As the title suggests , I'm having problems working out the integration of some basic signal functions. Shown below are the questions and my workings. For some reason I'm just unable to get the answer.

For a) and c) , I know that for u(t) is basically 0 before t=0 , thus I changed the limits for the integration. I did the same thing for u(t-1)

Thanks for the help!



  • $\begingroup$ Welcome to DSP.SE! I've added tags of homework and self-study. Please feel free to delete one or the other of them if they are inaccurate. $\endgroup$
    – Peter K.
    Oct 2, 2015 at 12:01
  • 1
    $\begingroup$ Any chance for TeX formatting? $\endgroup$
    – jojeck
    Oct 2, 2015 at 12:41

1 Answer 1


So you've already figured out the $u(t-T)$ trick: the unit step (delayed) just changes the limits of integration.

The other trick is that $\delta(t-T)$ just "samples" the value of the rest of the integral's argument. So, for example, $\int x(t) \delta(t-T) dt = x(T)$ where $T$ is somewhere within the integral's limits.

  • $\begingroup$ I tried using that for c) , assuming that T=0 , but I still am unable to get the correct answer. $\endgroup$
    – John
    Oct 2, 2015 at 12:20
  • $\begingroup$ Well, that's the other trick: note I said "where $T$ is somewhere within the integral's limits". If $T$ is not within the integral's limits then $\int x(t) \delta(t-T) dt = 0$. $\endgroup$
    – Peter K.
    Oct 2, 2015 at 12:25
  • $\begingroup$ But if I assume T=0 , then it is obviously within the integral limits. $\endgroup$
    – John
    Oct 2, 2015 at 13:02
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    $\begingroup$ Are you telling me that $T=0$ is within the integration limits of $\int_{1}^T \cos(t)\delta(t) dt$? $\endgroup$
    – Peter K.
    Oct 2, 2015 at 13:05
  • $\begingroup$ I'm confused , the integration limits for that question are from 1 to infinity. Knowing that T=0 , it obviously fits. $\endgroup$
    – John
    Oct 2, 2015 at 13:15

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