laplacetransform-t/4-3/8 sin(2t)

解答

拉普拉斯变换

- \frac{t}{4} - \frac{3}{8} sin (2 t)

- \frac{1}{4 s ^{2}} - \frac{3}{4 ( s ^{2} + 4 )}

求解步骤

L {- \frac{t}{4} - \frac{3}{8} sin (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}{4} L {t} - \frac{3}{8} L {sin (2 t)}

L {t} : \frac{1}{s ^{2}}

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

= - \frac{1}{4} \cdot \frac{1}{s ^{2}} - \frac{3}{8} \cdot \frac{2}{s ^{2} + 4}

整理

- \frac{1}{4} \frac{1}{s ^{2}} - \frac{3}{8} \frac{2}{s ^{2} + 4} : - \frac{1}{4 s ^{2}} - \frac{3}{4 ( s ^{2} + 4 )}

= - \frac{1}{4 s ^{2}} - \frac{3}{4 ( s ^{2} + 4 )}

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