5 傅里叶变换

傅里叶变换

$f_T(t)=f(t+T)$fT​(t)=f(t+T)

$f_T(t) = \sum_{-\infty}^{\infty}Cne^{inw_0t} \quad w_0=\frac{2\pi}{T}基频率$fT​(t)=∑−∞∞​Cneinw0​tw0​=T2π​基频率

$Cn=\frac{1}{T}\int_{0}^{T}f_T(t)e^{-inw_0t}dt$Cn=T1​∫0T​fT​(t)e−inw0​tdt

当函数是非周期
无限不重复 $T \rightarrow \infty$T→∞
$\Delta w =(n+1)w_0 - nw_0 = w_0 = \frac{2\pi}{T}$Δw=(n+1)w0​−nw0​=w0​=T2π​
当 $T \rightarrow \infty \quad \Delta w \rightarrow 0$T→∞Δw→0

$\begin{aligned} f_T(t) &= \sum_{n=-\infty}^{\infty}Cne^{inw_0t} \\ &= \sum_{n=-\infty}^{\infty} \frac{1}{T}\int_{0}^{T}f_T(t)e^{-inw_0t}dt e^{inw_0t} \\ &= \sum_{n=-\infty}^{\infty} \frac{\Delta w}{2\pi} \int_{0}^{T}f_T(t)e^{-inw_0t}dt e^{inw_0t} \\ \end{aligned}$fT​(t)​=n=−∞∑∞​Cneinw0​t=n=−∞∑∞​T1​∫0T​fT​(t)e−inw0​tdteinw0​t=n=−∞∑∞​2πΔw​∫0T​fT​(t)e−inw0​tdteinw0​t​

$w=\frac{\pi}{L}=\frac{2\pi}{T}$w=Lπ​=T2π​

$\frac{1}{T}=\frac{\Delta w}{2\pi}$T1​=2πΔw​

当 $T \rightarrow \infty$T→∞

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dt \rightarrow \int_{-\infty}^{\infty}dt$∫−2π​2π​​dt→∫−∞∞​dt

$nw_0 \rightarrow w$nw0​→w

$\sum_{n=-\infty}^{\infty} \Delta w \rightarrow \int_{-\infty}^{+\infty}dw$∑n=−∞∞​Δw→∫−∞+∞​dw

结论：

$f(t)=\frac{1}{2\pi} \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(t) e^{-iwt}dt \;e^{iwt}dw$f(t)=2π1​∫−∞+∞​∫−∞+∞​f(t)e−iwtdteiwtdw

傅里叶变换FT: $F(w)=\int_{-\infty}^{+\infty} f(t) e^{-iwt}dt$F(w)=∫−∞+∞​f(t)e−iwtdt

傅里叶变换逆变换： $f(t)=\frac{1}{2\pi} \int_{-\infty}^{+\infty} F(w) e^{iwt}dw$f(t)=2π1​∫−∞+∞​F(w)eiwtdw

原视频：
https://www.bilibili.com/video/av35810587/?spm_id_from=333.788.videocard.0