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Math Help - Unit step function and Laplace

  1. #1
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    Unit step function and Laplace

    I usually try to use Latex and make everything look pretty and easy to read, but Im in a bit of a rush so please excuse my ugly equations.

    Im doing one-sided Laplace transforms involving unit step functions and Im not sure how to deal with one of the form f(x) * u(t) * u(c-t)

    Explicitly my problem is:

    f(t) = [2t^2 - 10t] * u(t) * u(2-t)

    I know what the graph looks like, but Im not sure how to find F(s) in this situation.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by sixstringartist View Post
    I usually try to use Latex and make everything look pretty and easy to read, but Im in a bit of a rush so please excuse my ugly equations.

    Im doing one-sided Laplace transforms involving unit step functions and Im not sure how to deal with one of the form f(x) * u(t) * u(c-t)

    Explicitly my problem is:

    f(t) = [2t^2 - 10t] * u(t) * u(2-t)

    I know what the graph looks like, but Im not sure how to find F(s) in this situation.

    Thanks in advance.
    Does * represent multiplication or convolution?
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  3. #3
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    Quote Originally Posted by sixstringartist View Post
    I usually try to use Latex and make everything look pretty and easy to read, but Im in a bit of a rush so please excuse my ugly equations.

    Im doing one-sided Laplace transforms involving unit step functions and Im not sure how to deal with one of the form f(x) * u(t) * u(c-t)

    Explicitly my problem is:

    f(t) = [2t^2 - 10t] * u(t) * u(2-t)

    I know what the graph looks like, but Im not sure how to find F(s) in this situation.

    Thanks in advance.
    u(t)u(2-t)=u(2-t)

    (that is for the product to be non zero both u(t) and u(2-t) must both be non-zero
    and when they are they are both 1, ..)

    RonL
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  4. #4
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    they represent multiplication,


    but I dont see how u(t)u(2-t) = u(2-t) because that would include negative values. If anything it can be written as u(t)u(2-t) = u(t) - u(t-2) is that how I am supposed to evaluate this?
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  5. #5
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    Ive found what I was looking for:

    <br />
Au(t)u(T-t) = A[u(t) - u(t-T)]<br />

    Thanks for the help
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  6. #6
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    Quote Originally Posted by sixstringartist View Post
    they represent multiplication,


    but I dont see how u(t)u(2-t) = u(2-t) because that would include negative values. If anything it can be written as u(t)u(2-t) = u(t) - u(t-2) is that how I am supposed to evaluate this?
    CaptainBlack made a boo boo .... u(t)u(2-t) = u(2-t) is not correct. Note however that u(t)u(t-2) = u(t-2) is correct .....The good captain made a simple and easy-to-make slip-up.

    Well spotted and good for you in finding the correct expression.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    CaptainBlack made a boo boo .... u(t)u(2-t) = u(2-t) is not correct. Note however that u(t)u(t-2) = u(t-2) is correct .....The good captain made a simple and easy-to-make slip-up.

    Well spotted and good for you in finding the correct expression.
    Opps, and so:

    u(t)u(2-t) = \left\{ \begin{array} {ll} 1,& u \in [0,2]\\ 0, & \mbox{otherwise} \end{array} \right.

    (OP double check the end points yourself to see if the interval over which this is 1 is open or closed)

    RonL
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  8. #8
    shenky365
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    Unit step function and Laplace-dcnianhmappnhicocnopcoodak.gif

    HI

    I am new here and i am having a little trouble finding the constants A, B, C and t in this question. Please i need help.
    the question is :-

    f(t) = AH(t) + (-8t + B) H(t - 5) - (-8t - C) H(t - 22) where H(t) is the unit step.
    thanks

    Unit step function and Laplace-pkookmabcdgkcimbndhibpolmn.gif
    Last edited by shenky365; February 29th 2008 at 11:21 AM.
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