laplacetransform 1/2 e^t-1/2 cos(t)-1/2 sin(t)

解答

拉普拉斯变换

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

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

求解步骤

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

L {e^{t}} : \frac{1}{s - 1}

L {cos (t)} : \frac{s}{s ^{2} + 1}

L {sin (t)} : \frac{1}{s ^{2} + 1}

= \frac{1}{2} \cdot \frac{1}{s - 1} - \frac{1}{2} \cdot \frac{s}{s ^{2} + 1} - \frac{1}{2} \cdot \frac{1}{s ^{2} + 1}

化简

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

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

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