Aufgaben:Exercise 2.6Z: Magnitude and Phase - Revision history

Javier: Text replacement - "analyse" to "analyze"

2021-09-22T11:47:59Z

Text replacement - "analyse" to "analyze"

Guenter at 11:28, 16 April 2021

2021-04-16T11:28:59Z

Guenter at 11:24, 16 April 2021

2021-04-16T11:24:48Z

Guenter at 11:22, 16 April 2021

2021-04-16T11:22:36Z

Guenter at 11:20, 16 April 2021

2021-04-16T11:20:16Z

Noah at 09:27, 26 March 2021

2021-03-26T09:27:52Z

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

2021-03-23T12:39:19Z

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

Noah at 20:36, 16 January 2021

2021-01-16T20:36:55Z

Noah at 20:36, 16 January 2021

2021-01-16T20:36:37Z

Noah at 20:32, 16 January 2021

2021-01-16T20:32:13Z

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID348__Sig_Z_2_6.png|right|frame|Signal&nbsp; $x(t)$&nbsp; to be analysed]]
+[[File:P_ID348__Sig_Z_2_6.png|right|frame|Signal&nbsp; $x(t)$&nbsp; to be analyzed]]
 The aim is to show the connection between
 * the real Fourier coefficients&nbsp; $A_n$&nbsp; und&nbsp; $B_n$,

@@ Line 97: / Line 97: @@
 :$$\varphi_n=\arctan\left({B_n}/{A_n}\right),\hspace{0.5cm}C_n=\sqrt{A_n^2+B_n^2},\hspace{0.5cm}D_n={1}/{2} \cdot (A_n-{\rm j}\cdot B_n).$$
-*Because&nbsp; $B_1 = 0$&nbsp; we get&nbsp; $\varphi_1 \hspace{0.1cm}\underline{= 0}, \ C_1 = A_1 \hspace{0.1cm}\underline{= 2 \,{\rm  V}}$&nbsp; und&nbsp; $D_1 = A_1/2 \hspace{0.1cm}\underline{= 1 \,{\rm  V}}$.
+*Because&nbsp; $B_1 = 0$&nbsp; we get&nbsp; $\varphi_1 \hspace{0.1cm}\underline{= 0}, \ C_1 = A_1 \hspace{0.1cm}\underline{= 2 \,{\rm  V}}$&nbsp; and&nbsp; $D_1 = A_1/2 \hspace{0.1cm}\underline{= 1 \,{\rm  V}}$.
-'''(4)'''&nbsp; With&nbsp; $A_2 = 2\,{\rm  V}$&nbsp; und&nbsp; $B_2 = -1\,{\rm  V}$&nbsp; one obtains:
+'''(4)'''&nbsp; With&nbsp; $A_2 = 2\,{\rm  V}$&nbsp; and&nbsp; $B_2 = -1\,{\rm  V}$&nbsp; one obtains:
 :$$\varphi_2=\arctan(-0.5)\hspace{0.15cm}\underline{=-26.56^{\circ}},\hspace{0.5cm}C_2=\sqrt{A_2^2+B_2^2}\hspace{0.15cm}\underline{=2.236 \; \rm V},$$
 :$$D_2={1}/{2} \cdot (A_2-{\rm j}\cdot B_2)=1\;\rm V+{\rm j}\cdot 0.5\, {\rm  V}
@@ Line 109: / Line 109: @@
-'''(5)'''&nbsp; It is&nbsp; $\varphi_3 \hspace{0.15cm}\underline{=\hspace{0.1cm}-90^{\circ}}$&nbsp; und&nbsp; $C_3 = |B_3| \hspace{0.15cm}\underline{ = 1 \,{\rm  V}}$.
+'''(5)'''&nbsp; It is&nbsp; $\varphi_3 \hspace{0.15cm}\underline{=\hspace{0.1cm}-90^{\circ}}$&nbsp; and&nbsp; $C_3 = |B_3| \hspace{0.15cm}\underline{ = 1 \,{\rm  V}}$.

@@ Line 31: / Line 31: @@
 <quiz display=simple>
-{What are the values of the coefficients&nbsp; $A_0$,&nbsp; $D_0$,&nbsp; $C_0$ and&nbsp; $\varphi_0$?
+{What are the  coefficients&nbsp; $A_0$,&nbsp; $D_0$,&nbsp; $C_0$ and&nbsp; $\varphi_0$?
 |type="{}"}
 $A_0\ = \ $  { 1 3% } &nbsp;$\text{V}$
@@ Line 49: / Line 49: @@
-{What are the values of the coefficients&nbsp; $\varphi_1$,&nbsp; $C_1$&nbsp; und&nbsp; $D_1$?
+{What are the coefficients&nbsp; $\varphi_1$,&nbsp; $C_1$&nbsp; and&nbsp; $D_1$?
 |type="{}"}
 $\varphi_1\ = \ $ { 0. } &nbsp;$\text{deg}$
@@ Line 57: / Line 57: @@
-{What are the values of the coefficients&nbsp; $\varphi_2$,&nbsp; $C_2$&nbsp; und&nbsp; $D_2$?
+{What are the coefficients&nbsp; $\varphi_2$,&nbsp; $C_2$&nbsp; and&nbsp; $D_2$?
 |type="{}"}
 $\varphi_2\ = \ $  { -26.6--26.5 } &nbsp;$\text{deg}$
@@ Line 65: / Line 65: @@
-{What are the values of the coefficients&nbsp; $\varphi_3$&nbsp; und&nbsp; $C_3$?
+{What are the coefficients&nbsp; $\varphi_3$&nbsp; and&nbsp; $C_3$?
 |type="{}"}
 $\varphi_3\ = \ $  { -91--89 } &nbsp;$\text{deg}$
@@ Line 71: / Line 71: @@
-{What is the value of the complex Fourier coefficient&nbsp; $D_\text{–3}$?
+{What is the complex Fourier coefficient&nbsp; $D_\text{–3}$?
 |type="{}"}
 $\text{Re}[D_{-3}]\ = \ $ { 0. } &nbsp;$\text{V}$

@@ Line 4: / Line 4: @@
 [[File:P_ID348__Sig_Z_2_6.png|right|frame|Signal&nbsp; $x(t)$&nbsp; to be analysed]]
-The relationship between
+The aim is to show the connection between
 * the real Fourier coefficients&nbsp; $A_n$&nbsp; und&nbsp; $B_n$,
-* the complex coefficients&nbsp; $D_n$, sowie
+* the complex coefficients&nbsp; $D_n$, and
 * the magnitude or phase coefficients&nbsp; $(C_n$,&nbsp; $\varphi_n)$.

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID348__Sig_Z_2_6.png|right|frame|Signal &nbsp; $x(t)$ to be analysed]]
+[[File:P_ID348__Sig_Z_2_6.png|right|frame|Signal&nbsp; $x(t)$&nbsp; to be analysed]]
 The relationship between
-:* the real Fourier coefficients&nbsp; $A_n$&nbsp; und&nbsp; $B_n$,
+* the real Fourier coefficients&nbsp; $A_n$&nbsp; und&nbsp; $B_n$,
 * the complex coefficients&nbsp; $D_n$, sowie
-:* the complex coefficients&nbsp; $D_n$, sowie
+* the magnitude or phase coefficients&nbsp; $(C_n$,&nbsp; $\varphi_n)$.
@@ Line 17: / Line 15: @@
 :$$x(t)=1{\rm V+2V}\cdot\cos(\omega_0 t)  +{\rm 2V}\cdot\cos(2\omega_0 t)- \ {\rm 1V}\cdot\sin(2\omega_0 t)-{\rm 1V}\cdot\sin(3\omega_0 t).$$
-This signal is shown in the graph in the range from&nbsp; $–2T_0$&nbsp; bis&nbsp; $+2T_0$&nbsp; dargestellt.
+This signal is shown in the graph in the range from&nbsp; $–2T_0$&nbsp; to&nbsp; $+2T_0$.
@@ Line 26: / Line 22: @@
 ''Hints:''
-*This exercise belongs the the chapter&nbsp; [[Signal_Representation/Fourier_Series|Fourier Series]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Series|Fourier Series]].
 *You can find a compact summary of the topic in the two learning videos
-::[[Zur_Berechnung_der_Fourierkoeffizienten_(Lernvideo)|Zur Berechnung der Fourierkoeffizienten]],
+:[[Zur_Berechnung_der_Fourierkoeffizienten_(Lernvideo)|Zur Berechnung der Fourierkoeffizienten]]&nbsp; &rArr; &nbsp; "To calculate the Fourier coefficients",
-::[[Eigenschaften_der_Fourierreihendarstellung_(Lernvideo)|Eigenschaften der Fourierreihendarstellung]].
+: [[Eigenschaften_der_Fourierreihendarstellung_(Lernvideo)|Eigenschaften der Fourierreihendarstellung]] &nbsp; &rArr; &nbsp;   "Properties of the Fourier series representation".

← Older revision		Revision as of 12:39, 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^]]

← Older revision		Revision as of 20:36, 16 January 2021
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−	'''(6)'''  It is  $D_3 = -{\rm j} · B_3/2 ={\rm j}· 0.5 \,{\rm V}$  ~~und~~  $D_\text{–3} = D_3^{\star} ={\rm j}· B_3/2 = {- {\rm j} · 0.5 \,{\rm V}}$	+	'''(6)'''  It is  $D_3 = -{\rm j} · B_3/2 ={\rm j}· 0.5 \,{\rm V}$  and  $D_\text{–3} = D_3^{\star} ={\rm j}· B_3/2 = {- {\rm j} · 0.5 \,{\rm V}}$

	:$$\Rightarrow \hspace{0.3cm} \text{Re}[D_{-3}]\hspace{0.15cm}\underline{=0}, \hspace{0.5cm}\text{Im}[D_{-3}]\hspace{0.15cm}\underline{=\hspace{0.1cm}- 0.5 \,{\rm V}}.$$		:$$\Rightarrow \hspace{0.3cm} \text{Re}[D_{-3}]\hspace{0.15cm}\underline{=0}, \hspace{0.5cm}\text{Im}[D_{-3}]\hspace{0.15cm}\underline{=\hspace{0.1cm}- 0.5 \,{\rm V}}.$$

@@ Line 86: / Line 86: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Der Gleichsignalkoeffizient beträgt&nbsp; $A_0 = 1\,{\rm  V}$.
+'''(1)'''&nbsp;  The DC signal coefficient is&nbsp; $A_0 = 1\,{\rm  V}$.
-*Gleichzeitig gilt&nbsp; $C_0 = D_0 = A_0 \hspace{0.1cm}\Rightarrow  \hspace{0.1cm} C_0 \hspace{0.1cm}\underline{= 1\,{\rm  V}}, \varphi_0 \hspace{0.1cm}\underline{= 0}$.
+*At the same time,&nbsp; $C_0 = D_0 = A_0 \hspace{0.1cm}\Rightarrow  \hspace{0.1cm} C_0 \hspace{0.1cm}\underline{= 1\,{\rm  V}}, \varphi_0 \hspace{0.1cm}\underline{= 0}$.
-'''(2)'''&nbsp;  <u>Richtig sind die Antworten 1, 3, 4 und 6</u>:
+'''(2)'''&nbsp;  <u>The correct answers are 1, 3, 4 and 6</u>:
-*Es gibt keine Anteile mit&nbsp; $\sin(\omega_0t)$&nbsp; und&nbsp; $\cos(3\omega_0t)$.
+*There are no components with&nbsp; $\sin(\omega_0t)$&nbsp; and&nbsp; $\cos(3\omega_0t)$.
-*Daraus folgt direkt&nbsp; $B_1 = A_3 = 0$.
+*It follows directly that&nbsp; $B_1 = A_3 = 0$.
-*Alle anderen hier aufgeführten Koeffizienten sind ungleich Null.
+*All other coefficients listed here are non-zero.
-'''(3)'''&nbsp; Allgemein gilt:
+'''(3)'''&nbsp; In general:
 :$$\varphi_n=\arctan\left({B_n}/{A_n}\right),\hspace{0.5cm}C_n=\sqrt{A_n^2+B_n^2},\hspace{0.5cm}D_n={1}/{2} \cdot (A_n-{\rm j}\cdot B_n).$$
-*Wegen&nbsp; $B_1 = 0$&nbsp; erhält man&nbsp; $\varphi_1 \hspace{0.1cm}\underline{= 0}, \ C_1 = A_1 \hspace{0.1cm}\underline{= 2 \,{\rm  V}}$&nbsp; und&nbsp; $D_1 = A_1/2 \hspace{0.1cm}\underline{= 1 \,{\rm  V}}$.
+*Because&nbsp; $B_1 = 0$&nbsp; we get&nbsp; $\varphi_1 \hspace{0.1cm}\underline{= 0}, \ C_1 = A_1 \hspace{0.1cm}\underline{= 2 \,{\rm  V}}$&nbsp; und&nbsp; $D_1 = A_1/2 \hspace{0.1cm}\underline{= 1 \,{\rm  V}}$.
-'''(4)'''&nbsp; Mit&nbsp; $A_2 = 2\,{\rm  V}$&nbsp; und&nbsp; $B_2 = -1\,{\rm  V}$&nbsp; erhält man:
+'''(4)'''&nbsp; With&nbsp; $A_2 = 2\,{\rm  V}$&nbsp; und&nbsp; $B_2 = -1\,{\rm  V}$&nbsp; one obtains:
 :$$\varphi_2=\arctan(-0.5)\hspace{0.15cm}\underline{=-26.56^{\circ}},\hspace{0.5cm}C_2=\sqrt{A_2^2+B_2^2}\hspace{0.15cm}\underline{=2.236 \; \rm V},$$
 :$$D_2={1}/{2} \cdot (A_2-{\rm j}\cdot B_2)=1\;\rm V+{\rm j}\cdot 0.5\, {\rm  V}
@@ Line 113: / Line 114: @@
-'''(5)'''&nbsp; Es ist&nbsp; $\varphi_3 \hspace{0.15cm}\underline{=\hspace{0.1cm}-90^{\circ}}$&nbsp; und&nbsp; $C_3 = |B_3| \hspace{0.15cm}\underline{ = 1 \,{\rm  V}}$.
+'''(5)'''&nbsp; It is&nbsp; $\varphi_3 \hspace{0.15cm}\underline{=\hspace{0.1cm}-90^{\circ}}$&nbsp; und&nbsp; $C_3 = |B_3| \hspace{0.15cm}\underline{ = 1 \,{\rm  V}}$.
-'''(6)'''&nbsp; Es gilt&nbsp; $D_3 = -{\rm j} · B_3/2 ={\rm j}· 0.5 \,{\rm  V}$&nbsp; und&nbsp; $D_\text{–3} = D_3^{\star} ={\rm j}· B_3/2 =  {- {\rm j} · 0.5 \,{\rm  V}}$
+'''(6)'''&nbsp; It is&nbsp; $D_3 = -{\rm j} · B_3/2 ={\rm j}· 0.5 \,{\rm  V}$&nbsp; und&nbsp; $D_\text{–3} = D_3^{\star} ={\rm j}· B_3/2 =  {- {\rm j} · 0.5 \,{\rm  V}}$
 :$$\Rightarrow \hspace{0.3cm} \text{Re}[D_{-3}]\hspace{0.15cm}\underline{=0}, \hspace{0.5cm}\text{Im}[D_{-3}]\hspace{0.15cm}\underline{=\hspace{0.1cm}- 0.5 \,{\rm  V}}.$$