1 Answer

Maxim Tulupov · Answer 1 · 2023-10-18T18:16:30+0000

Decompose each given F(s) into partial fractions.

F(s) = (s+1)/(s(s-1)(s-3))

has partial fraction decomposition

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

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

(s+1)/(s(s-1)(s-3)) = (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

$\begin{cases}3a=1 \\ -4a-3b-c = 0 \\ a+b+c=0 \end{cases} \implies a=\frac13, b=-\frac12, c=\frac16$

Then

$F(s) = \frac13 * \frac1s - \frac12 * \frac1{s-1} + \frac16 * \frac1{s-3}$

Using the frequency-shifting property, the inverse transform is

$\boxed{f(t) = \frac13 - \frac{e^t}2 + \frac{e^(3t)}6}$

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

$F(s) = \frac1{(s-1)(s-2)(s-3)} = \frac a{s-1} + \frac b{s-2} + \frac c{s-3}$

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

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

$\implies \begin{cases}6 a + 3 b + 2 c=1 \\ -5a-4b-3c = 0 \\ a+b+c=0\end{cases} \implies a=\frac12, b=-1, c=\frac12$

$\implies F(s) = \frac12 * \frac1{s-1} - \frac1{s-2} + \frac12 * \frac1{s-3}$

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

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