统计动力学笔记（二）频谱密度与线性随机系统的动态准确性（自留用）

1. 频谱密度

频谱密度是对相关函数$R(t)$的傅里叶变换：
$$S(\omega )={\int }_{-\infty }^{\infty }R(\tau )e^{-j\omega \tau }\mathrm{d}\tau \text{\hspace{1em}(1)}$$显然，相关函数即为频谱密度的傅里叶逆变换：
$$R(\tau )=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }S(\omega )e^{j\omega \tau }\mathrm{d}\omega \text{\hspace{1em}(2)}$$替换$e^{-j\omega \tau }=\cos \omega \tau -j\sin \omega \tau$，并注意到奇函数$\sin \omega \tau$从$-\infty$到$+\infty$的积分为0，而偶函数$\cos \omega \tau$的积分对称，则式(1)可以化为：
$$S(\omega )={\int }_{-\infty }^{\infty }R(\tau )\cos \omega \tau \mathrm{d}\tau -j{\int }_{-\infty }^{\infty }R(\tau )\sin \omega \tau \mathrm{d}\tau =2{\int }_0^{\infty }R(\tau )\cos \omega \tau \mathrm{d}\tau \text{\hspace{1em}(3)}$$由此可见，频谱密度是偶函数：
$$\begin{gathered} S(-\omega )=2{\int }_0^{\infty }R(\tau )\cos (-\omega \tau )\mathrm{d}\tau =S(\omega ), \\ R(\tau )=\frac{1}{\pi }{\int }_0^{\infty }S(\omega )\cos \omega \tau \mathrm{d}\omega \text{\hspace{1em}(4)} \end{gathered}$$进而均方差为：
x 2 ‾ = D x = R x ( 0 ) = 1 2 π ∫ − ∞ ∞ S ( ω ) e j ω τ d ω ∣ τ = 0 = 1 π ∫ 0 ∞ S ( ω ) d ω (5) \overline{x^2} = D_x = R_x (0) = \frac{1}{2\pi} \int_{-\infty} ^\infty S(\omega) e^{ j \omega \tau} {\rm d} \omega \bigg\rvert_{\tau = 0} = \frac{1}{\pi} \int_{0} ^\infty S( \omega) {\rm d} \omega \tag{5} x2=Dx​=Rx​(0)=2π1​∫−∞∞​S(ω)ejωτdω ​τ=0​=π1​∫0∞​S(ω)dω(5)在物理意义上，$S(\omega )$表示了不同频率$\omega$下的功率分布，而其积分表示了平均功率。

复习一下$R(\tau )$的公式：
$$R(\tau )=\underset{T\to \infty }{\lim }\frac{1}{2T}{\int }_{-T}^{T}x(t)x(t+\tau )\mathrm{d}t$$代入定义式(1)：
$$S(\omega )={\int }_{-\infty }^{\infty }R(\tau )e^{-j\omega \tau }\mathrm{d}\tau =S(\omega )={\int }_{-\infty }^{\infty }{e}^{-j\omega \tau }\mathrm{d}\tau \underset{T\to \infty }{\lim }\frac{1}{2T}{\int }_{-T}^{T}x(t)x(t+\tau )\mathrm{d}t$$设$\theta =t+\tau$，则$\tau =\theta -t,\mathrm{d}\tau =\mathrm{d}\theta$，于是
$$\begin{array}{rl}S(\omega )& ={\int }_{-\infty }^{\infty }{e}^{-j\omega \tau }\mathrm{d}\tau \underset{T\to \infty }{\lim }\frac{1}{2T}{\int }_{-T}^{T}x(t)x(t+\tau )\mathrm{d}t\\ & ={\int }_{-\infty }^{\infty }{e}^{-j\omega \theta }{e}^{j\omega t}\mathrm{d}\theta \underset{T\to \infty }{\lim }\frac{1}{2T}{\int }_{-T}^{T}x(t)x(\theta )\mathrm{d}t\\ & ={\int }_{-\infty }^{\infty }x(\theta )e^{-j\omega \theta }\mathrm{d}\theta \underset{T\to \infty }{\lim }\frac{1}{2T}{\int }_{-T}^{T}x(t)e^{j\omega t}\mathrm{d}t\\ & =\underset{T\to \infty }{\lim }\frac{1}{2T}\left[{\int }_{-\infty }^{\infty }x(\theta )e^{-j\omega \theta }\mathrm{d}\theta {\int }_{-T}^{T}x(t)e^{j\omega t}\mathrm{d}t\right]\\ & =\underset{T\to \infty }{\lim }\frac{1}{2T}X(j\omega )X\left(-j\omega \right)\end{array}$$即：
$$S(\omega )=\underset{T\to \infty }{\lim }\frac{1}{2T}{\left|X(j\omega )\right|}^2\text{\hspace{1em}(6)}$$

例：设相关函数$R(\tau )=e^{-\alpha |\tau |}$，其中$\alpha >0$。则
S x ( ω ) = ∫ − ∞ ∞ e − α ∣ τ ∣ d τ = ∫ − ∞ 0 e α τ e − j ω τ d τ + ∫ 0 ∞ e − α τ e − j ω τ d τ = ∫ − ∞ 0 e ( α − j ω ) τ d τ + ∫ 0 ∞ e − ( α + j ω ) τ d τ = 1 α − j ω e α − j ω ∣ − ∞ 0 + − 1 α + j ω e α − j ω ∣ 0 ∞ = 1 α − j ω + 1 α + j ω = 2 α α 2 + ω 2 \begin{aligned} S_x (\omega) &= \int_{-\infty} ^\infty e^{- \alpha \lvert \tau \rvert} {\rm d} \tau \\ &= \int_{-\infty} ^0 e^{\alpha \tau } e^{-j \omega \tau } {\rm d} \tau + \int_{0} ^\infty e^{-\alpha \tau } e^{-j \omega \tau } {\rm d} \tau \\ &= \int_{-\infty} ^0 e^{ \left( \alpha - j \omega \right) \tau } {\rm d} \tau + \int_{0} ^\infty e^{- \left( \alpha + j \omega \right) \tau } {\rm d} \tau \\ &= \frac{1}{\alpha - j \omega} e^{ \alpha - j \omega} \bigg\rvert _{-\infty} ^0 + \frac{-1}{\alpha + j \omega} e^{ \alpha - j \omega} \bigg\rvert _{0} ^\infty \\ &= \frac{1}{\alpha - j \omega} + \frac{1}{\alpha + j \omega} \\ &= \frac{2\alpha}{\alpha^2 + \omega^2} \end{aligned} Sx​(ω)​=∫−∞∞​e−α∣τ∣dτ=∫−∞0​eατe−jωτdτ+∫0∞​e−ατe−jωτdτ=∫−∞0​e(α−jω)τdτ+∫0∞​e−(α+jω)τdτ=α−jω1​eα−jω ​−∞0​+α+jω−1​eα−jω ​0∞​=α−jω1​+α+jω1​=α2+ω22α​​

根据定义式(1)，很显然可以得到互频谱密度的表达式：
$$S_{xu}(\omega )={\int }_{-\infty }^{\infty }{R}_{xu}(\tau )e^{-j\omega \tau }\mathrm{d}\tau \text{\hspace{1em}(7)}$$及
$$R_{xu}(\tau )=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{S}_{xu}(\omega )e^{j\omega \tau }\mathrm{d}\omega \text{\hspace{1em}(8)}$$

2. 线性系统输出端的随机信号的频谱密度

在统计动力学笔记（一）动态系统随机信号在时域中的变换（自留用）一文的公式(3)中给出了输出端$x$的相关函数计算公式：
$$R_x(\tau )={\int }_{-\infty }^{\infty }K(\lambda )\mathrm{d}\lambda {\int }_{-\infty }^{\infty }{R}_u(\tau +\lambda -\eta )K(\eta )\mathrm{d}\eta$$则根据本文定义式(1)，输出端$x$的频谱密度为
$$\begin{array}{rl}{S}_x(\omega )& ={\int }_{-\infty }^{\infty }{e}^{-j\omega \tau }{R}_x(\tau )\mathrm{d}\tau \\ & ={\int }_{-\infty }^{\infty }{e}^{-j\omega \tau }\mathrm{d}\tau {\int }_{-\infty }^{\infty }K(\lambda )\mathrm{d}\lambda {\int }_{-\infty }^{\infty }{R}_u(\tau +\lambda -\eta )K(\eta )\mathrm{d}\eta \end{array}$$令$\xi =\tau +\lambda -\eta$，则$\mathrm{d}\tau =\mathrm{d}\xi$，上式为
S x ( ω ) = ∫ − ∞ ∞ e − j ω τ d τ ∫ − ∞ ∞ K ( λ ) d λ ∫ − ∞ ∞ R u ( τ + λ − η ) K ( η ) d η = ∫ − ∞ ∞ e − j ω ξ e − j ω η e j ω λ d ξ ∫ − ∞ ∞ K ( λ ) d λ ∫ − ∞ ∞ R u ( ξ ) K ( η ) d η = ∫ − ∞ ∞ e − j ω η K ( η ) d η ⏟ W ( j ω ) ∫ − ∞ ∞ e j ω λ K ( λ ) d λ ⏟ W ( − j ω ) ∫ − ∞ ∞ e − j ω ξ R u ( ξ ) d ξ ⏟ S u ( ω ) = W ( j ω ) W ( − j ω ) S u ( ω ) = ∣ W ( j ω ) ∣ 2 S u ( ω ) \begin{aligned} S_x(\omega) &= \int_{-\infty} ^\infty e^{- j \omega \tau} {\rm d} \tau \int_{-\infty} ^\infty K(\lambda) {\rm d} \lambda \int_{-\infty} ^\infty R_u (\tau + \lambda - \eta) K(\eta) {\rm d} \eta \\ &= \int_{-\infty} ^\infty e^{- j \omega \xi} e^{- j \omega \eta} e^{ j \omega \lambda} {\rm d} \xi \int_{-\infty} ^\infty K(\lambda) {\rm d} \lambda \int_{-\infty} ^\infty R_u (\xi) K(\eta) {\rm d} \eta \\ &= \underbrace{ \int_{-\infty} ^\infty e^{- j \omega \eta} K(\eta) {\rm d} \eta}_{W(j \omega)} \underbrace{ \int_{-\infty} ^\infty e^{ j \omega \lambda} K(\lambda) {\rm d} \lambda }_{W(-j \omega)} \underbrace{\int_{-\infty} ^\infty e^{- j \omega \xi} R_u (\xi) {\rm d} \xi}_{S_u (\omega)} \\ &= W(j \omega) W(-j \omega) S_u (\omega) \\ &= \big\lvert W ( j\omega) \big\rvert^2 S_u (\omega) \end{aligned} Sx​(ω)​=∫−∞∞​e−jωτdτ∫−∞∞​K(λ)dλ∫−∞∞​Ru​(τ+λ−η)K(η)dη=∫−∞∞​e−jωξe−jωηejωλdξ∫−∞∞​K(λ)dλ∫−∞∞​Ru​(ξ)K(η)dη=W(jω) ∫−∞∞​e−jωηK(η)dη​​W(−jω) ∫−∞∞​ejωλK(λ)dλ​​Su​(ω) ∫−∞∞​e−jωξRu​(ξ)dξ​​=W(jω)W(−jω)Su​(ω)= ​W(jω) ​2Su​(ω)​即
S x ( ω ) = ∣ W ( j ω ) ∣ 2 S u ( ω ) (9) S_x(\omega) = \big\lvert W ( j\omega) \big\rvert^2 S_u (\omega) \tag{9} Sx​(ω)= ​W(jω) ​2Su​(ω)(9)同理可以推出互频谱密度
$$S_{xu}(\omega )=W(j\omega )S_u(\omega )\text{\hspace{1em}(10)}$$
显然，对于强度$N^2=1$的白噪声来说，其$S_u(\omega )=1$，则 S x u ( ω ) = W ( j ω ) , S x ( ω ) = ∣ W ( j ω ) ∣ 2 S_{xu}(\omega) = W(j \omega), S_x(\omega) = \big\lvert W ( j\omega) \big\rvert^2 Sxu​(ω)=W(jω),Sx​(ω)= ​W(jω) ​2，由此可见，可以在输入端加载强度为1的白噪声，通过测量输出端的频谱密度$S_x(\omega )$，即可获得系统的传递函数$W(j\omega )$，具有很强的应用价值。