Question

Problem 0: Compute the inverse Laplace Transforms of:

Answer 1

Decompose each given F(s) into partial fractions.

$F(s) = \dfrac{s+1}{s(s-1)(s-3)}$

has partial fraction decomposition

$\dfrac{s+1}{s(s-1)(s-3)} = \dfrac as + \dfrac b{s-1} + \dfrac c{s-3}$

Combine the rational terms on the right and solve for the coefficients:

$\dfrac{s+1}{s(s-1)(s-3)} = \dfrac{a(s-1)(s-3) + b s(s-3) + c s(s-1)}{s (s-1) (s-3)}$

$1 = a(s-1)(s-3) + bs(s-3) + c s(s-1)$

$1 = 3 a + (-4 a - 3 b - c) s + (a + b + c) s^2$

$\begin{cases}3a=1 \\ -4a-3b-c = 0 \\ a+b+c=0 \end{cases} \implies a=\dfrac13, b=-\dfrac12, c=\dfrac16$

Then

$F(s) = \dfrac13 \times \dfrac1s - \dfrac12 \times \dfrac1{s-1} + \dfrac16 \times \dfrac1{s-3}$

Using the frequency-shifting property, the inverse transform is

$\boxed{f(t) = \dfrac13 - \dfrac{e^t}2 + \dfrac{e^{3t}}6}$

The other transform can be dealt with in the same manner.

$F(s) = \dfrac1{(s-1)(s-2)(s-3)} = \dfrac a{s-1} + \dfrac b{s-2} + \dfrac c{s-3}$

$\implies 1 = a(s-2)(s-3) + b(s-1)(s-3) + c(s-1)(s-2)$

$\implies 1 = 6 a + 3 b + 2 c + (-5 a - 4 b - 3 c) s + (a + b + c) s^2$

$\implies \begin{cases}6 a + 3 b + 2 c=1 \\ -5a-4b-3c = 0 \\ a+b+c=0\end{cases} \implies a=\dfrac12, b=-1, c=\dfrac12$

$\implies F(s) = \dfrac12 \times \dfrac1{s-1} - \dfrac1{s-2} + \dfrac12 \times \dfrac1{s-3}$

$\implies \boxed{f(t) = \dfrac{e^t}2 - e^{2t} + \dfrac{e^{3t}}2}$