数字通信基础仇佩亮课后答案18章

《数字通信基础仇佩亮课后答案18章》由会员上传分享，免费在线阅读，更多相关内容在教育资源-天天文库。

1、2008-3-211-1ABCD41/21/41/81/8[]1H(XE)logpX()1111log2log4log8log824881133228831(bits)41-301011/210[]XYXYp(X0)pX(1)0.50.5000.5p(Y0

2、X0)p(YX1

3、0)0.5111p(YX0

4、1)0p(YX1

5、1)1p(X0,Y0)p(XY0,1)0.25p(XY1,0)0p(XY1,1)0.5pY(0)0.25pY(1)0.751p(XY,)I(X;YE)log2p(X)pY()0.250.250.50.25log0.25log0.

6、5log2220.50.250.50.750.50.750.250.250.25log310.5log3221.50.75log3(bits)2I(X;Y)H(X)H(XY

7、)1p(Y1)H(XY

8、1)110.75h31.50.75log3(bits)2=0.3113(bit)h(x)xlogx(1xx)log(1)1-5125s63[]=1/(12510)=810Baud33=810log4=1610bits/s22-350Hzx()t1,40nx(nTn)1,04s0,(1)x(0.005)(2)(1)1x(t)x(kT)sinc2W()t

9、kTTs0.01()sssk2W4x(0.005)sinc(0.5kk)sinc(0.5)k1sinc(0.5)sinc(4.5)sin(0.5)sin(4.5)0.5660.54.5(2)sinc(tkTk),0,1,s2218Ex(t)dxx()kTs100k10022-11x(t)sinc(t)cos2fth(t)sinc(t)sin2ft00yt()x(t)sinc(t)cos2ftxˆ(t)sinc(t)sin2ft00x(tt)sinc()lh(t)sinc22(t)sin2fthˆ(t)sinc(t)cos2ft00j22h(t)

10、sinc()tel1y(t)Rey()tej2ft0ly(t)x(t)ht()lll1Y(f)X(f)Hf()lll211f2Xf()l10f2(1f)/jf01Hf()l(1f)/jf101(1f)/20jf2Yf()l1(1f)/2jf0212j2ftyll(t)1Y()fedf21112j22ft0jft(1)fedf1(1)fedf22jj02sint111sintt(cos)2222t2t22ttjj11sint(1cost)jsintt(cos1)22224t4t44tt11y(t)(1cost)sintsin(2)ft22044t

11、t2-19()t(tt)2cos(2)PP(0)(/2)1/2E[(t)]R(0,1)E[(t)]P(0)2cos(2t)Pt()2cos(2)22cos(2tt)sin(2)2RE(0,1)[2cos2cos(2)]5PP(0)4()4coscos22222-25N/2fB0cP2-25

12、H(f)

13、BB-fc0fcfP2-2512NB0ffcN0222P(f)Hf()N2B0ffc2jf2R()P()fedfNNBsinc(Bf)cos(2)0c20NB021nfn()expN22222-30()tR()P2-30[]Y(t)(t)()tT3

14、E[Y(t)Y(t)]E[((t)(tT))((t)(tT))]121122E[(t)(t)]E[(tTt)()]1212E[()t(tT)]E[(tT)(tT)]12122R()R(T)RT()P(f)[2R()R(T)RT()]Yj22fTjfTP(f)(2)ee2P(f)(1cos2)fT2-35Xt()Yt()Y(t)X(t)cos(2ft)Xˆ(t)sin(2)ft00f[0,2]X()tX()t0P2-35Yt()PX(f)A-B0BfP2-35Y(t)X(t)cosXtˆ()sincos2ft0X(t)sinXˆ(t)cossin

15、2ft0Z(t)X(t)cosXtˆ()sinZˆ(t)Xˆ(t)cosXˆˆˆ(t)sinX(t)sinXt()cosY(t)Z(t)cos2ftZˆ(t)sin2ft00R()EY(t)Yt()YR()cos2fRfˆ()sin2ZZ0022R()RR()cos()sinZXXEX(t)Xtˆˆ()cossinEX(t)Xt()cossinR()X4P(f)Pf()ZXP(f)FR()YYP(ff)P()ff(ff)()ffZZ0000jsgn(f)Pf()Z22jP()ffP()ffZZ001sgn()fff1sgn()f0022P(ff

16、)fffBZ00P(ff)BfffZ0000fff00PX(f)A-B0BfPY(f)A-f0-B-f00f0f0+B2-37X(t)=A+BtAB[