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

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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:

This task belongs to chapter Direct Current Signal - Limit Case of a Periodic Signal.

Questions

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:

$$z_0=\frac{1}{T_0/2}\int^{T_0/2}_0 x^2(t)\,{\rm d}t=\frac{4\rm V^2}{T_0/2}\int^{T_0/2}_0 ({2t}/{T_0})^2\, {\rm d}t={4}/{3}\rm \;V^2\hspace{0.15cm}\underline{\approx1.333\rm \;V^2}.$$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Signaldarstellung/Allgemeine Beschreibung
+{{quiz-Header|Buchseite=Signal Representation/General Description
 }}
-[[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&nbsp; ${x(t)}$&nbsp; gemäß der oberen Abbildung aus.
+We start from the triangular signal&nbsp; ${x(t)}$&nbsp; 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
 :$$z(t)=x^2(t).$$
-Die Gleichsignalanteile werden nachfolgend mit&nbsp; $x_0$,&nbsp; $y_0$&nbsp; bzw.&nbsp; $z_0$&nbsp; bezeichnet.
+The DC signal components are designated&nbsp; $x_0$,&nbsp; $y_0$&nbsp; and&nbsp; $z_0$&nbsp; in the following.
@@ Line 17: / Line 17: @@
+''Hint:''
+*This task belongs to chapter&nbsp; [[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&nbsp; [[Signal_Representation/Gleichsignal_-_Grenzfall_eines_periodischen_Signals|Gleichsignal - Grenzfall eines periodischen Signals]].
+===Questions===
-===Fragebogen===
 <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="{}"}
 $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="{}"}
 $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="{}"}
 $z_0\ = \ $ { 1.333 3% }&nbsp;  $\text{V}^2$
@@ Line 48: / Line 43: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{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:
 :$$x_0=\frac{1}{T_0}\int^{T_0}_0 x(t)\,{\rm d} t \hspace{0.15cm}\underline{=1\,\rm V}.$$
-'''(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}.$$
-\hspace{0.15cm}\underline{\approx1.333\rm \;V^2}.$$
 {{ML-Fuß}}
-[[Category:Aufgaben zu Signaldarstellung|^2. Periodische Signale^]]
+[[Category:Signal Representation: Exercises|^2.2 Direct Current Signal^]]
+[[de:Aufgaben:Aufgabe 2.2Z: Nichtlinearitäten]]