The correct answer is $\boxed{\left( {{1 \over a}} \right)\left( {1 – {e^{ – at}}} \right)}$.
The unit impulse response of a linear time invariant system is the unit step function $u(t)$. This means that if the system is excited with a unit impulse, $u(t)$, the output will be a unit step function, $u(t)$.
The excitation in this question is $e^{-at}u(t)$. This is a decaying exponential function that is multiplied by a unit step function. The output of the system to this excitation can be found using the convolution integral:
$$y(t) = \int_{-\infty}^t h(t-\tau)e^{-a\tau}d\tau$$
where $h(t)$ is the unit impulse response of the system.
The unit impulse response of the system is given as $h(t) = u(t)$. Substituting this into the convolution integral gives:
$$y(t) = \int_{-\infty}^t u(t-\tau)e^{-a\tau}d\tau$$
The convolution of $u(t)$ and $e^{-at}u(t)$ can be found using the following formula:
$$\int_{-\infty}^t u(t-\tau)e^{-a\tau}d\tau = \left( {{1 \over a}} \right)\left( {1 – {e^{ – at}}} \right)$$
Substituting this into the convolution integral gives:
$$y(t) = \left( {{1 \over a}} \right)\left( {1 – {e^{ – at}}} \right)$$
Therefore, the response of the system to an excitation $e^{-at}u(t)$, $a > 0$ will be $\left( {{1 \over a}} \right)\left( {1 – {e^{ – at}}} \right)$.