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LTI output for h(t)=e^(-2t), rectangular input

Short answer

By convolution, y(t)=0 for t<0; y(t)=(1-e^(-2t))/2 for 0≤t≤1; y(t)=((e^2-1)/2)e^(-2t) for t>1.

Find the output of the LTI system having impulse response h(t) = e^(-2t); t > 0 to the input x(t) = 0 for t<0, 1 for 0≤t≤1, 0 for 1<t.

Given h(t) = e^(-2t) for t > 0 and x(t) = 1 for 0 ≤ t ≤ 1, else 0.

y(t) = x(t) * h(t) = ∫ x(τ) h(t - τ) dτ, with h(t - τ) = e^(-2(t - τ)) for τ < t.

x(t): 1 |‾‾‾‾‾|        h(t): 1 |\
        |     |                |  \__
      --+-----+---> t          +-------> t
        0     1                0

Region 1: t < 0

No overlap, so y(t) = 0.

Region 2: 0 ≤ t ≤ 1

The overlap is from τ = 0 to τ = t:

y(t) = ∫₀^t e^(-2(t - τ)) dτ = e^(-2t) [e^(2τ)/2]₀^t = e^(-2t)(e^(2t) - 1)/2

y(t) = (1 - e^(-2t)) / 2

Region 3: t > 1

The overlap is from τ = 0 to τ = 1:

y(t) = e^(-2t) [e^(2τ)/2]₀^1 = e^(-2t)(e² - 1)/2

y(t) = (e² - 1) e^(-2t) / 2

At t = 1 both give y(1) = (1 - e^(-2))/2 = 0.4323, so the output is continuous.

Answer:

        0                          t < 0
y(t) =  (1 - e^(-2t))/2            0 <= t <= 1
        (e^2 - 1) e^(-2t)/2        t > 1

The output rises to 0.432 at t = 1 and then decays exponentially.

✓ Answer checked by Claude Opus 5.5 (high effort)

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