[足式机器人]Part2 Dr. CAN学习笔记-数学基础Ch0-5Laplace Transform of Convolution卷积的拉普拉斯变换

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Laplace Transform : $X\left(s\right)=\mathcal{L}\left[x\left(t\right)\right]={\int }_0^{\infty }x\left(t\right)e^{-st}\mathrm{d}t$

Convolution : $x\left(t\right)\ast g\left(t\right)={\int }_0^{t}x\left(\tau \right)g\left(t-\tau \right)\mathrm{d}\tau$

证明：$\mathcal{L}\left[x\left(t\right)\ast g\left(t\right)\right]=X\left(s\right)G\left(s\right)$
$\mathcal{L}\left[x\left(t\right)\ast g\left(t\right)\right]={\int }_0^{\infty }{\int }_0^{t}x\left(\tau \right)g\left(t-\tau \right)\mathrm{d}\tau e^{-st}\mathrm{d}t={\int }_0^{\infty }{\int }_{\tau }^{\infty }x\left(\tau \right)g\left(t-\tau \right)e^{-st}\mathrm{d}t\mathrm{d}\tau$
>令：$u=t-\tau ,t=u+\tau ,\mathrm{d}t=\mathrm{d}u+\mathrm{d}\tau ,t\in \left[\tau ,+\infty \right)\Rightarrow u\in \left[0,+\infty \right)$
$\mathcal{L}\left[x\left(t\right)\ast g\left(t\right)\right]={\int }_0^{\infty }{\int }_0^{\infty }x\left(\tau \right)g\left(u\right)e^{-s\left(u+\tau \right)}\mathrm{d}u\mathrm{d}\tau ={\int }_0^{\infty }x\left(\tau \right)e^{-s\tau }\mathrm{d}\tau {\int }_0^{\infty }g\left(u\right)e^{-su}\mathrm{d}u=X\left(s\right)G\left(s\right)$

$$\mathcal{L}\left[x\left(t\right)\ast g\left(t\right)\right]=\mathcal{L}\left[x\left(t\right)\right]\mathcal{L}\left[g\left(t\right)\right]=X\left(s\right)G\left(s\right)$$