Optimum Signal Processing: An Introduction

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"Optimum Signal Processing - An Introduction, Second Edition by Sophocles J. Orfanidis" 《最优信号处理》是Sophocles J. Orfanidis教授撰写的一本教材，适用于深入学习信号处理的读者。这本书的第二版于1988年由McGraw-Hill Publishing Company出版，并在1996年版权回归作者。书中详细介绍了信号处理领域的最优方法，对于理解如何在实际应用中优化信号处理技术具有重要价值。该书涵盖了阵列处理这一关键领域，阵列处理涉及到多个传感器或天线同时收集信号的情况，通过分析这些信号来提升检测、定位和通信系统的性能。在阵列处理中，最优信号处理方法可以帮助我们设计出能最大限度地提高信噪比、减小干扰或实现精确源定位的算法。书中可能包含了以下核心知识点： 1. **最优滤波理论**：讲解了如何设计滤波器以达到最佳性能，如最小均方误差（Minimum Mean Square Error, MMSE）滤波器和卡尔曼滤波器。 2. **阵列信号处理基础**：包括阵列配置、波达方向估计（Direction of Arrival, DOA）和空间谱估计等，这些都是阵列处理的关键概念。 3. **阵列处理算法**：如MUSIC（Multiple Signal Classification）算法和ESPRIT（Estimation of Signal Parameters via Rotational Invariance Techniques）算法，用于高效估计信号的方向。 4. **自适应信号处理**：讨论了自适应滤波器和LMS（Least Mean Squares）算法，它们能够根据输入信号的变化自动调整滤波器参数。 5. **优化理论与方法**：介绍了在信号处理中应用的优化问题，如凸优化和非凸优化，以及相关的求解策略。 6. **噪声和干扰抑制**：讲解了如何在噪声环境中提取有用信号，以及如何对抗干扰源。 7. **实际系统应用**：可能包含雷达、声纳、无线通信和遥感等领域的实例，展示最优信号处理在实际系统中的应用。此外，作者提供了与本书配套的网页（http://www.ece.rutgers.edu/~orfanidi/osp2e），可能包含额外的材料、习题解答和MATLAB代码示例，便于读者进行实践操作和深入学习。此书适合对信号处理有高级需求的工程师、科研人员和学生，它不仅提供理论知识，还强调了实用性和计算方法。通过阅读《最优信号处理》，读者可以掌握信号处理的前沿技术，并有能力解决实际工程问题。

2 1. Random Signals

p(x)=

√

2πσ

exp



−(x −m)

/2σ



Example 1.1.2: Uniform distribution

p(x)=

⎧

⎨

⎩

1/Q , for −Q/2 ≤ x ≤ Q/2

, otherwise

Its variance is σ

= Q

/12. 

Both the gaussian and the uniform distributions will prove to be important examples.

In typical signal processing problems of designing ﬁlters to remove or separate noise

from signal, it is often assumed that the noise interference is gaussian. This assumption

is justiﬁed on the grounds of the central limit theorem, provided that the noise arises

from many different noise sources acting independently of each other.

The uniform distribution is also important. In digital signal processing applications.

the quantization error arising from the signal quantization in the A/D converters, or the

roundoff error arising from the ﬁnite accuracy of the internal arithmetic operations in

digital ﬁlters, can often be assumed to be uniformly distributed.

Every computer provides system routines for the generation of random numbers.

For example, the routines RANDU and GAUSS of the IBM Scientiﬁc Subroutine Package

generate uniformly distributed random numbers over the interval

[0, 1], and gaussian-

distributed numbers, respectively. GAUSS calls RANDU twelve times, thus generating

twelve independent uniformly distributed random numbers

,...,x

. Then, their

sum

x = x

+···+x

, will be approximately gaussian, as guaranteed by the central

limit theorem. It is interesting to note that the variance of

x is unity, as it follows from

the fact that the variance of each

,is1/12:

= σ

+ σ

+···+σ

+···+

= 1

The mean of

x is 12/2 = 6. By shifting and scaling x, one can obtain a gaussian-

distributed random number of any desired mean and variance. Appendix A contains a

discussion of random number generators and methods to improve them, such as shuf-

ﬂing or using several generators in combination. A number of FORTRAN and C routines

for generating uniform and gaussian random numbers are included.

1.2 Chebyshev’s Inequality

The variance σ

of a random variable x is a measure of the spread of the x-values

about their mean. This intuitive interpretation of the variance is a direct consequence

of Chebyshev’s inequality, which states that the

x-values tend to cluster about their

mean in the sense that the probability of a value not occurring in the near vicinity of the

mean is small; and it is smaller the smaller the variance.

1.2. Chebyshev’s Inequality 3

More precisely, for any probability density

p(x) and any Δ>0, the probability that

x will fall outside the interval of values [m −Δ, m +Δ] is bounded by σ

/Δ

. Thus, for

ﬁxed

Δ, as the variance σ

becomes smaller, the x-values tend to cluster more narrowly

about the mean. In the extreme limiting case of a deterministic variable

x = m, the

density becomes inﬁnitely narrow,

p(x)= δ(x − m), and has zero variance.

Prob



|x − m|≥Δ



≤

(Chebyshev’s Inequality)

Chebyshev’s inequality is especially important in proving asymptotic convergence

results for sample estimates of parameters. For example, consider

N independent sam-

ples

,...,x

} drawn from a gaussian probability distribution of mean m and

variance

. The sample estimate of the mean is

m =

+ x

+···+x

) (1.2.1)

Being a sum of

N gaussian random variables,

m will itself be a gaussian random

variable. Its probability density is completely determined by the corresponding mean

and variance. These are found as follows.

m]=



E[x

]+E[x

]+···+E[x

]



(m + m +···+m)= m

Therefore,

is an unbiased estimator of m. However, the goodness of

as an es-

timator must be judged by how small its variance is—the smaller the better, by Cheby-

shev’s inequality. By the assumption of independence, we have

var

(

m)= E



(

m − m)





+ σ

+···+σ





Nσ



(1.2.2)

Thus,

m is also a consistent estimator of m in the sense that its variance tends to

zero as the number of samples

N increases. The values of

m will tend to cluster more

and more closely about the true value of

m as N becomes larger. Chebyshev’s inequality

implies that the probability of

m falling outside any ﬁxed neighborhood of m will tend

to zero for large

N. Equivalently,

will converge to m with probability one. This can

also be seen from the probability density of

m, which is the gaussian

m)=

1/2

(2π)

1/2

exp



−

2σ

(

m − m)



In the limit of large N, this density tends to the inﬁnitely narrow delta function

density

m)= δ(

m − m). In addition to the sample mean, we may also compute

sample estimates of the variance



i=1

−

(1.2.3)

It is easily shown [2,3] that this estimator is slightly biased. But for large

N,itis

asymptotically unbiased and consistent as can be seen from its mean and variance:

N −

, var(

N − 1

2σ

(1.2.4)

An unbiased and consistent estimator of

is the standard deviation deﬁned by

4 1. Random Signals

N − 1



i=1

−

(1.2.5)

It has

E[s

]= σ

and var(s

)= 2σ

/(N − 1) . In addition to the requirements of

asymptotic unbiasedness and consistency, a good estimator of a parameter must also be

judged in terms of its efﬁciency [2,3], which determines how closely the estimator meets

its Cram

er-Rao bound. This is discussed in Section 1.17. We will see there that the es-

timators (1.2.1) and (1.2.3)—being maximum likelihood estimators—are asymptotically

efﬁcient.

1.3 Joint and Conditional Densities, and Bayes’ Rule

Next, we discuss random vectors. A pair of two different random variables x = (x

)

may be thought of as a vector-valued random variable. Its statistical description is more

complicated than that of a single variable and requires knowledge of the joint probability

density

p(x

). The two random variables may or may not have any dependence on

each other. It is possible, for example, that if

assumes a particular value, then this

fact may inﬂuence, or restrict, the possible values that

can then assume.

A quantity that provides a measure for the degree of dependence of the two variables

on each other is the conditional density

p(x

) of x

given x

; and p(x

) of x

given

. These are related by Bayes’ rule

p(x

)= p(x

)p(x

)= p(x

)p(x

)

More generally, Bayes’ rule for two events A and B is

p(A, B)= p(A|B)p(B)= p(B|A)p(A)

The two random variables x

and x

are independent of each other if they do not

condition each other in any way, that is, if

p(x

)= p(x

) or p(x

)= p(x

)

In other words, the occurrence of x

does not in any way inﬂuence the variable x

When two random variables are independent, their joint density factors into the product

of single (marginal) densities:

p(x

)= p(x

)p(x

)

The converse is also true. The correlation between x

and x

is deﬁned by the expectation

value

E[x



p(x

)dx

When x

and x

are independent, the correlation also factors as E[x

]= E[x

]E[x

Example 1.3.1:

Suppose x

is related to x

= 5x

+ v

where v is a zero-mean, unit-variance, gaussian random variable assumed to be indepen-

dent of

. Determine the conditional density and conditional mean of x

given x

1.3. Joint and Conditional Densities, and Bayes’ Rule 5

Solution:

The randomness of x

arises both from the randomness of x

and the randomness of

v. But if x

takes on a particular value, then the randomness of x

will arise only from v.

Identifying elemental probabilities we have

p(x

)dx

= p(v)dv = (2π)

−1/2

exp



−



But, dx

= dv and v = x

− 5x

. Therefore,

p(x

)= (2π)

−1/2

exp



−

− 5x

)



The conditional mean is the mean of x

with respect to the density p(x

). It is evident

from the above gaussian expression that the conditional mean is

E[x

]= 5x

. This can

also be found directly as follows.

E[x

]= E



(5x

+ v)|x



+ E[v|x

]= 5x

where we used the independence of v and x

to replace the conditional mean of v with its

unconditional mean, which was given to be zero, that is,

E[v|x

]= E[v]= 0. 

The concept of a random vector generalizes to any dimension. A vector of N random

variables

⎡

⎢

⎣

⎤

⎥

⎦

requires knowledge of the joint density

p(x)= p(x

,...,x

)

(1.3.1)

for its complete statistical description. The second-order statistics of x are its mean, its

correlation matrix, and its covariance matrix, deﬁned by

= E[x], R= E[xx

], Σ= E



(x − m)(x − m)



(1.3.2)

where the superscript

T denotes transposition, and the expectation operations are de-

ﬁned in terms of the joint density (1.3.1); for example,

E[x]=



x p(x)d

where

x = dx

···dx

denotes the corresponding N-dimensional volume ele-

ment. The

ijth matrix element of the correlation matrix R is the correlation between

the

ith random variable x

with the jth random variable x

, that is, R

= E[x

] .Itis

easily shown that the covariance and correlation matrices are related by

Σ = R − mm

When the mean is zero, R and Σ coincide. Both R and Σ are symmetric positive semi-

deﬁnite matrices.

Example 1.3.2:

The probability density of a gaussian random vector x = [x

,...,x

]

completely speciﬁed by its mean m and covariance matrix

Σ, that is,

p(x)=

(2π)

N/2

(det Σ)

1/2

exp



−

(x −m)

−1

(x −m)



6 1. Random Signals

Example 1.3.3:

Under a linear transformation, a gaussian random vector remains gaussian. Let

x be a gaussian random vector of dimension

N, mean m

, and covariance Σ

. Show that

the linearly transformed vector

ξ = B x where B is a nonsingular N×N matrix

is gaussian-distributed with mean and covariance given by

= B m

,Σ

= BΣ

(1.3.3)

The relationships (1.3.3) are valid also for non-gaussian random vectors. They are easily

derived as follows:

E[ξ

ξ]= E[B x]= BE[x], E[ξ

ξξ

]= E[B x(B x)

]= BE[xx

The probability density p

(ξ

ξ) is related to the density p

(x) by the requirement that,

under the above change of variables, they both yield the same elemental probabilities:

(ξ

ξ) d

ξ = p

(x)d

x (1.3.4)

Since the Jacobian of the transformation from x to

ξ is d

ξ =|det B|d

x, we obtain

(ξ

ξ)= p

)/|det B|. Noting the invariance of the quadratic form

(ξ

ξ −m

)

−1

(ξ

ξ −m

) = (x − m

)

(BΣ

)

−1

B(x −m

)

= (

− m

)

−1

(x −m

)

and that det

= det(BΣ

)= (det B)

det Σ

, we obtain

(ξ

ξ)=

(2π)

N/2

(det

)

1/2

exp



−

(ξ

ξ −m

)

−1

(ξ

ξ −m

)



Example 1.3.4: Consider two zero-mean random vectors x and y of dimensions N and M, re-

spectively. Show that if they are uncorrelated and jointly gaussian, then they are also

independent of each other. That x and y are jointly gaussian means that the

(N+M)-

dimensional joint vector z





is zero-mean and gaussian, that is,

p(z)=

(2π)

(N+M)/2

(det R

)

1/2

exp



−

−1



where the correlation (covariance) matrix R

= E





, y

]





]E[xy

]

]E[yy

]







If x and y are uncorrelated, that is,

= E[xy

]= 0, then the matrix R

becomes block

diagonal and the quadratic form of the joint vector becomes the sum of the individual

quadratic forms:

−1

z = [x

, y

]



−1





−1

x +y

−1

Since R

= 0 also implies that det R

= (det R

)(det R

), it follows that the joint

density

p(z)= p(x, y) factors into the marginal densities:

p(x, y)= p(x)p(y)

which shows the independence of x and y.

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