Find the inverse transform of [latex]F(s) = \frac{4(s + 3)}{s(s + 2)²}[/latex]

The given transform has a simple pole at s = 0 and a double pole at s = −2. Factoring out one of the multiple poles and expanding the remainder by partial fractions yields [latex]F(s) = \frac{1}{s + 2}\left[\frac{4(s + 3)}{s(s + 2)}\right] = \frac{1}{s + 2} \left[\frac{6}{s} - \frac{2}{s + 2}\right] [/latex] Multiplying through by the removed factor and expanding again by partial fractions produces [latex]F(s) = \frac{6}{s(s + 2)} - \frac{2}{(s + 2)^{2}} = \frac{3}{s} - \frac{3}{s + 2} - \frac{2}{(s + 2)^{2}}[/latex] The last expansion on the right yields the inverse transform as [latex]f(t) = [3 − 3e^{−2t} −2te^{−2t}] u(t)[/latex]

Question 9.15: Find the inverse transform of F(s) = 4(s + 3)/s(s + 2)²...

The Analysis and Design of Linear Circuits [1078619]

Find the inverse transform of

$F(s) = \frac{4(s + 3)}{s(s + 2)²}$

Step-by-Step

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The given transform has a simple pole at s = 0 and a double pole at s = −2. Factoring out one of the multiple poles and expanding the remainder by partial fractions yields

$F(s) = \frac{1}{s + 2}\left[\frac{4(s + 3)}{s(s + 2)}\right] = \frac{1}{s + 2} \left[\frac{6}{s} – \frac{2}{s + 2}\right]$

Multiplying through by the removed factor and expanding again by partial fractions produces

$F(s) = \frac{6}{s(s + 2)} – \frac{2}{(s + 2)^{2}} = \frac{3}{s} – \frac{3}{s + 2} – \frac{2}{(s + 2)^{2}}$

The last expansion on the right yields the inverse transform as

$f(t) = [3 − 3e^{−2t} −2te^{−2t}] u(t)$