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Math Help - Laplace Transform

  1. #1
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    Laplace Transform

    How do we prove that L{f(2t)} = 1/2F(S/2).
    I know to prove L{f(t)} = 1/S

    Thanks,
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  2. #2
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    Re: Laplace Transform

    $F(s)=\int f(t) e^{-s t}~dt$

    $G(s)=\int f(2t) e^{-s t}~dt$

    $u=2t, du=2dt,~t=u/2~,dt=du/2$

    $G(s)=\frac 1 2 \int f(u) e^{-s u/2}~du=\frac 1 2 \in f(u) e^{-(s/2)u}~du=\frac 1 2 F(s/2)$
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  3. #3
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    Re: Laplace Transform

    Uh, you know that "L(f)= 1/s" is only true for the specific function, f(t)= 1 for all t, don't you? Saying that "L{f(t)}= 1/s" is meaningless.

    I hope, at least, that you know that L(f)= \int_0^\infty f(t)e^{-st}dt.
    So that L{f(2t)}= \int_0^\infty f(2t)e^{-st}dt. Now make the substitution u= 2x in that integral.
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  4. #4
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    Re: Laplace Transform

    Quote Originally Posted by romsek View Post
    $F(s)=\int f(t) e^{-s t}~dt$

    $G(s)=\int f(2t) e^{-s t}~dt$

    $u=2t, du=2dt,~t=u/2~,dt=du/2$

    $G(s)=\frac 1 2 \int f(u) e^{-s u/2}~du=\frac 1 2 \in f(u) e^{-(s/2)u}~du=\frac 1 2 F(s/2)$
    this should be

    $G(s)=\frac 1 2 \int f(u) e^{-s u/2}~du=\frac 1 2 \int f(u) e^{-(s/2)u}~du=\frac 1 2 F(s/2)$
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  5. #5
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    Re: Laplace Transform

    Thank you Romsek
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