Thread: Tricky inverse Laplace

G'day,

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



Thanks

Originally Posted by Jimmy_W

G'day,

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

Split it into partial fractions . 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



So for s = 1,

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



So for s = 1,

and so A = 4

but when i cant get B and C.

You should have had . 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