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Convolution of e^(-at)u(t) with u(t)

Short answer

An LTI system is linear and time-invariant; its output is the convolution of input and impulse response. For x(t)=e^(-at)u(t) and h(t)=u(t), y(t) = (1/a)(1 - e^(-at)) u(t).

Define LTI system. Find the convolution between x(t) = e^(-at)u(t), (a > 0) and h(t) = u(t).

LTI system: a system that is linear and time-invariant; its output is y(t) = x(t) * h(t) = ∫ x(τ) h(t - τ) dτ, where h(t) is its impulse response.

Convolution

x(t) = e^(-at) u(t), h(t) = u(t), a > 0.

y(t) = ∫ x(τ) h(t - τ) dτ = ∫ e^(-aτ) u(τ) u(t - τ) dτ

  • u(τ) restricts τ ≥ 0.
  • u(t - τ) restricts τ ≤ t.

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

t ≥ 0: y(t) = ∫₀^t e^(-aτ) dτ = [-e^(-aτ)/a]₀^t = (1 - e^(-at))/a

Answer: y(t) = (1/a)(1 - e^(-at)) u(t)

y(t)
1/a |       __________
    |    _-'
    |  /
    | /
----+------------------> t

The output rises from 0 and settles at 1/a as t → ∞ (the step response of the system, i.e. the integral of the exponential).

✓ Answer checked by Claude Opus 5.5 (high effort)

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