Aufgaben:Exercise 2.6: Complex Fourier Series - Revision history

Guenter at 09:49, 16 April 2021

2021-04-16T09:49:16Z

Guenter at 09:42, 16 April 2021

2021-04-16T09:42:38Z

Guenter at 09:41, 16 April 2021

2021-04-16T09:41:14Z

Noah at 09:26, 26 March 2021

2021-03-26T09:26:57Z

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

2021-03-23T12:38:57Z

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

Noah at 20:18, 16 January 2021

2021-01-16T20:18:31Z

Noah at 19:57, 16 January 2021

2021-01-16T19:57:30Z

Noah at 19:47, 16 January 2021

2021-01-16T19:47:20Z

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 2.6: Komplexe Fourierreihe to Aufgaben:Exercise 2.6: Complex Fourier Series

2020-11-02T14:21:13Z

Oezdemir moved page Aufgabe 2.6: Komplexe Fourierreihe to Exercise 2.6: Complex Fourier Series

Oezdemir at 14:15, 2 November 2020

2020-11-02T14:15:14Z

@@ Line 73: / Line 73: @@
 :$${\rm Im}[D_n] =\frac{T_0/T_1}{(2\pi n)^2} \cdot \sin(2\pi nT_1/T_0)-\frac{1}{2\pi n}.$$
-*With the approximation&nbsp; $\text{sin}(\alpha ) \approx \alpha$&nbsp; valid for small&nbsp; $\alpha$-values one obtains for the imaginary part:
+*With the approximation&nbsp; $\text{sin}(\alpha ) \approx \alpha$&nbsp; valid for small&nbsp; $\alpha$&ndash;values one obtains for the imaginary part:
 :$${\rm Im}[D_n] =\frac{T_0/T_1}{(2\pi n)^2}\cdot(2\pi nT_1/T_0)-\frac{1}{2\pi n}=0.$$
@@ Line 98: / Line 98: @@
 :$$D_n^{(y)}=\frac{-\rm j}{2\pi n}={1}/{2} \cdot (A_n- {\rm j} \cdot B_n).$$
-*The coefficient comparison yields&nbsp; $A_n^{(y)} = 0$&nbsp; and&nbsp; $B_n^{(y)} = 1/(\pi n)$. In particular&nbsp; $A_1^{(y)} \hspace{0.1cm}\underline{= 0}$&nbsp; und&nbsp; $B_1^{(y)}\hspace{0.1cm}\underline{ \approx 0.318}$.
+*The coefficient comparison yields&nbsp; $A_n^{(y)} = 0$&nbsp; and&nbsp; $B_n^{(y)} = 1/(\pi n)$.&nbsp; In particular&nbsp; $A_1^{(y)} \hspace{0.1cm}\underline{= 0}$&nbsp; und&nbsp; $B_1^{(y)}\hspace{0.1cm}\underline{ \approx 0.318}$.
 *As expected, $B_{-n}^{(y)} = -B_n^{(y)}$ always holds.
@@ Line 122: / Line 122: @@
 '''(4)'''&nbsp; The correct <u>solutions are 2, 4 and 5</u>:
-*The signal&nbsp; $x(t)$&nbsp;  is equal to the difference between&nbsp; $y(t)$&nbsp; and&nbsp; $z(t)$. Since&nbsp; $z(t)$&nbsp; is an even and&nbsp; $y(t)$&nbsp; an odd function, the cosine coefficients&nbsp; $A_n$&nbsp; are determined by the coefficients of the signal&nbsp; $z(t)$&nbsp; alone, but with negative signs.
+*The signal&nbsp; $x(t)$&nbsp;  is equal to the difference between&nbsp; $y(t)$&nbsp; and&nbsp; $z(t)$.&nbsp; Since&nbsp; $z(t)$&nbsp; is an even and&nbsp; $y(t)$&nbsp; an odd function, the cosine coefficients&nbsp; $A_n$&nbsp; are determined by the coefficients of the signal&nbsp; $z(t)$&nbsp; alone, but with negative signs.
-*The sine coefficients $B_n$ completely agree with those of&nbsp; $y(t)$&nbsp; überein.
+*The sine coefficients&nbsp; $B_n$&nbsp; completely agree with those of&nbsp; $y(t)$.
 *The DC component of&nbsp; $x(t)$&nbsp; results from the difference of the two DC components of&nbsp; $y(t)$&nbsp; und&nbsp; $z(t)$: &nbsp;
 :$$A_0 = 0.5 - 0.25 = 0.25.$$

← Older revision		Revision as of 09:42, 16 April 2021
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	''Hints:''		''Hints:''
−	*This exercise refers to the page  [[Signal_Representation/Fourier_Series#~~Complex_Fourier_Series~~\|Complex Fourier series]].	+	*This exercise refers to the page  [[Signal_Representation/Fourier_Series#Complex_Fourier_series\|Complex Fourier series]].

@@ Line 5: / Line 5: @@
 [[File:P_ID312__Sig_A_2_6.png|right|frame|Various periodic triangular signals]]
-We consider the signal&nbsp; $x(t)$, defined by the two parameters&nbsp; $T_0$&nbsp; and&nbsp; $T_1$&nbsp; where&nbsp; $T_1 \leq T_0$&nbsp; should always apply. For the complex Fourier coefficients
+We consider the signal&nbsp; $x(t)$,&nbsp; defined by the two parameters&nbsp; $T_0$&nbsp; and&nbsp; $T_1$&nbsp; where&nbsp; $T_1 \leq T_0$&nbsp; should always apply.&nbsp; For the complex Fourier coefficients
 :$$D_n=\frac{1}{T_0} \cdot \int_0^{T_0}x(t)\cdot\rm e^{-\rm j\it n\omega_0t}\,{\rm d} \it t$$
@@ Line 11: / Line 11: @@
 of this signal are obtained after mathematical transformations:
-:$$D_n=\frac{T_0/T_1} {(2\pi n)^2} \cdot  \bigg(1-{\rm e}^{-{\rm j} 2\pi nT_1/T_0}\bigg)-\frac{\rm j}{2\pi n}.$$
+:$$D_n=\frac{T_0/T_1} {(2\pi n)^2} \cdot  \big(1-{\rm e}^{-{\rm j} 2\pi nT_1/T_0}\big)-\frac{\rm j}{2\pi n}.$$
 *The parameter set dealt with in subtasks&nbsp; '''(1)'''&nbsp; and&nbsp; '''(3)'''&nbsp; &nbsp; $($with $T_1 = T_0/2)$&nbsp; is represented as the signal&nbsp; $x(t)$&nbsp;.
 *For&nbsp; $T_1 = T_0$&nbsp; &rArr; &nbsp; subtask&nbsp; '''(2)'''&nbsp; the function&nbsp; $y(t)$ results.
-*In subtask&nbsp; '''(4)'''&nbsp; the signal&nbsp; $z(t)$&nbsp; is considered. Its Fourier coefficients are:
+*In subtask&nbsp; '''(4)'''&nbsp; the signal&nbsp; $z(t)$&nbsp; is considered.&nbsp; Its Fourier coefficients are:
 :$$A_0=1/4,\hspace{1cm}
@@ Line 28: / Line 28: @@
 ''Hints:''
-*This exercise refers to the page&nbsp; [[Signal_Representation/Fourier_Series#Complex_Fourier_Series|Complex Fourier Series]].
+*This exercise refers to the page&nbsp; [[Signal_Representation/Fourier_Series#Complex_Fourier_Series|Complex Fourier series]].
@@ Line 39: / Line 39: @@
 <quiz display=simple>
-{Calculate the coefficient&nbsp; $D_0$&nbsp; and show that it is always real. What value results for&nbsp; $T_1 = T_0/2$, i.e. for the signal&nbsp; $x(t)$?
+{Calculate the coefficient&nbsp; $D_0$&nbsp; and show that it is always real.&nbsp; What value results for&nbsp; $T_1 = T_0/2$, i.e. for the signal&nbsp; $x(t)$?
 |type="{}"}
 $D_0^{(x)}\ = \ $  { 0.25 3% }
-{Calculate the complex Fourier coefficients&nbsp; $D_n^{(y)}$&nbsp; for&nbsp; $n \neq 0$ for the special case &nbsp; $T_1 = T_0$&nbsp; corresponding to the signal&nbsp; $y(t)$&nbsp;. <br>What are the coefficients&nbsp; $A_n^{(y)}$&nbsp; and&nbsp; $B_n^{(y)}$, especially for&nbsp; $n = 1$?
+{Calculate the complex Fourier coefficients&nbsp; $D_n^{(y)}$&nbsp; for&nbsp; $n \neq 0$&nbsp; for the special case &nbsp; $T_1 = T_0$&nbsp; corresponding to the signal&nbsp; $y(t)$&nbsp;. <br>What are the coefficients&nbsp; $A_n^{(y)}$&nbsp; and&nbsp; $B_n^{(y)}$,&nbsp; especially for&nbsp; $n = 1$?
 |type="{}"}
 $A_1^{(y)}\ = \ $ { 0. }
 $B_1^{(y)}\ = \ $ { 0.318 3% }
-{Now calculate the coefficients&nbsp; $A_n^{(x)}$&nbsp; and&nbsp; $B_n^{(x)}$&nbsp; for the signal&nbsp; $x(t)$&nbsp; with&nbsp; $T_1 = T_0/2$&nbsp; for&nbsp; $n \neq 0$. What are the values for&nbsp; $A_1^{(x)}$&nbsp; and&nbsp; $B_1^{(x)}$?
+{Now calculate the coefficients&nbsp; $A_n^{(x)}$&nbsp; and&nbsp; $B_n^{(x)}$&nbsp; for the signal&nbsp; $x(t)$&nbsp; with&nbsp; $T_1 = T_0/2$&nbsp; for&nbsp; $n \neq 0$.&nbsp; What are the values for&nbsp; $A_1^{(x)}$&nbsp; and&nbsp; $B_1^{(x)}$?
 |type="{}"}
 $A_1^{(x)}\ = \ $ { 0.203 3% }

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID312__Sig_A_2_6.png|right|frame|Verschiedene periodische Dreiecksignale]]
+[[File:P_ID312__Sig_A_2_6.png|right|frame|Various periodic triangular signals]]
 We consider the signal&nbsp; $x(t)$, defined by the two parameters&nbsp; $T_0$&nbsp; and&nbsp; $T_1$&nbsp; where&nbsp; $T_1 \leq T_0$&nbsp; should always apply. For the complex Fourier coefficients

← Older revision		Revision as of 12:38, 23 March 2021
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	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:~~Exercises for~~ Signal Representation\|^2.4 Fourier Series^]]	+	[[Category:Signal Representation: Exercises\|^2.4 Fourier Series^]]

@@ Line 18: / Line 18: @@
 :$$A_0=1/4,\hspace{1cm}
-A_n=\left\{ \begin{array}{cl} {\frac{\displaystyle-2}{\displaystyle(\pi  n)^2}} & {\rm   f\ddot{u}r\; geradzahliges\; \it n \rm ,} \\ 0 &  {\rm f\ddot{u}r\; ungeradzahliges\; \it n,} \end{array}\right. $$
+A_n=\left\{ \begin{array}{cl} {\frac{\displaystyle-2}{\displaystyle(\pi  n)^2}} & {\rm   for\; even\; \it n \rm ,} \\ 0 &  {\rm for\; odd\; \it n,} \end{array}\right. $$
 :$$B_n=0\; \;\;          \rm{ f\ddot{u}r\; alle\; \it n.}$$
@@ Line 67: / Line 67: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Mit dem Eulerschen Satz ist der komplexe Fourierkoeffizient&nbsp; $D_n$&nbsp; wie folgt darstellbar:
+'''(1)'''&nbsp; With Euler's theorem, the complex Fourier coefficient&nbsp; $D_n$ can be represented as follows:
 :$${\rm Re} [D_n] =\frac{T_0/T_1}{(2\pi n)^2}\cdot(1-\cos(2\pi nT_1/T_0)),$$
@@ Line 73: / Line 73: @@
 :$${\rm Im}[D_n] =\frac{T_0/T_1}{(2\pi n)^2} \cdot \sin(2\pi nT_1/T_0)-\frac{1}{2\pi n}.$$
-*Mit der für kleine&nbsp; $\alpha$-Werte gültigen Näherung&nbsp; $\text{sin}(\alpha ) \approx \alpha$&nbsp; erhält man für den Imaginärteil:
+*With the approximation&nbsp; $\text{sin}(\alpha ) \approx \alpha$&nbsp; valid for small&nbsp; $\alpha$-values one obtains for the imaginary part:
 :$${\rm Im}[D_n] =\frac{T_0/T_1}{(2\pi n)^2}\cdot(2\pi nT_1/T_0)-\frac{1}{2\pi n}=0.$$
-*Für den Realteil erhält man mit&nbsp; $\text{cos}(\alpha) \approx 1 – \alpha^{2}/2$:
+*For the real part one obtains with&nbsp; $\text{cos}(\alpha) \approx 1 – \alpha^{2}/2$:
 :$${\rm Re}[D_n] =\frac{T_0/T_1}{(2\pi n)^2}\frac{(2\pi nT_1/T_0)^2}{2}=\frac{T_1/T_0}{2}.$$
-*Für&nbsp; $T_1 = T_0/2$&nbsp; folgt daraus der Gleichsignalkoeffizient&nbsp; $D_0^{(x)} \hspace{0.1cm}\underline{= 0.25}$.
+*For&nbsp; $T_1 = T_0/2$&nbsp; it follows that the DC signal coefficient&nbsp; $D_0^{(x)} \hspace{0.1cm}\underline{= 0.25}$.
-*Mit&nbsp; $T_1 = T_0$&nbsp; ergibt sich&nbsp; $D_0^{(y)} = 0.5$.
+*With&nbsp; $T_1 = T_0$&nbsp; it results in&nbsp; $D_0^{(y)} = 0.5$.
-*Ein Vergleich mit den Signalen&nbsp; $x(t)$&nbsp; und&nbsp; $y(t)$&nbsp; auf der Angabenseite zeigen die Richtigkeit dieser Ergebnisse.
+*A comparison with the signals &nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp;on the data page show the correctness of these results.
-'''(2)'''&nbsp; Es wird nun&nbsp; $n \neq 0$&nbsp; vorausgesetzt. Mit&nbsp; $T_1 = T_0$&nbsp; erhält man für den Realteil wegen&nbsp; $\text{cos}(2\pi n) = 1$:
+'''(2)'''&nbsp; It is now assumed&nbsp; $n \neq 0$&nbsp;. With&nbsp; $T_1 = T_0$&nbsp; one obtains for the real part because of&nbsp; $\text{cos}(2\pi n) = 1$:
 :$${\rm Re}[D_n^{(y)}] =\frac{1}{(2\pi n)^2}\cdot(1-\cos(2\pi n))=0.$$
-*Der Imagnärteil lautet:
+*The imagnary part is:
 :$${\rm Im}[D_n^{(y)}] =\frac{1}{(2\pi n)^2}\cdot(\sin(2\pi n))-\frac{1}{2\pi n}.$$
-*Wegen&nbsp; $\text{sin}(2\pi n) = 0$&nbsp; folgt daraus &nbsp; ${\rm Im}[D_n] =-{1}/({2\pi n}).$ Somit ist
+*Because&nbsp; $\text{sin}(2\pi n) = 0$&nbsp; it follows that &nbsp; ${\rm Im}[D_n] =-{1}/({2\pi n}).$ Thus
 :$$D_n^{(y)}=\frac{-\rm j}{2\pi n}={1}/{2} \cdot (A_n- {\rm j} \cdot B_n).$$
-*Der Koeffizientenvergleich liefert&nbsp; $A_n^{(y)} = 0$&nbsp; und&nbsp; $B_n^{(y)} = 1/(\pi n)$. Insbesondere sind&nbsp; $A_1^{(y)} \hspace{0.1cm}\underline{= 0}$&nbsp; und&nbsp; $B_1^{(y)}\hspace{0.1cm}\underline{ \approx 0.318}$.
+*The coefficient comparison yields&nbsp; $A_n^{(y)} = 0$&nbsp; and&nbsp; $B_n^{(y)} = 1/(\pi n)$. In particular&nbsp; $A_1^{(y)} \hspace{0.1cm}\underline{= 0}$&nbsp; und&nbsp; $B_1^{(y)}\hspace{0.1cm}\underline{ \approx 0.318}$.
-*Wie zu erwarten war, gilt stets $B_{-n}^{(y)} = -B_n^{(y)}$.
+*As expected, $B_{-n}^{(y)} = -B_n^{(y)}$ always holds.
-'''(3)'''&nbsp; Aus der in der Teilaufgabe&nbsp; '''(1)'''&nbsp; berechneten allgemeinen Gleichung folgt mit&nbsp; $T_1/T_0 = 1/2$:
+'''(3)'''&nbsp; From the general equation calculated in subtask&nbsp; '''(1)'''&nbsp; it follows with&nbsp; $T_1/T_0 = 1/2$:
 :$$D_n^{(x)}=\frac{2}{(2\pi n)^2}(1-\cos(\pi n))+{\rm j}\cdot \left[\frac{2\sin(\pi n)}{(2\pi n)^2}-\frac{1}{(2\pi n)}\right].$$
-*Daraus erhält man die Cosinuskoeffizienten
+*From this one obtains the cosine coefficients
-:$$A_n^{(x)}={2}\cdot{\rm Re}[D_n] =\left\{ \begin{array}{cl} {\frac{\displaystyle 2}{\displaystyle(\pi n)^2}} & {\rm   f\ddot{u}r\; ungeradzahliges\; \it n ,} \\ 0 &  {\rm f\ddot{u}r\; geradzahliges\;\it n.} \end{array}\right. $$
+:$$A_n^{(x)}={2}\cdot{\rm Re}[D_n] =\left\{ \begin{array}{cl} {\frac{\displaystyle 2}{\displaystyle(\pi n)^2}} & {\rm   for\; odd\; \it n ,} \\ 0 &  {\rm for\; even\;\it n.} \end{array}\right. $$
-*Die Sinuskoeffizienten lauten:
+*The sine coefficients are:
 :$$B_n^{(x)}=-2\cdot{\rm Im}[D_n] =\frac{1}{\pi n}.$$
-*Hierbei ist berücksichtigt, dass für alle ganzzahligen Werte von&nbsp; $n$&nbsp; die Funktion&nbsp; $\text{sin}(n\pi ) = 0$&nbsp; ist. Die jeweils ersten reellen Koeffizienten lauten
+*Here it is taken into account that for all integer values of&nbsp; $n$&nbsp; the function&nbsp; $\text{sin}(n\pi ) = 0$&nbsp;. The first real coefficients are as follows
 :$$A_1^{(x)} = 2/\pi^{2} \hspace{0.1cm}\underline{\approx 0.203},$$
 :$$B_1 = 1/\pi \hspace{0.1cm}\underline{\approx 0.318}.$$
-'''(4)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 2, 4 und 5</u>:
+'''(4)'''&nbsp; The correct <u>solutions are 2, 4 and 5</u>:
-*Das Signal&nbsp; $x(t)$&nbsp; ist gleich der Differenz zwischen&nbsp; $y(t)$&nbsp; und&nbsp; $z(t)$. Da&nbsp; $z(t)$&nbsp; eine gerade und&nbsp; $y(t)$&nbsp; eine ungerade Funktion ist, werden die Cosinuskoeffizienten&nbsp; $A_n$&nbsp; allein durch die Koeffizienten des Signals&nbsp; $z(t)$&nbsp; bestimmt, allerdings mit negativen Vorzeichen.
+*The signal&nbsp; $x(t)$&nbsp;  is equal to the difference between&nbsp; $y(t)$&nbsp; and&nbsp; $z(t)$. Since&nbsp; $z(t)$&nbsp; is an even and&nbsp; $y(t)$&nbsp; an odd function, the cosine coefficients&nbsp; $A_n$&nbsp; are determined by the coefficients of the signal&nbsp; $z(t)$&nbsp; alone, but with negative signs.
-*Die Sinuskoeffizienten $B_n$ stimmen vollständig mit denen von&nbsp; $y(t)$&nbsp; überein.
+*The sine coefficients $B_n$ completely agree with those of&nbsp; $y(t)$&nbsp; überein.
-*Der Gleichsignalanteil von&nbsp; $x(t)$&nbsp; ergibt sich aus der Differenz der beiden Gleichanteile von&nbsp; $y(t)$&nbsp; und&nbsp; $z(t)$: &nbsp;
+*The DC component of&nbsp; $x(t)$&nbsp; results from the difference of the two DC components of&nbsp; $y(t)$&nbsp; und&nbsp; $z(t)$: &nbsp;
 :$$A_0 = 0.5 - 0.25 = 0.25.$$
 {{ML-Fuß}}

@@ Line 28: / Line 28: @@
 ''Hints:''
-*This exercise refers to the page&nbsp; [[Signal_Representation/Fourier_Series#Komplexe_Fourierreihe|Komplexe Fourierreihe]].
+*This exercise refers to the page&nbsp; [[Signal_Representation/Fourier_Series#Complex_Fourier_Series|Complex Fourier Series]].
@@ Line 35: / Line 35: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Berechnen Sie den Koeffizienten&nbsp; $D_0$&nbsp; und zeigen Sie, dass dieser stets reell ist. Welcher Wert ergibt sich für&nbsp; $T_1 = T_0/2$, also für das Signal&nbsp; $x(t)$?
+{Calculate the coefficient&nbsp; $D_0$&nbsp; and show that it is always real. What value results for&nbsp; $T_1 = T_0/2$, i.e. for the signal&nbsp; $x(t)$?
 |type="{}"}
 $D_0^{(x)}\ = \ $  { 0.25 3% }
-{Berechnen Sie für den Sonderfall&nbsp; $T_1 = T_0$&nbsp; entsprechend dem Signal&nbsp; $y(t)$&nbsp; die komplexen Fourierkoeffizienten&nbsp; $D_n^{(y)}$&nbsp; für&nbsp; $n \neq 0$. <br>Wie lauten die Koeffizienten&nbsp; $A_n^{(y)}$&nbsp; und&nbsp; $B_n^{(y)}$, insbesondere für&nbsp; $n = 1$?
+{Calculate the complex Fourier coefficients&nbsp; $D_n^{(y)}$&nbsp; for&nbsp; $n \neq 0$ for the special case &nbsp; $T_1 = T_0$&nbsp; corresponding to the signal&nbsp; $y(t)$&nbsp;. <br>What are the coefficients&nbsp; $A_n^{(y)}$&nbsp; and&nbsp; $B_n^{(y)}$, especially for&nbsp; $n = 1$?
 |type="{}"}
 $A_1^{(y)}\ = \ $ { 0. }
 $B_1^{(y)}\ = \ $ { 0.318 3% }
-{Berechnen Sie nun für das Signal&nbsp; $x(t)$&nbsp; mit&nbsp; $T_1 = T_0/2$&nbsp; die Koeffizienten&nbsp; $A_n^{(x)}$&nbsp; und&nbsp; $B_n^{(x)}$&nbsp; für&nbsp; $n \neq 0$. Welche Werte ergeben sich für&nbsp; $A_1^{(x)}$&nbsp; und&nbsp; $B_1^{(x)}$?
+{Now calculate the coefficients&nbsp; $A_n^{(x)}$&nbsp; and&nbsp; $B_n^{(x)}$&nbsp; for the signal&nbsp; $x(t)$&nbsp; with&nbsp; $T_1 = T_0/2$&nbsp; for&nbsp; $n \neq 0$. What are the values for&nbsp; $A_1^{(x)}$&nbsp; and&nbsp; $B_1^{(x)}$?
 |type="{}"}
 $A_1^{(x)}\ = \ $ { 0.203 3% }
 $B_1^{(x)}\ = \ $ { 0.318 3% }
-{Welche der folgenden Aussagen treffen bezüglich&nbsp; $x(t)$,&nbsp; $y(t)$&nbsp; und&nbsp; $z(t)$&nbsp; zu?
+{Which of the following statements are true regarding&nbsp; $x(t)$,&nbsp; $y(t)$&nbsp; and&nbsp; $z(t)$&nbsp;?
 |type="[]"}
-- Es gilt&nbsp; $x(t) = y(t) + z(t)$.
+- It is true that&nbsp; $x(t) = y(t) + z(t)$.
-+ Es gilt&nbsp; $x(t) = y(t) - z(t)$.
++ It is true that&nbsp; $x(t) = y(t) - z(t)$.
-- Die Cosinuskoeffizienten&nbsp; $A_n$&nbsp; von&nbsp; $x(t)$&nbsp; und&nbsp; $z(t)$&nbsp; sind identisch.
+- The cosine coefficients&nbsp; $A_n$&nbsp; of&nbsp; $x(t)$&nbsp; und&nbsp; $z(t)$&nbsp; are identical.
-+ Die Cosinuskoeffizienten&nbsp; $A_n$&nbsp; von&nbsp; $x(t)$&nbsp; und&nbsp; $z(t)$&nbsp; sind betragsgleich.
++ The cosine coefficients&nbsp; $A_n$&nbsp; of&nbsp; $x(t)$&nbsp; und&nbsp; $z(t)$&nbsp; are equal in magnitude.
-+ Die Sinuskoeffizienten&nbsp; $B_n$&nbsp; von&nbsp; $y(t)$&nbsp; und&nbsp; $z(t)$&nbsp; sind identisch.
++ The sine coefficients&nbsp; $B_n$&nbsp; of&nbsp; $y(t)$&nbsp; und&nbsp; $z(t)$&nbsp; are identical.
@@ Line 65: / Line 65: @@
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; Mit dem Eulerschen Satz ist der komplexe Fourierkoeffizient&nbsp; $D_n$&nbsp; wie folgt darstellbar:

← Older revision		Revision as of 14:15, 2 November 2020
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	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:Exercises for Signal Representation\|^2. ~~Periodische Signale~~^]]	+	[[Category:Exercises for Signal Representation\|^2.4 Fourier Series^]]