step Laplace

**pberardi** · April 18th 2009, 04:45 PM

Hi,
I am bit confused by the step function. I was under the impression that it is basically a plug and chug but I think I am doing something wrong and need some clarification and some guidance.

Find the L[usub2(t)e^(-t)cos(3t)
So I get e^(-2s)L[e^(-2t-4)cos(3t+6)]

I am almost sure that this is wrong but really don't have any examples that are similar. Can someone please help me with this?

**CaptainBlack** · April 18th 2009, 11:34 PM

Originally Posted by pberardi

Hi,
I am bit confused by the step function. I was under the impression that it is basically a plug and chug but I think I am doing something wrong and need some clarification and some guidance.

Find the L[usub2(t)e^(-t)cos(3t)
So I get e^(-2s)L[e^(-2t-4)cos(3t+6)]

I am almost sure that this is wrong but really don't have any examples that are similar. Can someone please help me with this?

$\mathcal{L} u_2(t) e^{-t}\cos(3t) =\int_2^\infty e^{-t}\cos(3t) e^{-st} dt$

Now change the variable to $\tau=t-2$ then we have:

$\mathcal{L} u_2(t) e^{-t}\cos(3t) =\int_0^\infty e^{-\tau-2}\cos(3\tau + 6) e^{-s\tau-2s} d\tau$

Simplify:

$\mathcal{L} u_2(t) e^{-t}\cos(3t) =e^{-2s-2}\int_0^\infty e^{-\tau}\cos(3\tau + 6) e^{-s\tau} d\tau=e^{-2s-2}\left[\mathcal{L}e^{-t}\cos(3t + 6)\right]$

Or you may prefereint in the form:

$\mathcal{L} u_2(t) e^{-t}\cos(3t)=e^{-2s}\left[\mathcal{L}e^{-t-2}\cos(3t + 6)\right]$

Alternativly fiddle with the translation theorem to get the same result.

CB

**pberardi** · April 19th 2009, 04:51 AM

Well the step after that is actually the confusing part. How does the exponential with the t mesh with the exponential with the s? Also are you taking the Laplace of the exp with the t? How does that come out? Thank you for your assistance but it is from this part on that I get confused about.

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