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Laplace transform with delta function

  1. Jun 14, 2005 #1
    I am sort of stuck on this one:
    compute Laplace trasnform of this signal (directly by evaluating the integral)

    f(t) = cos(pi*t + theta)*delta(t-2);

    I know what the LT integral looks like, but I don't think I'm evaluating it right.
    Would the answer be just: cos(pi*t + theta)*e^(-2s) ?

    Any help is much appreciated.
    Last edited: Jun 14, 2005
  2. jcsd
  3. Jun 15, 2005 #2


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    There are numerous examples of the Laplace Transform in the Mathematics > Calculus & Analysis forum, e.g. https://www.physicsforums.com/showthread.php?t=49248 , or just search the forum for "Lapalce Transform".

    There is no time variable 't' in the transform.
  4. Jun 15, 2005 #3


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    Since [tex]\delta (t-2)[/tex] exists only at t = 2, your f(t) is:
    [tex]f(t) = cos(2\pi + \theta)*\delta (t-2) = cos(\theta)*\delta (t-2)[/tex] and
    [tex]F(s) = cos(\theta)*e^{-2s}[/tex].
  5. Jun 15, 2005 #4
    Agree, I should have searched at other places on these forums.
    Thanks everyone for explanation.
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