1
$\begingroup$

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?

Improve this question
$\endgroup$
3
$\begingroup$

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...

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.