Exercise 2.1: Linear? Or Non-Linear?

Interconnected system

We consider the sketched arrangement with input $x(t)$ and output $z(t)$:

The system $S_1$ is describable by the following equation:

$$y(t) = x(t) + {1 \, \rm V}^{\rm -1} \cdot x^2(t) .$$

Nothing else is known about the system $S_2$ with input $y(t)$ and output $z(t)$ .

The system $S_3$ is the interconnection of $S_1$ and $S_2$.

An oscillation with frequency $f_0 = 5 \ \rm kHz$ is applied to the input:

$$x(t) = {2 \, \rm V} \cdot {\rm cos}(2\pi f_0 t ) .$$

Hence, at the output of the overall system $S_3$ the following is obtained:

$$z(t) = {1 \, \rm V} \cdot {\rm sin}(2\pi f_0 t ) .$$

Please note:

The exercise belongs to the chapter Classification of Distortions.

The following trigonometric relation is given:

$$\cos^2(\alpha) = {1}/{2} \cdot \big[ 1 + \cos(2\alpha)\big].$$

Questions

Solution

(1) The following holds due to the characteristic curve with linear and quadratic components:

$$y(t) = {2 \, \rm V} \cdot {\rm cos}(2\pi f_0 t ) + {1 \, \rm V}^{\rm -1} \cdot ({2 \, \rm V})^2 \cdot {\rm cos}^2(2\pi f_0 t ) = {2 \, \rm V} \cdot \big[ 1 + {\rm cos}(2\pi \cdot f_0 \cdot t ) +{\rm cos}(2\pi \cdot 2f_0 \cdot t ) \big].$$

Therefore, the signal value 6 V occurs at time $t= 0$ .

(2) Alternatives 2 snd 3 are possible:

An ideal system is out of the question because of $z(t) ≠ x(t)$ .
With only one input frequency $(f_0 = 5 \ \rm kHz)$ in the test signal it is not possible to make a statement whether a second frequency component with $f \ne f_0$ would also be attenuated by $\alpha = 0.5$ and delayed by $\tau = T_0/4 = 50 \ µ\rm s$ .
If for the second frequency component $\alpha = 0.5$ and $\tau = T_0/4 = 50 \ µ \rm s$ arose as a result, a distortion-free system could exist.
If for the second frequency component $\alpha \ne 0.5$ and/or $\tau \ne T_0/4$ arose as a result, then the system would be linearly distorting.
The last alternative would have to be logically denied – although partially – by the observer.

(3) Approach 2 is correct:

Der Beobachter würde erkennen, dass $S_2$ ein linear verzerrendes System ist.
Bei einem verzerrungsfreien System müsste $z(t)$ zusätzlich noch eine Gleichkomponente und eine $10 \ \rm kHz$–Komponente beinhalten,
bei einem nichtlinear verzerrenden System noch größere Frequenzanteile $($bei Vielfachen von $10 \ \rm kHz)$.

(4) In diesem Fall würde gelten:

$$y(t) = {2 \, \rm V} \cdot \big[ 1 + {\rm cos}(2\pi \cdot 10 \ {\rm kHz} \cdot t ) +{\rm cos}(2\pi \cdot 20 \ {\rm kHz} \cdot t ) \big].$$

Das heißt: $Y(f)$ würde Spektrallinien bei $f = 0$, $10 \ \rm kHz$ und $20 \ \rm kHz$ aufweisen.

Die auf der Angabenseite beschriebene Messung mit $f_0 = 5 \ \rm kHz$ hat aber gezeigt, dass $H_2(f = 0) = H_2(f = 10 \ {\rm kHz}) = 0$ gelten muss.
Die einzig mögliche Signalform ist somit

$$z(t) = {2 \, \rm V} \cdot H_2 (f = {20 \, \rm kHz})\cdot {\rm cos}(2\pi \cdot {20 \, \rm kHz} \cdot t ) .$$

Möglich sind also die Lösungsvorschläge 1 and 3, je nachdem, ob das System $S_2$ die Frequenz $20 \ {\rm kHz}$ unterdrückt oder durchlässt.

	$S_3$ is an ideal system.
	$S_3$ is a distortion-free system.
	$S_3$ is a linearly distorting system.
	$S_3$ is a non-linearly distorting system.

	$S_2$ is a distortion-free system.
	$S_2$ is a linearly distorting system.
	$S_2$ is a non-linearly distorting system.

	The signal $z(t)$ is zero for all times.
	A signal of the form $z(t) = A \cdot {\rm cos}(2\pi \cdot 10 \ {\rm kHz} \cdot t ) ,$ with $A \ne 0.$
	A signal of the form $z(t) = A \cdot {\rm cos}(2\pi \cdot 20 \ {\rm kHz} \cdot t ) ,$ with $A \ne 0.$