Not for sale 1 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. Not for sale 2 Chapter 9 9.1 We obtain the solutions by using Eq. (9.3) and Eq. (9.4). (a) p=110 p/20 =1100.24 /20= 0.0273,s=10s /20 =1049/20= 0.0035. (b) p=110 p/20 =1100.14 /20= 0.016,s=10s /20 =1068/20= 0.000398. 9.2 We obtain the solutions by using Eqs. (9.3) and (9.4). (a) p= 20log101p ()= 20log10(10.04) = 0.3546dB, s= 20log10s( )= 20log100.08()= 21.9382 dB. (b) p= 20log101p ()= 20log10(10.015) = 0.1313dB, s= 20log10s( )= 20log100.04()= 27.9588 dB. 9.3 G(z) = H2(z), or equivalently, G(e j) = H2(ej) = H(ej)2. Let pand s denote the passband and stopband ripples of H(e j), respectively. Also, let p,2= 2p, and s,2 denote the passband and stopband ripples of G(e j), respectively. Then p,2=1(1p)2, and s,2= (s)2. For a cascade of sections, p,M=1(1p)M,and s,M= (s)M. 9.4 HLP(ej) p s ps s 1+p 1 p 0 HHP(ej) s 1+ p 1 p p s( s)( p) 0 Therefore, the passband edge and the stopband edge of the highpass filter are given by p,HP= p, and s,HP= s, respectively. 9.5 Note that G(z) is a complex bandpass filter with a passband in the range 0 . Its passband edges are at p,BP=op,and stopband edges at s,BP=os. A real coefficient bandpass transfer function can be generated according to GBP(z) = HLP(e joz)+ HLP(e joz) which will have a passband in the range 0 Not for sale 3 and another passband in the range 0. However because of the overlap of the two spectra a simple formula for the bandedges cannot be derived. HLP(ej) p s ps s 1+p 1 p 0 G(ej) s 1+ p 1 p 0o o+s o+po p o s 9.6 (a) hp(t) = ha(t) p(t) where p(t) =(t nT). n= Thus, hp(t) =ha(nT) n= (t nT) We also have, gn = ha(nT). Now, Ha(s) =ha(t)est dt and Hp(s) =hp(t)est dt =ha(nT)(t nT)est dt n= =ha(nT)esnT n= . Comparing the above expression with G(z) =gnzn n= =h(nT)zn n= , we conclude that G(z) = Hp(s) s= 1 T lnz . We can also show that a Fourier series expansion of p(t) is given by p(t) = 1 T ej(2kt /T) k= . Therefore, hp(t) = 1 T e j(2kt /T) k= ha(t) = 1 T ha(t)ej(2kt /T) k= . Hence, Hp(s) = 1 T Has+ j 2kt T k= . As a result, we have G(z) = 1 T Has+ j 2kt T k= s= 1 T lnz. (7-1) (b) The transformation from the s-plane to z-plane is given by z = esT. If we express s =o+ jo, then we can write z = re j = eoTe joT . Therefore, Not for sale 4 z = 1,for o1. Or in other words, a point in the left-half -plane is mapped onto 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 j-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. (c) However, all points in the s-plane defined by s =o+ jo j 2k T , k = 0, 1, 2, , a are mapped onto a single point in the z-plane as z = eoTe j o2k T T = eoTe joT . The mapping is illustrated in the figure below 1 1 jzIm zRe z-plane -planes T 3 T T 3 T Note that the strip of width 2/T in the s-plane for values of s in the range T T is mapped into the entire z-plane, and so are the adjacent strips of width 2/T. The mapping is many-to-one with infinite number of such strips of width 2/T. It follows from the above figure and also from Eq. (7-1) that if the frequency response Ha(j) = 0 for T , then G(e j) =1 T Ha(j T ) for , and there is no aliasing. (d) For z = e j = e jT , or equivalently, = T. 9.7 Assume ha(t) is causal. Now, ha(t) =Ha(s)estds. Hence, gn = ha(nT) =Ha(s)esnTds. Therefore, Not for sale 5 G(z) =gnzn n=0 =Ha(s)esnTzn n=0 ds =Ha(s)zn n=0 esnTds = Ha(s) 1esTz1 ds . Hence G(z) =Residues Ha(s) 1esTz1 all poles of Ha(s) . 9.8 Ha(s) = A s+ . The transfer function has a pole at s = . Now G(z) = Residue at s= A