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I am studying for an exam and have the following question:

$$x(t) = u(t)$$ $$h(t) = [e^{-t}-e^{-2t}]u(t)$$

where u(t) is a unit-step function. I need to find the convolution x(t)*h(t).

So: $$ x(t)*h(t) = \int_{-\infty}^\infty u(t)[e^{-t}-e^{-2t}]u(t-\tau)d\tau\ $$ $$ x(t)*h(t) = u(t)[e^{-t}-e^{-2t}]\int_{-\infty}^\infty u(t-\tau)d\tau\ $$

The question I have is how do I take the integral of $$ \int_{-\infty}^\infty u(t-\tau)d\tau\ $$

And how would I graph this convolution? TIA!

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  • $\begingroup$ You should check your definition of convolution; you might have swapped a $\tau$ with a $t$ somewhere :). $\endgroup$
    – user14717
    Feb 8, 2015 at 18:47
  • $\begingroup$ @NickThompson would I also put $$(t-\tau)$$ for $$e^{-t} - e^{-2t}$$ $\endgroup$
    – Ryan_W4588
    Feb 8, 2015 at 18:52
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    $\begingroup$ Nope: Check the definition of convolution: en.wikipedia.org/wiki/Convolution $\endgroup$
    – user14717
    Feb 8, 2015 at 18:53
  • $\begingroup$ @NickThompson Oh I think I see what you're saying now. I think this would be my correct equation then: $$ x(t)*h(t) = [e^{-t}-e^{-2t}]\int_{-\infty}^\infty u(\tau)u(t-\tau)d\tau\ $$ $\endgroup$
    – Ryan_W4588
    Feb 8, 2015 at 18:55
  • $\begingroup$ Not quite! Your exponents have the wrong argument. $\endgroup$
    – user14717
    Feb 8, 2015 at 19:01

1 Answer 1

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using your notation, \begin{align*} (x\ast h)(t) &= \int_\mathbb{R} u(\tau)h(t-\tau)d\tau \overset{\textrm{u is step function}}{=} \int_0^\infty h(t-\tau)d\tau \\ &= \int_0^\infty (e^{-(t-\tau)}-e^{-2(t-\tau)})u(t-\tau) d\tau \\ &= \int_0^t (e^{-(t-\tau)}-e^{-2(t-\tau)}) d\tau, \qquad t\geq 0 \\ &= \left( \frac{1}{2} - e^{-t} + \frac{1}{2}e^{-2t} \right) u(t) \end{align*} the key point is that $u(t-\tau)$ as a function of $\tau$ is $1$ for $\tau \leq t$ and zero otherwise (a graph with these transformations - first shift to the right, then invert sign- may help).

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  • $\begingroup$ Awesome, this was super helpful! Thanks bbn! $\endgroup$
    – Ryan_W4588
    Feb 8, 2015 at 19:09

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