数字信号处理-基于计算机的方法(第四版)第九章答案

发布时间：2024-09-01

数字信号处理基于计算机方法第四版第九章答案

SOLUTIONS MANUAL

to accompany

Digital Signal Processing: A Computer-Based

Approach

Fourth Edition

Sanjit K. Mitra

Prepared by

Chowdary Adsumilli, John Berger, Marco Carli,

Hsin-Han Ho, Rajeev Gandhi, Martin Gawecki, Chin Kaye Koh, Luca Lucchese, Mylene Queiroz de Farias, and Travis Smith

Copyright © 2011 by Sanjit K. Mitra. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of Sanjit K. Mitra, including, but not limited to, in any network or other electronic Storage or transmission, or broadcast for distance learning.

数字信号处理基于计算机方法第四版第九章答案

Chapter 9

9.1 We obtain the solutions by using Eq. (9.3) and Eq. (9.4).

(a) δp=1 10

αp/20

=1 10 0.24/20=0.0273,δs=10 αs/20=10 49/20=0.0035.

=1 10 0.14/20=0.016,δs=10 αs/20=10 68/20=0.000398. (b) δp=1 10p

9.2 We obtain the solutions by using Eqs. (9.3) and (9.4). (a)αp= 20log101 δp= 20log10(1 0.04)=0.3546dB,

α/20

αs= 20log10(δs)= 20log10(0.08)=21.9382dB. (b)αp= 20log101 δp= 20log10(1 0.015)=0.1313dB,

()

αs= 20log10(δs)= 20log10(0.04)=27.9588dB.

G(ejω)=H2(ejω)=H(ejω).

jω

()

9.3 G(z)=H2(z), or equivalently,

Let δpand δs denote the passband and stopband ripples of H(ejω), respectively. Also, let

δp,2=2δp, and δs,2 denote the passband and stopband ripples of G(eThen δp,2=1 (1 δp)2, and δs,2=(δs)2. For a cascade of

δp,M=1 (1 δp)M,and δs,M=(δs).

), respectively.

sections,

–(πpsω

Therefore, the passband edge and the stopband edge of the highpass filter are given by ωp,HP=π ωp, and ωs,HP=π ωs, respectively.

9.5 Note that G(z) is a complex bandpass filter with a passband in the range 0≤ω≤π. Its passband edges are at ωp,BP=ωo±ωp,and stopband edges at ωs,BP=ωo±ωs. A real

coefficient bandpass transfer function can be generated according to

GBP(z)=HLP(ejωoz)+HLP(e–jωoz) which will have a passband in the range 0≤ω≤π

数字信号处理基于计算机方法第四版第九章答案

and another passband in the range –π≤ω≤0. However because of the overlap of the two spectra a simple formula for the bandedges cannot be derived.

∞

osωs

opop

9.6 (a) hp(t)=ha(t) p(t) where p(t)=

∑δ(t nT). Thus, hp(t)=∑ha(nT)δ(t nT)..

n= ∞

∞

n= ∞

We also have, g[n]=ha(nT). Now, Ha(s)=

∞

Hp(s)=

∞

∫hp(t)e

dt=

n= ∞ ∞

∑∫ha(nT)δ(t nT)e

∞

∞∞

∞

∫ha(t)e stdt and

dt=

∑ha(nT)e snT.

n= ∞

∞n= ∞

∞

Comparing the above expression with G(z)=conclude that G(z)=Hp(s)

∞

∑g[n]z n=∑h(nT)z n, we

n= ∞

We can also show that a Fourier series expansion of p(t) is given by

. 1lnzT

1p(t)=

∑e j(2πkt/T). Therefore,

∞

1 j(2πkt/T) h(t)=h(t)e. Hence, ∑e j(2πkt/T) ∑aa T k= ∞k= ∞

1hp(t)= T

k= ∞ ∞

Hp(s)=

∑

k= ∞∞

∞

2πkt Ha s+j . As a result, we have T

(7-1)

G(z)=

2πkt Hs+jlnz. ∑a

T 1

k= ∞

s=T

(b) The transformation from the s-plane to z-plane is given by z=e

jωσoTjΩoT

. If we express

s=σo+jΩo, then we can write z=re

=ee. Therefore,

数字信号处理基于计算机方法第四版第九章答案

<1,for σo<1, z= =1,for σo=1, Or in other words, a point in the left-half -plane is mapped onto

>1,for σo>1.

a point inside the unit circle in the z-plane, a point in the right-half -plane is mapped onto a point outside the unit circle in the z-plane, and a point on the jω-axis in the s-plane is mapped onto a point on the unit circle in the z-plane. As a result, the mapping has the desirable properties enumerated in Section 9.1.3.

2πk

T are mapped onto a single point in the z-plane as z

mapping is illustrated in the figure below

€

2πk j Ωo± T σoT T=ee

=eσoTejΩoT. The

Rez

s-planez-plane

Note that the strip of width 2π/T in the s-plane for values of s in the range ππ

≤Ω≤ is mapped into the entire z-plane, and so are the adjacent strips of width TT

2π/T. The mapping is many-to-one with infinite number of such strips of width 2π/T. It follows from the above figure and also from Eq. (7-1) that if the frequency response

π1ω

Ha(jΩ)=0 for Ω≥, then G(ejω)=Ha(j) for ≤π, and there is no aliasing.

TTT(d) For z=ejω=ejΩT, or equivalently, ω=ΩT.

9.7 Assume ha(t) is causal. Now, ha(t)=

g[n]=ha(nT)=

Ha(s)estds. Hence,

Ha(s)esnTds. Therefore,

数字信号处理基于计算机方法第四版第九章答案

∞

G(z)=

n=0

∑g[n]z n=∑Ha(s)esnTz nds=Ha(s)∑z nesnTds= H(s)

Residues ∑ . sT 1 1 ez allpolesofHa(s)

n=0

H(s)1 e

Hence G(z)=

9.8 Ha(s)=

. The transfer function has a pole at s= α. Now s+α

AAA

G(z)=Residue==. ats=–α (s+α)(1 ez) 1 ezs=–α1 ez

10.5 0.5j0.5+0.5j

++ s+3(s+2 j)(s+2+j)(s+3)(s+4s+5)

1s+3 1s+21=+=++. s+3(s+2)+12s+3(s+2)+12(s+2)+12

Using Eq (9.71), we get

11 z 1e 2Tcos(T)z 1e 2Tsin(T)

Ga(z)=++.1 ez1 2zecos(T)+ez1 2zecos(T)+ezSince T = 0.25, we get

11 0.4376z 1

Ga(z)=+.. 1 1 2

1 0.4724z1 1.1754z+0.3679z

2s2+s 11.50.25+0.75j0.25 0.75j

=++(b) Hb(s)= s+4(s+1 3j)(s+1+3j)(s+4)(s+2s+10)

1.5s 8 1s+13=+0.5=+0.5 0.53. ()222s+4s+3(s+1)+3(s+1)+3(s+1)+3

2(s+2)

9.9 (a) Ha(s)=

Using Eq (9.71), we get

1.51 z 1e Tcos(3T)z 1e Tsin(3T)

Gb(z)=+0.5 1.5.1 ez1 2zecos(3T)+ez1 2zecos(3T)+ezSince T = 0.25, we get

Gb(z)=

1.51 0.3679z

0.51 2.1624z 11 1.1397z

+0.6065z

s2+2s+11

(s+2s+5)(s1.5+j1.5 j=+(s+1 2j)(s

+1+2j)

数字信号处理基于计算机方法第四版第九章答案

s 1/3

(s+1)+4

s+1

2+3 2/3

3 ()22(s+1)+2(s+1)+2

(

Using Eq (9.71), we get

1 z 1e Tcos(2T)z 1e Tsin(2T)

. Gc(z)=3 21 2zecos(2T)+ez1 2zecos(2T)+ez

Since T = 0.25, we get

Gc(z)=

.. 1 1.3669z+0.6065z1 1.5622z+0.7788z

2zz e

+AzAz

=+. 12z ez ez e5z

31 0.9324z 1

31 0.4629z 1

9.10 (a) Ga(z)=

Since T = 0.5, α1=2.6, α2=4, A1=2, A2=5, it follows Ha(s)=

ze 1.4sin(1.6)

ze βTsin(λT)

+. s+2.6s+4

(b) Gb(z)==. z 2zecos(1.6)+ez 2zecos(λT)+e

3.2

Since T = 0.5, λ=3.2, β=2.6, it follows Hb(s)=. 2

(s+2.6)+3.2

9.11 (a)Ha(s)=Ga(z)z=4 1+s =

1 s

(b)Hb(s)=Gb(z)

1+s =z=4

1 s

45s2+18s+9

75s+154s+91

105s3+385s2+467s+195

(13s+11)27s+46s+23

9.12 For the impulse invariance design:

ωp=ΩpT=2πFpT=2π0.88×1030.25×10 3=0.44π.

For the bilinear transformation method:

1ΩpT 1 13 3

ωp=2tan =2tanFpTπ=2tan0.88×10 0.25×10 π=0.385π.

()()

()

9.13 For the impulse invariance method: 2πFp=

ωp

0.45π0.4×10

Fp=562.5 Hz.

For the bilinear transformation method:

ωp 1 0.45π 1

Fp=tan =tan =679.7Hz.

2 π0.4×10 3 2 πT

数字信号处理基于计算机方法第四版第九章答案

9.14 The passband and the stopband edges of the analog lowpass filter are assumed to

Ωp=0.25π and Ωs=0.55π. The requirements to be satisfied by the analog lowpass filter are thus 20log10Ha(j0.25π)≥ 0.5 dB and 20log10Ha(j0.55π)≤ 15 dB.

From αp=20log10

=0.5 we obtain ε2=0.1220184543. From

αs=10log10(A2)=15 we obtain A. From Eq. (A.6), the inverse

1==15.841979 and from Eq. (A.5) the inverse

k11Ω

transition ratio is given by ==2.2. Substituting these values in Eq. (A.9) we obtain

kΩp

log(1/k)log(15.841979) N===3.503885. We choose N = 4.

log10(1/k)log10(2.2)

2N Ωp

From Eq. (A.7) we have =ε2. Substituting the values of Ωp, N, and ε2 we get

Ωc

Ωc=1.3007568(Ωp)=1.021612.

Using the statement [z,p,k] = buttap(4) we get the poles of the 4-th order Butterworth analog filter with a 3-dB cutoff at 1 rad/s as p1= 0.3827+j0.9239, p2= 0.3827 j0.9239, p3= 0.9239+j0.3827, and p4= 0.9239 j0.3827. Therefore,

Han(s)=

=..

(s p1)(s p2)(s p3)(s p4)(s+0.7654s+1)(s+1.8478s+1)

Next we expand Han(s) in a partial-fraction expansion using the M-file residue and

0.9238729s 0.70713230.9238729s+1.7071323

+. We next s+0.7654s+1s+1.8478s+1

denormalizeHan(s) to move the 3-dB cutoff frequency to Ωc=1.021612 using the M-file

lp2lp resulting in Ha(s)=Han 1.021612

0.943847s 0.7380390.943847s+1.78174665

s+0.781947948s+1.0437074244s+1.887749436s+1.0437074244

0.943847s 0.7380390.943847s+1.78174665=+ . (s+0.390974)+(0.9438467)(s+0.94387471)+(0.39090656)arrive at Han(s)=

数字信号处理基于计算机方法第四版第九章答案

Making use of the M-file bilinear we finally arrive at

0.943847z2+0.68178386z0.943847z2 0.25640047z

+ G(z)= .

z 1.363567724z+0.4575139z 0.77823439z+0.1514122

9.15 The mapping is given by s=

(1 z 1) or equivalently, by z=

. Therefore, z2=. Hence, z<1 for σo<0. As 1 σoT jΩoT(1 σoT)+(ΩoT)

a result, a stable Ha(s) results in a stable H(z) after the transformation. However, for

which is equal to 1 only for Ωo=0. Hence, only the point Ωo=0 σo=0,z2=1+(ΩoT)

on the jΩ-axis in the s-plane is mapped onto the point z=1 on the unit circle.

Consequently, this mapping is not useful for the design of digital filters via analog filter transformation.

. For s=σo+jΩo,1 sT

9.16 For no aliasing T≤

H1(z)

. Figure below shows the magnitude responses of the digital filters Ωc

(a) The magnitude responses of the digital filters G1(z) and G2(z)

are shown below:

(b) As can be seen from the above G1(z) is a multi-passband filter, whereas, G2(z) is ahighpass filter.

T 1 z

1 z 1 9.17 Ha(s) is causal and stable and Ha(s) ≤ 1, s, Now, G(z)=Ha(s)s=2 . Thus,

G(z)is causal and stable. Now,

G(ejω)=Ha(s)

2 1 e

s= T 1+e

jω =Ha(s)s=j2tan(ω/2)=Ha(jtan(ω/2).

数字信号处理基于计算机方法第四版第九章答案

Therefore, G(ejω)=Ha(jtan(ω/2)≤1 for all values of ω. Hence, G(z) is a BR

Tfunction.

s+2/T(σ+2/T)+jΩ2 1+z 1 1s 2/T z==. 9.18 (a) s= , or z= s 2/Tσ 2/T+jΩT s+2/T1 z

(σ+2/T)2+Ω222

z=1z<1. The (b) z=, so and 2σ+jΩ,σ=0s=σ+jΩ,σ<0(σ 2/T)+Ω

first proves that a point on the jΩ axis is mapped to a point on the unit circle, and

the second proves that a point in the left-half s-plane is mapped to a point inside the unit circle (stability is preserved). This mapping does indeed have all the desirable properties.

bilinear transformation in Eq. (9.72), then s(f2(z))=s( f1(z)).

21+e jω 1 ΩT ω= 2cot .

(d) jΩ=

2 T1 e

(e) G(z) is a high-pass filter because small Ω (large ω) frequencies passed while

large Ω (small ω) frequencies are attenuated.

9.19 G(z)=HLP(s)s=

21+z 1T1 z

1 α 1 z 1

, where α is defined in Eq. (9.25). G(z) = 2 1+αz

is a high-pass filter. G(z)=GLP( z).G(z)=GLP( z If β = α

, then

数字信号处理基于计算机方法第四版第九章答案

9.20

, where α is defined in Eq. (9.25).

G(z) is a low-pass filter. GLPβ= α, then

1 β 1+z 1

G(z)= =GLP(z).. 2 1 βz

9.21 Let y(t)=

dx(nT) x(nT T)

. Then x(t) be approximated by y(nT)≈

dtT1 z 11 z 1

, or Y(z)=X(z), which suggests the mapping s=

TT111z==. z2=, so 21 Ts(1 Tσ) j(TΩ)(1 Tσ)+(TΩ)

z+jΩ,σ=0 z 0.5=0.5 and z

s=σ+jΩ,σ<0

<1. This the imaginary axis in the

s-domain is mapped to the circle of radius 0.5 centered at z = 0.5, but the

transformation preserves stability. An analog high-pass filter cannot be mapped to a digital high-pass filter because the poles of the digital filter do not lie in the left-half of the z-plane.

1+α1 2βz 1+z 2

9.22 G(z)= . For β=cosωo, the numerator of G(z) becomes 21 β(1+α)z+αz

1 2cos(ωo)z 1+z 2=(1 ejωoz 1)(1 e jωoz 1) which has roots at z=e±jωo. The

numerator of G(zN) is then given by (1 ejωoz N)(1 e jωoz N) whose roots are obtained by solving the equation zN=e±jωo, and are given by

z=ej(2πn±ωo)/N, 0≤n≤N 1. Hence G(zN) has N complex conjugate zero-pairs

2πn±ωlocated on the unit circle at angles of radians, 0≤n≤N 1. For ωo=π/2,

there are 2N equally spaced zeros on the unit circle starting at ω=π/2N.

a+as+as2+ +asN

9.23 Let Ha(s)= denote the analog transfer function and b0+b1s+b2s+ +bNs

q+qz 1+qz 2+ +qz N

denote the digital transfer function G(z)=d0+d1z+d2z+ +dNz

z 1€

obtained after applying the bilinear transformation s=c where c=2/T. All

z+1

transfer function coefficients are assumed to be real numbers. The numerator and

€

the denominator coefficients of Ha(s) and G(z) can be represented in vector form as shown below: A=[a0,a1,a2,…,aN], B=[b0,b1,b2,…,bN],

数字信号处理基于计算机方法第四版第九章答案

€

Q=[q0,q1,q2,…,qN], D=[d0,d1,d2,…,dN].

To determine the elements of Q and D from A and B, we compare the numerator and denominator coefficients of both transfer functions. It can be shown that for N

q+qz 1+qz 2

= 2, we have G(z)= d0+d1z+d2z

(a+ac+ac2)+(2a 2ac2)z 1+(a ac+ac2)z 2(b0+b1c+b2c)+(2b0 2b2c)z

By comparing the coefficients of like powers of z 1 we get

q 0 1 11 a0

q=20 2ac 1 1 .

a2c2 q 2 1 11

A similar relation between the elements of B and D can be derived. In the general case, these matrix equations can be compactly represented in the form

Q=PNAc and D=PNBc where PN is the Pascal matrix of order N+1 and

Ac=a0,a1c,a2c2,…,aNcN and Bc=b0,b1c,b2c2,…,bNcN.

In the general case, the elements of PN are given as follows: (1) The elements of the first row are all ones, (2) The elements of the last column are given by

€ n!i 1

pi,N+1=( 1), 1≤i≤N+1,

(n i+1)!(i 1)!

(3) The remaining elements are given by

pi,j=pi 1,j+pi 1,j+1+pi,j+1,

where 2≤i≤N+1, N≥j≥1.

[1+A4(z)]=2

+(b0 b1c+b2c)z

[][]

€

9.24 (a) H(z)=

z 2D(z 1)z 2D(z 1)N(z)

, where . We can write A4(z)=

D1(z)D2(z)D(z)

D1(z)=1 β1(1+α1)z 1+α1z 2 and D2(z)=1 β2(1+α2)z 1+α2z 2. Therefore,

N(z)=

D1(z)D2(z)+z 4D1(z 1)D2(z 1). Now, 2

1 4

zD1(z 1)D2(z 1)+D1(z)D2(z)=N(z). Hence, N(z) is a symmetric 2

[]

z 4N(z 1)=

polynomial. It follows then

P(z)=

[]

[(α1 β1(1+α1)z 1+z 2)(α2 β2(1+α2)z 1+z 2)

+1 β1(1+α1)z 1+α1z 21 β2(1+α2)z 1+zα2 2]

()()

数字信号处理基于计算机方法第四版第九章答案

1+αα(1+α)(1+α)(β+β) 12[α+α+ββ(1+α)(1+α)] 2

[1 z+z 21+α1α21+α1α2(1+α)(1+α)(β+β) 3 4

z+z]=a(1+b1z 1+b2z 2+b1z 3+z 3),

1+α1α2=

where

b1=

(1+α)(1+α)(β+β)

1+α1α2

2[α+α+ββ(1+α)(1+α)]b2=,

1+α1α2

(7-a) (7-b) (7-c)

(b) a=

1+αα. 2

(c) for z=ejω, we can write N(ejω)=a(1+b1e jω+b2e j2ω+b1e j3ω+e j4ω) =ae j2ω(b2+2b1cosω+2cos2ω). Now, N(ejω)=0 for i=1, 2. For i=1, we get b2+2b1cosω1+2cos2ω1=0, (7-d) for i=2, we get b2+2b1cosω2+2cos2ω2=0,

(7-e)

Solving Eqs. (7-d) and (7-e) we get b1= 2(cosω1+cosω2), (7-f) and b2=2(2cosω1cosω2+1). (7-g)

(1+α)(1+α)(β+β)

=2(cosω1+cosω2), (7-h)

1+α1α2

and from Eqs. (7-b) and (7-g) we have

2[α+α+ββ(1+α)(1+α)]

=2(2cosω1cosω2+1). (7-i)

1+α1α2

1 tan(B/2)1 tan(B/2)

, and after rearrangement we get Substituting α1= and α2=

1+tan(B1/2)1+tan(B2/2)

From Eqs. (7-a) and (7-f) we have

and

β1+β2=(cosω1+cosω2)[1+tan(B1/2)tan(B2/2)]=θ1, β1β2=[1+tan(B1/2)tan(B2/2)]cosω1cosω2=

θ2Δ

(7-j)

(7-k) The

above two nonlinear equations can be solved yielding β1=β2=.

θ1(d) For the double notch filter with the following specifications: ω1=0.2π, ω2=0.6π, B1=0.2π,andB2=0.25π we get the following values for the parameters of the notch filter transfer function:

α1=0.5095, α2=0.4142, θ1=0.5673, θ2= 0.1491, β1=0.7628, and

β2= 0.1955. H(z)=

[1+A4(z)]:

数字信号处理基于计算机方法第四版第九章答案

9.25 A zero (pole) of HLP(z) is given by the factor (z zk). After applying the lowpass-to- αz

zk, and hence the new location of the lowpass transformation, this factor becomes

1 αz

zero (pole) is given by the roots of the equation

α+z or z =. For zk= 1, z α zk+αzkz =(1+αzk)z (α+zk)=0k

1+αzk

a 1 z == 1. k

1 a

z 2 b+az 2αβ

a=9.26 The lowpass-to-bandpass transformation is given by z→ where

+bz β+11 az

β 1

and b=. A zero (pole) of HLP(z) is given by the factor (z zk). After applying the

β+1

z 2 b+az

zk, and hence, the lowpass-to-bandpass transformation, this factor becomes

+bz 21 az

new location of the zero (pole) of the bandpass transfer function is given by the roots of the

a(1+z)

b+zequation (1+bzk)z2 a(1+zk)z+(b+zk)=0, or z2 z+=0, whose

1+bzk

a(1+z)solution is given by z ±zk= 1, z k=k=±1.2(1+bzk)

9.27 GLP(z)=

0.34041+z

1+0.1842z+0.1776z

c=0.27π HLP(z) for ω

, with ωc=0.55π.

数字信号处理基于计算机方法第四版第九章答案

tan 0.55π

α= 2 tan 0.27π 2 sin 0.55π 0.27π tan 0.55π0.27π=2

0.55π=0.4434

2 +tan 2 sin +0.27π2

0.3404 1 +z

1 α 2 HLP(z)=GLP(z)z 1=z

1 α= 1 αz

1 αz

1+0.1842 z 1 α z 1 α 1 αz +0.1776 1 αz

=0.1055+0.2109z 1+0.1055z

20.9532 0.8239z

+0.2925z

0.55π+0.45π 9.28 α cos = 2

cos 0.55π 0.45π 2

=0.

HHP(z)

=G0.3404 0.6808z 1+0.3404z

2LP(z)z 1= z 1 +α = 1

+αz

1 0.1842z +0.1776z

9.29 ωc=ω

c2 ω c1, and α=cos(ω c)=0.0628

数字信号处理基于计算机方法第四版第九章答案

1 2 α 0.0628z z

. = z = 1 0.0628z 1 αz 1 z 1

1 2 3 4 0.6075 0.0389z 0.5930z 0.0389z +0.6075z

HBP(z)=.

+2.2169z 0.2480z +0.8781z 3.1250 0.5302z

0.52π 0.48π

sin 2 p=0.48π, and ωp=0.52π. λ=9.30 ω=0.0628. 0.52π+0.48πsin 2

1 2 3 0.3766 0.6803z +0.6803z 0.3766z 1 0.0628=HHP(z)=GHP(z)z=z .

1.3954+0.0705z +0.9783z +0.1892z 1 0.0628z

2 1 z

. 9.31 Eq. (7.79):HBP(z)=0.136728736 1 0.53353098z+0.726542528z

数字信号处理基于计算机方法第四版第九章答案

0.4π 0.6π sin 2

0=0.6π, and ω0=0.4π. λ== 0.3090 ω

0.4π+0.6πsin 2

2 0.1013 0.1013z 1 +0.3090=GBP(z)=HBP(z)z 1=z .

+0.5381z 0.7406+0.3951z 1+0.3090z

B 1 tan 120 5 2

9.32 ω0=2π =0.48π, Bω=2π =0.06π. α==0.8273

B 500 400 1+tan 2

β=cos(ω0)=0.0628.

(1+α) 2β(1+α)z 1+(1+α)z 20.9137 0.1148z 1+0.9137z 2

GBS(z)==. 1 β(1+α)z+αz1 0.1148z+0.8273z

ω ω sin 40 2

0= 2ω=0.5706 π=0.16π. α= ω+ω 500 sin2

1 2 0.3192 0.5594z +0.3192z 1 α=. HBS(z)=GBS(z)z 1=z

0.3337 0.5594z +0.3046z 1 αz

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