Aufgaben:Exercise 2.2Z: Non-Linearities

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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

$$y(t)=\left\{ {x(t)\atop \rm 1V}{\hspace{0.5cm} {\rm for}\quad x(t)\le \rm 1V \atop {\rm else}}\right..$$

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:

• This exercise belongs to the chapter  Direct Current Signal - Limit Case of a Periodic Signal.

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::

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