Aufgaben:Exercise 2.3Z: Oscillation Parameters: Difference between revisions
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[[File:P_ID316__Sig_Z_2_3.png|right| | [[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 | |||
:$$x(t)=C\cdot\cos\bigg(2\pi \cdot \frac{t-\tau}{T_0}\bigg)$$ | :$$x(t)=C\cdot\cos\bigg(2\pi \cdot \frac{t-\tau}{T_0}\bigg).$$ | ||
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 $\varphi$: | |||
:$$x(t)=C \cdot\cos(2\pi f_0t-\varphi).$$ | :$$x(t)=C \cdot\cos(2\pi f_0t-\varphi).$$ | ||
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:'' | ||
* | *This exercise belongs to the chapter [[ Signal_Representation/Harmonic_Oscillation|Harmonic Oscillation]]. | ||
=== | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
What is the period duration $T_0$ and the base frequency $f_0$? | |||
|type="{}"} | |type="{}"} | ||
$T_0$ | $T_0\hspace{0.2cm} = \ $ { 12 3% } $\text{ms}$ | ||
$f_0$ | $f_0\hspace{0.2cm} = \ $ { 83.33 3% } $\text{Hz}$ | ||
{ | {What is the value of the shift $\tau$ and the phase $\varphi$ $($in $\text{degrees})$ ? | ||
|type="{}"} | |type="{}"} | ||
$\tau$ | $\tau\hspace{0.25cm} = \ $ { 2 3% } $\text{ms}$ | ||
$\varphi$ | $\varphi\hspace{0.2cm} = \ $ { 60 3% } $\text{deg}$ | ||
{ | {What is the amplitude of the harmonic oscillation? | ||
|type="{}"} | |type="{}"} | ||
${C}$ | ${C}\ = \ $ { 6 3% } $\text{V}$ | ||
{ | {What is the spectrum $X(f)$? What is the weight of the spectral line at $+f_0$ ? | ||
|type="{}"} | |type="{}"} | ||
$\text{Re}[X(f = f_0)]$ | $\text{Re}\big[X(f = f_0)\big]\ = \ $ { 1.5 3% } $\text{V}$ | ||
$\text{Im}[X(f = f_0)]$ | $\text{Im}\big[X(f = f_0)\big] \ = \ $ { -2.65--2.55 } $\text{V}$ | ||
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</quiz> | </quiz> | ||
=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''1 | '''(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 ${C}/2 \cdot {\rm e}^{–\text{j}\varphi} = 3 \,\text{V} \cdot \cos(60^\circ)- 3 \,\text{V} \cdot \sin(60^\circ)\hspace{0.05cm}\approx \underline{1.5 \,\text{V} - \text{j} \cdot 2.6 \,\text{V}}$. | |||
{{ML-Fuß}} | {{ML-Fuß}} | ||
[[Category: | [[Category:Signal Representation: Exercises|^2.3 Harmonic Oscillation^]] | ||
[[de:Aufgaben:Aufgabe 2.3Z: Schwingungsparameter]] | |||
Latest revision as of 17:53, 16 March 2026

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:
- This exercise belongs to the chapter Harmonic Oscillation.
Questions
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 ${C}/2 \cdot {\rm e}^{–\text{j}\varphi} = 3 \,\text{V} \cdot \cos(60^\circ)- 3 \,\text{V} \cdot \sin(60^\circ)\hspace{0.05cm}\approx \underline{1.5 \,\text{V} - \text{j} \cdot 2.6 \,\text{V}}$.