I need to simplify the following convolution

$$x(t)\star [e^{-2t} u(t)]$$

where $u(t)$ is the unit step function. I'm very confused with this. Using the definition of convolution of continuous-time signals, I obtained

$$\int\limits_{-\infty}^t {x(\tau)e^{-2(t-\tau)}u(t-\tau)}{d\tau} $$

However, I have no idea what to do at this step. I'm stuck. Is there any property / theorem / trick that can help me simplify this?


Indeed you have reached what can be reached, may be the following additional line can be obtained by moving the $t$ function $e^{-2t}$ out of the integral and replacing the $u(t-\tau)$ by $1$ as:

$$ y(t) = e^{-2t} \int\limits_{-\infty}^t {x(\tau)e^{2\tau}{}d\tau} $$

In the general case you cannot simplify it any further...

  • $\begingroup$ How can you replace $u(t-\tau)$ by 1? Is it always 1 from - infinity to t? $\endgroup$ – anon Dec 6 '17 at 20:30
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    $\begingroup$ Indeed the better question is how did you determined the upper limit of the intgral as $t$ in the first place? i.e., $$ \int_{-\infty}^{\infty} f(\tau)u(t-\tau) d\tau = \int_{-\infty}^{t} f(\tau) u(t-\tau) d\tau $$. If you reach this line then you implicitly say that $$ u(t-\tau) = \begin{cases} 1 &, \text{ for } -\infty < \tau < t \\ 0 &, \text{ for } ~~~~~ t < \tau < \infty \\ \end{cases} $$ I think now it's clear... $\endgroup$ – Fat32 Dec 6 '17 at 20:37
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    $\begingroup$ Yes sorry my bad. Should have thought of that. Thank you for the explanation. Unfortunately it seems that there is no way x(t) can get out of the integration right? $\endgroup$ – anon Dec 6 '17 at 20:43
  • $\begingroup$ you welcome. In the general case you cannot get $x(\tau)$ out of integration... $\endgroup$ – Fat32 Dec 6 '17 at 20:46

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