Laplace Transform

Question

Laplace Transform

anonymous · Answer

Finding the inverse transform of:

$$F(s) = \frac{ 3(s^3 + 2s^2 +4s + 1) }{ s(s+3)^2 }$$

Answer is:

$$f(t) = 3 \delta (t) + 1/3  + [20t - (37/3)]e^{-t}$$

anonymous · Answer

I don't know where the delta dirac function came from.

I do notice that the partial fraction numerator is one order lower than F(s)'s numerator.

anonymous · Answer

@KingGeorge  ^_^

kinggeorge · Answer

I think the dirac delta comes from the 3 in front. I haven't worked out the whole thing, but I think that's where it comes from.

kinggeorge · Answer

Or maybe not....

kinggeorge · Answer

Well, with a bit of help from wolfram, I think it's the fact that you have an $s^3$ in the numerator and the denominator. And the inverse laplace transform of a constant, is the dirac delta.

kinggeorge · Answer

If I write the whole thing out, and do polynomial division, I get$$1+\frac{3(s+1)}{s(s+3)^2}$$

anonymous · Answer

oh. i see. I should always divide when partial fraction numerator < F(s)'s numerator? 

because otherwise terms will be omitted by partial fractions

anonymous · Answer

thank you very much. :D

kinggeorge · Answer

I don't know if you should always do that, but it certainly made things clearer in this case, and you're welcome.

anonymous · Answer

I think I do need to.

I'm comparing a second order polynomial (partial fraction forced numerator) to a 3rd order polynomial (F(s)'s numerator).