Inverse Laplace transform (Initial Value Problem)

  • #1

Homework Statement



I'm stuck trying to find out the inverse Laplace of F(s) to get y(t) (the solution for the differential equation):

Y(s) = 1 / [ (s-1)^2 + 1 ]^2



The Attempt at a Solution



I tried using a translation theorem and then apply the sine formula, but the denominator is still all squared. I also tried partial fractions to expand Y(s) but I didn't get it right...

Any suggestions please?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



I'm stuck trying to find out the inverse Laplace of F(s) to get y(t) (the solution for the differential equation):

Y(s) = 1 / [ (s-1)^2 + 1 ]^2



The Attempt at a Solution



I tried using a translation theorem and then apply the sine formula, but the denominator is still all squared. I also tried partial fractions to expand Y(s) but I didn't get it right...

Any suggestions please?
Yes: show us what you did and what you got. How can we possibly help if we have no idea of what your problem is?
 
  • #3
96
3
[itex][(s-1)+1]^2=[(s-1+i)(s-1-i)]^2[/itex]

Therefore, [tex]\left[\frac{1}{ (s-1-i)(s-1+i) }\right]^2=\left[\frac{1}{2i}\left(\frac{1}{s-1-i}-\frac{1}{s-1+i} \right ) \right ]^2[/tex]

Can you continue it from there?
 
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  • #4
Yes: show us what you did and what you got. How can we possibly help if we have no idea of what your problem is?

My friend, that's the thing, I can't get past that. I stated the problem clearly: "I'm stuck trying to find out the INVERSE LAPLACE of the given Y(s) to get y(t)". There's no point in showing what I did before, it's not needed for what's after the equation I posted.

Anyway, I got it. It was just applying one of the formulas from the Laplace transform tables. I thought partial fractions or something else had to be done.

Thanks for your help
 

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