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Math Help - Inverse Laplace of this function

  1. #1
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    Inverse Laplace of this function

    Hi,

    I've looked through the transform tables, and haven't been able to figure out how to find the inverse:

    Thanks for any help.
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  2. #2
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    (s+a)e^{-(a+b)t}
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by algorithm View Post
    Hi,

    I've looked through the transform tables, and haven't been able to figure out how to find the inverse:

    Thanks for any help.
    What exactly are you trying to find the inverse transform of? What's the X(s) doing there?

    If the problem is to find the inverse transform of
    \frac{s + a}{s + a + b}

    Note that
    \frac{s + a}{s + a + b} = 1 - \frac{b}{s + a + b}

    The second term is simple enough. You may need to look up the inverse transform of 1. It is \delta (t).

    -Dan
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  4. #4
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    Hi

    Thanks.

    Quote Originally Posted by lilaziz1 View Post
    (s+a)e^{-(a+b)t}
    Should the "s" be "n(t)"?

    Quote Originally Posted by topsquark
    What exactly are you trying to find the inverse transform of? What's the X(s) doing there?
    X(s) is an arbitrary function. I want to see how it is affected in the time domain by this multiplication. For example, if I had "sX(s)", then in the time domain it would be d/dt of x(t).
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by algorithm View Post
    Hi,

    I've looked through the transform tables, and haven't been able to figure out how to find the inverse:

    Thanks for any help.
    If that is the cases you can use the convolution theorem to bring this back to the time domain. Using Topsquark's idea gives

    X(s)\frac{s+a}{s+a+b}=X(s)\frac{s+a+b-b}{s+a+b}=X(s)-X(s)\frac{b}{s+a+b}

    So now taking the inverse transform gives

    x(t) - \int_{0}^{t}e^{-(a+b)(t-\tau)}X(\tau)d\tau=x(t) - be^{-(a+b)t}\int_{0}^{t}e^{(a+b)\tau}X(\tau)d\tau
    Last edited by TheEmptySet; May 8th 2011 at 03:19 PM.
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