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y(t) = x(t)*h(t) | 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. | I used the integral y(t) = <math>\int_{-\infty}^\infty h(\tau)x(t-\tau)\,d\tau</math> for simplicity. | ||
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Then, for t<3 y(t)=0 | Then, for t<3 y(t)=0 | ||
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 | ||
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| + | ==> Comment: I think your lower bound should be 0 rather than 3 so that the answer to part 1 is <math>y(t) =\frac{1-e^{-3(t-3)}}{3}</math> | ||
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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 | ||
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'''B''' | '''B''' | ||
y(t) = dx(t)/dt*h(t) = [delta(t-3) - delta(t-5)]*[e^-3t u(t)] | y(t) = dx(t)/dt*h(t) = [delta(t-3) - delta(t-5)]*[e^-3t u(t)] | ||
For t < 3, y(t) = 0. | For t < 3, y(t) = 0. | ||
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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>, | 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. | so y(t) = [e^-9 - e^-3(t-3)] / 3. | ||
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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>, | 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 | so y(t) = [e^-3(t-5) - e^-3(t-3)] / 3 | ||
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| + | I don't think that this is right, but I'm not sure what part I need to change. | ||
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| + | ==>Comment: I'm not completly sure but I think since the derivative is two impulse functions you can just get <math>y(\tau)</math> by the equation <math>-h(\tau-5) + h(\tau-3) = - e^{-3(\tau-5)} + e^{-3(\tau-3)}</math>. I think you then may need to add the step functions to the answer so that it is general for all cases: <math>y(\tau) = e^{-3(\tau-3)}u(\tau-3) - e^{-3(\tau-5)}u(\tau-5)</math> | ||
Latest revision as of 22:53, 17 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
==> Comment: I think your lower bound should be 0 rather than 3 so that the answer to part 1 is $ y(t) =\frac{1-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
I don't think that this is right, but I'm not sure what part I need to change.
==>Comment: I'm not completly sure but I think since the derivative is two impulse functions you can just get $ y(\tau) $ by the equation $ -h(\tau-5) + h(\tau-3) = - e^{-3(\tau-5)} + e^{-3(\tau-3)} $. I think you then may need to add the step functions to the answer so that it is general for all cases: $ y(\tau) = e^{-3(\tau-3)}u(\tau-3) - e^{-3(\tau-5)}u(\tau-5) $
