Getting stuck on a simple inverse Laplace transform. How do I massage this fraction?

I'm having some difficulty with what should be a simple (one sided) inverse Laplace transform:

$\displaystyle F(s) = \frac{2s}{s + 6}$

Now, I know the answer is $\displaystyle 2 \delta (t) - 12 e^{-6t} u(t)$

I'm just not sure how to get there. I'm not looking to apply the definition - I'm looking to massage the fraction into standard transform pairs or use standard properties of the transform.

It seems to me that I need to accomplish this:

$\displaystyle \frac{2s}{s + 6} = 2 - \frac{12}{s + 6}$

I'll probably kick myself for this, but what am I missing?

Re: Getting stuck on a simple inverse Laplace transform. How do I massage this fract

you can get it like this easily,

$\displaystyle \frac{2s}{s+6}=\frac{2(s+6)-12}{s+6}=\frac{2(s+6)}{s+6}-\frac{12}{s+6}=2-\frac{12}{s+6}$

then remember that

$\displaystyle L[\delta(t)]=1$ and $\displaystyle L[u(t)]=\frac{1}{s} $and$\displaystyle L[e^{at}f(t)]=F(s-a)$ where $\displaystyle L[f(t)]=F(s)$

Re: Getting stuck on a simple inverse Laplace transform. How do I massage this fract

Or "long division". s divides into 2s twice. 2 times s+ 6 is 2s+ 12 and subtracting: (2s+ 0)- (2s+ 12)= -12. s+ 6 divides into 2s twice with remainder -12:

$\displaystyle \frac{2s}{s+ 6}= 2+ \frac{-12}{s+ 6}$.

Re: Getting stuck on a simple inverse Laplace transform. How do I massage this fract

Quote:

Originally Posted by

**BAdhi**

you can get it like this easily,

$\displaystyle \frac{2s}{s+6}=\frac{2(s+6)-12}{s+6}=\frac{2(s+6)}{s+6}-\frac{12}{s+6}=2-\frac{12}{s+6}$

Yup, I was right. Kicking myself. This part of Laplace transforms is an art. Dammit. :)

Thanks for showing me what I was missing. :)