L{e^t} = 1/(s − 1),收敛域 s > 1,且 F(2) = 1 指数函数对:∫₀^∞ e^(−st) e^t dt = ∫₀^∞ e^(−(s−1)t) dt = 1/(s−1),仅当 s > 1 收敛。在 s = 2 处取值恰为 1。Laplace transform (table lookup)
f(t) = e^(t)
F(s) = 1/(s - 1)
Region of convergence: s > a (here s > 1)
Derivation: ∫₀^∞ e^(-(s-a)t) dt = 1/(s - a), the first-shift theorem with e^(at).
Evaluation: F(2) = 1
L{sin(2t)} = 2/(s² + 4)——对一切 s 收敛 正弦对:两次分部积分得 k/(s² + k²);k = 2 时变换为 2/(s² + 4)。由于 sin 有界,积分对一切 s 收敛。Laplace transform (table lookup)
f(t) = sin(2t)
F(s) = 2/(s^2 + 4)
Region of convergence: all s (sin is bounded)
Derivation: two integrations by parts (or Euler's formula on e^(ikt)).
Evaluation: F(3) = 0.1538