Determine the inverse Laplace transform of the given function.

$\frac{\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{16}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{s}\mathbf{+}\mathbf{13}}$.

• For a given transfer function H, the Inverse Laplace Transform takes the output Y(s) and determines what X(s) it is in terms of (s).
• Consider a function F(s), if there is a function f(t) that is continuous on$[0,\infty )$ and satisfies$\mathcal{L}\left\{f\right\}=F$ then we say that f(t) is the inverse Laplace transform of F(s) and employ the notation
• $f={\mathcal{L}}^{-1}\left\{F\right\}$
• ${\mathcal{L}}^{-1}\left\{\frac{n!}{{(s-a)}^{n+1}}\right\}=e^{at}{t}^n,n=1,2,\dots$

The given function is $\frac{2s+16}{s^2+4s+13}$.

Simplify $\frac{2s+16}{s^2+4s+13}$as follows:

$\begin{array}{rcl}\frac{2s+16}{s^2+4s+4+9}& =& \frac{2(s+8)}{(s^2+4s+4)+9}\\ & =& \frac{2(s+2)+12}{{(s+2)}^2+{\left(3\right)}^2}\\ & =& \frac{2(s+2)}{{(s+2)}^2+{\left(3\right)}^2}+\frac{12}{{(s+2)}^2+{\left(3\right)}^2}\\ & & \end{array}$

Find the inverse laplace transform of $\frac{2s+16}{s^2+4s+4+9}=\frac{2(s+2)}{{(s+2)}^2+{\left(3\right)}^2}+\frac{12}{{(s+2)}^2+{\left(3\right)}^2}$using ${\mathcal{L}}^{-1}\left\{\frac{b}{{(s-a)}^2+{\left(b\right)}^2}\right\}=e^{at}\sin bt$and ${\mathcal{L}}^{-1}\left\{\frac{b}{{(s-a)}^2+{\left(b\right)}^2}\right\}=e^{at}\cos bt$as:

$\begin{array}{rcl}{\mathcal{L}}^{-1}\left\{\frac{2s+16}{s^2+4s+4+9}\right\}& =& {\mathcal{L}}^{-1}\left\{\frac{2(s+2)}{{(s+2)}^2+{\left(3\right)}^2}+\frac{12}{{(s+2)}^2+{\left(3\right)}^2}\right\}\\ & =& 2{\mathcal{L}}^{-1}\left\{\frac{s+2}{{(s+2)}^2+{\left(3\right)}^2}\right\}+4{\mathcal{L}}^{-1}\left\{\frac{3}{{(s+2)}^2+{\left(3\right)}^2}\right\}\\ & =& 2e^{-2t}\cos 3t+4e^{-2t}\sin 3t\\ & & \end{array}$

Therefore, the inverse laplace transform of the given function is $2e^{-2t}\cos 3t+4e^{-2t}\sin 3t$.