Question

Find the inverse transform using partial fractions and the shifting theorem:

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

Answer 1

Laplace transform*

Answer 2

Well, quick, concise, right to the point -- these are the Laplace transform problems I like.$$\frac1{(s+3)^2(s-2)^2}=\frac{A}{s+3}+\frac{B}{(s+3)^2}+\frac{C}{s-2}+\frac{D}{(s-2)^2}$$

Answer 3

hold on I hate typing math. I'm doing it on a separate sheet of paper.

Answer 4

I'm counting on you man. Anything you can help me with would be appreciated.

Answer 5

\[\frac{1}{(s+3)^2(s-2)^2}=\frac{2}{125(s+3)}+\frac{1}{25(s+3)^2}-\frac{2}{125(s-2)}+\frac{1}{25(s-2)^2}\]

Answer 6

@SithsAndGiggles well yes all of us can use WolframAlpha do to it for us...

Answer 7

Yep, I prefer leaving the work to the askers.

Answer 8

I think the reason it hasn't been working for me for the past hour is that I'm collecting my coefficients incorrectly... the system I've ended up with I have correctly solved but they aren't the coefficients.

Answer 9

\(1=A(s+3)(s-2)^2+B(s-2)^2+C(s+3)^2(s-2)+D(s+3)^2\)

Letting \(s=-3\), you'll find that \(B=\dfrac{1}{25}\), and letting \(s=2\) gives you \(D=\dfrac{1}{25}\) as well.

Next, I'd try letting \(s=0\):
\(1=12A+\dfrac{4}{25}-18C+\dfrac{9}{25}\\
\color{red}{\dfrac{2}{25}=2A-3C}\)

Letting \(s=1\):
\(1=4A+\dfrac{1}{25}-16C+\dfrac{16}{25}\\
\color{blue}{\dfrac{2}{25}=A-4C}\)

Which tells you \(A-4C=2A-3C~\Rightarrow~ A=-C\). Substituting into the red equation:
\(\dfrac{2}{25}=5A\\
A=\dfrac{2}{125}~\Rightarrow~C=-\dfrac{2}{125}\)

And thus the result I got from WA.

Answer 10

@SithsAndGiggles for some reason I convinced myself that wouldn't work in this case... ah, I did eventually finish it using Heaviside's cover-up method.

Answer 11

From here, it looks like the inverse Laplace transform has two purely exponential terms and two exponential/trig terms.

@oldrin.bataku, I usually use the method that involves matching up coefficients of like-powered terms, but I don't really like multiplying polynomials.