Laplace transform of triangle pulse. Step function confusion

Discussion in 'Math' started by RolfRomeo, Dec 7, 2009.

  1. RolfRomeo

    Thread Starter Member

    Apr 27, 2009
    Exams are approaching, and I'm working through some old assignments. One of which is a cause of great confusion:

    I'm supposed to find the time-domain description of a symmetric triangle pulse with halfperiod T=1, and maximum amplitude A=1, starting at t=0 and returning to 0 at t=2T, and then Laplace-transform it. In of itself this is not very difficult, and I think I know how to do it correctly, BUT:

    Take a look at the attached plot. The blue curve correctly describes the signal, but the red curve is what gives the correct Laplace-transform. Can someone please explain to me why this is, as I can't seem to wrap my head around it.

    The only thing I can imagine is if the delayed unit step function u(t-d) has some additional operator-like properties, so that u(t-d)*f(t) becomes u(t-d)*f(t-d). Or something...

    Any thoughts are appreciated.

    edit: a = A/T
  2. RolfRomeo

    Thread Starter Member

    Apr 27, 2009
    Replying to myself here, but since I'm partially answering my own question I figured it was called for.

    After digging through the web some more I found this:
    Time-displacement theorem
    L[u(t-a)*g(t-a)] = exp(-as)*G(s)

    I have only found it referenced in a table, with no proof or discussion, but that may be what I'm looking for.

    Eureka! :)
    By looking at the line segments as functions of t: f(t) = a*t I am able to arrive at the correct transform A/(T*s^2) * (1 - exp(-Ts))^2 from the time-description that looks right. What still puzzles me is that the shortcut from the not-right-looking time-description works. Or maybe rather the extra s-terms the correct looking one gives in the laplace domain, when not using the above theorem.
    Last edited: Dec 7, 2009