(New page: 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) = <math>\int_{-\infty}^\infty h(\tau)x(t-\tau)\,d\tau</math> for simplicity. Then, for t...) |
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For 3 < t < 5 y(t) = (e^-9 - e^-3(t-3)) / 3 | 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 | 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) = <math>\int_{-\infty}^\infty h(\tau)x(t-\tau)\,d\tau</math> = <math>\int_0^{t-3} h(\tau)x(t-\tau)\,d\tau</math>, | ||
| + | so y(t) = [e^-9 - e^-3(t-3)] / 3. | ||
| + | For t > 5 y(t) = <math>\int_{-\infty}^\infty h(\tau)x(t-\tau)\,d\tau</math> = <math>\int_{t-5}^{t-3} h(\tau)x(t-\tau)\,d\tau</math>, | ||
| + | so y(t) = [e^-3(t-5) - e^-3(t-3)] / 3 | ||
Revision as of 16:43, 16 June 2008
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
