傅立葉變換
筆記來源:b站:BV1St41117fH
1. 三角函數的正交性
三角函數系:集合
{ sin n x cos n x n = 0 , 1 , 2... \begin{cases} \sin nx \\ \cos nx \qquad n = 0, 1, 2 ... \end{cases} {sinnxcosnxn=0,1,2...
有正交性:
∫ − π π sin n x cos m x d x = 0 n ≠ m \int_{-\pi}^{\pi} \sin nx \cos mx \mathrm{d}x = 0 \qquad \qquad n \neq m ∫−ππsinnxcosmxdx=0n=m
同樣:
∫ − π π cos n x cos m x d x = 0 \int_{-\pi}^{\pi}\cos nx \cos mx \mathrm{d}x = 0 ∫−ππcosnxcosmxdx=0
∫ − π π sin n x sin m x d x = 0 \int_{-\pi}^{\pi}\sin nx \sin mx \mathrm{d}x = 0 ∫−ππsinnxsinmxdx=0
若n = m:
∫ − π π cos m x cos m x d x = π \int_{-\pi}^{\pi}\cos mx\cos mx \mathrm{d}x = \pi ∫−ππcosmxcosmxdx=π
2. 周期為“2π”的函數展開為傅立葉級數
有了正交性就可以把周期函數展開為傅立葉級數;
把 f ( x ) = f ( x + 2 π ) = ∑ n = 0 ∞ a n cos n x + ∑ n = 0 ∞ b n sin n x = a 0 + ∑ n = 1 ∞ a n cos n x + ∑ n = 1 ∞ b n sin n x 把f(x)= f(x+ 2\pi)= \sum_{n=0}^{\infty}a_{n}\cos nx+ \sum_{n=0}^{\infty}b_{n}\sin nx = a_0 + \sum_{n=1}^{\infty}a_{n}\cos nx + \sum_{n=1}^{\infty}b_{n}\sin nx 把f(x)=f(x+2π)=∑n=0∞ancosnx+∑n=0∞bnsinnx=a0+∑n=1∞ancosnx+∑n=1∞bnsinnx
求 a 0 : 求a_0: 求a0:
∫ − π π f ( x ) d x = ∫ − π π a 0 d x + ∫ − π π ∑ n = 1 ∞ a n cos n x d x + ∫ − π π ∑ n = 1 ∞ b n sin n x d x = a 0 ∫ − π π d x = 2 π a 0 \int_{-\pi}^{\pi} f(x) \mathrm{d}x = \int_{-\pi}^{\pi} a_0 \mathrm{d}x + \int_{-\pi}^{\pi} \sum_{n=1}^{\infty}a_{n}\cos nx \mathrm{d}x + \int_{-\pi}^{\pi} \sum_{n=1}^{\infty}b_{n}\sin nx \mathrm{d}x = a_0 \int_{-\pi}^{\pi} \mathrm{d}x = 2\pi a_0 ∫−ππf(x)dx=∫−ππa0dx+∫−ππn=1∑∞ancosnxdx+∫−ππn=1∑∞bnsinnxdx=a0∫−ππdx=2πa0
a 0 = 1 2 π ∫ − π π f ( x ) d x a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) \mathrm{d}x a0=2π1∫−ππf(x)dx
使 用 a 0 2 就 得 到 a 0 = 1 π ∫ − π π f ( x ) d x 使用\frac{a_0}{2}就得到a_0= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \mathrm{d}x 使用2a0就得到a0=π1∫−ππf(x)dx
找 a n : 找a_{n}: 找an:
∫ − π π f ( x ) cos m x d x = ⋯ = ∫ − π π ∑ n = 1 ∞ a n cos n x cos m x d x \int_{-\pi}^{\pi} f(x)\cos mx \mathrm{d}x = \cdots = \int_{-\pi}^{\pi} \sum_{n=1}^{\infty}a_{n}\cos nx \cos mx \mathrm{d}x ∫−ππf(x)cosmxdx=⋯=∫−ππn=1∑∞ancosnxcosmxdx
∫ − π π f ( x ) cos n x d x = a n ∫ − π π ∑ n = 1 ∞ cos 2 n x d x = a n π \int_{-\pi}^{\pi} f(x)\cos nx \mathrm{d}x = a_{n}\int_{-\pi}^{\pi} \sum_{n=1}^{\infty}\cos^2 nx \mathrm{d}x= a_{n}\pi ∫−ππf(x)cosnxdx=an∫−ππn=1∑∞cos2nxdx=anπ
a n = 1 π ∫ − π π f ( x ) cos n x d x a_{n}= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos nx \mathrm{d} x an=π1∫−ππf(x)cosnxdx
同 理 b n : 同理b_{n}: 同理bn:
b n = 1 π ∫ − π π f ( x ) sin n x d x b_{n}= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin nx \mathrm{d} x bn=π1∫−ππf(x)sinnxdx
這 樣 就 能 寫 出 : 這樣就能寫出: 這樣就能寫出:
f ( x ) = a 0 2 + ∑ n = 1 ∞ ( a n cos n x + b n sin n x ) f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}(a_{n}\cos nx + b_{n}\sin nx) f(x)=2a0+n=1∑∞(ancosnx+bnsinnx)
a 0 = 1 π ∫ − π π f ( x ) d x a n = 1 π ∫ − π π f ( x ) cos n x d x b n = 1 π ∫ − π π f ( x ) sin n x d x \begin{aligned} &a_0= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \mathrm{d}x\\ &a_n= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos nx \mathrm{d}x\\ &b_n= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin nx \mathrm{d}x \end{aligned} a0=π1∫−ππf(x)dxan=π1∫−ππf(x)cosnxdxbn=π1∫−ππf(x)sinnxdx
3.周期為“2L”的函數展開為傅立葉級數
f ( t ) = f ( t + 2 L ) f(t) = f(t+ 2L) f(t)=f(t+2L)
為了利用原來的方法, 可以進行換元:
x = π L t ⇒ t = L π x x = \frac{\pi}{L}t \Rightarrow t = \frac{L}{\pi}x x=Lπt⇒t=πLx
然 後 : f ( t ) = f ( L π x ) = 變 成 g ( x ) 然後:f(t)= f(\frac{L}{\pi}x) \stackrel{變成}{=} g(x) 然後:f(t)=f(πLx)=變成g(x)
因 此 : g ( x ) = g ( x + 2 π ) 是以:g(x) = g(x+ 2\pi) 是以:g(x)=g(x+2π)
然後同上部分解得:
g ( x ) = a 0 2 + ∑ n = 1 ∞ ( a n cos n x + b n sin n x ) g(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}(a_{n}\cos nx + b_{n}\sin nx) g(x)=2a0+n=1∑∞(ancosnx+bnsinnx)
a 0 = 1 π ∫ − π π g ( x ) d x a n = 1 π ∫ − π π g ( x ) cos n x d x b n = 1 π ∫ − π π g ( x ) sin n x d x \begin{aligned} &a_0= \frac{1}{\pi}\int_{-\pi}^{\pi} g(x) \mathrm{d}x\\ &a_n= \frac{1}{\pi}\int_{-\pi}^{\pi} g(x)\cos nx \mathrm{d}x\\ &b_n= \frac{1}{\pi}\int_{-\pi}^{\pi} g(x)\sin nx \mathrm{d}x \end{aligned} a0=π1∫−ππg(x)dxan=π1∫−ππg(x)cosnxdxbn=π1∫−ππg(x)sinnxdx
因 為 x = π L t : 因為x= \frac{\pi}{L}t: 因為x=Lπt:
cos n x = cos n π L t sin n x = sin n π L t g ( x ) = f ( t ) ∫ − π π d x = ∫ − L L d ( π L t ) 1 π d x = 1 π π L ∫ − L L d t = 1 L ∫ − L L d t \begin{aligned} &\cos nx= \cos \frac{n\pi}{L}t\\ &\sin nx= \sin \frac{n\pi}{L}t\\ &g(x)= f(t)\\ &\int_{-\pi}^{\pi} \mathrm{d}x= \int_{-L}^{L} \mathrm{d}(\frac{\pi}{L}t)\\ &\frac{1}{\pi} \mathrm{d}x= \frac{1}{\pi}\frac{\pi}{L}\int_{-L}^{L} \mathrm{d}t= \frac{1}{L}\int_{-L}^{L} \mathrm{d}t \end{aligned} cosnx=cosLnπtsinnx=sinLnπtg(x)=f(t)∫−ππdx=∫−LLd(Lπt)π1dx=π1Lπ∫−LLdt=L1∫−LLdt
代 入 後 得 到 : 代入後得到: 代入後得到:
f ( x ) = a 0 2 + ∑ n = 1 ∞ ( a n cos n π L t + b n sin n π L t ) f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}(a_{n}\cos n\frac{\pi}{L}t + b_{n}\sin n\frac{\pi}{L}t) f(x)=2a0+n=1∑∞(ancosnLπt+bnsinnLπt)
a 0 = 1 L ∫ − L L f ( t ) d t a n = 1 L ∫ − L L f ( t ) cos n π L t d t b n = 1 L ∫ − L L f ( t ) sin n π L t d t \begin{aligned} &a_0= \frac{1}{L}\int_{-L}^{L} f(t) \mathrm{d}t\\ &a_n= \frac{1}{L}\int_{-L}^{L} f(t)\cos n\frac{\pi}{L}t \mathrm{d}t\\ &b_n= \frac{1}{L}\int_{-L}^{L} f(t)\sin n\frac{\pi}{L}t \mathrm{d}t \end{aligned} a0=L1∫−LLf(t)dtan=L1∫−LLf(t)cosnLπtdtbn=L1∫−LLf(t)sinnLπtdt
工程中: t 從 0 開 始 , 周 期 T = 2 L , ω = π L = 2 π T t 從0開始,周期T= 2L,\omega= \frac{\pi}{L}= \frac{2\pi}{T} t從0開始,周期T=2L,ω=Lπ=T2π
∫ − L L d t ⇒ ∫ 0 2 L d t ⇒ ∫ 0 T d t \int_{-L}^{L} \mathrm{d}t \Rightarrow \int_{0}^{2L} \mathrm{d}t \Rightarrow \int_{0}^{T} \mathrm{d}t ∫−LLdt⇒∫02Ldt⇒∫0Tdt
得 到 : 得到: 得到:
f ( x ) = a 0 2 + ∑ n = 1 ∞ ( a n cos n ω t + b n sin n ω t ) f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}(a_{n}\cos n\omega t + b_{n}\sin n\omega t) f(x)=2a0+n=1∑∞(ancosnωt+bnsinnωt)
a 0 = 2 T ∫ T 0 f ( t ) d t a n = 2 T ∫ T 0 f ( t ) cos n ω t d t b n = 2 T ∫ T 0 f ( t ) sin n ω t d t \begin{aligned} &a_0= \frac{2}{T}\int_{T}^{0} f(t) \mathrm{d}t\\ &a_n= \frac{2}{T}\int_{T}^{0} f(t)\cos n\omega t \mathrm{d}t\\ &b_n= \frac{2}{T}\int_{T}^{0} f(t)\sin n\omega t \mathrm{d}t \end{aligned} a0=T2∫T0f(t)dtan=T2∫T0f(t)cosnωtdtbn=T2∫T0f(t)sinnωtdt
若T -> ∞ , f(t)就不再為周期函數,就要傅立葉變換了
4. 傅立葉級數的複數形式
歐拉公式:
e i θ = cos θ + i sin θ e^{i\theta}= \cos \theta + i\sin \theta eiθ=cosθ+isinθ
可得到:
f ( t ) = a 0 2 + ∑ n = 1 ∞ [ 1 2 a n ( e i n ω t + e − i n ω t ) − 1 2 i b n ( e i n ω t − e − i n ω t ) ] = a 0 2 + ∑ n = 1 ∞ [ a n − i b n 2 e i n ω t + a n + i b n 2 e − i n ω t ] = a 0 2 + ∑ n = 1 ∞ a n − i b n 2 e i n ω t + ∑ n = − ∞ − 1 a − n + i b − n 2 e i n ω t ∑ n = 1 ∞ a n + i b n 2 e − i n ω t = ∑ n = 0 0 a 0 2 e i n ω t + ∑ n = 1 ∞ a n − i b n 2 e i n ω t + ∑ n = − ∞ − 1 a − n + i b − n 2 e i n ω t = ∑ − ∞ ∞ C n e i n ω t \begin{aligned} f(t) = &\frac{a_0}{2} + \sum_{n=1}^{\infty}[\frac{1}{2}a_{n}(e^{in \omega t}+ e^{-i n \omega t})- \frac{1}{2}ib_{n}(e^{in\omega t}- e^{-in \omega t})]\\ = &\frac{a_0}{2}+ \sum_{n=1}^{\infty}[\frac{a_{n}- ib_{n}}{2}e^{in \omega t} + \frac{a_{n}+ ib_{n}}{2}e^{-in \omega t}]\\ = &\frac{a_0}{2}+ \sum_{n=1}^{\infty}\frac{a_{n}- ib_{n}}{2}e^{in \omega t} + \stackrel{\sum_{n=1}^{\infty}\frac{a_{n}+ ib_{n}}{2}e^{-in \omega t}}{\sum_{n=-\infty}^{-1}\frac{a_{-n}+ ib_{-n}}{2}e^{in \omega t}}\\ = &\sum_{n=0}^{0}\frac{a_{0}}{2}e^{in \omega t} + \sum_{n=1}^{\infty}\frac{a_{n}- ib_{n}}{2}e^{in \omega t} + \sum_{n=-\infty}^{-1}\frac{a_{-n}+ ib_{-n}}{2}e^{in \omega t}\\ = &\sum_{-\infty}^{\infty}C_{n}e^{in \omega t} \end{aligned} f(t)=====2a0+n=1∑∞[21an(einωt+e−inωt)−21ibn(einωt−e−inωt)]2a0+n=1∑∞[2an−ibneinωt+2an+ibne−inωt]2a0+n=1∑∞2an−ibneinωt+n=−∞∑−12a−n+ib−neinωt∑n=1∞2an+ibne−inωtn=0∑02a0einωt+n=1∑∞2an−ibneinωt+n=−∞∑−12a−n+ib−neinωt−∞∑∞Cneinωt
看 系 數 C n : 看系數C_{n}: 看系數Cn:
C n = { a 0 2 = 1 T ∫ 0 T f ( t ) d t , n = 0 a n − i b n 2 = 1 T ∫ 0 T f ( t ) e − i n ω t d t n = 1 , 2 , 3 , ⋯ a − n + i b − n 2 = 1 T ∫ 0 T f ( t ) e − i n ω t d t n = − 1 , − 2 , − 3 , ⋯ C_{n}= \begin{cases} \frac{a_0}{2} = \frac{1}{T}\int_{0}^{T} f(t)\mathrm{d}t, \qquad n= 0\\ \frac{a_{n}- ib_{n}}{2}= \frac{1}{T}\int_{0}^{T} f(t)e^{-in \omega t} \mathrm{d}t \qquad n= 1, 2, 3, \cdots\\ \frac{a_{-n}+ ib_{-n}}{2} = \frac{1}{T}\int_{0}^{T} f(t)e^{-in \omega t} \mathrm{d}t \qquad n= -1, -2, -3, \cdots \end{cases} Cn=⎩⎪⎨⎪⎧2a0=T1∫0Tf(t)dt,n=02an−ibn=T1∫0Tf(t)e−inωtdtn=1,2,3,⋯2a−n+ib−n=T1∫0Tf(t)e−inωtdtn=−1,−2,−3,⋯
驚喜的發現:
f ( t ) = ∑ − ∞ ∞ C n e i n ω t f(t) = \sum_{-\infty}^{\infty}C_{n}e^{in \omega t} f(t)=−∞∑∞Cneinωt
可以變為:
C n = 1 T ∫ 0 T f ( t ) e − i n ω t d t C_{n}= \frac{1}{T}\int_{0}^{T} f(t)e^{-in \omega t} \mathrm{d}t Cn=T1∫0Tf(t)e−inωtdt
傅立葉變換,FT
有 f T ( t ) = f ( t + T ) 有f_T(t)= f(t+ T) 有fT(t)=f(t+T)
f T ( t ) = ∑ − ∞ ∞ C n e i n ω 0 t ( 1 ) ω 0 = 2 π T 基 頻 率 f_T(t)= \sum_{-\infty}^{\infty}C_{n}e^{in\omega_0 t} \qquad (1) \qquad \omega_0 = \frac{2\pi}{T} \quad 基頻率 fT(t)=−∞∑∞Cneinω0t(1)ω0=T2π基頻率
C n = 1 T ∫ − T 2 T 2 f T ( t ) e − i n ω 0 t d t ( 2 ) ( 周 期 積 分 ) C_{n}= \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} f_T(t)e^{-in \omega_0 t}\mathrm{d}t \qquad (2) \qquad (周期積分) Cn=T1∫−2T2TfT(t)e−inω0tdt(2)(周期積分)
真正區分開不同函數的是Cn, 它定義了函數。
非周期,一般形式:
lim T → ∞ f T ( t ) = f ( t ) \lim_{T \to \infty}f_T(t)= f(t) limT→∞fT(t)=f(t)
Δ ω = ( n + 1 ) ω 0 − n ω 0 = ω 0 = 2 π T \Delta \omega = (n+ 1)\omega_0 - n \omega_0 = \omega_0 = \frac{2\pi}{T} Δω=(n+1)ω0−nω0=ω0=T2π
T → ∞ , Δ ω → 0 , 離 散 → 連 續 T \to \infty, \Delta \omega \to 0 , 離散 \to 連續 T→∞,Δω→0,離散→連續
将(2)代入(1)且代入 1 T = Δ ω 2 π \frac{1}{T}= \frac{\Delta \omega}{2\pi} T1=2πΔω:
f T ( t ) = ∑ n = − ∞ ∞ Δ ω 2 π ∫ − T 2 T 2 f T ( t ) e − i n ω 0 t d t e i n ω 0 t f_T(t) = \sum_{n=-\infty}^{\infty}\frac{\Delta \omega}{2\pi}\int_{-\frac{T}{2}}^{\frac{T}{2}} f_T(t)e^{-in\omega_0 t}\mathrm{d}t e^{in\omega_0 t} fT(t)=n=−∞∑∞2πΔω∫−2T2TfT(t)e−inω0tdteinω0t
當 T → ∞ : 當T\to \infty: 當T→∞:
∫ − T 2 T 2 d t → ∫ − ∞ ∞ d t \int_{-\frac{T}{2}}^{\frac{T}{2}} \mathrm{d}t \to \int_{-\infty}^{\infty} \mathrm{d}t ∫−2T2Tdt→∫−∞∞dt
n ω 0 → ω n\omega_0 \to \omega nω0→ω
∑ n = − ∞ ∞ Δ ω → ∫ − ∞ ∞ d ω \sum_{n=-\infty}^{\infty}\Delta \omega \to \int_{-\infty}^{\infty} \mathrm{d}\omega n=−∞∑∞Δω→∫−∞∞dω
代入得:
f ( t ) = 1 2 π ∫ − ∞ ∞ ∫ − ∞ ∞ f ( t ) e − i ω t d t e i ω t d ω f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(t)e^{-i \omega t}\mathrm{d}t e^{i\omega t}\mathrm{d}\omega f(t)=2π1∫−∞∞∫−∞∞f(t)e−iωtdteiωtdω
我們取出中間的積分式得到了FT:
F ( ω ) = ∫ − ∞ ∞ f ( t ) e − i ω t d t F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-i\omega t}\mathrm{d}t F(ω)=∫−∞∞f(t)e−iωtdt
FT逆變換則為:
f ( t ) = 1 2 π ∫ − ∞ ∞ F ( ω ) e i ω t d ω f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(\omega)e^{i\omega t}\mathrm{d}\omega f(t)=2π1∫−∞∞F(ω)eiωtdω