梦开始的地方——Fourier级数与变换

2021-12-02 23:43 作者:子瞻Louis 0人读过 | 我要投稿

这篇文章肯定是写不下太多内容的，因此我会写一些以后会用到的一些东西，因为能力有限所以文章可能会稍微有些简略……

Euler公式

对 $e%5E%7Bix%7D$ 展开为Maclaurin级数，有

$%5Cbegin%7Baligned%7De%5E%7Bix%7D%26%3D1%2Bix-%5Cfrac%7Bx%5E2%7D%7B2!%7D-i%5Cfrac%7Bx%5E2%7D%7B3!%7D%2B%5Cfrac%7Bx%5E4%7D%7B4!%7D-i%5Cfrac%7Bx%5E5%7D%7B5!%7D%2B%E2%80%A6%5C%5C%26%3D1-%5Cfrac%7Bx%5E2%7D%7B2!%7D%2B%5Cfrac%7Bx%5E4%7D%7B4!%7D-%E2%80%A6%2Bi%5Cleft(x-%5Cfrac%7Bx%5E3%7D%7B3!%7D%2B%5Cfrac%7Bx%5E5%7D%7B5!%7D-%E2%80%A6%5Cright)%5Cend%7Baligned%7D$

根据正弦余弦的Maclaurin级数，便有

$e%5E%7Bix%7D%3D%5Ccos%20x%2Bi%5Csin%20x$

这只是及其简略且不严谨的证明，不严谨在于这里并没有指出它满足交换求和次序的条件

它的详细证明以及推广什么的这里就不讨论了，互联网一搜就是一堆

Fourier级数

众所周知，正弦函数和余弦函数都是周期为2π的连续函数，因此多个正弦函数和余弦函数的和仍然是周期函数

$%5Csin%20x%2B2%5Csin%203x%3D%5Csin%20(x%2B2n%5Cpi)%2B2%5Csin%203(x%2B2n%5Cpi)%0A$

$%5Ccos%202x%2B%5Csin%204x%3D%5Ccos%202(x%2Bn%5Cpi)%2B%5Csin%204(x%2Bn%5Cpi)$

$%E2%80%A6%E2%80%A6$

那么反过来，是不是周期函数都可以写成正弦函数与余弦函数的和呢？

Fourier就正好注意到了这点，他认为，任何在一个周期内只有有限个第一类间断点的周期函数都可表为正余弦函数之和，即对周期函数f(t)，有

$f(t)%3D%5Cfrac%7Ba_%7B0%7D%7D%7B2%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7Da_%7Bn%7D%5Ccos%20%5Cleft(%5Cfrac%7B2%5Cpi%20nt%7D%7BT%7D%5Cright)%20%2Bb_%7Bn%7D%5Csin%20%5Cleft(%5Cfrac%7B2%5Cpi%20nt%7D%7BT%7D%5Cright)$

其中T为f(t)的最小正周期，此即为经典的Fourier三角级数

假设f(t)是处处连续的，且Fourier级数收敛于f(t)，我们来求解他的Fourier系数，利用欧拉公式，有

$%5Ccos%20x%3D%5Cfrac%7Be%5E%7Bix%7D%2Be%5E%7B-ix%7D%7D%7B2%7D%2C%5Csin%20x%3D%5Cfrac%7Be%5E%7Bix%7D-e%5E%7B-ix%7D%7D%7B2i%7D%3D-i%5Cfrac%7Be%5E%7Bix%7D-e%5E%7B-ix%7D%7D%7B2%7D$

代入到其中

$%5Cbegin%7Baligned%7Df(t)%26%3D%5Cfrac%7Ba_%7B0%7D%7D%7B2%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Ba_%7Bn%7D%7D%7B2%7D%5Cleft(e%5E%7B%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D%2Be%5E%7B-%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D%5Cright)-i%5Cfrac%7Bb_%7Bn%7D%7D%7B2%7D%5Cleft(e%5E%7B%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D-e%5E%7B-%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D%5Cright)%5C%5C%26%3D%5Cfrac%7Ba_%7B0%7D%7D%7B2%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Ba_%7Bn%7D-ib_%7Bn%7D%7D%7B2%7De%5E%7B%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D%2B%5Cfrac%7Ba_%7Bn%7D%2Bib_%7Bn%7D%7D%7B2%7De%5E%7B-%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D%5Cend%7Baligned%7D$

看起来越变越复杂了，但我们可以做如下变换

$c_%7Bn%7D%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%0A%5Cfrac%7Ba_%7Bn%7D-ib_%7Bn%7D%7D%7B2%7D%20%26%20%5Cmbox%7Bfor%7D%0A%26%20n%EF%BC%9E0%20%5C%5C%20%5Cfrac%7Ba_%7B0%7D%7D%7B2%7D%20%26%20%5Cmbox%7Bfor%7D%20%26%20n%3D0%20%5C%5C%0A%5Cfrac%7Ba_%7B-n%7D%2Bb_%7B-n%7D%7D%7B2%7D%20%26%20%5Cmbox%7Bfor%7D%20%26%20n%EF%BC%9C0%0A%5Cend%7Barray%7D%5Cright.$

则

$f(t)%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D$

到这一步后就看上去似乎无从下手了，

但如果我们用一指数函数乘以f(t)再对其在一个周期上进行积分（这里选[-T/2,T/2]）

$%5Cbegin%7Baligned%7Df(t)e%5E%7B-%5Cfrac%7B2%5Cpi%20ikt%7D%7BT%7D%7D%26%3D%5Cleft(%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D%5Cright)e%5E%7B-%5Cfrac%7B2%5Cpi%20ikt%7D%7BT%7D%7D%5C%5C%26%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20i(n-k)t%7D%7BT%7D%7D%5Cend%7Baligned%7D$

$%5Cbegin%7Baligned%7D%5CRightarrow%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(t)e%5E%7B-%5Cfrac%7B2%5Cpi%20ikt%7D%7BT%7D%7Ddt%26%3D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20i(n-k)t%7D%7BT%7D%7Ddt%5C%5C%26%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20i(n-k)t%7D%7BT%7D%7Ddt%5Cend%7Baligned%7D$

其中，积分与和式可以交换次序是因为该级数的和收敛

假设该积分绝对值收敛，这样我们便可以对其每一项讨论

n=k时

$%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20i(n-k)t%7D%7BT%7D%7Ddt%3D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Dc_%7Bn%7De%5E%7B0%7Ddt%3Dc_%7Bn%7DT$

n≠k时

$%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20i(n-k)t%7D%7BT%7D%7Ddt%3Dc_%7Bn%7D%5Cfrac%7BT%7D%7B2%5Cpi%20i(n-k)t%7De%5E%7B%5Cfrac%7B2%5Cpi%20i(n-k)t%7D%7BT%7D%7D%5Cvert_%7B-T%2F2%7D%5E%7BT%2F2%7D%3D0$

于是我们得到

$%5Cbegin%7Baligned%7Dc_%7Bk%7D%26%3D%5Cfrac%7B1%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(t)e%5E%7B-%5Cfrac%7B2%5Cpi%20ikt%7D%7BT%7D%7Ddt%5C%5C%26%3D%5Cfrac%7B1%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(t)%5Ccos%5Cleft(%7B%5Cfrac%7B2%5Cpi%20kt%7D%7BT%7D%7D%5Cright)dt-i%5Cfrac%7B1%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(t)%5Csin%5Cleft(%7B%5Cfrac%7B2%5Cpi%20kt%7D%7BT%7D%7D%5Cright)dt%5Cend%7Baligned%7D$

根据 $c_n$ 的定义对比实部与虚部又可得：

$a_%7Bk%7D%3Dc_n%2Bc_%7B-n%7D%3D%5Cfrac%7B2%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(t)%5Ccos%5Cleft(%7B%5Cfrac%7B2%5Cpi%20kt%7D%7BT%7D%7D%5Cright)dt$

$b_%7Bk%7D%3D-i(c_n-c_%7B-n%7D)%3D%5Cfrac%7B2%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(t)%5Csin%5Cleft(%7B%5Cfrac%7B2%5Cpi%20kt%7D%7BT%7D%7D%5Cright)dt$

这样就能得到周期函数的Fourier级数展开了，

$f(t)%3D%5Cfrac%7Ba_%7B0%7D%7D%7B2%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7Da_%7Bn%7D%5Ccos%5Cleft(%7B%5Cfrac%7B2%5Cpi%20nt%7D%7BT%7D%7D%5Cright)%2Bb_%7Bn%7D%5Csin%20%5Cleft(%7B%5Cfrac%7B2%5Cpi%20nt%7D%7BT%7D%7D%5Cright)$

为 $f(t)$ 的Fourier三角级数，同样有一下更优美形式的三角级数

$f(t)%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cleft(%5Cfrac%7B1%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(x)e%5E%7B%7B-%5Cfrac%7B2%5Cpi%20nx%7D%7BT%7D%7D%7Ddx%5Cright)e%5E%7B%7B%5Cfrac%7B2%5Cpi%20nx%7D%7BT%7D%7D%7D$

Fourier积分

看到这里的同学不难发现Fourier级数仅仅局限于对周期函数展开，那么非周期函数要怎么办呢？

对此，我们可以将一个非周期连续函数看为周期为无穷大的周期函数

令 $%5Cfrac%20nT%3D%5Comega_n$ ，则 $%5Cfrac1T%3D%5Cfrac%7Bn%2B1%7DT-%5Cfrac%20nT%3D%5CDelta%5Comega$

$%5Cbegin%7Baligned%7D%5CRightarrow%20f(t)%26%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cleft(%5Cfrac%7B1%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(x)e%5E%7B-%5Cfrac%7B2%5Cpi%20inx%7DT%7Ddx%5Cright)e%5E%7B%5Cfrac%7B2%5Cpi%20in%20t%7DT%7D%5C%5C%26%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cleft(%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(x)e%5E%7B-2%5Cpi%20i%5Comega_n%20x%7Ddx%5Cright)e%5E%7B2%5Cpi%20i%5Comega_n%20t%7D%5CDelta%20%5Comega%5Cend%7Baligned%7D$

令 $T%5Crightarrow%20%2B%5Cinfty$ ，则 $%5CDelta%20%5Cxi%20%5Crightarrow%200%5E%7B%2B%7D$

$f(t)%3D%5Clim_%7B%5CDelta%5Comega%20%5Cto0%5E%7B%2B%7D%7D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cleft(%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7Df(x)e%5E%7B2%5Cpi%20i%5Comega_n%20x%7Ddx%5Cright)e%5E%7B2%5Cpi%20i%5Comega_n%20t%7D%5CDelta%20%5Comega$

运用小学二年级学过的微积分知识，可知右边为一黎曼和，于是可以将级数写成积分

$f(t)%3D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cleft(%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7Df(t)e%5E%7B-2%5Cpi%20i%5Comega%20t%7Ddt%5Cright)e%5E%7B2%5Cpi%20i%5Comega%20t%7Dd%5Comega$

这样，我们就得到了Fourier积分，其中

$%5Cmathcal%20F%5C%7Bf(t)%5C%7D(%5Comega)%3D%5Chat%20f(%5Comega)%3D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%7Bf%7D(t)e%5E%7B-2%5Cpi%20i%5Comega%20t%7Ddt$

称为f(t)的Fourier变换，且其逆变换为

$%5Cmathcal%20F%5E%7B-1%7D%5C%7B%5Chat%20f(%5Comega)%5C%7D(t)%3Df(t)%3D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Chat%7Bf%7D(%5Comega)e%5E%7B2%5Cpi%20i%5Comega%20t%7D%5Cmathrm%20d%5Comega$