inverselaplace 1/(7s^2+37.4s+19990)

解答

的拉普拉斯逆变换

\frac{1}{7 s ^{2} + 37.4 s + 19990}

0.00267 \dots e^{- 2.67142 \dots t} sin (53.37206 \dots t)

求解步骤

L^{- 1} {\frac{1}{7 s ^{2} + 37.4 s + 19990}}

展开

\frac{1}{7 s ^{2} + 37.4 s + 19990} : \frac{1}{7} \cdot \frac{1}{( s + 2.67142 \dots ) ^{2} + 2848.57775 \dots}

= L^{- 1} {\frac{1}{7} \cdot \frac{1}{( s + 2.67142 \dots ) ^{2} + 2848.57775 \dots}}

利用逆拉普拉斯变换的常数乘法特性:

对于函数

f (t)

和常数

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{1}{7} L^{- 1} {\frac{1}{( s + 2.67142 \dots ) ^{2} + 2848.57775 \dots}}

L^{- 1} {\frac{1}{( s + 2.67142 \dots ) ^{2} + 2848.57775 \dots}} : e^{- 2.67142 \dots t} \frac{1}{53.37206 \dots} sin (53.37206 \dots t)

= \frac{1}{7} e^{- 2.67142 \dots t} \frac{1}{53.37206 \dots} sin (53.37206 \dots t)

化简

\frac{1}{7} e^{- 2.67142 \dots t} \frac{1}{53.37206 \dots} sin (53.37206 \dots t) : 0.00267 \dots e^{- 2.67142 \dots t} sin (53.37206 \dots t)

= 0.00267 \dots e^{- 2.67142 \dots t} sin (53.37206 \dots t)

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