inverselaplace (6s+2)/(s(9s^2+6s+2))

解答

的拉普拉斯逆变换

\frac{6 s + 2}{s ( 9 s ^{2} + 6 s + 2 )}

H (t) - e^{- \frac{t}{3}} (cos (\frac{t}{3}) - sin (\frac{t}{3}))

求解步骤

L^{- 1} {\frac{6 s + 2}{s ( 9 s ^{2} + 6 s + 2 )}}

将

\frac{6 s + 2}{s ( 9 s ^{2} + 6 s + 2 )}

用部份分式展开:

\frac{1}{s} - \frac{9 s}{9 s ^{2} + 6 s + 2}

= L^{- 1} {\frac{1}{s} - \frac{9 s}{9 s ^{2} + 6 s + 2}}

展开

\frac{9 s}{9 s ^{2} + 6 s + 2} : \frac{s + \frac{1}{3}}{( s + \frac{1}{3} ) ^{2} + \frac{1}{9}} - \frac{1}{3} \cdot \frac{1}{( s + \frac{1}{3} ) ^{2} + \frac{1}{9}}

= L^{- 1} {\frac{1}{s} - (\frac{s + \frac{1}{3}}{( s + \frac{1}{3} ) ^{2} + \frac{1}{9}} - \frac{1}{3} \cdot \frac{1}{( s + \frac{1}{3} ) ^{2} + \frac{1}{9}})}

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{1}{s}} - L^{- 1} {\frac{s + \frac{1}{3}}{( s + \frac{1}{3} ) ^{2} + \frac{1}{9}} - \frac{1}{3} \cdot \frac{1}{( s + \frac{1}{3} ) ^{2} + \frac{1}{9}}}

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

L^{- 1} {\frac{s + \frac{1}{3}}{( s + \frac{1}{3} ) ^{2} + \frac{1}{9}} - \frac{1}{3} \cdot \frac{1}{( s + \frac{1}{3} ) ^{2} + \frac{1}{9}}} : e^{- \frac{t}{3}} (cos (\frac{t}{3}) - sin (\frac{t}{3}))

= H (t) - e^{- \frac{t}{3}} (cos (\frac{t}{3}) - sin (\frac{t}{3}))

\frac{d}{d x} (\frac{2 x}{125})

t an g e n t \frac{6}{x ^{2} + 2}

\frac{d}{d x} (\frac{e ^{x} ( x - 2 )}{x ^{3}})

d er i v a t i v e f (x) = \frac{3}{4 x + 1}

x \to 1 - lim (- \frac{1}{x - 1})