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!

Please, ive been stuck on this inverse laplace for awhile

  1. Oct 14, 2005 #1
    I have to find the laplace inverse of a function y(s) which has repeated complex roots.
    Y(s)=s^2 / (s^2+4)^2
    so s=2i, s=2i, s=-2i, s=-2i.

    My partial fraction is as follows:
    A/(s-2i) + B/(s-2i)^2 + C/(s+2j) + D/(s+2j)^2

    I use the standard method for finding regular repeated roots but I get stuck trying to calculating C and D. My values are undefined. My work is below...

    A= d/ds[(s-2i)^2*Y(s)]=8s(3s^2-4)/(s^2+4)^3 + 4s^2(s^2-12)*i/(s^2+4)^3-->then you set s=2i which then results in A=-6i.

    And B=1

    But now for C, when I use the same process as A but instead of 2i, I use -2i, my answer is a number over 0 which results in undefined.

    Am I even doing this problem correctly? Any help would be appreciated...

    Thanks!
     
  2. jcsd
  3. Oct 14, 2005 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    You didn't do your partial fraction decomposition correctly. Because the polynomials are quadratic you need terms like (As + B) in the numerators.

    BTW - if you are familiar with the Cauchy integral formula you could save yourself a lot of work and just evaluate the residues directly to evaluate the integral.
     
  4. Oct 14, 2005 #3
    I gave it a go.

    The values I have are:
    A=undefined
    B=1
    C=undefined
    D=1/4

    Checked with maple 10
    A=0
    B=1
    C=0
    D=-4

    However this was not done in complex numbers (for maple)

    For my results I followed the method of residues. When evaulating for the Unknowns A and C (where you use d/ds) both resulted in undefined which I chose to interpret as just leave it alone:approve:

    Obviously, you could evaluate this another way which would be advantageous right now but on a test where class lasts fifty minutes you need to know which method works most accurately. Since I am a novice at this I cannot be of much more help.
     
  5. Oct 15, 2005 #4
    Actually I just realized my Y(s)=s^2 / [(s^2+4)^2*(s^2+2)^2]
    so s=2i, s=-2i, s=sqrt(2)i, s=-sqrt(2)i
    Thus, resulting in no repeating roots.

    But I still have problems coming up with the coefficients.

    I think you setup the partial fraction like this:

    As+2B/(s^2+4)^2 + Cs+sqrt(2)D/(s^2+2)^2

    But I don't know where to go from there....
     
  6. Oct 15, 2005 #5
    Here is what I got

    Ok, given:

    Y(s) = s^2
    [ (s-j2)^2*(s+j2)^2*(s-jsqrt(2))^2*(s+jsqrt(2))^2]

    which equates to:

    =K1/(s-j2)^2 + K2/(s-j2) + {plus the conjugates of K1 and K2}

    Q1/(s-jsqrt(2))^2 + Q2/(s-jsqrt2) + {plus the conjugates of Q1 and Q2}

    Solving for K1, K2, Q1 ,Q2 using the method of residues.

    K1=6.1
    K2=0
    Q1=1/16
    Q2=0

    I do not know if this is correct but this is what I got. I really can't stand complex partial fraction decomposition.
     
  7. Oct 15, 2005 #6

    See, I haven't learned the method of residues...This problem is for my signals and systems class and I'm only allowed to use inverse laplace methods.
     
  8. Oct 15, 2005 #7
    If you you've taken circuit analysis 2 then you should have seen the method of residues. Maybe you should refer to it. If not google it. If you are interested I can explain it to you quickly.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Please, ive been stuck on this inverse laplace for awhile
Loading...