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Dr. CAN学习笔记-Ch03 傅里叶级数与变换
- [1. 三角函数的正交性](#1. 三角函数的正交性)
- [2. 周期为 2 π 2\pi 2π的函数展开为傅里叶级数](#2. 周期为 2 π 2\pi 2π的函数展开为傅里叶级数)
- [3. 周期为 2 L 2L 2L的函数展开](#3. 周期为 2 L 2L 2L的函数展开)
- [4. 傅里叶级数的复数形式](#4. 傅里叶级数的复数形式)
- [5. 从傅里叶级数推导傅里叶变换FT](#5. 从傅里叶级数推导傅里叶变换FT)
- [6. 总结](#6. 总结)
1. 三角函数的正交性
三角函数系 ： 集合 { sin n x , cos n x } n = 0 , 1 , 2 , ⋯ \left\{ \sin nx,\cos nx \right\} n=0,1,2,\cdots {sinnx,cosnx}n=0,1,2,⋯
正交 ：
∫ − π π sin n x sin m x d x = 0 , n ≠ m ∫ − π π sin n x cos m x d x = 0 , n ≠ m ∫ − π π cos n x sin m x d x = 0 , n ≠ m \int_{-\pi}^{\pi}{\sin nx\sin mx}\mathrm{d}x=0,n\ne m \\ \int_{-\pi}^{\pi}{\sin nx\cos mx}\mathrm{d}x=0,n\ne m \\ \int_{-\pi}^{\pi}{\cos nx\sin mx}\mathrm{d}x=0,n\ne m ∫−ππsinnxsinmxdx=0,n=m∫−ππsinnxcosmxdx=0,n=m∫−ππcosnxsinmxdx=0,n=m
积化和差： ⇒ ∫ − π π 1 2 [ cos ( n − m ) x + cos ( n + m ) x ] d x = 1 2 1 ( n − m ) sin ( n − m ) x ∣ − π π + 1 2 1 ( n + m ) sin ( n + m ) x ∣ − π π \Rightarrow \int_{-\pi}^{\pi}{\frac{1}{2}\left[ \cos \left( n-m \right) x+\cos \left( n+m \right) x \right]}\mathrm{d}x=\frac{1}{2}\frac{1}{\left( n-m \right)}\sin \left( n-m \right) x\mid_{-\pi}^{\pi}+\frac{1}{2}\frac{1}{\left( n+m \right)}\sin \left( n+m \right) x\mid_{-\pi}^{\pi} ⇒∫−ππ21[cos(n−m)x+cos(n+m)x]dx=21(n−m)1sin(n−m)x∣−ππ+21(n+m)1sin(n+m)x∣−ππ
∫ − 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. 周期为 2 π 2\pi 2π的函数展开为傅里叶级数
T = 2 π : f ( x ) = f ( x + 2 π ) T=2\pi :f\left( x \right) =f\left( x+2\pi \right) T=2π:f(x)=f(x+2π)
f ( x ) = ∑ n = 0 ∞ a n cos n x + ∑ n = 0 ∞ b n sin n x = a 0 cos o x + ∑ n = 1 ∞ a n cos n x + b 0 sin 0 x + ∑ n = 1 ∞ b n sin n x = a 0 + ∑ n = 1 ∞ a n cos n x + ∑ n = 1 ∞ b n sin n x f\left( x \right) =\sum_{n=0}^{\infty}{a_{\mathrm{n}}\cos nx}+\sum_{n=0}^{\infty}{b_{\mathrm{n}}\sin nx}=a_0\cos ox+\sum_{n=1}^{\infty}{a_{\mathrm{n}}\cos nx}+b_0\sin 0x+\sum_{n=1}^{\infty}{b_{\mathrm{n}}\sin nx}=a_0+\sum_{n=1}^{\infty}{a_{\mathrm{n}}\cos nx}+\sum_{n=1}^{\infty}{b_{\mathrm{n}}\sin nx} f(x)=n=0∑∞ancosnx+n=0∑∞bnsinnx=a0cosox+n=1∑∞ancosnx+b0sin0x+n=1∑∞bnsinnx=a0+n=1∑∞ancosnx+n=1∑∞bnsinnx
- 找 a 0 a_0 a0:
∫ − π π f ( x ) d x = ∫ − π π a 0 d x + ∫ − π π 1 ⋅ ∑ n = 1 ∞ a n cos n x d x + ∫ − π π 1 ⋅ ∑ n = 1 ∞ b n sin n x d x = a 0 ∫ − π π d x + 0 + 0 = a 0 ⋅ 2 π \int_{-\pi}^{\pi}{f\left( x \right)}\mathrm{d}x=\int_{-\pi}^{\pi}{a_0}\mathrm{d}x+\int_{-\pi}^{\pi}{1\cdot \sum_{n=1}^{\infty}{a_{\mathrm{n}}\cos nx}}\mathrm{d}x+\int_{-\pi}^{\pi}{1\cdot \sum_{n=1}^{\infty}{b_{\mathrm{n}}\sin nx}}\mathrm{d}x \\ =a_0\int_{-\pi}^{\pi}{}\mathrm{d}x+0+0=a_0\cdot 2\pi ∫−ππf(x)dx=∫−ππa0dx+∫−ππ1⋅n=1∑∞ancosnxdx+∫−ππ1⋅n=1∑∞bnsinnxdx=a0∫−ππdx+0+0=a0⋅2π
⇒ a 0 = 1 2 π ∫ − π π f ( x ) d x \Rightarrow a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}{f\left( x \right)}\mathrm{d}x ⇒a0=2π1∫−ππf(x)dx - 找 a n a_n an:
∫ − π π f ( x ) cos m x d x = ∫ − π π a 0 cos m x ⋅ 1 d x + ∫ − π π ∑ n = 1 ∞ a n cos n x cos m x d x + ∫ − π π ∑ n = 1 ∞ b n sin n x cos m x d x = ∫ − π π a n cos n x cos n x d x = a n π \int_{-\pi}^{\pi}{f\left( x \right) \cos mx}\mathrm{d}x=\int_{-\pi}^{\pi}{a_0}\cos mx\cdot 1\mathrm{d}x+\int_{-\pi}^{\pi}{\sum_{n=1}^{\infty}{a_{\mathrm{n}}\cos nx\cos mx}}\mathrm{d}x+\int_{-\pi}^{\pi}{\sum_{n=1}^{\infty}{b_{\mathrm{n}}\sin nx\cos mx}}\mathrm{d}x=\int_{-\pi}^{\pi}{a_{\mathrm{n}}\cos nx\cos nx}\mathrm{d}x=a_{\mathrm{n}}\pi ∫−ππf(x)cosmxdx=∫−ππa0cosmx⋅1dx+∫−ππn=1∑∞ancosnxcosmxdx+∫−ππn=1∑∞bnsinnxcosmxdx=∫−ππancosnxcosnxdx=anπ
⇒ a n = 1 π ∫ − π π f ( x ) cos n x d x \Rightarrow a_{\mathrm{n}}=\frac{1}{\pi}\int_{-\pi}^{\pi}{f\left( x \right) \cos nx}\mathrm{d}x ⇒an=π1∫−ππf(x)cosnxdx - 找 b n b_n bn:
∫ − π π f ( x ) sin . m x d x ⇒ b n = 1 π ∫ − π π f ( x ) sin n x d x \int_{-\pi}^{\pi}{f\left( x \right) \sin .mx}\mathrm{d}x\Rightarrow b_{\mathrm{n}}=\frac{1}{\pi}\int_{-\pi}^{\pi}{f\left( x \right) \sin nx}\mathrm{d}x ∫−ππf(x)sin.mxdx⇒bn=π1∫−ππf(x)sinnxdx
⇒ f ( x ) = f ( x + 2 π ) , T = 2 π { f ( x ) = a 0 2 + ∑ n = 0 ∞ a n cos n x + ∑ n = 0 ∞ b n sin n x a 0 = 1 2 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 \Rightarrow f\left( x \right) =f\left( x+2\pi \right) ,T=2\pi \begin{cases} f\left( x \right) =\frac{a_0}{2}+\sum_{n=0}^{\infty}{a_{\mathrm{n}}\cos nx}+\sum_{n=0}^{\infty}{b_{\mathrm{n}}\sin nx}\\ a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}{f\left( x \right)}\mathrm{d}x\\ a_{\mathrm{n}}=\frac{1}{\pi}\int_{-\pi}^{\pi}{f\left( x \right) \cos nx}\mathrm{d}x\\ b_{\mathrm{n}}=\frac{1}{\pi}\int_{-\pi}^{\pi}{f\left( x \right) \sin nx}\mathrm{d}x\\ \end{cases} ⇒f(x)=f(x+2π),T=2π⎩ ⎨ ⎧f(x)=2a0+∑n=0∞ancosnx+∑n=0∞bnsinnxa0=2π1∫−ππf(x)dxan=π1∫−ππf(x)cosnxdxbn=π1∫−ππf(x)sinnxdx
3. 周期为 2 L 2L 2L的函数展开
f ( t ) = f ( t + 2 L ) f\left( t \right) =f\left( t+2L \right) f(t)=f(t+2L) , 换元： x = π L t , t = L π x x=\frac{\pi}{L}t,t=\frac{L}{\pi}x x=Lπt,t=πLx
f ( t ) = f ( L π x ) = g ( x ) , g ( x + 2 π ) = f ( L π ( x + 2 π ) ) = f ( L π x + 2 L ) = f ( L π x ) = g ( x ) f\left( t \right) =f\left( \frac{L}{\pi}x \right) =g\left( x \right) ,g\left( x+2\pi \right) =f\left( \frac{L}{\pi}\left( x+2\pi \right) \right) =f\left( \frac{L}{\pi}x+2L \right) =f\left( \frac{L}{\pi}x \right) =g\left( x \right) f(t)=f(πLx)=g(x),g(x+2π)=f(πL(x+2π))=f(πLx+2L)=f(πLx)=g(x)
g ( x ) = a 0 2 + ∑ n = 1 ∞ [ a n cos n x + b n sin n x ] a 0 = 1 2 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 g\left( x \right) =\frac{a_0}{2}+\sum_{n=1}^{\infty}{\left[ a_{\mathrm{n}}\cos nx+b_{\mathrm{n}}\sin nx \right]} \\ a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}{f\left( x \right)}\mathrm{d}x,a_{\mathrm{n}}=\frac{1}{\pi}\int_{-\pi}^{\pi}{f\left( x \right) \cos nx}\mathrm{d}x,b_{\mathrm{n}}=\frac{1}{\pi}\int_{-\pi}^{\pi}{f\left( x \right) \sin nx}\mathrm{d}x g(x)=2a0+n=1∑∞[ancosnx+bnsinnx]a0=2π1∫−ππf(x)dx,an=π1∫−ππf(x)cosnxdx,bn=π1∫−ππf(x)sinnxdx
→ x = π L t ⇒ cos n x = cos n π L t , sin n x = sin n π L t , g ( x ) = f ( t ) ∫ − π π d x = ∫ − π π d π L t ⇒ 1 π ∫ − π π d x = 1 L ∫ − L L d t \rightarrow x=\frac{\pi}{L}t\Rightarrow \cos nx=\cos \frac{n\pi}{L}t,\sin nx=\sin \frac{n\pi}{L}t,g\left( x \right) =f\left( t \right) \\ \int_{-\pi}^{\pi}{}\mathrm{d}x=\int_{-\pi}^{\pi}{}\mathrm{d}\frac{\pi}{L}t\Rightarrow \frac{1}{\pi}\int_{-\pi}^{\pi}{}\mathrm{d}x=\frac{1}{L}\int_{-L}^L{}\mathrm{d}t →x=Lπt⇒cosnx=cosLnπt,sinnx=sinLnπt,g(x)=f(t)∫−ππdx=∫−ππdLπt⇒π1∫−ππdx=L1∫−LLdt
⇒ f ( t ) = a 0 2 + ∑ n = 1 ∞ [ a n cos n π L t + b n sin n π L t ] , 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 \Rightarrow f\left( t \right) =\frac{a_0}{2}+\sum_{n=1}^{\infty}{\left[ a_{\mathrm{n}}\cos \frac{n\pi}{L}t+b_{\mathrm{n}}\sin \frac{n\pi}{L}t \right]},a_0=\frac{1}{L}\int_{-L}^L{f\left( t \right)}\mathrm{d}t,a_{\mathrm{n}}=\frac{1}{L}\int_{-L}^L{f\left( t \right)}\cos \frac{n\pi}{L}t\mathrm{d}t,b_{\mathrm{n}}=\frac{1}{L}\int_{-L}^L{f\left( t \right)}\sin \frac{n\pi}{L}t\mathrm{d}t ⇒f(t)=2a0+n=1∑∞[ancosLnπt+bnsinLnπt],a0=L1∫−LLf(t)dt,an=L1∫−LLf(t)cosLnπtdt,bn=L1∫−LLf(t)sinLnπtdt
4. 傅里叶级数的复数形式
f ( t ) = a 0 2 + ∑ n = 1 ∞ [ a n 1 2 ( e i n w t + e − i n w t ) − b n 1 2 ( e i n w t − e − i n w t ) ] = a 0 2 + ∑ n = 1 ∞ [ a n − i b n 2 e i n w t + a n + i b n 2 e − i n w t ] = ∑ n = 0 0 a 0 2 e i n w t + ∑ n = 1 ∞ a n − i b n 2 e i n w t + ∑ n = − ∞ − 1 a n + i b n 2 e i n w t = ∑ n = − ∞ ∞ C n e i n w t , C n = { a 0 2 n = 0 a n − i b n 2 n = 1 , 2 , 3 , ⋯ a n + i b n 2 n = − 1 , − 2 , − 3 , ⋯ → 1 T ∫ 0 T f ( t ) d t → 1 T ∫ 0 T f ( t ) ( cos n w t − i sin n w t ) d t = 1 T ∫ 0 T f ( t ) ( cos ( − n w t ) + i sin ( − n w t ) ) d t = 1 T ∫ 0 T f ( t ) e − i n w t d t → 1 T ∫ 0 T f ( t ) e − i n w t d t ⇒ f ( t ) = ∑ − ∞ ∞ C n e i n w t , C n = 1 T ∫ 0 T f ( t ) e − i n w t d t f\left( t \right) =\frac{a_0}{2}+\sum_{n=1}^{\infty}{\left[ a_{\mathrm{n}}\frac{1}{2}\left( e^{inwt}+e^{-inwt} \right) -b_{\mathrm{n}}\frac{1}{2}\left( e^{inwt}-e^{-inwt} \right) \right]}=\frac{a_0}{2}+\sum_{n=1}^{\infty}{\left[ \frac{a_{\mathrm{n}}-ib_{\mathrm{n}}}{2}e^{inwt}+\frac{a_{\mathrm{n}}+ib_{\mathrm{n}}}{2}e^{-inwt} \right]} \\ =\sum_{n=0}^0{\frac{a_0}{2}e^{inwt}}+\sum_{n=1}^{\infty}{\frac{a_{\mathrm{n}}-ib_{\mathrm{n}}}{2}e^{inwt}}+\sum_{n=-\infty}^{-1}{\frac{a_{\mathrm{n}}+ib_{\mathrm{n}}}{2}e^{inwt}} \\ =\sum_{n=-\infty}^{\infty}{C_{\mathrm{n}}e^{inwt}},C_{\mathrm{n}}=\begin{cases} \frac{a_0}{2}\,\,n=0\\ \frac{a_{\mathrm{n}}-ib_{\mathrm{n}}}{2}\,\,n=1,2,3,\cdots\\ \frac{a_{\mathrm{n}}+ib_{\mathrm{n}}}{2}\,\,n=-1,-2,-3,\cdots\\ \end{cases}\begin{array}{c} \rightarrow \frac{1}{T}\int_0^T{f\left( t \right)}\mathrm{d}t\\ \rightarrow \frac{1}{T}\int_0^T{f\left( t \right)}\left( \cos nwt-i\sin nwt \right) \mathrm{d}t=\frac{1}{T}\int_0^T{f\left( t \right)}\left( \cos \left( -nwt \right) +i\sin \left( -nwt \right) \right) \mathrm{d}t=\frac{1}{T}\int_0^T{f\left( t \right)}e^{-inwt}\mathrm{d}t\\ \rightarrow \frac{1}{T}\int_0^T{f\left( t \right) e^{-inwt}}\mathrm{d}t\\ \end{array} \\ \Rightarrow f\left( t \right) =\sum_{-\infty}^{\infty}{C_{\mathrm{n}}e^{inwt}},C_{\mathrm{n}}=\frac{1}{T}\int_0^T{f\left( t \right) e^{-inwt}}\mathrm{d}t f(t)=2a0+n=1∑∞[an21(einwt+e−inwt)−bn21(einwt−e−inwt)]=2a0+n=1∑∞[2an−ibneinwt+2an+ibne−inwt]=n=0∑02a0einwt+n=1∑∞2an−ibneinwt+n=−∞∑−12an+ibneinwt=n=−∞∑∞Cneinwt,Cn=⎩ ⎨ ⎧2a0n=02an−ibnn=1,2,3,⋯2an+ibnn=−1,−2,−3,⋯→T1∫0Tf(t)dt→T1∫0Tf(t)(cosnwt−isinnwt)dt=T1∫0Tf(t)(cos(−nwt)+isin(−nwt))dt=T1∫0Tf(t)e−inwtdt→T1∫0Tf(t)e−inwtdt⇒f(t)=−∞∑∞Cneinwt,Cn=T1∫0Tf(t)e−inwtdt
- Euler's Formula
5. 从傅里叶级数推导傅里叶变换FT
f T ( t ) = f ( t + T ) f_{\mathrm{T}}\left( t \right) =f\left( t+T \right) fT(t)=f(t+T)
f T ( t ) = ∑ − ∞ ∞ C n e i n w 0 t f_{\mathrm{T}}\left( t \right) =\sum_{-\infty}^{\infty}{C_{\mathrm{n}}e^{inw_0t}} fT(t)=∑−∞∞Cneinw0t, 基频率： w 0 = 2 π T w_0=\frac{2\pi}{T} w0=T2π, 定义函数: C n = 1 T ∫ − T 2 T 2 f T ( t ) e − i n w t d t C_{\mathrm{n}}=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}{f_{\mathrm{T}}\left( t \right) e^{-inwt}}\mathrm{d}t Cn=T1∫−2T2TfT(t)e−inwtdt
非周期，一般形式： T → ∞ T\rightarrow \infty T→∞
lim T → ∞ f T ( t ) = f ( t ) , Δ w = ( n + 1 ) w 0 − n w 0 = w 0 = 2 π T T ↗ Δ w ↘ \underset{T\rightarrow \infty}{\lim}f_{\mathrm{T}}\left( t \right) =f\left( t \right) ,\varDelta w=\left( n+1 \right) w_0-nw_0=w_0=\frac{2\pi}{T}\,\,T\nearrow \varDelta w\searrow T→∞limfT(t)=f(t),Δw=(n+1)w0−nw0=w0=T2πT↗Δw↘
f T ( t ) = ∑ − ∞ ∞ ( 1 T ∫ − T 2 T 2 f T ( t ) e − i n w 0 t d t ) e i n w 0 t , 1 T = Δ w 2 π ⇒ f T ( t ) = ∑ − ∞ ∞ ( Δ w 2 π ∫ − T 2 T 2 f T ( t ) e − i n w 0 t d t ) e i n w 0 t , T → ∞ : { ∫ − T 2 T 2 d t → ∫ − ∞ ∞ d t n w 0 → w ∑ − ∞ ∞ Δ w → ∫ − ∞ ∞ d w ⇒ f ( t ) = 1 2 π ∫ − ∞ ∞ ( ∫ − ∞ ∞ f ( t ) e − i w t d t ) e i w t d w , ∫ − ∞ ∞ f ( t ) e − i w t d t = F ( w ) ⇒ f ( t ) = 1 2 π ∫ − ∞ ∞ F ( w ) e i w t d w f_{\mathrm{T}}\left( t \right) =\sum_{-\infty}^{\infty}{\left( \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}{f_{\mathrm{T}}\left( t \right) e^{-inw_0t}}\mathrm{d}t \right) e^{inw_0t}},\frac{1}{T}=\frac{\varDelta w}{2\pi} \\ \Rightarrow f_{\mathrm{T}}\left( t \right) =\sum_{-\infty}^{\infty}{\left( \frac{\varDelta w}{2\pi}\int_{-\frac{T}{2}}^{\frac{T}{2}}{f_{\mathrm{T}}\left( t \right) e^{-inw_0t}}\mathrm{d}t \right) e^{inw_0t}},T\rightarrow \infty :\begin{cases} \int_{-\frac{T}{2}}^{\frac{T}{2}}{}\mathrm{d}t\rightarrow \int_{-\infty}^{\infty}{}\mathrm{d}t\\ nw_0\rightarrow w\\ \sum_{-\infty}^{\infty}{\varDelta w}\rightarrow \int_{-\infty}^{\infty}{}\mathrm{d}w\\ \end{cases} \\ \Rightarrow f\left( t \right) =\frac{1}{2\pi}\int_{-\infty}^{\infty}{\left( \int_{-\infty}^{\infty}{f\left( t \right) e^{-iwt}}\mathrm{d}t \right)}e^{iwt}\mathrm{d}w,\int_{-\infty}^{\infty}{f\left( t \right) e^{-iwt}}\mathrm{d}t=F\left( w \right) \\ \Rightarrow f\left( t \right) =\frac{1}{2\pi}\int_{-\infty}^{\infty}{F\left( w \right)}e^{iwt}\mathrm{d}w fT(t)=−∞∑∞(T1∫−2T2TfT(t)e−inw0tdt)einw0t,T1=2πΔw⇒fT(t)=−∞∑∞(2πΔw∫−2T2TfT(t)e−inw0tdt)einw0t,T→∞:⎩ ⎨ ⎧∫−2T2Tdt→∫−∞∞dtnw0→w∑−∞∞Δw→∫−∞∞dw⇒f(t)=2π1∫−∞∞(∫−∞∞f(t)e−iwtdt)eiwtdw,∫−∞∞f(t)e−iwtdt=F(w)⇒f(t)=2π1∫−∞∞F(w)eiwtdw
F ( w ) = ∫ − ∞ ∞ f ( t ) e − i w t d t F\left( w \right) =\int_{-\infty}^{\infty}{f\left( t \right) e^{-iwt}}\mathrm{d}t F(w)=∫−∞∞f(t)e−iwtdt : FT 傅里叶变换
f ( t ) = 1 2 π ∫ − ∞ ∞ F ( w ) e i w t d w f\left( t \right) =\frac{1}{2\pi}\int_{-\infty}^{\infty}{F\left( w \right)}e^{iwt}\mathrm{d}w f(t)=2π1∫−∞∞F(w)eiwtdw : 逆变换
6. 总结
三角函数的正交性：
[ 0 , 1 , sin x , cos x , sin 2 x , cos 2 x , ⋯ , sin n x , cos n x ] , n = 0 , 1 , 2 , ⋯ \left[ 0,1,\sin x,\cos x,\sin 2x,\cos 2x,\cdots ,\sin nx,\cos nx \right] ,n=0,1,2,\cdots [0,1,sinx,cosx,sin2x,cos2x,⋯,sinnx,cosnx],n=0,1,2,⋯
∫ − π π sin n x sin m x d x = 0 , n ≠ m ∫ − π π sin n x sin m x d x = 0 , n ≠ m ∫ − π π sin n x cos m x d x = 0 , n ≠ m \int_{-\pi}^{\pi}{\sin nx\sin mx}\mathrm{d}x=0,n\ne m \\ \int_{-\pi}^{\pi}{\sin nx\sin mx}\mathrm{d}x=0,n\ne m \\ \int_{-\pi}^{\pi}{\sin nx\cos mx}\mathrm{d}x=0,n\ne m ∫−ππsinnxsinmxdx=0,n=m∫−ππsinnxsinmxdx=0,n=m∫−ππsinnxcosmxdx=0,n=m
周期 2 π 2\pi 2π :
f ( x ) = f ( x + 2 π ) f\left( x \right) =f\left( x+2\pi \right) f(x)=f(x+2π)
f ( x ) = ∑ n = 0 ∞ a n cos n x + ∑ n = 0 ∞ b n sin n x ← f ( x ) = a 0 2 + ∑ n = 0 ∞ a n cos n x + ∑ n = 0 ∞ b n sin n x f\left( x \right) =\sum_{n=0}^{\infty}{a_{\mathrm{n}}\cos nx}+\sum_{n=0}^{\infty}{b_{\mathrm{n}}\sin nx}\gets f\left( x \right) =\frac{a_0}{2}+\sum_{n=0}^{\infty}{a_{\mathrm{n}}\cos nx}+\sum_{n=0}^{\infty}{b_{\mathrm{n}}\sin nx} f(x)=n=0∑∞ancosnx+n=0∑∞bnsinnx←f(x)=2a0+n=0∑∞ancosnx+n=0∑∞bnsinnx
a 0 = 1 2 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 a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}{f\left( x \right)}\mathrm{d}x,a_{\mathrm{n}}=\frac{1}{\pi}\int_{-\pi}^{\pi}{f\left( x \right) \cos nx}\mathrm{d}x,b_{\mathrm{n}}=\frac{1}{\pi}\int_{-\pi}^{\pi}{f\left( x \right) \sin nx}\mathrm{d}x a0=2π1∫−ππf(x)dx,an=π1∫−ππf(x)cosnxdx,bn=π1∫−ππf(x)sinnxdx
周期 2 L 2L 2L :
T = 2 L , f ( t ) = f ( t + 2 L ) , x = π L t T=2L,f\left( t \right) =f\left( t+2L \right) ,x=\frac{\pi}{L}t T=2L,f(t)=f(t+2L),x=Lπt
f ( t ) = a 0 2 + ∑ n = 1 ∞ [ a n cos n π L t + b n sin n π L t ] f\left( t \right) =\frac{a_0}{2}+\sum_{n=1}^{\infty}{\left[ a_{\mathrm{n}}\cos \frac{n\pi}{L}t+b_{\mathrm{n}}\sin \frac{n\pi}{L}t \right]} f(t)=2a0+n=1∑∞[ancosLnπt+bnsinLnπt]
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 a_0=\frac{1}{L}\int_{-L}^L{f\left( t \right)}\mathrm{d}t,a_{\mathrm{n}}=\frac{1}{L}\int_{-L}^L{f\left( t \right)}\cos \frac{n\pi}{L}t\mathrm{d}t,b_{\mathrm{n}}=\frac{1}{L}\int_{-L}^L{f\left( t \right)}\sin \frac{n\pi}{L}t\mathrm{d}t a0=L1∫−LLf(t)dt,an=L1∫−LLf(t)cosLnπtdt,bn=L1∫−LLf(t)sinLnπtdt
复指数：
f ( t ) = ∑ − ∞ ∞ C n e i n w 0 t , w 0 = 2 π T , C n = 1 T ∫ 0 T f ( t ) e − i n w 0 t d t f\left( t \right) =\sum_{-\infty}^{\infty}{C_{\mathrm{n}}e^{inw_0t}},w_0=\frac{2\pi}{T},C_{\mathrm{n}}=\frac{1}{T}\int_0^T{f\left( t \right) e^{-inw_0t}}\mathrm{d}t f(t)=−∞∑∞Cneinw0t,w0=T2π,Cn=T1∫0Tf(t)e−inw0tdt
FT ：
f ( t ) = f ( t + T ) , T → ∞ f\left( t \right) =f\left( t+T \right) ,T\rightarrow \infty f(t)=f(t+T),T→∞ , f ( t ) = 1 2 π ∫ − ∞ ∞ ( ∫ − ∞ ∞ f ( t ) e − i w t d t ) e i w t d w f\left( t \right) =\frac{1}{2\pi}\int_{-\infty}^{\infty}{\left( \int_{-\infty}^{\infty}{f\left( t \right) e^{-iwt}}\mathrm{d}t \right)}e^{iwt}\mathrm{d}w f(t)=2π1∫−∞∞(∫−∞∞f(t)e−iwtdt)eiwtdw
F T → F ( w ) = ∫ − ∞ ∞ f ( t ) e − i w t d t I F T → ( t ) = 1 2 π ∫ − ∞ ∞ F ( w ) e i w t d w \begin{array}{c} FT\rightarrow F\left( w \right) =\int_{-\infty}^{\infty}{f\left( t \right) e^{-iwt}}\mathrm{d}t\\ IFT\rightarrow \left( t \right) =\frac{1}{2\pi}\int_{-\infty}^{\infty}{F\left( w \right)}e^{iwt}\mathrm{d}w\\ \end{array} FT→F(w)=∫−∞∞f(t)e−iwtdtIFT→(t)=2π1∫−∞∞F(w)eiwtdw
Laplace : F ( s ) : ∫ − ∞ ∞ f ( t ) e − s t d t F\left( s \right) :\int_{-\infty}^{\infty}{f\left( t \right) e^{-st}}\mathrm{d}t F(s):∫−∞∞f(t)e−stdt