工程數(shù)學(xué)附錄_傅里葉級數(shù)與傅里葉變換

2023-02-25 20:27 作者:sky92曇 0人讀過 | 我要投稿

引

這里要首先證明三角函數(shù)的正交性然后，既然證明了正交，利用加權(quán)求和的形式得到關(guān)于周期為 $2%5Cpi$ 函數(shù)的傅里葉級數(shù),再把周期擴(kuò)展到任意L
為了簡化公式，使用歐拉公式把復(fù)數(shù)域和三角函數(shù)混合的形式區(qū)別統(tǒng)合起來，于是有了周期函數(shù)的復(fù)指數(shù)表達(dá),最終的周期復(fù)指表達(dá)中會算得一個系數(shù)項，整個展開的矛盾就轉(zhuǎn)移到了該系數(shù)中,然后我們將整個函數(shù)推廣到非周期函數(shù)中，即將周期推到無限的函數(shù),無限導(dǎo)致的就是級數(shù)求和成為連續(xù)積分，最后得到規(guī)整的式子，就是傅里葉變換和逆變換；

三角函數(shù)的正交性

三角函數(shù)正交性是傅里葉級數(shù)的基礎(chǔ)

我們有個三角函數(shù)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，向量內(nèi)積為0是正交，函數(shù)則這里有個定義：
$%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Csin%20nx%20%5Ccos%20mx%20dx%20%3D0%20%20$
$%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20mx%20%5Ccos%20nx%20dx%20%3D0%20%20%2C%20n%5Cneq%20m%20$

那么上述的三角函數(shù)正交的情況是怎么來的？
簡單來說，其實正交就是垂直，也就是兩個向量的內(nèi)積為0的時候就是正交；

$%5Cvec%20a%20%5Ccdot%20%5Cvec%20b%20%3D%20%7C%5Cvec%20a%7C%7C%5Cvec%20b%7C%5Ccos%5Cphi%20%3D%20%7C%5Cvec%20a%7C%7C%5Cvec%20b%7C%20%5Ccdot%200%3D0$ 就平面上來說正交是這樣，那么如果用向量表達(dá)出來，假設(shè)兩者在n維度場：
$%5Cvec%20a%20%5Ccdot%20%5Cvec%20b%20%3D%20a_1%20b_1%2Ba_2%20b_2%2B...%2Ba_n%20b_n%3D%5Csum%5Climits%5E%5Cinfty_%7Bi%3D1%7Da_i%20b_i%3D0$
如果上述求和公式并非取整數(shù)，而是連續(xù)實數(shù)，那么上述的求和就成為了積分；
于是當(dāng)我們將向量轉(zhuǎn)為無限維的函數(shù)，意味著其向量內(nèi)元素是無限且稠密的，但某個元素總是可用函數(shù)自變量表達(dá)

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

由此我們定義出了函數(shù)之間的內(nèi)積以及函數(shù)的正交；

于是我們要在三角函數(shù)上證明，，利用三角積化和差公式，以及奇函數(shù)性質(zhì)：
$%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20nx%20%5Ccos%20mx%20dx%20%2C%20n%5Cneq%20m%20%5Crightarrow%20%20%5C%5C%20%0A%5Cfrac%7B1%7D%7B2%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20(n-m)x%20%5Ccos%20(n%2Bm)x%20dx%20%5C%5C%3D%0A%5Cfrac%7B1%7D%7B2%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20(n-m)x%20dx%20%5Cfrac%7B1%7D%7B2%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Ccos%20(n%2Bm)x%20dx%20%5C%5C%3D%0A%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B1%7D%7Bn-m%7D%20%5Csin%20(n-m)x%20%7C%5E%5Cpi_%7B-%5E%5Cpi%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B1%7D%7Bn-m%7D%5Csin%20(n%2Bm)x%7C%5E%5Cpi_%7B-%5E%5Cpi%7D%20%3D%0A0%2B0%3D0$

同樣我們也可驗證：
$%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Ccos%20nx%20%5Csin%20mx%20dx%20%3D%200%2C%20n%5Cneq%20m%20%5C%5C%20%5Cint%5E%5Cpi_%7B-%5Cpi%7D%20%5Csin%20nx%20%5Csin%20mx%20dx%20%3D%200%2C%20n%5Cneq%20m%20$

那么如果n = m ,就有?

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

那么我們將上述所有三角函數(shù)正交的情況羅列出來：

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

周期為 $2%5Cpi$ 的函數(shù)展開為傅里葉級數(shù)

拿到一個周期為 $T%20%3D%202%5Cpi$ 的函數(shù) ? $f(x)%3Df(x%2B2%5Cpi)%20$ ?

將其展開為三角函數(shù)的加和，那么兩個三角函數(shù)作為基底進(jìn)行加權(quán)組合，兩種表達(dá)方式：

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

我們需要求出這里的 $a_0$

對上述第二式兩邊積分

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

于是可以得到? $a_0%3D%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)dx%20$ ，這個公式有時為了后續(xù)計算方便通常兩側(cè)都乘2 得到

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

接下來求 $a_n$

我們對等式兩側(cè)乘以 $%5Ccos%20mx$ ，然后兩側(cè)進(jìn)行積分

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

此時我們觀察到上式僅在n=m的時候出現(xiàn)非零項

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

得到

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

接下來求解 $b_n$

我們對等式兩側(cè)乘以 $%5Csin%20mx$ 然后對兩側(cè)進(jìn)行積分

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

此時我們觀察到上式僅在n=m的時候出現(xiàn)非零項 $%5Cint%5E%5Cpi_%7B-%5Cpi%7Df(x)%5Ccos%20nx%20dx%20%3Db_n%5Cint%5E%5Cpi_%7B-%5Cpi%7D%5Csin%5E2%20nx%20dx%3Db_n%5Cpi%20$

得到

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

那么我們得到了周期為 $2%5Cpi$ 函數(shù)的完整傅里葉級數(shù)的展開

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

周期為2L的函數(shù)展開為傅里葉級數(shù)

拿到一個周期為 $T%20%3D%202L$ 的函數(shù)? $f(x)%3Df(x%2B2L)$

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

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

那么我們可得到

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

顯然其實形式并沒有太大的變化

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

在工程上時間通常不會是負(fù)數(shù)? $t%3E0$ ?, 周期為 $T%3D2L%09%2C%5Comega%20%3D%5Cfrac%7B%5Cpi%7D%7BL%7D%20%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D$

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

于是我們得到傅里葉工程表達(dá)

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

那么如果此時T 變?yōu)闊o限大，即函數(shù)已經(jīng)不是周期函數(shù)了，或者說全局只有一個周期的函數(shù) ；

傅里葉級數(shù)的復(fù)數(shù)表達(dá)形式

以上述工程表達(dá)形式為例 ?

連接復(fù)數(shù)以及三角可使用歐拉公式作中介

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

帶入上述的工程表達(dá)得到傅里葉級數(shù)的復(fù)數(shù)表達(dá)

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

觀察上述式子，將第三項的n的范圍改變符號得到 $%5Cfrac%7B1%7D%7B2%7D%5Csum%5Climits%5E%7B-1%7D_%7Bn%3D-%5Cinfty%7D(a_%7B-n%7D%2Bib_%7B-n%7D)e%5E%7Bin%5Comega%20t%7D$

此時可以發(fā)現(xiàn)n的取值成了 $n%5Cin(-%5Cinfty%2C%5Cinfty)$ ，出現(xiàn)了可以合并的項 $e%5E%7Bin%5Comega%20t%7D$ ，最后式子就變?yōu)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=%5Csum%5Climits%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7DC_n%20e%5E%7Bin%5Comega%20t%7D" alt="%5Csum%5Climits%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7DC_n%20e%5E%7Bin%5Comega%20t%7D">

于是

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

然后將原先傅里葉級數(shù)代入，我們就會有驚奇的發(fā)現(xiàn)

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

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

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

因此完全可以使用一個式子來表達(dá)這系數(shù)內(nèi)涵的三項，此時復(fù)數(shù)形式的傅里葉級數(shù)變得非常簡單；

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

傅里葉變換 FT

我們已經(jīng)得到了傅里葉級數(shù)的復(fù)數(shù)表達(dá)；

上述函數(shù)原表達(dá)，求和式，以及系數(shù)式

其求和式中 $%5Csum%5Climits%5E%5Cinfty_%7B-%5Cinfty%7D$ 和 $e%5E%7Bin%5Comega_0%20t%7D$ 兩者對于任意的傅里葉級數(shù)都是一樣，已經(jīng)是一種固定規(guī)則了，僅僅由 $C_n$ 來決定不同樣式的傅里葉級數(shù)，這系數(shù)是一個復(fù)數(shù)；

其實把求和式展開， $%20...%20%2B%20c_%7B-1%7De%5E%7B-i(-1)%5Comega_0%20t%7D%2B%20c_%7B0%7De%5E0%2B%20c_1%20e%5E%7Bi(1)%5Comega_0%20t%7D%2B%20c_2e%5E%7B-i(2)%5Comega_0%20t%7D%20%2B...$ , 表現(xiàn)為如下右圖；

就上圖來說，左邊在工程中總是稱為時域，畢竟和事件有關(guān)，右側(cè)表達(dá)的是系數(shù)在不同頻率下的大小不同，即每次都以某個頻率作為基礎(chǔ)，作加權(quán)，并對其各頻率加權(quán)結(jié)果之總和就是原函數(shù)；

上述是周期函數(shù)的情況；

那么如果一個函數(shù)不是周期函數(shù)，或者說整個函數(shù)就一個周期？

那么其周期就趨于無窮，此時就成了一般函數(shù) $%5Clim%5Climits_%7BT%5Cto%5Cinfty%7Df_T(t)-f(t)%20$

對基頻率來說 $T%20%5Cto%20%5Cinfty%20$ ,此時 $%5CDelta%5Comega%20%3D%20(n%2B1)%5Comega_0-n%5Comega_0%3D%5Comega_0%20%3D%20%5Cfrac%7B2%5Cpi%7DT%20$ 周期增加導(dǎo)致頻率間隔變小，當(dāng)趨于無窮則頻率間隔無限??；此時我們可以將傅里葉級數(shù)頻率的離散情況，轉(zhuǎn)為連續(xù)情況，即各個有微小差異的頻率稠密的順序排在一起組成了頻率函數(shù)（離散級數(shù)成連續(xù)函數(shù)）；

把系數(shù)式代入求和式 ?得到混合式：

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

可見當(dāng)周期無限時會出現(xiàn)如下情況：

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

將這些變化代入上述混合式得到：

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

而整體拿出來就是傅里葉逆向變換 IFT

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

傅里葉變換簡化寫法就是拉普拉斯變換 Lplace-Transform ? LT

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

標(biāo)簽：拉普拉斯變換傅里葉逆變換數(shù)學(xué)推導(dǎo)傅里葉級數(shù)傅里葉變換