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The problem at hand: enter image description here

Where I'm currently stuck:

enter image description here

I'm not entirely sure about how to move on from this point, I'm trying to find the superposition of the responses of the two individual signals so I can later check if that's equal to the response of their (the two signals') superpositions. The intervals in each response involve different signals and thus I'm unable to find the superposition of responses.

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    $\begingroup$ Do ponder on the difference between a signal and a system. A signal just is -- it's a bunch of values as a function of time. A system is something that operates on a signal. In this case, your system is unnamed; it's the expression that maps $x(t)$ onto $y(t)$. $\endgroup$
    – TimWescott
    Apr 10, 2021 at 16:08
  • $\begingroup$ Ouch, I actually meant system, let me fix the title. $\endgroup$
    – Essam
    Apr 10, 2021 at 16:09

3 Answers 3

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First, you are not trying to find the linearity of a signal, but the linearity of a system defined by the transform above between input and output signals.

If you suspect your system is not linear (the different expressions depending on the sign of $x(t)$ is a big red flag), then you only need a counterexample rather than a general proof of linearity.

In this case, if you scale your input by a negative number, the output is not scaled accordingly. That is: $T{\alpha x} \neq \alpha T{x}$ if $\alpha$ is negative.

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  • $\begingroup$ Well, Oppenheimer's solution manual says that it's linear. If I could work out the superposition of the responses and compare that to the response of the superposition of the inputs it would have solved the problem. $\endgroup$
    – Essam
    Apr 10, 2021 at 16:07
  • $\begingroup$ But what you said actually makes sense, so I might just assume that it's a mistake. I'm still interested in finding the superposition nonetheless. $\endgroup$
    – Essam
    Apr 10, 2021 at 16:13
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    $\begingroup$ Oppenheimer's solution manual is wrong! If they aren't showing a worked example, then it could just be a typo. $\endgroup$
    – TimWescott
    Apr 10, 2021 at 16:14
  • $\begingroup$ And you're really not finding the superposition -- you're disproving that it exists. $\endgroup$
    – TimWescott
    Apr 10, 2021 at 16:14
  • $\begingroup$ Yes, I successfully do that whenever I find that it's not equal to the response of the superposition of the inputs. But I need to find it first. $\endgroup$
    – Essam
    Apr 10, 2021 at 16:15
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As said by @Juancho, the expression let us suspect that the system could be non-linear. So we could look for a counter-example, yet it is not evident how to find a good one. Anyway, lt us try to better understand the system. Since there is something around sign change.

The insight is there to have a first signal that has a sign change, and a second that does not. Let us keep simple, and try with $x_1(t) = t$, and $x_2(t) = 1$. Intuitively, adding $x_2$ to $x_1$ will also affect where the sign changes. Grip paper is your friend, like in Understanding LTI systems graphically.

Now, let's get graphical! And by the way, you get a clearer understanding of the system, and a (family of) counter-example.

Graphical representation of the non-linearity of the system

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If you did want to grind through this, you could define $x_3(t) = A x_1(t) + B x_2(t)$, then substitute it into the left side of either of your top two system definitions above.

like this: $$x_3(t) = \begin{cases}0 & x_3(t) < 0 \\ x_3(t)+x_3(t-2) & x_3(t) \ge 0\end{cases} = \\ \begin{cases}0 & A x_1(t) + B x_2(t) < 0 \\ A x_1(t) + B x_2(t)+A x_1(t-2) + B x_2(t-2) & A x_1(t) + B x_2(t) \ge 0\end{cases}$$

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  • $\begingroup$ Yes, that would be my next step (finding the response of the superposition of inputs), however, I'm still stuck in the step of finding the superposition of the individual responses (the last step would be to check that they're equal for linearity) $\endgroup$
    – Essam
    Apr 10, 2021 at 16:14
  • $\begingroup$ Edited. It really is just plugging the long thing in wherever you see the short thing. $\endgroup$
    – TimWescott
    Apr 10, 2021 at 16:19
  • $\begingroup$ Yes, that was my next step as I said. To show that it's not linear I need to show that this expression isn't equal to the one I'm trying to find in the image. $\endgroup$
    – Essam
    Apr 10, 2021 at 16:21

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