1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inverse Laplace Help

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the inverse laplace transform of the equation listed:

    [tex](\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}})[/tex]


    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. jcsd
  3. Oct 9, 2011 #2
     
  4. Oct 9, 2011 #3
    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.
     
  5. Oct 9, 2011 #4
    So you want to transform:

    [tex]\frac{s+k}{s^2+s+c}[/tex]

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

    [tex]s^2+s+c=s^2+s+1/4-1/4+c=(s+1/2)^2+c-1/4[/tex]

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

    [tex]\frac{s+k}{(s+1/2)^2+f}[/tex]

    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.
     
  6. Oct 9, 2011 #5
    Well, after completing the square and then subtracting out what I added to complete the square, I get this (It's a bit messy):

    [tex]\frac{RC(s+\frac{R}{L}-1)}{(s+\frac{R}{2L})^2-\frac{R^2C^2-4LC}{4L^2C^2}}[/tex]
     
  7. Oct 9, 2011 #6
    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:

    [tex]\frac{s+k}{(s+1/2)^2+f}[/tex]

    I now write that as:

    [tex]\frac{(s+1/2)+(k-1/2)}{(s+1/2)^2+f}[/tex]

    Now we can use the shifting theorem:

    [tex]L^{-1}\{f(s-a)\}=e^{at}F(t)[/tex]

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

    You can figure that out.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Inverse Laplace Help
Loading...