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I'm currently working on this problem from Oppenheimer's book: input and output to the system

Given $x_{1}(t)$ and $y_{1}(t)$ I should figure out $y_{2}(t)$ and $y_{3}(t)$ given that the system is LTI. My progress so far: enter image description here I'm specifically asking about the graph for $y_{3}(t)$ because it looks like my solution doesn't agree with what's in the solution manual as for $0<t<1$ it's $y_{3}(t)=2$ instead of $y_{3}(t)=4t$ so am I missing anything?

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    $\begingroup$ Should have thought about that, will try it out now. Thank you. $\endgroup$
    – Essam
    Apr 10, 2021 at 18:00
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    $\begingroup$ I find that this exercise is very pedagogic in the progression of difficulties. Some other exercises of the book might have the same gradation, don't hesitate to look back before jumping into computations at first $\endgroup$
    – Laurent Duval
    Apr 10, 2021 at 18:31

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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 that this exercise is very pedagogic in the progression of difficulties. Some other exercises of the book might have the same gradation, don't hesitate to look back before jumping into computations at first.

Graphical solution

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    $\begingroup$ I was able to write it down as you said and things worked out, the schematic is incredibly insightful, thank you. $\endgroup$
    – Essam
    Apr 10, 2021 at 18:16
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    $\begingroup$ This doesn't not work always in exercises, however using simple bricks is never a waste of time in analyzing a system. $\endgroup$
    – Laurent Duval
    Apr 10, 2021 at 18:20

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