[足式机器人]Part2 Dr. CAN学习笔记 - Ch03 傅里叶级数与变换

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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
  • 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
  • 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
  • 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=mππsinnxcosmxdx=0,n=mππcosnxsinmxdx=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(nm)x+cos(n+m)x]dx=21(nm)1sin(nm)xππ+21(n+m)1sin(n+m)xππ
∫ − 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=0ancosnx+n=0bnsinnx=a0cosox+n=1ancosnx+b0sin0x+n=1bnsinnx=a0+n=1ancosnx+n=1bnsinnx

  1. 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+ππ1n=1ancosnxdx+ππ1n=1bnsinnxdx=a0ππdx+0+0=a02π
    ⇒ 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=2π1ππf(x)dx
  2. 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=ππa0cosmx1dx+ππn=1ancosnxcosmxdx+ππn=1bnsinnxcosmxdx=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=π1ππf(x)cosnxdx
  3. 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.mxdxbn=π1ππf(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=0ancosnx+n=0bnsinnxa0=2π1ππf(x)dxan=π1ππf(x)cosnxdxbn=π1ππf(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=2π1ππf(x)dx,an=π1ππf(x)cosnxdx,bn=π1ππf(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=Lπtcosnx=cosLt,sinnx=sinLt,g(x)=f(t)ππdx=ππdLπtπ1ππdx=L1LLdt

⇒ 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[ancosLt+bnsinLt],a0=L1LLf(t)dt,an=L1LLf(t)cosLtdt,bn=L1LLf(t)sinLtdt
[足式机器人]Part2 Dr. CAN学习笔记 - Ch03 傅里叶级数与变换_第1张图片

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+einwt)bn21(einwteinwt)]=2a0+n=1[2anibneinwt+2an+ibneinwt]=n=002a0einwt+n=12anibneinwt+n=12an+ibneinwt=n=Cneinwt,Cn= 2a0n=02anibnn=1,2,3,2an+ibnn=1,2,3,T10Tf(t)dtT10Tf(t)(cosnwtisinnwt)dt=T10Tf(t)(cos(nwt)+isin(nwt))dt=T10Tf(t)einwtdtT10Tf(t)einwtdtf(t)=Cneinwt,Cn=T10Tf(t)einwtdt

  • 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=T12T2TfT(t)einwtdt

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 TlimfT(t)=f(t),Δw=(n+1)w0nw0=w0=T2πTΔw

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)=(T12T2TfT(t)einw0tdt)einw0t,T1=2πΔwfT(t)=(2πΔw2T2TfT(t)einw0tdt)einw0t,T: 2T2Tdtdtnw0wΔwdwf(t)=2π1(f(t)eiwtdt)eiwtdw,f(t)eiwtdt=F(w)f(t)=2π1F(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)eiwtdt : 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)=2π1F(w)eiwtdw : 逆变换

[足式机器人]Part2 Dr. CAN学习笔记 - Ch03 傅里叶级数与变换_第2张图片

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=mππsinnxsinmxdx=0,n=mππsinnxcosmxdx=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=0ancosnx+n=0bnsinnxf(x)=2a0+n=0ancosnx+n=0bnsinnx
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=2π1ππf(x)dx,an=π1ππf(x)cosnxdx,bn=π1ππf(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[ancosLt+bnsinLt]
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=L1LLf(t)dt,an=L1LLf(t)cosLtdt,bn=L1LLf(t)sinLtdt

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=T10Tf(t)einw0tdt

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)eiwtdt)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} FTF(w)=f(t)eiwtdtIFT(t)=2π1F(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)estdt

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