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Math Help - Laplace Transforms

  1. #1
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    Laplace Transforms

    The question and answer is attached though I need some help understanding the following:

    1) When finding the laplace transform of the function, how do you know which order to do it in? For example, I could do it the order given in the answer or I could transform this way:

    First, I transform u(t) to 1/s
    Then I transform e^-at * sin wt to w/(((s+a)^2)+w^2)

    Hence, multiplying the two transforms I get,
    1/s . (w)/(s+a)^2 + w^2

    Which is different from the given answer, so does the order matter and if so, how do u decide which order to do it in?

    2) I think the solution is wrong because when completing the square i got (s+3/2)^2+(7/4) rather than what was given. Is it possible the answer is wrong?

    Thanks for your time!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by blueturkey View Post
    The question and answer is attached though I need some help understanding the following:

    1) When finding the laplace transform of the function, how do you know which order to do it in? For example, I could do it the order given in the answer or I could transform this way:

    First, I transform u(t) to 1/s
    Then I transform e^-at * sin wt to w/(((s+a)^2)+w^2)

    Hence, multiplying the two transforms I get,
    1/s . (w)/(s+a)^2 + w^2
    The Laplace transform of the product of two functions is not the product of their Laplace transforms.

    (Are you using the one-sided LT?, If so ask yourself how the LT of u(t)f(t) might differ from that of f(t))

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    (Are you using the one-sided LT?, If so ask yourself how the LT of u(t)f(t) might differ from that of f(t))

    CB
    What do you mean by one-sided? I thought about your question - the thing is I tried another question, using laplace transforms to solve ode, and when converting back to the f(t) I found that an extra u(t) popped up in the model answer but not in the answer that I got. Is this because the transform of u(t) = 1? Hence, just assume that the u(t) is there?

    Thanks for clearing up that its not the product though.

    PS.
    To others having some trouble with this topic,

    I found this video quite helpful.
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  4. #4
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    Quote Originally Posted by blueturkey View Post
    What do you mean by one-sided? I thought about your question - the thing is I tried another question, using laplace transforms to solve ode, and when converting back to the f(t) I found that an extra u(t) popped up in the model answer but not in the answer that I got. Is this because the transform of u(t) = 1? Hence, just assume that the u(t) is there?
    It is because the one-sided LT of a function does not know anything about the behaviour of the function at negative arguments, which is why the u(t) is irrelevant in the forward direction. It may appear for an inverse transform related to the solution to an ODE if the initial conditions dictate it.

    CB
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  5. #5
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    thanks for the clear reply
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