Determine the inverse Laplace transform of the given function.

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

• 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 \(\mathcal{L}\{ f\} = F\)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,

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

We Know that

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

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

Rewriting the function as:

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

Finding the inverse Laplace transform.

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

Therefore, the inverse Laplace transform of the given function is \({\rm{sin}}(2t)\).