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Visual Studio 圈复杂度评估

秀妮_5519 2024-11-19 阅读 28

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(t)+bnsin(t)]=T2t0t0+Tf(t)cos(t)dt=T2t0t0+Tf(t)sin(t)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(t+ϕn)anbn=A0+n=1Ansin(t+ϕn)=Ansinϕncost+Ancosϕnsint=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(kn)x]sinkxsinnx=21[cos(kn)xcos(k+n)x]coskxcosnx=21[cos(k+n)x+cos(kn)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(t)+bnsin(t)]

两边求积分:
∫ − π π f ( t ) d t = ∫ − π π A 0 d t + ∑ n = 1 ∞ a n ∫ − π π cos ⁡ ( n ω t ) d t + b n ∫ − π π s i n ( n ω t ) d t   = 2 π A 0 A 0 = ∫ − π π f ( t ) d t 2 π \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=1anππcos(t)dt+bnππsin(t)dt=2πA0=2πππf(t)dt
计算 a n : a_n: an:
f ( t ) cos ⁡ n ω t = ∑ n = 1 ∞ a n ∫ − T 2 T 2 cos ⁡ k ω t cos ⁡ n ω t d t + b n ∫ − T 2 T 2 cos ⁡ k ω t sin ⁡ n ω t d t = ∑ n = 1 ∞ a n ∫ − T 2 T 2 cos ⁡ k ω t cos ⁡ n ω t d t = ∑ n = 1 ∞ a n ∫ − T 2 T 2 cos ⁡ k ω t cos ⁡ n ω t d t = a n ∫ − T 2 T 2 cos ⁡ 2 ( n ω t ) d t ( n = k ) = a n 2 ( ∫ − T 2 T 2 1 d t + ∫ − T 2 T 2 cos ⁡ ( 2 n ω t ) ) d t = a n 2 × T = T a n 2 = π a n \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)cost=n=1an2T2Tcostcostdt+bn2T2Tcostsintdt=n=1an2T2Tcostcostdt=an2T2Tcos2(t)dt(n=k)=2an(2T2T1dt+2T2Tcos(2t))dt=2an×T=2Tan=πan=n=1an2T2Tcostcostdt
因此 a n = 2 T ∫ − T 2 T 2 cos ⁡ ( n ω t ) f ( t ) d t = 1 π ∫ − π π cos ⁡ ( n ω t ) f ( t ) d t 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=T22T2Tcos(t)f(t)dt=π1ππcos(t)f(t)dt;\
同理可得 b n = 2 T ∫ − T 2 T 2 sin ⁡ ( n ω t ) f ( t ) d t = 1 π ∫ − π π sin ⁡ ( n ω t ) f ( t ) d t 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=T22T2Tsin(t)f(t)dt=π1ππsin(t)f(t)dt。\
又有 A 0 = 2 T ∫ − T 2 T 2 f ( t ) d t = ∫ − π π f ( t ) d t π A_0 = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t)dt = \frac{ \int_{-\pi}^{\pi}f(t)dt}{\pi} A0=T22T2Tf(t)dt=πππf(t)dt。\
将求得的项代入得到: f ( t ) = a 0 2 + ∑ n = 1 ∞ a n cos ⁡ ( n ω t ) + b n sin ⁡ ( n ω t ) 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=1ancos(t)+bnsin(t)

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+isinθeiθ=cos(θ)+isin(θ)=cosθisinθcosθ=2eiθ+eiθsinθ=2eiθeiθ

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:
f ( t ) = A 0 + ∑ n = 1 ∞ [ a n cos ⁡ ( n ω t ) + b n sin ⁡ ( n ω t ) ] = A 0 + ∑ i = 1 ∞ a n e n i ω t + e − n i ω t 2 + b n e n i ω t − e − n i ω t 2 = A 0 + ∑ i = 1 ∞ [ a n − i b n 2 e n i ω t + a n + i b n 2 e − n i ω t ] \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(t)+bnsin(t)]=A0+i=1an2eniωt+eniωt+bn2eniωteniωt=A0+i=1[2anibneniωt+2an+ibneniωt]
a n = 2 T ∫ − T 2 T 2 f ( t ) cos ⁡ ( n ω t ) d t a_n = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t) \cos (n \omega t) dt an=T22T2Tf(t)cos(t)dt\ , b n = 2 T ∫ − T 2 T 2 f ( t ) sin ⁡ ( n ω t ) d t b_n = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t) \sin (n \omega t) dt bn=T22T2Tf(t)sin(t)dt可得:\
a n − i b n 2 = 1 T ∫ − T 2 T 2 f ( t ) [ cos ⁡ ( n ω t ) − i sin ⁡ ( n ω t ) ] d t = 1 T ∫ − T 2 T 2 f ( t ) e − n i ω t d t \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 2anibn=T12T2Tf(t)[cos(t)isin(t)]dt=T12T2Tf(t)eniωtdt\
a n + i b n 2 = 1 T ∫ − T 2 T 2 f ( t ) [ cos ⁡ ( n ω t ) + i sin ⁡ ( n ω t ) ] d t = 1 T ∫ − T 2 T 2 f ( t ) e n i ω t d t \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=T12T2Tf(t)[cos(t)+isin(t)]dt=T12T2Tf(t)eniωtdt\
再将其代入到 f ( t ) f(t) f(t)的表达式中可得: \
f ( t ) = A 0 + 1 T ∑ n = 1 ∞ ∫ − T 2 T 2 f ( t ) e − n i ω t d t   e n i ω t + 1 T ∑ n = 1 ∞ ∫ − T 2 T 2 f ( t ) e n i ω t d t   e − n i ω t 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=12T2Tf(t)eniωtdt eniωt+T1n=12T2Tf(t)eniωtdt eniωt\
将上式中第三项的 n n n换为 − n -n n可得:\
f ( t ) = A 0 + 1 T ∑ n = 1 ∞ ∫ − T 2 T 2 f ( t ) e − n i ω t d t   e n i ω t + 1 T ∑ n = − ∞ − 1 ∫ − T 2 T 2 f ( t ) e − n i ω t d t   e n i ω t 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=12T2Tf(t)eniωtdt eniωt+T1n=12T2Tf(t)eniωtdt eniωt\
又有 A 0 = 1 2 2 T ∫ − T 2 T 2 f ( t ) d t = 1 T ∫ − T 2 T 2 f ( t ) e − 0   i ω t d t   e 0 i ω t 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=21T22T2Tf(t)dt=T12T2Tf(t)e0 tdt e0t\
综上可得 f ( t ) = 1 T ∑ n = − ∞ ∞ ∫ − T 2 T 2 f ( t ) e − n i ω t d t   e n i ω t 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)eniωtdt 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 → ∞ T \to \infty T就可以得到非周期的了。\
T → ∞ , Δ ω → d W , ω 0 = 2 π T , Δ W = ω 0 = ( n + 1 ) ω 0 − n ω 0 , n ω 0 → W ; ∑ n = − ∞ ∞ Δ W = ∫ − ∞ ∞ d W 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,ΔωdW,ω0=T2π,ΔW=ω0=(n+1)ω0nω0,nω0W;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 ( t ) = ∑ n = − ∞ ∞ ( Δ W 2 π ∫ − ∞ ∞ f ( t ) e − i W t )   e i W t = 1 2 π ∑ n = − ∞ ∞ ∫ − ∞ ∞ f ( t ) e − i W t d t   e i W t Δ W = 1 2 π ∫ − ∞ ∞ F ( W ) e i W t d W \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=(2πΔWf(t)eiWt) eiWt=2π1n=f(t)eiWtdt eiWtΔW=2π1F(W)eiWtdW
其中 F ( W ) = ∫ − ∞ ∞ f ( t ) e − i W t d t F(W) = \int_{-\infty}^{\infty} f(t) e^{-i Wt}dt F(W)=f(t)eiWtdt称为傅立叶变换。

3. 参考

leinlin
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