Output for x = e^(−at)u(t), h = u(t)

For an LTI system the output is the convolution of input and impulse response:

$$y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau$$

Given: $x(t) = e^{-at}u(t)$ (take $a > 0$), $h(t) = u(t)$.

$$y(t) = \int_{-\infty}^{\infty} e^{-a\tau}u(\tau)\,u(t-\tau)\,d\tau$$

Limits: $u(\tau) = 1$ for $\tau \ge 0$ and $u(t-\tau) = 1$ for $\tau \le t$. The overlap is $0 \le \tau \le t$, which exists only for $t \ge 0$.

For $t < 0$: no overlap, $y(t) = 0$.

For $t \ge 0$:

$$\begin{aligned} y(t) &= \int_{0}^{t} e^{-a\tau}\,d\tau \\ &= \left[\frac{e^{-a\tau}}{-a}\right]_0^{t} \\ &= \frac{1}{a}\left(1 - e^{-at}\right) \end{aligned}$$

Answer:

$$y(t) = \frac{1}{a}\left(1 - e^{-at}\right)u(t)$$

 y(t)
 1/a |        .-------------
     |     .-'
     |   .'
     |  /
     | /
   0 +/----------------------> t
     0

The output starts at 0 at $t = 0$ and rises to the final value $1/a$ as $t \to \infty$. This is expected, because $h(t) = u(t)$ is an ideal integrator, so $y(t) = \int_{-\infty}^{t} x(\tau)\,d\tau$, the running area under $e^{-at}$, whose total area is $1/a$.