Unit Step Function

aka the Heaviside function

$$u(t)=\{\begin{array}{ll}0& t<0\\ 1& t>0\end{array}$$

Laplace Transform

Alone

$$\mathcal{L}\{u(t-a)\}=\frac{e^{-as}}{s},\text{ for }s>0$$

With Function

$$\mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(t)$$

Extension

$$\begin{array}{rl}& g(t)=f(t-a),f(t)=g(t+a)\\ & \\ & \mathcal{L}\{g(t)u(t-a)\}=e^{-as}\mathcal{L}\{g(t+a)\}\end{array}$$

Plug $t+a$ in before the transform

Inverse

$${\mathcal{L}}^{-1}\{e^{-as}F(s)\}=u(t-a){\mathcal{L}}^{-1}\{F(s)\}(t-a)$$

Note:

$${\mathcal{L}}^{-1}\{F(s)\}(t-a)={{\mathcal{L}}^{-1}\{F(s)\}|}_{t\to t-a}$$

Plug $t-a$ in after the transform