文章目录
1. 傅立叶级数
1.1 定义
f ( t ) = a 0 2 + ∑ n = 1 ∞ [ a n cos ( n ω t ) + b n sin ( n ω t ) ] a n = 2 T ∫ t 0 t 0 + T f ( t ) cos ( n ω t ) d t b n = 2 T ∫ t 0 t 0 + T f ( t ) sin ( n ω t ) d t \begin{align*} f(t) &= \frac{a_0}{2} + \sum_{n=1}^{\infty}[a_n\cos(n \omega t) + b_n \sin (n \omega t)]\\ a_n &= \frac{2}{T} \int_{t_0 }^{t_0 + T} f(t) \cos (n \omega t) dt \\ b_n &= \frac{2}{T} \int_{t_0}^{t_0 + T} f(t) \sin (n \omega t) dt \\ \end{align*} f(t)anbn=2a0+n=1∑∞[ancos(nωt)+bnsin(nωt)]=T2∫t0t0+Tf(t)cos(nωt)dt=T2∫t0t0+Tf(t)sin(nωt)dt
1.2 推导
1.2.1 三角函数逼近
f ( t ) = A 0 + ∑ n = 1 ∞ A n sin ( n ω t + ϕ n ) A n sin ( n ω t + ϕ n ) = A n sin ϕ n cos n ω t + A n cos ϕ n sin n ω t a n = A n sin ϕ n b n = A n cos ϕ n \begin{align*} f(t) &= A_0 + \sum_{n=1}^{\infty} A_n \sin(n \omega t + \phi _n)\\ A_n \sin(n \omega t + \phi_n) &= A_n \sin \phi_n \cos n\omega t + A_n \cos \phi_n \sin n \omega t\\ a_n &= A_n \sin \phi_n \\ b_n &= A_n \cos \phi_n \end{align*} f(t)Ansin(nωt+ϕn)anbn=A0+n=1∑∞Ansin(nωt+ϕn)=Ansinϕncosnωt+Ancosϕnsinnωt=Ansinϕn=Ancosϕn
1.2.2 多项式展开
f ( x ) = A + B x + C x 2 + D x 3 + ⋯ f ′ ( x ) = B + 2 C x + 3 D x 2 + ⋯ f ′ ′ ( x ) = 2 C + 6 D x + ⋯ A = f ( 0 ) , B = f ′ ( 0 ) , C = f ′ ′ ( 0 ) 2 , D = f ′ ′ ′ ( 0 ) 6 N = f ( n ) ( 0 ) n ! \begin{align*} f(x) &= A + Bx + Cx^2 + Dx^3 + \cdots \\ f'(x) &= B+2Cx + 3Dx^2 + \cdots \\ f''(x) &= 2C + 6Dx + \cdots\\ A & = f(0), B = f'(0), C = \frac{f''(0)}{2}, D = \frac{f'''(0)}{6}\\ N &= \frac{f^{(n)}(0)}{n!} \end{align*} f(x)f′(x)f′′(x)AN=A+Bx+Cx2+Dx3+⋯=B+2Cx+3Dx2+⋯=2C+6Dx+⋯=f(0),B=f′(0),C=2f′′(0),D=6f′′′(0)=n!f(n)(0)
1.2.3 三角函数的正交性
∫ − π π cos ( n x ) d x = 0 ( n = 1 , 2 , 3 , ⋯ ) ∫ − π π sin ( n x ) d x = 0 ( n = 1 , 2 , 3 , ⋯ ) ∫ − π π sin ( k x ) cos ( n x ) d x = 0 ( k , n = 1 , 2 , 3 , ⋯ , k ≠ n ) ∫ − π π cos ( k x ) cos ( n x ) d x = 0 ( k , n = 1 , 2 , 3 , ⋯ , k ≠ n ) ∫ − π π sin ( k x ) sin ( n x ) d x = 0 ( k , n = 1 , 2 , 3 , ⋯ , k ≠ n ) \begin{align*} \int_{-\pi}^{\pi} \cos (nx) dx = 0 \quad (n = 1,2,3,\cdots) \\ \int_{-\pi}^{\pi} \sin (nx) dx = 0 \quad (n = 1,2,3, \cdots) \\ \int_{-\pi}^{\pi} \sin (kx) \cos (nx) dx = 0 \quad (k,n = 1,2,3,\cdots, k \ne n) \\ \int_{-\pi}^{\pi} \cos (kx) \cos (nx) dx = 0 \quad (k,n = 1,2,3, \cdots, k \ne n) \\ \int_{-\pi}^{\pi} \sin (kx) \sin (nx) dx = 0 \quad (k,n = 1,2,3, \cdots, k \ne n)\\ \end{align*} ∫−ππcos(nx)dx=0(n=1,2,3,⋯)∫−ππsin(nx)dx=0(n=1,2,3,⋯)∫−ππsin(kx)cos(nx)dx=0(k,n=1,2,3,⋯,k=n)∫−ππcos(kx)cos(nx)dx=0(k,n=1,2,3,⋯,k=n)∫−ππsin(kx)sin(nx)dx=0(k,n=1,2,3,⋯,k=n)
1.2.4 积化和差
sin k x cos n x = 1 2 [ sin ( k + n ) x + sin ( k − n ) x ] sin k x sin n x = 1 2 [ cos ( k − n ) x − cos ( k + n ) x ] cos k x cos n x = 1 2 [ cos ( k + n ) x + cos ( k − n ) x ] \begin{align*} \sin kx \cos nx = \frac{1}{2} [\sin(k + n)x + \sin(k -n)x] \\ \sin kx \sin nx = \frac{1}{2}[\cos(k-n)x - \cos(k+n)x]\\ \cos kx \cos nx = \frac{1}{2}[ \cos(k+n)x + \cos(k-n)x] \end{align*} sinkxcosnx=21[sin(k+n)x+sin(k−n)x]sinkxsinnx=21[cos(k−n)x−cos(k+n)x]coskxcosnx=21[cos(k+n)x+cos(k−n)x]
1.2.5 实数下的傅立叶级数
f ( t ) = A 0 + ∑ n = 1 ∞ [ a n cos ( n ω t ) + b n sin ( n ω t ) ] f(t) = A_0 + \sum_{n=1}^{\infty}[a_n \cos(n \omega t) + b_n \sin(n \omega t)] f(t)=A0+∑n=1∞[ancos(nωt)+bnsin(nωt)]
两边求积分:
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\begin{align*} \int_{-\pi}^{\pi} f(t) dt &= \int_{-\pi}{\pi} A_0 dt + \sum_{n=1}^{\infty} a_n \int_{-\pi}^{\pi} \cos (n \omega t) dt + b_n \int_{-\pi}^{\pi}sin (n \omega t) dt\\ \ &= 2 \pi A_0 \\ A_0 &= \frac{ \int_{-\pi}^{\pi}f(t)dt}{2\pi} \end{align*}
∫−ππf(t)dt A0=∫−ππA0dt+n=1∑∞an∫−ππcos(nωt)dt+bn∫−ππsin(nωt)dt=2πA0=2π∫−ππf(t)dt
计算
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a_n:
an:
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\begin{align*} f(t)\cos n\omega t &= \sum_{n=1}^{\infty} a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} \cos k \omega t \cos n \omega t dt + b_n \int_{-\frac{T}{2}}^{\frac{T}{2}} \cos k\omega t \sin n \omega t dt &= \sum_{n=1}^{\infty} a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} \cos k \omega t \cos n \omega t dt \\ &= \sum_{n=1}^{\infty} a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} \cos k\omega t \cos n \omega t dt \\ &= a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} \cos^2 (n\omega t) dt \quad (n = k) \\ &= \frac{a_n}{2}( \int_{-\frac{T}{2}}^{\frac{T}{2}} 1 dt + \int_{-\frac{T}{2}}^{\frac{T}{2}} \cos(2n\omega t))dt \\ &= \frac{a_n}{2} \times T \\ &= \frac{Ta_n}{2}\\ &= \pi a_n \end{align*}
f(t)cosnωt=n=1∑∞an∫−2T2Tcoskωtcosnωtdt+bn∫−2T2Tcoskωtsinnωtdt=n=1∑∞an∫−2T2Tcoskωtcosnωtdt=an∫−2T2Tcos2(nωt)dt(n=k)=2an(∫−2T2T1dt+∫−2T2Tcos(2nωt))dt=2an×T=2Tan=πan=n=1∑∞an∫−2T2Tcoskωtcosnωtdt
因此
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cos
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f
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a_n = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} \cos ( n \omega t) f(t) dt = \frac{1}{\pi} \int_{-\pi}^{\pi} \cos (n \omega t) f(t) dt
an=T2∫−2T2Tcos(nωt)f(t)dt=π1∫−ππcos(nωt)f(t)dt;\
同理可得
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b_n = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} \sin ( n \omega t) f(t) dt = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin (n \omega t) f(t) dt
bn=T2∫−2T2Tsin(nωt)f(t)dt=π1∫−ππsin(nωt)f(t)dt。\
又有
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A_0 = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t)dt = \frac{ \int_{-\pi}^{\pi}f(t)dt}{\pi}
A0=T2∫−2T2Tf(t)dt=π∫−ππf(t)dt。\
将求得的项代入得到:
f
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2
+
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1
∞
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cos
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+
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sin
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n
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f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty}a_n \cos (n \omega t) + b_n \sin (n \omega t )
f(t)=2a0+∑n=1∞ancos(nωt)+bnsin(nωt)
至此,傅立叶级数实数上推导完成。
1.3 复数形式上的傅立叶级数
1.3.1 欧拉公式
e i θ = cos θ + i sin θ e i − θ = cos ( − θ ) + i sin ( − θ ) = cos θ − i sin θ cos θ = e i θ + e − i θ 2 sin θ = e i θ − e − i θ 2 \begin{align*} e^{i \theta} = \cos \theta + i \sin \theta \\ e^{i -\theta} = \cos(-\theta) + i \sin( -\theta) = \cos \theta - i \sin \theta \\ \cos \theta = \frac {e^{i \theta} + e^{-i\theta}}{2} \\ \sin \theta = \frac{e^{i \theta} - e^{-i\theta}}{2} \\ \end{align*} eiθ=cosθ+isinθei−θ=cos(−θ)+isin(−θ)=cosθ−isinθcosθ=2eiθ+e−iθsinθ=2eiθ−e−iθ
1.3.2 复数上的傅立叶级数
代入到傅立叶级数的实数形式:
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\begin{align*} f(t) &= A_0 + \sum_{n=1}^{\infty}[a_n \cos (n \omega t) + b_n \sin (n \omega t)]\\ &= A_0 + \sum_{i=1}^{\infty} a_n \frac{e^{ni\omega t} + e^{-ni\omega t}}{2} + b_n \frac{e^{ni\omega t} - e^{-ni\omega t}}{2}\\ &=A_0 + \sum_{i=1}^{\infty} [\frac{a_n - ib_n}{2} e^{ni\omega t} +\frac{a_n + ib_n}{2}e^{-ni\omega t}] \end{align*}
f(t)=A0+n=1∑∞[ancos(nωt)+bnsin(nωt)]=A0+i=1∑∞an2eniωt+e−niωt+bn2eniωt−e−niωt=A0+i=1∑∞[2an−ibneniωt+2an+ibne−niωt]
由
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a_n = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t) \cos (n \omega t) dt
an=T2∫−2T2Tf(t)cos(nωt)dt\ ,
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b_n = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t) \sin (n \omega t) dt
bn=T2∫−2T2Tf(t)sin(nωt)dt可得:\
a
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[
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\frac{a_n - ib_n}{2} = \frac{1}{T} \int_{ -\frac{T}{2}}^{\frac{T}{2}} f(t)[\cos (n\omega t) - i \sin (n \omega t)]dt =\frac{1}{T} \int_{ -\frac{T}{2}}^{\frac{T}{2}}f(t) e^{-ni\omega t}dt
2an−ibn=T1∫−2T2Tf(t)[cos(nωt)−isin(nωt)]dt=T1∫−2T2Tf(t)e−niωtdt\
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\frac{a_n + ib_n}{2} = \frac{1}{T} \int_{ -\frac{T}{2}}^{\frac{T}{2}} f(t)[\cos (n\omega t) + i \sin (n \omega t)]dt =\frac{1}{T} \int_{ -\frac{T}{2}}^{\frac{T}{2}}f(t) e^{ni\omega t}dt
2an+ibn=T1∫−2T2Tf(t)[cos(nωt)+isin(nωt)]dt=T1∫−2T2Tf(t)eniωtdt\
再将其代入到
f
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f(t)
f(t)的表达式中可得: \
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f(t)= A_0 + \frac{1}{T} \sum\limits_{n=1}^{\infty} \int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-ni\omega t}dt\ e^{ni \omega t} + \frac{1}{T} \sum\limits_{n=1}^{\infty} \int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{ni\omega t}dt\ e^{-ni \omega t}
f(t)=A0+T1n=1∑∞∫−2T2Tf(t)e−niωtdt eniωt+T1n=1∑∞∫−2T2Tf(t)eniωtdt e−niωt\
将上式中第三项的
n
n
n换为
−
n
-n
−n可得:\
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f(t)= A_0 + \frac{1}{T} \sum\limits_{n=1}^{\infty} \int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-ni\omega t}dt\ e^{ni \omega t} + \frac{1}{T} \sum\limits_{n=-\infty}^{-1} \int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-ni\omega t}dt\ e^{ni \omega t}
f(t)=A0+T1n=1∑∞∫−2T2Tf(t)e−niωtdt eniωt+T1n=−∞∑−1∫−2T2Tf(t)e−niωtdt eniωt\
又有
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A_0 = \frac{1}{2} \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t)dt = \frac{1}{T} \int_{ -\frac{T}{2}}^{\frac{T}{2}} f(t) e^{-0\ i\omega t}dt\ e^{0i\omega t}
A0=21T2∫−2T2Tf(t)dt=T1∫−2T2Tf(t)e−0 iωtdt e0iωt\
综上可得
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f(t) = \frac{1}{T} \sum\limits_{n=-\infty}^{\infty} \int_{ -\frac{T}{2}}^{\frac{T}{2}} f(t)e^{-ni\omega t} dt\ e^{ni\omega t}
f(t)=T1n=−∞∑∞∫−2T2Tf(t)e−niωtdt eniωt
2. 傅立叶变换
上面的傅立叶级数是周期为
T
T
T的形式的,我们让
T
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∞
T \to \infty
T→∞就可以得到非周期的了。\
T
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T \to \infty, \Delta \omega \to dW, \omega_0 = \frac{2 \pi}{T},\\ \Delta W= \omega_0 = (n+1)\omega_0 - n \omega_0, n\omega_0 \to W;\\ \sum\limits_{n=-\infty}^{\infty} \Delta W = \int_{-\infty}^{\infty}dW
T→∞,Δω→dW,ω0=T2π,ΔW=ω0=(n+1)ω0−nω0,nω0→W;n=−∞∑∞ΔW=∫−∞∞dW。
代入到上面的周期性傅立叶变换函数中去,可以得到:
f
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∞
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2
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∞
F
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\begin{align*} f(t) &= \sum_{n=-\infty}^{\infty}(\frac{\Delta W}{2\pi} \int_{-\infty}^{\infty}f(t)e^{-iWt})\ e^{iWt} \\ &= \frac{1}{2\pi} \sum_{n=-\infty}^{\infty} \int_{-\infty}^{\infty} f(t) e^{-iWt}dt\ e^{iWt} \Delta W\\ &=\frac{1}{2\pi} \int_{-\infty}^{\infty} F(W)e^{iWt}dW \end{align*}
f(t)=n=−∞∑∞(2πΔW∫−∞∞f(t)e−iWt) eiWt=2π1n=−∞∑∞∫−∞∞f(t)e−iWtdt eiWtΔW=2π1∫−∞∞F(W)eiWtdW
其中
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F(W) = \int_{-\infty}^{\infty} f(t) e^{-i Wt}dt
F(W)=∫−∞∞f(t)e−iWtdt称为傅立叶变换。
3. 参考
leinlin
kaizhao