Inverse Laplace Transform: Solve Homework Equation

[polarmystery]

Homework Statement

Find the inverse laplace transform of the equation listed:

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

Homework Equations

a, R, C, and L are constants.

The Attempt at a Solution

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

 

[jackmell]

You can start by starting small then building back up. First consider:

$$(1-e^{-s})(\frac{1}{s^2}-\frac{1}{s}+\frac{s+k}{s^2+s+c})$$

You can do that one right? Just break it up into individual pieces and compute the inverse transform of each piece. Get that working, then start adding all those constants. Won't change it much right since they're just constants.

 

[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.

 

[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.

 

[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):

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

 

[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.