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We are currently learning about convolution in my signals and systems class, and one of our textbook problems is to compute the convolution of $x(t) = e^{-\alpha t} u(t)$ and $y(t) = e^{-\beta t} u(t)$. After working out the integral, I obtained the following result

$$ x(t) * h(t) = \frac{e^{-\alpha t} - e^{-\beta t}}{\beta - \alpha}. $$

The solutions manual for my textbook has the same answer, except their result is multiplied by $u(t)$. I am unsure of where this is coming from, since I know that I computed the integral correctly. I used the unit step functions to simplify the bounds on the integral, so I do not see how they are still around in the final answer.

Can anyone explain what is going on here?

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A step by step derivation of this convolution would start with the following:

$$ \begin{align} f(t) &= \int_{-\infty}^\infty x(\tau) h(t-\tau) \,d\tau \\ &= \int_{-\infty}^\infty e^{-\alpha\tau} u(\tau) e^{-\beta(t-\tau)} u(t-\tau) \,d\tau \end{align} $$ And this is probably where you probably went on and tried to simplified the bounds on the integral with: $$ \begin{align} f(t) &= \int_0^t e^{-\alpha\tau} e^{-\beta(t-\tau)} \, d\tau \\ \end{align} $$ forgetting that this is only true if $t \geq 0$. Indeed for $t < 0$ the product $u(\tau) u(t-\tau)$ is 0 for all values of $\tau$, and the result of the integral is thus 0.

So, the correct expression would be: $$ \begin{align} f(t) &= u(t) \int_0^t e^{-\alpha\tau} e^{-\beta(t-\tau)} \, d\tau \\ &= \cdots \\ &= u(t) \frac{e^{-\alpha t} - e^{-\beta t}}{\beta-\alpha} \end{align} $$

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