laplacetransform 5(t-8)-10(t-4)

解答

拉普拉斯变换

5 \cdot (t - 8) - 10 \cdot (t - 4)

- \frac{5}{s ^{2}}

求解步骤

L {5 (t - 8) - 10 (t - 4)}

展开

5 (t - 8) - 10 (t - 4) : - 5 t

= L {- 5 t}

利用拉普拉斯变换的常数乘法特性:

对于函数

f (t)

和常数

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= - 5 L {t}

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

= - 5 \cdot \frac{1}{s ^{2}}

整理

- 5 \frac{1}{s ^{2}} : - \frac{5}{s ^{2}}

= - \frac{5}{s ^{2}}

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