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Homework Help: How would I find the inverse laplace transform

  1. Oct 8, 2005 #1
    Hi How would I find the inverse laplace transform of this?

    [tex] I(s) = \left( \frac{1}{s(1+e^{-s})}\right) \left( \frac{1}{Ls+R}\right)[/tex]

    L, R are constants. I recognize the first term to be a geometric progression (square-wave function). With an infinite number of terms in that progression I don't think I could use convolution here. I could also try partial fractions but don't know how to do it with that exponential in the denominator. Someone please help.
  2. jcsd
  3. Oct 8, 2005 #2


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    Well I don't know if I've become too reliant on Mathematica but what I would do would be first to see what it reports, then work backward to see how it was figured out. But Mathematica can't solve it; suppose I could review all the techniques I can think of but well, curious if it's a simple matter that just the math jocks at Wolfram can't.:smile:
  4. Oct 8, 2005 #3
    The original question was to solve a differential equation for i(t) for LR series electrical circuit...

    [tex]L \frac{di(t)}{dt} + Ri(t) = E(t)[/tex]

    Given the condition i(0) = 0 and E(t) is the square wave function.

    So I looked up the square wave function and got E(t) = 1 - H(t-1) + H(t-2) - H(t-3)... where H(t) is the heavisdie function.And taking the LT of that, I get [tex] \frac{1}{s(1+e^{-s})}[/tex]

    So when I take the LT of the differential equation I got what I wrote above in post 1.

    Ya just the trouble is now getting i(t).

    I don't have any of the math programs so it's all lecture notes, pencil and paper i have to work with.
  5. Oct 8, 2005 #4


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    And so why not a convolution? When I convolve them, I get:

    [tex]i(t)=\frac{1}{L}\int_0^t E(\beta)e^{-R/L(t-\beta)}d\beta[/tex]
  6. Oct 8, 2005 #5
    But doesn't that integral have to be evaluated to get the answer? That's what I can't evaluate. Hence, I went looking to try partial fractions which I can't get either.
  7. Oct 8, 2005 #6


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    So we got:

    [tex]i(t)=\frac{1}{L}\int_0^t E(\beta)e^{-R/L(t-\beta)}d\beta[/tex]

    with E(t) being a square-wave.

    For now, let's just let L and R both be 1:

    What is i(t) in the interval [0,1]? Wouldn't that just be:

    [tex]i(t)=\int_0^t e^{-(t-\beta)}d\beta\quad\text{for}\quad t\in[0,1][/tex]

    What about in the interval [1,2]? So that would be:

    [tex]i(t)=\int_0^1 E(\beta)e^{-(t-\beta)}d\beta+\int_1^t E(\beta)e^{-(t-\beta)}d\beta[/tex]

    but the second integral is zero because E(t) is zero in that region so:

    [tex]i(t)=\int_0^1 e^{-(t-\beta)}d\beta\quad\text{for}\quad t\in[1,2][/tex]

    Can you figure out what i(t) would be for the next interval?
    Last edited: Oct 8, 2005
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