Output of LTI with h(t) = e^(-at)u(t) and e^(at)u(-t)

For an LTI system the output is y(t) = x(t) * h(t). No input is given, so take the input as a unit impulse, x(t) = δ(t): the output is then h(t) itself. Below, each h(t) is sketched, and the output for a unit step input is also found as an extra illustration.

Case 1: h(t) = e^(-at)u(t), a > 0 (causal)

Impulse response (output for δ(t)):

h(t)
1 |*
  | *
  |  **
  |    ****_____
  +-----------------> t
  0

It decays from 1 at t = 0 towards 0. It is causal and stable, since ∫₀^∞ e^(-at) dt = 1/a < ∞.

Step response, s(t) = ∫₀^t e^(-aτ) dτ = (1 - e^(-at))/a for t ≥ 0:

s(t)
1/a |        _______
    |     _--
    |   /
    | /
    +-----------------> t

Case 2: h(t) = e^(at)u(-t), a > 0 (anticausal)

h(t)
          * 1
        * |
     **   |
_****     |
----------+--------> t
          0

The impulse response exists only for t ≤ 0 and grows towards 1 at t = 0. It is anticausal (non-causal) and stable, since ∫_{-∞}^0 e^(at) dt = 1/a.

Step response: s(t) = ∫_{-∞}^{t} e^(aτ)u(-τ) dτ.

• For t ≤ 0: s(t) = e^(at)/a (rising to 1/a at t = 0).
• For t > 0: s(t) = 1/a (constant).

s(t)
1/a |           ______________
    |        /
    |    __/
    +--------+--------------> t
             0

The output begins before the input is applied (at t < 0), which shows that the system is non-causal and cannot be realised in real time.