Aufgaben:Exercise 2.4: Rectified Cosine - Revision history

Guenter at 12:24, 15 April 2021

2021-04-15T12:24:54Z

Guenter at 12:14, 15 April 2021

2021-04-15T12:14:12Z

Noah at 09:23, 26 March 2021

2021-03-26T09:23:54Z

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

2021-03-23T12:40:01Z

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

Noah at 22:21, 13 January 2021

2021-01-13T22:21:28Z

Noah at 22:16, 13 January 2021

2021-01-13T22:16:26Z

Noah at 21:55, 13 January 2021

2021-01-13T21:55:55Z

Noah at 21:49, 13 January 2021

2021-01-13T21:49:20Z

Noah at 21:38, 13 January 2021

2021-01-13T21:38:46Z

Noah at 21:29, 13 January 2021

2021-01-13T21:29:44Z

@@ Line 72: / Line 72: @@
 '''(1)'''&nbsp; All solutions are correct except the fourth:
 *From the signal frequency&nbsp; $f_0= 10\,\rm{kHz}$&nbsp; follows&nbsp; $T_0 = 1/f_0 = 100\,&micro;\text{s}$.
-*The cosine signal is free of equal signals&nbsp; $(A_0 = 0)$&nbsp; and is completely described by a single cosine coefficient – namely&nbsp; $A_1$&nbsp;.
+*The cosine signal is mean&ndash;free&nbsp; $(A_0 = 0)$&nbsp; and it is completely described by a single cosine coefficient – namely&nbsp; $A_1$&nbsp;.
 *All sine coefficients are&nbsp; $B_n \equiv 0$, since&nbsp; $x(t)$&nbsp; is an even function.
 *The Fourier series representation&nbsp; $x_3(t)$&nbsp; reproduces&nbsp; $x(t)$&nbsp; without error.
@@ Line 83: / Line 83: @@
-'''(3)'''&nbsp; In the range from&nbsp; $–T_0/2$&nbsp; to&nbsp; $+T_0/2 \ (–25\,&micro;\text{s} \ \text{...}  +25\,&micro;\text{s})$&nbsp; ist&nbsp; $y(t) = x(t)$. Mit&nbsp; $f_x= 10\,\rm{kHz} = 1/(2T_0)$&nbsp; therefore applies to this section:
+'''(3)'''&nbsp; In the range from&nbsp; $–T_0/2$&nbsp; to&nbsp; $+T_0/2 \ (–25\,&micro;\text{s} \ \text{...}  +25\,&micro;\text{s})$&nbsp; is&nbsp; $y(t) = x(t)$. With&nbsp; $f_x= 10\,\rm{kHz} = 1/(2T_0)$&nbsp; therefore applies to this section:
 :$$y(t)={\rm 1V}\cdot\cos(2{\pi} f_0\hspace{0.05cm}t)={\rm 1V}\cdot\cos(\pi \cdot {t}/{T_0}).$$
-*This results in the following for the DC signal component:
+*This results in the following for the DC coefficient:
 :$$A_0=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}y(t)\,{\rm d} t=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}{\rm 1V}\cdot\cos(\pi\cdot {t}/{T_0})\,{\rm d}t.$$
@@ Line 96: / Line 96: @@
-'''(4)'''&nbsp; Since&nbsp; $y(–t) = y(t)$&nbsp; holds, all sine coefficients&nbsp; $B_n = 0$. Thus&nbsp; $B_2 \hspace{0.1cm}\underline{= 0}$ also holds.
+'''(4)'''&nbsp; Since&nbsp; $y(–t) = y(t)$&nbsp; holds, all sine coefficients&nbsp; $B_n = 0$.&nbsp; Thus&nbsp; $B_2 \hspace{0.1cm}\underline{= 0}$&nbsp; also holds.
@@ Line 111: / Line 111: @@
 '''(6)'''&nbsp; For the finite Fourier series with&nbsp; $N = 3$&nbsp; the following applies in general:
 :$$y_3(t)=\frac{2{\rm V}}{\pi} \cdot \left [ 1+{2}/{3} \cdot \cos(\omega_0t)-{2}/{15}\cdot \cos(2\omega_0t)+{2}/{35}\cdot \cos(3\omega_0t) \right ].$$
-At time&nbsp; $t = 0$&nbsp; &nbsp; $y_3(0) \approx 1.0125 \ \rm V$; thus the error is&nbsp; $\varepsilon_3(t = 0) \hspace{0.15cm}\underline{= 0.0125 \,\text{V}}$ .
+At time&nbsp; $t = 0$:&nbsp; &nbsp; $y_3(0) \approx 1.0125 \ \rm V$; thus the error is&nbsp; $\varepsilon_3(t = 0) \hspace{0.15cm}\underline{= 0.0125 \,\text{V}}$ .
-'''(7)'''&nbsp; The time&nbsp; $t = 25\,&micro;\text{s}$&nbsp;  corresponds to half the period of the signal&nbsp; $y(t)$. The following applies here because of&nbsp; $\omega_0 \cdot T_0 = 2\pi$:
+'''(7)'''&nbsp; The time&nbsp; $t = 25\,&micro;\text{s}$&nbsp;  corresponds to half the period of the signal&nbsp; $y(t)$.&nbsp; The following applies here because of&nbsp; $\omega_0 \cdot T_0 = 2\pi$:
 :$$y_3(T_0/2)  = \frac{2{\rm V}}{\pi} \left [1+\frac{2}{3} \cdot \cos({\pi}) -\frac{2}{15}\cdot \cos(2\pi)+\frac{2}{35}\cdot \cos(3\pi)\right ]=   \frac{2{\rm V}}{\pi}\left [1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}\right ] =  \frac{2{\rm V}}{7\pi}\approx 0.091{\rm V}.$$
 *Since&nbsp; $y(T_0/2) = 0$&nbsp; this also results in&nbsp; $\varepsilon_3(T_0/2) \hspace{0.15cm}\underline{\approx 0.091\,{\rm V}}$.
-*This error is larger than the error at&nbsp; $t = 0$  by more than a factor of$7$, since&nbsp; $y(t)$&nbsp; has more high-frequency components at&nbsp; $t = T_0/2$&nbsp; (peak-shaped course).
+*This error is larger than the error at&nbsp; $t = 0$  by more than a factor of &nbsp;$7$, since&nbsp; $y(t)$&nbsp; has more high-frequency components at&nbsp; $t = T_0/2$&nbsp; $($peak-shaped course$)$.
 *If it is required that the error&nbsp; $\varepsilon_3(T_0/2)$&nbsp; be smaller than&nbsp; $0.01$&nbsp; then at least&nbsp; $32$&nbsp; Fourier coefficients would have to be taken into account.

@@ Line 3: / Line 3: @@
 }}
-[[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&nbsp; $x(t)$&nbsp; with the amplitude&nbsp; $1\,\rm{V}$&nbsp; and the frequency&nbsp; $f_0= 10\,\rm{kHz}$&nbsp; is applied to the input of a half-wave rectifier. At its output, the signal&nbsp; $y(t)$ results, which is shown in the graph below.
+A cosine signal&nbsp; $x(t)$&nbsp; with amplitude&nbsp; $1\,\rm{V}$&nbsp; and frequency&nbsp; $f_0= 10\,\rm{kHz}$&nbsp; is applied to the input of a half-wave rectifier.&nbsp; At its output, the signal&nbsp; $y(t)$ results, which is shown in the graph below.
-In subtasks&nbsp; '''(6)'''&nbsp; and&nbsp; '''(7)'''&nbsp; the error signal&nbsp; $\varepsilon_3(t) = y_3(t) - y(t)$&nbsp; is also used. This describes the difference between the Fourier series&nbsp; ⇒ &nbsp; $y_3(t)$ &nbsp; limited to only&nbsp; $N = 3$&nbsp; coefficients and the actual output signal&nbsp; $y(t)$.
+In subtasks&nbsp; '''(6)'''&nbsp; and&nbsp; '''(7)'''&nbsp; the error signal&nbsp; $\varepsilon_3(t) = y_3(t) - y(t)$&nbsp; is also used.&nbsp; This describes the difference between the Fourier series&nbsp; ⇒ &nbsp; $y_3(t)$ &nbsp; limited to only&nbsp; $N = 3$&nbsp; coefficients and the actual output signal&nbsp; $y(t)$.
@@ Line 18: / Line 18: @@
 *This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Series|Fourier Series]].
-*To solve the problem, you can use the following definite integral &nbsp; (let $n$ be an integer$)$:
+*To solve the problem, you can use the following definite integral &nbsp; $($let $n$ be an integer$)$:
 :$$\int ^{\pi /2}_{-\pi /2}\cos(u)\cdot\cos(2nu)\,{\rm d}u  =  (-1)^{n+1}\cdot\frac{2}{4n^2-1}.$$
-*You can find a compact summary of the topic in the learning video&nbsp; [[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> &nbsp; &nbsp; &nbsp;[[Zur_Berechnung_der_Fourierkoeffizienten_(Lernvideo)|Zur Berechnung der Fourierkoeffizienten]] &nbsp; &rArr; &nbsp; "To calculate the Fourier coefficients".
@@ Line 54: / Line 56: @@
-{Specify the Fourier series&nbsp; $y_3(t)$&nbsp; analytically (limit to&nbsp; $N = 3$&nbsp; sine and cosine coefficients each).
+{Specify the Fourier series&nbsp; $y_3(t)$&nbsp; analytically&nbsp; $($limit to&nbsp; $N = 3$&nbsp; sine and cosine coefficients each$)$.
 <br>How large is the error between this finite Fourier series and the actual signal value at&nbsp; $t = 0$?
 |type="{}"}
 $\varepsilon_3(t= 0)\ = \ $  { 0.0125 3% } ${\rm V}$
-{Now calculate the error&nbsp; $\varepsilon_3(t= 25 \,&micro;{\rm s})$. Interpret this value in comparison to the result from&nbsp; '''(6)'''.
+{Now calculate the error&nbsp; $\varepsilon_3(t= 25 \,&micro;{\rm s})$.&nbsp; Interpret this value in comparison to the result from&nbsp; '''(6)'''.
 |type="{}"}
 $\varepsilon_3(t= 25 \,&micro;{\rm s})\ = \ $  { 0.091 3% } ${\rm V}$

@@ Line 3: / Line 3: @@
 }}
-[[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&nbsp; $x(t)$&nbsp; with the amplitude&nbsp; $1\,\rm{V}$&nbsp; and the frequency&nbsp; $f_0= 10\,\rm{kHz}$&nbsp; is applied to the input of a half-wave rectifier. At its output, the signal&nbsp; $y(t)$ results, which is shown in the graph below.
-In subtasks&nbsp; '''(6)'''&nbsp; und&nbsp; '''(7)'''&nbsp; the error signal&nbsp; $\varepsilon_3(t) = y_3(t) - y(t)$&nbsp; is also used. This describes the difference between the Fourier series&nbsp; ⇒ &nbsp; $y_3(t)$ &nbsp; limited to only&nbsp; $N = 3$&nbsp; coefficients and the actual output signal&nbsp; $y(t)$.
+In subtasks&nbsp; '''(6)'''&nbsp; and&nbsp; '''(7)'''&nbsp; the error signal&nbsp; $\varepsilon_3(t) = y_3(t) - y(t)$&nbsp; is also used. This describes the difference between the Fourier series&nbsp; ⇒ &nbsp; $y_3(t)$ &nbsp; limited to only&nbsp; $N = 3$&nbsp; coefficients and the actual output signal&nbsp; $y(t)$.
@@ Line 28: / Line 28: @@
 <quiz display=simple>
-{Which of the following statements are true for the signal&nbsp; $x(t)$&nbsp; zutreffend?
+{Which of the following statements are true for the signal&nbsp; $x(t)$&nbsp;?
 |type="[]"}
 + The period duration is&nbsp;  $T_0 = 100 \,&micro;{\rm s}$.

← Older revision		Revision as of 12:40, 23 March 2021
Line 127:		Line 127:

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

@@ Line 116: / Line 116: @@
-'''(7)'''&nbsp; At time&nbsp; $t = 25\,&micro;\text{s}$&nbsp;  entspricht der halben Periodendauer des Signals&nbsp; $y(t)$. Hierfür gilt wegen&nbsp; $\omega_0 \cdot T_0 = 2\pi$:
+'''(7)'''&nbsp; The time&nbsp; $t = 25\,&micro;\text{s}$&nbsp;  corresponds to half the period of the signal&nbsp; $y(t)$. The following applies here because of&nbsp; $\omega_0 \cdot T_0 = 2\pi$:
 :$$y_3(T_0/2)  = \frac{2{\rm V}}{\pi} \left [1+\frac{2}{3} \cdot \cos({\pi}) -\frac{2}{15}\cdot \cos(2\pi)+\frac{2}{35}\cdot \cos(3\pi)\right ]=   \frac{2{\rm V}}{\pi}\left [1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}\right ] =  \frac{2{\rm V}}{7\pi}\approx 0.091{\rm V}.$$
-*Da&nbsp; $y(T_0/2) = 0$&nbsp; ist, ergibt sich auch&nbsp; $\varepsilon_3(T_0/2) \hspace{0.15cm}\underline{\approx 0.091\,{\rm V}}$.
+*Since&nbsp; $y(T_0/2) = 0$&nbsp; this also results in&nbsp; $\varepsilon_3(T_0/2) \hspace{0.15cm}\underline{\approx 0.091\,{\rm V}}$.
-*Dieser Fehler ist um mehr als den Faktor $7$ größer als der Fehler bei&nbsp; $t = 0$, da&nbsp; $y(t)$&nbsp; bei&nbsp; $t = T_0/2$&nbsp; mehr hochfrequente Anteile besitzt (spitzförmiger Verlauf).
+*This error is larger than the error at&nbsp; $t = 0$  by more than a factor of$7$, since&nbsp; $y(t)$&nbsp; has more high-frequency components at&nbsp; $t = T_0/2$&nbsp; (peak-shaped course).
-*Wird gefordert, dass der Fehler&nbsp; $\varepsilon_3(T_0/2)$&nbsp; kleiner als&nbsp; $0.01$&nbsp; sein soll, dann müssten mindestens&nbsp; $32$&nbsp; Fourierkoeffizienten berücksichtigt werden.
+*If it is required that the error&nbsp; $\varepsilon_3(T_0/2)$&nbsp; be smaller than&nbsp; $0.01$&nbsp; then at least&nbsp; $32$&nbsp; Fourier coefficients would have to be taken into account.
 {{ML-Fuß}}

@@ Line 81: / Line 81: @@
-'''(3)'''&nbsp; Im Bereich von&nbsp; $–T_0/2$&nbsp; bis&nbsp; $+T_0/2 \ (–25\,&micro;\text{s} \ \text{...}  +25\,&micro;\text{s})$&nbsp; ist&nbsp; $y(t) = x(t)$. Mit&nbsp; $f_x= 10\,\rm{kHz} = 1/(2T_0)$&nbsp; gilt deshalb für diesen Abschnitt:
+'''(3)'''&nbsp; In the range from&nbsp; $–T_0/2$&nbsp; to&nbsp; $+T_0/2 \ (–25\,&micro;\text{s} \ \text{...}  +25\,&micro;\text{s})$&nbsp; ist&nbsp; $y(t) = x(t)$. Mit&nbsp; $f_x= 10\,\rm{kHz} = 1/(2T_0)$&nbsp; therefore applies to this section:
 :$$y(t)={\rm 1V}\cdot\cos(2{\pi} f_0\hspace{0.05cm}t)={\rm 1V}\cdot\cos(\pi \cdot {t}/{T_0}).$$
-*Daraus ergibt sich für den Gleichsignalanteil:
+*This results in the following for the DC signal component:
 :$$A_0=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}y(t)\,{\rm d} t=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}{\rm 1V}\cdot\cos(\pi\cdot {t}/{T_0})\,{\rm d}t.$$
-*Mit der Substitution&nbsp; $u = \pi \cdot t/T_0$&nbsp; erhält man schließlich:
+*With the substitution&nbsp; $u = \pi \cdot t/T_0$&nbsp; one obtains:
 :$$A_0=\left. \frac{ {\rm 1V}}{\pi}\int_{-\pi /2}^{\pi/2}\cos(u)\,{\rm d}u=\frac{ {\rm 1V}}{\pi}\sin(u)\; \right| _{-\pi/2}^{\pi/2}=\frac{ {\rm 1V}\cdot 2}{\pi} \hspace{0.15cm}\underline{\approx 0.637\;{\rm V}}.$$
-'''(4)'''&nbsp; Da&nbsp; $y(–t) = y(t)$&nbsp; gilt, sind alle Sinuskoeffizienten&nbsp; $B_n = 0$. Damit ist auch&nbsp; $B_2 \hspace{0.1cm}\underline{= 0}$.
+'''(4)'''&nbsp; Since&nbsp; $y(–t) = y(t)$&nbsp; holds, all sine coefficients&nbsp; $B_n = 0$. Thus&nbsp; $B_2 \hspace{0.1cm}\underline{= 0}$ also holds.
-'''(5)'''&nbsp;  Für die Koeffizienten&nbsp; $A_n$&nbsp; gilt mit der Substitution&nbsp; $u = \pi \cdot t/T_0$&nbsp; entsprechend dem angegebenen Integral:
+'''(5)'''&nbsp;  For the coefficients&nbsp; $A_n$&nbsp; applies with the substitution&nbsp; $u = \pi \cdot t/T_0$&nbsp; according to the given integral:
 :$$A_n  = \frac{2{\rm V}}{T_0}\int_{-T_0/2}^{T_0/2}\cos(\pi\frac{t}{T_0})\cdot \cos(n\cdot 2\pi\frac{t}{T_0})\,{\rm d}t  = \frac{2{\rm V}}{\pi}\int_{-\pi/2}^{\pi/2}\cos(u)\cdot \cos(2n u)\,{\rm d}u \quad
 \Rightarrow  \quad A_n  = \left( { - 1} \right)^{n + 1} \frac{{4\;{\rm{V}}}}{{{\rm{\pi }}\left( {4n^2  - 1} \right)}}.$$
-Der Koeffizient&nbsp; $A_2$&nbsp; ist damit gleich&nbsp; $-4 \,\text{V}/(15\pi) \hspace{0.1cm}\underline{\approx -\hspace{0.05cm}0.085 \, \text{V}}$.
+The coefficient&nbsp; $A_2$&nbsp; is thus equal to&nbsp; $-4 \,\text{V}/(15\pi) \hspace{0.1cm}\underline{\approx -\hspace{0.05cm}0.085 \, \text{V}}$.
-'''(6)'''&nbsp; Für die endliche Fourierreihe mit&nbsp; $N = 3$&nbsp; gilt allgemein:
+'''(6)'''&nbsp; For the finite Fourier series with&nbsp; $N = 3$&nbsp; the following applies in general:
 :$$y_3(t)=\frac{2{\rm V}}{\pi} \cdot \left [ 1+{2}/{3} \cdot \cos(\omega_0t)-{2}/{15}\cdot \cos(2\omega_0t)+{2}/{35}\cdot \cos(3\omega_0t) \right ].$$
-Zum Zeitpunkt&nbsp; $t = 0$&nbsp; ist&nbsp; $y_3(0) \approx 1.0125 \ \rm V$; damit ergibt sich der Fehler zu&nbsp; $\varepsilon_3(t = 0) \hspace{0.15cm}\underline{= 0.0125 \,\text{V}}$ .
+At time&nbsp; $t = 0$&nbsp; &nbsp; $y_3(0) \approx 1.0125 \ \rm V$; thus the error is&nbsp; $\varepsilon_3(t = 0) \hspace{0.15cm}\underline{= 0.0125 \,\text{V}}$ .
-'''(7)'''&nbsp; Die Zeit&nbsp; $t = 25\,&micro;\text{s}$&nbsp;  entspricht der halben Periodendauer des Signals&nbsp; $y(t)$. Hierfür gilt wegen&nbsp; $\omega_0 \cdot T_0 = 2\pi$:
+'''(7)'''&nbsp; At time&nbsp; $t = 25\,&micro;\text{s}$&nbsp;  entspricht der halben Periodendauer des Signals&nbsp; $y(t)$. Hierfür gilt wegen&nbsp; $\omega_0 \cdot T_0 = 2\pi$:
 :$$y_3(T_0/2)  = \frac{2{\rm V}}{\pi} \left [1+\frac{2}{3} \cdot \cos({\pi}) -\frac{2}{15}\cdot \cos(2\pi)+\frac{2}{35}\cdot \cos(3\pi)\right ]=   \frac{2{\rm V}}{\pi}\left [1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}\right ] =  \frac{2{\rm V}}{7\pi}\approx 0.091{\rm V}.$$

@@ Line 59: / Line 59: @@
 $\varepsilon_3(t= 0)\ = \ $  { 0.0125 3% } ${\rm V}$
-{Berechnen Sie nun den Fehler&nbsp; $\varepsilon_3(t= 25 \,&micro;{\rm s})$. Interpretieren Sie diesen Wert im Vergleich zum Ergebnis aus&nbsp; '''(6)'''.
+{Now calculate the error&nbsp; $\varepsilon_3(t= 25 \,&micro;{\rm s})$. Interpret this value in comparison to the result from&nbsp; '''(6)'''.
 |type="{}"}
 $\varepsilon_3(t= 25 \,&micro;{\rm s})\ = \ $  { 0.091 3% } ${\rm V}$
@@ Line 66: / Line 66: @@
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Richtig sind hier alle Lösungsvorschläge außer dem Vierten:
+'''(1)'''&nbsp; All solutions are correct except the fourth:
-*Aus der Signalfrequenz&nbsp; $f_0= 10\,\rm{kHz}$&nbsp; folgt&nbsp; $T_0 = 1/f_0 = 100\,&micro;\text{s}$.
+*From the signal frequency&nbsp; $f_0= 10\,\rm{kHz}$&nbsp; follows&nbsp; $T_0 = 1/f_0 = 100\,&micro;\text{s}$.
-*Das Cosinussignal ist gleichsignalfrei&nbsp; $(A_0 = 0)$&nbsp; und wird durch einen einzigen Cosinuskoeffizienten – nämlich&nbsp; $A_1$&nbsp; – vollständig beschrieben.
+*The cosine signal is free of equal signals&nbsp; $(A_0 = 0)$&nbsp; and is completely described by a single cosine coefficient – namely&nbsp; $A_1$&nbsp;.
-*Alle Sinuskoeffizienten  sind&nbsp; $B_n \equiv 0$, da&nbsp; $x(t)$&nbsp; eine gerade Funktion ist.
+*All sine coefficients are&nbsp; $B_n \equiv 0$, since&nbsp; $x(t)$&nbsp; is an even function.
-*Die Fourierreihendarstellung&nbsp; $x_3(t)$&nbsp; bildet&nbsp; $x(t)$&nbsp; fehlerfrei nach.
+*The Fourier series representation&nbsp; $x_3(t)$&nbsp; reproduces&nbsp; $x(t)$&nbsp; without error.
-'''(2)'''&nbsp; Aufgrund der Doppelweggleichrichtung ergibt sich für die Periodendauer nunmehr der halbe Wert:&nbsp; $T_0 \hspace{0.1cm}\underline{= 50\,&micro;\text{s}}$.
+'''(2)'''&nbsp; Due to the double path rectification, the period duration is now half the value:&nbsp; $T_0 \hspace{0.1cm}\underline{= 50\,&micro;\text{s}}$.
-*Bei allen nachfolgenden Punkten bezieht sich die Angabe&nbsp; $T_0$&nbsp; auf diesen Wert, also auf die Periodendauer des Signals&nbsp; $y(t)$.
+*For all subsequent points, the specification&nbsp; $T_0$&nbsp; refers to this value, i.e. to the period of the signal&nbsp; $y(t)$.

@@ Line 36: / Line 36: @@
 + The Fourier series&nbsp; $x_3(t)$&nbsp; does not deviate from the actual signal&nbsp; $x(t)$&nbsp;.
-{Wie groß ist die Periodendauer des Signals&nbsp; $y(t)$?
+{What is the period duration of the signal&nbsp; $y(t)$?
 |type="{}"}
 $T_0\ = \ $ { 50 3% } &nbsp; ${\rm &micro;s}$
-{Berechnen Sie den Gleichsignalanteil des Signals&nbsp; $y(t)$.
+{Calculate the DC component of the signal&nbsp; $y(t)$.
 |type="{}"}
 $A_0\ = \ $  { 0.637 3% } &nbsp; ${\rm V}$
-{Wie lauten die Sinuskoeffizienten&nbsp; $B_n$? Begründen Sie Ihr Ergebnis. Geben Sie zur Kontrolle den Koeffizienten&nbsp; $B_2$&nbsp; ein.
+{What are the sine coefficients&nbsp; $B_n$?  Justify your result. Enter the coefficient&nbsp; $B_2$&nbsp; as a check.
 |type="{}"}
 $B_2\ = \ $  { 0. } &nbsp; ${\rm V}$
-{Berechnen Sie nun die Cosinuskoeffizienten&nbsp; $A_n$. Geben Sie zur Kontrolle den Koeffizienten&nbsp; $A_2$&nbsp; ein.
+{Now calculate the cosine coefficient&nbsp; $A_n$. Enter the coefficient&nbsp; $A_2$&nbsp; as a check.
 |type="{}"}
 $A_2\ = \ $  { -0.087--0.083 } &nbsp; ${\rm V}$
-{Geben Sie die Fourierreihe&nbsp; $y_3(t)$&nbsp; analytisch an (Begrenzung auf je&nbsp; $N = 3$&nbsp; Sinus– bzw. Cosinuskoeffizienten).
+{Specify the Fourier series&nbsp; $y_3(t)$&nbsp; analytically (limit to&nbsp; $N = 3$&nbsp; sine and cosine coefficients each).
-<br>Wie groß ist der Fehler zwischen dieser endlichen Fourierreihe und dem tatsächlichen Signalwert bei&nbsp; $t = 0$?
+<br>How large is the error between this finite Fourier series and the actual signal value at&nbsp; $t = 0$?
 |type="{}"}
 $\varepsilon_3(t= 0)\ = \ $  { 0.0125 3% } ${\rm V}$

@@ Line 15: / Line 15: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum Kapitel&nbsp; [[Signal_Representation/Fourier_Series|Fourierreihe]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Series|Fourier Series]].
-*Zur Lösung der Aufgabe können Sie das folgende bestimmte Integral benutzen&nbsp; $(n$ sei ganzzahlig$)$:
+*To solve the problem, you can use the following definite integral &nbsp; (let $n$ be an integer$)$:
 :$$\int ^{\pi /2}_{-\pi /2}\cos(u)\cdot\cos(2nu)\,{\rm d}u  =  (-1)^{n+1}\cdot\frac{2}{4n^2-1}.$$
-*Eine kompakte Zusammenfassung der Thematik finden Sie im Lernvideo&nbsp; [[Zur_Berechnung_der_Fourierkoeffizienten_(Lernvideo)|Zur Berechnung der Fourierkoeffizienten]].
+*You can find a compact summary of the topic in the learning video&nbsp; [[Zur_Berechnung_der_Fourierkoeffizienten_(Lernvideo)|Zur Berechnung der Fourierkoeffizienten]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welche der folgenden Aussagen sind für das Signal&nbsp; $x(t)$&nbsp; zutreffend?
+{Which of the following statements are true for the signal&nbsp; $x(t)$&nbsp; zutreffend?
 |type="[]"}
-+ Die Periodendauer ist&nbsp;  $T_0 = 100 \,&micro;{\rm s}$.
++ The period duration is&nbsp;  $T_0 = 100 \,&micro;{\rm s}$.
-+ Der Gleichsignalkoeffizient ist&nbsp; $A_0 = 0$.
++ The DC signal coefficient is&nbsp; $A_0 = 0$.
-+ Von allen Cosinuskoeffizienten&nbsp; $A_n$&nbsp; ist genau einer ungleich Null.
++ Of all cosine coefficients&nbsp; $A_n$&nbsp; exactly one is not equal to zero.
-- Von allen Sinuskoeffizienten&nbsp; $B_n$&nbsp; ist genau einer ungleich Null.
+- Of all the sine coefficients&nbsp; $B_n$&nbsp; exactly one is not equal to zero.
-+ Die Fourierreihe&nbsp; $x_3(t)$&nbsp; weicht nicht vom tatsächlichen Signal&nbsp; $x(t)$&nbsp; ab.
++ The Fourier series&nbsp; $x_3(t)$&nbsp; does not deviate from the actual signal&nbsp; $x(t)$&nbsp;.
 {Wie groß ist die Periodendauer des Signals&nbsp; $y(t)$?