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Laplace Transformation and unit step func

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data

    I am given the following

    [tex]f(t)= e^t[/tex] where 0<t<2

    express the function f(t) in the terms of the unitstep func.


    2. Relevant equations

    I am told that the f(t) can be expressed

    [tex]\mathcal{L}(f(t-a)u(t-a)) = e^{-as}F(s) [/tex]

    3. The attempt at a solution


    where [tex]F(S) = \mathcal{L}(f(t))[/tex]

    then

    [tex]\mathcal{L}(f(t-0)u(t-0)) = \frac{1}{s} [/tex]

    and

    [tex]\mathcal{L}(f(t-2)u(t-2)) = e^{-2s} \frac{1}{s-1} [/tex]

    I am told that I need to add the two together (what is my motivation for doing this?)

    and if I do this I get

    [tex]\mathcal{L}(F(s)) =\frac{1+e^{-2s}}{s(s-1)}[/tex]

    what am I doing wrong here?

    Best Regards
    Susanne

    edit: I have discovered that if I take the place integral

    [tex]\mathcal{L}(f(t)) = \int _{0}^{2} e^{-st} \cdot e^{t} dt = \frac{e^{t-st}}{1-s}|_{t=0}^{2} = \frac{1-e^{2-2s}}{1-s}[/tex]

    which is desired result according to the textbook. But how do I show that this can the found using the original method above?
     
    Last edited: Oct 20, 2009
  2. jcsd
  3. Oct 20, 2009 #2

    LCKurtz

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    When you write your f(t) in terms of unit step functions you should get:

    f(t) = et(u(t-2) - u(t))

    = etu(t-2) - etu(t)

    [Edit] signs wrong here, should be:

    etu(t)-etu(t-2)


    The first term is not in the form f(t-a)u(t-a). It is of the form f(t)u(t-a).

    Have you had this form of the shift theorem:

    L(f(t)u(t-a)) = e-asL(f(t+a)) ?

    It is what you generally want for this type of transform. Try it and you will get your correct answer.
     
    Last edited: Oct 20, 2009
  4. Oct 20, 2009 #3
    Hi maybe this is stupid question the reason why you subtract is it because because it can be seen as an integral? And that being the reason why what I get with the integral is the same result?
     
  5. Oct 20, 2009 #4

    LCKurtz

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    "Hi maybe this is stupid question the reason why you subtract is it because because it can be seen as an integral? And that being the reason why what I get with the integral is the same result?"

    No. etu(t-2) is zero up until t = 2, then it is et.
    etu(t) agrees with et for t > 0.

    Look at the graphs of those. You need to subtract the part for t > 2 to get 0 for t > 2. That is why you subtract the u(t-2) part.

    For that reason, if b > a, u(t-a) - u(t-b) is sometimes called a filter function. Because if you multiply any f(t) by that what you get is just the portion of f(t) on [a,b] and zero elsewhere.

    They are very useful in LaPlace transforms for taking transforms of piecewise defined functions.

    You got the correct result when you integrated because you computed the transform directly and correctly. But you should still learn to do it with the shift theorem. Did you try what I suggested?
     
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