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Laplace transform of the dirac delta function

  1. Nov 27, 2012 #1
    1. The problem statement, all variables and given/known data

    L[[itex]t^{2}[/itex] - [itex]t^{2}[/itex]δ(t-1)]

    2. Relevant equations

    L[ [itex]t^{n}[/itex]f(t)] = ([itex]-1^{n}[/itex]) [itex]\frac{d^{n}}{ds^{n}}[/itex] L[f(t)]

    L[δ-t] = e^-ts

    3. The attempt at a solution

    My teacher wrote [itex]\frac{2}{s^{3}}[/itex] [itex]-e^{s}[/itex] as the answer.

    I got [itex]\frac{2}{s^{3}}[/itex] + [itex]\frac{e^-s}{s}[/itex] + 2 [itex]\frac{e^-s}{s^2}[/itex] + [itex]\frac{2e^-s}{s^3}[/itex]
     
    Last edited: Nov 27, 2012
  2. jcsd
  3. Nov 27, 2012 #2

    Mute

    User Avatar
    Homework Helper

    You should show us how you derived your answer. That way we can best help you see where you went wrong. Right now we have to guess.

    My guess is that you tried to use the identity under "Relevant equations" on the Dirac delta term, which likely made your problem more complicated than it needed to be.

    Are you aware of the rule ##f(t)\delta(t-t_0) = f(t_0)\delta(t-t_0)##? This will simplify your calculation.
     
  4. Nov 27, 2012 #3
    I don't see that formula in my textbook.

    When I plug in the numbers I get: [itex]\frac{2}{s^3}[/itex] + [itex]\frac{e^-s}{s}[/itex]

    This is what I did to get my first answer.
    L[t^2 - t^2 δ(t-1))]


    [itex]\frac{2}{s^3}[/itex] + [itex]\frac{d}{ds}[/itex] ([itex]\frac{d}{ds}[/itex] ([itex]\frac{e^(-s)}{s}[/itex])




    [itex]\frac{2}{s^3}[/itex] +[itex]\frac{d}{ds}[/itex] ([itex]\frac{e^-s * s + e^-s}{s^2}[/itex] + [itex]\frac{e^-s*s^2 + 2s*e^-s}{s^2}[/itex])

    [itex]\frac{2}{s^{3}}[/itex] + [itex]\frac{e^-s}{s}[/itex] + 2 [itex]\frac{e^-s}{s^2}[/itex] + [itex]\frac{2e^-s}{s^3}[/itex]


    Okay I just checked the answer on wolframalpha and what my teacher had was the correct answer. Why is it that L[t^n*d(t-a)] = e^-as

    Edit:
    Nevermind, I've found my mistake.
     
    Last edited: Nov 27, 2012
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