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January 8th, 2018, 03:41 AM   #1
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Inverse laplace transform, formula help me a little ;)

I have to find y, but I don't have a formula to solve that one, if anyone have an idea...
y=L−1[(e−2πs*s−2)/(s2+4)]

Last edited by skipjack; January 8th, 2018 at 06:05 AM.
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January 8th, 2018, 04:00 AM   #2
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Just to make sure, you are talking about
$$\mathcal{L}^{-1}\left(\frac{e - 2\pi s^2 - 2}{s^2 + 4}\right)$$

Last edited by Micrm@ss; January 8th, 2018 at 04:03 AM.
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January 8th, 2018, 04:30 AM   #3
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oops no, it is not that, it is hard to write it here but that one is correct
L−1[(e^(-2πs-2s))/(s^2-4)]
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January 8th, 2018, 06:25 AM   #4
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Use the entries numbered 17 and 27 in this list of Laplace transforms.
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January 8th, 2018, 08:31 AM   #5
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Quote:
Originally Posted by Justinetea View Post
oops no, it is not that, it is hard to write it here but that one is correct
L−1[(e^(-2πs-2s))/(s^2-4)]
Is −1[(e^(-2πs-2s))/(s^2-4)] a multiplication by -1?
If so, why do you show the "1"?
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January 8th, 2018, 08:57 AM   #6
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L-1 is an inverse of laplace transformation it should be L to the power of -1

Last edited by Justinetea; January 8th, 2018 at 09:02 AM.
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January 8th, 2018, 09:18 AM   #7
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Well, post it this way: L^(-1) or 1/L
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January 8th, 2018, 10:09 AM   #8
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Quote:
Originally Posted by Denis View Post
Well, post it this way: L^(-1) or 1/L
I'm afraid I have to disagree with you Sir Denis.

Writing the inverse of a transform $\mathscr{L}$ as $\dfrac {1}{\mathscr{L}}$ gets you sent to the corner.

If I was writing in charcoal in my cave I would write it as L^(-1){} (your first choice)

Using LaTex I'd write it as $\mathscr{L}^{-1}()$
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January 8th, 2018, 11:19 AM   #9
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Quote:
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using latex i'd write it as $\mathscr{l}^{-1}()$
yuk!!
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January 8th, 2018, 03:20 PM   #10
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Quote:
Originally Posted by Denis View Post
yuk!!
hrm.. you didn't see the script capital L?

what do you see here $\mathscr{L}$
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