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Math Help - Laplace Transform and Heaviside Function

  1. #1
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    Laplace Transform and Heaviside Function

    Hello, This is my first post here, after seeing that a lot of great help is on hand, so i thought i would give it a try as i am having difficulty.

    question: Find the Laplace Transform of the following from first principles;

    <br /> <br />
f(x)=(x+2)^{2}H(x-3)<br /> <br />

    After looking at some other examples on the forum, I have tried letting  g(t)=(t+5) in order to make the two expressions inside the brackets the same, ie  (x-3) so that standard rules can be applied, though I am not quite sure if this is right or where to go from here. ANy help would be much appreciated!

    Thanks in advance.
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    Progress so far

    In order to give you a batter idea of where my train of thought is, this is the answer I have got, through the method below;

    let  g(t)=(t^{2}+4t+1)

    therefore  f(x)=(x-3)H(x-3)

    so, from first principles, although, using rule:  <br /> <br />
\begin{aligned}L\left\{f(x)\right\}&=e^{-3s}\bigg[L\left\{x-3\right\}\bigg]\\ &=e^{-3s}\left(1/s^2-3/s\right) \\ &=\frac{e^{-3s}}{s^2}-\frac{3e^{-3s}}{s}\end{aligned}<br /> <br />

    The question also asks to "Indicate for what values of 's' is the transform valid?" I am not sure what this means, is it something involving the solution of the  g(t) function?

    Again, thanks a lot for any help or pointers or letting me know I am barking completely up the wrong tree!
    Last edited by arickard1988; October 19th 2009 at 02:51 AM.
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  3. #3
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    Quote Originally Posted by arickard1988 View Post
    Hello, This is my first post here, after seeing that a lot of great help is on hand, so i thought i would give it a try as i am having difficulty.

    question: Find the Laplace Transform of the following from first principles;

    <br /> <br />
f(x)=(x+2)^{2}H(x-3)<br /> <br />

    After looking at some other examples on the forum, I have tried letting  g(t)=(t+5) in order to make the two expressions inside the brackets the same, ie  (x-3) so that standard rules can be applied, though I am not quite sure if this is right or where to go from here. ANy help would be much appreciated!

    Thanks in advance.
    Form first principles:

    \mathcal{L}f(s)=\int_{x=0}^{\infty}(x+2)^2H(x-3)e^{-sx}\;dx=\int_{x=3}^{\infty}(x+2)^2e^{-sx}\;dx

    which you will now need to evaluate.

    CB
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    Form first principles:

    \mathcal{L}f(s)=\int_{x=0}^{\infty}(x+2)^2H(x-3)e^{-sx}\;dx=\int_{x=3}^{\infty}(x+2)^2e^{-sx}\;dx

    which you will now need to evaluate.

    CB
    And integration of the final integral found by CB can be avoided by making the substitution u = x - 3, take out the 'constant' of e^{-3s}, expand (u + 5)^2 and then express the integral as three seperate integrals (each of which is recognised as the Laplace transform of a simple function).


    Edit: Addition in italics (added to clarify my remark in light of CB's quite correct subsequent post).
    Last edited by mr fantastic; October 18th 2009 at 11:07 PM.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    And integration can be avoided by making the substitution u = x - 3, take out the 'constant' of e^{-3s}, expand (u + 5)^2 and then express the integral as three seperate integrals (each of which is recognised as the Laplace transform of a simple function).
    To take the LT from first principles requires that he construct the integrals given in my post, I don't care how he then evaluates them (but he cannot look them up in a table of LTs as that would not be doing this from first principles).

    CB
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    Quote Originally Posted by CaptainBlack View Post
    Form first principles:

    \mathcal{L}f(s)=\int_{x=0}^{\infty}(x+2)^2H(x-3)e^{-sx}\;dx=\int_{x=3}^{\infty}(x+2)^2e^{-sx}\;dx

    which you will now need to evaluate.

    CB
    CB,

    Many thanks! Its now crystal, and has helped with the rest of the paper.

    Alex
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    And integration of the final integral found by CB can be avoided by making the substitution u = x - 3, take out the 'constant' of e^{-3s}, expand (u + 5)^2 and then express the integral as three seperate integrals (each of which is recognised as the Laplace transform of a simple function).


    Edit: Addition in italics (added to clarify my remark in light of CB's quite correct subsequent post).
    Fantastic,

    Thankyou for the add-on advice, has wrapped it up nicely, seems trivial now, as always!

    I look forward to using this forum again, a GREAT tool in helping me learn.

    Alex
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