What is the Laplace Transform of cos(3t)*u(t)?

AI Thread Summary
The Laplace transform of h(t) = cos(3t) * u(t) is derived using the standard formula for the Laplace transform of cosine functions. The function f(t) is identified as cos(3t), leading to F(s) = s / (s^2 + 9). Since the unit step function u(t) does not affect the transform when evaluated at t = 0, the final result for H(s) remains H(s) = F(s) = s / (s^2 + 9). This confirms the correct application of the Laplace transform for the given function. The discussion emphasizes the consistency of the Laplace transform with established mathematical principles.
JonForbes
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Homework Statement



I need to know what the laplace transform is for f(t) * u(t)

More specifically, h(t) = cos(3t)*u(t)

Homework Equations



I know that f(t-to) * u(t-to) = e-tos * F(s)

The Attempt at a Solution



So, f(t) = cos(3t)

F(s) = s/(s2 + 9)

In the formula I gave in (2), I would then think that to = 0, because we just are multiplying f(t) by u(t - 0).

So H(s) = e-0*s * F(s)

H(S) = 1* F(s) = s/(s2 + 9)

I think this is right but could someone confirm this?

Thanks!
 
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H(s)=*L(cos(3t)(s))=s/(s^2+9)
because L(cos(omega*t)(s)=s/(s^2+omega^2) for real s and s/(s^2+a) for complex s
 
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