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

From LNTwww
 
(7 intermediate revisions by 3 users not shown)
Line 3: Line 3:
 
}}
 
}}
  
−
[[File:P_ID322__Sig_Z_2_2.png|right|frame|Gleichanteil nach Nichtlinearitäten]]
+
[[File:P_ID322__Sig_Z_2_2.png|right|frame|DC component after non-linearities]]
−
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
−
:$$y(t)=\left\{ {x(t)\atop \rm 1V}{\hspace{0.5cm} {\rm f\ddot{u}r}\quad x(t)\le \rm 1V \atop {\rm sonst}}\right..$$
+
:$$y(t)=\left\{ {x(t)\atop \rm 1V}{\hspace{0.5cm} {\rm for}\quad x(t)\le \rm 1V \atop {\rm else}}\right..$$
−
Eine zweite Nichtlinearität liefert das Signal
+
*Another 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.  
  
  
Line 17: Line 17:
  
  
−
 
+
''Hint:''  
−
 
+
*This task belongs to chapter  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal|Direct Current Signal - Limit Case of a Periodic Signal]].
−
''Hinweis:''  
 
−
*Die Aufgabe gehört zum Kapitel  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal|Gleichsignal - Grenzfall eines periodischen Signals]].
 
 
   
 
   
  
  
−
 
+
===Questions===
−
 
 
−
 
 
−
===Fragebogen===
 
  
 
<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;  $\text{V}^2$
Line 48: Line 43:
 
</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 of signal&nbsp; ${x(t)}$.&nbsp; Averaging over a period duration&nbsp; $T_0 = 1 \, \text{ms}$ is sufficient.&nbsp; 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 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}.$$
Line 67: Line 62:
  
  
−
[[Category:Exercises for Signal Representation|^2.2 Direct Current Signal - Limit Case of a Periodic Signal^]]
+
[[Category:Signal Representation: Exercises|^2.2 Direct Current Signal^]]

Latest revision as of 12:24, 14 April 2021

DC component after non-linearities

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
$$y(t)=\left\{ {x(t)\atop \rm 1V}{\hspace{0.5cm} {\rm for}\quad x(t)\le \rm 1V \atop {\rm else}}\right..$$
  • Another 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 of 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 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:
$$z_0=\frac{1}{T_0/2}\int^{T_0/2}_0 x^2(t)\,{\rm d}t=\frac{4\rm V^2}{T_0/2}\int^{T_0/2}_0 ({2t}/{T_0})^2\, {\rm d}t={4}/{3}\rm \;V^2 \hspace{0.15cm}\underline{\approx1.333\rm \;V^2}.$$

Retrieved from "http://en.lntwww.de/index.php?title=Aufgaben:Exercise_2.2Z:_Non-Linearities&oldid=37579"
  • Powered by MediaWiki