Difference between revisions of "Aufgaben:Exercise 2.1: Linear? Or Non-Linear?"
From LNTwww
(2 intermediate revisions by one other user not shown) | |||
Line 27: | Line 27: | ||
''Please note:'' | ''Please note:'' | ||
− | *The exercise belongs to the chapter [[Linear_and_Time_Invariant_Systems/Classification_of_the_Distortions|Classification of Distortions]]. | + | *The exercise belongs to the chapter [[Linear_and_Time_Invariant_Systems/Classification_of_the_Distortions|Classification of the Distortions]]. |
*The following trigonometric relation is given: | *The following trigonometric relation is given: | ||
Line 49: | Line 49: | ||
− | {What conclusions would the observer have to draw if all the information from the information | + | {What conclusions would the observer have to draw if all the information from the information section is known to them? |
|type="[]"} | |type="[]"} | ||
- $S_2$ is a distortion-free system. | - $S_2$ is a distortion-free system. | ||
Line 89: | Line 89: | ||
'''(3)''' <u>Approach 2</u> is correct: | '''(3)''' <u>Approach 2</u> is correct: | ||
*The observer would realise that $S_2$ is a linearly distorting system. | *The observer would realise that $S_2$ is a linearly distorting system. | ||
− | *For a | + | *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 $10 \ \rm kHz)$. | *for a non-linearly distorting system, even larger frequency components $($at multiples of $10 \ \rm kHz)$. | ||
Line 99: | Line 99: | ||
*That is: $Y(f)$ would have spectral lines at $f = 0$, $10 \ \rm kHz$ and $20 \ \rm kHz$ . | *That is: $Y(f)$ would have spectral lines at $f = 0$, $10 \ \rm kHz$ and $20 \ \rm kHz$ . | ||
− | *However, the measurement described | + | *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 | *The only possible signal form is therefore | ||
:$$z(t) = {2 \, \rm V} \cdot H_2 (f = {20 \, \rm kHz})\cdot {\rm | :$$z(t) = {2 \, \rm V} \cdot H_2 (f = {20 \, \rm kHz})\cdot {\rm |
Latest revision as of 13:45, 17 November 2022
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 the 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 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 $20 \ {\rm kHz}$ .