# Time Shift Property of The Laplace Transform

$$\bbox[10px,border:1px solid grey]{\mathcal{L}[\cos w t]=\frac{1}{2}\left(\frac{1}{s-j \omega}+\frac{1}{s+j \omega}\right)=\frac{s}{s^{2}+\omega^{2}} }\tag{A}$$
If $F(s)$ is the Laplace transform of $f(t)$, then
$$\begin{array}{r}\mathcal{L}[f(t-a) u(t-a)]=\int_{0}^{\infty} f(t-a) u(t-a) e^{-s t} d t \\a \geq 0\end{array}$$
But $u(t-a)=0$ for $t < a$ and $u(t-a)=1$ for $t > a$. Hence, $$\mathcal{L}[f(t-a) u(t-a)]=\int_{a}^{\infty} f(t-a) e^{-s t} d t$$ If we let $x=t-a$, then $d x=d t$ and $t=x+a$. As $t \rightarrow a, x \rightarrow 0$ and as $t \rightarrow \infty, x \rightarrow \infty$. Thus,
\begin{aligned}\mathcal{L}[f(t-a) u(t-a)] &=\int_{0}^{\infty} f(x) e^{-s(x+a)} d x \\&=e^{-a s} \int_{0}^{\infty} f(x) e^{-s x} d x=e^{-a s} F(s)\end{aligned}
or
$$\bbox[10px,border:1px solid grey]{\mathcal{L}[f(t-a) u(t-a)]=e^{-a s} F(s)} \tag{1}$$
In other words, if a function is delayed in time by $a$, the result in the $s$ domain is multiplying the Laplace transform of the function (without the delay) by $e^{-a s}$. This is called the time-delay or time-shift property of the Laplace transform. As an example, we know from Eq. (A) that
$$\mathcal{L}[\cos \omega t]=\frac{s}{s^{2}+\omega^{2}}$$
Using the time-shift property in Eq. (1),
$$\mathcal{L}[\cos \omega(t-a) u(t-a)]=e^{-a s} \frac{s}{s^{2}+\omega^{2}}$$