Question 6.11: Computing an Inverse Laplace Transform Find the inverse Lapl...

Known Quantities: Function to be inverse Laplace-transformed.

Find: f(t)=\mathcal{L}^{-1}\{F(s)\}.

Assumptions: None.

Analysis: A direct entry for the function cannot be found in Table 6.1. In such cases, one must compute a partial fraction expansion of the function F(s), and then individually transform each term in the expansion. A partial fraction expansion is the inverse operation of obtaining a common denominator, and is illustrated below.

F(s)=\frac{2 s+5}{s^{2}+5 s+6}=\frac{A}{s+2}+\frac{B}{s+3}

To obtain the constants A and B, we multiply the above expression by each of the denominator terms:

\begin{aligned}&(s+2) F(s)=A+\frac{(s+2) B}{s+3} \\&(s+3) F(s)=\frac{(s+3) A}{s+2}+B\end{aligned}

From the above two expressions, we can compute A and B as follows:

\begin{aligned}&A=\left.(s+2) F(s)\right|_{s=-2}=\left.\frac{2 s+5}{s+3}\right|_{s=-2}=1 \\&B=\left.(s+3) F(s)\right|_{s=-3}=\left.\frac{2 s+5}{s+2}\right|_{s=-3}=1\end{aligned}

Finally,

F(s)=\frac{2 s+5}{s^{2}+5 s+6}=\frac{1}{s+2}+\frac{1}{s+3}

and 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 compute

f(t)=\left(e^{-2 t}+e^{-3 t}\right) u(t)