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Math Help - Simplifying Laplace transform

  1. #1
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    Simplifying Laplace transform

    Hi,

    I need help simplifying the following so that i can use the Laplace transforms table

    LT[y] = (2/3s^2 -5s -2)/ (s^3 -3s^2 -1)

    Thank in advance,
    ArTiCk
    Last edited by ArTiCK; October 9th 2009 at 04:16 PM.
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  2. #2
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    Quote Originally Posted by ArTiCK View Post
    Hi,

    I need help simplifying the following so that i can use the Laplace transforms table

    LT[y] = (2/3s^2 -5s -2)/ (s^3 -3s^2 -1)

    Thank in advance,
    ArTiCk
    Are you sure this is the correct expression?
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  3. #3
    MHF Contributor chisigma's Avatar
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    If You write the LT as...

    \mathcal {L} \{y(t)\} = L(s)=\frac{N(s)}{D(s)} (1)

    ... where N(s) and D(s) are both polynomial in s, and You 'factorize' D(s) as follows ...

    D(s) = (s-\alpha_{1})\cdot (s-\alpha_{2})\dots (s-\alpha_{n}) (2)

    ... the LT can be written as...

    L(s) = \frac{\rho_{1}}{s-\alpha_{1}} + \frac{\rho_{2}}{s-\alpha_{2}} + \dots + \frac{\rho_{n}}{s-\alpha_{n}} (3)

    ... where...

    \rho_{n} = \lim_{s \rightarrow \alpha_{n}} (s-\alpha_{n})\cdot L(s) (4)

    In your case D(s) is a polynomial of order 3 and its roots are...

    \alpha_{1} = 3.1038034...

    \alpha_{2} = -.051901701... - i \cdot .56523585...

    \alpha_{2} = -.051901701... + i \cdot .56523585...

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by chisigma View Post
    If You write the LT as...

    \mathcal {L} \{y(t)\} = L(s)=\frac{N(s)}{D(s)} (1)

    ... where N(s) and D(s) are both polynomial in s, and You 'factorize' D(s) as follows ...

    D(s) = (s-\alpha_{1})\cdot (s-\alpha_{2})\dots (s-\alpha_{n}) (2)

    ... the LT can be written as...

    L(s) = \frac{\rho_{1}}{s-\alpha_{1}} + \frac{\rho_{2}}{s-\alpha_{2}} + \dots + \frac{\rho_{n}}{s-\alpha_{n}} (3)

    ... where...

    \rho_{n} = \lim_{s \rightarrow \alpha_{n}} (s-\alpha_{n})\cdot L(s) (4)

    In your case D(s) is a polynomial of order 3 and its roots are...

    \alpha_{1} = 3.1038034...

    \alpha_{2} = -.051901701... - i \cdot .56523585...

    \alpha_{2} = -.051901701... + i \cdot .56523585...

    Kind regards

    \chi \sigma
    IF the given expression is correct (and I doubt that it is) then the technique would be to factorise the denominator in a linear factor and an irreducible quadratic factor and then use a partical fraction decomposition. The difficulty of doing this convinces me that the given expression, wherever it has come from, is wrong.
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  5. #5
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    Quote Originally Posted by ArTiCK View Post
    Hi,

    I need help simplifying the following so that i can use the Laplace transforms table

    LT[y] = (2/3s^2 -5s -2)/ (s^3 -3s^2 -1)

    Thank in advance,
    ArTiCk
    Please repost with brackets so deployed to make the meaning of this explicit. As it stands you have (probably) posted:

    \mathcal{L}[y] = \frac{\frac{2}{3}s^2 -5s -2}{s^3 -3s^2 -1}

    or may-be:

    \mathcal{L}[y] = \frac{\frac{2}{3s^2} -5s -2}{s^3 -3s^2 -1}

    If you do not make yourself clear you are wasting our time

    CB
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  6. #6
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    Hi all,

    I would like to apologise for causing so much trouble. I tried to delete this post but i couldn't find a button to do so. As Mr. F has mentioned i have made a mistake earlier on in my calculations. I appreciate all the help that all you guys have given.

    Sorry once again,
    ArTiCk
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