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April 20th, 2017, 04:35 PM   #1
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Inverse Laplace of 1.5s/(s+100). s in the numerator makes me crazy!

Hello,

Inverse Laplace of 1.5s/(s+100). i would like to know the steps of solving this problem. s in numerator makes me confused.. which theorem i should use? any information that could help, is highly appreciated thanks in advance!
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April 20th, 2017, 04:37 PM   #2
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Should be something in the tables for $\dfrac{s}{s + a}$. The 1.5 just multiples the whole expression.
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Last edited by Joppy; April 20th, 2017 at 04:46 PM.
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April 20th, 2017, 04:48 PM   #3
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there is nothing in the tables with first order equations and the s in the numerator... Could you write integral that i need to solve?
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April 24th, 2017, 07:05 PM   #4
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Forgot about this sorry.

Note that,

$\dfrac{1.5s}{s + 100} = 1.5 \left[ 1 - \dfrac{100}{s + 100} \right]$

Since the Laplace transform is a linear operator, you can inverse term by term,

$1.5 \left[ \mathcal{L}^{-1} \left[ 1 \right] - \mathcal{L}^{-1} \left[ \dfrac{100}{s + 100} \right] \right] = 1.5(\delta(t) - 100e^{-100t}).$

It is very often the case with inverse Laplace exercises that you need to first split up the terms through partial fractions or other methods.
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