Revision as of 16:44, 16 June 2008 by Kseeger (Talk)

Let x(t) = u(t-3) - u(t-5) and h(t) = e^-3tu(t)

A) y(t) = x(t)*h(t) I used the integral y(t) = $ \int_{-\infty}^\infty h(\tau)x(t-\tau)\,d\tau $ for simplicity.

Then, for t<3 y(t)=0 For 3 < t < 5 y(t) = (e^-9 - e^-3(t-3)) / 3

For t>5 y(t) = (e^-3(t-5) - e^-3(t-3)) / 3

B y(t) = dx(t)/dt*h(t) = [delta(t-3) - delta(t-5)]*[e^-3t u(t)] For t < 3, y(t) = 0.

For 3 < t < 5, y(t) = $ \int_{-\infty}^\infty h(\tau)x(t-\tau)\,d\tau $ = $ \int_0^{t-3} h(\tau)x(t-\tau)\,d\tau $, so y(t) = [e^-9 - e^-3(t-3)] / 3.

For t > 5 y(t) = $ \int_{-\infty}^\infty h(\tau)x(t-\tau)\,d\tau $ = $ \int_{t-5}^{t-3} h(\tau)x(t-\tau)\,d\tau $, so y(t) = [e^-3(t-5) - e^-3(t-3)] / 3

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