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Math Help - Step function?

  1. #1
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    Step function?

    I have a laplace transform of F(P) = [1-exp(-p)]/[p(1+exp(-2p)] and it tells me to find the laplace transform of y(t) and it gives me a graph where:

    y(0) = 0
    y(1) = 1
    y(2) = 1
    y(3) = 0
    y(4) = 0
    y(5) = 1
    y(6) = 1
    and the pattern continues

    I think this is come kind of heaviside step function but how do i find out the equation to it?
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  2. #2
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    Quote Originally Posted by CookieC View Post
    I have a laplace transform of F(P) = [1-exp(-p)]/[p(1+exp(-2p)] and it tells me to find the laplace transform of y(t) and it gives me a graph where:

    y(0) = 0
    y(1) = 1
    y(2) = 1
    y(3) = 0
    y(4) = 0
    y(5) = 1
    y(6) = 1
    and the pattern continues

    I think this is come kind of heaviside step function but how do i find out the equation to it?
    You need to post your questions more clearly.

    Are you trying to find f(t) given that its laplace transform is F(P) = [1-exp(-p)]/[p(1+exp(-2p)]?
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  3. #3
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    yes I am and the graph I was given gives the answer of f(t) but I just don't know how to find the equation of that graph
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  4. #4
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    Quote Originally Posted by CookieC View Post
    yes I am and the graph I was given gives the answer of f(t) but I just don't know how to find the equation of that graph
    Here are two things (the first you should already have been taught):

    Theorem: If f(t) is a periodic function of period T then \displaystyle LT[f(t)] = \frac{1}{1 - e^{-pt}} \int_0^T f(t) e^{-pt} \, dt.

    In light of the above theorem, you need to find \displaystyle LT^{-1} \left[ \frac{1 - e^{-p}}{p}\right] and then define f(t) as an appropriate periodic function.

    Now you need to think hard about these two things and then make a strong attempt at the question before asking for more help.
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  5. #5
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    Is it still included as a periodic function if the lines are straight lines? as in from 0 to 1 it is a straight line with y = x then a flat line from 1 to 2 then from 3 to 4 is is a line from one to 0 thats downward sloping..not sure if that made sense
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Here are two things (the first you should already have been taught):

    Theorem: If f(t) is a periodic function of period T then \displaystyle LT[f(t)] = \frac{1}{1 - e^{-pt}} \int_0^T f(t) e^{-pt} \, dt.

    In light of the above theorem, you need to find \displaystyle LT^{-1} \left[ \frac{1 - e^{-p}}{p}\right] and then define f(t) as an appropriate periodic function.

    Now you need to think hard about these two things and then make a strong attempt at the question before asking for more help.
    My mistake. The above is not directly relevant to the question. I misread the denominator of the expression in post #1 as having a minus instead of a plus. If you re-arrange the form of the expression by multiplying numerator and denominator by \displaystyle 1 - e^{-2p} then the ideas I have suggested will work I think.
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