Laplace Time Shifting Property

Note

\mathcal{L}\{f(t-c)u(t-c)\} = e^{-cs}F(s)

Derivation:

u(t) = \begin{cases} 0 & t < 0 \\ 1 & t \geq 0 \end{cases}

u(t-c) = \begin{cases} 0 & t < c \\ 1 & t \geq c \end{cases}

\begin{align*} \mathcal{L}\{f(t)\} &= \int_{0}^{\infty} e^{-st}f(t)\, dt \\ \mathcal{L}\{f(t-c)u(t-c)\} &= \int_{0}^{\infty} e^{-st}f(t-c)u(t-c)\, dt \\ &= \underbrace{\int_{0}^{c} e^{-st}f(t-c)u(t-c)\, dt}_{0} + \int_{c}^{\infty} e^{-st}f(t-c)u(t-c)\, dt \\ &= \int_{c}^{\infty} e^{-st}f(t-c)\, dt \\ \text{let}\, u = t-c \\ (t \to c, u \to 0)\\ (t \to \infty, u \to \infty)\\ &= \int_{0}^{\infty} e^{-s(u+c)}f(u)\, du \\ &= e^{-cs}\int_{0}^{\infty} e^{-su}f(u)\, du \\ &= e^{-cs}F(s) \end{align*}