Second Shifting Property | Laplace Transform

Second Shifting Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   and   $g(t) = \begin{cases}{f(t - a) & t > a \\ 0 & t  

then,

$\mathcal{L} \left\{ g(t) \right\} = e^{-as} F(s)$

 

Proof of Second Shifting Property
$g(t) = \begin{cases}{f(t - a) & t > a \\ 0 & t  

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-st} g(t) \, dt$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^a e^{-st} (0) \, dt + \int_a^\infty e^{-st} f(t - a) \, dt$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_a^\infty e^{-st} f(t - a) \, dt$
 

Let
$z = t - a$

$t = z + a$

$dt = dz$
 

when   $t = a, \, z = 0$

when   $t = \infty, \, z = \infty$
 

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-s(z + a)} f(z) \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-sz - sa} f(z) \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-sz} e^{-sa} f(z) \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-sa} \int_0^\infty e^{-sz} f(z) \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-as} \mathcal{L} \left\{ f(z) \right\}$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-as} \mathcal{L} \left\{ f(t - a) \right\}$

$\mathcal{L} \left\{ g(t) \right\} = e^{-as} F(s)$       okay
 

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