1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Solving Transcendental Equations (and Laplace Transforms)

  1. Apr 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Given the equation [tex]H'(t) + u H(t - T) = 0[/tex] u > 0
    Look for solutions of the form [tex]e^{rt}[/tex]

    Show that these solutions are exponentially damped if [tex]e^{-1} > uT > 0[/tex]
    Find uT for which these solutions for r complex are oscillatory with growing, decaying, or constant amplitude.

    The book also hints that a Laplace Transform would be helpful (albeit not necessary), but I'm not sure how to do these.

    2. Relevant equations
    [tex]r = - u e^{-rT}[/tex]
    Let [tex]r = x + yi[/tex]
    [tex]x = -u e^{-xT} cos(yT)[/tex]
    [tex]y = u e^{-xT} sin(yT)[/tex]

    3. The attempt at a solution
    I find found the real solutions.
    Set y = 0, then cos(yT) = 1, so
    [tex]x = -u e^{-xT} [/tex]

    define [tex]F(x) = x + u e^{-xT}[/tex]
    [tex]F(0) = u >0[/tex]
    [tex]F(-1/T) = \frac{-1 + uTe}{T} < \frac{-1 + 1}{T} = 0[/tex]
    when [tex]e^{-1} > uT > 0[/tex]

    For oscillatory, I just set [tex]x = 0[/tex], [tex]so cos(yT) = 0[/tex], which means [tex]sin(yT) = +-1[/tex]
    So [tex]y = +-u[/tex], which means [tex]cos(uT) = 0 sin(uT) = +-1[/tex], which we know happens when uT = pi/2, 3pi/2, etc...

    I'm not sure how to do the rest.

    It says a Laplace Transform would make it faster, well, I took the transform from the DE and got
    [tex]Y(s) = \frac{H(0)}{s + e^{-sT}}[/tex]

    How do I interpret this?
  2. jcsd
  3. Apr 11, 2008 #2
    Although I don't like bumping threads, I want to make sure everyone sees this.

    In particular, I'm really curious to how I would be using Laplace Transforms to solve this problem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook