Aufgaben:Exercise 2.4: Rectified Cosine - Revision history

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	[[Category:Signal Representation: Exercises\|^2.4 Fourier Series^]]		[[Category:Signal Representation: Exercises\|^2.4 Fourier Series^]]
	[[de:Aufgaben:2.~~4_Geichgerichteter_Cosinus~~]]		[[de:Aufgaben:Aufgabe 2.4: Gleichgerichteter Cosinus]]

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	'''(5)'''  For the coefficients  $A_n$  applies with the substitution  $u = \pi \cdot t/T_0$  according to the given integral:		'''(5)'''  For the coefficients  $A_n$  applies with the substitution  $u = \pi \cdot t/T_0$  according to the given integral:

	:$$A_n = \frac{2{\rm V}}{T_0}\int_{-T_0/2}^{T_0/2}\cos(\pi\frac{t}{T_0})\cdot \cos(n\cdot 2\pi\frac{t}{T_0})\,{\rm d}t = \frac{2{\rm V}}{\pi}\int_{-\pi/2}^{\pi/2}\cos(u)\cdot \cos(2n u)\,{\rm d}u \quad		:$$A_n = \frac{2{\rm V}}{T_0}\int_{-T_0/2}^{T_0/2}\cos(\pi\frac{t}{T_0})\cdot \cos(n\cdot 2\pi\frac{t}{T_0})\,{\rm d}t = \frac{2{\rm V}}{\pi}\int_{-\pi/2}^{\pi/2}\cos(u)\cdot \cos(2n u)\,{\rm d}u \quad\Rightarrow \quad A_n = \left( { - 1} \right)^{n + 1} \frac{{4\;{\rm{V}}}}{{{\rm{\pi }}\left( {4n^2 - 1} \right)}}.$$
	\Rightarrow \quad A_n = \left( { - 1} \right)^{n + 1} \frac{{4\;{\rm{V}}}}{{{\rm{\pi }}\left( {4n^2 - 1} \right)}}.$$

	The coefficient  $A_2$  is thus equal to  $-4 \,\text{V}/(15\pi) \hspace{0.1cm}\underline{\approx -\hspace{0.05cm}0.085 \, \text{V}}$.		The coefficient  $A_2$  is thus equal to  $-4 \,\text{V}/(15\pi) \hspace{0.1cm}\underline{\approx -\hspace{0.05cm}0.085 \, \text{V}}$.

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	[[Category:Signal Representation: Exercises\|^2.4 Fourier Series^]]		[[Category:Signal Representation: Exercises\|^2.4 Fourier Series^]]
			[[de:Aufgaben:2.4_Geichgerichteter_Cosinus]]

Guenter at 12:24, 15 April 2021

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Guenter at 12:14, 15 April 2021

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Noah at 09:23, 26 March 2021

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	[[File:P_ID300__Sig_A_2_4.png\|right\|frame\|~~Gleichgerichteter Cosinus~~]]		[[File:P_ID300__Sig_A_2_4.png\|right\|frame\|Rectified cosine]]

	A cosine signal  $x(t)$  with the amplitude  $1\,\rm{V}$  and the frequency  $f_0= 10\,\rm{kHz}$  is applied to the input of a half-wave rectifier. At its output, the signal  $y(t)$ results, which is shown in the graph below.		A cosine signal  $x(t)$  with the amplitude  $1\,\rm{V}$  and the frequency  $f_0= 10\,\rm{kHz}$  is applied to the input of a half-wave rectifier. At its output, the signal  $y(t)$ results, which is shown in the graph below.

	In subtasks  '''(6)'''  ~~und~~  '''(7)'''  the error signal  $\varepsilon_3(t) = y_3(t) - y(t)$  is also used. This describes the difference between the Fourier series  ⇒   $y_3(t)$   limited to only  $N = 3$  coefficients and the actual output signal  $y(t)$.		In subtasks  '''(6)'''  and  '''(7)'''  the error signal  $\varepsilon_3(t) = y_3(t) - y(t)$  is also used. This describes the difference between the Fourier series  ⇒   $y_3(t)$   limited to only  $N = 3$  coefficients and the actual output signal  $y(t)$.


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	<quiz display=simple>		<quiz display=simple>
	{Which of the following statements are true for the signal  $x(t)$  ~~zutreffend~~?		{Which of the following statements are true for the signal  $x(t)$ ?
	\|type="[]"}		\|type="[]"}
	+ The period duration is  $T_0 = 100 \,µ{\rm s}$.		+ The period duration is  $T_0 = 100 \,µ{\rm s}$.

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

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	[[Category:~~Exercises for~~ Signal Representation\|^2.4 Fourier Series^]]		[[Category:Signal Representation: Exercises\|^2.4 Fourier Series^]]

Noah at 22:21, 13 January 2021

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	'''(7)'''  At time  $t = 25\,µ\text{s}$  ~~entspricht der halben Periodendauer des Signals~~  $y(t)$. ~~Hierfür gilt wegen~~  $\omega_0 \cdot T_0 = 2\pi$:		'''(7)'''  The time  $t = 25\,µ\text{s}$  corresponds to half the period of the signal  $y(t)$. The following applies here because of  $\omega_0 \cdot T_0 = 2\pi$:

	:$$y_3(T_0/2) = \frac{2{\rm V}}{\pi} \left [1+\frac{2}{3} \cdot \cos({\pi}) -\frac{2}{15}\cdot \cos(2\pi)+\frac{2}{35}\cdot \cos(3\pi)\right ]= \frac{2{\rm V}}{\pi}\left [1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}\right ] = \frac{2{\rm V}}{7\pi}\approx 0.091{\rm V}.$$		:$$y_3(T_0/2) = \frac{2{\rm V}}{\pi} \left [1+\frac{2}{3} \cdot \cos({\pi}) -\frac{2}{15}\cdot \cos(2\pi)+\frac{2}{35}\cdot \cos(3\pi)\right ]= \frac{2{\rm V}}{\pi}\left [1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}\right ] = \frac{2{\rm V}}{7\pi}\approx 0.091{\rm V}.$$

	*Da  $y(T_0/2) = 0$  ~~ist, ergibt sich auch~~  $\varepsilon_3(T_0/2) \hspace{0.15cm}\underline{\approx 0.091\,{\rm V}}$.		*Since  $y(T_0/2) = 0$  this also results in  $\varepsilon_3(T_0/2) \hspace{0.15cm}\underline{\approx 0.091\,{\rm V}}$.
	*~~Dieser Fehler ist um mehr als den Faktor $7$ größer als der Fehler bei~~  $t = 0$, da  $y(t)$  ~~bei~~  $t = T_0/2$  ~~mehr hochfrequente Anteile besitzt~~ (~~spitzförmiger Verlauf~~).		*This error is larger than the error at  $t = 0$ by more than a factor of$7$, since  $y(t)$  has more high-frequency components at  $t = T_0/2$  (peak-shaped course).
	*~~Wird gefordert, dass der Fehler~~  $\varepsilon_3(T_0/2)$  ~~kleiner als~~  $0.01$  ~~sein soll, dann müssten mindestens~~  $32$  ~~Fourierkoeffizienten berücksichtigt werden~~.		*If it is required that the error  $\varepsilon_3(T_0/2)$  be smaller than  $0.01$  then at least  $32$  Fourier coefficients would have to be taken into account.

	{{ML-Fuß}}		{{ML-Fuß}}

Noah at 22:16, 13 January 2021

2021-01-13T22:16:26Z

Noah at 21:55, 13 January 2021

2021-01-13T21:55:55Z

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	$\varepsilon_3(t= 0)\ = \ $ { 0.0125 3% } ${\rm V}$		$\varepsilon_3(t= 0)\ = \ $ { 0.0125 3% } ${\rm V}$

	{~~Berechnen Sie nun den Fehler~~  $\varepsilon_3(t= 25 \,µ{\rm s})$. ~~Interpretieren Sie diesen Wert im Vergleich zum Ergebnis aus~~  '''(6)'''.		{Now calculate the error  $\varepsilon_3(t= 25 \,µ{\rm s})$. Interpret this value in comparison to the result from  '''(6)'''.
	\|type="{}"}		\|type="{}"}
	$\varepsilon_3(t= 25 \,µ{\rm s})\ = \ $ { 0.091 3% } ${\rm V}$		$\varepsilon_3(t= 25 \,µ{\rm s})\ = \ $ { 0.091 3% } ${\rm V}$
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	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Richtig sind hier alle Lösungsvorschläge außer dem Vierten~~:		'''(1)'''  All solutions are correct except the fourth:
	*~~Aus der Signalfrequenz~~  $f_0= 10\,\rm{kHz}$  ~~folgt~~  $T_0 = 1/f_0 = 100\,µ\text{s}$.		*From the signal frequency  $f_0= 10\,\rm{kHz}$  follows  $T_0 = 1/f_0 = 100\,µ\text{s}$.
	*~~Das Cosinussignal ist gleichsignalfrei~~  $(A_0 = 0)$  ~~und wird durch einen einzigen Cosinuskoeffizienten~~ – ~~nämlich~~  $A_1$  ~~– vollständig beschrieben~~.		*The cosine signal is free of equal signals  $(A_0 = 0)$  and is completely described by a single cosine coefficient – namely  $A_1$ .
	*~~Alle Sinuskoeffizienten sind~~  $B_n \equiv 0$, da  $x(t)$  ~~eine gerade Funktion ist~~.		*All sine coefficients are  $B_n \equiv 0$, since  $x(t)$  is an even function.
	*~~Die Fourierreihendarstellung~~  $x_3(t)$  ~~bildet~~  $x(t)$  ~~fehlerfrei nach~~.		*The Fourier series representation  $x_3(t)$  reproduces  $x(t)$  without error.



	'''(2)'''  ~~Aufgrund der Doppelweggleichrichtung ergibt sich für die Periodendauer nunmehr der halbe Wert~~:  $T_0 \hspace{0.1cm}\underline{= 50\,µ\text{s}}$.		'''(2)'''  Due to the double path rectification, the period duration is now half the value:  $T_0 \hspace{0.1cm}\underline{= 50\,µ\text{s}}$.
	*~~Bei allen nachfolgenden Punkten bezieht sich die Angabe~~  $T_0$  ~~auf diesen Wert~~, ~~also auf die Periodendauer des Signals~~  $y(t)$.		*For all subsequent points, the specification  $T_0$  refers to this value, i.e. to the period of the signal  $y(t)$.

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	'''(1)'''  All solutions are correct except the fourth:		'''(1)'''  All solutions are correct except the fourth:
	*From the signal frequency  $f_0= 10\,\rm{kHz}$  follows  $T_0 = 1/f_0 = 100\,µ\text{s}$.		*From the signal frequency  $f_0= 10\,\rm{kHz}$  follows  $T_0 = 1/f_0 = 100\,µ\text{s}$.
	*The cosine signal is free ~~of equal signals~~  $(A_0 = 0)$  and is completely described by a single cosine coefficient – namely  $A_1$ .		*The cosine signal is mean–free  $(A_0 = 0)$  and it is completely described by a single cosine coefficient – namely  $A_1$ .
	*All sine coefficients are  $B_n \equiv 0$, since  $x(t)$  is an even function.		*All sine coefficients are  $B_n \equiv 0$, since  $x(t)$  is an even function.
	*The Fourier series representation  $x_3(t)$  reproduces  $x(t)$  without error.		*The Fourier series representation  $x_3(t)$  reproduces  $x(t)$  without error.
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	'''(3)'''  In the range from  $–T_0/2$  to  $+T_0/2 \ (–25\,µ\text{s} \ \text{...} +25\,µ\text{s})$  ~~ist~~  $y(t) = x(t)$. ~~Mit~~  $f_x= 10\,\rm{kHz} = 1/(2T_0)$  therefore applies to this section:		'''(3)'''  In the range from  $–T_0/2$  to  $+T_0/2 \ (–25\,µ\text{s} \ \text{...} +25\,µ\text{s})$  is  $y(t) = x(t)$. With  $f_x= 10\,\rm{kHz} = 1/(2T_0)$  therefore applies to this section:

	:$$y(t)={\rm 1V}\cdot\cos(2{\pi} f_0\hspace{0.05cm}t)={\rm 1V}\cdot\cos(\pi \cdot {t}/{T_0}).$$		:$$y(t)={\rm 1V}\cdot\cos(2{\pi} f_0\hspace{0.05cm}t)={\rm 1V}\cdot\cos(\pi \cdot {t}/{T_0}).$$

	*This results in the following for the DC ~~signal component~~:		*This results in the following for the DC coefficient:

	:$$A_0=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}y(t)\,{\rm d} t=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}{\rm 1V}\cdot\cos(\pi\cdot {t}/{T_0})\,{\rm d}t.$$		:$$A_0=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}y(t)\,{\rm d} t=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}{\rm 1V}\cdot\cos(\pi\cdot {t}/{T_0})\,{\rm d}t.$$
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	'''(4)'''  Since  $y(–t) = y(t)$  holds, all sine coefficients  $B_n = 0$. Thus  $B_2 \hspace{0.1cm}\underline{= 0}$ also holds.		'''(4)'''  Since  $y(–t) = y(t)$  holds, all sine coefficients  $B_n = 0$.  Thus  $B_2 \hspace{0.1cm}\underline{= 0}$  also holds.


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	'''(6)'''  For the finite Fourier series with  $N = 3$  the following applies in general:		'''(6)'''  For the finite Fourier series with  $N = 3$  the following applies in general:


	:$$y_3(t)=\frac{2{\rm V}}{\pi} \cdot \left [ 1+{2}/{3} \cdot \cos(\omega_0t)-{2}/{15}\cdot \cos(2\omega_0t)+{2}/{35}\cdot \cos(3\omega_0t) \right ].$$		:$$y_3(t)=\frac{2{\rm V}}{\pi} \cdot \left [ 1+{2}/{3} \cdot \cos(\omega_0t)-{2}/{15}\cdot \cos(2\omega_0t)+{2}/{35}\cdot \cos(3\omega_0t) \right ].$$

	At time  $t = 0$    $y_3(0) \approx 1.0125 \ \rm V$; thus the error is  $\varepsilon_3(t = 0) \hspace{0.15cm}\underline{= 0.0125 \,\text{V}}$ .		At time  $t = 0$:    $y_3(0) \approx 1.0125 \ \rm V$; thus the error is  $\varepsilon_3(t = 0) \hspace{0.15cm}\underline{= 0.0125 \,\text{V}}$ .



	'''(7)'''  The time  $t = 25\,µ\text{s}$  corresponds to half the period of the signal  $y(t)$. The following applies here because of  $\omega_0 \cdot T_0 = 2\pi$:		'''(7)'''  The time  $t = 25\,µ\text{s}$  corresponds to half the period of the signal  $y(t)$.  The following applies here because of  $\omega_0 \cdot T_0 = 2\pi$:

	:$$y_3(T_0/2) = \frac{2{\rm V}}{\pi} \left [1+\frac{2}{3} \cdot \cos({\pi}) -\frac{2}{15}\cdot \cos(2\pi)+\frac{2}{35}\cdot \cos(3\pi)\right ]= \frac{2{\rm V}}{\pi}\left [1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}\right ] = \frac{2{\rm V}}{7\pi}\approx 0.091{\rm V}.$$		:$$y_3(T_0/2) = \frac{2{\rm V}}{\pi} \left [1+\frac{2}{3} \cdot \cos({\pi}) -\frac{2}{15}\cdot \cos(2\pi)+\frac{2}{35}\cdot \cos(3\pi)\right ]= \frac{2{\rm V}}{\pi}\left [1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}\right ] = \frac{2{\rm V}}{7\pi}\approx 0.091{\rm V}.$$

	*Since  $y(T_0/2) = 0$  this also results in  $\varepsilon_3(T_0/2) \hspace{0.15cm}\underline{\approx 0.091\,{\rm V}}$.		*Since  $y(T_0/2) = 0$  this also results in  $\varepsilon_3(T_0/2) \hspace{0.15cm}\underline{\approx 0.091\,{\rm V}}$.
	*This error is larger than the error at  $t = 0$ by more than a factor of$7$, since  $y(t)$  has more high-frequency components at  $t = T_0/2$  (peak-shaped course).		*This error is larger than the error at  $t = 0$ by more than a factor of  $7$, since  $y(t)$  has more high-frequency components at  $t = T_0/2$  $($peak-shaped course$)$.
	*If it is required that the error  $\varepsilon_3(T_0/2)$  be smaller than  $0.01$  then at least  $32$  Fourier coefficients would have to be taken into account.		*If it is required that the error  $\varepsilon_3(T_0/2)$  be smaller than  $0.01$  then at least  $32$  Fourier coefficients would have to be taken into account.

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	}}		}}

	[[File:P_ID300__Sig_A_2_4.png\|right\|frame\|~~Rectified~~ cosine]]		[[File:P_ID300__Sig_A_2_4.png\|right\|frame\|Cosine and rectified cosine]]

	A cosine signal  $x(t)$  with ~~the~~ amplitude  $1\,\rm{V}$  and ~~the~~ frequency  $f_0= 10\,\rm{kHz}$  is applied to the input of a half-wave rectifier. At its output, the signal  $y(t)$ results, which is shown in the graph below.		A cosine signal  $x(t)$  with amplitude  $1\,\rm{V}$  and frequency  $f_0= 10\,\rm{kHz}$  is applied to the input of a half-wave rectifier.  At its output, the signal  $y(t)$ results, which is shown in the graph below.

	In subtasks  '''(6)'''  and  '''(7)'''  the error signal  $\varepsilon_3(t) = y_3(t) - y(t)$  is also used. This describes the difference between the Fourier series  ⇒   $y_3(t)$   limited to only  $N = 3$  coefficients and the actual output signal  $y(t)$.		In subtasks  '''(6)'''  and  '''(7)'''  the error signal  $\varepsilon_3(t) = y_3(t) - y(t)$  is also used.  This describes the difference between the Fourier series  ⇒   $y_3(t)$   limited to only  $N = 3$  coefficients and the actual output signal  $y(t)$.


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	*This exercise belongs to the chapter  [[Signal_Representation/Fourier_Series\|Fourier Series]].		*This exercise belongs to the chapter  [[Signal_Representation/Fourier_Series\|Fourier Series]].

	*To solve the problem, you can use the following definite integral   (let $n$ be an integer$)$:		*To solve the problem, you can use the following definite integral   $($let $n$ be an integer$)$:

	:$$\int ^{\pi /2}_{-\pi /2}\cos(u)\cdot\cos(2nu)\,{\rm d}u = (-1)^{n+1}\cdot\frac{2}{4n^2-1}.$$		:$$\int ^{\pi /2}_{-\pi /2}\cos(u)\cdot\cos(2nu)\,{\rm d}u = (-1)^{n+1}\cdot\frac{2}{4n^2-1}.$$

	*You can find a compact summary of the topic in the learning video  [[Zur_Berechnung_der_Fourierkoeffizienten_(Lernvideo)\|Zur Berechnung der Fourierkoeffizienten]].		*You can find a compact summary of the topic in the (German language) learning video<br>      [[Zur_Berechnung_der_Fourierkoeffizienten_(Lernvideo)\|Zur Berechnung der Fourierkoeffizienten]]   ⇒   "To calculate the Fourier coefficients".




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	{Specify the Fourier series  $y_3(t)$  analytically (limit to  $N = 3$  sine and cosine coefficients each).		{Specify the Fourier series  $y_3(t)$  analytically  $($limit to  $N = 3$  sine and cosine coefficients each$)$.
	<br>How large is the error between this finite Fourier series and the actual signal value at  $t = 0$?		<br>How large is the error between this finite Fourier series and the actual signal value at  $t = 0$?
	\|type="{}"}		\|type="{}"}
	$\varepsilon_3(t= 0)\ = \ $ { 0.0125 3% } ${\rm V}$		$\varepsilon_3(t= 0)\ = \ $ { 0.0125 3% } ${\rm V}$

	{Now calculate the error  $\varepsilon_3(t= 25 \,µ{\rm s})$. Interpret this value in comparison to the result from  '''(6)'''.		{Now calculate the error  $\varepsilon_3(t= 25 \,µ{\rm s})$.  Interpret this value in comparison to the result from  '''(6)'''.
	\|type="{}"}		\|type="{}"}
	$\varepsilon_3(t= 25 \,µ{\rm s})\ = \ $ { 0.091 3% } ${\rm V}$		$\varepsilon_3(t= 25 \,µ{\rm s})\ = \ $ { 0.091 3% } ${\rm V}$

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	'''(3)'''  ~~Im Bereich von~~  $–T_0/2$  ~~bis~~  $+T_0/2 \ (–25\,µ\text{s} \ \text{...} +25\,µ\text{s})$  ist  $y(t) = x(t)$. Mit  $f_x= 10\,\rm{kHz} = 1/(2T_0)$  ~~gilt deshalb für diesen Abschnitt~~:		'''(3)'''  In the range from  $–T_0/2$  to  $+T_0/2 \ (–25\,µ\text{s} \ \text{...} +25\,µ\text{s})$  ist  $y(t) = x(t)$. Mit  $f_x= 10\,\rm{kHz} = 1/(2T_0)$  therefore applies to this section:

	:$$y(t)={\rm 1V}\cdot\cos(2{\pi} f_0\hspace{0.05cm}t)={\rm 1V}\cdot\cos(\pi \cdot {t}/{T_0}).$$		:$$y(t)={\rm 1V}\cdot\cos(2{\pi} f_0\hspace{0.05cm}t)={\rm 1V}\cdot\cos(\pi \cdot {t}/{T_0}).$$

	*~~Daraus ergibt sich für den Gleichsignalanteil~~:		*This results in the following for the DC signal component:

	:$$A_0=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}y(t)\,{\rm d} t=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}{\rm 1V}\cdot\cos(\pi\cdot {t}/{T_0})\,{\rm d}t.$$		:$$A_0=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}y(t)\,{\rm d} t=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}{\rm 1V}\cdot\cos(\pi\cdot {t}/{T_0})\,{\rm d}t.$$

	*~~Mit der Substitution~~  $u = \pi \cdot t/T_0$  ~~erhält man schließlich~~:		*With the substitution  $u = \pi \cdot t/T_0$  one obtains:

	:$$A_0=\left. \frac{ {\rm 1V}}{\pi}\int_{-\pi /2}^{\pi/2}\cos(u)\,{\rm d}u=\frac{ {\rm 1V}}{\pi}\sin(u)\; \right\| _{-\pi/2}^{\pi/2}=\frac{ {\rm 1V}\cdot 2}{\pi} \hspace{0.15cm}\underline{\approx 0.637\;{\rm V}}.$$		:$$A_0=\left. \frac{ {\rm 1V}}{\pi}\int_{-\pi /2}^{\pi/2}\cos(u)\,{\rm d}u=\frac{ {\rm 1V}}{\pi}\sin(u)\; \right\| _{-\pi/2}^{\pi/2}=\frac{ {\rm 1V}\cdot 2}{\pi} \hspace{0.15cm}\underline{\approx 0.637\;{\rm V}}.$$


	'''(4)'''  Da  $y(–t) = y(t)$  ~~gilt~~, ~~sind alle Sinuskoeffizienten~~  $B_n = 0$. ~~Damit ist auch~~  $B_2 \hspace{0.1cm}\underline{= 0}$.		'''(4)'''  Since  $y(–t) = y(t)$  holds, all sine coefficients  $B_n = 0$. Thus  $B_2 \hspace{0.1cm}\underline{= 0}$ also holds.



	'''(5)'''  ~~Für die Koeffizienten~~  $A_n$  ~~gilt mit der Substitution~~  $u = \pi \cdot t/T_0$  ~~entsprechend dem angegebenen Integral~~:		'''(5)'''  For the coefficients  $A_n$  applies with the substitution  $u = \pi \cdot t/T_0$  according to the given integral:

	:$$A_n = \frac{2{\rm V}}{T_0}\int_{-T_0/2}^{T_0/2}\cos(\pi\frac{t}{T_0})\cdot \cos(n\cdot 2\pi\frac{t}{T_0})\,{\rm d}t = \frac{2{\rm V}}{\pi}\int_{-\pi/2}^{\pi/2}\cos(u)\cdot \cos(2n u)\,{\rm d}u \quad		:$$A_n = \frac{2{\rm V}}{T_0}\int_{-T_0/2}^{T_0/2}\cos(\pi\frac{t}{T_0})\cdot \cos(n\cdot 2\pi\frac{t}{T_0})\,{\rm d}t = \frac{2{\rm V}}{\pi}\int_{-\pi/2}^{\pi/2}\cos(u)\cdot \cos(2n u)\,{\rm d}u \quad
	\Rightarrow \quad A_n = \left( { - 1} \right)^{n + 1} \frac{{4\;{\rm{V}}}}{{{\rm{\pi }}\left( {4n^2 - 1} \right)}}.$$		\Rightarrow \quad A_n = \left( { - 1} \right)^{n + 1} \frac{{4\;{\rm{V}}}}{{{\rm{\pi }}\left( {4n^2 - 1} \right)}}.$$

	~~Der Koeffizient~~  $A_2$  ~~ist damit gleich~~  $-4 \,\text{V}/(15\pi) \hspace{0.1cm}\underline{\approx -\hspace{0.05cm}0.085 \, \text{V}}$.		The coefficient  $A_2$  is thus equal to  $-4 \,\text{V}/(15\pi) \hspace{0.1cm}\underline{\approx -\hspace{0.05cm}0.085 \, \text{V}}$.



	'''(6)'''  ~~Für die endliche Fourierreihe mit~~  $N = 3$  ~~gilt allgemein~~:		'''(6)'''  For the finite Fourier series with  $N = 3$  the following applies in general:


	:$$y_3(t)=\frac{2{\rm V}}{\pi} \cdot \left [ 1+{2}/{3} \cdot \cos(\omega_0t)-{2}/{15}\cdot \cos(2\omega_0t)+{2}/{35}\cdot \cos(3\omega_0t) \right ].$$		:$$y_3(t)=\frac{2{\rm V}}{\pi} \cdot \left [ 1+{2}/{3} \cdot \cos(\omega_0t)-{2}/{15}\cdot \cos(2\omega_0t)+{2}/{35}\cdot \cos(3\omega_0t) \right ].$$

	~~Zum Zeitpunkt~~  $t = 0$  ~~ist~~  $y_3(0) \approx 1.0125 \ \rm V$; ~~damit ergibt sich der Fehler zu~~  $\varepsilon_3(t = 0) \hspace{0.15cm}\underline{= 0.0125 \,\text{V}}$ .		At time  $t = 0$    $y_3(0) \approx 1.0125 \ \rm V$; thus the error is  $\varepsilon_3(t = 0) \hspace{0.15cm}\underline{= 0.0125 \,\text{V}}$ .



	'''(7)'''  ~~Die Zeit~~  $t = 25\,µ\text{s}$  entspricht der halben Periodendauer des Signals  $y(t)$. Hierfür gilt wegen  $\omega_0 \cdot T_0 = 2\pi$:		'''(7)'''  At time  $t = 25\,µ\text{s}$  entspricht der halben Periodendauer des Signals  $y(t)$. Hierfür gilt wegen  $\omega_0 \cdot T_0 = 2\pi$:

	:$$y_3(T_0/2) = \frac{2{\rm V}}{\pi} \left [1+\frac{2}{3} \cdot \cos({\pi}) -\frac{2}{15}\cdot \cos(2\pi)+\frac{2}{35}\cdot \cos(3\pi)\right ]= \frac{2{\rm V}}{\pi}\left [1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}\right ] = \frac{2{\rm V}}{7\pi}\approx 0.091{\rm V}.$$		:$$y_3(T_0/2) = \frac{2{\rm V}}{\pi} \left [1+\frac{2}{3} \cdot \cos({\pi}) -\frac{2}{15}\cdot \cos(2\pi)+\frac{2}{35}\cdot \cos(3\pi)\right ]= \frac{2{\rm V}}{\pi}\left [1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}\right ] = \frac{2{\rm V}}{7\pi}\approx 0.091{\rm V}.$$