Question 6.10: Computing an Inverse Laplace Transform Find the inverse Lapl...
Computing an Inverse Laplace Transform
Find the inverse Laplace transform of
F(s)=\frac{2}{s+3}+\frac{4}{s^{2}+4}+\frac{4}{s}
Known Quantities: Function to be inverse Laplace-transformed.
Find: f(t)=\mathcal{L}^{-1}\{F(s)\}.
Schematics, Diagrams, Circuits, and Given Data:
F(s)=\frac{2}{s+3}+\frac{4}{s^{2}+4}+\frac{4}{s}=F_{1}(s)+F_{2}(s)+F_{3}(s)
Assumptions: None.
Analysis: Using Table 6.1,
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}
we can individually inverse-transform each of the elements of F(s) :
\begin{aligned}&f_{1}(t)=2 \mathcal{L}^{-1}\left(\frac{1}{s+3}\right)=2 e^{-3 t} u(t) \\&f_{2}(t)=2 \mathcal{L}^{-1}\left(\frac{2}{s^{2}+2^{2}}\right)=2 \sin (2 t) u(t) \\&f_{3}(t)=4 \mathcal{L}^{-1}\left(\frac{1}{s}\right)=4 u(t)\end{aligned}
Thus
f (t) = f_1(t) + f_2(t) + f_3(t) = (2e^{−3t} + 2 \sin(2t) + 4)u(t).