91Ó°ÊÓ

• 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,

\(F(S) = {\rm{ }}\frac{3}{{{{(2s + 5)}^3}}}\)

Identify the form of given inverted Laplace, in this case the form matches the Laplace of \({e^{at}}{t^n}.\)

Simplify the transform by making the denominator match the denominator of the transform given in the first step:

\({\mathcal{L}^{ - 1}}\left\{ {\frac{3}{{{{(2s + 5)}^3}}}} \right\} = {\mathcal{L}^{ - 1}}\left\{ {\frac{3}{{{{\left( {2\left( {s + \frac{5}{2}} \right)} \right)}^3}}}} \right\}\)

Find \(a{\rm{ and }}n\).

\(\begin{array}{c}n + 1 = 3\\n = 2\end{array}\)

\(s + \frac{5}{2} \Rightarrow a = - \frac{5}{2}\)

Solve the equation by substituting the values for \(a\) and \(n\)in the equation from the first step, then finding the coefficient.

\(\begin{array}{c}{\mathcal{L}^{ - 1}}\left\{ {\frac{3}{{{{\left( {2\left( {s + \frac{5}{2}} \right)} \right)}^3}}}} \right\} = \frac{3}{8} \cdot \frac{1}{2}{\mathcal{L}^{ - 1}}\left\{ {\frac{2}{{{{\left( {s - \left( { - \frac{5}{2}} \right)} \right)}^3}}}} \right\}\\ = \frac{3}{{16}}{\mathcal{L}^{ - 1}}\left\{ {\frac{{2!}}{{{{\left( {s - \left( { - \frac{5}{2}} \right)} \right)}^3}}}} \right\}\\ = \frac{3}{{16}}{e^{ - \frac{5}{2}t}}{t^2}\end{array}\)

Therefore the inverse laplace transform of the given function is \(\frac{3}{{16}}{e^{ - \frac{5}{2}t}}{t^2}\).