Determine the inverse Laplace transform of the given function.

$\frac{\mathbf{1}}{{\mathbf{s}}^{\mathbf{5}}}$.

• 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
• localid="1662725690417" $\mathrm{f}={\mathcal{L}}^{-1}\left\{\mathrm{F}\right\}$
• localid="1662725700279" ${\mathcal{L}}^{-1}\left\{\frac{\mathrm{n}!}{{(\mathrm{s}-\mathrm{a})}^{\mathrm{n}+1}}\right\}={\mathrm{e}}^{\mathrm{at}}{\mathrm{t}}^{\mathrm{n}},\mathrm{n}=1,2,\dots$
• Constant multiplication property of inverse laplace transform, for function and constant.
• localid="1662725707425" ${\mathrm{L}}^{-1}\{\mathrm{a}·\mathrm{f}(\mathrm{t}\left)\right\}=\mathrm{a}·{\mathrm{L}}^{-1}\left\{\mathrm{f}\right(\mathrm{t}\left)\right\}$

The given function is $\frac{1}{s^5}$.

Find the inverse laplace transform of $\frac{1}{s^5}$using localid="1662725728091" ${\mathrm{L}}^{-1}\{\mathrm{a}·\mathrm{f}(\mathrm{t}\left)\right\}=\mathrm{a}·{\mathrm{L}}^{-1}\left\{\mathrm{f}\right(\mathrm{t}\left)\right\}$and $L^{-1}\left\{\frac{\left(n\right)!}{s^{n+1}}\right\}=t^n$as:

localid="1662725718604" $\begin{array}{rcl}{\mathcal{L}}^{-1}\left\{\frac{1}{s^5}\right\}& =& {\mathcal{L}}^{-1}\left\{\frac{1}{24}·\frac{24}{s^5}\right\}\\ & =& \frac{1}{24}{\mathcal{L}}^{-1}\left\{\frac{24}{s^5}\right\}\\ & =& \frac{t^4}{24}\end{array}$

Therefore, the inverse laplace transform of the given function is$\frac{t^4}{24}$.