Aufgaben:Exercise 2.2Z: Non-Linearities: Difference between revisions

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[[File:P_ID322__Sig_Z_2_2.png|right|frame|Gleichanteil nach Nichtlinearitäten]]
[[File:P_ID322__Sig_Z_2_2.png|right|frame|Gleichanteil nach Nichtlinearitäten]]
Wir gehen von dem dreieckförmigen Signal  ${x(t)}$  gemäß der oberen Abbildung aus.  
We start from the triangular signal  ${x(t)}$  according to the figure above.  


Gibt man dieses Signal auf einen Amplitudenbegrenzer, so entsteht das Signal
If we apply this signal to an amplitude limiter, we get the signal
:$$y(t)=\left\{ {x(t)\atop \rm 1V}{\hspace{0.5cm} {\rm f\ddot{u}r}\quad x(t)\le \rm 1V \atop {\rm sonst}}\right..$$
:$$y(t)=\left\{ {x(t)\atop \rm 1V}{\hspace{0.5cm} {\rm f\ddot{u}r}\quad x(t)\le \rm 1V \atop {\rm sonst}}\right..$$
Eine zweite Nichtlinearität liefert das Signal
A second non-linearity provides the signal
:$$z(t)=x^2(t).$$
:$$z(t)=x^2(t).$$
Die Gleichsignalanteile werden nachfolgend mit  $x_0$,  $y_0$  bzw.  $z_0$  bezeichnet.  
The DC signal components are designated  $x_0$,  $y_0$  and  $z_0$  in the following.  




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''Hinweis:''  
''Hint:''  
*Die Aufgabe gehört zum Kapitel  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal|Gleichsignal - Grenzfall eines periodischen Signals]].
*This exercise belongs to the chapter  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal|Direct Current Signal - Limit Case of a Periodic Signal.]].
   
   


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===Fragebogen===
===Questions===


<quiz display=simple>
<quiz display=simple>
{Ermitteln Sie den Gleichsignalanteil&nbsp; $x_0$&nbsp; des Signals&nbsp; ${x(t)}$.
{Determine the DC signal component&nbsp; $x_0$&nbsp; of the signal&nbsp; ${x(t)}$.
|type="{}"}
|type="{}"}
$x_0\ = \ $  { 1 3% } &nbsp; $\text{V}$
$x_0\ = \ $  { 1 3% } &nbsp; $\text{V}$




{Ermitteln Sie den Gleichsignalanteil&nbsp; $y_0$&nbsp; des Signals&nbsp; ${y(t)}$.
{Determine the DC signal component&nbsp; $y_0$&nbsp; of the signal&nbsp; ${y(t)}$.
|type="{}"}
|type="{}"}
$y_0\ = \ $ { 0.75 3% } &nbsp; $\text{V}$
$y_0\ = \ $ { 0.75 3% } &nbsp; $\text{V}$




{Ermitteln Sie den Gleichsignalanteil&nbsp; $z_0$&nbsp; des Signals&nbsp; ${z(t)}$.
{Determine the DC signal component&nbsp; $z_0$&nbsp; of the signal&nbsp; ${z(t)}$.
|type="{}"}
|type="{}"}
$z_0\ = \ $ { 1.333 3% }&nbsp;  $\text{V}^2$
$z_0\ = \ $ { 1.333 3% }&nbsp;  $\text{V}^2$
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</quiz>
</quiz>


===Musterlösung===
===Solution===
{{ML-Kopf}}
{{ML-Kopf}}
'''(1)'''&nbsp;  Der Gleichsignalanteil&nbsp; $x_0$&nbsp; ist der Mittelwert des Signals&nbsp; ${x(t)}$. Es genügt die Mittelung über eine Periodendauer&nbsp; $T_0 = 1 \, \text{ms}$, und man erhält:
'''(1)'''&nbsp;  The DC signal&nbsp; $x_0$&nbsp; is the mean value of the signal&nbsp; ${x(t)}$. Averaging over a period duration&nbsp; $T_0 = 1 \, \text{ms}$ is sufficient. One obtains:
:$$x_0=\frac{1}{T_0}\int^{T_0}_0 x(t)\,{\rm d} t \hspace{0.15cm}\underline{=1\,\rm V}.$$
:$$x_0=\frac{1}{T_0}\int^{T_0}_0 x(t)\,{\rm d} t \hspace{0.15cm}\underline{=1\,\rm V}.$$






'''(2)'''&nbsp;  In der Hälfte der Zeit ist&nbsp; ${y(t)} = 1\, \text{V}$, in der anderen Hälfte liegt es zwischen&nbsp; $0$&nbsp; und&nbsp; $1\, \text{V}$&nbsp; mit dem Mittelwert bei&nbsp; $0.5 \,\text{V}$&nbsp; &rArr; &nbsp;  $y_0 \hspace{0.15cm}\underline{= 0.75 \,\text{V}}$.
'''(2)'''&nbsp;  In half the time&nbsp; ${y(t)} = 1\, \text{V}$, in the other half is is between&nbsp; $0$&nbsp; and&nbsp; $1\, \text{V}$&nbsp; with the mean value at&nbsp; $0.5 \,\text{V}$&nbsp; &rArr; &nbsp;  $y_0 \hspace{0.15cm}\underline{= 0.75 \,\text{V}}$.






'''(3)'''&nbsp; Aufgrund der Periodizität und der Symmetrie genügt die Mittelung im Bereich von&nbsp; $0$&nbsp; bis&nbsp; $T_0/2$.
'''(3)'''&nbsp; Due to the periodicity and symmetry, averaging in the range from&nbsp; $0$&nbsp; bis&nbsp; $T_0/2$ is sufficient.
* Mit der entsprechenden Kennlinie gilt dann:
* With the corresponding characteristic curve, the following then applies::
:$$z_0=\frac{1}{T_0/2}\int^{T_0/2}_0 x^2(t)\,{\rm d}t=\frac{4\rm V^2}{T_0/2}\int^{T_0/2}_0 ({2t}/{T_0})^2\, {\rm d}t={4}/{3}\rm \;V^2
:$$z_0=\frac{1}{T_0/2}\int^{T_0/2}_0 x^2(t)\,{\rm d}t=\frac{4\rm V^2}{T_0/2}\int^{T_0/2}_0 ({2t}/{T_0})^2\, {\rm d}t={4}/{3}\rm \;V^2
\hspace{0.15cm}\underline{\approx1.333\rm \;V^2}.$$
\hspace{0.15cm}\underline{\approx1.333\rm \;V^2}.$$

Revision as of 04:14, 4 January 2021

Gleichanteil nach Nichtlinearitäten

We start from the triangular signal  ${x(t)}$  according to the figure above.

If we apply this signal to an amplitude limiter, we get the signal

$$y(t)=\left\{ {x(t)\atop \rm 1V}{\hspace{0.5cm} {\rm f\ddot{u}r}\quad x(t)\le \rm 1V \atop {\rm sonst}}\right..$$

A second non-linearity provides the signal

$$z(t)=x^2(t).$$

The DC signal components are designated  $x_0$,  $y_0$  and  $z_0$  in the following.





Hint:




Questions

1 Determine the DC signal component  $x_0$  of the signal  ${x(t)}$.

$x_0\ = \ $   $\text{V}$

2 Determine the DC signal component  $y_0$  of the signal  ${y(t)}$.

$y_0\ = \ $   $\text{V}$

3 Determine the DC signal component  $z_0$  of the signal  ${z(t)}$.

$z_0\ = \ $   $\text{V}^2$


Solution

Solution

(1)  The DC signal  $x_0$  is the mean value of the signal  ${x(t)}$. Averaging over a period duration  $T_0 = 1 \, \text{ms}$ is sufficient. One obtains:

$$x_0=\frac{1}{T_0}\int^{T_0}_0 x(t)\,{\rm d} t \hspace{0.15cm}\underline{=1\,\rm V}.$$


(2)  In half the time  ${y(t)} = 1\, \text{V}$, in the other half is is between  $0$  and  $1\, \text{V}$  with the mean value at  $0.5 \,\text{V}$  ⇒   $y_0 \hspace{0.15cm}\underline{= 0.75 \,\text{V}}$.


(3)  Due to the periodicity and symmetry, averaging in the range from  $0$  bis  $T_0/2$ is sufficient.

  • With the corresponding characteristic curve, the following then applies::
$$z_0=\frac{1}{T_0/2}\int^{T_0/2}_0 x^2(t)\,{\rm d}t=\frac{4\rm V^2}{T_0/2}\int^{T_0/2}_0 ({2t}/{T_0})^2\, {\rm d}t={4}/{3}\rm \;V^2

\hspace{0.15cm}\underline{\approx1.333\rm \;V^2}.$$

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