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

发布时间:2024年01月08日

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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 ?ππ?sinnxsinmxdx=0,n=m?ππ?sinnxcosmxdx=0,n=m?ππ?cosnxsinmxdx=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} ??ππ?21?[cos(n?m)x+cos(n+m)x]dx=21?(n?m)1?sin(n?m)x?ππ?+21?(n+m)1?sin(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 ?ππ?cosmxcosmxdx=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?an?cosnx+n=0?bn?sinnx=a0?cosox+n=1?an?cosnx+b0?sin0x+n=1?bn?sinnx=a0?+n=1?an?cosnx+n=1?bn?sinnx

  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 ?ππ?f(x)dx=?ππ?a0?dx+?ππ?1?n=1?an?cosnxdx+?ππ?1?n=1?bn?sinnxdx=a0??ππ?dx+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??ππ?f(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 ?ππ?f(x)cosmxdx=?ππ?a0?cosmx?1dx+?ππ?n=1?an?cosnxcosmxdx+?ππ?n=1?bn?sinnxcosmxdx=?ππ?an?cosnxcosnxdx=an?π
    ? 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??ππ?f(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 ?ππ?f(x)sin.mxdx?bn?=π1??ππ?f(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?an?cosnx+n=0?bn?sinnxa0?=2π1??ππ?f(x)dxan?=π1??ππ?f(x)cosnxdxbn?=π1??ππ?f(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=Lπ?t,t=πL?x
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(πL?x)=g(x),g(x+2π)=f(πL?(x+2π))=f(πL?x+2L)=f(πL?x)=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?[an?cosnx+bn?sinnx]a0?=2π1??ππ?f(x)dx,an?=π1??ππ?f(x)cosnxdx,bn?=π1??ππ?f(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=Lπ?t?cosnx=cosL?t,sinnx=sinL?t,g(x)=f(t)?ππ?dx=?ππ?dLπ?t?π1??ππ?dx=L1??LL?dt

? 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?[an?cosL?t+bn?sinL?t],a0?=L1??LL?f(t)dt,an?=L1??LL?f(t)cosL?tdt,bn?=L1??LL?f(t)sinL?tdt
在这里插入图片描述

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?[an?21?(einwt+e?inwt)?bn?21?(einwt?e?inwt)]=2a0??+n=1?[2an??ibn??einwt+2an?+ibn??e?inwt]=n=00?2a0??einwt+n=1?2an??ibn??einwt+n=??1?2an?+ibn??einwt=n=??Cn?einwt,Cn?=? ? ??2a0??n=02an??ibn??n=1,2,3,?2an?+ibn??n=?1,?2,?3,??T1?0T?f(t)dtT1?0T?f(t)(cosnwt?isinnwt)dt=T1?0T?f(t)(cos(?nwt)+isin(?nwt))dt=T1?0T?f(t)e?inwtdtT1?0T?f(t)e?inwtdt??f(t)=??Cn?einwt,Cn?=T1?0T?f(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)=??Cn?einw0?t, 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??2T?2T??fT?(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 Tlim?fT?(t)=f(t),Δw=(n+1)w0??nw0?=w0?=T2π?TΔw

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??2T?2T??fT?(t)e?inw0?tdt)einw0?t,T1?=2πΔw??fT?(t)=??(2πΔw??2T?2T??fT?(t)e?inw0?tdt)einw0?t,T:? ? ???2T?2T??dt??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

在这里插入图片描述

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 ?ππ?sinnxsinmxdx=0,n=m?ππ?sinnxsinmxdx=0,n=m?ππ?sinnxcosmxdx=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?an?cosnx+n=0?bn?sinnxf(x)=2a0??+n=0?an?cosnx+n=0?bn?sinnx
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??ππ?f(x)dx,an?=π1??ππ?f(x)cosnxdx,bn?=π1??ππ?f(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=Lπ?t
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?[an?cosL?t+bn?sinL?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 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??LL?f(t)dt,an?=L1??LL?f(t)cosL?tdt,bn?=L1??LL?f(t)sinL?tdt

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)=??Cn?einw0?t,w0?=T2π?,Cn?=T1?0T?f(t)e?inw0?tdt

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} FTF(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

文章来源:https://blog.csdn.net/LiongLoure/article/details/135453891
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