How do i find the inverse laplace?

Question

$$ F(s) = \frac{2s-1}{s^2(s+1)^3} $$

If I try to use partial fractions, I end up with 8 constants to solve for! Is there some shortcut I'm not seeing? Am I supposed to simplify it first? Am I even doing the partial decomposition right?

Answer 1

You should have five.

$$\displaystyle F(s) = \frac{2s-1}{s^2(s+1)^3} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1} + \frac{D}{(s+1)^2} + \frac{E}{(s+1)^3}$$

This will produce:

• $A = 5$
• $B = -1$
• $C = -5$
• $D = -4$
• $E = -3$

The inverse Laplace transform, using this table is:

$$-\dfrac{1}{2} e^{-t} (3~ t^2+2~ t~ e^t+ 8~t-10~ e^t+10)$$

Answer 2

Alternatively, if you know some complex analysis, the ILT is the sum of the residues of $F(s) e^{s t}$ at the poles of $F$. There is a double pole at $s=0$ and a triple pole at $s=-1$, so that

$$f(t) = \left [ \frac{d}{ds} \frac{(2 s-1) e^{s t}}{(s+1)^3} \right ]_{s=0} + \left [ \frac12 \frac{d^2}{ds^2} \frac{(2 s-1) e^{s t}}{s^2} \right ]_{s=-1}$$

I leave the details to the reader; the result is

$$-\frac{1}{2} e^{-t} \left(3 t^2+8 t+10\right)-t+5$$

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{{\rm F}\pars{s} = {2s - 1 \over s^{2}\pars{s + 1}^{3}}.\quad {\cal F}\pars{t} = \int_{\gamma - \ic\infty}^{\gamma + \ic\infty}{\rm F}\pars{s}\expo{st}\, {\dd s \over 2\pi\ic}\quad\mbox{where}\quad \Re\gamma > 0}$

\begin{align} {\cal F}\pars{t} &= \int_{\gamma - \ic\infty}^{\gamma + \ic\infty}\,{2s - 1 \over s^{2} \pars{s + 1}^{3}}\,\expo{st}\,{\dd s \over 2\pi\ic} \\[3mm]&= \lim_{s \to 0}\partiald{}{s}\bracks{{\pars{2s - 1}\expo{st} \over \pars{s + 1}^{3}}} + {1 \over 2}\,\lim_{s \to\ -1}\partiald[2]{}{s} \bracks{{\pars{2s - 1}\expo{st} \over s^{2}}}\tag{1} \end{align}

\begin{align} &\lim_{s \to 0}\partiald{}{s}\bracks{{\pars{2s - 1}\expo{st} \over \pars{s + 1}^{3}}} = \lim_{s \to 0}\bracks{% {2\expo{st} + \pars{2s - 1}\expo{st}t \over \pars{s + 1}^{3}} - 3\,{\pars{2s - 1}\expo{st} \over \pars{s + 1}^{4}}} \\[3mm]&= \pars{2 - t} + 3 = 5 - t\tag{2} \\[3mm]& {1 \over 2}\,\lim_{s \to\ -1}\partiald[2]{}{s} \bracks{{\pars{2s - 1}\expo{st} \over s^{2}}} = {1 \over 2}\,\lim_{s \to 0}\partiald[2]{}{s} \braces{{\bracks{2\pars{s - 1} - 1}\expo{\pars{s - 1}t} \over \pars{s - 1}^{2}}} \\[3mm]&= {1 \over 2}\,\expo{-t}\lim_{s \to 0}\partiald[2]{}{s} \bracks{{\pars{2s - 3}\expo{st} \over \pars{s - 1}^{2}}} = {1 \over 2}\,\expo{-t}\underbrace{\pars{-3t^{2} - 8t - 10}}_{\mbox{From WA}}\tag{3} \end{align} By replacing $\pars{2}$ and $\pars{3}$ in $\pars{1}$, we get: $$\color{#0000ff}{\large% \int_{\gamma - \ic\infty}^{\gamma + \ic\infty}\,{2s - 1 \over s^{2} \pars{s + 1}^{3}}\,\expo{st}\,{\dd s \over 2\pi\ic} = -\,{3 \over 2}\expo{-t}\pars{t^{2} + {8 \over 3}\,t + {10 \over 3}} - t + 5}\,, \quad \Re\gamma > 0 $$