# Frequency Shift Property of The Laplace Transform

If $F(s)$ is the Laplace transform of $f(t)$, then
\begin{aligned}\mathcal{L}\left[e^{-a t} f(t)\right] &=\int_{0}^{\infty} e^{-a t} f(t) e^{-s t} d t \\&=\int_{0}^{\infty} f(t) e^{-(s+a) t} d t=F(s+a)\end{aligned}
or
$$\bbox[10px,border:1px solid grey]{\mathcal{L}\left[e^{-a t} f(t)\right]=F(s+a) }\tag{1}$$
That is, the Laplace transform of $e^{-a t} f(t)$ can be obtained from the Laplace transform of $f(t)$ by replacing every $s$ with $s+a$. This is known as frequency shift or frequency translation. As an example, we know that and
$$\cos \omega t \quad \Longleftrightarrow \quad \frac{s}{s^{2}+\omega^{2}}$$
$$\sin \omega t \quad \Longleftrightarrow \quad \frac{\omega}{s^{2}+\omega^{2}}$$
Using the shift property in Eq. (1), we obtain the Laplace transform of the damped sine and damped cosine functions as
\begin{aligned}\mathcal{L}\left[e^{-a t} \cos \omega t\right] &=\frac{s+a}{(s+a)^{2}+\omega^{2}} \\\mathcal{L}\left[e^{-a t} \sin \omega t\right] &=\frac{\omega}{(s+a)^{2}+\omega^{2}}\end{aligned}