The Laplace Transform of Common Functions

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

Transforms of some common functions

\(\delta(t)\)

\[ \mathcal{L}[\delta(t)]=\int_{0^{-}}^{\infty} \delta(t) e^{-s t} d t \]

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

\(u(t)\)

To find the Laplace transform of \(u(t)\), recall that \(u(t)=1\) for \(t \geq 0\). Therefore, \[ \begin{aligned} \mathcal{L}[u(t)] & =\int_{0^{-}}^{\infty} u(t) e^{-s t} d t=\int_{0^{-}}^{\infty} e^{-s t} d t=-\left.\frac{1}{s} e^{-s t}\right|_{0^{-}} ^{\infty} \\ & =\frac{1}{s} \quad \operatorname{Re} s>0 \end{aligned} \]

\(\cos \omega_0 t u(t)\)

Because \[ \cos \omega_0 t u(t)=\frac{1}{2}\left[e^{j \omega_0 t}+e^{-j \omega_0 t}\right] u(t) , \] we know that \[ \mathcal{L}\left[\cos \omega_0 t u(t)\right]=\frac{1}{2} \mathcal{L}\left[e^{j \omega_0 t} u(t)+e^{-j \omega_0 t} u(t)\right] \]

From \(\eqref{eq4_6}\), it follows that

\[ \begin{aligned} \mathcal{L}\left[\cos \omega_0 t u(t)\right] & =\frac{1}{2}\left[\frac{1}{s-j \omega_0}+\frac{1}{s+j \omega_0}\right] \quad \operatorname{Re}(s \pm j \omega)=\operatorname{Re} s>0 \\ & =\frac{s}{s^2+\omega_0^2} \quad \operatorname{Re} s>0 \end{aligned} \]