工程数学附录_傅里叶级数与傅里叶变换

2023-02-25 20:27 作者:sky92昙 0人读过 | 我要投稿

引

这里要首先证明三角函数的正交性然后，既然证明了正交，利用加权求和的形式得到关于周期为 $2%5Cpi$ 函数的傅里叶级数,再把周期扩展到任意L
为了简化公式，使用欧拉公式把复数域和三角函数混合的形式区别统合起来，于是有了周期函数的复指数表达,最终的周期复指表达中会算得一个系数项，整个展开的矛盾就转移到了该系数中,然后我们将整个函数推广到非周期函数中，即将周期推到无限的函数,无限导致的就是级数求和成为连续积分，最后得到规整的式子，就是傅里叶变换和逆变换；

三角函数的正交性

三角函数正交性是傅里叶级数的基础

我们有个三角函数系集合

$%5C%7B%20%5Csin%200x%3D0%20%2C%20%5Ccos%200x%3D1%2C%20%5Csin%20x%20%5Ccos%20x%2C%5Csin%202x%20%5Ccos%202x%2C...%2C%5Csin%20nx%20%5Ccos%20nx%2C...%20%5C%7D$

即

$%5C%7B%20%5Csin%20nx%20%2C%5Ccos%20nx%20%5C%7D%2Cn%20%3D%200%2C1%2C2%2C...$

那么什么是正交，向量内积为0是正交，函数则这里有个定义：
$%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Csin%20nx%20%5Ccos%20mx%20dx%20%3D0%20%20$
$%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20mx%20%5Ccos%20nx%20dx%20%3D0%20%20%2C%20n%5Cneq%20m%20$

那么上述的三角函数正交的情况是怎么来的？
简单来说，其实正交就是垂直，也就是两个向量的内积为0的时候就是正交；

$%5Cvec%20a%20%5Ccdot%20%5Cvec%20b%20%3D%20%7C%5Cvec%20a%7C%7C%5Cvec%20b%7C%5Ccos%5Cphi%20%3D%20%7C%5Cvec%20a%7C%7C%5Cvec%20b%7C%20%5Ccdot%200%3D0$ 就平面上来说正交是这样，那么如果用向量表达出来，假设两者在n维度场：
$%5Cvec%20a%20%5Ccdot%20%5Cvec%20b%20%3D%20a_1%20b_1%2Ba_2%20b_2%2B...%2Ba_n%20b_n%3D%5Csum%5Climits%5E%5Cinfty_%7Bi%3D1%7Da_i%20b_i%3D0$
如果上述求和公式并非取整数，而是连续实数，那么上述的求和就成为了积分；
于是当我们将向量转为无限维的函数，意味着其向量内元素是无限且稠密的，但某个元素总是可用函数自变量表达

$a%20%5Ccdot%20b%20%3D%5Cint%5E%7Bx_1%7D_%7Bx_0%7Df(x)g(x)dx%20%5C%5Ca%3Df(x)%2Cb%3Dg(x)%20%2Cx%5Cin(x_0%2Cx_1)$

由此我们定义出了函数之间的内积以及函数的正交；

于是我们要在三角函数上证明，，利用三角积化和差公式，以及奇函数性质：
$%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20nx%20%5Ccos%20mx%20dx%20%2C%20n%5Cneq%20m%20%5Crightarrow%20%20%5C%5C%20%0A%5Cfrac%7B1%7D%7B2%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20(n-m)x%20%5Ccos%20(n%2Bm)x%20dx%20%5C%5C%3D%0A%5Cfrac%7B1%7D%7B2%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20(n-m)x%20dx%20%5Cfrac%7B1%7D%7B2%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Ccos%20(n%2Bm)x%20dx%20%5C%5C%3D%0A%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B1%7D%7Bn-m%7D%20%5Csin%20(n-m)x%20%7C%5E%5Cpi_%7B-%5E%5Cpi%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B1%7D%7Bn-m%7D%5Csin%20(n%2Bm)x%7C%5E%5Cpi_%7B-%5E%5Cpi%7D%20%3D%0A0%2B0%3D0$

同样我们也可验证：
$%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20nx%20%5Csin%20mx%20dx%20%3D%200%2C%20n%5Cneq%20m%20%5C%5C%20%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Csin%20nx%20%5Csin%20mx%20dx%20%3D%200%2C%20n%5Cneq%20m%20$

那么如果n = m ,就有

$%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20nx%20%5Ccos%20mx%20dx%20%2C%20n%3Dm%5Cneq0%20%20%5Crightarrow%20%5C%5C%0A%20%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Cfrac%7B1%7D%7B2%7D%5B1%2B%5Ccos2mx%5D%20dx%20%5C%5C%3D%20%0A%5Cfrac%7B1%7D%7B2%7D%20%5B%5Cint%5E%5Cpi_%7B-%5Cpi%7D%201%20dx%2B%20%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Ccos2mx%20dx%20%5D%20%20%5C%5C%3D%0A%5Cfrac%7B1%7D%7B2%7D%20%5B%5Cint%5E%5Cpi_%7B-%5Cpi%7D%201%20dx%2B%20%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Ccos0x%5Ccos2mx%20dx%20%5D%20%20%5C%5C%3D%0A%5Cfrac%7B1%7D%7B2%7D%5B%5Cpi%2B0%5D%3D%5Cpi$

那么我们将上述所有三角函数正交的情况罗列出来：

$%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Csin%20nx%20%5Ccos%20mx%20dx%20%3D0%20%5C%5C%20%0A%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Csin%20nx%20%5Csin%20mx%20dx%20%3D%200%20%2Cn%5Cneq%20m%20%5C%5C%20%0A%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Csin%20nx%20%5Csin%20mx%20dx%20%3D%20%5Cpi%20%2Cn%3Dm%5Cneq0%20%5C%5C%20%0A%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Csin%20nx%20%5Csin%20mx%20dx%20%3D%202%5Cpi%20%2Cn%3Dm%3D0%20%5C%5C%0A%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20nx%20%5Ccos%20mx%20dx%20%3D%200%20%2Cn%5Cneq%20m%20%5C%5C%20%0A%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20nx%20%5Ccos%20mx%20dx%20%3D%20%5Cpi%20%2Cn%3Dm%5Cneq0%20%5C%5C%20%0A%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20nx%20%5Ccos%20mx%20dx%20%3D%202%5Cpi%20%2Cn%3Dm%3D0%0A$

周期为 $2%5Cpi$ 的函数展开为傅里叶级数

拿到一个周期为 $T%20%3D%202%5Cpi$ 的函数 $f(x)%3Df(x%2B2%5Cpi)%20$

将其展开为三角函数的加和，那么两个三角函数作为基底进行加权组合，两种表达方式：

$f(x)%3D%5Csum%5Climits%5E%5Cinfty_%7Bn%3D0%7Da_n%5Ccos%20nx%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D0%7Db_n%5Csin%20nx%20%5C%5C%0Af(x)%3Da_0%5Ccos0x%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Da_n%5Ccos%20nx%2Bb_0%5Csin0x%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Db_n%5Csin%20nx%3Da_0%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Da_n%5Ccos%20nx%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Db_n%5Csin%20nx$

我们需要求出这里的 $a_0$

对上述第二式两边积分

$%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)dx%3D%5Cint%5E%5Cpi_%7B-%5Cpi%7Da_0dx%2B%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Da_n%5Ccos%20nx%20dx%2B%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Db_n%5Csin%20nx%20dx%20%3D%0Aa_0%5Cint%5E%5Cpi_%7B-%5Cpi%7Ddx%2B0%2B0%3Da_0x%7C%5E%5Cpi_%7B-%5Cpi%7D%3D2%5Cpi%20a_0$

于是可以得到 $a_0%3D%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)dx%20$ ，这个公式有时为了后续计算方便通常两侧都乘2 得到

$a_0%3D%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)dx%20%2C%20%20a'_0%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)dx$

接下来求 $a_n$

我们对等式两侧乘以 $%5Ccos%20mx$ ，然后两侧进行积分

$%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)%5Ccos%20mx%20dx%20%5C%5C%0A%3D%5Cint%5E%5Cpi_%7B-%5Cpi%7Da_0%5Ccos%20mxdx%2B%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Da_n%5Ccos%20nx%5Ccos%20mx%20dx%2B%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Db_n%5Csin%20nx%20%5Ccos%20mxdx%5C%5C%0A%3D0%2B%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Da_n%5Ccos%20nx%5Ccos%20mx%20dx%2B0$

此时我们观察到上式仅在n=m的时候出现非零项

$f(x)%3Df(x%2B2L)$

得到

$a_n%20%3D%20%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)%5Ccos%20nx%20dx$

接下来求解 $b_n$

我们对等式两侧乘以 $%5Csin%20mx$ 然后对两侧进行积分

$%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)%5Csin%20mx%20dx%20%5C%5C%0A%3D%5Cint%5E%5Cpi_%7B-%5Cpi%7Da_0%5Csin%20mxdx%2B%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Da_n%5Ccos%20nx%5Csin%20mx%20dx%2B%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Db_n%5Csin%20nx%20%5Csin%20mxdx%5C%5C%0A%3D0%2B0%2B%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Db_n%5Csin%20nx%5Csin%20mx%20dx$

此时我们观察到上式仅在n=m的时候出现非零项 $%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)%5Ccos%20nx%20dx%20%3Db_n%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Csin%5E2%20nx%20dx%3Db_n%5Cpi%20$

得到

$b_n%20%3D%20%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)%5Csin%20nx%20dx$

那么我们得到了周期为 $2%5Cpi$ 函数的完整傅里叶级数的展开

$f(x)%3Df(x%2B2%5Cpi)%2CT%3D2%5Cpi%20%5C%5C%0Af(x)%0A%3D%20%5Cfrac%7Ba'_0%7D%7B2%7D%20%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Da_n%5Ccos%20nx%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Db_n%5Csin%20nx%20%0A%3D%20a_0%20%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Da_n%5Ccos%20nx%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Db_n%5Csin%20nx%20%5C%5C%20%20%0Aa'_0%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)dx%2C%20a_0%3D%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)dx%20%5C%5C%20%20%0Aa_n%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)%5Ccos%20nxdx%20%5C%5C%20%20%0Ab_n%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)%5Csin%20nxdx$

周期为2L的函数展开为傅里叶级数

拿到一个周期为 $T%20%3D%202L$ 的函数 $f(x)%3Df(x%2B2L)$

可以直接使用换元方法 $%5Cfrac%7B%5Cpi%7D%7BL%7D%3D%5Cfrac%7Bx%7D%7Bt%7D%5Crightarrow%20x%3D%5Cfrac%7B%5Cpi%7D%7BL%7Dt%20%5Crightarrow%20t%20%3D%20%5Cfrac%7BL%7D%7B%5Cpi%7Dx$

$f(t)%3Df(%5Cfrac%7BL%7D%7B%5Cpi%7Dx)%5Ctriangleq%20g(x)$

那么我们可得到

$%20x%20%3D%20%5Cfrac%7B%5Cpi%7D%7BL%7Dt%2C%5Ccos%20nx%3D%5Ccos%5Cfrac%7Bn%5Cpi%7D%7BL%7Dt%2C%5Csin%20nx%3D%5Csin%20%5Cfrac%7Bn%5Cpi%7D%7BL%7Dt%2C%5C%5C%0A%5Cint%5E%5Cpi_%7B-%5Cpi%7Ddx%3D%5Cint%5EL_%7B-L%7Dd%5Cfrac%7B%5Cpi%7D%7BL%7Dt%20%5Crightarrow%20%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7Ddx%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cfrac%7B%5Cpi%7D%7BL%7D%5Cint%5EL_%7B-L%7Ddt%3D%20%5Cfrac%7B1%7D%7BL%7D%5Cint%5EL_%7B-L%7Ddt$

显然其实形式并没有太大的变化

$f(t)%3Df(t%2B2L)%2CT%3D2L%20%5C%5C%0Af(t)%0A%3D%20%5Cfrac%7Ba'_0%7D%7B2%7D%20%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Da_n%5Ccos%20%5Cfrac%7Bn%5Cpi%7D%7BL%7Dt%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Db_n%5Csin%20%5Cfrac%7Bn%5Cpi%7D%7BL%7Dt%20%20%20%5C%5C%20%0Aa'_0%3D%5Cfrac%7B1%7D%7BL%7D%5Cint%5EL_%7B-L%7Df(t)dt%20%5C%5C%0Aa_n%3D%5Cfrac%7B1%7D%7BL%7D%5Cint%5EL_%7B-L%7Df(t)%5Ccos%20%5Cfrac%7Bn%5Cpi%7D%7BL%7Dtdt%5C%5C%20%20%0Ab_n%3D%5Cfrac%7B1%7D%7BL%7D%5Cint%5EL_%7B-L%7Df(t)%5Csin%20%5Cfrac%7Bn%5Cpi%7D%7BL%7Dtdt$

在工程上时间通常不会是负数 $t%3E0$ , 周期为 $T%3D2L%09%2C%5Comega%20%3D%5Cfrac%7B%5Cpi%7D%7BL%7D%20%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D$

$%5Cint%5EL_%7B-L%7D%20%5Crightarrow%20%5Cint_0%5E%7B2L%7Ddt%5Crightarrow%20%5Cint%5ET_0%20t$

于是我们得到傅里叶工程表达

$f(t)%3Df(t%2B2L)%2CT%3D2L%20%2C%5Comega%20%3D%5Cfrac%7B%5Cpi%7D%7BL%7D%20%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D%5C%5C%0Af(t)%0A%3D%20%5Cfrac%7Ba'_0%7D%7B2%7D%20%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Da_n%5Ccos%20n%5Comega%20t%2B%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Db_n%5Csin%20n%5Comega%20t%20%0A%20%5C%5C%20%20%0Aa'_0%3D%5Cfrac%7B2%7D%7BT%7D%5Cint%5ET_0f(t)dt%20%5C%5C%20%20%0Aa_n%3D%5Cfrac%7B2%7D%7BT%7D%5Cint%5ET_0f(t)%5Ccos%20n%5Comega%20tdt%5C%5C%20%20%0Ab_n%3D%5Cfrac%7B2%7D%7BT%7D%5Cint%5ET_0f(t)%5Csin%20n%5Comega%20tdt$

那么如果此时T 变为无限大，即函数已经不是周期函数了，或者说全局只有一个周期的函数；

傅里叶级数的复数表达形式

以上述工程表达形式为例

连接复数以及三角可使用欧拉公式作中介

$e%5E%7Bi%5Ctheta%7D%20%3D%20%5Ccos%5Ctheta%20%2Bi%5Csin%5Ctheta%20%5C%5C%0A%5Ccos%5Ctheta%3D%5Cfrac%7B1%7D%7B2%7D(e%5E%7Bi%5Ctheta%7D%2Be%5E%7B-i%5Ctheta%7D)%20%5C%5C%0A%5Csin%5Ctheta%3D-%5Cfrac%7B1%7D%7B2%7Di(e%5E%7Bi%5Ctheta%7D-e%5E%7B-i%5Ctheta%7D)$

带入上述的工程表达得到傅里叶级数的复数表达

$f(t)%20%5C%5C%3D%0A%5Cfrac%7Ba'_0%7D%7B2%7D%2B%5Cfrac%7B1%7D%7B2%7D%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Da_n(e%5E%7Bin%5Comega%20t%7D%2Be%5E%7B-in%5Comega%20t%7D)%2B%5Cfrac%7B-1%7D%7B2%7Di%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7Db_n(e%5E%7Bin%5Comega%20t%7D-e%5E%7B-in%5Comega%20t%7D)%20%5C%5C%3D%0A%5Cfrac%7Ba'_0%7D%7B2%7D%2B%5Cfrac%7B1%7D%7B2%7D%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7D%20(a_n-ib_n)e%5E%7Bin%5Comega%20t%7D%20%2B%5Cfrac%7B1%7D%7B2%7D%5Csum%5Climits%5E%5Cinfty_%7Bn%3D1%7D(a_n%2Bib_n)e%5E%7B-in%5Comega%20t%7D$

观察上述式子，将第三项的n的范围改变符号得到 $%5Cfrac%7B1%7D%7B2%7D%5Csum%5Climits%5E%7B-1%7D_%7Bn%3D-%5Cinfty%7D(a_%7B-n%7D%2Bib_%7B-n%7D)e%5E%7Bin%5Comega%20t%7D$

此时可以发现n的取值成了 $n%5Cin(-%5Cinfty%2C%5Cinfty)$ ，出现了可以合并的项 $e%5E%7Bin%5Comega%20t%7D$ ，最后式子就变为 $%5Csum%5Climits%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7DC_n%20e%5E%7Bin%5Comega%20t%7D$

于是

$f(t)%3D%5Csum%5Climits%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7DC_n%20e%5E%7Bin%5Comega%20t%7D%20%20%20%0A%5C%5C%20%20%0A%5Cbegin%7Bequation%7D%0AC_n%3D%0A%5Cbegin%7Bcases%7D%0A%5Cfrac%7Ba_0%7D%7B2%7D%20%2C%20n%3D0%20%5C%5C%0A%5Cfrac12(a_n-ib_n)%2Cn%3D1%2C2%2C3%2C...%20%5C%5C%0A%5Cfrac12(a_%7B-n%7D%2Bib_%7B-n%7D)%2Cn%3D-1%2C-2%2C3%2C...%20%5C%5C%0A%5Cend%7Bcases%7D%0A%5Cend%7Bequation%7D$

然后将原先傅里叶级数代入，我们就会有惊奇的发现

$n%3D0%20%5Crightarrow%20%20C_0%20%3D%5Cfrac%7Ba_0%7D%7B2%7D%20%3D%20%5Cfrac1T%20%5Cint%5Et_0f(t)dt%20%3D%5Cfrac1T%5Cint%5ET_0f(t)e%5E%7B-i0%5Comega%20t%7Ddt%20%20$

$%0An%3D1%2C2%2C...%20%5Crightarrow%20C_n%20%3D%5Cfrac12(a_n-ib_n)%20%5C%5C%3D%0A%5Cfrac12%5B%20%20%5Cfrac2T%20%5Cint%5ET_0%20f(t)%5Ccos%20n%5Comega%20t%20dt%20-%20i%20%5Cfrac2T%20%5Cint%5ET_0%20f(t)%5Csin%20n%5Comega%20t%20dt%20%5D%20%5C%5C%3D%0A%5Cfrac1T%5Cint%5ET_0f(t)%5B%5Ccos%20n%5Comega%20t-i%5Csin%20n%5Comega%20t%5Ddt%20%5C%5C%3D%0A%5Cfrac1T%5Cint%5ET_0f(t)%5B%5Ccos%20(-n%5Comega%20t)%2Bi%5Csin%20(-n%5Comega%20t)%5Ddt%20%5C%5C%3D%0A%5Cfrac1T%5Cint%5ET_0f(t)e%5E%7B-in%5Comega%20t%7Ddt%20%0A$

$%0An%3D-1%2C-2%2C...%20%5Crightarrow%20C_n%20%3D%5Cfrac12(a_%7B-n%7D%2Bib_%7B-n%7D)%20%5C%5C%3D%0A%5Cfrac12%5B%20%20%5Cfrac2T%20%5Cint%5ET_0%20f(t)%5Ccos%20(-n%5Comega%20t%20)dt%20%2B%20i%20%5Cfrac2T%20%5Cint%5ET_0%20f(t)%5Csin%20(-n%5Comega%20t)%20dt%20%5D%20%5C%5C%3D%0A%5Cfrac1T%5Cint%5ET_0f(t)%5B%5Ccos%20n%5Comega%20t-i%5Csin%20n%5Comega%20t%5Ddt%20%5C%5C%3D%0A%5Cfrac1T%5Cint%5ET_0f(t)%5B%5Ccos%20(-n%5Comega%20t)%2Bi%5Csin%20(-n%5Comega%20t)%5Ddt%20%5C%5C%3D%0A%5Cfrac1T%5Cint%5ET_0f(t)e%5E%7B-in%5Comega%20t%7Ddt%20%0A$

因此完全可以使用一个式子来表达这系数内涵的三项，此时复数形式的傅里叶级数变得非常简单；

$f(t)%3Df(t%2BT)%2C%5Comega%20%3D%20%5Cfrac%7B2%5Cpi%7DT%5C%5C%0Af(t)%3D%5Csum%5Climits%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7DC_n%20e%5E%7Bin%5Comega%20t%7D%20%20%5C%5C%20%0AC_n%3D%5Cfrac1T%20%5Cint%5ET_0f(t)e%5E%7B-in%5Comega%20t%7Ddt$

傅里叶变换 FT

我们已经得到了傅里叶级数的复数表达；

上述函数原表达，求和式，以及系数式

其求和式中 $%5Csum%5Climits%5E%5Cinfty_%7B-%5Cinfty%7D$ 和 $e%5E%7Bin%5Comega_0%20t%7D$ 两者对于任意的傅里叶级数都是一样，已经是一种固定规则了，仅仅由 $C_n$ 来决定不同样式的傅里叶级数，这系数是一个复数；

其实把求和式展开， $%20...%20%2B%20c_%7B-1%7De%5E%7B-i(-1)%5Comega_0%20t%7D%2B%20c_%7B0%7De%5E0%2B%20c_1%20e%5E%7Bi(1)%5Comega_0%20t%7D%2B%20c_2e%5E%7B-i(2)%5Comega_0%20t%7D%20%2B...$ , 表现为如下右图；

就上图来说，左边在工程中总是称为时域，毕竟和事件有关，右侧表达的是系数在不同频率下的大小不同，即每次都以某个频率作为基础，作加权，并对其各频率加权结果之总和就是原函数；

上述是周期函数的情况；

那么如果一个函数不是周期函数，或者说整个函数就一个周期？

那么其周期就趋于无穷，此时就成了一般函数 $%5Clim%5Climits_%7BT%5Cto%5Cinfty%7Df_T(t)-f(t)%20$

对基频率来说 $T%20%5Cto%20%5Cinfty%20$ ,此时 $%5CDelta%5Comega%20%3D%20(n%2B1)%5Comega_0-n%5Comega_0%3D%5Comega_0%20%3D%20%5Cfrac%7B2%5Cpi%7DT%20$ 周期增加导致频率间隔变小，当趋于无穷则频率间隔无限小；此时我们可以将傅里叶级数频率的离散情况，转为连续情况，即各个有微小差异的频率稠密的顺序排在一起组成了频率函数（离散级数成连续函数）；

把系数式代入求和式得到混合式：

$f_T(t)%3D%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%3D-%5Cinfty%7D%5Cfrac1T%20%5Cint%5E%7B%5Cfrac%20T2%7D_%7B-%5Cfrac%20T2%7Df_T(t)e%5E%7B-in%5Comega_0%20t%7Ddt%20e%5E%7Bin%5Comega_0%20t%7D%2C%5Cfrac1T%3D%5Cfrac%7B%5CDelta%5Comega%7D%7B2%5Cpi%7D%EF%BC%8CT%20%5Cto%20%5Cinfty$

可见当周期无限时会出现如下情况：

$%20%20%5Cint%5E%7B%5Cfrac%20T2%7D_%7B-%5Cfrac%20T2%7D%20dt%20%5Cto%20%5Cint%5E%5Cinfty_%7B-%5Cinfty%7D%20dt%20%5C%5C%0A%20%20%20n%5Comega_0%20%5Cto%20%5Comega%20%20%5C%5C%0A%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%3D-%5Cinfty%7D%5CDelta%5Comega%5Cto%20%5Cint%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7Dd%5Comega$

将这些变化代入上述混合式得到：

$%5Cfrac1%7B2%5Cpi%7D%20%5Cint%5E%5Cinfty_%7B-%5Cinfty%7D%5Cint%5E%5Cinfty_%7B-%5Cinfty%7Df(t)e%5E%7B-i%5Comega%20t%7D%20dt%20%5C%20%20e%5E%7Bi%5Comega%20t%7D%20d%5Comega$

而整体拿出来就是傅里叶逆向变换 IFT

$f(t)%3D%5Cfrac1%7B2%5Cpi%7D%20%5Cint%5E%5Cinfty_%7B-%5Cinfty%7DF(%5Comega)%20e%5E%7Bi%5Comega%20t%7D%20d%5Comega$

傅里叶变换简化写法就是拉普拉斯变换 Lplace-Transform LT

$F(S)%3D%5Cint%5E%5Cinfty_%7B-%5Cinfty%7Df(t)e%5E%7B-S%20t%7D%20dt$

标签：拉普拉斯变换傅里叶逆变换数学推导傅里叶级数傅里叶变换