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Laplace transform with abs value in exponential

  1. Aug 27, 2005 #1
    Hi everyone,
    I have this problem and just need someone to check or correct:
    [tex] f(t) = -4e^{-3|t|}(u(t + 3) - u(t-1))[/tex]
    where u(t) is step function: u(t) = 1 for t >= 1 and 0 elsewhere;
    so, i guess I need to break abs value into 2 cases and have 2 different equations? anyway, here's what I have if someone would be so kind and check my work (at first I applied linearity property and distributed [tex]-4e^{-3|t|}[/tex]:
    1. for t > 0:
    [tex] L[f(t)] = \frac{-4}{s+3} + \frac{4e^{-(s+3)}}{s+3} [/tex]

    2. for t < 0:
    [tex] L[f(t)] = \frac{-4}{s-3} + \frac{4e^{-(s-3)}}{s-3} [/tex]

    thanks in advance!

    ps: i guess one thing I should explain is that by definition of unilateral laplace transform, even though first part of signal starts at -3 we do not consider it, what we are doing is one-sided Laplace transform, so I started integrating from 0- the first part of the expression.

    edit: to (hopefully) increase chances that someone looks at this here's the Lapl. trnsf. that are relevant:
    [tex] L[u(t-k)] = \frac{e^{-sk}}{s}[/tex]

    [tex]L[e^{at}f(t)] = F(s-a)[/tex]
    but you probably know this anyway :shy:
    Last edited: Aug 27, 2005
  2. jcsd
  3. Aug 27, 2005 #2


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    Do you mean u(t)=0, for t<0 and u(t)=1 fot t>0? That's what the standard (Heavyside) step function does.

    I think the easiest way would be direct integration. If you look at the expression u(t+3)-u(t-1), you notice it is 1 inside the interval [-3,1] and zero elsewhere. This makes the integral pretty easy to evaluate.
  4. Aug 27, 2005 #3
    ooops, sorry about the typo, u(t) = 1 for t >= 0, you're right.
    so I would have to integrate from 0 to 1 (for the one-sided laplace trnsfm)...
  5. Aug 27, 2005 #4


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    Yeah, and you would get the same answer you already had :)
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