Question

Determine the inverse Laplace transform of the given function.

$\frac{\mathbf{6}}{{\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{4}}}$.

Short answer

The inverse laplace transform of the given function is$e^t{t}^3$.

Step-by-step solution

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

Find inverse laplace transform for the given function

The given function is $\frac{6}{{(s-1)}^4}$.

Find the inverse laplace transform of $\frac{6}{{(s-1)}^4}$using as ${\mathcal{L}}^{-1}\left\{\frac{n!}{{(s-a)}^{n+1}}\right\}=e^{at}{t}^n,n=1,2,\dots$as

$\begin{array}{rcl}{\mathcal{L}}^{-1}\left\{\frac{6}{{(s-1)}^4}\right\}& =& {\mathcal{L}}^{-1}\left\{\frac{3!}{{(s-1)}^{3+1}}\right\}\\ & =& e^t{t}^3\\ & & \end{array}$

Therefore, the inverse laplace transform of the given function is$e^t{t}^3$.