laplacetransform cos(4t)cos(6t)

解答

拉普拉斯变换

cos (4 t) cos (6 t)

\frac{s}{2 ( s ^{2} + 100 )} + \frac{s}{2 ( s ^{2} + 4 )}

求解步骤

L {cos (4 t) cos (6 t)}

使用三角恒等式改写

L {\frac{1}{2} cos (10 t) + \frac{1}{2} cos (- 2 t)}

利用拉普拉斯变换的线性特性:

对于函数

f (t), g (t)

和常数

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= \frac{1}{2} L {cos (10 t)} + \frac{1}{2} L {cos (- 2 t)}

L {cos (10 t)} : \frac{s}{s ^{2} + 100}

L {cos (- 2 t)} : \frac{s}{s ^{2} + 4}

= \frac{1}{2} \cdot \frac{s}{s ^{2} + 100} + \frac{1}{2} \cdot \frac{s}{s ^{2} + 4}

整理

\frac{1}{2} \frac{s}{s ^{2} + 100} + \frac{1}{2} \frac{s}{s ^{2} + 4} : \frac{s}{2 ( s ^{2} + 100 )} + \frac{s}{2 ( s ^{2} + 4 )}

= \frac{s}{2 ( s ^{2} + 100 )} + \frac{s}{2 ( s ^{2} + 4 )}

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