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首页第五版《自适应滤波原理》英文课后习题答案

第五版《自适应滤波原理》英文课后习题答案

自适应滤波器

课后答案

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《自适应滤波器原理（第五版）》英文版课后答案解析，由Simon Haykin和Kelvin Hall两位作者共同创作，他们是加拿大麦克马斯特大学的学者。本书是针对赫金教授的经典教材，专门探讨自适应滤波器的理论与实践。这本Solution Manual涵盖了第五版的所有章节，旨在帮助学生理解和解决书中习题，深入理解自适应滤波器的设计、学习算法如Least Mean Squares (LMS)和Levenberg-Marquardt Backpropagation (LMBP)等。书中特别提到了对复杂数据处理的Wirtiger Calculus的支持，这是由马里兰大学的Tulai Adali教授引入的，它在处理部分微分问题时具有重要作用。Ashique Rupam Mahomood博士也贡献了大量计算机实验，尤其是在第十三章中应用AutoStep方法及其解决方案，以及该章末尾的相关问题，他的帮助使读者能够通过实际操作加深理解。 Erkan Baser的研究生项目也被收录在附录中，该项目回顾了第六章关于LMS算法的自适应均衡实验，但这次使用的是IDB算法。这部分不仅展示了理论在实际项目中的应用，也提供了实际问题求解的实例，有助于学生掌握如何在实际环境中设计和优化自适应滤波器。这本Solution Manual不仅是学习者解决作业难题的宝贵工具，还包含了作者们对自适应滤波技术不断探索和发展的最新成果，对于那些希望在信号处理、通信工程或机器学习领域深入研究的读者来说，是一份不可多得的学习资料。无论是理论概念的理解还是实践经验的积累，都能从中获益匪浅。

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PROBLEM 1.9. CHAPTER 1.

Equivalently, except for the scaling factor r(0),

, a

, . . . , a

}  {r(1), r(2), . . . , r(M)} (2)

Combining Equation (1) and Equation (2):

, a

, . . . , a

}  {r(0), r(1), r(2), . . . , r(M)} (3)

Problem 1.9

The transfer function of the MA model of Fig. 1.3 is

H(z) = 1 + b

∗

−1

+ b

∗

−2

+ . . . + b

∗

−K

The transfer function of the ARMA model of Fig. 1.4 is

H(z) =

+ b

∗

−1

+ b

∗

−2

+ . . . + b

∗

−K

1 + a

∗

−1

+ a

∗

−2

+ . . . + a

∗

−M

The ARMA model reduces to an AR model when

= b

= . . . = b

= 0

The ARMA model reduces to MA model when

= a

= . . . = a

= 0

PROBLEM 1.10. CHAPTER 1.

Problem 1.10

∗

Taking the z-transform of both sides of the correct equation:

X(z) = (1 + 0.75z

−1

+ 0.25z

−2

)V (z)

Hence, the transfer function of the MA model is:

X(z)

V (z)

=1 + 0.75z

−1

+ 0.75z

−1

(1 + 0.75z

−1

+ 0.75z

−1

)

−1

(1)

Using long division we may perform the following expansion of the denominator in Equa-

tion (1):

(1 + 0.75z

−1

+ 0.75z

−1

)

−1

= 1 −

−1

−2

−

−3

−

256

−4

−

1024

−5

−

4096

−6

16283

−7

−

65536

−8

−

627

262144

−9

1541

1048576

−10

+ . . .

≈ 1 − 0.75z

−1

+ 0.3125z

−2

− 0.0469z

−3

− 0.043z

−4

− 0.0439z

−5

− 0.0222z

−6

+ 0.0057z

−7

− 0.0013z

−8

− 0.0024z

−9

+ 0.0015z

−10

(2)

M = 2

Retaining terms in Equation (2) up to z

−2

, we may approximate the MA model with an

AR model of order two as follows:

X(z)

V (z)

≈

1 − 0.75z

−1

+ 0.3125z

−2

∗

Correction: the question was meant to ask the reader to consider an MA process x(n) of order two

described by the difference equation

x(n) = ν(n) + 0.75ν(n − 1) + 0.25ν(n − 2)

not the equation

x(n) = ν(n) + 0.75ν(n − 1) + 0.75ν(n − 2)

PROBLEM 1.11. CHAPTER 1.

M = 5

Retaining terms in Equation (2) up to z

−5

, we may approximate the MA model with an

AR model of order two as follows:

X(z)

V (z)

≈

1 − 0.75z

−1

+ 0.3125z

−2

− 0.0469z

−3

− 0.043z

−4

+ 0.0439z

−5

M = 10

Retaining terms in Equation (2) up to z

−10

, we may approximate the MA model with an

AR model of order two as follows:

X(z)

V (z)

≈

D(z)

where D(z) is given by the polynomial on the right-hand side of Equation (2).

Problem 1.11

The ﬁlter output is

x(n) = w

u(n)

where u(n) is the tap-input vector. The average power of the ﬁlter output is therefore

E[|x(n)|

] = E[w

u(n)u

(n)w]

= w

E[u(n)u

(n)]w

= w

If u(n) is extracted from a zero-mean white noise with variance σ

, then

R = σ

where I is the identity matrix. Hence,

E[|x(n)|

] = σ

PROBLEM 1.12. CHAPTER 1.

Problem 1.12

The process u(n) is a linear combination of Gaussian samples. Hence, u(n) is Gaussian.

From inverse ﬁltering, we recognize that ν(n) may also be expressed as a linear combina-

tion of samples relating to u(n). Hence, if u(n) is Gaussian, then ν(n) is also Gaussian.

Problem 1.13

From the Gaussian moment factoring theorem:

E[(u

∗

)

] =E[u

∗

. . . u

∗

. . . u

]

=k!E[u

∗

] . . . E[u

∗

]

=k!(E[u

∗

])

(1)

By allowing u

= u

= u, Equation (1) reduces to:

E[|u|

] = k!(E[|u|

])

Problem 1.14

It is not permissible to interchange the order of expectation and limiting operation in Equa-

tion (1.113). The reason is that the expectation is a linear operation, whereas the limiting

operation with respect to the number of samples N is nonlinear.

Problem 1.15

The ﬁlter output is

y(n) =

h(i)u(n − i)

PROBLEM 1.16. CHAPTER 1.

Similarly, we may write

y(m) =

h(k)u(m − k)

Hence,

(n, m) = E[y(n)y

∗

(m)]

= E

h(i)u(n − i)

∗

(k)u

∗

(m − k)

h(i)h

∗

(k)E [u(n − i)u

∗

(m − k)]

h(i)h

∗

(k)r

(n − i, m − k)

Problem 1.16

The mean-square value of the ﬁlter output response to white noise input is

2σ

∆ω

The value P

is linearly proportional to the ﬁlter bandwidth ∆ω. This relation holds irre-

spective of how small ∆ω is compared to the mid-band frequency of the ﬁlter.

Problem 1.17

The variance of the ﬁlter output is

2σ

∆ω

It has been stated that

= 0.1 volts

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