Output of transfer function given impulse response and input the

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To find the output of a transfer function given an impulse response and input, one must first compute the Laplace transforms of both x(t) and h(t), then multiply these transforms to obtain Y(s). The inverse Laplace transform is then used to find y(t). A key point of confusion is why certain terms, like e^5t, disappear; this is clarified by understanding that the one-sided Laplace transform assumes x(t) = 0 for t < 0, making it inapplicable for all time. Alternatives such as the Fourier transform and two-sided Laplace transform can also be used, with convolution being preferred for its effectiveness. Ultimately, convolution is highlighted as a superior method compared to the Fourier approach.
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Homework Statement


upload_2018-2-9_10-5-47.png


Homework Equations


Find laplace of x(t) and h(t)
Multiply the laplaced values to get Y(s)
Find inverse laplace to get y(t)

The Attempt at a Solution


upload_2018-2-9_10-17-34.png

In book the circled term isn't there. Why does that go away?
Book gets y(t) as
upload_2018-2-9_10-8-58.png

In laplace why does the e5t term go away?
 

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You will fin that the answer derives from the convolution integral. That method is valid for all time -∞ < t < ∞. The Laplace transform you're familiar with is most likely the one-sided Laplace transform which assumes x(t) = 0 for t < 0, so not applicable here.

There are two other alternatives: the Fourier transform and the two-sided Laplace transform. The latter is seldom encountered (see footnote) while the Fourier is appropriate and gives the same answer.

[Footnote: Then Brooklyn Poly's Professor John G. Truxal's venerable "Automatic Feedback Control System Synthesis" invokes it in preference to the Fourier].
 
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Thanks. Convolution is better than the Fourier route.
 

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