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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Originally Posted by Jimmy_W 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)} $ Split it into partial fractions $\displaystyle F(s) = \frac A{s-1} + \frac B{s+1} + \frac C{(s+1)^2}$. Then use your table of inverse Laplace transforms to find the inverse transform of each fraction.
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.
Originally Posted by Jimmy_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. You should have had $\displaystyle 6s^2 + 9s+1 \equiv A(s+1)^2 + B(s-1)(s+1) + C(s-1)$. Try it now.
I get A = 4, C=-2, B = 1. Is this correct?
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
Last edited by Calculus26; Aug 27th 2009 at 08:18 PM.
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