Consider the signal $x(t)=e^{-t}u(t)$
where $u(t)=\mathbb{1}(t\geq0)$, i.e. the Heaviside function.
Find the signal $y(t)=x(t)*x(-t)$
My attempt:
$y(t)=x(t)*x(-t)$
$=e^{-t}u(t)*e^{t}u(-t)$
$=e^{-t}u(t)*e^{t}u(-t)$
$= \int_{-\infty}^{\infty}[e^{-\tau}u(\tau)][e^{t-\tau}u(-t+\tau)]d\tau $
$= \int_{-\infty}^{\infty}e^{t}e^{-2\tau}u(\tau)u(-t+\tau)d\tau $
$= e^{t}\int_{-\infty}^{\infty}e^{-2\tau}u(\tau)u(-t+\tau)d\tau $
$= e^{t}\int_{0}^{t}e^{-2\tau}d\tau $ (I think)
$=e^{t}[(-\frac{1}{2}e^{-2\tau}]_0^t$
$=-\frac{1}{2}e^{t}[e^{-2t}-1]$
$=\frac{1}{2}[e^{t}-e^{-t}]$
The solution and Wolfram Alpha give $y(t)=\frac{1}{2}e^{-|t|}$. Did I evaulate the step functions wrong? Guidance is appreciated.
Best Answer
You need $u(\tau)$ and $u(-t+\tau)$ to both be $1$, so $\tau > 0$ and $\tau > t$. So, if $t > 0$, then this becomes $\tau > t$, and your integral (in the line where you say "I think") should actually have limits $t$ to $\infty$. This gives $y(t) = \frac{1}{2}e^{-t}$ for $t > 0$.
On the other hand, if $t < 0$, this becomes $\tau > 0$, so your integral should have limits $0$ to $\infty$. This gives $y(t) = \frac{1}{2}e^{t}$ for $t < 0$. Putting this together with the solution in the $t > 0$ case, you can see that $y(t) = \frac{1}{2}e^{-\left\lvert t\right\rvert}$ covers both cases in a single expression.