The Laplace Transform of Common Functions

For more results, refer to Table of the Laplace Transform Pairs.

Transforms of some common functions

$\delta (t)$

$$\mathcal{L}[\delta (t)]={\int }_{0^{-}}^{\mathrm{\infty }}\delta (t)e^{-st}dt$$

Using the sampling property, we obtain $$\mathcal{L}[\delta (t)]=1\text{ for all }s$$ that is, $$\delta (t)⟺1\text{ for all }s$$

$u(t)$

To find the Laplace transform of $u(t)$, recall that $u(t)=1$ for $t\ge 0$. Therefore, $$\begin{array}{rl}\mathcal{L}[u(t)]& ={\int }_{0^{-}}^{\mathrm{\infty }}u(t)e^{-st}dt={\int }_{0^{-}}^{\mathrm{\infty }}{e}^{-st}dt=-{\frac{1}{s}{e}^{-st}|}_{0^{-}}^{\mathrm{\infty }}\\ & =\frac{1}{s}\mathrm{Re}s>0\end{array}$$

$\cos {\omega }_{0}tu(t)$

Because $$\cos {\omega }_{0}tu(t)=\frac{1}{2}[e^{j{\omega }_{0}t}+e^{-j{\omega }_{0}t}]u(t),$$ we know that $$\mathcal{L}[\cos {\omega }_{0}tu(t)]=\frac{1}{2}\mathcal{L}[e^{j{\omega }_{0}t}u(t)+e^{-j{\omega }_{0}t}u(t)]$$

From $\text{(???)}$, it follows that

$$\begin{array}{rl}\mathcal{L}[\cos {\omega }_{0}tu(t)]& =\frac{1}{2}[\frac{1}{s-j{\omega }_0}+\frac{1}{s+j{\omega }_0}]\mathrm{Re}(s\pm j\omega )=\mathrm{Re}s>0\\ & =\frac{s}{s^2+{\omega }_0^2}\mathrm{Re}s>0\end{array}$$