Inverse Laplace transform question:
Find Y
$$L(y)=\frac{s-1}{(s^2-1)((s-1)^2+1)}$$

I'm hitting a road block here

Question

Inverse Laplace transform question: 
Find Y
$$L(y)=\frac{s-1}{(s^2-1)((s-1)^2+1)}$$

I'm hitting a road block here

fibonaccichick666 · Answer

@zepdrix

fibonaccichick666 · Answer

wolfram is telling me it should be immaginary

fibonaccichick666 · Answer

@Kainui, @ganeshie8 not sure if this is your territory

hartnn · Answer

have you considered frequency shifting property ? 
$e^{at}f(t) <->   F(s-a)$

fibonaccichick666 · Answer

I'm not sure how that would work here

fibonaccichick666 · Answer

(I've never heard that particualr phrase either)

hartnn · Answer

$\Large L^{-1}[\dfrac{s-1}{(s^2-1)((s-1)^2+1)}] = e^t L^{-1}[\dfrac{s}{s(s+2)((s)^2+1)}]$

hartnn · Answer

after that normal partial fraction would work :)

fibonaccichick666 · Answer

ahh ok Thanks, I think that will work! I'll check back once I give it a shot

hartnn · Answer

$\Large \dfrac{A}{s+2} + \dfrac{Bs+C}{s^2+1}$

hartnn · Answer

sure, ask if any more doubts, now or later :)

fibonaccichick666 · Answer

Thank you! that was it!  Quick question if you shift it a times would it be $e^{at}$?

hartnn · Answer

thats correct.

if you see s-a in the s domain, you can plug out e^(at) in t domain

fibonaccichick666 · Answer

thank you!!!!!  I totally forgot about this