Partial fractions and residue theorem

Question

I need to find the inverse Laplace transform of this function:

$$F(s) = \frac{50(s+1)}{s(s^2+20s+116)(0.8s+1)} $$

$$ \frac{50(s+1)}{s(s^2+20s+116)(0.8s+1)} = \frac{K_1}{s} + \frac{K_2}{0.8s+1} + \frac{K_3s + K_4}{s^2+20s+116} $$

I know that $ K_1 = 0.431$ and $ K_2 = 0.135 $ so I rewrite it as follows

$$ \frac{50(s+1)}{s(s^2+20s+116)(0.8s+1)} = \frac{0.431}{s} + \frac{0.135}{0.8s+1} + \frac{K_3s + K_4}{s^2+20s+116} $$

Now, if I understood correctly, to find $K_3$ and $K_4$ I can assign two random values to $ s $, ending up with a system of two equations. I rewrite everything as follows to make it somewhat faster:

$$ 50s + 50 = 0.431(s^2+20s+116)(0.8s+1) + 0.135s(s^2+20s+116) + s(0.8s+1)(K_3s + K_4)$$

For $ s = 1 $

$$ 100 = 0.431(137(1.8) + 0.135(137) + 1.8K_3 + 1.8K_4 $$

$$ K_3 = -13.77 - K_4$$

For $ s = 2 $

$$150 = 0.431(160)(2.6)+0.27(160) + 5.2(2K_3+K_4) $$

$$ K_4 = 2K_3 - 13.94$$

$$ \left\{ \begin{array}{c} K_3 = -13.77 - K_4 \\ K_4 = 2K_3 - 13.94 \\ \end{array} \right. $$

ending up with $K_3 = 2.05$ and $K_4 = -15.83$ which are both wrong. The correct result should be $K_3 = 0.566 $ and $K_4 = -11.15$

I don't know where the mistake is.

Answer 1

Lets use the grind it out method of Example 3.

We have

$$\dfrac{50(s+1)}{s(.8s + 1)(s^2+20s+116)} = \dfrac{250(s+1)}{s(4 s + 5)(s^2+20s+116)} = \dfrac{A}{s} + \dfrac{B}{4s+5} + \dfrac{Cs+D}{s^2+20s+115}$$

Cross multiplying $$250(s+1) = A(4s+5)(s^2+20s+116) + B(s)(s^2+20s+116) + (Cs+D)(s)(4s+5)$$

Expanding

• $A(4s+5)(s^2+20s+116) = 4As^3 + 85 A S^2 + 564 A s + 580A$

• $B(s)(s^2+20s+116) = Bs^3 + 20 B s^2 + 116 B s$

• $(Cs+D)(s)(4s+5) = 4Cs^3 + 4D s^2 + 5C s^2 + 5D s$

Equating terms, we have the system

$$\begin{align} 4A + B + 4C + 0D &= 0 \\ 85 A +20 B + 5C + 4 D &= 0 \\ 564 A + 116 B + 0 C + 5D &= 250\\580 A &= 250\end{align}$$

Solving

$$A = \dfrac{25}{58},~~ B = \dfrac{800}{1481}, ~~C = - \dfrac{48625}{85898}, ~~D = - \dfrac{479000}{42949}$$

Next, try this approach to see where you made a slight error.

Answer 2

Working with whole numbers, $$F(s)=\frac{250 (s+1)}{s (s^2+20s+116)(4 s+5) }=\frac A s+\frac {B s+C}{s^2+20s+116}+\frac D{4s+5}$$ $$250 (s+1)=A(4s+5)(s^2+20s+116)+(Bs+C)s(4s+5)+Ds(s^2+20s+116)$$

$$s=0 \implies 250=A \times 5\times 116=580 \implies A=\frac{25}{58}$$ $$s=-\frac 54 \implies -\frac{125}{2}=-\frac{7405}{64} D\implies D=\frac{800}{1481}$$

Now, make $s=\pm 1$ to obtain $B+C$ and $B-C$ (this is simpler than using $1$ and $2$).

I think that $K_2=0.135$ is not correct.