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[parent] delay theorem (Theorem)

Theorem. If $ f(t) \equiv 0$ for $ t < 0$, then

$\displaystyle \mathcal{L}\{f(t-t_0)\} = e^{-t_0s}\mathcal{L}\{f(t)\}.$

Proof. Since $ f(t-t_0) \equiv 0$ for $ t < t_0$, the definition of Laplace transform at first gives

$\displaystyle \mathcal{L}\{f(t-t_0)\} = \int_{t_0}^\infty e^{-st}f(t-t_0)\,dt.$
The substitution $ t-t_0 := u$ yields
$\displaystyle \mathcal{L}\{f(t-t_0)\} = \int_{0}^\infty e^{-s(u+t_0)}f(u)\,du = e^{-t_0s}\int_{0}^\infty e^{-su}f(u)\,du = e^{-t_0s}\mathcal{L}\{f(t)\}.$



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See Also: Heaviside step function

Other names:  delay theorem of Laplace transform
Keywords:  Laplace transform

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Cross-references: Laplace transform

This is version 3 of delay theorem, born on 2008-05-07, modified 2008-05-07.
Object id is 10570, canonical name is DelayTheorem.
Accessed 86 times total.

Classification:
AMS MSC: 44A10 (Integral transforms, operational calculus :: Laplace transform)
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