# Tricky inverse Laplace

• Aug 27th 2009, 07:40 AM
Jimmy_W
Tricky 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 AM
Opalg
Quote:

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.
• Aug 27th 2009, 09:38 AM
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.
• Aug 27th 2009, 10:59 AM
halbard
Quote:

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.
• Aug 27th 2009, 07:16 PM
Jimmy_W
I get A = 4, C=-2, B = 1. Is this correct?
• Aug 27th 2009, 08:01 PM
Calculus26
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