G'day,

I struggle with Laplace transforms and calculating their inverses. Could someone please help me with the following?

$\displaystyle F(s) = \frac{6s^2 + 9s +1}{(s+1)^2 (s-1)} $

Thanks

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- Aug 27th 2009, 07:40 AMJimmy_WTricky inverse Laplace
G'day,

I struggle with Laplace transforms and calculating their inverses. Could someone please help me with the following?

$\displaystyle F(s) = \frac{6s^2 + 9s +1}{(s+1)^2 (s-1)} $

Thanks - Aug 27th 2009, 08:05 AMOpalg
- Aug 27th 2009, 09:38 AMJimmy_W
I havent gotten far, even with your help.

So from your post I have

$\displaystyle 6s^2 + 9s+1 = A(s+1)^2 + B(s-1)(s+1)^2 + C (s-1)(s+1)^2 $

So for s = 1,

$\displaystyle 16 = 4A $ and so A = 4

but when i cant get B and C. - Aug 27th 2009, 10:59 AMhalbard
- Aug 27th 2009, 07:16 PMJimmy_W
I get A = 4, C=-2, B = 1. Is this correct?

- Aug 27th 2009, 08:01 PMCalculus26
http://www.mathhelpforum.com/math-he...e009cbca-1.gif

6s^2 + 9s +1 = 6(s+1)^2 - 3(s+1) -2

F(s) = 6/(s-1) -3/[(s+1)(s-1)] -2/[(s+1)^2(s-1)]

The first 2 transforms are standards the last could be done fairly simple with convolution

f(t) = -2te^(-t) g(t) = e^t