Difference between revisions of "Aufgaben:Exercise 2.3Z: Oscillation Parameters"

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[[File:P_ID316__Sig_Z_2_3.png|right|frame|Definition von  $x_0$,  $t_1$  und  $t_2$]]
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[[File:P_ID316__Sig_Z_2_3.png|right|frame|Definitions of  $x_0$,  $t_1$  und  $t_2$]]
Jede harmonische Schwingung kann auch in der Form
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Every harmonic oscillation can also be written in the form
:$$x(t)=C\cdot\cos\bigg(2\pi \cdot \frac{t-\tau}{T_0}\bigg)$$
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:$$x(t)=C\cdot\cos\bigg(2\pi \cdot \frac{t-\tau}{T_0}\bigg).$$
geschrieben werden. Die Schwingung ist somit durch drei Parameter vollständig bestimmt:
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The oscillation is thus completely determined by three parameters:
  
:* die Amplitude  $C$,
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:* the amplitude  $C$,
  
:* die Periodendauer  $T_0$,
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:* the period duration  $T_0$,
  
:* die Verschiebung  $\tau$  gegenüber einem Cosinussignal.
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:* the shift  $\tau$  with respect to a cosine signal.
  
  
Eine zweite Darstellungsform lautet mit der Grundfrequenz  $f_0$  und der Phase  $\varphi$:
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A second form of representation is with the basic frequency  $f_0$  and the phase  $\varphi$:
 
:$$x(t)=C \cdot\cos(2\pi f_0t-\varphi).$$
 
:$$x(t)=C \cdot\cos(2\pi f_0t-\varphi).$$
Von einer harmonischen Schwingung ist nun bekannt, dass
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From a harmonic oscillation it is now known that
 +
:* the first signal maximum occurs at  $t_1 = 2 \,\text{ms}$,
  
:* das erste Signalmaximum bei  $t_1 = 2 \,\text{ms}$  auftritt,
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:* the second signal maximum occurs at  $t_2 = 14 \,\text{ms}$,
  
:* das zweite Signalmaximum bei  $t_2 = 14 \,\text{ms}$  auftritt,
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:* the value  $x_0 ={x(t = 0)} = 3 \,\text{V}$.
  
:* der Wert  $x_0 ={x(t = 0)} = 3 \,\text{V}$  ist.
 
  
  
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''Hint:''  
''Hinweis:''  
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*This exercise belongs to the chapter  [[ Signal_Representation/Harmonic_Oscillation|Harmonic Oscillation]].
*Die Aufgabe gehört zum Kapitel  [[Signaldarstellung/Harmonische_Schwingung|Harmonische Schwingung]].
 
 
   
 
   
  
  
  
===Fragebogen===
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===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Wie groß ist die Periodendauer&nbsp; $T_0$&nbsp; und die Grundfrequenz&nbsp; $f_0$?
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What is the period duration&nbsp; $T_0$&nbsp; and the base frequency&nbsp; $f_0$?
 
|type="{}"}
 
|type="{}"}
 
$T_0\hspace{0.2cm} = \ $  { 12 3% } &nbsp;$\text{ms}$
 
$T_0\hspace{0.2cm} = \ $  { 12 3% } &nbsp;$\text{ms}$
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{Welchen Wert haben hier die Verschiebung&nbsp; $\tau$&nbsp; und die Phase&nbsp; $\varphi$&nbsp; $($in&nbsp; $\text{Grad})$&nbsp;?
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{What is the value of the shift&nbsp; $\tau$&nbsp; and the phase&nbsp; $\varphi$&nbsp; $($in&nbsp; $\text{degrees})$&nbsp;?
 
|type="{}"}
 
|type="{}"}
 
$\tau\hspace{0.25cm} = \ $  { 2 3% } &nbsp;$\text{ms}$
 
$\tau\hspace{0.25cm} = \ $  { 2 3% } &nbsp;$\text{ms}$
$\varphi\hspace{0.2cm} = \ $  { 60 3% } &nbsp;$\text{Grad}$
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$\varphi\hspace{0.2cm} = \ $  { 60 3% } &nbsp;$\text{deg}$
  
  
{Wie groß ist die Amplitude der harmonischen Schwingung?
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{What is the amplitude of the harmonic oscillation?
 
|type="{}"}
 
|type="{}"}
 
${C}\ = \ $  { 6 3% } &nbsp;$\text{V}$
 
${C}\ = \ $  { 6 3% } &nbsp;$\text{V}$
  
  
{Wie lautet das Spektrum&nbsp; $X(f)$?&nbsp; Welches Gewicht hat die Spektrallinie bei&nbsp; $+f_0$&nbsp;?
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{What is the spectrum&nbsp; $X(f)$?&nbsp; What is the weight of the spectral line at&nbsp; $+f_0$&nbsp;?
 
|type="{}"}
 
|type="{}"}
 
$\text{Re}\big[X(f = f_0)\big]\ = \ $  { 1.5 3% } &nbsp;$\text{V}$
 
$\text{Re}\big[X(f = f_0)\big]\ = \ $  { 1.5 3% } &nbsp;$\text{V}$
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</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp;  Es gilt $T_0 = t_2 - t_1 = 12\, \text{ms}$ und $f_0 = 1/T_0 \hspace{0.15cm} \underline{\approx 83.33\, \text{Hz}}$.
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'''(1)'''&nbsp;  It is&nbsp; $T_0 = t_2 - t_1 = 12\, \text{ms}$&nbsp; and&nbsp; $f_0 = 1/T_0 \hspace{0.15cm} \underline{\approx 83.33\, \text{Hz}}$.
 +
 
 +
 
  
 +
'''(2)'''&nbsp; The shift is&nbsp; $\tau \hspace{0.1cm} \underline{= 2\, \text{ms}}$&nbsp; and the phase is&nbsp; $\varphi = 2\pi \cdot \tau/T_0 = \pi/3$&nbsp; corresponding to&nbsp; $\varphi =\hspace{0.15cm} \underline{60^{\circ}}$.
  
'''(2)'''&nbsp; 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 $\varphi =\hspace{0.15cm} \underline{60^{\circ}}$.
 
  
  
'''(3)'''&nbsp; Aus dem Wert zum Zeitpunkt $t = 0$ folgt für die Amplitude ${C}$:
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'''(3)'''&nbsp; From the value at timet&nbsp; $t = 0$&nbsp; it follows for the amplitude&nbsp; ${C}$:
 
:$$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)'''&nbsp; Die dazugehörige Spektralfunktion lautet:
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 +
'''(4)'''&nbsp; 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 &nbsp; ${C}/2 \cdot {\rm e}^{–\text{j}\varphi} = 3 \,\text{V} \cdot \cos(60^\circ)- 3 \,\text{V} \cdot \sin(60^\circ)\hspace{0.05cm}\approx \underline{1.5 \,\text{V} - \text{j} \cdot 2.6 \,\text{V}}$.
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*The weight of the Dirac delta line at &nbsp; $f = f_0$&nbsp; (first term) is &nbsp; ${C}/2 \cdot {\rm e}^{–\text{j}\varphi} = 3 \,\text{V} \cdot \cos(60^\circ)- 3 \,\text{V} \cdot \sin(60^\circ)\hspace{0.05cm}\approx \underline{1.5 \,\text{V} - \text{j} \cdot 2.6 \,\text{V}}$.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
[[Category:Aufgaben zu Signaldarstellung|^2. Periodische Signale^]]
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[[Category:Signal Representation: Exercises|^2.3 Harmonic Oscillation^]]

Latest revision as of 15:14, 18 January 2023

Definitions of  $x_0$,  $t_1$  und  $t_2$

Every harmonic oscillation can also be written in the form

$$x(t)=C\cdot\cos\bigg(2\pi \cdot \frac{t-\tau}{T_0}\bigg).$$

The oscillation is thus completely determined by three parameters:

  • the amplitude  $C$,
  • the period duration  $T_0$,
  • the shift  $\tau$  with respect to a cosine signal.


A second form of representation is with the basic frequency  $f_0$  and the phase  $\varphi$:

$$x(t)=C \cdot\cos(2\pi f_0t-\varphi).$$

From a harmonic oscillation it is now known that

  • the first signal maximum occurs at  $t_1 = 2 \,\text{ms}$,
  • the second signal maximum occurs at  $t_2 = 14 \,\text{ms}$,
  • the value  $x_0 ={x(t = 0)} = 3 \,\text{V}$.




Hint:



Questions

1

What is the period duration  $T_0$  and the base frequency  $f_0$?

$T_0\hspace{0.2cm} = \ $

 $\text{ms}$
$f_0\hspace{0.2cm} = \ $

 $\text{Hz}$

2

What is the value of the shift  $\tau$  and the phase  $\varphi$  $($in  $\text{degrees})$ ?

$\tau\hspace{0.25cm} = \ $

 $\text{ms}$
$\varphi\hspace{0.2cm} = \ $

 $\text{deg}$

3

What is the amplitude of the harmonic oscillation?

${C}\ = \ $

 $\text{V}$

4

What is the spectrum  $X(f)$?  What is the weight of the spectral line at  $+f_0$ ?

$\text{Re}\big[X(f = f_0)\big]\ = \ $

 $\text{V}$
$\text{Im}\big[X(f = f_0)\big] \ = \ $

 $\text{V}$


Solution

(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)  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)  From the value at timet  $t = 0$  it follows for the amplitude  ${C}$:

$$x_0=x(t=0)=C\cdot\cos(-60\,^\circ)={C}/{2}=\rm 3\,V \hspace{0.3 cm} \Rightarrow \hspace{0.3 cm}\hspace{0.15cm}\underline{\it C=\rm 6\,V}.$$


(4)  The corresponding spectral function is:

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