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Math Help - series RL analysis using Laplace

  1. #1
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    series RL analysis using Laplace

    Hi Guys,
    Could anyone offer some assistance please?

    I have a question which states the following:
    L[v(t)] = V(s) = 1/(1+s^s)(1-e^(pi*s)
    if an inductor (1 H) and a resistor (1 Ohm) are connected in series then show that the resulting current is,

    i(t)=sigma(n=0,infinity).f(t-n*pi)
    where f(t) =(sin(t)-cos(t)+e^(-t))u(t)

    i am able to analysis the circuit to give the following:
    L[di/dt] +L[Ri]=L[v(t)]
    giving I(s)=1/(s+1)(s^2+1)(1-e^(pi*s))

    using partial fractions it becomes:
    I(s)=(1/(1-e^(-pi*s)))*0.5*((-s/(s^2+1))+(1/(s^2+1))+(1/(s+1))

    I am really not sure where to go next, I am sure that 1/(1-e^(-s*pi)) is some kind of geometric series but other than that I am stumped!<br><br>I would really appreciate it if someone could point me in the right direction.
    Regards,

    Mckiseldeck
    Last edited by Mckiseldeck; March 23rd 2013 at 10:11 AM.
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  2. #2
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    Re: series RL analysis using Laplace

    Hey Mckiseldeck.

    Have you considered using repeated applications of the convolution theorem?
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  3. #3
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    Re: series RL analysis using Laplace

    Quote Originally Posted by chiro View Post
    Hey Mckiseldeck.

    Have you considered using repeated applications of the convolution theorem?
    Hi Thanks for reply,

    Please could you elaborate?

    I know L-1{F(s)G(s)}=(f*g)(t) but I don not know what you mean in this instance.

    Regards,

    Mckiseldeck
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  4. #4
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    Re: series RL analysis using Laplace

    Basically I mean this (note that Laplace transforms satisfy having this theorem too):

    Convolution theorem - Wikipedia, the free encyclopedia
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