数字信号处理-基于计算机的方法(第四版)第九章答案

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,

2

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

M

δp,M=1 (1 δp)M,and δs,M=(δs).

), respectively.

sections,

s

p

p

ω

–(πpsω

sp

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.

s

p

p

∞

ω

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

st

dt=

n= ∞ ∞

∑∫ha(nT)δ(t nT)e

∞

∞∞

∞

∫ha(t)e stdt and

st

dt=

∑ha(nT)e snT.

n= ∞

∞n= ∞

∞

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

∞

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)=

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= ∞ ∞

1

Hp(s)=

T

∑

k= ∞∞

∞

2πkt Ha s+j . As a result, we have T

(7-1)

G(z)=

1

T

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

sT

. 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

(c) However, all points in the s-plane defined by s=σo+jΩo±j, k=0, 1, 2, …,, a

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

n=0

H(s)1 e

z

.

Hence G(z)=

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