3
$\begingroup$

I have a problem with the convolution of two signals:

$$x_{1}(t) = e^{2t}*u(-t)$$ $$x_{2}(t) = u(t-3)$$

$$x_1 \mathbin{\mathrm{(conv)}} x_2 = \int_{-\infty}^{+\infty} x_2(\tau) * x_1(t-\tau) \, d\tau$$

My usual way is, as the integral says, leave one function as it is and mirror the other one. In my previous task there was no intersection after the mirroring and I could shift the mirrored function until they both intersect for the first time and this was my integration border. But in this task, if I mirror $x_{2}$, I get a function, that is already intersecting $x_{1}$, because it runs exponentially to $0$ for large $\tau$.

How can I do this convolution?

enter image description here

$\endgroup$
  • 1
    $\begingroup$ $x_1(t) = 0$ when $t > 0$ and $x_2(t) = 0$ when $t < 3$ so $x_1 \ast x_2 (t) = \int_{-\infty}^\infty x_2(\tau) x_1(t-\tau) d\tau = \int_{\max(t,3)}^\infty x_2(\tau) x_1(t-\tau) d\tau = \int_{\max(t,3)}^\infty x_1(t-\tau) d\tau$ $\endgroup$ – reuns Jan 2 '16 at 21:53
  • $\begingroup$ @user1952009 Thanks for the quick response, but could you explain why the your integral starts at max(t,3) (not even sure what that means)? $\endgroup$ – JavaForStarters Jan 2 '16 at 21:58
  • $\begingroup$ Never mind, I think I got it now. So it's basically $\int_{3}^{\infty} e^{2(t-3)}dt $ ? $\endgroup$ – JavaForStarters Jan 2 '16 at 22:04
  • 1
    $\begingroup$ yes, the result should be a continuous function, constant for $t < 3$ and diminishing exponentially for $t > 3$ $\endgroup$ – reuns Jan 2 '16 at 22:09

Your Answer

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

Browse other questions tagged or ask your own question.