L⁻¹{1/(s² + 3s + 2)} = e^(−t) − e^(−2t) 分母分解为 (s + 1)(s + 2),故 1/(s² + 3s + 2) = 1/(s + 1) − 1/(s + 2),两项分别反变换为指数:f(t) = e^(−t) − e^(−2t)。Inverse Laplace transform via partial fractions
F(s) = 1/(s^2 + 3 s + 2)
Step 1 — Denominator roots: r = -1 (×1), r = -2 (×1)
Step 2 — Partial fractions: F(s) = 1/(s + 1) - 1/(s + 2)
Step 3 — Invert term by term: A/(s - r) → A·e^(rt)
f(t) = e^(-t) - e^(-2t)
L⁻¹{(3s + 5)/(s² + 4)} = 3cos(2t) + 2.5sin(2t) 分母 s² + 4 有共轭根 ±2i,分解保持二次形式 (3s + 5)/(s² + 4)。移位为 ((s)² + 2²) 后反变换该二次对,得 f(t) = 3·cos(2t) + (5/2)·sin(2t)。Inverse Laplace transform via partial fractions
F(s) = (3 s + 5)/(s^2 + 4)
Step 1 — Denominator roots: conjugate pair α = 0, β = 2
Step 2 — Partial fractions: F(s) = (3 s + 5)/(s^2 + 4)
Step 3 — Invert the quadratic pair: e^(αt)·[B·cos(βt) + ((C + Bα)/β)·sin(βt)]
f(t) = 3 cos(2t) + 2.5 sin(2t)