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Can someone help me understand how did we get from the second line to the third one?

$$\begin{align} u(t-1)*u(t) &= \int_{-\infty}^{\infty}u(\tau-1)u(t-\tau)d\tau \\ &=\int_{1}^{t}u(t-\tau)d\tau \\ &=(t-1)u(t-1) \end{align}$$

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  • $\begingroup$ This question isn't really about signal processing; it's just an a question about how integrals work which is off-topic for this forum. $\endgroup$
    – Peter K.
    Sep 17, 2015 at 20:24
  • $\begingroup$ @PeterK. Is there a more suitable stack exchange site for these type of questions? I thought unit step fumction was related to this site. $\endgroup$ Sep 17, 2015 at 21:01

1 Answer 1

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The integration variable is $\tau$ and the integration limits are determined by the range of $\tau$ over which both step functions under the integral are non-zero.

For the first step functions that's $\tau > 1$ and for the second one it's $\tau < t$ . This determines the integration interval. Within this interval the function to integrate is simply 1, i.e. $u(t- \tau) = 1$ for $ \tau < t$ . Integral from a to b over one is simply a-b.

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  • $\begingroup$ I think you're almost there... I found this [pdf ](courses.washington.edu/bioen316/Assignments/316_SCP.pdf) that explains why the upper limit is bounded by t and since r>1 and r<t --> t>1 and that means for t<1 we get 0 output, so the answer that was t-1 should be reformed to (t-1)u(t-1) to reflect the complete answer. $\endgroup$ Sep 17, 2015 at 19:19
  • $\begingroup$ @user2692669 "I think you're almost there..." What a snide and ungracious comment on an answer that was written to help you and which you found acceptable! Hilmar's answer is all there. +1 to him and a -1 to your question. $\endgroup$ Sep 19, 2015 at 1:55
  • $\begingroup$ @DilipSarwate I think you read more into that comment than you are supposed to. I encouraged him to add some things that would make his answer something for others to read by and be like "Yes, that answer is very informative and covers all my concerns about the problem.". This is a community, I didn't intend this to be a question only for myself, others will stumble upon it, too. Marking a question -1 for personal feelings is just offensive and disgraceful. $\endgroup$ Sep 19, 2015 at 6:17
  • $\begingroup$ Guys: no point in fighting over this: no offense was meant and none was taken $\endgroup$
    – Hilmar
    Sep 20, 2015 at 11:46

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