Exercise 2.1: Linear? Or Non-Linear?

From LNTwww

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 following trigonometric relation is given:
$$\cos^2(\alpha) = {1}/{2} \cdot \big[ 1 + \cos(2\alpha)\big].$$


Questions

1

What is the signal $y(t)$? What is the signal value at time zero?

$y(t = 0) \ = \ $

$\ \rm V$

2

What correct conclusions could be drawn by an observer who only knows the signals $x(t)$  and  $z(t)$  but not the structure of  $S_3$?

$S_3$  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.

3

What conclusions would the observer have to draw if all the information from the information section is known to them?

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

4

What signal $z(t)$  could arise as a result with the input frequency $f_0 = 10 \ \rm kHz$ ?

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 $A \ne 0.$


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 about 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 true – by the observer.


(3)  Approach 2 is correct:

  • The observer would realise that  $S_2$  is a linearly distorting system.
  • For a distortion-free system,  $z(t)$  would additionally have to include a direct (DC) component and a $10 \ \rm kHz$–component,
  • for a non-linearly distorting system, even larger frequency components  $($at multiples of  $10 \ \rm kHz)$.


(4)  In this case the following would hold:

$$y(t) = {2 \, \rm V} \cdot \big[ 1 + {\rm cos}(2\pi \cdot 10 \ {\rm kHz} \cdot t ) +{\rm cos}(2\pi \cdot 20 \ {\rm kHz} \cdot t ) \big].$$
  • That is:   $Y(f)$  would have spectral lines at  $f = 0$,  $10 \ \rm kHz$  and  $20 \ \rm kHz$ .
  • However, the measurement described in the information section with  $f_0 = 5 \ \rm kHz$  has shown that  $H_2(f = 0) = H_2(f = 10 \ {\rm kHz}) = 0$  must hold.
  • The only possible signal form is therefore
$$z(t) = {2 \, \rm V} \cdot H_2 (f = {20 \, \rm kHz})\cdot {\rm cos}(2\pi \cdot {20 \, \rm kHz} \cdot t ) .$$
  • So, approaches 1 and 3 are possible depending on whether the system  $S_2$  suppresses or passes the frequency  $20 \ {\rm kHz}$ .