Q7.4 - 5E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the inverse Laplace transform of the given function.

$\frac{\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{s}\mathbf{+}\mathbf{8}}$.

Step-by-Step Solution

Answer

The inverse laplace transform of the given function is $\frac{1}{2}{e}^{-2t}\sin 2t$.

Determining the inverse laplace transform

• 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 
•  $f={\mathcal{L}}^{-1}\left\{F\right\}$
•  ${\mathcal{L}}^{-1}\left\{\frac{\mathrm{n}!}{{(\mathrm{s}-\mathrm{a})}^{\mathrm{n}+1}}\right\}={\mathrm{e}}^{\mathrm{at}}{\mathrm{t}}^{\mathrm{n}},\mathrm{n}=1,2,\dots$

Find inverse laplace transform for the given function

The given function is $\frac{1}{s^2+4s+8}$.

Find the inverse laplace transform of $\frac{1}{s^2+4s+8}$using ${\mathcal{L}}^{-1}\left\{\frac{b}{{(s-a)}^2+{\left(b\right)}^2}\right\}=e^{at}\sin bt$ as:

$\begin{array}{rcl}{\mathcal{L}}^{-1}\left\{\frac{1}{{(s+2)}^2+{\left(2\right)}^2}\right\}& =& \frac{1}{2}{\mathcal{L}}^{-1}\left\{\frac{b}{{[s-(-2\left)\right]}^2+{\left(2\right)}^2}\right\}\\ & =& \frac{1}{2}{e}^{-2t}\sin 2t\end{array}$

Therefore, the inverse laplace transform of the given function is $\frac{1}{2}{e}^{-2t}\sin 2t$.