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Laplace transform

  1. May 12, 2009 #1
    http://users.on.net/~rohanlal/laplacetrans5.jpg [Broken]
    In the third line, what is S+4 the laplace transform of?
     
    Last edited by a moderator: May 4, 2017
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  3. May 12, 2009 #2

    gabbagabbahey

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    It doesn't matter. Just isolate Y(s) by dividing both sides by (s+4). Then use partial fractions to compute the inverse Laplace Transform of the RHS.
     
  4. May 13, 2009 #3
    it does matter because this is part of the solution to the problem and i need to understand how its done.
     
  5. May 13, 2009 #4

    HallsofIvy

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    No, "the function whose Laplace transform is s+ 4" is not part of the solution, the function whose Laplace transform is 1/(s+ 4) is.

    As Gabbagabbahey said, you need to divide both sides by s+ 4 to get
    [tex]Y(s)= \frac{17}{(s^2+1)(s+4)}+ \frac{1}{s+4}[/tex]
    Partial fractions will give something of the form
    [tex]Y(s)= \frac{As}{s^2+1}+ \frac{B}{s^2+1}+ \frac{C}{s+4}[/tex]
    and then you can look up the inverse transform of each of those.
     
  6. May 13, 2009 #5
    so is the laplace transform of 4y = 1/s+4? then how does it end up being a factor of y(s)?
     
  7. May 13, 2009 #6

    gabbagabbahey

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    I'm not exactly sure what you are trying to ask here. If you are asking whether [itex]4Y(s)=\frac{1}{s+4}[/itex], then the answer is no.

    You have correctly gotten to this point:

    [tex](s+4)Y(s)=\frac{17}{s^2+1}+1[/tex]

    However, for this to be useful you need to divide both sides of the equation by [itex]s+4[/itex] in order to isolate [itex]Y(s)[/itex]. The reason why simply taking the inverse Laplace transform of both sides in the current form is useless is because [itex]\mathcal{L}^{-1}[(s+4)Y(s)]\neq(\mathcal{L}^{-1}[(s+4)])\cdot(\mathcal{L}^{-1}[Y(s)])[/itex]

    But if you divide by [itex]s+4[/itex] first, you get

    [tex]Y(s)=\frac{17}{(s^2+1)(s+4)}+\frac{1}{s+4} \implies y(t)=\mathcal{L}^{-1}[Y(s)]=\mathcal{L}^{-1}[\frac{17}{(s^2+1)(s+4)}+\frac{1}{s+4}][/tex]

    which allows you to find [itex]y(t)[/itex].
     
  8. May 13, 2009 #7
    The working i attached wasn't carried out by me. I havent gotten as far as
    [tex]
    (s+4)Y(s)=\frac{17}{s^2+1}+1
    [/tex]

    someone else did. i want to know how they got there.
    im unsure as to how they got y(s)[s+4]
    i also dont know what the laplace transform of 4y is.
     
  9. May 13, 2009 #8

    gabbagabbahey

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    Oh! Now I see what you were asking.:bugeye:

    The workings in your attached image should be fairly clear: [itex]\mathcal{L}[4y(t)]=4\mathcal{L}[y(t)]=4Y(s)[/itex] and [itex]\mathcal{L}[y'(t)]=sY(s)-y(0)=sY(s)-1[/itex]. The reasons for this should be in your textbook/notes, and if you are not comfortable with such basic Laplace transforms you need to study your text/notes and perhaps get a tutor.
     
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