\[\widetilde{P}(w,t) = e^{- iwU_{0}e^{\lambda t} - \frac{\lambda}{2B}w^{2}(e^{2\lambda t} - 1)}\]对w进行傅里叶逆变换
时间: 2023-07-24 09:33:53 浏览: 45
傅里叶逆变换公式为:
$$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\widetilde{f}(w)e^{iwt}dw$$
将给定的式子带入傅里叶逆变换公式中,即可得到:
$$\begin{aligned} P(w,t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}\widetilde{P}(u,t)e^{-iuw}du\\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{- iuU_{0}e^{\lambda t} - \frac{\lambda}{2B}u^{2}(e^{2\lambda t} - 1)}e^{-iuw}du\\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{- iuU_{0}e^{\lambda t}}e^{- \frac{\lambda}{2B}u^{2}(e^{2\lambda t} - 1)}e^{-iuw}du\\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{- iuU_{0}e^{\lambda t}}e^{- \frac{\lambda}{2B}(u(e^{\lambda t} - iwB\lambda) - iwB\lambda)^{2} + \frac{\lambda}{2B}w^{2}B^{2}\lambda^{2}(e^{2\lambda t} - 1)}du\\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{- iuU_{0}e^{\lambda t}}e^{- \frac{\lambda}{2B}(u(e^{\lambda t} - iwB\lambda) - iwB\lambda)^{2}}du \cdot e^{\frac{\lambda}{2}w^{2}(e^{2\lambda t} - 1)}\end{aligned}$$
其中,我们对第三行到第四行进行了平方配方法,将指数部分化为二次型的形式,再通过高斯积分将u的部分积掉。因此,傅里叶逆变换后的式子为:
$$\boxed{P(w,t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{- iuU_{0}e^{\lambda t}}e^{- \frac{\lambda}{2B}(u(e^{\lambda t} - iwB\lambda) - iwB\lambda)^{2}}du \cdot e^{\frac{\lambda}{2}w^{2}(e^{2\lambda t} - 1)}}$$
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