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Laplace transforms

  1. May 24, 2013 #1
    I have two questions:

    I had to find the Laplace transform of:
    [itex]t \cdot sin(t)[/itex]
    Not by definition, using a table of transforms and the properties.
    I did:
    [itex]sin(t) = i \cdot sinh(it) = i \cdot \frac{e^{t}}{2}-i \cdot \frac{e^{-t}}{2} [/itex]
    Then
    [itex]t \cdot sin(t) = it \cdot \frac{e^{t}}{2}-it \cdot \frac{e^{-t}}{2}[/itex]
    And
    [itex]t^{n}e^{-at}=\frac{n!}{(s+a)^{n+1}}[/itex]
    So
    [itex]\frac{i}{2} t^{1}e^{-it}=\frac{i}{2} \frac{1}{(s+i)^{2}}=\frac{i}{2} \frac{1}{s^{2}+2si-1}[/itex] and [itex]\frac{i}{2} t^{1}e^{it}=\frac{i}{2} \frac{1}{(s-i)^{2}}=\frac{i}{2} \frac{1}{s^{2}-2si-1}[/itex]
    [itex]\frac{i}{2} \frac{s^{2}-2si-1-(s^{2}+2si-1)}{(s^2+1)^2}=\frac{2s}{(s^2+1)^2}[/itex]
    I want to know other way of calculate this without using the definition, because I don't know if I'm meant to use complex numbers.

    The other problem is:
    [itex]f(t)=\frac{sin(t)}{t}[/itex] if [itex]t \neq 0[/itex]
    [itex]f(t)=1[/itex] if [itex]t = 0[/itex]
    Find the Mac Laurin serie of the function and check that [itex]L\left\{f(t)\right\}=arctan(\frac{1}{s})[/itex] s >1
    I find the following serie:
    [itex]\sum_{n=0}^{\infty} (-1)^n (\frac{1}{s})^{2n+1}\frac{1}{2n+1}[/itex]
    But it diverges for s>1, so it's useless to my purpose.
    And I had problems trying to compute a Taylor Serie of [itex]arctan(\frac{1}{s})[/itex], I don't know what point to use.

    Sorry if I made some gramatical mistakes, I don't speak English very well.
     
    Last edited: May 24, 2013
  2. jcsd
  3. May 24, 2013 #2

    Ray Vickson

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    What is the function ##sen(t)##? I have never heard of it. Do you mean ##\sin(t)##? I have also never heard of ##senh(t)##, but I do know about ##\sinh(t)##. Is that what you mean?
     
  4. May 24, 2013 #3
    Yes, I forget to translate them when I originally created the thread.
     
  5. May 25, 2013 #4
    It isn't, I don't know why I confused the divergence of t with the diverge of s, it has nothing to do with it.
    So, I guess all I have to do is to find a good point to calculate the serie.
    It's a good idea to calculate the McLaurin serie of arctan(x) then use it to arctan(1/x) changing x by 1/x and restricting the x values?
     
    Last edited: May 25, 2013
  6. May 25, 2013 #5

    Ray Vickson

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    Your very first step is incorrect: you wrote
    [tex]\sin(t)=i⋅\sinh(it)=i⋅\frac{e^t}{2}−i⋅\frac{e^{−t}}{2} \; \leftarrow \text{ wrong!} [/tex]
    It should be
    [tex] \sin(t) = \frac{1}{2i} \left( e^{it} - e^{-it} \right)
    = - \frac{i}{2} e^{it} + \frac{i}{2} e^{-it}[/tex]
     
  7. May 26, 2013 #6

    vela

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    Your table should list a property for finding the Laplace transform of t f(t) in terms of F(s), the transform of f(t).
     
  8. May 26, 2013 #7

    vela

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    Use the fact that the derivative of arctan x is ##\frac{1}{1+x^2}##. Expand the latter as a series and then integrate it term by term.
     
  9. May 26, 2013 #8

    Thanks.

    I guess it's it.
    [itex]t^n f(t) \leftrightarrow (-1)^n F^{(n)}(s)[/itex]
    It was named "derivative of a transform", so I didn't noticed it the first time because I didn't pay attention to this propertie (perhaps we didn't see it in the lectures, perhaps we saw it was the day I was sick).
    Thank you again, it clarified a lot and it was a lot shorter than my previous approach xD
     
    Last edited: May 26, 2013
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