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Second shift theorem

  1. Feb 2, 2016 #1
    1. The problem statement, all variables and given/known data
    for the alternative form of second shift property (4.8) , why he integral of (e^-sp) g(p+a) dp isn't equal to integral of (e^-sp) g(t) dp ? why it will become L{ g(t+a) } ?

    2. Relevant equations


    3. The attempt at a solution
     

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  3. Feb 2, 2016 #2

    RUber

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    Would you mind showing more work explaining your reasoning?
    I imagine the answer is either in the endpoints of your integrals or the substitution.

    Straight off, from your question, for ##t = p+a##
    ##\int e^{-sp}g(p+a) dp = \int e^{-sp}g(t) dp##
    but you can't do much with that...complete the substitution to change the whole integral into one with only dependence on t and no p's.
     
  4. Feb 2, 2016 #3
    ya , why not the final ans not L {g(p+a)} ?
     
  5. Feb 2, 2016 #4

    RUber

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    Write out the steps with the limits of integration. Do the substitutions where appropriate to make sure the new integrals are from zero to infinity.
     
  6. Feb 2, 2016 #5
    is my working until here correct ?
     

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  7. Feb 3, 2016 #6

    RUber

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    It looks like you are treating g(p+a) as a multiplication. g(x) is a function.
    Also, when you do substitutions within integrals, you have to change the limits as well.

    For example:
    Substituting ##t = p+a\quad p = t-a\quad p=0 \implies t = a##
    ##\int_0^\infty e^{-sp}g(p+a) dp = \int_{a}^\infty e^{-s(t-a)}g(t) dt##
    Does that make sense?
     
  8. Feb 3, 2016 #7
    i gt stucked here , how to continue?
     

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  9. Feb 4, 2016 #8

    RUber

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    You are looking to show that ##\mathcal{L}\{ g(t) \mathcal{H}(t-a) \} = e^{-as}\mathcal{L}\{ g(t+a)\}.##
    Start by expanding the transform:
    ##\mathcal{L}\{ g(t) \mathcal{H}(t-a) \} =\int_0^\infty e^{-st}g(t) \mathcal{H}(t-a)dt##
    Apply the Heaviside function:
    ##\qquad \qquad =\int_a^\infty e^{-st}g(t) dt##
    Because this integral does not start at 0, we need to shift it with a substitution.
    Substitute t = p+a ... p = t-a ... dt = dp ...and where t = a, p = 0 (for the limits of integration).
    ##\qquad \qquad =\int_0^\infty e^{-s(p+a)}g(p+a) dp=\int_0^\infty e^{-sp-sa}g(p+a) dp##
    Since a and s are not variables inside the integral, you can move them to the outside:
    ##\qquad \qquad =e^{-as} \int_0^\infty e^{-sp}g(p+a) dp.##
    In this form you can see that this is the same as
    ## e^{-as}\mathcal{L}\{ g(p+a)\}## but since p could be any letter, you can name it t.
    This will give you what you want to show:
    ##\mathcal{L}\{ g(t) \mathcal{H}(t-a) \} = e^{-as}\mathcal{L}\{ g(t+a)\}.##
     
  10. Feb 4, 2016 #9

    RUber

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    Two things I noticed in your post #7...
    1) There is no need to do integration by parts on this.
    2) You did not distribute properly in your exponential: -s(t-a) = -st+sa.
     
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