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Laplace nightmare

  1. Mar 20, 2006 #1
    ** disclaimer - not homework ** =)

    Is the following correct?

    I have a simple series RL circuit with a square wave input (periodic) defined by:

    Code (Text):

    e(t) = {10 0 <= t < 2
           { 0 2 <= t < 4
    With a period of 4

    E(s) = 1/(1-exp(-2as) * int (0 -> 2) exp(-st) f(t) dt
    Here the period = 2a thus a = 2

    E(s) = 1/(1-exp(-4s) * int (0->2) exp(-st) (10) dt
    E(s) = 1/(1-exp(-4s) * 10 * exp(-st)/-s | 0 -> 2
    E(s) = 10/s * (1-exp(-2s)/(1-exp(-4s)

    Since this is an RL circuit, the equation is

    E(s) = I(s) (R + sL)
    I(s) = E(s) / (R+sL)
    = 10/s * (1-exp(-2s)/(1-exp(-4s)(R+sL)
    = 10 / (s * (R+sL) * (1+exp(-2s))

    Now, I don't have a transform where the exponent is on the bottom - how do I get the inverse laplace transform?
  2. jcsd
  3. Mar 22, 2006 #2


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    Homework Helper

    I don't know that a closed form exists for the inverse transform but you can try one of the following.

    (i) Multiply numerator and denominator by exp(s) giving an exponential in the numerator and cosh(s) in the denominator - then scour the tables!

    (ii) Try developing the asymptotic behavior of the inverse transform using the stationary phase approximation.

    (iii) Try to find the inverse by numerical means.

    Good luck!
  4. Mar 22, 2006 #3
    i have doubts about your (II) quote in fact if we make a change of variable s=c+ix we have the next integral (but a factor 2pi):

    [tex] \int_{R}dxF(c+ix)e^{ixt} [/tex] with R=(-oo,oo) how do you apply

    in that case the "Asymptotic" behavior?...note that exp(ixt) is complex and Laplace method won,t work at all.
  5. Mar 23, 2006 #4


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    Science Advisor
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    How does that rule out stationary phase? We are free to deform the contour of integration in any manner we choose and find the portion that gives the greatest contribution to the integral.
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