What is the Inverse Laplace Transform of (2s^2 + 20s + 36) / (s^3 + 6s^2 + 9s)?

minuswhale
Messages
2
Reaction score
0

Homework Statement



F(s) = (2s2 + 20s + 36) / (s3 + 6s2 + 9s)

Homework Equations



L[e-at] = 1 / (s + a)
L-1[(s+a) / ((s+a)2 + w2)] = e-atcos(wt)
L-1[w / ((s+a)2 + w2)] = e-atsin(wt)

The Attempt at a Solution



Factored the top by 2 and the bottom by s, but the rest of the top seems unfactorable. Stuck after that :frown:
]
 
Physics news on Phys.org
I would try factoring the denominator and then applying partial fractions, if you can.
 
Somehow I got from partial fraction 4 unknowns with 3 equations?

A/s + B/(s+3) + (Cs+D) / (s+3)2
 
You don't need the C in the last term. Your expansion should be

\frac{A}{s} + \frac{B}{s+3} + \frac{C}{(s+3)^2}

You use Cs+D in the numerator when the denominator is irreducible, e.g. x2+1.
 
^c=0
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top