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Laplace Transforms

by adamwitt
Tags: laplace, transforms
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adamwitt
#1
Jul23-11, 06:35 AM
P: 25
Ok, so I have uploaded/attached the question and the solution. I just need help understanding the solution please. I understand how to calculate the initial inverse transform, but I included it as the reference to the second part of the question regarding the y'' + 4y' = H(t-3)

Can someone please explain the full steps to invert the laplace transform Y to f, like the solution shows in the last step?
I have got to the following point, but I think I may be forgetting some Laplace Transform identities needed to make my life easier?

I can split the equation into parts where I recognise a few Laplace Transforms but not sure about the rest, cheers...

Y(s) = (1/s).e-3s.[1/(s2+22)] + [s/(s2+22)] - [2/s2+22]


I recognise a few inverse laplace transforms there but without adding my confusion to the mess can someone please clarify how they got the answer? many thanks in advance.
hopefully I didnt make this post toooo convoluted with my thoughts!
Attached Thumbnails
laplace transforms.png  
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icystrike
#2
Jul23-11, 11:48 AM
P: 436
U are confused because they have skipped a step.

I bet u must have recognised this part: [s/(s2+22)] - [2/s2+22]

For this part: (1/s).e-3s.[1/(s2+22)] :
(1/(s(s+2)))
=(1/4)((1/s)-(s/(s+2)))
=(1/4)(H(t-3)-cos(2(t-3)))
adamwitt
#3
Jul23-11, 12:27 PM
P: 25
Quote Quote by icystrike View Post
U are confused because they have skipped a step.

I bet u must have recognised this part: [s/(s2+22)] - [2/s2+22]

For this part: (1/s).e-3s.[1/(s2+22)] :
(1/(s(s+2)))
=(1/4)((1/s)-(s/(s+2)))
=(1/4)(H(t-3)-cos(2(t-3)))

How did you get rid of the e^(-3s) ?

thanks for helping.

icystrike
#4
Jul23-11, 11:01 PM
P: 436
Laplace Transforms

You are most welcome adam :)

[tex]L^{-1}(e^{-cs}F(s))=f(t-c)[/tex]
Given :
[tex] L^{-1}(F(s))=f(t) [/tex]



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