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Laplace transform and region of convergence

  1. Oct 30, 2014 #1
    Find the LT and specify ROC of:
    x(t) = e-at, 0 ≤ t ≤ T
    = 0, elsewhere
    where a > 0

    Attempt:
    X(s) = - 1/(s+a)*e-(s+a) integrated from 0 to T
    => -1/(s+a)[e-(s+a) + 1]
    Converges to X(s) = 1/(s+a) , a ⊂ R, if Re{s} > -a for 0≤t≤T
    Elsewhere ROC is empty (LT doesn't exist).

    Is this correct?
     
  2. jcsd
  3. Nov 1, 2014 #2
    The Laplace transform of ##f\left(t\right)## is:
    $$ \mathscr{L}\left\{f\left(t\right)\right\} = F\left(s\right) = \int\limits_{0}^{\infty} e^{-st}f\left(t\right) dt$$
    Hence, the Laplace transform of your function is:
    $$ X\left(s\right) = \int\limits_{0}^{T} e^{-at}e^{-st}dt + \int\limits_{T}^{\infty} 0 \cdot e^{-st}dt= \int\limits_{0}^{T} e^{-\left(a+s\right)t}dt = -\dfrac{1}{s+a}e^{-\left(s+a\right)t}\bigg|_{0}^{T}$$
    $$ X\left(s\right) = \dfrac{1}{s+a}\left(1-e^{-\left(s+a\right)T}\right)$$
    The ROC is as you stated: ##\Re\left\{s\right\}>-a##
     
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