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Heaviside function

  1. Jul 16, 2009 #1
    1. The problem statement, all variables and given/known data

    how to i take the laplace transform of this ,

    -tH(t-1)

    so we need to get thr right shift so is it -(t-1 + 1 ) so do i take the laplace transform of
    -(t+1) so would it be -(1/s^2 + 1/s ) *e^(-s)
     
  2. jcsd
  3. Jul 16, 2009 #2

    CompuChip

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    You just need to calculate
    [tex]- \int_0^\infty e^{-s t} t H(t - 1) \, \mathrm dt[/tex]

    You can break up the integral in two parts: 0 < t < 1 and t > 1.
     
  4. Jul 16, 2009 #3
    we cant just force the shift , if we had t^2H(t-1)
    then to shift the quadratic we would (t-2+2)^ then we would expand
    (t+2)^2 to t^2+4t+4 then take the laplace transform of that .
     
  5. Jul 16, 2009 #4

    CompuChip

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    What do you want to shift? H(t - 1) is zero for t < 1, so

    [tex]
    \int_0^\infty f(t) H(t - 1) \, \mathrm dt
    =
    \int_0^1 0 \cdot f(t) \, \mathrm dt + \int_1^\infty 1 \cdot f(t) \, \mathrm dt
    =
    \int_1^\infty f(t) \, \mathrm dt
    [/tex]
    or am I really stupid?
     
    Last edited: Jul 16, 2009
  6. Jul 16, 2009 #5
    um i dont know my teacher never really talked about it like that,
    ur prolly right , and another thing i cant see what ever u typed in those black boxes
    my computer wont let me so its hard for me to tell what ur doing .

    im sure ur right , but my teacher told us to take the laplace tranform
    of a function time the heaviside function
    he said the function needed to have the same shift as the heaviside
    function in order for the formula to work .
    f(t)H(t-1) = e^(-s)*L(f(t))
     
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