zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

inverselaplace ((s^2+s))/(2s^2+2s+1)

解答

的拉普拉斯逆变换

\frac{( s ^{2} + s )}{2 s ^{2} + 2 s + 1}

解答

\frac{1}{2} δ (t) - \frac{1}{2} e^{- \frac{t}{2}} sin (\frac{t}{2})

求解步骤

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

将

\frac{s ^{2} + s}{2 s ^{2} + 2 s + 1}

用部份分式展开:

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

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

展开

\frac{1}{2 ( 2 s ^{2} + 2 s + 1 )} : \frac{1}{4} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}

= L^{- 1} {\frac{1}{2} - \frac{1}{4} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}}

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}{2}} - \frac{1}{4} L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}}

L^{- 1} {\frac{1}{2}} : \frac{1}{2} δ (t)

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}} : e^{- \frac{t}{2}} \cdot 2 sin (\frac{t}{2})

= \frac{1}{2} δ (t) - \frac{1}{4} e^{- \frac{t}{2}} \cdot 2 sin (\frac{t}{2})

整理

\frac{1}{2} δ (t) - \frac{1}{4} e^{- \frac{t}{2}} 2 sin (\frac{t}{2}) : \frac{1}{2} δ (t) - \frac{1}{2} e^{- \frac{t}{2}} sin (\frac{t}{2})

= \frac{1}{2} δ (t) - \frac{1}{2} e^{- \frac{t}{2}} sin (\frac{t}{2})

流行的例子

derivative f(x)= 1/((3x+8)^2)

d er i v a t i v e f (x) = \frac{1}{( 3 x + 8 ) ^{2}}

derivative of (-x^2+2x+1/((x^2+1)^2))

\frac{d}{d x} (\frac{- x ^{2} + 2 x + 1}{( x ^{2} + 1 ) ^{2}})

integral of cos(2pinx)

\int cos (2 πn x) d x

sum from n=0 to infinity of ((x-3)^n)/n

n = 0 \sum \infty \frac{( x - 3 ) ^{n}}{n}

derivative of 1/(e^{1/x})

\frac{d}{d x} (\frac{1}{e ^{\frac{1}{x}}})