第4章 高阶统计量在随机信号建模中应用
4.1 引 言
·建模:根据采集的有限个含有噪声的观测信号找到一个合适的模型与观测值或其统计量相匹配,这就是随机信号建模。
·建模的关键:
(1) 模型的假设
(2) 参数的估计
·模型

·线性模型
(1)
参数模型:

其中

(4.1)
若

;
若

;
(2) 非参数模型

(4.2)
(3) 二者关系
·建模的目的:
由观测值

确定系统的频响


的幅度与相位,或

·用自相关函数建模:
系统输入必须为白噪声;系统为最小相位系统。
·用高阶累积量建模
输入可为非高斯有色或白色噪声; 系统可以是非最小相位系统
4.2 封闭型递推方法—

模型建模方法
(1)

是iid.过程,(非高斯白噪声)
(2)

是高斯噪声(有色或无色),方差

未知,且

与

互相独立。
(3)

4.2.1

与

的辨识
设

,则上式

, 又


, [

只能取0];

,[

只能取0];
设


(4.3)

(4.4)

(4.5)
4.2.2 封闭型递推方法
设

(归一化)
则

递推公式如下:

(4.6)

(4.7)
如果

为偶数,则

(4.8)

(4.9)
说明:
同理,

同理可证(4.7)式。
4.2.3 递推方法
(1)
根据观测值
(2)
计算
,

,

[
根据式(4.3)—(4.5)]
。
(3)
求
,

设

再由式
(4.6)及式(4.7),得
,

设

再根据式
(4.6)及式(4.7),得
,

,依此类推。
优点:简单。
缺点:有累计误差,且较严重。
4.3

公式—

模型建模方法二
设

系统单位冲击响应为

则

(4.10)
称此为

公式。
证明:

设


时,

上式变为

(4.11)
再设


(4.12)
(4.11)/ (4.12),得
同理:

(4.13)
还可得:

(4.14)

(4.15)
·

公式计算方法:
(1)
根据观测值

(2) 估计:

4.4

切片法
设

是

模型单位冲激响应(IR)
根据冲激响应定义:

(4.16)
(此公式今后经常见到)。
则

(4.17)
证明: 设

(4.18)
设

将上式代入(4.18)式,得

(4.19)
由(4.16)式得 (设:

)
所以(4.19)式成为:
设

,则

(4.20)
设

,则(4.20)式变为:
再设

,则(4.20)式变为:
则

讨论:
(1) 若为

模型,即

模型变为:

于是:此公式与

公式等价。
(2) 若是

模型,求

已知

和

。
4.5 线性系统中二阶与高阶统计量之间的关系
4.5.1

的双谱与

维谱之间的关系

一维对角线切片

(4.21)
证明:


变换:


复卷积定理
曲线积分
设

,

则

(4.22)

, 第2章得出的
由此得出
代入(4.22)式,原式得证。
4.5.2 功率谱与

维谱之间的关系

(4.23)

—

方程 (4.24)
证明:

前面提到过 (4.25)
而

(第2章讲过) (4.26)

(4.27)
将(4.27)式代入(4.25)中,得:
·推广:

·时域

方程:
设系统为

系统
则时域

方程为:

(4.28)
证明:

两边取

反变换:
根据复卷积定理
·推广

(4.29)
三、 三阶累计量之间的关系

第2章而来
对于

系统(过程)
设

,

,则

(4.30)
设

,则(4.30)式成为:

(4.31)
设


(4.32)

(4.33)
(4.32)/ (4.33),得:

域乘积
时域解:
时域卷积(此为三阶累积量之间的关系)
设

,

,则

(4.34)
―称此为累积量

方程
看如何转换:
设

,根据(4.34)式得:

取0,1,2,

表达成矩阵形式:

(4.35)

用最小二乘法,准确度高,取出的是

,符号没了。
建议:用

方法识别符号。

(4.36)
4.6 非最小相位

模型建模
4.6.1 模型假设
设

表示一非高斯信号过程,它可以用一个

模型来描述:

(4.37)
这里,如果

,则退化为

模型。
假设
(1) 阶次

已知;
(2) 输入过程

是一个不可观测的、零均值
i.i.d.非高斯过程,它至少存在一个有限的非零高阶累积量
,

,

。
(3) 系统是因果的、指数稳定的且可以是非最小相位的,即

的根在单位圆内,而

的根可以在单位圆内,也可以在单位圆外。
(4) 观测噪声

是一个零均值有色高斯过程,其能谱密度未知,且

与

相互独立。
这样,非最小相位

信号建模的实质就是怎样由观测值

的高阶累积量估计信号模型的

参数和

参数。
4.6.2

参数估计
1. 基于高阶累积量的高阶Yule-Walker方程
设系统

的单位冲激响应为

,则根据单位冲激响应的定义,有

(4.38)
又由第1章式(1.38)可知
综合上面两式,有
当

时,

,于是

(4.39)
方程式(4.39)称作基于高阶累积量的高阶Yule-Walker方程。
2. 参数的唯一可识别性定理
众所周知,一个非高斯过程的

模型给定的话,则其

阶累积量

就唯一确定。因此,从累积量匹配的观点出发,

参数

是由所有的

阶累积量对所有

求解式
(2)而唯一确定。由于方程组的个数是无穷大,这样的方程组的求解显然是不现实的。于是,必须选择少量合适的高阶累积量切片构造具有唯一解的方程组。
在式
(4.39)中,取
并固定

,且记


,得到下列由一个切片构造的方程组
(4.40)
或

(4.41)
若矩阵

为满矩,则由式
(4.40)可以得到唯一的
参数;然而,矩阵

不一定为满秩,这样,式
(4.40)不能保证
参数的唯一可识别性。
尽管单独某个切片构造的矩阵可能非满秩,
Giannakis and Mendel建议利用
个一维切片构造的矩阵则是满秩的,并提出了

参数的唯一可识别性定理。
定理1:
在假设(1)~(4)下,当且仅当模型式(4.37)不存在着零、极点相消时,
参数可由

个一维累积量切片构造的下列线性方程组唯一确定
(4.42)
其中,

。
令式
(4.39)中
,得到下列矩阵方程
(4.43)
3.

参数估计的
SVD-TLS方法
实际上,只能由观测值

得到高阶累积量的估计

,同时阶次

也未知,因此必须确定方程式
(4.43)的实用性问题。
1) 用奇异值分解
(SVD)法确定
阶次

设模型阶次

的上限为

(通常可以预先选定一组较大的值),取

;

,则矩阵
的秩为

,若用

代替

,则构造的新矩阵

的有效秩为

,这样,通过对

进行奇异值分解可以确定

的有效秩,从而确定模型

阶次

。
2) 用SVD-TLS法估计参数
式(4.43)是忽略噪声影响的一种近似。通常有两种方法来补偿这种噪声扰动的影响。第一种方法是最小二乘(LS)法,这种方法假定在式(4.43)右边的矢量中有一扰动项;第二种方法是主特征矢量(PE)法,这种方法则假定式(4.43)左边的系数矩阵中有一扰动项。显然,这两种方法都是不够的,因为式(4.43)中的系数矩阵和累积量矢量中元素均由观测值估计得到,因此都含有噪声扰动项。我们采用同时考虑这两种噪声扰动项的整体最小二乘(TLS)法来解决这一问题。由于TLS法通常采用SVD来实现,因而又称作SVD-TLS法。
SVD-TLS法步骤如下:
(1) 选择阶次上限

,取

,并
(2) 对

进行奇异值分解

(4.44)
其中

和

分别为

和

的特征向量矩阵,

为由奇异值

所构成的对角阵。
(3) 取

,计算

(4.45)
(4) 计算矩阵

(4.46)
其中,

表示由矩阵

中第

列向量

中的

个元素组成的

维向量,且
(5) 求解线性方程组

(4.47)
其中,

和

均为

维向量,且

,

,

的选取是以

中第一个元素为
1为原则,若
为奇异矩阵,则以

的零特征值对应的归一化特征向量当作

参数估计

。
4.6.3 MA参数估计
我们采用残差时间序列法来估计

参数。
设估计的

参数为

,则定义残差时间序列为

(4.48)
由式

,代入得

(4.49)
并考虑到式(4.37),有
如果

,则
由于假设
(4)可知
是高斯有色噪声,故

也是高斯有色噪声,于是残差时间序列可以看成纯非高斯

过程与高斯有色噪声的合成,因此参数

的估计转化成高斯有色噪声中的非高斯

过程的参数估计问题,我们将在下一节非最小相位

模型建模中一起讨论这个问题。
4.7 非高斯相位

模型建模
4.7.1 模型假设
设非最小相位信号模型为

模型,即
观测模型为
其中,动态噪声不可观测,并假定:
(1) 为零均值、独立地服从同一分布的非高斯白噪声;
(2) 为零均值、高斯分布噪声;
(3) 与相互独立,因而与也相互独立;
(4) 阶次已知且。
有关及更进一步的假设将在研究不同估计算法中给出。由于系统是非最小相位的,故传递函数
的零点可以在单位圆外。
我们感兴趣的问题是,怎样利用观测值的统计量信息来估计模型参数。
目前,基于高阶统计量解决上述问题的方法有两大类:非线性方法和线性方法。线性方法运算简单,且能保证估计值收敛于整体极值,同时线性方法的估计值通常可以当作非线性优化方法的估计初值;非线性方法则运算复杂,且估计值有时收敛于局部极值,因此线性方法的研究引人注目。至今,线性方法主要包括封闭型解和线性代数解。本文也是基于线性方法的基本思想着重研究了几种新的封闭型解和线性代数解。