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Math Help - Inverse Laplace transform

  1. #1
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    Inverse Laplace transform

    Could anybody please point me in the right direction with this inverse laplace transform?

    L-1 {(e^(-0.165s)) / ((1+0.63s)s)

    i think its a heaviside function

    so H(t-a)= H(t-0.165)

    which leaves me with

    1 / ((1+0.63s)s)

    and this where i'm stuck (No doubt the solution is staring me in the face!)
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  2. #2
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    Quote Originally Posted by al2308 View Post
    Could anybody please point me in the right direction with this inverse laplace transform?

    L-1 {(e^(-0.165s)) / ((1+0.63s)s)

    i think its a heaviside function

    so H(t-a)= H(t-0.165)

    which leaves me with

    1 / ((1+0.63s)s)

    and this where i'm stuck (No doubt the solution is staring me in the face!)
    Let LT^{-1}[F(s)] = f(t).

    Then LT^{-1}[e^{-as} F(s)] = f(t - a) H(t - a).

    In your case you have the form F(s) = \frac{1}{s(bs + 1)} so you need to find LT^{-1} \left[\, \frac{1}{s(bs + 1)} \, \right].

    To do this, express \frac{1}{s(bs + 1)} in the partial fraction form \frac{A_1}{s} + \frac{A_2}{bs + 1}.

    Note that \frac{A_2}{bs + 1} = \frac{A_2/b}{s - \left( -\frac{1}{b} \right)}.
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  3. #3
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    I thought this one looked simple but i was wrong again!

    I've split it down into the partial fractions

    A/s + B/(bs+1)

    so

    1=A(bs+1) + Bs

    let S=0

    A=1

    expanding the equation

    1=Abs+A+Bs

    equating the coefficients

    0=Ab+B

    as A=1

    then B=-b

    i think this is right so far (but i'm often wrong )

    this leaves me with the following to transform

    (L-1) 1/s - b/(bs+1)

    or

    (L-1) 1/s - 1/(s-(-1/b))

    = 1- e^-((1/b)t)


    is this near the mark?
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  4. #4
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    Quote Originally Posted by al2308 View Post
    I thought this one looked simple but i was wrong again!

    I've split it down into the partial fractions

    A/s + B/(bs+1)

    so

    1=A(bs+1) + Bs

    let S=0

    A=1

    expanding the equation

    1=Abs+A+Bs

    equating the coefficients

    0=Ab+B

    as A=1

    then B=-b

    i think this is right so far (but i'm often wrong )

    this leaves me with the following to transform

    (L-1) 1/s - b/(bs+1)

    or

    (L-1) 1/s - 1/(s-(-1/b))

    = 1- e^-((1/b)t)


    is this near the mark?
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