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Hello all I am having some trouble with the following convolution integral $$h(t) = e^{-t} u(t) ~,~x(t) = e^{-2t}u(t-3)$$

I see that to evaluate that I need to first shift x(t) by t-3 $$e^{-2(t-3)}u(t-3) $$ After which I can perform the convolution of the by modifying h(t) since it is a simpler function

$$\int_{3}^{t} e^{-2(\tau-3)}u(\tau-3)~e^{\tau+t}u(t-\tau)d\tau$$

After integrating I end up with the following answer:

$$(e^{9-t}-e^{6})u(t-3)$$ I dont feel that this answer is correct, can someone point out where my mistake is? Thanks in advanced

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$$\int_{-\infty}^{\infty} x(\tau) h(t-\tau)d\tau$$

$$=\int_{-\infty}^{\infty} e^{-2\tau}u(\tau-3)~e^{-t+\tau}u(t-\tau)d\tau$$

$$=\int_{-\infty}^{\infty} e^{-(t+\tau)}u(\tau-3)u(t-\tau)d\tau$$

There are two cases:

Case $1$: If $t < 3$, then the answer is $0$.

Case $2$: If $t \geq 3$, then:

$$=\int_{3}^{t} e^{-(t+\tau)}d\tau=e^{-t}(e^{-3}-e^{-t})$$

In a simple way, the answer is $e^{-t}(e^{-3}-e^{-t})u(t-3).$

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  • $\begingroup$ $e^{-t}u(t-3) \neq e^{-(t-3)} u(t-3)$, just take $t=3$. $\endgroup$ – Rajat Oct 9 '17 at 19:30
  • $\begingroup$ thank you for the clarification $\endgroup$ – james Oct 9 '17 at 22:25

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