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Homework Help: 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

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    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.
     
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