inverselaplace (7s+10)/(7s^2+26s+25)

解答

的拉普拉斯逆变换

\frac{7 s + 10}{7 s ^{2} + 26 s + 25}

e^{- \frac{13 t}{7}} cos (\frac{6 t}{7}) - \frac{3 e ^{- \frac{13 t}{7}} sin ( \frac{6 t}{7} )}{2}

求解步骤

L^{- 1} {\frac{7 s + 10}{7 s ^{2} + 26 s + 25}}

展开

\frac{7 s + 10}{7 s ^{2} + 26 s + 25} : \frac{s + \frac{13}{7}}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}} - \frac{3}{7} \cdot \frac{1}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}}

= L^{- 1} {\frac{s + \frac{13}{7}}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}} - \frac{3}{7} \cdot \frac{1}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}}}

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{s + \frac{13}{7}}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}}} - \frac{3}{7} L^{- 1} {\frac{1}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}}}

L^{- 1} {\frac{s + \frac{13}{7}}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}}} : e^{- \frac{13 t}{7}} cos (\frac{6 t}{7})

L^{- 1} {\frac{1}{( s + \frac{13}{7} ) ^{2} + \frac{6}{49}}} : e^{- \frac{13 t}{7}} \frac{7}{6} sin (\frac{6 t}{7})

= e^{- \frac{13 t}{7}} cos (\frac{6 t}{7}) - \frac{3}{7} e^{- \frac{13 t}{7}} \frac{7}{6} sin (\frac{6 t}{7})

整理

e^{- \frac{13 t}{7}} cos (\frac{6 t}{7}) - \frac{3}{7} e^{- \frac{13 t}{7}} \frac{7}{6} sin (\frac{6 t}{7}) : e^{- \frac{13 t}{7}} cos (\frac{6 t}{7}) - \frac{3 e ^{- \frac{13 t}{7}} sin ( \frac{6 t}{7} )}{2}

= e^{- \frac{13 t}{7}} cos (\frac{6 t}{7}) - \frac{3 e ^{- \frac{13 t}{7}} sin ( \frac{6 t}{7} )}{2}

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