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Laplace Transform help needed

  1. Apr 17, 2012 #1
    Hi,

    I've been asked to find the Laplace transform of a function and I have not the slightest clue where to begin. My professor derived the basic Laplace transforms in class(sin, cos, delta function, step function, etc), all of which I understood perfectly. However, he never really gave us an example of how to use those to find other Laplace transforms, let alone discuss how to approach the homework problem.

    rc34wp.png

    i made a few attempts. The first was by saying that the transform of a summation is the summation of the transforms. And then trying to take the transform of that, which seems ugly. I also tried graphing it and re-writing it as a series of step functions to get an idea, but I obviously can't do that until the end of time, so I'm stuck. Could somebody lead me in the right direction? Thanks!
     
  2. jcsd
  3. Apr 17, 2012 #2

    Mark44

    Staff: Mentor

    Try sketching a graph of the summation - it's just the sum of some step functions, with each multiplied by either 1 or -1.
     
  4. Apr 17, 2012 #3
    Right, I've done that already, but how do I get the Laplace transform from the graph? My graph starts at 1 when n=0, drops to -1 a n=1, goes to 1 at n=2, -1 at n=3, etc.

    So the series I computed from that was u(t)-2u(t-1)+2u(t-2)-2u(t-3)+2u(t-4)...

    But now what?
     
  5. Apr 17, 2012 #4

    Mark44

    Staff: Mentor

    Where are the 2's coming from?

    If you expand the series, don't you get just u(t) - u(t - 1) + u(t - 2) -+ ... + (-1)nu(t - n) + ... ?

    Now, what's the Laplace transform of u(t - a)?
     
  6. Apr 17, 2012 #5
    Well isn't a step function minus another step function zero (1-1=0)? so I need to subtract it by another step function, 2u(t-n), to make it reach -1.

    And the laplace transform of u(t-a) = e^(-as)/s
     
  7. Apr 17, 2012 #6

    Mark44

    Staff: Mentor

    Yes
    ???
    Why do you think you need to reach -1?

    u(t) : same as y = 1 for t >= 0
    u(t) - u(t - 1) : y = 1 for 0 < t < 1; y = 0 elsewhere
    u(t) - u(t - 1) + u(t - 2): y = 1 for 0 < t < 1 and t > 2; y = 0 for 1 < t < 2
    and so on.
    Yes.
     
  8. Apr 17, 2012 #7
    I must've been mixing myself up with a plot from the book.

    Anyway, now that I have the Laplace transform of u(t-a), then is the answer I'm looking for the summation of that Laplace transform times (-1)^n from 0 to infinity?
     
  9. Apr 17, 2012 #8

    Mark44

    Staff: Mentor

    That's what I get.
     
  10. Apr 17, 2012 #9
    Now is there any way to simplify that even further, like get rid of the summation?
     
  11. Apr 17, 2012 #10

    Mark44

    Staff: Mentor

    I don't think so.
     
  12. Apr 17, 2012 #11
    ok, thanks
     
  13. Apr 17, 2012 #12
    wish you were able to get it worked out!http://www.infoocean.info/avatar1.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  14. Apr 17, 2012 #13

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    [QUOTYE=audifanatic51;3870693]Now is there any way to simplify that even further, like get rid of the summation?[/QUOTE]

    Yes. You just have a geometric series.

    RGV
     
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