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

1. ## 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:

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

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

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:

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

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

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

you can get it like this easily,

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

then remember that

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

3. ## 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:
$\frac{2s}{s+ 6}= 2+ \frac{-12}{s+ 6}$.

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

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