inverse laplace transform

Discussion in 'Homework Help' started by ironmike828, Jun 10, 2010.

  1. ironmike828

    Thread Starter Member

    Jan 29, 2010
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    I've been stuck on this problem for a few days now and I'm not sure where im going wrong.

    Using Inv Laplace Transform determine the time domain equation.
    Y(s) = ((4s + 6)/(3s^2 + 6s + 51))

    1) I factored out the 3 from the denominator which resulted in (1.3s + 2)/(s^2 + 2s + 17)

    2) from there I solved for the roots resulting in an imaginary number
    >>s = -1 + j4 ; s = -1 - j4

    3) using partial fraction method I get 2 fractions
    (A)/(s+1-j4) ; B/(s+1+j4)

    Can I set up A(S+1+j4) + B(s + 1 -j4) = 1.3s + 2 ??
    this is where I get lost

    any help would be appreciated
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Are you familiar with the use of tables of Laplace transforms?
     
  3. ironmike828

    Thread Starter Member

    Jan 29, 2010
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    Yes I am familiar but I figured it out later last night. My problem was in setting up the partial fractions. I wasn't setting the common denominator correctly so that was throwing off my values by quite a bit. But thank you any way
     
  4. Georacer

    Moderator

    Nov 25, 2009
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    There is nothing wrong with your last equation. Just expand the parentheses and then regroup the factors. You should have one coefficient for "s", and once constant coefficient. Equate each one of these with zero and you get a 2x2 equation system which can be solved relatively easily, and will yield the values for A and B.
    I hope this helps.
     
  5. BenjaminSweet

    New Member

    Jun 4, 2010
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    I prefer not to solve complex Laplace transforms with complex partial fractions, I prefer to use the "Completed Square" form:

    F(s) = (A*s + B*wd)/([s + r]^2 + (wd)^2), where:
    * "r" is the real part of the complex pole, and
    * "wd" (Omega-d, the damped oscillation frequency) is the imaginary part of the complex pole.

    So in your case:
    Y(s) = (1.3s + 2)/(s^2 + 2s + 17) = (A*s + B*4)/([s + 1]^2 + (4)^2)

    To solve for the coefficients, put everything over a common denominator (which they already are...) and equate the numerators.

    The inverse Laplace transform pairs for this form (under-damped response) are the exponentially decaying Sine and Cosine.

    I hope this is helpful...
     
  6. Georacer

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    Nov 25, 2009
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    What is the reasoning while inputing the coefficients and the powers of s in front of A and B, in the right side of the equation?
     
  7. Georacer

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    Nov 25, 2009
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    To BenjaminSweet:
    There must be a fault in your line of thought. You clearly demonstrate that A=1.3 and B=0.5 which is wrong. By solving the system I proposed you get A=0.65-0.875j and B=0.65+0.0875j. These are the same results Matlab gives through the function "residue".

    To ironmike828:
    Partial Fractions actually are no help to you in this exercise, as at least one of A and B are imaginary numbers, and you cannot extract an inverse Laplace from them. You are far better off trying to bring the initial transfer function (your factored simplified version works best) to a sum of two of the standard forms found in this list:http://www.vibrationdata.com/Laplace.htm
    Try to guess the resulting time-domain function and the table will hint you about the needed form of the s-domain funtion.

    If you still have trouble finding it, ask further and I will help you.
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Actually if you go back to the original post we were given

    Y(s) = ((4s + 6)/(3s^2 + 6s + 51))

    In partial fraction form this is more accurately given by

    Y(s)=\frac{0.667+j0.0833}{s+1+j4}+\frac{0.667-j0.0833}{s+1-j4}

    I think BenjaminSweet was trying to find another different pair of unknown variables A & B which would fit the completed square form of Y(s) - he wasn't looking for the partial fraction coefficients.
     
  9. Georacer

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    Nov 25, 2009
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    Well, the two pairs of A and B you and I found give identical solutions. I checked it on Matlab. Knowing the solution of the problem though, I think what BenjaminSweet was looking for was the equation
    (1.3s + 2)=(A(s+1)+4B)
    in order to simlify the s-plane expressions for an easy inverse Laplace transformation.
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Sure - although your answers for A and B terms don't actually agree with mine - perhaps you had a typo for your value A ....? My other point was that the approximation of 4/3=1.3 by ironmike828 seemed a little rough. That's why I referred back to his original equation.

    Re-reading BenjaminSweet's post.

    He had .....

    Y(s) = (1.3s + 2)/(s^2 + 2s + 17) = (A*s + B*4)/([s + 1]^2 + (4)^2)

    - it seemed pretty clear he was trying re-cast the numerator to satisfy his desire to used the alternate form. His re-casting of the denominator form appears correct.
     
  11. Georacer

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    Nov 25, 2009
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    Duh! Obviously my set of A and B are wrong! I took ironmike828's sipmlification of the transfer function for granted and didn't notice that he equated 4/3 with 1.3. So that settles this matter.
    On the other hand, as it concerns BenjaminSweet's efforts, I see what he has done with the denumerator, and agree with it, but the numerator still eludes me. It's not that he has done anything forbidden, but I can't see how it helps us solve the problem. BenjaminSweet, could you clarify this for us, please?
     
  12. BenjaminSweet

    New Member

    Jun 4, 2010
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    Georacer, t_n_k,

    Yes, I was trying to put the partial fractions in a form form for an easy inverse transform. By "pre-multiplying" A by (s+1) and B by 4, once the values of A and B are found they will be the coefficients of the exponentially decaying Cosine and Sine respectively (rows 2.25 and 2.26 on the link provided by Georacer.)

    Note that the A and B value found this way will be different than the A and B values using the complex partial fractions, as t_n_k pointed out.

    If you put the complex partial fractions over a common denominator, I suspect that you will end-up with the original equation with (1.3s + 2) in the numerator (just a hunch, It's too late at night for me to try that right now...)

    I believe that this would lead to A=1.3 and B=0.5, as Georacer pointed out, and the inverse Laplace transform could be read directly off the table:
    y(t) = exp(-1t)*{1.3*cos(4t) + 0.5*sin(4t)}
     
  13. Georacer

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    Nov 25, 2009
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    You almost got it. It actually was <br />
e^{\small{-t}}\cdot(1.3\cdot\cos(4t)+0.175\cdot\sin(4t))<br />
     
  14. ericyeoh

    Active Member

    Aug 2, 2009
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    is the answer (4/3).exp^-t .cos 4t + (1/6) . exp^-t sin 4t ?
     
  15. t_n_k

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    Mar 6, 2009
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    That's correct.

    Alternatively .....

    1.34371e^{-t}sin(4t+82.875^{o})
     
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