Aufgaben:Exercise 2.3: Cosine and Sine Components - Revision history

Guenter at 16:33, 17 May 2021

2021-05-17T16:33:01Z

Guenter at 11:33, 14 April 2021

2021-04-14T11:33:44Z

Guenter at 14:18, 13 April 2021

2021-04-13T14:18:23Z

Guenter at 14:09, 13 April 2021

2021-04-13T14:09:07Z

Noah at 09:22, 26 March 2021

2021-03-26T09:22:43Z

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

2021-03-23T12:38:46Z

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

Javier: Javier moved page Aufgaben:Exercises 2.3: Cosine and Sine Components to Aufgaben:Exercise 2.3: Cosine and Sine Components

2021-01-04T09:47:59Z

Javier moved page Exercises 2.3: Cosine and Sine Components to Exercise 2.3: Cosine and Sine Components

Noah at 03:00, 4 January 2021

2021-01-04T03:00:17Z

Noah at 02:37, 4 January 2021

2021-01-04T02:37:48Z

Noah at 02:31, 4 January 2021

2021-01-04T02:31:03Z

@@ Line 3: / Line 3: @@
 }}
-[[File: P_ID278_Sig_A_2_3neu.png|right|frame|Spectra of cosine and sine components]]
+[[File: P_ID278_Sig_A_2_3neu.png|right|frame|Spectra of DC, cosine and sine components]]
 Given is the amplitude spectrum&nbsp; $X(f)$&nbsp; of a signal&nbsp; $x(t)$&nbsp; according to the graph.
@@ Line 61: / Line 61: @@
 :$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$
-[[File:P_ID293__Sig_A_2_3_a.png|right|frame|Sum signal of cosine and sine components]]
+[[File:P_ID293__Sig_A_2_3_a.png|right|frame|Sum signal of DC, cosine and sine components]]
 *Here&nbsp; $\omega_1 = 2\pi f_1$&nbsp; denotes the circular frequency of the cosine component.

@@ Line 57: / Line 57: @@
 ===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; The time signal has the following form:
 :$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$
 *Here&nbsp; $\omega_1 = 2\pi f_1$&nbsp; denotes the circular frequency of the cosine component.
 *At time&nbsp; $t = 0$&nbsp; the signal has the value&nbsp; $x(t=0)\hspace{0.15 cm}\underline{=1\,\rm V}$.
@@ Line 74: / Line 74: @@
 From this follows&nbsp; $f_0 = 2{\,\rm kHz}$ &nbsp; &rArr; &nbsp;  period duration $T_0 = 1/f_0 \hspace{0.1cm}\underline{= 0.5 {\,\rm ms}}$.
 <br clear=all>
 '''(3)'''&nbsp; The following applies to the output signal $y(t)$ of the differentiatior:
 :$$y(t)=\frac{1}{\omega_1}\cdot\frac{ {\rm d}x(t)}{{\rm d}t}=\frac{ {\rm -2V}}{\omega_1}\cdot\omega_1 \cdot (-\sin(\omega_1 t))+\frac{\rm 4V}{\omega_1}\cdot 2.5\omega_1\cdot {\rm cos}(2.5\omega_1t).$$
 *This leads to the solution:

@@ Line 62: / Line 62: @@
 :$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$
-*Here&nbsp; $\omega_1 = 2\pi f_1$&nbsp; denotes the angular frequency of the cosine component.
+*Here&nbsp; $\omega_1 = 2\pi f_1$&nbsp; denotes the circular frequency of the cosine component.
 *At time&nbsp; $t = 0$&nbsp; the signal has the value&nbsp; $x(t=0)\hspace{0.15 cm}\underline{=1\,\rm V}$.
-'''(2)'''&nbsp; The base frequency&nbsp; $f_0$&nbsp; is the least common divisor
+'''(2)'''&nbsp; The basic frequency&nbsp; $f_0$&nbsp; is the greatest common divisor
 *of $f_1 = 4{\,\rm kHz}$
 *and $2.5 · f_1 = 10{\,\rm kHz}$.

@@ Line 3: / Line 3: @@
 }}
-[[File: P_ID278_Sig_A_2_3neu.png|right|frame|Spectrum of cosine and sine components]]
+[[File: P_ID278_Sig_A_2_3neu.png|right|frame|Spectra of cosine and sine components]]
 Given is the amplitude spectrum&nbsp; $X(f)$&nbsp; of a signal&nbsp; $x(t)$&nbsp; according to the graph.
 *Let&nbsp; $f_1 = 4\,\text{kHz}$ be the normalisation frequency.
-*Thus the actual frequencies of the signal components are&nbsp; $0\,\text{kHz}$,&nbsp; $4\,\text{kHz}$&nbsp; and&nbsp; $10\,\text{kHz}$.
+*Thus the frequencies of the signal components are&nbsp; $0\,\text{kHz}$,&nbsp; $4\,\text{kHz}$&nbsp; and&nbsp; $10\,\text{kHz}$.
@@ Line 41: / Line 41: @@
 $T_0\ = \ $ { 0.5 3% } &nbsp; ${\rm ms}$
-{Calculate the output signal&nbsp; $y(t)$&nbsp; of the differentiator&nbsp; What is the signal value at time&nbsp; $t = 0$?
+{Calculate the output signal&nbsp; $y(t)$&nbsp; of the differentiator.&nbsp; What is the signal value at time&nbsp; $t = 0$?
 |type="{}"}
 $y(t=0)\ = \ $ { 10 3% } &nbsp; ${\rm V}$

@@ Line 3: / Line 3: @@
 }}
-[[File: P_ID278_Sig_A_2_3neu.png|right|frame|Spektrum von Cosinus- und Sinusanteilen]]
+[[File: P_ID278_Sig_A_2_3neu.png|right|frame|Spectrum of cosine and sine components]]
 Given is the amplitude spectrum&nbsp; $X(f)$&nbsp; of a signal&nbsp; $x(t)$&nbsp; according to the graph.
@@ Line 57: / Line 57: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID293__Sig_A_2_3_a.png|right|frame|Summensignal aus Cosinus- und Sinusanteilen]]
+[[File:P_ID293__Sig_A_2_3_a.png|right|frame|Sum signal of cosine and sine components]]
 '''(1)'''&nbsp; The time signal has the following form:
@@ Line 74: / Line 74: @@
 From this follows&nbsp; $f_0 = 2{\,\rm kHz}$ &nbsp; &rArr; &nbsp;  period duration $T_0 = 1/f_0 \hspace{0.1cm}\underline{= 0.5 {\,\rm ms}}$.
 <br clear=all>
-[[File:P_ID294__Sig_A_2_3_d_neu.png|right|300px|frame|Spektrum mit diskreten Anteilen]]
+[[File:P_ID294__Sig_A_2_3_d_neu.png|right|300px|frame|Spectrum with discrete components]]
 '''(3)'''&nbsp; The following applies to the output signal $y(t)$ of the differentiatior:

← Older revision		Revision as of 12:38, 23 March 2021
Line 97:		Line 97:

	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:~~Exercises for~~ Signal Representation\|^2.3 Harmonic Oscillation^]]	+	[[Category:Signal Representation: Exercises\|^2.3 Harmonic Oscillation^]]

← Older revision	Revision as of 09:47, 4 January 2021
(No difference)

@@ Line 21: / Line 21: @@
 ''Hint:''
-*This exercise belongs to the chpater&nbsp; [[ Signal_Representation/Harmonic_Oscillation|Harmonic Oscillation]].
+*This exercise belongs to the chapter&nbsp; [[ Signal_Representation/Harmonic_Oscillation|Harmonic Oscillation]].
@@ Line 58: / Line 58: @@
 {{ML-Kopf}}
 [[File:P_ID293__Sig_A_2_3_a.png|right|frame|Summensignal aus Cosinus- und Sinusanteilen]]
-'''(1)'''&nbsp; Das Zeitsignal hat die folgende Form:
+'''(1)'''&nbsp; The time signal has the following form:
 :$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$
-*Hierbei bezeichnet&nbsp; $\omega_1 = 2\pi f_1$&nbsp; die Kreisfrequenz des Cosinusanteils.
+*Here&nbsp; $\omega_1 = 2\pi f_1$&nbsp; denotes the angular frequency of the cosine component.
-*Zum Zeitpunkt&nbsp; $t = 0$&nbsp; hat das Signal den Wert&nbsp; $x(t=0)\hspace{0.15 cm}\underline{=1\,\rm V}$.
+*At time&nbsp; $t = 0$&nbsp; the signal has the value&nbsp; $x(t=0)\hspace{0.15 cm}\underline{=1\,\rm V}$.
-'''(2)'''&nbsp; Die Grundfrequenz&nbsp; $f_0$&nbsp; ist der kleinste gemeinsame Teiler
+'''(2)'''&nbsp; The base frequency&nbsp; $f_0$&nbsp; is the least common divisor
-*von $f_1 = 4{\,\rm kHz}$
+*of $f_1 = 4{\,\rm kHz}$
-*und $2.5 · f_1 = 10{\,\rm kHz}$.
+*and $2.5 · f_1 = 10{\,\rm kHz}$.
-Daraus folgt&nbsp; $f_0 = 2{\,\rm kHz}$ &nbsp; &rArr; &nbsp;  Periodendauer $T_0 = 1/f_0 \hspace{0.1cm}\underline{= 0.5 {\,\rm ms}}$.
+From this follows&nbsp; $f_0 = 2{\,\rm kHz}$ &nbsp; &rArr; &nbsp;  period duration $T_0 = 1/f_0 \hspace{0.1cm}\underline{= 0.5 {\,\rm ms}}$.
 <br clear=all>
 [[File:P_ID294__Sig_A_2_3_d_neu.png|right|300px|frame|Spektrum mit diskreten Anteilen]]
-'''(3)'''&nbsp; Für das Ausgangssignal $y(t)$ des Differenzierers gilt:
+'''(3)'''&nbsp; The following applies to the output signal $y(t)$ of the differentiatior:
 :$$y(t)=\frac{1}{\omega_1}\cdot\frac{ {\rm d}x(t)}{{\rm d}t}=\frac{ {\rm -2V}}{\omega_1}\cdot\omega_1 \cdot (-\sin(\omega_1 t))+\frac{\rm 4V}{\omega_1}\cdot 2.5\omega_1\cdot {\rm cos}(2.5\omega_1t).$$
-*Dies führt zum Ergebnis:
+*This leads to the solution:
 :$$y(t)={\rm 2V}\cdot\sin(\omega_1 t)+{\rm 10V}\cdot\cos(2.5\omega_1 t).$$
-*Für&nbsp; $t = 0$&nbsp; ergibt sich der Wert&nbsp; $y(t=0)\hspace{0.15cm}\underline{=10\,\rm V}$.
+*For&nbsp; $t = 0$&nbsp; the value&nbsp; $y(t=0)\hspace{0.15cm}\underline{=10\,\rm V}$ follows.
-*Rechts ist das Spektrum&nbsp; $Y(f)$&nbsp; dargestellt.
+*The spectrum&nbsp; $Y(f)$&nbsp; is shown on the right.
-'''(4)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 4</u>:
+'''(4)'''&nbsp; The <u>solutions 1 and 4</u> are correct:
-*Die Periodendauer $T_0$ wird durch die Amplitude und die Phase der beiden Anteile nicht verändert.
+*The period duration $T_0$ is not changed by the amplitude and phase of the two components.
-*Das bedeutet, dass weiterhin&nbsp; $T_0 = 0.5 {\,\rm ms}$&nbsp;  gilt.
+*This means, that&nbsp; $T_0 = 0.5 {\,\rm ms}$&nbsp;  still applies.
-*Der Gleichanteil verschwindet aufgrund der Differentiation.
+*The DC component disappears due to the differentiation.
-*Der Anteil bei&nbsp; $f_1$&nbsp; ist sinusförmig. Somit hat&nbsp; $X(f)$&nbsp; einen (imaginären) Dirac bei&nbsp; $f = f_1$, jedoch mit negativem Vorzeichen.
+*The component&nbsp; $f_1$&nbsp; is sinusoidal. Thus&nbsp; $X(f)$&nbsp; has an (imaginary) Dirac at&nbsp; $f = f_1$, but with a negative sign.
-*Der Cosinusanteil mit der Amplitude&nbsp; ${10\,\rm V}$&nbsp; hat die beiden Diracfunktionen bei&nbsp; $\pm 2.5 \cdot f_1$&nbsp; zur Folge, jeweils mit dem Gewicht&nbsp; ${5\,\rm V}$ .
+*The cosine component with amplitude&nbsp; ${10\,\rm V}$&nbsp; results in the two Dirac functions at&nbsp; $\pm 2.5 \cdot f_1$&nbsp;, each with weight&nbsp; ${5\,\rm V}$ .
 {{ML-Fuß}}
 __NOEDITSECTION__
 [[Category:Exercises for Signal Representation|^2.3 Harmonic Oscillation^]]

@@ Line 45: / Line 45: @@
 $y(t=0)\ = \ $ { 10 3% } &nbsp; ${\rm V}$
-{Welche der folgenden Aussagen sind bezüglich des Signals&nbsp; $y(t)$&nbsp; bzw. seines Spektrums&nbsp; $Y(f)$&nbsp; zutreffend?
+{Which of the following statements are true regarding the signal&nbsp; $y(t)$&nbsp; or its spectrum&nbsp; $Y(f)$&nbsp;?
 |type="[]"}
-+ $y(t)$&nbsp; hat die gleiche Periodendauer wie das Signal&nbsp; $x(t)$.
++ $y(t)$&nbsp; has the same period duration as the signal&nbsp; $x(t)$.
-- $Y(f)$&nbsp; beinhaltet eine Diracfunktion bei der Frequenz&nbsp; $f = 0$.
+- $Y(f)$&nbsp; contains a Dirac function at the frequency&nbsp; $f = 0$.
-- $Y(f)$&nbsp; beinhaltet eine Diracfunktion bei&nbsp; $+f_1$&nbsp; mit dem Gewicht&nbsp; $\rm{j} · 1\,{\rm V}$.
+- $Y(f)$&nbsp; contains a Dirac function at&nbsp; $+f_1$&nbsp; with weight&nbsp; $\rm{j} · 1\,{\rm V}$.
-+ $Y(f)$&nbsp; beinhaltet eine Diracfunktion bei&nbsp; $–\hspace{-0.1cm}2.5 \cdot f_1$&nbsp; mit dem Gewicht&nbsp; $5\,{\rm V}$.
++ $Y(f)$&nbsp; contains a Dirac function at&nbsp; $–\hspace{-0.1cm}2.5 \cdot f_1$&nbsp; with weight&nbsp; $5\,{\rm V}$.
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 [[File:P_ID293__Sig_A_2_3_a.png|right|frame|Summensignal aus Cosinus- und Sinusanteilen]]

@@ Line 5: / Line 5: @@
 [[File: P_ID278_Sig_A_2_3neu.png|right|frame|Spektrum von Cosinus- und Sinusanteilen]]
-Gegeben ist das Amplitudenspektrum&nbsp; $X(f)$&nbsp; eines Signals&nbsp; $x(t)$&nbsp; entsprechend der Grafik.
+Given is the amplitude spectrum&nbsp; $X(f)$&nbsp; of a signal&nbsp; $x(t)$&nbsp; according to the graph.
-*Die Normierungsfrequenz sei&nbsp; $f_1 = 4\,\text{kHz}$.
+*Let&nbsp; $f_1 = 4\,\text{kHz}$ be the normalisation frequency.
-*Damit liegen die tatsächlichen Frequenzen der Signalanteile bei&nbsp; $0\,\text{kHz}$,&nbsp; $4\,\text{kHz}$&nbsp; und&nbsp; $10\,\text{kHz}$.
+*Thus the actual frequencies of the signal components are&nbsp; $0\,\text{kHz}$,&nbsp; $4\,\text{kHz}$&nbsp; and&nbsp; $10\,\text{kHz}$.
-Dieses Signal&nbsp; $x(t)$&nbsp; liegt am Eingang eines linearen Differenzierers, dessen Ausgang mit&nbsp; $\omega_1 = 2\pi f_1$&nbsp; wie folgt dargestellt werden kann:
+This signal&nbsp; $x(t)$&nbsp;  is at the input of a linear differentiator whose output can be represented with&nbsp; $\omega_1 = 2\pi f_1$&nbsp; as follows:
 :$$y(t)=\frac{1}{\omega_1}\cdot\frac{ {\rm d}  x(t)}{{\rm d}  t}.$$
@@ Line 20: / Line 20: @@
-''Hinweis:''
+''Hint:''
-*Die Aufgabe gehört zum Kapitel&nbsp; [[ Signal_Representation/Harmonic_Oscillation|Harmonische Schwingung]].
+*This exercise belongs to the chpater&nbsp; [[ Signal_Representation/Harmonic_Oscillation|Harmonic Oscillation]].
@@ Line 29: / Line 29: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Geben Sie&nbsp; $x(t)$&nbsp; analytisch an.&nbsp; Wie groß ist der Signalwert bei&nbsp; $t = 0$?
+{Give&nbsp; $x(t)$&nbsp; analytically.&nbsp; What is the signal value at&nbsp; $t = 0$?
 |type="{}"}
 $x(t=0)\ = \ $ { 1 3% } &nbsp; ${\rm V}$
-{Wie groß ist die Periodendauer des Signals&nbsp; $x(t)$?
+{What is the period duration of the signal&nbsp; $x(t)$?
 |type="{}"}
 $T_0\ = \ $ { 0.5 3% } &nbsp; ${\rm ms}$
-{Berechnen Sie das Ausgangssignal&nbsp; $y(t)$&nbsp; des Differenzierers.&nbsp; Wie groß ist der Signalwert zum Zeitpunkt&nbsp; $t = 0$?
+{Calculate the output signal&nbsp; $y(t)$&nbsp; of the differentiator&nbsp; What is the signal value at time&nbsp; $t = 0$?
 |type="{}"}
 $y(t=0)\ = \ $ { 10 3% } &nbsp; ${\rm V}$