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Inverse Laplace Tranform

  • Thread starter chocok
  • Start date
  • #1
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<Inverse Laplace Transform>

Hi, I was given this: F(z) = 1/(z^2)(z^2+1) and was asked to use its residues to compute f(t). [z is a complex number]

The answer I got is -sin(t). But the answer on the book says t-sin(t).
I double checked and the residue of z=0 is 0, I don't get where the t is from.
Please, can anyone help? Thanks
 

Answers and Replies

  • #2
benorin
Homework Helper
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[tex]\frac{1}{z^2(z^2+1)}=\frac{1}{z^2}-\frac{1}{z^2+1}[/tex]

hence f(t)=t-sin(t).
 

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