I have following convolution:
$$e^{-at}u(t)*e^{at}u(-t);a>0$$
$u(t)$ is the unit step function.
I have tried the following:
$$\begin{align*}e^{-at}u(t)*e^{at}u(-t)&=\int_{-\infty}^{\infty}e^{-a\tau}u(\tau)\ e^{a(t-\tau)}u(-(t-\tau))\mathrm{d}\tau=e^{at} \int_{-\infty}^{\infty}e^{-2a\tau}u(\tau)\ u(\tau-t)\mathrm{d}\tau \\ &=e^{at}\int_{t}^{\infty}e^{-2a\tau}\mathrm{d}\tau=-\frac{e^{at}}{2a}[e^{-2a\tau}]_t^{\infty}=\frac{e^{-at}}{2a} \\ \end{align*}$$
But the given answer is $\displaystyle \frac{e^{-at}}{2a}u(t)+\frac{e^{at}}{2a}u(-t)$.
Can anyone please tell me what am I doing wrong over here.
Best Answer
By definition, \begin{equation} f(t) = \int_{-\infty}^{\infty} e^{-a \tau}u(\tau) e^{a(t-\tau)}u(\tau - t) \ d \tau \end{equation} that is \begin{equation} f(t) = e^{at}\int_{-\infty}^{\infty} e^{-2a \tau} u(\tau) u(\tau - t) \ d \tau \end{equation} Integral is non zero when $\tau > 0$ or $\tau - t > 0$ ($\tau >t$).
Assuming $t>0$ \begin{equation} f(t) = e^{at}\int_{t}^{\infty} e^{-2a \tau} \ d \tau = \frac{e^{at}}{2a} [ e^{-2at} ] = \frac{e^{-at}}{2a} \end{equation}
Assuming $t<0$
If $t < 0$, then the above integral must start from $0$, \begin{equation} f(t) = e^{at}\int_{0}^{\infty} e^{-2a \tau} \ d \tau = \frac{e^{at}}{2a} [ 1 ] = \frac{e^{at}}{2a} \end{equation}
Combining both results
We get $$f(t) = \frac{e^{-at}}{2a}u(t) + \frac{e^{at}}{2a} u(-t)$$ i.e. the first term is non zero when $t >0$ and the second term is non zero when $t < 0$. Note that when one is zero, the other is zero.