Aufgaben:Exercise 2.3Z: Oscillation Parameters - 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.~~3Z_Schwingungsparameter~~]]		[[de:Aufgaben:Aufgabe 2.3Z: Schwingungsparameter]]

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	'''(3)'''  From the value at timet  $t = 0$  it follows for the amplitude  ${C}$:		'''(3)'''  From the value at timet  $t = 0$  it follows for the amplitude  ${C}$:
	:$$x_0=x(t=0)=C\cdot\cos(-60\,^\circ)={C}/{2}=\rm 3\,V		:$$x_0=x(t=0)=C\cdot\cos(-60\,^\circ)={C}/{2}=\rm 3\,V\hspace{0.3 cm} \Rightarrow \hspace{0.3 cm}\hspace{0.15cm}\underline{\it C=\rm 6\,V}.$$
	\hspace{0.3 cm} \Rightarrow \hspace{0.3 cm}\hspace{0.15cm}\underline{\it C=\rm 6\,V}.$$

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	[[Category:Signal Representation: Exercises\|^2.3 Harmonic Oscillation^]]		[[Category:Signal Representation: Exercises\|^2.3 Harmonic Oscillation^]]
			[[de:Aufgaben:2.3Z_Schwingungsparameter]]

Hwang at 13:14, 18 January 2023

2023-01-18T13:14:53Z

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	:$$X(f)={C}/{2}\cdot{\rm e}^{-{\rm j}\varphi}\cdot\delta(f-f_0)+{C}/{2}\cdot{\rm e}^{{\rm j}\varphi}\cdot\delta(f+f_0).$$		:$$X(f)={C}/{2}\cdot{\rm e}^{-{\rm j}\varphi}\cdot\delta(f-f_0)+{C}/{2}\cdot{\rm e}^{{\rm j}\varphi}\cdot\delta(f+f_0).$$

	*The weight of the Dirac line at   $f = f_0$  (first term) is   ${C}/2 \cdot {\rm e}^{–\text{j}\varphi} = 3 \,\text{V} \cdot \cos(60^\circ)- 3 \,\text{V} \cdot \sin(60^\circ)\hspace{0.05cm}\approx \underline{1.5 \,\text{V} - \text{j} \cdot 2.6 \,\text{V}}$.		*The weight of the Dirac delta line at   $f = f_0$  (first term) is   ${C}/2 \cdot {\rm e}^{–\text{j}\varphi} = 3 \,\text{V} \cdot \cos(60^\circ)- 3 \,\text{V} \cdot \sin(60^\circ)\hspace{0.05cm}\approx \underline{1.5 \,\text{V} - \text{j} \cdot 2.6 \,\text{V}}$.
	{{ML-Fuß}}		{{ML-Fuß}}


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

Guenter at 14:31, 13 April 2021

2021-04-13T14:31:27Z

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	\|type="{}"}		\|type="{}"}
	$\tau\hspace{0.25cm} = \ $ { 2 3% }  $\text{ms}$		$\tau\hspace{0.25cm} = \ $ { 2 3% }  $\text{ms}$
	$\varphi\hspace{0.2cm} = \ $ { 60 3% }  $\text{~~degree~~}$		$\varphi\hspace{0.2cm} = \ $ { 60 3% }  $\text{deg}$

Guenter at 14:27, 13 April 2021

2021-04-13T14:27:21Z

Noah at 09:23, 26 March 2021

2021-03-26T09:23:02Z

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	}}		}}

	[[File:P_ID316__Sig_Z_2_3.png\|right\|frame\|Definition ~~von~~  $x_0$,  $t_1$  und  $t_2$]]		[[File:P_ID316__Sig_Z_2_3.png\|right\|frame\|Definition of  $x_0$,  $t_1$  und  $t_2$]]
	Every harmonic oscillation can also be written in the form		Every harmonic oscillation can also be written in the form
	:$$x(t)=C\cdot\cos\bigg(2\pi \cdot \frac{t-\tau}{T_0}\bigg).$$		:$$x(t)=C\cdot\cos\bigg(2\pi \cdot \frac{t-\tau}{T_0}\bigg).$$

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

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

Noah at 21:23, 13 January 2021

2021-01-13T21:23:45Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Es gilt~~  $T_0 = t_2 - t_1 = 12\, \text{ms}$  ~~und~~  $f_0 = 1/T_0 \hspace{0.15cm} \underline{\approx 83.33\, \text{Hz}}$.		'''(1)'''  It is  $T_0 = t_2 - t_1 = 12\, \text{ms}$  and  $f_0 = 1/T_0 \hspace{0.15cm} \underline{\approx 83.33\, \text{Hz}}$.



	'''(2)'''  ~~Die Verschiebung beträgt~~  $\tau \hspace{0.1cm} \underline{= 2\, \text{ms}}$  ~~und die Phase ist~~  $\varphi = 2\pi \cdot \tau/T_0 = \pi/3$  ~~entsprechend~~  $\varphi =\hspace{0.15cm} \underline{60^{\circ}}$.		'''(2)'''  The shift is  $\tau \hspace{0.1cm} \underline{= 2\, \text{ms}}$  and the phase is  $\varphi = 2\pi \cdot \tau/T_0 = \pi/3$  corresponding to  $\varphi =\hspace{0.15cm} \underline{60^{\circ}}$.



	'''(3)'''  ~~Aus dem Wert zum Zeitpunkt~~  $t = 0$  ~~folgt für die Amplitude~~  ${C}$:		'''(3)'''  From the value at timet  $t = 0$  it follows for the amplitude  ${C}$:
	:$$x_0=x(t=0)=C\cdot\cos(-60\,^\circ)={C}/{2}=\rm 3\,V		:$$x_0=x(t=0)=C\cdot\cos(-60\,^\circ)={C}/{2}=\rm 3\,V
	\hspace{0.3 cm} \Rightarrow \hspace{0.3 cm}\hspace{0.15cm}\underline{\it C=\rm 6\,V}.$$		\hspace{0.3 cm} \Rightarrow \hspace{0.3 cm}\hspace{0.15cm}\underline{\it C=\rm 6\,V}.$$


	'''(4)'''  ~~Die dazugehörige Spektralfunktion lautet~~:		'''(4)'''  The corresponding spectral function is:
	:$$X(f)={C}/{2}\cdot{\rm e}^{-{\rm j}\varphi}\cdot\delta(f-f_0)+{C}/{2}\cdot{\rm e}^{{\rm j}\varphi}\cdot\delta(f+f_0).$$		:$$X(f)={C}/{2}\cdot{\rm e}^{-{\rm j}\varphi}\cdot\delta(f-f_0)+{C}/{2}\cdot{\rm e}^{{\rm j}\varphi}\cdot\delta(f+f_0).$$

	*~~Das Gewicht der Diraclinie bei~~  $f = f_0$  (~~erster Term~~) ~~ist~~   ${C}/2 \cdot {\rm e}^{–\text{j}\varphi} = 3 \,\text{V} \cdot \cos(60^\circ)- 3 \,\text{V} \cdot \sin(60^\circ)\hspace{0.05cm}\approx \underline{1.5 \,\text{V} - \text{j} \cdot 2.6 \,\text{V}}$.		*The weight of the Dirac line at   $f = f_0$  (first term) is   ${C}/2 \cdot {\rm e}^{–\text{j}\varphi} = 3 \,\text{V} \cdot \cos(60^\circ)- 3 \,\text{V} \cdot \sin(60^\circ)\hspace{0.05cm}\approx \underline{1.5 \,\text{V} - \text{j} \cdot 2.6 \,\text{V}}$.
	{{ML-Fuß}}		{{ML-Fuß}}


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

Noah at 21:00, 13 January 2021

2021-01-13T21:00:27Z

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	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Es gilt  $T_0 = t_2 - t_1 = 12\, \text{ms}$  und  $f_0 = 1/T_0 \hspace{0.15cm} \underline{\approx 83.33\, \text{Hz}}$.		'''(1)'''  Es gilt  $T_0 = t_2 - t_1 = 12\, \text{ms}$  und  $f_0 = 1/T_0 \hspace{0.15cm} \underline{\approx 83.33\, \text{Hz}}$.

← Older revision		Revision as of 16:27, 13 April 2021
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	}}		}}

	[[File:P_ID316__Sig_Z_2_3.png\|right\|frame\|~~Definition~~ of  $x_0$,  $t_1$  und  $t_2$]]		[[File:P_ID316__Sig_Z_2_3.png\|right\|frame\|Definitions of  $x_0$,  $t_1$  und  $t_2$]]
	Every harmonic oscillation can also be written in the form		Every harmonic oscillation can also be written in the form
	:$$x(t)=C\cdot\cos\bigg(2\pi \cdot \frac{t-\tau}{T_0}\bigg).$$		:$$x(t)=C\cdot\cos\bigg(2\pi \cdot \frac{t-\tau}{T_0}\bigg).$$
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	:* the amplitude  $C$,		:* the amplitude  $C$,

	:* the period duration   $T_0$,		:* the period duration  $T_0$,

	:* the shift  $\tau$  with respect to a cosine signal.		:* the shift  $\tau$  with respect to a cosine signal.


	A second form of representation is with the ~~base~~ frequency  $f_0$  and the phase  $\varphi$:		A second form of representation is with the basic frequency  $f_0$  and the phase  $\varphi$:
	:$$x(t)=C \cdot\cos(2\pi f_0t-\varphi).$$		:$$x(t)=C \cdot\cos(2\pi f_0t-\varphi).$$
	From a harmonic oscillation it is now known that		From a harmonic oscillation it is now known that
	:* the first signal maximum occurs at  $t_1 = 2 \,\text{ms}$~~  auftritt~~,		:* the first signal maximum occurs at  $t_1 = 2 \,\text{ms}$,

	:* the second signal maximum occurs at  $t_2 = 14 \,\text{ms}$~~  auftritt~~,		:* the second signal maximum occurs at  $t_2 = 14 \,\text{ms}$,

	:* the value  $x_0 ={x(t = 0)} = 3 \,\text{V}$~~ ~~.		:* the value  $x_0 ={x(t = 0)} = 3 \,\text{V}$.


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	\|type="{}"}		\|type="{}"}
	$\tau\hspace{0.25cm} = \ $ { 2 3% }  $\text{ms}$		$\tau\hspace{0.25cm} = \ $ { 2 3% }  $\text{ms}$
	$\varphi\hspace{0.2cm} = \ $ { 60 3% }  $\text{~~Grad~~}$		$\varphi\hspace{0.2cm} = \ $ { 60 3% }  $\text{degree}$


	{What is the amplitude of the harmonic oscillation??		{What is the amplitude of the harmonic oscillation?
	\|type="{}"}		\|type="{}"}
	${C}\ = \ $ { 6 3% }  $\text{V}$		${C}\ = \ $ { 6 3% }  $\text{V}$