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LTI output for h(t)=e^(-at)u(t), x(t)=u(t)-u(t-2)

Short answer

Output: y(t)=0 (t<0), y(t)=(1-e^(-at))/a (0≤t<2), y(t)=(e^(2a)-1)e^(-at)/a (t≥2).

Sketch the output of Linear Time invariant system having impulse response h(t) = e^(-at)u(t), for a > 0 and input signal x(t) = u(t) - u(t-2).

h(t) = e^(-at) u(t), a > 0; x(t) = u(t) - u(t - 2), a rectangular pulse of unit height from 0 to 2.

By linearity and time invariance, the output is the step response s(t) minus its 2-second delayed copy. The step response is

s(t) = h(t) * u(t) = (1 - e^(-at))/a for t ≥ 0 (and 0 for t < 0).

y(t) = s(t) - s(t - 2)

Region 1: t < 0

y(t) = 0.

Region 2: 0 ≤ t < 2

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

Region 3: t ≥ 2

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

Answer:

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

Sketch

y(t)
 (1-e^-2a)/a |        .-'‾'-.
             |      /         '-.
             |    /               '-.__
             |  /                       '-.___
 ------------+------------+------------------> t
             0            2

The output charges exponentially while the pulse is on, peaks at t = 2 with y(2) = (1 - e^(-2a))/a, then decays exponentially.

✓ Answer checked by Claude Opus 5.5 (high effort)

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