The Inverse Laplace Transform (example #3)

Determine f(t) given:

$$ F(s) = \frac{8}{s(s+1)^3} $$ Begin by expanding F(s) using partial fraction decomposition:

$$ F(s) = \frac{8}{s(s+1)^3} = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+1)^2} + \frac{D}{(s+1)^3} \qquad,\;(Eqn\;1) $$

We will now examine two different approaches to obtaining the expansion coefficients A, B, C and D.

Determine coefficient "A"

Recall the definition of the Heaviside theorem (from the intro to inverse Laplace Transform page): $$ k_i = (s+p_i) \; F(s) \;\Big|_{s=-p_i} $$ Therefore: $$ \begin{align} A &= (s+0) \; F(s) \;\Big|_{s=0} \\\\ &= \frac{8}{(s+1)^3} \;\Big|_{s=0} \\\\ &= \frac{8}{(0+1)^3} \\\\ A &= 8 \end{align} $$

Determine coefficient "D"

Again, we use the definition of the Heaviside theorem: $$ k_i = (s+p_i) \; F(s) \Big|_{s=-p_i} $$ Therefore: $$ \begin{align} D &= (s+1)^3 \; F(s) \;\Big|_{s=-1} \\\\ &= \frac{8}{s} \;\Big|_{s=-1} \\\\ &= \frac{8}{-1} \\\\ D &= -8 \end{align} $$

Determine coefficient "C"

Since we are dealing with a repeated root, let us recall the Residue Method: $$ k_{n-m} = \Big(\frac{1}{m!}\Big) \frac{d^m}{ds^m}[(s+p)^n \; F(s)]\;\Big|_{s=-p} $$ If we refer back to equation #1, we see that for the coefficient "C", we have the following: $$ n=3, \; m=1 $$ Therefore:

$$ \begin{align} k_{(n-m)} = k_2 = C &= \Big(\frac{1}{1!}\Big) \frac{d}{ds} [(s+1)^3 \; F(s)] \;\Big|_{s=-1} \\\\ &= \frac{d}{ds} \Big[ \frac{8}{s} \Big] \;\Big|_{s=-1} \\\\ &= \frac{s(0)-8(1)}{s^2} \;\Big|_{s=-1} \\\\ &= \frac{-8}{(-1)^2} \;\Big|_{s=-1} \\\\ C &= -8 \end{align} $$

Determine coefficient "B"

Once again, this is a repeated root and we will use the Residue Method to find "B". As a reminder, the residue method is defined as: $$ k_{n-m} = \Big(\frac{1}{m!}\Big) \frac{d^m}{ds^m}[(s+p)^n \; F(s)]\;\Big|_{s=-p} $$ Refering back to equation #1, we see that for the coefficient "B", we have the following: $$ n=3, \; m=2 $$ Therefore:

$$ \begin{align} k_{(n-m)} = k_1 = B &= \Big(\frac{1}{2!}\Big) \frac{d^2}{ds^2} [(s+1)^3 \; F(s)] \;\Big|_{s=-1} \\\\ &= \frac{1}{2}\frac{d}{ds} \Big[ \frac{-8}{s^2} \Big] \;\Big|_{s=-1} \\\\ &= \frac{1}{2} \Big[ \frac{s^2(0)-8(2)(s)}{s^4}\Big] \;\Big|_{s=-1} \\\\ &= \frac{1}{2} \Big[ \frac{16s}{s^4} \Big] \;\Big|_{s=-1} \\\\ &= \frac{8}{s^3} \;\Big|_{s=-1} \\\\ &= \frac{1}{(-1)^3} \\\\ B &= -8 \end{align} $$ Now that we have determined the expansion coefficients, go ahead and substitute them into equation #1 and obtain the following:

$$ F(s) = \frac{8}{s} - \frac{8}{s+1} - \frac{8}{(s+1)^2} - \frac{8}{(s+1)^3} \qquad,(Eqn\;2) $$

Recall equation #1: $$ F(s) = \frac{8}{s(s+1)^3} = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+1)^2} + \frac{D}{(s+1)^3} \qquad,\;(Eqn\;1) $$ Now multiply both sides of equation #1 by the common denominator:

$$ \begin{align} 8 &= A(s+1)^3 + Bs(s+1)^2 + Cs(s+1) + Ds \\\\ &= A(s+1)(s^2+2s+1) + Bs(s^2+2s+1) + Cs^2 + Cs + Ds \\\\ &= A(s^3+2s^2+s+s^2+2s+1) + Bs^3 + 2Bs^2 + Bs + Cs^2 + Cs + Ds \\\\ &= As^3 + 3As^2 + 3As + A + Bs^3 + 2Bs^2 + Bs + Cs^2 + Cs + Ds \\\\ 8 &= s^3(A+B) + s^2(3A+2B+C) + s(3A+B+C+D) + s(3A+B+C+D) + A \\\\ \end{align} $$

Now equate coefficients: $$ for \; s^3: \;\;\; A+B=0 $$ $$ for \; s^2: \;\;\; 3A+2B+C = 0 $$ $$ for \; s: \;\;\; 3A+B+C+D=0 $$ $$ for \; constants: \;\;\; A=8 $$ Now solve the system of equations. Plugging (A=8) into the equation for "s^3" gives us: $$ 8+B=0 $$ $$ B=-8 $$ Plugging (A=8, B=-8) into the equation for "s^2" gives us: $$ 3(8)+2(-8)+C=0 $$ $$ C=-8 $$ Plugging (A=8, B=-8, C=-8) into the equation for "s" gives us: $$ 3(8)-8-8+D=0 $$ $$ D=-8 $$ If we substitute these values into equation #1, we get the same result as we did with the Residue Method:

$$ F(s) = \frac{8}{s} - \frac{8}{s+1} - \frac{8}{(s+1)^2} - \frac{8}{(s+1)^3} \qquad,(Eqn\;2) $$

Rearrange equation #2 so the terms resemble properties in our table

$$ F(s) = \frac{8}{s} - \frac{8}{s+1} - 8\frac{1!}{(s+1)^2} - 4\frac{2!}{(s+1)^3} $$ The form above will allow us to use property #1 in the 1st term, property #2 on the 2nd term and property #23 on the 3rd and 4th terms. Now lets determine the inverse Laplace transform of the above expression:

$$ \begin{align} f(t) &= \mathcal{L^{-1}}[F(s)] \\\\ &= 8\mathcal{L^{-1}} \Big[ \frac{1}{s} \Big] - 8\mathcal{L^{-1}} \Big[ \frac{1}{s+1} \Big] - 8\mathcal{L^{-1}}\Big[ \frac{1!}{(s+1)^2} \Big] - 4\mathcal{L^{-1}} \Big[ \frac{2!}{(s+1)^3} \Big] \end{align} $$

$$ f(t) = 8 - 8e^{-t} - 8te^{-t} - 4t^2e^{-t} $$