop {\lim }\limits_{x \to 0 } \frac{ {\sin \left( x \right)}}{x} = 1.$$
-*Daraus folgt:
+*From this follows:
 :$$\left| {X( {f = 0} )} \right| = A \cdot T \hspace{0.15 cm}\underline{= 0.1 \;{\rm{mV/Hz}}}{\rm{.}}$$
-*Dieses Ergebnis ist einsichtig, da nach dem ersten Fourierintegral der Spektralwert bei&nbsp; $f = 0$&nbsp; genau der Fläche unter der Zeitfunktion entspricht.
+*This result is obvious because, according to the first Fourier integral, the spectral value at&nbsp; $f = 0$&nbsp; corresponds exactly to the area under the time function.
 [[File:P_ID563__Sig_A_3_3_c.png|right|frame|Betragsspektrum des Rechteckimpulses]]
-'''(3)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 3</u>:
+'''(3)'''&nbsp; Proposed<u> solutions 1 and 3</u> are correct::
-*Entsprechend dem Ergebnis zur Teilaufgabe&nbsp; '''(2)'''&nbsp; treten die Nullstellen im Abstand&nbsp; $f_0 = 1/T$&nbsp; auf.
+*According to the result of subtask&nbsp; '''(2)'''&nbsp; the zeros occur at the distance&nbsp; $f_0 = 1/T$&nbsp;.
-*Bei&nbsp; $f_0 = 1/(2T) = f = 10 \;{\rm{kHz}}$&nbsp;  ist zwar der Realteil&nbsp; $0$, aber nicht der Imaginärteil.
+*With&nbsp; $f_0 = 1/(2T) = f = 10 \;{\rm{kHz}}$&nbsp;  the real part is&nbsp; $0$, but not the imaginary part.
-*Bei den Argumenten&nbsp; $f \cdot T = 0.5, 1.5, 2.5,\hspace{0.05cm}\text{ ... }$&nbsp; ist die Sinusfunktion jeweils betragsmäßig gleich&nbsp; $1$, und es gilt:
+*With the arguments&nbsp; $f \cdot T = 0.5, 1.5, 2.5,\hspace{0.05cm}\text{ ... }$&nbsp; the sine function is in each case equal in magnitude to&nbsp; $1$, and it holds:
 :$$\left| {X( f )} \right| = \frac{A}{ {{\rm{\pi }}\left| f \right|}} = X_{\rm S} ( f ).$$
-*Bei anderen Frequenzen dient&nbsp; $X_{\rm S}(f)$&nbsp; als obere Schranke, das heißt, es gilt stets&nbsp; $|Xf)| \leq X_{\rm S}(f)$.
+*At other frequencies,&nbsp; $X_{\rm S}(f)$&nbsp; als obere Schranke, serves as an upper bound, i.e.&nbsp; $|Xf)| \leq X_{\rm S}(f)$ always applies.
-*In der Skizze ist diese Schranke zusätzlich zu&nbsp; $|X(f)|$&nbsp; als violette Kurve eingezeichnet.
+*In the sketch, this bound is drawn as a violet curve in addition to&nbsp; $|X(f)|$&nbsp;.
-'''(4)'''&nbsp; Nach der Definition auf der Angabenseite kann man die Phasenfunktion wie folgt berechnen:
+'''(4)'''&nbsp; According to the definition on the information page, one can calculate the phase function as follows:
 :$$\varphi ( f ) =  - \arctan \frac{ { {\mathop{\rm Im}\nolimits} ( f )}}{ { {\mathop{\rm Re}\nolimits} ( f )}}.$$
-*Mit den Ergebnissen aus Teilaufgabe&nbsp; '''(1)'''&nbsp; gilt somit:
+*With the results from subtask&nbsp; '''(1)'''&nbsp; the following thus applies:
 :$$\varphi ( f ) = \arctan \left( {\frac{ {1 - \cos ( {\omega T} )}}{ {\sin ( {\omega T} )}}} \right).$$
-*Das Argument dieser Funktion ist entsprechend der Angabe gleich&nbsp; $\tan(\omega T/2) = \tan(\pi fT)$. Daraus folgt ein mit der Frequenz linear ansteigender Verlauf:
+*The argument of this function is equal to&nbsp; $\tan(\omega T/2) = \tan(\pi fT)$ according to the specification. From this follows a linearly increasing course with the frequency:
 :$$\varphi ( f ) = \arctan \left( {\tan ( { {\rm{\pi }}fT} )} \right) = {\rm{\pi }}fT.$$
-*Mit&nbsp; $f = 10\,\text{kHz}$&nbsp;  und&nbsp; $T = 50\,\text{µs}$&nbsp;  erhält man daraus den Phasenwinkel&nbsp; $\pi /2$&nbsp; entsprechend&nbsp; $\underline{90^{\circ}}$ .
+*With&nbsp; $f = 10\,\text{kHz}$&nbsp;  and&nbsp; $T = 50\,\text{µs}$&nbsp;  one obtains from this the phase angle&nbsp; $\pi /2$&nbsp; corresponding to&nbsp; $\underline{90^{\circ}}$ .
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 [[Category:Exercises for Signal Representation|^3.2 Special Cases of Impulse Signals^]]

@@ Line 54: / Line 54: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Mit der Abkürzung&nbsp; $\omega = 2\pi f$&nbsp; lautet die Spektralfunktion gemäß dem ersten Fourierintegral:
+'''(1)'''&nbsp; With the abbreviation&nbsp; $\omega = 2\pi f$&nbsp; the spectral function is according to the first Fourier integral:
 :$$X( f ) = \int_0^T {A \cdot {\rm{e}}^{-{\rm{j}}\omega t} \hspace{0.05cm}{\rm d}t = } \int_0^T {A \cdot \cos \left( {\omega t} \right)\hspace{0.05cm}{\rm d}t  }\hspace{0.1cm}-\hspace{0.1cm} {\rm{j}} \cdot \int_{\rm{0}}^T {A \cdot \sin ( {\omega t} )} \hspace{0.05cm}{\rm d}t.$$
-*Nach Integration und Einsetzen der Grenzen folgt daraus:
+*After integration and insertion of the limits, it follows:
 :$${\mathop{\rm Re}\nolimits} [ {X( f )} ] = \frac{A}{\omega } \cdot \sin( {\omega T} ),$$
@@ Line 66: / Line 66: @@
 :$${\mathop{\rm Im}\nolimits} [ {X( f)} ] = \frac{A}{\omega } \cdot \left( {\cos ( {\omega T}) - 1} \right) =  - \frac{A}{\omega } \cdot\left( {1 - \cos ( {\omega T} )} \right).$$
-*Für die Frequenz&nbsp; $f = 1/(2T) = 10\, \text{kHz}$  &nbsp; &rArr; &nbsp;  $\omega \cdot T = \pi$&nbsp;  erhält man:
+*For the frequency&nbsp; $f = 1/(2T) = 10\, \text{kHz}$  &nbsp; &rArr; &nbsp;  $\omega \cdot T = \pi$&nbsp;  we get::
 :$${\mathop{\rm Re}\nolimits}[{X( {f = 10 \;{\rm{kHz}}} )}] = \frac{A}{ {2{\rm{\pi }}f}} \cdot \sin ( {\rm{\pi }} ) \hspace{0.15 cm}\underline{= 0},$$
@@ Line 73: / Line 73: @@
-'''(2)'''&nbsp; Das Betragsquadrat ist die Summe von Real- und Imaginärteil, jeweils quadriert:
+'''(2)'''&nbsp; The magnitude square is the sum of the real and imaginary parts, squared in each case:
 :$$\left| {X( f )} \right|^2  = \frac{ {A^2 }}{ {\omega ^2 }}\big[ {\sin ^2 ( {\omega T} ) + 1 - 2 \cdot \cos ( {\omega T}) + \cos ^2 ( {\omega T} )} \big].$$
-*Wegen&nbsp; $\sin^2(\alpha) + \cos^2(\alpha) = 1$&nbsp; kann hierfür auch geschrieben werden:
+*Because of&nbsp; $\sin^2(\alpha) + \cos^2(\alpha) = 1$&nbsp;, this can also be written:
 :$$\left| {X( f )} \right|^2  = \frac{ {2A^2 }}{ {\omega ^2 }} \cdot \big( {1 - \cos ( {\omega T} )} \big) = \frac{ {4A^2 }}{ {\omega ^2 }} \cdot \sin ^2( {\omega T/2} ).$$
-*Setzt man für&nbsp; $\omega = 2\pi f$&nbsp; und zieht die Wurzel, so erhält man unter der Voraussetzung&nbsp; $A > 0$:
+*Setting &nbsp; $\omega = 2\pi f$&nbsp; and taking the root, we obtain, under the condition&nbsp; $A > 0$:
 :$$\left| {X( f )} \right| =  \left| \frac{A}{ {\rm\pi }f} \cdot \sin ( {\rm \pi }fT ) \right| = A \cdot T \cdot \left| \frac{\sin ( {\rm\pi }fT )}{ {\rm \pi}fT} \right|.$$
-*Mit der Abkürzung&nbsp; $\text{si}(x) = \sin(x)/x$&nbsp; lautet das Ergebnis:
+*With the shortcut&nbsp; $\text{si}(x) = \sin(x)/x$&nbsp; the result is:
 :$$\left| {X( f)} \right| = A \cdot T \cdot\left|{\rm si} ( { {\rm{\pi }}fT} ) \right|.$$
-*Der Spektralwert bei der Frequenz&nbsp; $f = 1/T = \text{20 kHz}$&nbsp; ergibt sich zu
+*The spectral value at the frequency&nbsp; $f = 1/T = \text{20 kHz}$&nbsp; is given by
 :$$\left| {X( {f = 20\;{\rm{kHz}}} )} \right| = \frac{ {A \cdot T}}{ {\rm{\pi }}} \cdot \sin ( {\rm{\pi }} ) \hspace{0.15 cm}\underline{= 0}.$$
-*Bei der Berechnung des Wertes für&nbsp; $f = 0$&nbsp; erscheint der Quotient&nbsp; $\text{0 durch 0}$. Durch Anwendung der&nbsp; [https://de.wikipedia.org/wiki/Regel_von_de_l%E2%80%99Hospital l'Hospitalschen Regel]&nbsp; kann der Grenzwert berechnet werden:
+*When calculating the value for&nbsp; $f = 0$&nbsp; the quotient&nbsp; $\text{0 divided by 0}$. By applying&nbsp; [https://de.wikipedia.org/wiki/Regel_von_de_l%E2%80%99Hospital l'Hospitalschen Regel]&nbsp; the limiting value can be calculated:
 :$$\mathop {\lim }\limits_{x \to 0 } \frac{ {\sin \left( x \right)}}{x} = 1.$$

@@ Line 5: / Line 5: @@
 [[File:EN_Sig_A_3_3.png|250px|right|frame|Rechteckimpuls und zugehöriges Spektrum]]
-Betrachtet wird ein Rechteckimpuls&nbsp; $x(t)$&nbsp; der Dauer&nbsp; $T = 50\,\text{µs}$&nbsp; und der Höhe&nbsp; $A = 2\,\text{V}$. An den Sprungstellen bei&nbsp; $t = 0$&nbsp; und&nbsp; $t = T$&nbsp; ist der Signalwert jeweils&nbsp; $A/2$, was aber für die Lösung der Aufgabe keinen Einfluss hat.
+A rectangular pulse&nbsp; $x(t)$&nbsp; with a duration of&nbsp; $T = 50\,\text{µs}$&nbsp; and  the height of&nbsp; $A = 2\,\text{V}$ is considered. At the jump points at&nbsp; $t = 0$&nbsp; and&nbsp; $t = T$&nbsp; the signal value is&nbsp; $A/2$, in each case, but this has no influence on the solution of the task.
-In der unteren Grafik ist die dazugehörige Spektralfunktion nach Betrag und Phase qualitativ skizziert. Es gilt:
+In the graphic below, the corresponding spectral function is sketched qualitatively according to magnitude and phase. It is valid:
 :$$X( f ) = \left| {X( f )} \right| \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \varphi ( f )} .$$
-Der analytische Funktionsverlauf von&nbsp; $X(f)$&nbsp; soll ermittelt werden.
+The analytical function progression of&nbsp; $X(f)$&nbsp; is to be determined.
@@ Line 19: / Line 19: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum Kapitel&nbsp;  [[Signal_Representation/Special_Cases_of_Impulse_Signals|Einige Sonderfälle impulsartiger Signale]].
+*This exercise belongs to the chapter&nbsp;  [[Signal_Representation/Special_Cases_of_Impulse_Signals|Special Cases of Impulse Signals]].
-*Gegeben sind weiterhin folgende trigonometrischen Umformungen:
+*The following trigonometric transformations are also given:
 :$$\sin ^2( \alpha ) = {1}/{2} \cdot \big( {1 - \cos ( {2\alpha } )} \big),\hspace{0.5cm} \tan( {\alpha /2} ) = \frac{ {1 - \cos ( \alpha  )}}{ {\sin ( \alpha  )}}.$$
@@ Line 28: / Line 28: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>