inverselaplace (s^2+3s+5)/(s^3(s+2))

解答

的拉普拉斯逆变换

\frac{s ^{2} + 3 s + 5}{s ^{3} ( s + 2 )}

\frac{3}{8} H (t) + \frac{t}{4} + \frac{5 t ^{2}}{4} - \frac{3}{8} e^{- 2 t}

求解步骤

L^{- 1} {\frac{s ^{2} + 3 s + 5}{s ^{3} ( s + 2 )}}

将

\frac{s ^{2} + 3 s + 5}{s ^{3} ( s + 2 )}

用部份分式展开:

\frac{3}{8 s} + \frac{1}{4 s ^{2}} + \frac{5}{2 s ^{3}} - \frac{3}{8 ( s + 2 )}

= L^{- 1} {\frac{3}{8 s} + \frac{1}{4 s ^{2}} + \frac{5}{2 s ^{3}} - \frac{3}{8 ( s + 2 )}}

利用逆拉普拉斯变换的线性特性:

对于函数

f (s), g (s)

和常数

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{3}{8 s}} + L^{- 1} {\frac{1}{4 s ^{2}}} + L^{- 1} {\frac{5}{2 s ^{3}}} - L^{- 1} {\frac{3}{8 ( s + 2 )}}

L^{- 1} {\frac{3}{8 s}} : \frac{3}{8} H (t)

L^{- 1} {\frac{1}{4 s ^{2}}} : \frac{t}{4}

L^{- 1} {\frac{5}{2 s ^{3}}} : \frac{5 t ^{2}}{4}

L^{- 1} {\frac{3}{8 ( s + 2 )}} : \frac{3}{8} e^{- 2 t}

= \frac{3}{8} H (t) + \frac{t}{4} + \frac{5 t ^{2}}{4} - \frac{3}{8} e^{- 2 t}

\frac{d y}{d x} = x

n = 0 \sum \infty \frac{2}{n ^{2}}

\int_{- 4}^{2} (x + 4) d x

a re a y^{2} = 4 x, y = x^{2}

t an g e n t 8 x^{2} - 2, at x = - 4