Inverse Transforms of Rational Functions

Let $F(s)$ be a rational function of $s$ . We give a few examples to explain the methods for calculating the inverse Laplace transform of $F(s).$

EXAMPLE 10.3.16  

1. DENOMINATOR OF MATHEND000# HAS DISTINCT REAL ROOTS:
   $${\mbox{If }}\;\; F(s) = \frac{(s+1)(s+3)}{s(s+2)(s+8)} \;\; {\mbox{ find }} f(t).$$
   
   Solution: $F(s) = \displaystyle\frac{3}{16 s} + \frac{1}{ 12 (s + 2)} + \frac{35}{ 48 (s+8)}.$ Thus,

   $f(t) = \displaystyle\frac{3}{16} + \frac{1}{12} e^{-2t}+ \frac{35 }{48} e^{-8t}.$

2. DENOMINATOR OF MATHEND000# HAS DISTINCT COMPLEX ROOTS:
   $${\mbox{ If }} \;\; F(s) = \frac{4s+3 }{s^2 + 2 s + 5} \;\; {\mbox{ find }} \;\; f(t).$$
   
   Solution: $F(s) = \displaystyle 4 \frac{s+1}{(s+1)^2 + 2^2} - \frac{1}{2} \cdot \frac{2}{ (s+1)^2 + 2^2}.$ Thus,

   $f(t) = \displaystyle 4 e^{-t} \cos (2 t) - \frac{1}{2} e^{-t} \sin (2t).$

3. DENOMINATOR OF MATHEND000# HAS REPEATED REAL ROOTS:
   $${\mbox{ If }} \;\; F(s) = \frac{3s+4}{(s+1)(s^2 + 4 s + 4)} \;\; {\mbox{ find }} \;\; f(t).$$
   
   Solution: Here,
   $$F(s) = \frac{3s+4}{(s+1)(s^2 + 4 s + 4)} = \frac{3s+4}{(s+1)(s+2)^2} = \frac{a}{s+1} + \frac{b}{s+2} + \frac{c}{(s+2)^2}.$$
   Solving for $a, b$ and $c$ , we get $F(s) = \frac{1}{s+1} - \frac{1}{s+2} + \frac{2}{ (s+2)^2} = \frac{1}{s+1} - \frac{1}{s+2} + 2 \frac{d}{ds} \left( - \frac{1}{ (s+2)}\right).$ Thus, $f(t) = e^{-t} - e^{-2t} + 2 te^{-2t}.$