Inverse laplace transform question

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  • #1
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How do i find the inverse laplace transform of s/(s-2)^2
 

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  • #2
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take the transform of [tex] f(t)=(1+t)e^{2t} [/tex]

using the "shift" property so..

[tex] F(s+a)=L[e^{-at}f(t)] [/tex] using

[tex] \frac{s+2}{s^{2}}=s^{-1}+2s^{-2} [/tex]

the inverse of above is just 1+t then multiply it by exp(2t) and you get it without recalling "residue theorem"...:tongue2:
 
  • #3
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Use partial fractions:

[tex]\frac{s}{(s-2)^2} = \frac{A}{(s-2)^2} + \frac{B}{s-2}[/tex]

To find A, multiply both sides by [tex]s-2[/tex] and evaluate at [tex]s=2[/tex]:

[tex]s = A[/tex]

[tex]A = 2[/tex]

Now to find B, go back to the original expression again, and multiply (again) both sides by [tex]s-2[/tex],
then differentiate with respect to s and evaluate at [tex]s=2[/tex]:

[tex]\frac{d}{ds}s = \frac{d}{ds}[B(s-2)][/tex]

[tex]B = 1[/tex]

Now plug A and B into the original expression:

[tex]\frac{2}{(s-2)^2} + \frac{1}{s-2}[/tex]

So the inverse Laplace would be:

[tex]L^{-1} = [2te^{2t}+e^{2t}] u(t)[/tex]
 

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