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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 ...


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Suppose that you could write $x_.$ as $x_.(t) = a_1x_1(t-n_1)+a_2x_1(t-n_2)$, then by LTI hypothesis, you can derive $y_.(t) = a_1y_1(t-n_1)+a_2y_1(t-n_2)$ with very little computation. You can get a first insight to that method by looking at the shape of $x_2$ which is self-evident, and a little more imagination gives you the structure for $x_3$. I find ...


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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 ...


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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)+...


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A 5-point moving average can be performed in different ways. The two principal options consist in: causal: take the current point, and average it with the four most recent past samples, or sum it and divide by the length of the average span (which seems to be your choiice, regarding your for bounds $$ y[n] = \left(\sum^n_{k=n-4} x[k]\right)/5\,,$$ ...


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HINT (really an answer but not providing code so its a hint): In your setup, a $M$-point moving average filter uses the average of the previous $M-1$ samples and the current sample to compute the current output. So for $n=5$, the output will be $\frac{1}{5}(x[1]+x[2]+x[3]+x[4]+x[5])$. In general, for $n=k$ the output of this $M$-point MA filter is: $$ \frac{...


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