Second Shifting Property | Laplace Transform | Advance Engineering Mathematics Review at MATHalino

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

then,

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

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

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

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

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

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

Second Shifting Property | Laplace Transform