1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

DIFFEQ - Discontinuous Forcing Functions (should be an easy question)

  1. Oct 16, 2005 #1
    Ok, we just started this chapter, and I am slightly confused with one specific aspect of the info... I'll just go through an example, it's the best way to explain it IMHO.
    I have to find the Laplace transform of the following function.
    The table of transforms that I can use are (sorry about the formatting, I know they are not equal to each other):
    [tex] u_c(t) = \frac{e^{-cs}}{s} [/tex]
    [tex] u_c(t)f(t-c) = e^{-cs}F(s) [/tex]
    [tex] t^n = \frac{n!}{s^{n+1}} [/tex]

    [tex] f(t)= [/tex] is defined as a system of equations (sorry I don't know the LaTeX formatting for it).
    [tex] f(t)=0|t<1 [/tex]
    [tex] f(t)=t^2-2t+2|t\geq1 [/tex]

    So [tex] f(t) [/tex] can be rewritten as:
    [tex] f(t) = u_1(t)(t^2-2t+2) [/tex]

    Ok, so now this is where I get confused. I have to do the Laplace transform of [tex] f(t) = u_1(t)(t^2-2t+2 [/tex]. But the only table value I have is:
    [tex] u_c(t)f(t-c) = e^{-cs}F(s) [/tex]

    But, this doesn't actually match what I have. Since, f(t) is not of the form f(t-c). So if anyone could just explain this part better to me... that would be awesome. My thought process here is that I have to change f(t-c) to be f(t).
    So:
    [tex] (t-1)^2 = t^2-2t+1 [/tex]
    [tex] (t-1)^2 +1 = f(t) [/tex]
    This would allow me to use the rule right?

    So I would then have:

    [tex] F(s) = e^{-cs}/s L((t-1)^2+1) = e^{-cs}/s [L(t^2)+L(-2t)+L(2)][/tex]

    Is this idea even right? I guess I just don't understand what is really going on here.
    On a second note what the hell is going on with the latex formatting? Is anyone else having troubles previewing their changes?
     
    Last edited: Oct 16, 2005
  2. jcsd
  3. Oct 16, 2005 #2
    Nevermind

    Nevermind... I figured it out. Thanks though :)
    If anyone wants to elaborate, be my guest.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: DIFFEQ - Discontinuous Forcing Functions (should be an easy question)
  1. DiffEq question (Replies: 1)

  2. DiffEq Question (Replies: 5)

Loading...