北邮通信原理课后习题答案汇总

2第三章13456789第4章〔2〕s(t)=m(t)c(t)=sin(2pfmt)Acsin(2pfct)fc-ft-cos2p(fc+fm)t]S(f)={d[f+(fc-fd[f-(fc-f4-Ac{d[f+(fc+fm)]+d[f-(fc+fm)]}4r(t)理想低通滤波器输出y0(t)Cos(Wct)与发端相干解调相干解调：将接收信号与载波信号sin(2pfct)相乘，得到r(t)sin(2pfct)=Acm(t)sin(2pfct)sin(2pfct)=m(t)[1-cos(4pfct)]通过低通滤波器抑制载频的二倍频分量，得到解调信号为2解：(1)s(t)=cos(2p´104t)+4cos(2p´1.1´104t)+cos(2p´1.2´104t)=4cos(2p´1.1´104t)[1+0.5cos(2p´0.1´104t)](2)S(f)=[d(ff-104)]+2[d(ff-1.1´104)]d(f+1.2´104)+d(f-1.2´104)]S(f)5/223/21-12000-1000001000012000f(Hz)-11000(3)包络检波器(2)理想低通滤波器载波提取电路wctw通过低通滤波器抑制载频的二倍频分量，得到解调信号(3)如上图：在信号上加上导频，在接收时就可以提取导频作为解调波fc+fm)t]S(f)=d(f+fc+fm)+d(f-fc-fS(f)S(f)21-(fc+fm)fc+fm0f(2)调制方式为上边带调制。(3)r(t)理想低通滤波器cos(2*pi*ft)载波提取电路(1)m(t)=cos2000pt+2sin2000pt(t)=sin2000pt-2cos2000pty(t)(2)调制信号：m(t)载波：c(t)=Accos(2p´8´105´t)s下(t)=m(t)cos(2p´8´105´t)+(t)sin(2p´8´105´t)=50(cos2000pt+2sin2000pt)cos(2p´8´105´t)+50(sin2000pt-2cos2000pt)sin(2p´8´105´t)=50[cos2000ptcos(2p´8´105´t)+sin2000ptsin(2p´8´105´t)]+100[sin2000ptcos(2p´8´105´t)-cos2000ptsin(2p´8´105´t)]=50[cos(2p´799´103t)-2sin(2p´799´103t)]S下(f)=50[2d(ff-79900)-jd(+jd(f-79900)]=50[(--j)d(f+79900)+(-+j)d(f-S下(f)=255[d(ff-79900)](3)S(f)0f(Hz)79900-799000f(Hz)799006(1)A:s上(t)=Acm(t)cos2pfct-Ac(t)sin2pfct=Acm(t)cos(2p´455´103´t)-Ac(t)sin(2p´455´103´t)sB(t)=Acm(t)cos(2pfct)cos(2pfct)-Ac(t)sin(2pfct)cos(2pfct)=2[Acm(t)+Acm(t)cos(4pfct)-Ac(t)sin(4pfct)]sD(t)=Acm(t)cos(2pfct)sin(2pfct)-Acmˆ(t)sin(2pfct)sin(2pfct)=2[Acm(t)sin(4pfct)+Acm(t)cos(4pfct)-F:G:sG(t)=Acm(t)(2)只需最末端的相加改为相减即可，如图：coswcoswctLPF震荡器GA-90度-FLPFDE希尔伯特滤波器sinwcts(t)+CB下(t)=Acm(t)cos2pfct+Ac(t)sin2pfctF:=Acm(t)cos(2p´455´103´t)+Ac(t)sin(2p´455´103´t)sB(t)=Acm(t)cos(2pfct)cos(2pfct)+Acmˆ(t)sin(2pfct)cos(2pfct)=2[Acm(t)+Acm(t)cos(4pfct)+Ac(t)sin(4pfct)]sD(t)=Acm(t)cos(2pfct)sin(2pfct)+Ac(t)sin(2pfct)sin(2pfct)=2[Acm(t)sin(4pfct)+Acm(t)-Acm(t)cos(4psG(t)=Acm(t)r(t)理想低通滤波器输出y0(t)Cos(Wct)与发端相干解调A2R(f)=-[M(f+fc)+M(f-fcA2f+fc)+d(f-fc)]A1Y1(f)=2[f+fc)+d(f-fc)]A1==-[M(f+2fc)H(f+fc)+M(f)H(f+fc)+M(f)H(f-f4+M(f-2fc)H(f-fA4通过低通滤波器后Y0(f)=-M(f)[H(f+fc)+H(f-fcA4由图可看出H(f+fc)+H(f-fc)在[]内为常数，所以不失真。H(f)W/2W0f-WW/2W0f(2)调制指数：bf=4fmax=bf´fm=4´100Hz=400Hz(4)调频信号带宽：B=2(1+bf)W=2´5´100Hz=1000Hz9解：(1)调制指数：bf=4，信号带宽：B=2(1+bf)fm=2´5´1000=10000Hz(2)调制指数假设加倍，那么bf=2信号带宽：B=2(1+bf)fm=2´3´2000=12000Hz(3)调制指数：bp=4，信号带宽：B=2(1+bf)fm=2´5´1000=10000Hz(4)调制指数：bp=Kpmaxm(t)，假设加倍，bp=4信号带宽：B=2(1+bf)fm=2´5´2000=20000Hz10解：(1)因为K>m(t)max，s(t)=Kcoswct[1+]，所以调制为有离散大载波的双边带调制。(2)s(t)=coswct-sinwctò-t¥m(t)dt，为窄带调制。(3)s下(t)=m(t)coswct+(t)sinwct，为下边带调制。(4)调制。(5)调制。11解：已调双边带调幅波()信号：s(t)=m(t)cos(2pfct)解调接收信号：r(t)=m(t)cos(2pfct)+nc(t)cos(2pfct)-ns(t)sin(2pfct)已调双边带调幅波()信号的功率：Ps=m(t)2=PM=´4W=2P0W那么PM=4P0W输入加性白高斯噪声的功：Pn=1nc(t)2+1ns(t)2=n(t)2=N04W=解调后信号：解调后信号功率：P0=1m(t)2=1PM=P0W44解调后噪声功率：1N0S0N0WN1N0S0N0W2r(t)=Ac[m(t)coswct-(t)sinwct]+nc(t)coswct-ns(t)sinwct解调输入信号的平均功率：P=Ac2PM由得，所以PR=4´10-4W解调输入噪声功率：Pn=2W=N0W=2´10-6W解调输出信号的平均功率：解调输出噪声功率：2AcS0N0WN2AcS0N0W4(2)解调输入信号：r(t)=Ac[1+amn(t)]coswct+nc(t)coswct-ns(t)sinwct-Ac22PAc222Ac2a2PMn由得，所以PR=4´10-4W解调输入噪声功率：Pn=2´´2W=2N0W=4´10-6W解调信号：s0(t)=1Acamn(t)+1nc(2Ac22P0=-m(2Ac2244解调输出噪声功率：2NP04aPMnAc2a2PMnS2NP04aPMnAc2a2PMnS0Pn0N0W2N0Wa2PMna2PMn2(1+aPMn)a2PMnPR1+a2PMnN0W1+a2PMnN0W因为PMn所以(N)a2PMnPRPRS)所以(N)a2PMnPRPRfm48KHzfm48KHz(2)调频解调器带通带通低通低通带通低通(3)鉴频器输出的噪声功率谱为=(F-1)T〔上P120〕为等效噪声系数T为热噪声源温度F为K。(2)馈线与前置射频放大器级联的噪声系数或等效噪声温度；〔上P121〕为级联等效噪声温度1为馈线的等效噪声温度2为前放的等效噪声温度Kpa1为馈线的增益系数对于无源对应的等效噪声系数(3)超外差接收机〔含天线、馈线〕的噪声单边功率普密度N0(W/Hz)。由NJ/K为波尔兹曼常数为天线等效噪声温度为馈线和前放级联的等效噪声温度在由条件和前面的结果N0=1.38´10-23J/K[55K+177.26K]=0.32´10-20W/Hz23456789235461解：由图可知y(t)=x(t)+x(t-t)，所以信道的冲击响应为h(t)=δ(t)+δ(t-t)，传输j负t，可知该信道为恒参信道。e-j(ej+e-j)=2cos，(2n间的距离为2πt。H(l[c-C2][-c-C2]c)+M(c)，因而输出为R(w)=S(w)H(w)=w+wc)e-j(w+w)tw-wc)e-j(w-w)t，r(t)=-t0)ejwt-t0)e-jwt=m(t-t0)coswct，比照输入s(t)=m(t)coswct，信号整体有失真，但包络m(t)无失真。1jwCjwC