傅立葉級數

  • 法國數學家 Joseph Fourier在研究熱傳導問題的時候,發現的一種三角級數的應用。可以將複雜的週期性訊號,用sine和cosine的組合形式來描述。其應用相當廣泛包括訊號處理和影像分析等。
  • 適用於滿足以下條件:
    • f(x)
      是單一值
    • f(x)
      為有限個不連續的點
    • f(x)
      為有限個最大值和最小值
    • f(x)
      是可積分函數,在一個週期內的積分值是有限的

基本思想:
任何週期函數都可以拆解為無窮多個不同頻率的正弦波和餘弦波的加總。

基本架構

  • 傅立葉級數的公式並不好解釋,原因是涉及很多抽象的數學概念,並且範圍很廣主要的三個部分週期函數、諧波、基底都是很過去幾乎沒有接觸的數學概念。

週期函數

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  • 一個函數
    f(x)
    存在一個常數
    T
    T>0
    ,這時候可以注意到只要
    T
    值以正倍數變化,其數值都會相等。
    f(x)=f(x+T)=f(x+3T)=....=f(x+nT)
  • 如果
    f(x)=sin(x)
    T
    2π
    的正整數倍(在傅立葉分析中,常見的週期函數多以 2π 為基本週期)
    sin(x)=sin(x+T)=sin(x+nT)

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  • 如果一個週期函數
    f(x)
    有一個週期
    T
    且在任何有限區間內都是可積分的,且 a, b 為任意的實數。可以藉由圖表得知一個週期函數對其相同長度的區間做積分結果都是相同的。
    aa+Tf(x)dx=bb+Tf(x)dx

諧波

  • 諧波是指頻率為基波頻率整數倍的波,任何複雜的週期性訊號都可以分解成一系列不同頻率和振幅的弦波。

  • 基波頻率:

    f0=1T

  • 也可以是 1,2,3,4

    n 個諧波
    fn=nf0

  • ω=2πf=2π/T 所以
    T=2π/ω
    ,延續前面的正弦的週期函數改寫為:
    y=Asin[w(x+2πw)+ϕ]=Asin[(wx+ϕ)+2π]=Asin(wx+ϕ)

    |A|:諧波的振幅
    w
    :諧波的頻率
    ϕ
    :諧波初始的相位

  • 對此周函數做複角公式:

    y=A(cos(ωx)sin(ϕ)+sin(ωx)cos(ϕ))

    • a=Asin(ϕ), b=Acos(ϕ)
      整理為:
      y=acos(ωx)+bsin(ωx)

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  • ω=2π/T ,如果
    T=2l
    那麼
    ω=π/l
    帶入剛剛整理的公式得到:
    y=acos(xπl)+bsin(xπl)

  • 加入諧波的概念得到:

    y=ancos(nxπl)+bnsin(nxπl)n=0,1,2,3.....

  • 嘗試以級數的型式表達:

    f(x)=a0+Σn=1(ancos(nxπl)+bnsin(nxπl))

  • 在查找相關資料時我注意到一個疑惑點,在工數課本強調為

    a0/2。但是在我的想法
    a0
    應該單純是級數帶入
    n=0
    計算得到的數值,為何需要平均。回想到電路學直流成分的數值是交流成分端點的平均這樣的結果。在級數描述時都是交流成分所以
    n=0
    也是,但是結果我們希望
    a0
    是直流成分所以取平均。修正公式得到:
    f(x)=a02+Σn=1(ancos(nxπl)+bnsin(nxπl))

正交

  • 在數學中,正交的概念不僅限於幾何中的 90 度垂直,而是指兩個函數的內積為 0。這個情況下彼此都不會相互影響,這個情況就是正交。
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  • 對於函數 f(x) 和 g(x)做積分:
    ππf(x)g(x)dx=0
  • 對於 sin(x), cos(x)做積分:
    ππcos(nx)dx=sin(nx)n|ππ=0ππsin(nx)dx=cos(nx)n|ππ=0
  • 對於 sin(x), cos(x)的平方做積分:
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    所以對於
    f2(x)
    做積分結果為:
    ππf2(x)dx=π

係數

  • 依據傅立葉級數的公式使其乘上一個
    x0
    後做積分:
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    結合正交性質可以知道在
    n>0
    的項結果皆為0,所以
    a0
    結果為:
    ππf(x)dx=a0πa0=1πππf(x)dx
  • 接續分別乘上
    cos(x),sin(x)
    以找出
    an,bn
    :
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    因為
    a0
    的項視為
    <1,cos(x)>
    所以為0, 而
    sin(x)cos(x)
    也為0所以只要專注於
    cos(kx)cos(nx)
    的項恰好對應到
    an
    :
    ππcos(kx)cos(nx)dx={0,if n  kπ,if n=kann=k

    整合上面的結果:
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    an=1πππf(x)cos(nx)dxn=1,2,3,4.....

    在工程數學課本則是這樣描述的:
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sin(x) 則是:
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結合對於 cos(x) 運算的經驗可以得到:
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bn=1πππf(x)sin(nx)dxn=1,2,3,4.....

在工程數學課本描述則是這樣:
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