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

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	[[Category:Signal Representation: Exercises\|^2.3 Harmonic Oscillation^]]		[[Category:Signal Representation: Exercises\|^2.3 Harmonic Oscillation^]]
	[[de:Aufgaben:2.~~3_cos~~-~~_und_sin~~-Anteil]]		[[de:Aufgaben:Aufgabe 2.3: cos- und sin-Anteil]]

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	[[Category:Signal Representation: Exercises\|^2.3 Harmonic Oscillation^]]		[[Category:Signal Representation: Exercises\|^2.3 Harmonic Oscillation^]]
			[[de:Aufgaben:2.3_cos-_und_sin-Anteil]]

Guenter at 16:33, 17 May 2021

2021-05-17T16:33:01Z

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	[[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  $X(f)$  of a signal  $x(t)$  according to the graph.		Given is the amplitude spectrum  $X(f)$  of a signal  $x(t)$  according to the graph.
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	:$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$		:$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$

	[[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  $\omega_1 = 2\pi f_1$  denotes the circular frequency of the cosine component.		*Here  $\omega_1 = 2\pi f_1$  denotes the circular frequency of the cosine component.

Guenter at 11:33, 14 April 2021

2021-04-14T11:33:44Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	~~[[File:P_ID293__Sig_A_2_3_a.png\|right\|frame\|Sum signal of cosine and sine components]]~~
	'''(1)'''  The time signal has the following form:		'''(1)'''  The time signal has the following form:

	:$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$		:$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$

			[[File:P_ID293__Sig_A_2_3_a.png\|right\|frame\|Sum signal of cosine and sine components]]

	*Here  $\omega_1 = 2\pi f_1$  denotes the circular frequency of the cosine component.		*Here  $\omega_1 = 2\pi f_1$  denotes the circular frequency of the cosine component.
	*At time  $t = 0$  the signal has the value  $x(t=0)\hspace{0.15 cm}\underline{=1\,\rm V}$.		*At time  $t = 0$  the signal has the value  $x(t=0)\hspace{0.15 cm}\underline{=1\,\rm V}$.



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	From this follows  $f_0 = 2{\,\rm kHz}$   ⇒   period duration $T_0 = 1/f_0 \hspace{0.1cm}\underline{= 0.5 {\,\rm ms}}$.		From this follows  $f_0 = 2{\,\rm kHz}$   ⇒   period duration $T_0 = 1/f_0 \hspace{0.1cm}\underline{= 0.5 {\,\rm ms}}$.
	<br clear=all>		<br clear=all>
	~~[[File:P_ID294__Sig_A_2_3_d_neu.png\|right\|300px\|frame\|Spectrum with discrete components]]~~
	'''(3)'''  The following applies to the output signal $y(t)$ of the differentiatior:		'''(3)'''  The following applies to the output signal $y(t)$ of the differentiatior:

	:$$y(t)=\frac{1}{\omega_1}\cdot\frac{ {\rm d}x(t)}{{\rm d}t}=\frac{ {\rm -2V}}{\omega_1}\cdot\omega_1 \cdot (-\sin(\omega_1 t))+\frac{\rm 4V}{\omega_1}\cdot 2.5\omega_1\cdot {\rm cos}(2.5\omega_1t).$$		:$$y(t)=\frac{1}{\omega_1}\cdot\frac{ {\rm d}x(t)}{{\rm d}t}=\frac{ {\rm -2V}}{\omega_1}\cdot\omega_1 \cdot (-\sin(\omega_1 t))+\frac{\rm 4V}{\omega_1}\cdot 2.5\omega_1\cdot {\rm cos}(2.5\omega_1t).$$

			[[File:P_ID294__Sig_A_2_3_d_neu.png\|right\|300px\|frame\|Spectrum with discrete components]]

	*This leads to the solution:		*This leads to the solution:

Guenter at 14:18, 13 April 2021

2021-04-13T14:18:23Z

← Older revision		Revision as of 16:18, 13 April 2021
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	:$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$		:$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$

	*Here  $\omega_1 = 2\pi f_1$  denotes the ~~angular~~ frequency of the cosine component.		*Here  $\omega_1 = 2\pi f_1$  denotes the circular frequency of the cosine component.
	*At time  $t = 0$  the signal has the value  $x(t=0)\hspace{0.15 cm}\underline{=1\,\rm V}$.		*At time  $t = 0$  the signal has the value  $x(t=0)\hspace{0.15 cm}\underline{=1\,\rm V}$.



	'''(2)'''  The ~~base~~ frequency  $f_0$  is the ~~least~~ common divisor		'''(2)'''  The basic frequency  $f_0$  is the greatest common divisor
	*of $f_1 = 4{\,\rm kHz}$		*of $f_1 = 4{\,\rm kHz}$
	*and $2.5 · f_1 = 10{\,\rm kHz}$.		*and $2.5 · f_1 = 10{\,\rm kHz}$.

Guenter at 14:09, 13 April 2021

2021-04-13T14:09:07Z

← Older revision		Revision as of 16:09, 13 April 2021
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	}}		}}

	[[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  $X(f)$  of a signal  $x(t)$  according to the graph.		Given is the amplitude spectrum  $X(f)$  of a signal  $x(t)$  according to the graph.
	*Let  $f_1 = 4\,\text{kHz}$ be the normalisation frequency.		*Let  $f_1 = 4\,\text{kHz}$ be the normalisation frequency.
	*Thus the ~~actual~~ frequencies of the signal components are  $0\,\text{kHz}$,  $4\,\text{kHz}$  and  $10\,\text{kHz}$.		*Thus the frequencies of the signal components are  $0\,\text{kHz}$,  $4\,\text{kHz}$  and  $10\,\text{kHz}$.


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	$T_0\ = \ $ { 0.5 3% }   ${\rm ms}$		$T_0\ = \ $ { 0.5 3% }   ${\rm ms}$

	{Calculate the output signal  $y(t)$  of the differentiator  What is the signal value at time  $t = 0$?		{Calculate the output signal  $y(t)$  of the differentiator.  What is the signal value at time  $t = 0$?
	\|type="{}"}		\|type="{}"}
	$y(t=0)\ = \ $ { 10 3% }   ${\rm V}$		$y(t=0)\ = \ $ { 10 3% }   ${\rm V}$

Noah at 09:22, 26 March 2021

2021-03-26T09:22:43Z

← Older revision		Revision as of 11:22, 26 March 2021
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	}}		}}

	[[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  $X(f)$  of a signal  $x(t)$  according to the graph.		Given is the amplitude spectrum  $X(f)$  of a signal  $x(t)$  according to the graph.
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{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)'''  The time signal has the following form:		'''(1)'''  The time signal has the following form:

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	From this follows  $f_0 = 2{\,\rm kHz}$   ⇒   period duration $T_0 = 1/f_0 \hspace{0.1cm}\underline{= 0.5 {\,\rm ms}}$.		From this follows  $f_0 = 2{\,\rm kHz}$   ⇒   period duration $T_0 = 1/f_0 \hspace{0.1cm}\underline{= 0.5 {\,\rm ms}}$.
	<br clear=all>		<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)'''  The following applies to the output signal $y(t)$ of the differentiatior:		'''(3)'''  The following applies to the output signal $y(t)$ of the differentiatior:

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"

← Older revision		Revision as of 14:38, 23 March 2021
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	[[Category:~~Exercises for~~ Signal Representation\|^2.3 Harmonic Oscillation^]]		[[Category:Signal Representation: Exercises\|^2.3 Harmonic Oscillation^]]

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 Aufgaben:Exercises 2.3: Cosine and Sine Components to Aufgaben:Exercise 2.3: Cosine and Sine Components

← Older revision	Revision as of 11:47, 4 January 2021
(No difference)

Noah at 03:00, 4 January 2021

2021-01-04T03:00:17Z

← Older revision		Revision as of 05:00, 4 January 2021
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	''Hint:''		''Hint:''
	*This exercise belongs to the ~~chpater~~  [[ Signal_Representation/Harmonic_Oscillation\|Harmonic Oscillation]].		*This exercise belongs to the chapter  [[ Signal_Representation/Harmonic_Oscillation\|Harmonic Oscillation]].


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	{{ML-Kopf}}		{{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\|Summensignal aus Cosinus- und Sinusanteilen]]
	'''(1)'''  ~~Das Zeitsignal hat die folgende Form~~:		'''(1)'''  The time signal has the following form:

	:$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$		:$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$

	*~~Hierbei bezeichnet~~  $\omega_1 = 2\pi f_1$  ~~die Kreisfrequenz des Cosinusanteils~~.		*Here  $\omega_1 = 2\pi f_1$  denotes the angular frequency of the cosine component.
	*~~Zum Zeitpunkt~~  $t = 0$  ~~hat das Signal den Wert~~  $x(t=0)\hspace{0.15 cm}\underline{=1\,\rm V}$.		*At time  $t = 0$  the signal has the value  $x(t=0)\hspace{0.15 cm}\underline{=1\,\rm V}$.



	'''(2)'''  ~~Die Grundfrequenz~~  $f_0$  ~~ist der kleinste gemeinsame Teiler~~		'''(2)'''  The base frequency  $f_0$  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~~  $f_0 = 2{\,\rm kHz}$   ⇒   ~~Periodendauer~~ $T_0 = 1/f_0 \hspace{0.1cm}\underline{= 0.5 {\,\rm ms}}$.		From this follows  $f_0 = 2{\,\rm kHz}$   ⇒   period duration $T_0 = 1/f_0 \hspace{0.1cm}\underline{= 0.5 {\,\rm ms}}$.
	<br clear=all>		<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\|Spektrum mit diskreten Anteilen]]
	'''(3)'''  ~~Für das Ausgangssignal~~ $y(t)$ ~~des Differenzierers gilt~~:		'''(3)'''  The following applies to the output signal $y(t)$ of the differentiatior:

	:$$y(t)=\frac{1}{\omega_1}\cdot\frac{ {\rm d}x(t)}{{\rm d}t}=\frac{ {\rm -2V}}{\omega_1}\cdot\omega_1 \cdot (-\sin(\omega_1 t))+\frac{\rm 4V}{\omega_1}\cdot 2.5\omega_1\cdot {\rm cos}(2.5\omega_1t).$$		:$$y(t)=\frac{1}{\omega_1}\cdot\frac{ {\rm d}x(t)}{{\rm d}t}=\frac{ {\rm -2V}}{\omega_1}\cdot\omega_1 \cdot (-\sin(\omega_1 t))+\frac{\rm 4V}{\omega_1}\cdot 2.5\omega_1\cdot {\rm cos}(2.5\omega_1t).$$

	*~~Dies führt zum Ergebnis~~:		*This leads to the solution:

	:$$y(t)={\rm 2V}\cdot\sin(\omega_1 t)+{\rm 10V}\cdot\cos(2.5\omega_1 t).$$		:$$y(t)={\rm 2V}\cdot\sin(\omega_1 t)+{\rm 10V}\cdot\cos(2.5\omega_1 t).$$

	*~~Für~~  $t = 0$  ~~ergibt sich der Wert~~  $y(t=0)\hspace{0.15cm}\underline{=10\,\rm V}$.		*For  $t = 0$  the value  $y(t=0)\hspace{0.15cm}\underline{=10\,\rm V}$ follows.
	*~~Rechts ist das Spektrum~~  $Y(f)$  ~~dargestellt~~.		*The spectrum  $Y(f)$  is shown on the right.



	'''(4)'''  ~~Richtig sind die~~ <u>~~Lösungsvorschläge~~ 1 ~~und~~ 4</u>:		'''(4)'''  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~~  $T_0 = 0.5 {\,\rm ms}$  ~~gilt~~.		*This means, that  $T_0 = 0.5 {\,\rm ms}$  still applies.
	*~~Der Gleichanteil verschwindet aufgrund der Differentiation~~.		*The DC component disappears due to the differentiation.
	*~~Der Anteil bei~~  $f_1$  ~~ist sinusförmig~~. ~~Somit hat~~  $X(f)$  ~~einen~~ (~~imaginären~~) Dirac ~~bei~~  $f = f_1$, ~~jedoch mit negativem Vorzeichen~~.		*The component  $f_1$  is sinusoidal. Thus  $X(f)$  has an (imaginary) Dirac at  $f = f_1$, but with a negative sign.
	*~~Der Cosinusanteil mit der Amplitude~~  ${10\,\rm V}$  ~~hat die beiden Diracfunktionen bei~~  $\pm 2.5 \cdot f_1$  ~~zur Folge~~, ~~jeweils mit dem Gewicht~~  ${5\,\rm V}$ .		*The cosine component with amplitude  ${10\,\rm V}$  results in the two Dirac functions at  $\pm 2.5 \cdot f_1$ , each with weight  ${5\,\rm V}$ .
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Exercises for Signal Representation\|^2.3 Harmonic Oscillation^]]		[[Category:Exercises for Signal Representation\|^2.3 Harmonic Oscillation^]]