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

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Homework Help Overview

The discussion revolves around finding the inverse Laplace transform of the function F(s) = (2s^2 + 20s + 36) / (s^3 + 6s^2 + 9s), which involves concepts from Laplace transforms and partial fraction decomposition.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss factoring the numerator and denominator, with some attempting partial fraction decomposition. There are questions about the correct form of the partial fraction expansion and the necessity of certain terms.

Discussion Status

The discussion is ongoing, with participants providing suggestions on factoring and partial fractions. There is a recognition of confusion regarding the number of unknowns and equations in the partial fraction setup, indicating a productive exploration of the problem.

Contextual Notes

Participants are navigating the complexities of the partial fraction decomposition, with some noting the irreducibility of certain terms and the implications for the expansion.

minuswhale
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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:
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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

[tex]\frac{A}{s} + \frac{B}{s+3} + \frac{C}{(s+3)^2}[/tex]

You use Cs+D in the numerator when the denominator is irreducible, e.g. x2+1.
 
^c=0
 

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