# Inverse Laplace Help

1. Oct 9, 2011

### polarmystery

1. The problem statement, all variables and given/known data

Find the inverse laplace transform of the equation listed:

$$(\frac{1-e^{-sa}}{a})*(\frac{1}{s^2}-\frac{RC}{s}+\frac{sRC+R^2\frac{C}{L}-1}{s^2+s\frac{R}{L}+\frac{1}{LC}})$$

2. Relevant equations

a, R, C, and L are constants.

3. The attempt at a solution

Completely lost, not sure how to even begin to reduce this massive function.

2. Oct 9, 2011

3. Oct 9, 2011

### polarmystery

If I complete the square term on the bottom (s^2 + s + c) => (s + x)^2 + c, where does the additional x^2 term go that I added? Does it get added to the numerator also? It's been a while since I've done this stuff before.

4. Oct 9, 2011

### jackmell

So you want to transform:

$$\frac{s+k}{s^2+s+c}$$

I'd have to review that also but as far as completing the square, you'd write:

$$s^2+s+c=s^2+s+1/4-1/4+c=(s+1/2)^2+c-1/4$$

or to make it a little easier, just let f=c-1/4 so that we need to invert now:

$$\frac{s+k}{(s+1/2)^2+f}$$

Yeah, well I don't know how to do that one either. I'd have to review. I did it in Mathematica and it looks pretty messy but the more you work on these, the easier they get.

5. Oct 9, 2011

### polarmystery

Well, after completing the square and then subtracting out what I added to complete the square, I get this (It's a bit messy):

$$\frac{RC(s+\frac{R}{L}-1)}{(s+\frac{R}{2L})^2-\frac{R^2C^2-4LC}{4L^2C^2}}$$

6. Oct 9, 2011

### jackmell

Ok, that's a good start although you're quick to jump right back to the complicated problem. So Let me work a simple one that's related using the exponential shifting theorem:

Suppose I have:

$$\frac{s+k}{(s+1/2)^2+f}$$

I now write that as:

$$\frac{(s+1/2)+(k-1/2)}{(s+1/2)^2+f}$$

Now we can use the shifting theorem:

$$L^{-1}\{f(s-a)\}=e^{at}F(t)$$

where F(t) is the inverse transform of f(s).

You can figure that out.