laplacetransform (t-1)cos(4(t-1))+t^2

解答

拉普拉斯变换

(t - 1) cos (4 (t - 1)) + t^{2}

- \frac{- cos ( 4 ) s ^{2} - 8 sin ( 4 ) s + 16 cos ( 4 )}{( s ^{2} + 16 ) ^{2}} - \frac{cos ( 4 ) s + 4 sin ( 4 )}{s ^{2} + 16} + \frac{2}{s ^{3}}

求解步骤

L {cos (4 (t - 1)) (t - 1) + t^{2}}

展开

(t - 1) cos (4 (t - 1)) + t^{2} : t cos (4 t - 4) - cos (4 t - 4) + t^{2}

= L {t cos (4 t - 4) - cos (4 t - 4) + t^{2}}

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)}

= L {t cos (4 t - 4)} - L {cos (4 t - 4)} + L {t^{2}}

L {t cos (4 t - 4)} : - \frac{- cos ( 4 ) s ^{2} - 8 sin ( 4 ) s + 16 cos ( 4 )}{( s ^{2} + 16 ) ^{2}}

L {cos (4 t - 4)} : \frac{cos ( 4 ) s + 4 sin ( 4 )}{s ^{2} + 16}

L {t^{2}} : \frac{2}{s ^{3}}

= - \frac{- cos ( 4 ) s ^{2} - 8 sin ( 4 ) s + 16 cos ( 4 )}{( s ^{2} + 16 ) ^{2}} - \frac{cos ( 4 ) s + 4 sin ( 4 )}{s ^{2} + 16} + \frac{2}{s ^{3}}

f (x) = \frac{x ^{2}}{4 - cos ( x )}

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