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Second shifting theorem of Laplace transform

  1. Aug 14, 2015 #1
    Can anybody direct me towards a proof of the second shifting theorem for Laplace transforms? I'm understanding how to use it but I can't figure out where it comes from. I've been learning from Boas, which doesn't offer much in way of proof for this theorem. Are there any good resources online where I can get a feel for where this theorem comes from? Or, if it's simple, can somebody explain the derivation?

    Thanks in advance.
     
  2. jcsd
  3. Aug 14, 2015 #2

    RUber

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    I'll give it a shot, but I don't work much with the Laplace Transform, so please forgive any mistakes.
    The transform is defined by:
    ##\mathcal{L} \{ f(t) \} = F(s) = \int_0^\infty e^{-ts} f(t) dt.##
    And the second shifting theorem is
    ##\mathcal{L} \{H(t-c) f(t-c) \} = e^{-sc} \mathcal{L} \{f(t) \}##
    So, let's start with the result and work backward.
    ##\mathcal{L} \{H(t-c) f(t-c) \} = e^{-sc} \mathcal{L} \{f(t) \} \\
    \qquad \qquad =e^{-sc} \int_0^\infty e^{-ts} f(t) dt \\
    \qquad \qquad = \int_0^\infty e^{-(t+c)s} f(t) dt##
    Now, the goal is to show that this is equivalent to:
    ##\int_0^\infty e^{ts} H(t-c) f(t-c)dt\\
    =\int_c^\infty e^{-ts} f(t-c)dt\\
    =\int_0^\infty e^{-(t+c)s} f(t)dt\\##
    So...there it is.
     
  4. Aug 15, 2015 #3
    Thanks for your reply. I think I am on the verge of understanding, but how do you justify that last step where you change the bottom limit of integration back to zero, and substitute (t+c) for t? Also correct me if I'm wrong but I think you left out a minus sign in the exponent on the first line after "is equivalent to".
     
  5. Aug 15, 2015 #4
    I just came across another proof of the second shifting theorem using the convolution integral and the Dirac delta function.

    We want to find the inverse transform of F(s) = e-saG(s), where G is the transform of some function g(t). The inverse Laplace transform of e-sa is δ(t-a), and the inverse of G(s) is g(t). To find the inverse of the product of e-sa and G(s), we take the convolution of their independent inverses:

    f(t) = ∫0tg(t-T)δ(T-a)dT = g(t-a), provided that t is greater than a, so as to include the "spike" of the delta function in the bounds of integration, otherwise the convolution will be zero. This result can now be expressed as the product g(t-a)H(t-a), where H(t-a) is the unit step function.

    Thus the Laplace transform of g(t-a)H(t-a) = e-saG(s).
     
  6. Aug 16, 2015 #5

    RUber

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    Yes, I did miss the negative exponent.
    Sometimes the substitution is easier to see if a different variable is used.
    Let u = t-c, t= u+c. du=dt, so with the substitution, the integral from c to infinity dt becomes an integral from 0 to infinity du.
     
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