laplacetransform f(t)=(1+e^{4t})^2

解答

拉普拉斯变换

f (t) = (1 + e^{4 t})^{2}

\frac{1}{s} + \frac{2}{s - 4} + \frac{1}{s - 8}

求解步骤

L {(1 + e^{4 t})^{2}}

展开

(1 + e^{4 t})^{2} : 1 + 2 e^{4 t} + e^{8 t}

= L {1 + 2 e^{4 t} + e^{8 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)}

= L {1} + 2 L {e^{4 t}} + L {e^{8 t}}

L {1} : \frac{1}{s}

L {e^{4 t}} : \frac{1}{s - 4}

L {e^{8 t}} : \frac{1}{s - 8}

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

整理

\frac{1}{s} + 2 \frac{1}{s - 4} + \frac{1}{s - 8} : \frac{1}{s} + \frac{2}{s - 4} + \frac{1}{s - 8}

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

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