Transforming Cosine with Unit Step?

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Homework Statement



Find the Laplace transform of cos(t)h(t)
if the LT of h(t) (unit step) is 1/s


Homework Equations



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The Attempt at a Solution



Since the unit step isn't shifting anything anywhere, can I just treat it as 1? Seems too simple, haha. Any ideas? thanks.
 
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Here's a similar one - Find the LT of (e^-3t + 2)h(t)
with h(t) the same as before.

Would the h(t) just act as a 'turn on' in these cases?
 
Yes, and as you noted, it's superfluous with the usual Laplace transform. If you are using the double-sided transform, the presence of the step function would effectively give you the one-sided transform.
 
Ok, thanks.

meb09JW said:
Here's a similar one - Find the LT of (e^-3t + 2)h(t)
with h(t) the same as before.

So would the above LT simply be-

1/(s+3) + 2/s

?
 
Yup.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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