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What does ROC Mean?

  1. Sep 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Find laplace of e-atu(t) where a > 0

    2. Relevant equations
    Laplace is integration from -inf to +inf f(t)e-stdt
    u(t) is 1 for t more than equal to 0.

    3. The attempt at a solution
    Well i got the answer as X(s) = 1/(s+a).

    But the book said something like ROC like region of convergence and for that s must be more than -a
    I understand that s more than minus a, gives positive 1/(s+a)
    but why?

    Why can't ROC be negative? What does ROC mean? I am studying Signals and Processing.
  2. jcsd
  3. Sep 23, 2016 #2


    Staff: Mentor

  4. Sep 23, 2016 #3


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    Your integral is ##\int_1^\infty e^{-st}e^{-at}~dt = \int_1^\infty e^{-(s+a)t}~dt##. That integral will only converge if the exponent is negative, meaning ##s+a>0##. ROC in this case means the region of convergence, the values of ##s## for which the transform is defined.
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