参考：DR_CAN
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动态系统的建模与分析 自动控制原理 Advanced控制理论 傅里叶级数与变换 工程数学三角函数系是一个集合： { 1 , sin x , cos x , sin 2 x , cos 2 x , ⋯ , sin n x , cos n x , ⋯ } \left \{ 1,\sin x, \cos x, \sin 2x,\cos 2x,\cdots,\sin nx, \cos nx,\cdots \right \} {1,sinx,cosx,sin2x,cos2x,⋯,sinnx,cosnx,⋯}
三角函数系更一般的形式： { sin 0 x , cos 0 x , sin x , cos x , sin 2 x , cos 2 x , ⋯ , sin n x , cos n x , ⋯ } \left \{ \sin 0x, \cos 0x,\sin x, \cos x, \sin 2x,\cos 2x,\cdots,\sin nx, \cos nx,\cdots \right \} {sin0x,cos0x,sinx,cosx,sin2x,cos2x,⋯,sinnx,cosnx,⋯}
集合中任意两个三角函数乘积的积分为0，即正交： ∫ − π π sin n x cos m x d x = 0 ∫ − π π cos n x cos m x d x = 0 , n ≠ m \begin{aligned} &\int_{-\pi}^{\pi} \sin nx \cos mx dx=0 \\ &\int_{-\pi}^{\pi} \cos nx \cos mx dx=0 \quad,n \ne m \end{aligned} ∫−ππsinnxcosmxdx=0∫−ππcosnxcosmxdx=0,n=m
证明： a ⃗ ⋅ b ⃗ = ( 2 , 1 ) ⋅ ( − 1 , 2 ) = 0 \vec a\cdot \vec b=(2,1)\cdot (-1,2)=0 a ⋅b =(2,1)⋅(−1,2)=0
a ⃗ = ( a 1 , a 2 , ⋯ , a n ) b ⃗ = ( b 1 , b 2 , ⋯ , b n ) a ⃗ ⋅ b ⃗ = a 1 b 1 + a 2 b 2 + ⋯ + a n b n = ∑ i = 1 n a i b i \vec a=(a_1,a_2,\cdots ,a_n)\\ \vec b=(b_1,b_2,\cdots,b_n)\\ \vec a \cdot \vec b=a_1b_1+a_2b_2+\cdots +a_nb_n=\sum_{i=1}^{n}a_ib_i a =(a1,a2,⋯,an)b =(b1,b2,⋯,bn)a ⋅b =a1b1+a2b2+⋯+anbn=∑i=1naibi 离散为求和，连续为积分
写成积分的形式，向量变为函数 a = f ( x ) , b = g ( x ) a ⋅ b = ∫ x 0 x 1 f ( x ) g ( x ) d x = 0 a=f(x),b=g(x)\\ a\cdot b=\int_{x_0}^{x_1}f(x)g(x)dx=0 a=f(x),b=g(x)a⋅b=∫x0x1f(x)g(x)dx=0
例证： ∫ − π π cos 0 x sin x d x = ∫ − π π sin x d x = 0 \int_{-\pi}^{\pi}\cos 0x \sin x dx=\int_{-\pi}^{\pi}\sin x dx=0 ∫−ππcos0xsinxdx=∫−ππsinxdx=0 sin x 是 奇 函 数 ， 因 此 对 称 区 间 面 积 抵 消 \sin x是奇函数，因此对称区间面积抵消 sinx是奇函数，因此对称区间面积抵消
∫ − π π cos n x cos m x = 1 2 [ ∫ − π π cos ( n − m ) x + ∫ − π π cos ( n + m ) x ] = 1 2 [ 1 n − m sin ( n − m ) x ∣ − π π + 1 n + m sin ( n + m ) x ∣ − π π ] = 0 \begin{aligned} &\int_{-\pi}^{\pi}\cos nx \cos mx\\ &=\frac{1}{2} \left [\int_{-\pi}^{\pi}\cos (n-m)x +\int_{-\pi}^{\pi}\cos (n+m)x \right ]\\ &=\frac{1}{2}\left [\left . \frac{1}{n-m} \sin (n-m)x \right |_{-\pi}^{\pi}+ \left . \frac{1}{n+m} \sin (n+m)x \right |_{-\pi}^{\pi}\right ]\\ &=0 \end{aligned} ∫−ππcosnxcosmx=21[∫−ππcos(n−m)x+∫−ππcos(n+m)x]=21[n−m1sin(n−m)x∣∣∣∣−ππ+n+m1sin(n+m)x∣∣∣∣−ππ]=0
∫ − π π cos n x sin m x d x = 0 , n ≠ m ∫ − π π sin n x sin m x d x = 0 , n ≠ m \int_{-\pi}^{\pi}\cos nx \sin mx dx=0,n \ne m\\ \int_{-\pi}^{\pi}\sin nx \sin mx dx=0,n \ne m ∫−ππcosnxsinmxdx=0,n=m∫−ππsinnxsinmxdx=0,n=m
如果n=m ∫ − π π cos n x cos m x d x = 1 2 ∫ − π π 1 d x + ∫ − π π cos 2 m x d x = 1 2 ∫ − π π 1 d x = π \begin{aligned} &\int_{-\pi}^{\pi}\cos nx \cos mx dx\\ &=\frac{1}{2}\int_{-\pi}^{\pi}1dx+\int_{-\pi}^{\pi}\cos 2mx dx\\ &=\frac{1}{2}\int_{-\pi}^{\pi}1dx=\pi \end{aligned} ∫−ππcosnxcosmxdx=21∫−ππ1dx+∫−ππcos2mxdx=21∫−ππ1dx=π
T = 2 π , f ( x ) = f ( x + 2 π ) T=2\pi,f(x)=f(x+2\pi) T=2π,f(x)=f(x+2π)
傅立叶级数展开式 f ( x ) = ∑ n = 0 ∞ a n cos n x + ∑ n = 0 ∞ b n sin n x = a 0 cos 0 x + ∑ n = 1 ∞ a n cos n x + b 0 sin 0 x + ∑ n = 1 ∞ b n sin n x = a 0 2 + ∑ n = 1 ∞ a n cos n x + ∑ n = 1 ∞ b n sin n x \begin{aligned} f(x) &=\sum_{n=0}^{\infty}a_n \cos nx+\sum_{n=0}^{\infty}b_n \sin nx\\ &=a_0 \cos 0x + \sum_{n=1}^{\infty}a_n \cos nx+b_0 \sin 0x+\sum_{n=1}^{\infty}b_n \sin nx\\ &=\frac{a_0}{2} + \sum_{n=1}^{\infty}a_n \cos nx+\sum_{n=1}^{\infty}b_n \sin nx \end{aligned} f(x)=n=0∑∞ancosnx+n=0∑∞bnsinnx=a0cos0x+n=1∑∞ancosnx+b0sin0x+n=1∑∞bnsinnx=2a0+n=1∑∞ancosnx+n=1∑∞bnsinnx
确定 a 0 a_0 a0 ∫ − π π f ( x ) d x = ∫ − π π a 0 d x = 2 π a 0 a 0 = 1 π ∫ − π π f ( x ) d x \int_{-\pi}^{\pi}f(x)dx = \int_{-\pi}^{\pi}a_0 dx=2\pi a_0\\ a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx ∫−ππf(x)dx=∫−ππa0dx=2πa0a0=π1∫−ππf(x)dx
确定 a n a_n an ∫ − π π f ( x ) cos m x d x = ∫ − π π ∑ ∞ n = 1 a n cos n x cos m x d x = π a n a n = 1 π ∫ − π π f ( x ) cos n x d x \int_{-\pi}^{\pi}f(x)\cos mxdx = \int_{-\pi}^{\pi}\sum_{\infty}^{n=1}a_n \cos nx \cos mx dx=\pi a_n\\ a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx dx ∫−ππf(x)cosmxdx=∫−ππ∞∑n=1ancosnxcosmxdx=πanan=π1∫−ππf(x)cosnxdx
确定b_n 同理可得 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 dx bn=π1∫−ππf(x)sinnxdx
f ( x ) = f ( x + 2 π ) T = 2 π f ( x ) = a 0 2 + ∑ n = 1 ∞ a n cos n x + ∑ n = 1 ∞ b n sin n x 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} f(x)&=f(x+2\pi) \quad T=2\pi \\ f(x)&=\frac{a_0}{2} + \sum_{n=1}^{\infty}a_n \cos nx+\sum_{n=1}^{\infty}b_n \sin nx\\ a_0&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx\\ a_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx dx\\ b_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin nx dx \end{aligned} f(x)f(x)a0anbn=f(x+2π)T=2π=2a0+n=1∑∞ancosnx+n=1∑∞bnsinnx=π1∫−ππf(x)dx=π1∫−ππf(x)cosnxdx=π1∫−ππf(x)sinnxdx
f ( t ) = f ( t + 2 L ) f(t)=f(t+2L) 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=Lπtt=πLx
t t t x x x002L2 π \pi π4L 4 π 4\pi 4π ⋮ \vdots ⋮ ⋮ \vdots ⋮用新的函数表示原函数 f ( t ) = f ( L π x ) = g ( x ) f(t)=f(\frac{L}{\pi}x)=g(x) f(t)=f(πLx)=g(x)
g ( x ) = a 0 2 + ∑ n = 1 ∞ a n cos n x + ∑ n = 1 ∞ b n sin n x 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} g(x)&=\frac{a_0}{2} + \sum_{n=1}^{\infty}a_n \cos nx+\sum_{n=1}^{\infty}b_n \sin nx\\ a_0&=\frac{1}{\pi}\int_{-\pi}^{\pi}g(x)dx\\ a_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}g(x)\cos nx dx\\ b_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}g(x)\sin nx dx \end{aligned} g(x)a0anbn=2a0+n=1∑∞ancosnx+n=1∑∞bnsinnx=π1∫−ππg(x)dx=π1∫−ππg(x)cosnxdx=π1∫−ππg(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 ) 1 π ∫ − π π d x = 1 π ∫ − L L d π L t = 1 L ∫ − L L d t x=\frac{\pi}{L}t\\ \cos nx=\cos \frac{n\pi}{L}t\\ \sin nx=\sin \frac{n\pi}{L}t\\ g(x)=f(t)\\ \frac{1}{\pi}\int_{-\pi}^{\pi}dx=\frac{1}{\pi}\int_{-L}^{L}d\frac{\pi}{L}t=\frac{1}{L}\int_{-L}^{L}dt x=Lπtcosnx=cosLnπtsinnx=sinLnπtg(x)=f(t)π1∫−ππdx=π1∫−LLdLπt=L1∫−LLdt
f ( t ) = a 0 2 + ∑ n = 1 ∞ a n cos n π L t + ∑ n = 1 ∞ 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 \begin{aligned} f(t)&=\frac{a_0}{2} + \sum_{n=1}^{\infty}a_n\cos \frac{n\pi}{L}t+\sum_{n=1}^{\infty}b_n \sin \frac{n\pi}{L}t\\ a_0&=\frac{1}{L}\int_{-L}^{L}f(t)dt\\ a_n&=\frac{1}{L}\int_{-L}^{L}f(t)\cos \frac{n\pi}{L}tdt\\ b_n&=\frac{1}{L}\int_{-L}^{L}f(t)\sin \frac{n\pi}{L}tdt \end{aligned} f(t)a0anbn=2a0+n=1∑∞ancosLnπt+n=1∑∞bnsinLnπt=L1∫−LLf(t)dt=L1∫−LLf(t)cosLnπtdt=L1∫−LLf(t)sinLnπtdt
工程中t从0开始，周期为T=2L， ω = π L = 2 π T \omega =\frac{\pi}{L}=\frac{2\pi}{T} ω=Lπ=T2π ∫ − L L d t → ∫ 0 2 L d t → ∫ 0 T d t \int_{-L}^{L}dt \to \int_{0}^{2L}dt \to \int_{0}^{T}dt ∫−LLdt→∫02Ldt→∫0Tdt
f ( t ) = a 0 2 + ∑ n = 1 ∞ a n cos n w t + ∑ n = 1 ∞ b n sin n w t a 0 = 2 T ∫ 0 T f ( t ) d t a n = 2 T ∫ 0 T f ( t ) cos n w t d t b n = 2 T ∫ 0 T f ( t ) sin n w t d t \begin{aligned} f(t)&=\frac{a_0}{2} + \sum_{n=1}^{\infty}a_n\cos nwt+\sum_{n=1}^{\infty}b_n \sin nwt\\ a_0&=\frac{2}{T}\int_{0}^{T}f(t)dt\\ a_n&=\frac{2}{T}\int_{0}^{T}f(t)\cos nwtdt\\ b_n&=\frac{2}{T}\int_{0}^{T}f(t)\sin nwtdt \end{aligned} f(t)a0anbn=2a0+n=1∑∞ancosnwt+n=1∑∞bnsinnwt=T2∫0Tf(t)dt=T2∫0Tf(t)cosnwtdt=T2∫0Tf(t)sinnwtdt
欧拉公式 e i θ = cos θ + i sin θ cos θ = 1 2 ( e i θ + e − i θ ) sin θ = − 1 2 i ( e i θ − e − i θ ) e^{i\theta}=\cos \theta + i \sin \theta\\ \cos \theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})\\ \sin \theta=-\frac{1}{2}i(e^{i\theta}-e^{-i\theta}) eiθ=cosθ+isinθcosθ=21(eiθ+e−iθ)sinθ=−21i(eiθ−e−iθ)
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 ] = a 0 2 + ∑ 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 = 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 = ∑ − ∞ ∞ C n e i n w t \begin{aligned} f(t)&=\frac{a_0}{2} + \sum_{n=1}^{\infty} \left [ a_n\frac{1}{2}(e^{inwt}+e^{-inwt})-b_n\frac{1}{2}(e^{inwt}+e^{-inwt}) \right ]\\ &=\frac{a_0}{2} + \sum_{n=1}^{\infty} \left [ \frac{a_n-ib_n}{2}e^{inwt}+\frac{a_n+ib_n}{2}e^{-inwt} \right ]\\ &=\frac{a_0}{2} + \sum_{n=1}^{\infty} \frac{a_n-ib_n}{2}e^{inwt}+\sum_{n=1}^{\infty} \frac{a_n+ib_n}{2}e^{-inwt}\\ &=\sum_{n=0}^{0}\frac{a_0}{2}e^{inwt} + \sum_{n=1}^{\infty} \frac{a_n-ib_n}{2}e^{inwt}+\sum_{n=-\infty}^{-1} \frac{a_n+ib_n}{2}e^{-inwt}\\ &=\sum_{-\infty}^{\infty}C_n e^{inwt} \end{aligned} f(t)=2a0+n=1∑∞[an21(einwt+e−inwt)−bn21(einwt+e−inwt)]=2a0+n=1∑∞[2an−ibneinwt+2an+ibne−inwt]=2a0+n=1∑∞2an−ibneinwt+n=1∑∞2an+ibne−inwt=n=0∑02a0einwt+n=1∑∞2an−ibneinwt+n=−∞∑−12an+ibne−inwt=−∞∑∞Cneinwt
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 , ⋯ C_n= \left \{ \begin{array}{l} \frac{a_0}{2} &,n=0\\ \frac{a_n-ib_n}{2} &,n=1,2,3,\cdots\\ \frac{a_n+ib_n}{2} &,n=-1,-2,-3,\cdots \end{array} \right . Cn=⎩⎨⎧2a02an−ibn2an+ibn,n=0,n=1,2,3,⋯,n=−1,−2,−3,⋯
n=0 C n = a 0 2 = 1 2 ⋅ 2 T ∫ 0 T f ( t ) d t = 1 T ∫ 0 T f ( t ) e 0 d t C_n=\frac{a_0}{2}=\frac{1}{2}\cdot \frac{2}{T}\int_{0}^{T}f(t)dt=\frac{1}{T}\int_{0}^{T}f(t)e^0 dt Cn=2a0=21⋅T2∫0Tf(t)dt=T1∫0Tf(t)e0dt
n=1,2,3, ⋯ \cdots ⋯ C n = 1 2 ( 2 T ∫ 0 T f ( t ) cos n w t d t − i 2 T ∫ 0 T f ( t ) 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 ) ( cos n w t + i sin ( − n w t ) ) d t = 1 T ∫ 0 T f ( t ) e − i n w t d t \begin{aligned} C_n &=\frac{1}{2}\left ( \frac{2}{T}\int_{0}^{T}f(t)\cos nwtdt - i \frac{2}{T}\int_{0}^{T}f(t)\sin nwtdt\right )\\ &=\frac{1}{T}\int_{0}^{T}f(t)(\cos nwt-i\sin nwt)dt \\ &=\frac{1}{T}\int_{0}^{T}f(t)(\cos nwt+i\sin (-nwt))dt \\ &=\frac{1}{T}\int_{0}^{T}f(t)e^{-inwt}dt \end{aligned} Cn=21(T2∫0Tf(t)cosnwtdt−iT2∫0Tf(t)sinnwtdt)=T1∫0Tf(t)(cosnwt−isinnwt)dt=T1∫0Tf(t)(cosnwt+isin(−nwt))dt=T1∫0Tf(t)e−inwtdt
n=-1,-2,-3, ⋯ \cdots ⋯ C n = 1 2 ( 2 T ∫ 0 T f ( t ) cos ( − n ) w t d t + i 1 T ∫ 0 T f ( t ) sin ( − n ) w t d t ) = 1 T ∫ 0 T f ( t ) e − i n w t d t \begin{aligned} C_n &=\frac{1}{2}\left ( \frac{2}{T}\int_{0}^{T}f(t)\cos (-n)wtdt + i \frac{1}{T}\int_{0}^{T}f(t)\sin (-n)wtdt\right )\\ &=\frac{1}{T}\int_{0}^{T}f(t)e^{-inwt}dt \end{aligned} Cn=21(T2∫0Tf(t)cos(−n)wtdt+iT1∫0Tf(t)sin(−n)wtdt)=T1∫0Tf(t)e−inwtdt
f ( t ) = f ( t + 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 \begin{aligned} &f(t)=f(t+T)\\ & f(t)=\sum_{-\infty}^{\infty}C_n e^{inwt}\\ & C_n=\frac{1}{T}\int_{0}^{T}f(t)e^{-inwt}dt \end{aligned} f(t)=f(t+T)f(t)=−∞∑∞CneinwtCn=T1∫0Tf(t)e−inwtdt
f ( t ) = f ( t + T ) f T ( t ) = ∑ − ∞ ∞ C n e i n w 0 t C n = 1 T ∫ T / 2 T / 2 f T ( t ) e − i n w 0 t d t w 0 = 2 π T 为 基 频 率 \begin{aligned} &f(t)=f(t+T)\\ &f_{T}(t)=\sum_{-\infty}^{\infty}C_n e^{inw_0t}\\ &C_n=\frac{1}{T}\int_{T/2}^{T/2}f_{T}(t)e^{-inw_0t}dt\\ &w_0=\frac{2\pi}{T}为基频率 \end{aligned} f(t)=f(t+T)fT(t)=−∞∑∞Cneinw0tCn=T1∫T/2T/2fT(t)e−inw0tdtw0=T2π为基频率
在工程中，信号一般都是非周期信号，即 T → ∞ T \to \infty T→∞ lim T → ∞ f T ( t ) = f ( t ) Δ w = ( n + 1 ) w 0 + n w 0 = w 0 = 2 π T 1 T = Δ 2 π \lim_{T\to \infty} f_T(t)=f(t)\\ \Delta w=(n+1)w_0+ nw_0=w_0=\frac{2\pi}{T}\\ \frac{1}{T}=\frac{\Delta}{2\pi} T→∞limfT(t)=f(t)Δw=(n+1)w0+nw0=w0=T2πT1=2πΔ 当 T → ∞ T \to \infty T→∞时 ， Δ w → 0 \Delta w \to 0 Δw→0，离散 → \to → 连续
则有 ∫ T / 2 T / 2 d t → ∫ − ∞ ∞ d t n w 0 → w ∑ n = − ∞ i n f t y Δ w → ∫ − ∞ ∞ \int_{T/2}^{T/2}dt\to \int_{-\infty}^{\infty}dt\\ nw_0 \to w\\ \sum_{n=-\infty}^{infty}\Delta w \to \int_{-\infty}^{\infty} ∫T/2T/2dt→∫−∞∞dtnw0→wn=−∞∑inftyΔw→∫−∞∞
f ( t ) = 1 2 π ∫ − ∞ ∞ ∫ − ∞ ∞ f ( t ) e − i w t d t e i w t d w f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(t)e^{-iwt}dt e^{iwt}dw f(t)=2π1∫−∞∞∫−∞∞f(t)e−iwtdteiwtdw 其中 F ( w ) = ∫ − ∞ ∞ f ( t ) e − i w t d t F(w)=\int_{-\infty}^{\infty}f(t)e^{-iwt}dt F(w)=∫−∞∞f(t)e−iwtdt为傅立叶变换， f ( t ) = 1 2 π ∫ − ∞ ∞ F ( w ) e i w t d w f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(w)e^{iwt}dw f(t)=2π1∫−∞∞F(w)eiwtdw为傅里叶逆变换。
