基础维纳滤波

模型

输入输出为平稳随机过程，求最优权向量使得输出估计结果的均方误差最小
$$\begin{array}{rl}& \hat{d}(n)=w^{H}u(n)\\ & min\{J(w)\\ & \{\begin{array}{rl}& e(n)=d(n)-\hat{d}(n)\\ & J(w)==E\{|e(n){|}^2\}\end{array}\end{array}$$

方法

最优权向量：
Wiener-Hopf方程：
R w o = p R = E { u ( n ) u H ( n ) } = [ r ( 0 ) r ( 1 ) ⋯ r ( M − 1 ) r ( − 1 ) r ( 0 ) ⋯ r ( M − 2 ) ⋮ ⋯ ⋱ ⋯ r ( − M + 1 ) ⋯ ⋯ r ( 0 ) ] p = E { u ( n ) d ∗ ( n ) } = [ E { u ( n ) d ∗ ( n ) } E { u ( n − 1 ) d ∗ ( n ) } ⋮ E { u ( n − M + 1 ) d ∗ ( n ) } ] = [ p ( 0 ) p ( − 1 ) ⋮ p ( − M + 1 ) ] \begin {align} &Rw_o=p\\ &R=E\{u(n)u^H(n)\}=\begin{bmatrix} r(0) & r(1) & \cdots & r(M-1) \\r(-1)&r(0) & \cdots & r(M-2) \\ \vdots &\cdots &\ddots &\cdots \\ r(-M+1)&\cdots&\cdots&r(0)\end{bmatrix}\\ &p=E\{u(n)d^*(n)\}=\begin{bmatrix} E\{u(n)d^*(n)\} \\ E\{u(n-1)d^*(n)\}\\ \vdots\\ E\{u(n-M+1)d^*(n)\} \end{bmatrix} =\begin{bmatrix} p(0)\\ p(-1)\\ \vdots\\ p(-M+1) \end{bmatrix} \end {align} ​Rwo​=pR=E{u(n)uH(n)}= ​r(0)r(−1)⋮r(−M+1)​r(1)r(0)⋯⋯​⋯⋯⋱⋯​r(M−1)r(M−2)⋯r(0)​ ​p=E{u(n)d∗(n)}= ​E{u(n)d∗(n)}E{u(n−1)d∗(n)}⋮E{u(n−M+1)d∗(n)}​ ​= ​p(0)p(−1)⋮p(−M+1)​ ​​​

算法

1.最陡下降算法

(1)核心

$$\Delta w=-\frac{1}{2}\mu \Delta J(w(n))$$

(2)计算

$$\begin{array}{r}\{\begin{array}{rl}& w(n+1)=w(n)+\mu [p-Rw(n)]\\ & 0<\mu <\frac{2}{{\lambda }_{max}}\end{array}\end{array}$$

(3)性能

当$0<\mu <\frac{2}{{\lambda }_{max}}$时，最陡下降算法收敛：
$$\underset{n->\infty }{\lim }w(n)=w_o$$

2.LMS算法

(1)核心

$$\begin{array}{rl}& \hat{R}=\frac{1}{N}\sum _{i=1}^{N}u(i)u^H(i)\\ & \hat{p}=\frac{1}{N}\sum _{i=1}^{N}u(i)d^{\ast }(i)\end{array}$$

(2)计算

$$\hat{w}(n+1)=\hat{w}(n)+\mu u(n)e^{\ast }(n)$$

(3)性能

滤波器权向量1阶收敛，但均方误差大于最小均方误差

1阶收敛

当$0<\mu <\frac{2}{{\lambda }_{max}}$时，最陡下降算法收敛：
$$\underset{n->\infty }{\lim }E\{\hat{w}(n)\}=w_o$$

均方误差的稳态性能

步长因子满足
$$0<\mu <\frac{2}{\sum _{i=1}^M{\lambda }_i}=\frac{2}{Mr(0)}$$

a.剩余均方误差$J_{ex}$

$$\begin{array}{rl}& J_{ex}(n)=E\hat{J}(n)-J_{min}\\ & J_{ex}(\infty )\approx \mu J_{min}\frac{\sum _{i=1}^M{\lambda }_i}{2-\mu \sum _{i=1}^M{\lambda }_i}\end{array}$$

b.失调参数$M$

$$\begin{array}{rl}& M=\frac{J_{ex}(\infty )}{J_{min}}\\ & M=\frac{\mu \sum _{i=1}^M{\lambda }_i}{2-\mu \sum _{i=1}^M{\lambda }_i}\approx \frac{\mu }{2}\sum _{i=1}^M{\lambda }_i=\frac{\mu }{2}Mr(0)=\frac{\mu }{2}M{\lambda }_{av}=\frac{M}{4{\tau }_{av}}\end{array}$$

c.平均时间常数${\tau }_{av}$

$${\tau }_{av}=\frac{1}{2\mu {\lambda }_{av}}{\lambda }_{av}=\frac{1}{M}\sum _{i=1}^M{\lambda }_i$$

d.特征值拓展/特征值比$\chi$

$$\chi (R)=\frac{{\lambda }_{max}}{{\lambda }_{min}}$$