Determine the inverse Laplace transform of the given function.

\(\frac{4}{{{s^2} + 9}}\)

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 \(\left[ {0,\infty } \right)\) 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

\(f = {\mathcal{L}^{ - 1}}\{ F\} \)

\({\mathcal{L}^{ - 1}}\left\{ {\frac{{n!}}{{{{(s - a)}^{n + 1}}}}} \right\} \Leftrightarrow {e^{at}}{t^n},n = 1,2, \ldots \)

Given the function,

\(F\left( S \right) = \frac{4}{{{s^2} + 9}}\)

We Know that

\(\mathcal{L}\left\{ {{\rm{sin}}bt} \right\} = \frac{b}{{{s^2} + {b^2}}}\)

Identify the form of given inverted Laplace, in this case the form matches the Laplace of \({\rm{sin}}bt\).

Rewritting the function

\({\mathcal{L}^{ - 1}}\left\{ {\frac{b}{{{s^2} + {b^2}}}} \right\} = {\mathcal{L}^{ - 1}}\left\{ {\frac{4}{{{s^2} + 9}}} \right\}\)

Let \({b^2} = 9 \Rightarrow b = 3\) and fill in the original equation.

Finding the inverse laplace transform.

\(\begin{array}{c}{\mathcal{L}^{ - 1}}\left\{ {\frac{4}{{{s^2} + 9}}} \right\} = \frac{4}{3}{\mathcal{L}^{ - 1}}\left\{ {\frac{3}{{{s^2} + 9}}} \right\}\\ = \frac{4}{3}{\rm{sin}}\left( {3t} \right)\end{array}\)

Therefore the inverse laplace transform of the given function is \(\frac{4}{3}{\rm{sin}}\left( {3t} \right)\).