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Laplace transform - step function

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data
    f(t)= 1 if 0≤t≤1 ; 0 is t>1
    find the laplace transform


    2. Relevant equations



    3. The attempt at a solution
    I know u(t)= 0 for t<0 and 1 for t≥0

    I know I have to shift it and get
    u_a(t)=u(t-a)= 1 if 0≤t≤a, 0 if a>1

    am I even going the right way?
    then I think I integrate it from 0 to inf with
    ∫e^(-st)u_a(t) dt = ?

    not sure what to do from here
     
  2. jcsd
  3. Feb 9, 2012 #2

    vela

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    What?
     
  4. Feb 9, 2012 #3
    I thought there was a step involving replacing the discontinuous point with a
     
  5. Feb 9, 2012 #4

    vela

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    There is, but your description doesn't make sense. When is ua(t) equal to 0 and when is it equal to 1?
     
  6. Feb 9, 2012 #5
    f(t) is 0 when t>1 and 1 when it is between 0 and 1, inclusive.

    so for ua(t) wouldnt I just replace the 1 with a?
     
  7. Feb 9, 2012 #6

    vela

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    I'm not following what you're getting at. Replace what 1 with a?
     
  8. Feb 9, 2012 #7
    You replace the discontinuous point with a right? Which happens to be one in this problem because it has a value of 0 whe greater than 1 and a value of 1 when between 0 and 1. So the discontinuous point would be at 1? Which you replace with a?
     
  9. Feb 9, 2012 #8

    vela

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    You have it backwards. You don't replace 1 with a. You set a to 1, i.e., u1(t) = u(t-1). That's the step function shifted to the right by 1.
    $$u_1(t) = u(t-1) = \begin{cases}
    0 & t<1 \\
    1 & t\ge 1
    \end{cases}$$
     
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