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Math Help - Laplace transforms

  1. #1
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    Laplace transforms

    Using only the definition of the Laplace transform, and showing your work, find the Laplace transform of:

    (t^2)u(t-(1/4) where u is the Heaviside function.



    I don't know what it means for u to be the Heaviside function and I have the definition of Laplace transforms right in front of me, and I am not sure how to use it to solve this. Any help would be greatly appreciated. Thanks!
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Hollysti View Post
    Using only the definition of the Laplace transform, and showing your work, find the Laplace transform of:

    (t^2)u(t-(1/4) where u is the Heaviside function.



    I don't know what it means for u to be the Heaviside function and I have the definition of Laplace transforms right in front of me, and I am not sure how to use it to solve this. Any help would be greatly appreciated. Thanks!
    The Heaviside function is defined as the following piecewise function:
    u(t) = {0 for t<0, 1 for t>0}
    u(t - a) = {0 for t<a, 1 for t>a} ... (t - a) represents a shift of the Heviside function to the right "a" units.

    When multiplying the Heaviside function by some other function f(t), we get:
    f(t)*u(t) = {0*f(t)=0 for t<0, 1*f(t)=f(t) for t>0}

    Now, taking the Laplace of this, where f(t) = t^2, and a = 1/4
    L{t^2*u(t - 1/4)} = INT{0,inf} e^(-st)*t^2*u(t - 1/4) dt

    Since u(t - 1/4) = {0 for t < 1/4, 1 for t > 1/4}, the limits of integration break up at t = 1/4 to become:
    INT{0,1/4} e^(-st)*t^2*0 dt + INT{1/4,inf} e^(-st)*t^2*1 dt
    = INT{1/4,inf} e^(-st)*t^2 dt

    Give me a second to finish this up.
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    = INT{1/4,inf} e^(-st)*t^2 dt
    Now, to change this function into a true Laplace integration, perform the following substitution:

    let t = v + 1/4 <--> dt = dv

    Notice that pluging the limits of integration into this result in:

    t = 1/4: 1/4 = v + 1/4 --> v = 0
    t -> inf: inf = v + 1/4 --> v -> inf

    INT{0,inf} e^[-s(v + 1/4)]*(v + 1/4)^2 dv
    = e^(-s/4) INT{0,inf} e^(-sv)*(v + 1/4)^2 dv
    = e^(-s/4) INT{0,inf} e^(-sv)*(v^2 + 1/2*v + 1/16) dv

    Notice that e^(-sv)*(v^2 + 1/2*v + 1/16) becomes three seperate terms being added, which becomes:
    e^(-s/4) * L{v^2 + 1/2*v + 1/16}
    = e^(-s/4) * [L{v^2} + L{1/2*v} + L{1/16}]
    = e^(-s/4) * [L{v^2} + 1/2 * L{v} + 1/16 * L{1}]

    I'm going to assume you know how to do this problem from here.
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  4. #4
    Senior Member ecMathGeek's Avatar
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    Hollysti, I'll answer your questions in here.

    Quote Originally Posted by Hollysti
    Thank you very much for your help... for the most part it all makes sense, but I have a couple more questions if you don't mind about that problem.

    1. After you helped me figure out what the heaviside function was, I looked it up on the internet, and Heaviside Step Function -- from Wolfram MathWorld

    declares it as 0 for x<0, 1/2 for x=0, and 1 for x>0. Do I need to put the x=0 part in there somehow???



    2. How did you know how to substitute (v + 1/4) in for t?



    3. Near the bottom, how did you eventually get rid of the integral?



    Thank you for all your help, but if you wouldn't mind answering these questions too to help me understand, that would be amazing!
    1. When taking the integration, we are conserned with the "limits" of integration, so if the value of the function does not match the limit of the function at a point (in this case, x = 0 and limit{x -> 0} are not equal), all we are concerned with is the limit of the function as x->0, so that f(x) = 1/2 for x = 0 doesn't matter.

    2. The Laplace Transform of f(t) is by definition the INT{0,inf} e^(-st)f(t) dt, not INT{a,inf} e^(-st)f(t) dt. In order to change the limits of integration from {1/4,inf} to {0,inf}, we needed to "shift" the function back 1/4 units. Using a substitution of t = v + 1/4 effectively shift the function back 1/4 units as needed.

    3. I didn't "get rid of the integral" I mearly noted that e^(-s/4)*INT{0,inf} e^(-sv)(v^2 + 1/2*v + 1/16) dv = e^(-s/4) of the Laplace Transformation of (v^2 + 1/2*v + 1/16). I rewrote the problem then as the Laplace of this polynomial to show that you can take the Laplace Transformation of each term individually.

    If you need me to clarify anything further, feel free to ask.
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