Solutions Manual For Digital Communications, 5th Edition ...

1 Solutions Manual For Digital Communications, 5th edition prepared by Kostas Stamatiou Solutions ManualforDigital Communications, 5th edition (Chapter 2)1 prepared byKostas StamatiouJanuary 11, 20081 PROPRIETARY The McGraw-Hill Companies, Inc. All rights reserved. No part of thisManual may be displayed, reproduced or distributed in any form or by any means, without the prior writtenpermission of the publisher, or used beyond the limited distribution to teachers and educators permitted byMcGraw-Hill for their individual course preparation. If you are a student using this Manual , you are usingit without x(t) =1 Z x(a)t adaHence : x( t) = 1 R x(a) t ada= 1 R x( b) t+b( db)= 1 R x(b) t+bdb=1 R x(b)t bdb= x(t)where we have made the change of variables :b= aand used the relationship :x(b) =x( b).

2 Exactly the same way as in part (a) we prove : x(t) = x( t) (t) = cos 0t, so its Fourier transform is :X(f) =12[ (f f0) + (f+f0)], f0= 2 the phase-shifting property (2-1-4) of the Hilbert transform : X(f) =12[ j (f f0) +j (f+f0)] =12j[ (f f0) (f+f0)] =F 1{sin 2 f0t}Hence, x(t) = sin a similar way to part (c) :x(t) = sin 0t X(f) =12j[ (f f0) (f+f0)] X(f) =12[ (f f0) (f+f0)] X(f) = 12[ (f f0) + (f+f0)] = F 1{cos 2 0t} x(t) = cos positive frequency content of the new signal will be : ( j)( j)X(f) = X(f), f >0,whilethe negative frequency content will be :j jX(f) = X(f), f < , since X(f) = X(f),we have : x(t) = x(t). the magnitude response of the Hilbert transformer is characterized by :|H(f)|= 1,wehave that : X(f) =|H(f)||X(f)|=|X(f)|.Hence :Z X(f) 2df=Z |X(f)|2dfPROPRIETARY The McGraw-Hill Companies, Inc.

3 All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual , you are using it without using Parseval s relationship :Z x2(t)dt=Z x2(t) parts (a) and (b) above, we note that ifx(t) is even, x(t) is odd and vice-versa. Therefore,x(t) x(t) is always odd and hence :R x(t) x(t)dt= Using relationsX(f) =12Xl(f f0) +12Xl( f f0)Y(f) =12Yl(f f0) +12Yl( f f0)and Parseval s relation, we haveZ x(t)y(t)dt=Z X(f)Y (f)dt=Z 12Xl(f f0) +12Xl( f f0) 12Yl(f f0) +12Yl( f f0) df=14Z Xl(f f0)Y l(f f0)df+14Z Xl( f f0)Yl( f f0)df=14Z Xl(u)Y l(u)du+14X l(v)Y(v)dv=12Re Z Xl(f)Y l(f)df =12Re Z xl(t)y l(t)dt where we have used the fact that sinceXl(f f0) andYl( f f0) do not overlap,Xl(f f0)Yl( f f0) = 0 and similarlyXl( f f0)Yl(f f0) = Puttingy(t) =x(t) we get the desired result from the result of part The McGraw-Hill Companies, Inc.

4 All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual , you are using it without well-known result in estimation theory based on the minimum mean-squared-error criterion statesthat the minimum ofEeis obtained when the error is orthogonal to each of the functions in theseries expansion. Hence :Z "s(t) KXk=1skfk(t)#f n(t)dt= 0, n= 1,2,..,K(1)since the functions{fn(t)}are orthonormal, only the term withk=nwill remain in the sum, so :Z s(t)f n(t)dt sn= 0, n= 1,2,..,Kor:sn=Z s(t)f n(t)dt n= 1,2.

5 ,KThe corresponding residual errorEeis :Emin=R hs(t) PKk=1skfk(t)ihs(t) PKn=1snfn(t)i dt=R |s(t)|2dt R PKk=1skfk(t)s (t)dt PKn=1s nR hs(t) PKk=1skfk(t)if n(t)dt=R |s(t)|2dt R PKk=1skfk(t)s (t)dt=Es PKk=1|sk|2where we have exploited relationship (1) to go from the second to the third step in the : Relationship (1) can also be obtained by simple differentiation of the residual error withrespect to the coefficients{sn}.Sincesnis, in general, complex-valuedsn=an+jbnwe have todifferentiate with respect to both real and imaginary parts :ddanEe=ddanR hs(t) PKk=1skfk(t)ihs(t) PKn=1snfn(t)i dt= 0 R anfn(t)hs(t) PKn=1snfn(t)i +a nf n(t)hs(t) PKn=1snfn(t)idt= 0 2anR Renf n(t)hs(t) PKn=1snfn(t)iodt= 0 R Renf n(t)hs(t) PKn=1snfn(t)iodt= 0, n= 1,2,..,KPROPRIETARY The McGraw-Hill Companies, Inc.

6 All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual , you are using it without we have exploited the identity : (x+x ) = 2Re{x}.Differentiation ofEewith respect tobnwill give the corresponding relationship for the imaginarypart; combining the two we get (1).Problem procedure is very similar to the one for the real-valued signals described in the book (pages33-37). The only difference is that the projections should conform to the complex-valued vectorspace :c12=Z s2(t)f 1(t)dtand, in general for thek-th function :cik=Z sk(t)f i(t)dt, i= 1,2.

7 ,k 1 Problem first basis function is :g4(t) =s4(t) E4=s4(t) 3= 1/ 3,0 t 30, Then, for the second basis function :c43=Z s3(t)g4(t)dt= 1/ 3 g 3(t) =s3(t) c43g4(t) = 2/3,0 t 2 4/3,2 t 30, Hence :g3(t) =g 3(t) E3= 1/ 6,0 t 2 2/ 6,2 t 30, whereE3denotes the energy ofg 3(t) :E3=R30(g 3(t))2dt= 8 the third basis function :c42=Z s2(t)g4(t)dt= 0 and c32=Z s2(t)g3(t)dt= 0 PROPRIETARY The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual , you are using it without :g 2(t) =s2(t) c42g4(t) c32g3(t) =s2(t)andg2(t) =g 2(t) E2= 1/ 2,0 t 1 1/ 2,1 t 20, where :E2=R20(s2(t))2dt= for the fourth basis function :c41=Z s1(t)g4(t)dt= 2/ 3,c31=Z s1(t)g3(t)dt= 2/ 6, c21= 0 Hence :g 1(t) =s1(t) c41g4(t) c31g3(t) c21g2(t) = 0 g1(t) = 0 The last result is expected, since the dimensionality of thevector space generated by these signalsis 3.

8 Based on the basis functions (g2(t),g3(t),g4(t)) the basis representation of the signals is :s4= 0,0, 3 E4= 3s3= 0,p8/3, 1/ 3 E3= 3s2= 2,0,0 E2= 2s1= 2/ 6, 2/ 3,0 E1= 2 Problem the set of signalse nl(t) =j nl(t),1 n N, then by definition of lowpass equivalentsignals and by Equations and , we see that n(t) s are 2 times the lowpass equivalentsof nl(t) s ande n(t) s are 2 times the lowpass equivalents ofe nl(t) s. We also note that since n(t) s have unit energy,h nl(t),e nl(t)i=h nl(t),j nl(t)i= jand since the inner product is pureimaginary, we conclude that n(t) ande n(t) are orthogonal. Using the orthonormality of the set nl(t), we haveh nl(t), j ml(t)i=j mnand using the result of problem we haveh n(t),e m(t)i= 0 for alln,mWe also haveh n(t), m(t)i= 0 for alln6=mPROPRIETARY The McGraw-Hill Companies, Inc.

9 All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual , you are using it without n(t),e m(t)i= 0 for alln6=mUsing the fact that the energy in lowpass equivalent signal is twice the energy in the bandpasssignal we conclude that the energy in n(t) s ande n(t) s is unity and hence the set of 2 Nsignals{ n(t),e n(t)}constitute an orthonormal set. The fact that this orthonormal set is sufficient forexpansion of bandpass signals follows from Equation (t) =m(t) cos 2 f0twherem(t) is real and lowpass with bandwidth less thanf0.

10 ThenF[ x(t)] = jsgn(f) 12M(f f0) +12M(f+f0) and henceF[ x(t)] = j2M(f f0) +j2M(f+f0)where we have used that fact thatM(f f0) = 0 forf <0 andM(f+f0) = 0 forf >0. Thisshows that x(t) =m(t) sin 2 f0t. Similarly we can show that Hilbert transform ofm(t) sin 2 f0tis m(t) cos 2 f0t. From above and Equation we haveH[ n(t)] = 2 ni(t) sin 2 f0t+ 2 nq(t) cos 2 f0t= e n(t)Problem real-valued signals the correlation coefficients are given by : km=1 EkEmR sk(t)sm(t)dtandthe Euclidean distances by :d(e)km= Ek+Em 2 EkEm km 1 the signals in this problem :E1= 2,E2= 2,E3= 3,E4= 3 12= 0 13=2 6 14= 2 6 23= 0 24= 0 34= 13and:d(e)12= 2d(e)13=q2 + 3 2 62 6= 1d(e)14=q2 + 3 + 2 62 6= 3d(e)23= 2 + 3 = 5d(e)24= 5d(e)34=q3 + 3 + 2 313= 2 2 PROPRIETARY The McGraw-Hill Companies, Inc. All rights reserved.