数字通信原理课后习题解答.pdf

《数字通信原理课后习题解答.pdf》由会员分享，可在线阅读，更多相关《数字通信原理课后习题解答.pdf（459页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、Solutions ManualforDigital Communications,5th Edition(Chapter 2)1Prepared byKostas StamatiouJanuary 11,20081PROPRIETARY MATERIAL.c?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 writtenp

2、ermission 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 permission.2Problem 2.1a.x(t)=1Zx(a)t adaHence:x(t)=1Rx(a)tada=1Rx(b)t+b(db)=1Rx

3、(b)t+bdb=1Rx(b)tbdb=x(t)where we have made the change of variables:b=a and used the relationship:x(b)=x(b).b.In exactly the same way as in part(a)we prove:x(t)=x(t)c.x(t)=cos0t,so its Fourier transform is:X(f)=12(f f0)+(f+f0),f0=20.Exploiting the phase-shifting property(2-1-4)of the Hilbert transfor

4、m:X(f)=12j(f f0)+j(f+f0)=12j(f f0)(f+f0)=F1sin2f0tHence,x(t)=sin0t.d.In a similar way to part(c):x(t)=sin0t X(f)=12j(f f0)(f+f0)X(f)=12(f f0)(f+f0)X(f)=12(f f0)+(f+f0)=F1cos20t x(t)=cos0te.The positive frequency content of the new signal will be:(j)(j)X(f)=X(f),f 0,whilethe negative frequency conten

5、t will be:j jX(f)=X(f),f 0.Hence,sinceX(f)=X(f),we have:x(t)=x(t).f.Since 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 MATERIAL.c?The McGraw-Hill Companies,Inc.All rights reserved.No part o

6、f 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

7、Manual,you are using it without permission.3and using Parsevals relationship:Z x2(t)dt=Zx2(t)dtg.From parts(a)and(b)above,we note that if x(t)is even,x(t)is odd and vice-versa.Therefore,x(t)x(t)is always odd and hence:Rx(t)x(t)dt=0.Problem 2.21.Using relationsX(f)=12Xl(f f0)+12Xl(f f0)Y(f)=12Yl(f f0

8、)+12Yl(f f0)and Parsevals relation,we haveZx(t)y(t)dt=ZX(f)Y(f)dt=Z?12Xl(f f0)+12Xl(f f0)?12Yl(f f0)+12Yl(f f0)?df=14ZXl(f f0)Yl(f f0)df+14ZXl(f f0)Yl(f f0)df=14ZXl(u)Yl(u)du+14Xl(v)Y(v)dv=12Re?ZXl(f)Yl(f)df?=12Re?Zxl(t)yl(t)dt?where we have used the fact that since Xl(f f0)and Yl(f f0)do not overla

9、p,Xl(f f0)Yl(f f0)=0 and similarly Xl(f f0)Yl(f f0)=0.2.Putting y(t)=x(t)we get the desired result from the result of part 1.PROPRIETARY MATERIAL.c?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

10、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 permission.4Problem 2.3A well-known result in estimation theo

11、ry based on the minimum mean-squared-error criterion statesthat the minimum of Eeis obtained when the error is orthogonal to each of the functions in theseries expansion.Hence:Zs(t)KXk=1skfk(t)#fn(t)dt=0,n=1,2,.,K(1)since the functions fn(t)are orthonormal,only the term with k=n will remain in the s

12、um,so:Zs(t)fn(t)dt sn=0,n=1,2,.,Kor:sn=Zs(t)fn(t)dtn=1,2,.,KThe corresponding residual error Eeis:Emin=Rhs(t)PKk=1skfk(t)ihs(t)PKn=1snfn(t)idt=R|s(t)|2dt RPKk=1skfk(t)s(t)dt PKn=1snRhs(t)PKk=1skfk(t)ifn(t)dt=R|s(t)|2dt RPKk=1skfk(t)s(t)dt=EsPKk=1|sk|2where we have exploited relationship(1)to go from

13、 the second to the third step in the abovecalculation.Note:Relationship(1)can also be obtained by simple differentiation of the residual error withrespect to the coefficients sn.Since snis,in general,complex-valued sn=an+jbnwe have todifferentiate with respect to both real and imaginary parts:ddanEe

14、=ddanRhs(t)PKk=1skfk(t)ihs(t)PKn=1snfn(t)idt=0 Ranfn(t)hs(t)PKn=1snfn(t)i+anfn(t)hs(t)PKn=1snfn(t)idt=0 2anRRenfn(t)hs(t)PKn=1snfn(t)iodt=0RRenfn(t)hs(t)PKn=1snfn(t)iodt=0,n=1,2,.,KPROPRIETARY MATERIAL.c?The McGraw-Hill Companies,Inc.All rights reserved.No part of this Manual may be displayed,reprod

15、uced 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 permi

16、ssion.5where we have exploited the identity:(x+x)=2Rex.Differentiation of Eewith respect to bnwill give the corresponding relationship for the imaginary part;combining the two we get(1).Problem 2.4The procedure is very similar to the one for the real-valued signals described in the book(pages33-37).

17、The only difference is that the projections should conform to the complex-valued vectorspace:c12=Zs2(t)f1(t)dtand,in general for the k-th function:cik=Zsk(t)fi(t)dt,i=1,2,.,k 1Problem 2.5The first basis function is:g4(t)=s4(t)E4=s4(t)3=1/3,0 t 30,o.w.Then,for the second basis function:c43=Zs3(t)g4(t

18、)dt=1/3 g3(t)=s3(t)c43g4(t)=2/3,0 t 24/3,2 t 30,o.wHence:g3(t)=g3(t)E3=1/6,0 t 22/6,2 t 30,o.wwhere E3denotes the energy of g3(t):E3=R30(g3(t)2dt=8/3.For the third basis function:c42=Zs2(t)g4(t)dt=0andc32=Zs2(t)g3(t)dt=0PROPRIETARY MATERIAL.c?The McGraw-Hill Companies,Inc.All rights reserved.No part

19、 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 thi

20、s Manual,you are using it without permission.6Hence:g2(t)=s2(t)c42g4(t)c32g3(t)=s2(t)andg2(t)=g2(t)E2=1/2,0 t 11/2,1 t 20,o.wwhere:E2=R20(s2(t)2dt=2.Finally for the fourth basis function:c41=Zs1(t)g4(t)dt=2/3,c31=Zs1(t)g3(t)dt=2/6,c21=0Hence:g1(t)=s1(t)c41g4(t)c31g3(t)c21g2(t)=0 g1(t)=0The last resu

21、lt is expected,since the dimensionality of the vector space generated by these signalsis 3.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=2Problem 2.6Consider the set of signalsenl(t)=jnl(t),1 n

22、 N,then by definition of lowpass equivalentsignals and by Equations 2.2-49 and 2.2-54,we see that n(t)s are2 times the lowpass equivalentsof nl(t)s anden(t)s are2 times the lowpass equivalents ofenl(t)s.We also note that sincen(t)s have unit energy,hnl(t),enl(t)i=hnl(t),jnl(t)i=j and since the inner

23、 product is pureimaginary,we conclude that n(t)anden(t)are orthogonal.Using the orthonormality of the setnl(t),we havehnl(t),jml(t)i=jmnand using the result of problem 2.2 we havehn(t),em(t)i=0for all n,mWe also havehn(t),m(t)i=0for all n 6=mPROPRIETARY MATERIAL.c?The McGraw-Hill Companies,Inc.All r

24、ights 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

25、are astudent using this Manual,you are using it without permission.7andhen(t),em(t)i=0for all n 6=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 anden(t)s is unity and hence the set of 2N signalsn(t),en(t)co

26、nstitute an orthonormal set.The fact that this orthonormal set is sufficient forexpansion of bandpass signals follows from Equation 2.2-57.Problem 2.7Let x(t)=m(t)cos2f0t where m(t)is real and lowpass with bandwidth less than f0.ThenF x(t)=j sgn(f)?12M(f f0)+12M(f+f0)?and hence F x(t)=j2M(f f0)+j2M(

27、f+f0)where we have used that fact that M(f f0)=0 for f 0.Thisshows that x(t)=m(t)sin2f0t.Similarly we can show that Hilbert transform of m(t)sin2f0t ism(t)cos2f0t.From above and Equation 2.2-54 we haveHn(t)=2ni(t)sin2f0t+2nq(t)cos2f0t=en(t)Problem 2.8For real-valued signals the correlation coefficie

28、nts are given by:km=1EkEmRsk(t)sm(t)dt andthe Euclidean distances by:d(e)km=?Ek+Em 2EkEmkm?1/2.For the signals in this problem:E1=2,E2=2,E3=3,E4=312=013=2614=2623=024=034=13and:d(e)12=2d(e)13=q2+3 2626=1d(e)14=q2+3+2626=3d(e)23=2+3=5d(e)24=5d(e)34=q3+3+2 313=22PROPRIETARY MATERIAL.c?The McGraw-Hill

29、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

30、preparation.If you are astudent using this Manual,you are using it without permission.8Problem 2.9We know from Fourier transform properties that if a signal x(t)is real-valued then its Fouriertransform satisfies:X(f)=X(f)(Hermitian property).Hence the condition under which sl(t)is real-valued is:Sl(

31、f)=Sl(f)or going back to the bandpass signal s(t)(using 2-1-5):S+(fc f)=S+(fc+f)The last condition shows that in order to have a real-valued lowpass signal sl(t),the positive fre-quency content of the corresponding bandpass signal must exhibit hermitian symmetry around thecenter frequency fc.In gene

32、ral,bandpass signals do not satisfy this property(they have Hermitiansymmetry around f=0),hence,the lowpass equivalent is generally complex-valued.Problem 2.10a.To show that the waveforms fn(t),n=1,.,3 are orthogonal we have to prove that:Zfm(t)fn(t)dt=0,m 6=nClearly:c12=Zf1(t)f2(t)dt=Z40f1(t)f2(t)d

33、t=Z20f1(t)f2(t)dt+Z42f1(t)f2(t)dt=14Z20dt 14Z42dt=14 2 14(4 2)=0Similarly:c13=Zf1(t)f3(t)dt=Z40f1(t)f3(t)dt=14Z10dt 14Z21dt 14Z32dt+14Z43dt=0and:c23=Zf2(t)f3(t)dt=Z40f2(t)f3(t)dt=14Z10dt 14Z21dt+14Z32dt 14Z43dt=0PROPRIETARY MATERIAL.c?The McGraw-Hill Companies,Inc.All rights reserved.No part of this

34、 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

35、,you are using it without permission.9Thus,the signals fn(t)are orthogonal.It is also straightforward to prove that the signals have unitenergy:Z|fi(t)|2dt=1,i=1,2,3Hence,they are orthonormal.b.We first determine the weighting coefficientsxn=Zx(t)fn(t)dt,n=1,2,3x1=Z40 x(t)f1(t)dt=12Z10dt+12Z21dt 12Z

36、32dt+12Z43dt=0 x2=Z40 x(t)f2(t)dt=12Z40 x(t)dt=0 x3=Z40 x(t)f3(t)dt=12Z10dt 12Z21dt+12Z32dt+12Z43dt=0As it is observed,x(t)is orthogonal to the signal wavaforms fn(t),n=1,2,3 and thus it can notrepresented as a linear combination of these functions.Problem 2.11a.As an orthonormal set of basis functi

37、ons we consider the setf1(t)=10 t 10o.wf2(t)=11 t 20o.wf3(t)=12 t 30o.wf4(t)=13 t 40o.wIn matrix notation,the four waveforms can be represented ass1(t)s2(t)s3(t)s4(t)=2111211011111222f1(t)f2(t)f3(t)f4(t)Note that the rank of the transformation matrix is 4 and therefore,the dimensionality of thewavef

38、orms is 4PROPRIETARY MATERIAL.c?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 permi

39、tted by McGraw-Hill for their individual course preparation.If you are astudent using this Manual,you are using it without permission.10b.The representation vectors ares1=h2111is2=h2110is3=h1111is4=h1222ic.The distance between the first and the second vector is:d1,2=p|s1 s2|2=r?h4221i?2=25Similarly

40、we find that:d1,3=p|s1 s3|2=r?h1020i?2=5d1,4=p|s1 s4|2=r?h1113i?2=12d2,3=p|s2 s3|2=r?h3201i?2=14d2,4=p|s2 s4|2=r?h3332i?2=31d3,4=p|s3 s4|2=r?h0133i?2=19Thus,the minimum distance between any pair of vectors is dmin=5.Problem 2.12As a set of orthonormal functions we consider the waveformsf1(t)=10 t 10

41、o.wf2(t)=11 t 20o.wf3(t)=12 t x)=Q(x)andp(X 104)=Q?104104?=Q(1)=.159p(X 4 104)=Q?4 104104?=Q(4)=3.17 105p(2 104 104?X 0)=p(X 104,X 0)p(X 0)=p(X 104)p(X 0)=.159.5=.318PROPRIETARY MATERIAL.c?The McGraw-Hill Companies,Inc.All rights reserved.No part of this Manual may be displayed,reproduced or distrib

42、uted 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 permission.14Problem

43、 2.201)y=g(x)=ax2.Assume without loss of generality that a 0.Then,if y 0 there are two solutions to the system,namelyx1,2=py/a.Hence,fY(y)=fX(x1)|g(x1)|+fX(x2)|g(x2)|=fX(py/a)2apy/a+fX(py/a)2apy/a=1ay22ey2a22)The equation y=g(x)has no solutions if y b.Thus FY(y)and fY(y)are zero for y b.Ifb y b,then

44、 for a fixed y,g(x)y if x b then g(x)b yfor every x;hence FY(y)=1.At the points y=b,FY(y)is discontinuous and the discontinuitiesequal toFY(b+)FY(b)=FX(b)andFY(b+)FY(b)=1 FX(b)The PDF of y=g(x)isfY(y)=FX(b)(y+b)+(1 FX(b)(y b)+fX(y)u1(y+b)u1(y b)=Q?b?(y+b)+(y b)+122ey222u1(y+b)u1(y b)3)In the case of

45、 the hard limiterp(Y=b)=p(X 0)=1 FX(0)=12Thus FY(y)is a staircase function andfY(y)=FX(0)(y b)+(1 FX(0)(y a)PROPRIETARY MATERIAL.c?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

46、 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 permission.154)The random variable y=g(x)takes the values yn=xnwith probabilit

47、yp(Y=yn)=p(an X an+1)=FX(an+1)FX(an)Thus,FY(y)is a staircase function with FY(y)=0 if y xN.The PDFis a sequence of impulse functions,that isfY(y)=NXi=1FX(ai+1)FX(ai)(y xi)=NXi=1hQ?ai?Q?ai+1?i(y xi)Problem 2.21For n odd,xnis odd and since the zero-mean Gaussian PDF is even their product is odd.Sincet

48、he integral of an odd function over the interval,is zero,we obtain EXn=0 for n odd.Let In=Rxnexp(x2/22)dx.Obviously Inis a constant and its derivative with respect to xis zero,i.e.,ddxIn=Z?nxn1ex22212xn+1ex222?dx=0which results in the recursionIn+1=n2In1This is true for all n.Now let n=2k 1,we will

49、have I2k=(2k 1)2I2k2,with the initialcondition I0=22.Substituting we haveI2=222I4=32I2=3422I6=5 32I4=5 3622I8=7 2I6=7 5 3822.=.and in general if I2k=(2k1)(2k3)(2k5)312k22,then I2k+2=(2k+1)2I2k=(2k+1)(2k1)(2k3)(2k5)312k+222.Using the fact that EX2k=I2k/22,we obtainIn=1 3 5 (n 1)nfor n even.PROPRIETAR

50、Y MATERIAL.c?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