Aufgaben:Exercise 3.6: Even and Odd Time Signals - Revision history

Guenter at 13:22, 27 April 2021

2021-04-27T13:22:40Z

Guenter at 13:12, 27 April 2021

2021-04-27T13:12:06Z

Guenter at 13:07, 27 April 2021

2021-04-27T13:07:33Z

Guenter: Guenter moved page Aufgaben:Exercise 3.6: Even/Odd Time Signal to Aufgaben:Exercise 3.6: Even and Odd Time Signals

2021-04-27T12:55:28Z

Guenter moved page Exercise 3.6: Even/Odd Time Signal to Exercise 3.6: Even and Odd Time Signals

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

2021-03-23T12:39:40Z

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

Noah at 17:36, 27 January 2021

2021-01-27T17:36:37Z

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 3.6: Gerades/ungerades Zeitsignal to Aufgaben:Exercise 3.6: Even/Odd Time Signal

2020-11-02T15:30:01Z

Oezdemir moved page Aufgabe 3.6: Gerades/ungerades Zeitsignal to Exercise 3.6: Even/Odd Time Signal

Oezdemir at 15:24, 2 November 2020

2020-11-02T15:24:30Z

Javier: Text replacement - "Signaldarstellung/Gesetzmäßigkeiten der Fouriertransformation" to "Signal_Representation/Fourier_Transform_Laws"

2020-09-01T09:49:54Z

Text replacement - "Signaldarstellung/Gesetzmäßigkeiten der Fouriertransformation" to "Signal_Representation/Fourier_Transform_Laws"

Javier: Text replacement - "Signal_Representation/Gesetzmäßigkeiten_der_Fouriertransformation" to "Signal_Representation/Fourier_Transform_Laws"

2020-09-01T09:48:59Z

Text replacement - "Signal_Representation/Gesetzmäßigkeiten_der_Fouriertransformation" to "Signal_Representation/Fourier_Transform_Laws"

@@ Line 57: / Line 57: @@
 :$$U( {f = 0.5\;{\rm{kHz}}} ) =  - {\rm{j}} \cdot \frac{ {A_u  \cdot T}}{ {\rm{\pi }}} \cdot {\mathop{\rm si}\nolimits} ( {{ {\rm{\pi }}}/{2}} ) =  - {\rm{j}} \cdot \frac{2}{ { {\rm{\pi }}^{\rm{2}} }} \cdot A_{\rm u}  \cdot T.$$
-*The imaginary part is numerically&nbsp;  ${\rm Im}[U(f=0.5 \,\text{kHz})]\; \underline{\approx 0.2 \,\text{mV/Hz}}$.
+*The imaginary part is&nbsp;  ${\rm Im}[U(f=0.5 \,\text{kHz})]\; \underline{\approx 0.2 \,\text{mV/Hz}}$.
-*In contrast, the si function at&nbsp; $f \cdot T = 1$&nbsp; yields the value zero, while the cosine is equal to&nbsp; $-1$&nbsp;. Thus, with&nbsp; $A_u = 1\,\text{V}$&nbsp; and&nbsp; $T = 1\,\text{ms}$ one obtains:
+*In contrast, the&nbsp; $\rm si$&ndash;function at&nbsp; $f \cdot T = 1$&nbsp; yields the value zero, while the cosine is equal to&nbsp; $-1$.&nbsp; Thus with&nbsp; $A_u = 1\,\text{V}$&nbsp; and&nbsp; $T = 1\,\text{ms}$&nbsp; one obtains:
 :$$U( {f = 1\;{\rm{kHz}}} ) = {\rm{j}} \cdot \frac{ {A_{\rm u} \cdot T}}{ { {\rm{2\pi }}}} \hspace{0.3 cm} \Rightarrow \hspace{0.3 cm} {\rm Re} [\text{...}] \hspace{0.15 cm}\underline{ =  0}, \hspace{0.3 cm}{\rm Im} [\text{...}] \hspace{0.15 cm}\underline{\approx  0.159 \;{\rm{mV/Hz}}}.$$
@@ Line 64: / Line 64: @@
-'''(2)'''&nbsp; According to the mapping theorem, an odd time function&nbsp; $u(t)$&nbsp; always has an imaginary and at the same time odd spectrum &nbsp;
+'''(2)'''&nbsp; According to the&nbsp; "Assignment Theorem", an odd time function&nbsp; $u(t)$&nbsp; always has an imaginary and at the same time odd spectrum&nbsp; $U( { - f} ) =  - U( f )$.&nbsp; With the boundary transition&nbsp; $f \rightarrow \infty$&nbsp; follows from the given equation
 :$$U( f ) =  - {\rm{j}} \cdot \frac{ {A_u  \cdot T}}{ {2{\rm{\pi }}fT}}\big[ { {\mathop{\rm si}\nolimits} ( {{\rm{\pi }}fT} ) - \cos ( { {\rm{\pi }}fT} )} \big]$$
-the result&nbsp; $U(f = 0) = 0$. Formally, one could confirm this result by applying l'Hospital's rule.
+the result&nbsp; $U(f = 0) = 0$.&nbsp; Formally, one could confirm this result by applying l'Hospital's rule.
 We proceed a little more pragmatically.
@@ Line 77: / Line 76: @@
 *For even smaller frequency values, the result also becomes smaller and smaller.
-*You get to the result&nbsp; $U(f = 0)\;\underline{ = 0}$, more quickly if you take into account that the integral over&nbsp; $u(t)$&nbsp; disappears.
+*You get to the result&nbsp; $U(f = 0)\;\underline{ = 0}$&nbsp; more quickly if you take into account that the integral over&nbsp; $u(t)$&nbsp; disappears.
 *So you don't have to calculate at all.
-'''(3)'''&nbsp; The signal&nbsp; $x(t)$&nbsp; can be divided into the even and the odd part, which lead to the even real part and the odd imaginary part of&nbsp; $X(f)$&nbsp;:
+'''(3)'''&nbsp; The signal&nbsp; $x(t)$&nbsp; can be divided into an even and an odd part, which lead to the even real part and the odd imaginary part of&nbsp; $X(f)$&nbsp;:
-*The even part is equal to the function&nbsp; $g(t)$&nbsp; with&nbsp; $A_g = 3\,\text{V}$. From this follows for the real part of the spectral value at&nbsp; $f \cdot T = 0.5$:
+*The even part is equal to the function&nbsp; $g(t)$&nbsp; with&nbsp; $A_g = 3\,\text{V}$.&nbsp; From this follows for the real part of the spectral value at&nbsp; $f \cdot T = 0.5$:
 :$${\mathop{\rm Re}\nolimits} \left[ {X( {f \cdot T = 0.5} )} \right] = A_{\rm g}  \cdot T \cdot {\mathop{\rm si}\nolimits} ( {{ {\rm{\pi }}}/{2}} ) \hspace{0.15 cm}\underline{= 1.91  \;{\rm{mV/Hz}}}{\rm{.}}$$
-*The imaginary part results from the spectral function&nbsp; $U(f)$ with $A_u = 1\,\text{V}$. This was already calculated in subtask&nbsp; '''(1)'''&nbsp;:
+*The imaginary part results from the spectral function&nbsp; $U(f)$&nbsp; with $A_u = 1\,\text{V}$.&nbsp; This was already calculated in subtask&nbsp; '''(1)''':
 :$${\mathop{\rm Im}\nolimits} \left[ {X( {f \cdot T = 0.5} )} \right] \hspace{0.15 cm}\underline{\approx  - 0.2  \;{\rm{mV/Hz}}}{\rm{.}}$$

@@ Line 23: / Line 23: @@
 *This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_Theorems|Fourier Transform Theorems]].
 *All the theorems presented here are illustrated with examples in the (German language) learning video<br> &nbsp; &nbsp; &nbsp;[[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)|Gesetzmäßigkeiten der Fouriertransformation]] &nbsp; &rArr; &nbsp;  "Regularities to the Fourier transform".
-*Solve this task with the help of the&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Mapping_theorem|mapping theorem]].
+*Solve this task with the help of the&nbsp; [[Signal_Representation/Fourier_Transform_Theorems#Assignment_Theorem|Assignment Theorem]].
-*Use the signal parameters&nbsp; $A_u = 1\,\text{ V}$&nbsp; and&nbsp; $T = 1\,\text{ ms}$ for the first two subtasks.
+*For the first two subtasks use the signal parameters&nbsp; $A_u = 1\,\text{V}$&nbsp; and&nbsp; $T = 1\,\text{ms}$.
@@ Line 32: / Line 32: @@
 <quiz display=simple>
-{Calculate the (purely imaginary) spectral values of the unbalanced signal&nbsp; $u(t)$&nbsp; at the frequencies&nbsp; $f = 0.5\,\text{kHz}$&nbsp; and&nbsp; $f = 1\,\text{kHz}$.
+{Calculate the (imaginary) spectral values of the odd signal&nbsp; $u(t)$&nbsp; at the frequencies&nbsp; $f = 0.5\,\text{kHz}$&nbsp; and&nbsp; $f = 1\,\text{kHz}$.
 |type="{}"}
 ${\rm Im}\big[U(f=0.5 \,\text{kHz})\big] \ = \ $ { -0.205--0.195 } &nbsp;$\text{mV/Hz}$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID516__Sig_A_3_6_neu.png|250px|right|frame|&bdquo;Keilfunktion&rdquo; sowie ein gerades und ein ungerades Zeitsignal]]
+[[File:P_ID516__Sig_A_3_6_neu.png|250px|right|frame|"Wedge function" as well as an even and an odd time signal]]
-We are looking for the spectrum&nbsp; $X(f)$&nbsp; of the pulse-shaped signal&nbsp; $x(t)$ sketched opposite, which rises linearly from&nbsp; $2\,\text{ V}$&nbsp; to&nbsp; $4\,\text{ V}$&nbsp; in the range from&nbsp; $–T/2$&nbsp; to&nbsp; $+T/2$&nbsp; and is zero outside.
+We are looking for the spectrum&nbsp; $X(f)$&nbsp; of the pulse&nbsp; $x(t)$ sketched opposite, which rises linearly from&nbsp; $2\hspace{0.05cm} \text{V}$&nbsp; to&nbsp; $4\hspace{0.05cm} \text{V}$&nbsp; in the range from&nbsp; $–T/2$&nbsp; to&nbsp; $+T/2$&nbsp; and is zero outside.
 The spectral functions of the signals&nbsp; $g(t)$&nbsp; and&nbsp; $u(t)$&nbsp; shown below are assumed to be known:
-*The even, rectangular time function&nbsp; $g(t)$&nbsp; has the spectrum
+*The even&nbsp;(German:&nbsp; gerade) rectangular time function&nbsp; $g(t)$&nbsp; has the spectrum
-:$$G( f ) = A_g  \cdot T \cdot {\mathop{\rm si}\nolimits}( { {\rm{\pi }}fT} ) \hspace{0.3cm} {\rm{mit}}\hspace{0.3cm} {\mathop{\rm si}\nolimits}( x ) =  {\sin ( x )}/{x}.$$
+:$$G( f ) = A_g  \cdot T \cdot {\mathop{\rm si}\nolimits}( { {\rm{\pi }}fT} ) \hspace{0.3cm} {\rm{with}}\hspace{0.3cm} {\mathop{\rm si}\nolimits}( x ) =  {\sin ( x )}/{x}.$$
-*The spectrum of the asymmetric function&nbsp; $u(t)$&nbsp; is:
+*The spectrum of the odd&nbsp;(German:&nbsp; ungerade) function&nbsp; $u(t)$&nbsp; is:
 :$$U( f ) =  - {\rm{j}} \cdot \frac{ {A_u  \cdot T}}{ {2{\rm{\pi }}fT}}\big[ {{\mathop{\rm si}\nolimits} ( { {\rm{\pi }}fT} ) - \cos ( { {\rm{\pi }}fT} )} \big].$$
@@ Line 24: / Line 21: @@
 ''Hints:''
-*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_Laws|Fourier Transform Laws]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_Theorems|Fourier Transform Theorems]].
-*All of these laws are illustrated with examples in the learning video&nbsp; [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)|Gesetzmäßigkeiten der Fouriertransformation]]&nbsp;.
+*All the theorems presented here are illustrated with examples in the (German language) learning video<br> &nbsp; &nbsp; &nbsp;[[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)|Gesetzmäßigkeiten der Fouriertransformation]] &nbsp; &rArr; &nbsp;  "Regularities to the Fourier transform".
 *Solve this task with the help of the&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Mapping_theorem|mapping theorem]].
 *Use the signal parameters&nbsp; $A_u = 1\,\text{ V}$&nbsp; and&nbsp; $T = 1\,\text{ ms}$ for the first two subtasks.

← Older revision		Revision as of 12:39, 23 March 2021
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−	[[Category:~~Exercises for~~ Signal Representation\|^3.3 Fourier Transform Theorems^]]	+	[[Category:Signal Representation: Exercises\|^3.3 Fourier Transform Theorems^]]

@@ Line 5: / Line 5: @@
 [[File:P_ID516__Sig_A_3_6_neu.png|250px|right|frame|&bdquo;Keilfunktion&rdquo; sowie ein gerades und ein ungerades Zeitsignal]]
-Gesucht ist das Spektrum&nbsp; $X(f)$&nbsp; des nebenstehend skizzierten impulsförmigen Signals&nbsp; $x(t)$, das im Bereich von&nbsp; $–T/2$&nbsp; bis&nbsp; $+T/2$&nbsp; linear von&nbsp; $2\,\text{ V}$&nbsp; auf&nbsp; $4\,\text{ V}$&nbsp;  ansteigt und außerhalb Null ist.
+We are looking for the spectrum&nbsp; $X(f)$&nbsp; of the pulse-shaped signal&nbsp; $x(t)$ sketched opposite, which rises linearly from&nbsp; $2\,\text{ V}$&nbsp; to&nbsp; $4\,\text{ V}$&nbsp; in the range from&nbsp; $–T/2$&nbsp; to&nbsp; $+T/2$&nbsp; and is zero outside.
-Die Spektralfunktionen der unten dargestellten Signale&nbsp; $g(t)$&nbsp; und&nbsp; $u(t)$&nbsp; werden als bekannt vorausgesetzt:
+The spectral functions of the signals&nbsp; $g(t)$&nbsp; and&nbsp; $u(t)$&nbsp; shown below are assumed to be known:
-*Die gerade, rechteckförmige Zeitfunktion&nbsp; $g(t)$&nbsp; hat das Spektrum
+*The even, rectangular time function&nbsp; $g(t)$&nbsp; has the spectrum
 :$$G( f ) = A_g  \cdot T \cdot {\mathop{\rm si}\nolimits}( { {\rm{\pi }}fT} ) \hspace{0.3cm} {\rm{mit}}\hspace{0.3cm} {\mathop{\rm si}\nolimits}( x ) =  {\sin ( x )}/{x}.$$
-*Das Spektrum der unsymmetrischen Funktion&nbsp; $u(t)$&nbsp; lautet:
+*The spectrum of the asymmetric function&nbsp; $u(t)$&nbsp; is:
 :$$U( f ) =  - {\rm{j}} \cdot \frac{ {A_u  \cdot T}}{ {2{\rm{\pi }}fT}}\big[ {{\mathop{\rm si}\nolimits} ( { {\rm{\pi }}fT} ) - \cos ( { {\rm{\pi }}fT} )} \big].$$
@@ Line 23: / Line 23: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Signal_Representation/Fourier_Transform_Laws|Gesetzmäßigkeiten der Fouriertransformation]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_Laws|Fourier Transform Laws]].
-*Alle diese Gesetzmäßigkeiten werden im Lernvideo&nbsp; [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)|Gesetzmäßigkeiten der Fouriertransformation]]&nbsp;  an Beispielen verdeutlicht.
+*All of these laws are illustrated with examples in the learning video&nbsp; [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)|Gesetzmäßigkeiten der Fouriertransformation]]&nbsp;.
-*Lösen Sie diese Aufgabe mit Hilfe des&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Zuordnungssatz|Zuordnungssatzes]].
+*Solve this task with the help of the&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Mapping_theorem|mapping theorem]].
-*Verwenden Sie für die beiden ersten Teilaufgaben die Signalparameter&nbsp; $A_u = 1\,\text{ V}$&nbsp; und&nbsp; $T = 1\,\text{ ms}$.
+*Use the signal parameters&nbsp; $A_u = 1\,\text{ V}$&nbsp; and&nbsp; $T = 1\,\text{ ms}$ for the first two subtasks.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Berechnen Sie die (rein imaginären) Spektralwerte des unsymmetrischen Signals&nbsp; $u(t)$&nbsp; bei den Frequenzen&nbsp; $f = 0.5\,\text{kHz}$&nbsp; und&nbsp; $f = 1\,\text{kHz}$.
+{Calculate the (purely imaginary) spectral values of the unbalanced signal&nbsp; $u(t)$&nbsp; at the frequencies&nbsp; $f = 0.5\,\text{kHz}$&nbsp; and&nbsp; $f = 1\,\text{kHz}$.
 |type="{}"}
 ${\rm Im}\big[U(f=0.5 \,\text{kHz})\big] \ = \ $ { -0.205--0.195 } &nbsp;$\text{mV/Hz}$
 ${\rm Im}\big[U(f=1.0 \,\text{kHz})\big]\ = \ $ { 0.159 3% } &nbsp;$\text{mV/Hz}$
-{Wie groß ist der Spektralwert von&nbsp; $u(t)$&nbsp; bei der Frequenz&nbsp; $f = 0$? &nbsp; &nbsp;
+{What is the spectral value of&nbsp; $u(t)$&nbsp; at the frequency&nbsp; $f = 0$? &nbsp; &nbsp;
-''Hinweis'': Lieber denken als rechnen.
+''Hint'': Think before you calculate.
 |type="{}"}
 ${\rm Im}\big[U(f=0)\big]\ = \ $ { 0. } &nbsp;$\text{mV/Hz}$
-{Berechnen Sie unter Verwendung des Ergebnisses aus&nbsp; '''(1)'''&nbsp; den Spektralwert des Signals&nbsp; $x(t)$&nbsp; bei der Frequenz&nbsp; $f=0.5 \,\text{kHz}$.
+{Using the result from&nbsp; '''(1)'''&nbsp; calculate the spectral value of the signal&nbsp; $x(t)$&nbsp; at the frequency&nbsp; $f=0.5 \,\text{kHz}$.
 |type="{}"}
 ${\rm Re}\big[X(f=0.5 \,\text{kHz})\big]\ = \ $ { 1.91 3% } &nbsp;$\text{mV/Hz}$
@@ Line 53: / Line 53: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Für&nbsp; $f \cdot T  = 0.5$&nbsp; erhält man aus der angegebenen Gleichung:
+'''(1)'''&nbsp;  For&nbsp; $f \cdot T  = 0.5$&nbsp; one obtains from the given equation:
 :$$U( {f = 0.5\;{\rm{kHz}}} ) =  - {\rm{j}} \cdot \frac{ {A_u  \cdot T}}{ {\rm{\pi }}} \cdot {\mathop{\rm si}\nolimits} ( {{ {\rm{\pi }}}/{2}} ) =  - {\rm{j}} \cdot \frac{2}{ { {\rm{\pi }}^{\rm{2}} }} \cdot A_{\rm u}  \cdot T.$$
-*Der Imaginärteil ist zahlenmäßig&nbsp;  ${\rm Im}[U(f=0.5 \,\text{kHz})]\; \underline{\approx 0.2 \,\text{mV/Hz}}$.
+*The imaginary part is numerically&nbsp;  ${\rm Im}[U(f=0.5 \,\text{kHz})]\; \underline{\approx 0.2 \,\text{mV/Hz}}$.
-*Dagegen liefert die si-Funktion bei&nbsp; $f \cdot T = 1$&nbsp; den Wert Null, während der Cosinus gleich&nbsp; $-1$&nbsp; ist. Damit erhält man mit&nbsp; $A_u = 1\,\text{V}$&nbsp; und&nbsp; $T = 1\,\text{ms}$:
+*In contrast, the si function at&nbsp; $f \cdot T = 1$&nbsp; yields the value zero, while the cosine is equal to&nbsp; $-1$&nbsp;. Thus, with&nbsp; $A_u = 1\,\text{V}$&nbsp; and&nbsp; $T = 1\,\text{ms}$ one obtains:
 :$$U( {f = 1\;{\rm{kHz}}} ) = {\rm{j}} \cdot \frac{ {A_{\rm u} \cdot T}}{ { {\rm{2\pi }}}} \hspace{0.3 cm} \Rightarrow \hspace{0.3 cm} {\rm Re} [\text{...}] \hspace{0.15 cm}\underline{ =  0}, \hspace{0.3 cm}{\rm Im} [\text{...}] \hspace{0.15 cm}\underline{\approx  0.159 \;{\rm{mV/Hz}}}.$$
@@ Line 67: / Line 67: @@
-'''(2)'''&nbsp; Eine ungerade Zeitfunktion&nbsp; $u(t)$&nbsp; besitzt nach dem Zuordnungssatz stets ein imaginäres und gleichzeitig ungerades Spektrum: &nbsp;
+'''(2)'''&nbsp; According to the mapping theorem, an odd time function&nbsp; $u(t)$&nbsp; always has an imaginary and at the same time odd spectrum &nbsp;
-$U( { - f} ) =  - U( f ).$ Mit dem Grenzübergang&nbsp; $f \rightarrow \infty$&nbsp; folgt aus der angegebenen Gleichung
+$U( { - f} ) =  - U( f ).$ With the boundary transition&nbsp; $f \rightarrow \infty$&nbsp; follows from the given equation
 :$$U( f ) =  - {\rm{j}} \cdot \frac{ {A_u  \cdot T}}{ {2{\rm{\pi }}fT}}\big[ { {\mathop{\rm si}\nolimits} ( {{\rm{\pi }}fT} ) - \cos ( { {\rm{\pi }}fT} )} \big]$$
-das Ergebnis&nbsp; $U(f = 0) = 0$. Formal könnte man dieses Ergebnis durch Anwendung der l'Hospitalschen Regel bestätigen.
+the result&nbsp; $U(f = 0) = 0$. Formally, one could confirm this result by applying l'Hospital's rule.
-Wir gehen etwas pragmatischer vor.
+We proceed a little more pragmatically.
-*Setzen wir zum Beispiel&nbsp; $f \cdot T = 0.01$, so erhält man:
+*For example, if we set&nbsp; $f \cdot T = 0.01$, we obtain:
-:$$U( {f \cdot T = 0.01}) = -{\rm{j}} \cdot \frac{ {A_{\rm u} \cdot T}}{{0.02{\rm{\pi }}}}\big[ {{\mathop{\rm si}\nolimits} ( {0.01{\rm{\pi }}} ) - \cos ( {0.01{\rm{\pi }}} )} \big ] =  - {\rm{j}} \cdot \frac{ {A{\rm u} \cdot T}}{{0.02{\rm{\pi }}}}( {0.999836 - 0.999507} ) \approx  - {\rm{j}} \cdot 5 \cdot 10^{ - 6} \;{\rm{V/Hz}}{\rm{.}}$$
+$$U( {f \cdot T = 0.01}) = -{\rm{j}} \cdot \frac{ {A_{\rm u} \cdot T}}{{0.02{\rm{\pi }}}}\big[ {{\mathop{\rm si}\nolimits} ( {0.01{\rm{\pi }}} ) - \cos ( {0.01{\rm{\pi }}} )} \big ] =  - {\rm{j}} \cdot \frac{ {A{\rm u} \cdot T}}{{0.02{\rm{\pi }}}}( {0.999836 - 0.999507} ) \approx  - {\rm{j}} \cdot 5 \cdot 10^{ - 6} \;{\rm{V/Hz}}{\rm{.}}$$
-*Für noch kleinere Frequenzwerte wird auch das Ergebnis immer kleiner.
+*For even smaller frequency values, the result also becomes smaller and smaller.
-*Schneller kommt man zum Ergebnis&nbsp; $U(f = 0)\;\underline{ = 0}$, wenn man berücksichtigt, dass das Integral über&nbsp; $u(t)$&nbsp; verschwindet.
+*You get to the result&nbsp; $U(f = 0)\;\underline{ = 0}$, more quickly if you take into account that the integral over&nbsp; $u(t)$&nbsp; disappears.
-*Man muss also gar nicht rechnen.
+*So you don't have to calculate at all.
-'''(3)'''&nbsp; Das Signal&nbsp; $x(t)$&nbsp; kann in den geraden und den ungeraden Anteil aufgeteilt werden, die zum geraden Realteil bzw. ungeraden Imaginärteil von&nbsp; $X(f)$&nbsp; führen:
+'''(3)'''&nbsp; The signal&nbsp; $x(t)$&nbsp; can be divided into the even and the odd part, which lead to the even real part and the odd imaginary part of&nbsp; $X(f)$&nbsp;:
-*Der gerade Anteil ist gleich der Funktion&nbsp; $g(t)$&nbsp; mit&nbsp; $A_g = 3\,\text{V}$. Daraus folgt für den Realteil des Spektralwertes bei&nbsp; $f \cdot T = 0.5$:
+*The even part is equal to the function&nbsp; $g(t)$&nbsp; with&nbsp; $A_g = 3\,\text{V}$. From this follows for the real part of the spectral value at&nbsp; $f \cdot T = 0.5$:
 :$${\mathop{\rm Re}\nolimits} \left[ {X( {f \cdot T = 0.5} )} \right] = A_{\rm g}  \cdot T \cdot {\mathop{\rm si}\nolimits} ( {{ {\rm{\pi }}}/{2}} ) \hspace{0.15 cm}\underline{= 1.91  \;{\rm{mV/Hz}}}{\rm{.}}$$
-*Der Imaginärteil ergibt sich aus der Spektralfunktion&nbsp; $U(f)$ mit $A_u = 1\,\text{V}$. Dieser wurde bereits in der Teilaufgabe&nbsp; '''(1)'''&nbsp; berechnet:
+*The imaginary part results from the spectral function&nbsp; $U(f)$ with $A_u = 1\,\text{V}$. This was already calculated in subtask&nbsp; '''(1)'''&nbsp;:
 :$${\mathop{\rm Im}\nolimits} \left[ {X( {f \cdot T = 0.5} )} \right] \hspace{0.15 cm}\underline{\approx  - 0.2  \;{\rm{mV/Hz}}}{\rm{.}}$$

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