# Laplace ^(-1) of (s^3)/(s-e^-1)^4 ?

• May 31st 2012, 01:07 AM
mariusrb
Laplace ^(-1) of (s^3)/(s-e^-1)^4 ?
Laplace ^(-1) of (s^3)/(s-e^-1)^4
• May 31st 2012, 05:21 AM
Prove It
Re: Laplace ^(-1) of (s^3)/(s-e^-1)^4 ?
Quote:

Originally Posted by mariusrb
Laplace ^(-1) of (s^3)/(s-e^-1)^4

I suggest you let \displaystyle \displaystyle \begin{align*} u = s - e^{-1} \end{align*}, then

\displaystyle \displaystyle \begin{align*} s &= u + e^{-1} \\ s^3 &= \left(u + e^{-1}\right)^3 \\ &= u^3 + 3e^{-1}u^2 + 3e^{-2}u + e^{-3} \\ &= \left(s - e^{-1}\right)^3 + 3e^{-1}\left(s - e^{-1}\right)^2 + 3e^{-2}\left(s - e^{-1}\right) + e^{-3} \end{align*}

then you can write

\displaystyle \displaystyle \begin{align*} \mathcal{L}^{-1}\left\{ \frac{s^3}{\left(s - e^{-1}\right)^4} \right\} &= \mathcal{L}^{-1}\left\{ \frac{\left(s - e^{-1}\right)^3 + 3e^{-1}\left(s - e^{-1}\right)^2 + 3e^{-2}\left(s - e^{-1}\right) + e^{-3}}{\left(s - e^{-1}\right)^4} \right\} \\ &= \mathcal{L}^{-1} \left\{ \frac{1}{s - e^{-1}} \right\} + 3e^{-1}\mathcal{L}^{-1}\left\{ \frac{1}{\left(s - e^{-1}\right)^2} \right\} + 3e^{-2}\mathcal{L}^{-1}\left\{ \frac{1}{\left(s - e^{-1}\right)^3} \right\} + e^{-3}\mathcal{L}^{-1}\left\{ \frac{1}{\left( s - e^{-1} \right)^4} \right\} \end{align*}

and you can now apply the shift theorem.
• May 31st 2012, 06:47 AM
mariusrb
Re: Laplace ^(-1) of (s^3)/(s-e^-1)^4 ?
Good idea.Thanks!