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I have solved a problem, Kindly help me in determing if I solved it correctly. I will post the question and my own working below.

Problem

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My working

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  • $\begingroup$ @MattL Why did you change the title of the question to "... in terms of its step response"? There is no mention of the step response of the LTI system anywhere in the problem statement or the OP's working; it is all about the impulse response. $\endgroup$ – Dilip Sarwate Jun 16 at 22:14
  • $\begingroup$ @DilipSarwate: If you look at the OP's working you see that he computed the step response $s(t)$ in order to express the response to the given input signal as $s(t)-s(t-1)$. $\endgroup$ – Matt L. Jun 17 at 9:42
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HINTS:

  1. $\cos(2)-\cos(2)=0$
  2. What remains is $y(t)=\cos(t-1)-\cos(t)$. Can this really be the response of a causal system to an input signal that starts at $t=0$? Maybe you forgot some important detail in your answer.
  3. The correct answer must consist of $3$ different expressions for the time intervals $t<2$, $2<t<3$, and $t>3$. Your answer corresponds to one of those three expressions, so you need to add the other two, and the corresponding time intervals.
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  • $\begingroup$ Sorry for not removing cos(2) as it subtracts out, i forgot to write it down.Understood your points 2 and 3, but can you tell if my methodology is correct ? Like the way I computed y(t) (via step-response and impulse response relationship and then applying transformation for the answer for u(t), is it correct ? $\endgroup$ – calculusnoob Jun 16 at 21:33
  • $\begingroup$ Also, for t >3, at what point 3 is defined ? $\endgroup$ – calculusnoob Jun 16 at 21:39
  • $\begingroup$ @calculusnoob: You expressed the response to the given input signal as the difference between the step response and its shifted version. If you go back to your calculations you'll see that the step response starts at $t=2$. So if you subtract from it a shifted version of the step response, the subtraction will have an effect for $t>3$ because that's where the shifted version of the step response starts. $\endgroup$ – Matt L. Jun 17 at 9:44

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