Question 6.8: Computing a Laplace Transform Find the Laplace transform of ...

Known Quantities: Function to be Laplace-transformed.

Find: F(s)=\mathcal{L}\{f(t)\}.

Schematics, Diagrams, Circuits, and Given Data: f(t)=e^{-a t} u(t).

Assumptions: None.

Analysis: From equation 6.51:

\mathcal{L}[f(t)]=F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t        (6.51)

F(s)=\int_{0}^{\infty} e^{-a t} e^{-s t} d t=\int_{0}^{\infty} e^{-(s+a) t} d t=\left.\frac{1}{s+a} e^{-(s+a) t}\right|_{0} ^{\infty}=\frac{1}{s+a}

Comments: Table 6.1 contains a list of common Laplace transform pairs.

Table 6.1 Laplace transform pairs
\begin{array}{ll} \hline {\boldsymbol{f}(\boldsymbol{t})} & \boldsymbol{F}(\boldsymbol{s}) \\\hline\delta(t) (\text{unit impulse}) & 1 \\\\u(t) (\text{unit step}) & \frac{1}{s}\\ \\e^{-a t} u(t) & \frac{1}{s+a}\\ \\\sin \omega t u(t) & \frac{\omega}{s^2+\omega^2}\\ \\\cos \omega t u(t) & \frac{s}{s^2+\omega^2} \\\\e^{-a t} \sin \omega t u(t) & \frac{\omega}{(s+a)^2+\omega^2}\\ \\e^{-a t} \cos \omega t u(t) & \frac{s+a}{(s+a)^2+\omega^2} \\\\t u(t) & \frac{1}{s^2} \\ \hline\end{array}