通信原理教程第二版答案
通信原理教程第二版答案
《通信原理》习题第一章
第一章习题
习题1.1 在英文字母中E出现的概率最大，等于0.105，试求其信息量。
1解:E的信息量: I,log,,logPE,，，,log0.105,3.25bE222，，PE
习题1.2 某信息源由A，B，C，D四个符号组成，设每个符号独立出现，其出现的概率分别为1/4，1/4，3/16，5/16。试求该信息源中每个符号的信息量。
解:
11 I,log,,logP(A),,log,2bA222P(A)4
335 I,,log,2.415bI,,log,2.415bI,,log,1.678bB2C2D2161616
习题1.3 某信息源由A，B，C，D四个符号组成，这些符号分别用二进制码组00，01，10，11表示。若每个二进制码元用宽度为5ms的脉冲传输，试分别求出在下列条件下的平均信息速率。
(1) 这四个符号等概率出现; (2)这四个符号出现概率如习题
1.2所示。
解:(1)一个字母对应两个二进制脉冲，属于四进制符号，故一个字母的持续时间为2×5ms。传送字母的符号速率为
1 R,,100BdB,32，5，10
等概时的平均信息速率为
R,RlogM,Rlog4,200bsbB2B2
(2)平均信息量为
11316516 H,log4，log4，log，log,1.977比特符号222244163165
则平均信息速率为 R,RH,100，1.977,197.7bsbB
习题1.4 试问上题中的码元速率是多少,
11解: R,,,200 BdB,3T5*10B
习题1.5 设一个信息源由64个不同的符号组成，其中16个符号的出现概率均为1/32，其余48个符号出现的概率为1/96，若此信息源每秒发出1000个独立的符号，试求该信息源的平均信息速率。
解:该信息源的熵为
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《通信原理》习题第一章
M6411H(X),,P(x)logP(x),,P(x)logP(x),16*log32，
48*log96,ii,ii22223296i,i,11
=5.79比特/符号
因此，该信息源的平均信息速率。 RmH,,,1000*5.795790 b/sb
习题1.6 设一个信息源输出四进制等概率信号，其码元宽度为125 us。试求码元速率和信息速率。
11解: ,,,R8000 BdB,6T125*10B
等概时， R,RlogM,8000*log4,16kb/sbB22
习题1.7 设一台接收机输入电路的等效电阻为600欧姆，输入电路的带宽为6 MHZ，环境温度为23摄氏度，试求该电路产生的热噪声电压的有效值。
,,23612解: V44*1.38*10*23*600*6*104.57*10 V,,,kTRB
习题1.8 设一条无线链路采用视距传输方式通信，其收发天线的架设高度都等于80 m，试求其最远的通信距离。
26解:由，得 Drh,8Drh,,,88*6.37*10*8063849 km
第二章习题
习题2.1 设随机过程X(t)可以表示成:
Xttt()2cos(2), ,，,,,,,,,
式中，是一个离散随机变量，它具有如下概率分布:P(=0)=0.5，
P(=/2)=0.5 ,,,,试求E[X(t)]和。 R(0,1)X
,解:E[X(t)]=P(=0)2+P(=/2) cos(2),t2cos(2)=cos(2)sin2ttt，,,,,,,2 cos,t
习题2.2 设一个随机过程X(t)可以表示成:
Xttt()2cos(2), ,，,,,,,,,
判断它是功率信号还是能量信号,并求出其功率谱密度或能量谱密度。
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《通信原理》习题第一章
解:为功率信号。
1T/2RXtXtdt,,,，()lim()(),,,XTT/2,T 1T/2,，，，
ttdtlim2cos(2)*2cos2(),,,,,,,,,,TT/2,T
jtjt22,,, ,,，2cos(2),,ee
,,,,,jfjtjtjf2222,,,,,,PfRedeeed()()(),,，,,,,,,,X,,
,,，，(1)(1)ff,,
习题2.3 设有一信号可表示为:
4exp() ,t0,,t Xt(){,0, t<0
试问它是功率信号还是能量信号,并求出其功率谱密度或能量谱密度。解:它是能量信号。X(t)的傅立叶变换为:
4，,,，,,,，,,，jttjtjt,,,(1) Xxtedteedtedt,,,,,()()44,,00,,,，j1, 24162则能量谱密度 G(f)== Xf(),22,,，，114jf
习题2.4 X(t)=，它是一个随机过程，其中和是相互统
xtxtcos2sin2,,,xx1212
2计独立的高斯随机变量，数学期望均为0，方差均为。试求: ,
2(1)E[X(t)]，E[];(2)X(t) 的概率分布密度;(3) Xt()Rtt(,)X12
解:(1) ,,,,，，,,，，EXt,Excos2,t,xsin2,t,cos2,t,Ex,sin2,t,Ex,01212因为相互独立，所以。 ,,,,,,Pf()x和xExx,Ex,Ex121212X
222222又因为，，所以。 ,,,,,,,,,,Ex,Ex,0,,,,Ex,ExEx,Ex,,121112
22222故 ,, ，，，，EXt,cos2,t，sin2,t,,,
(2)因为服从高斯分布，的线性组合，所以也服从高斯分，，，，x和xXt是x 和xXt1212
2,,1z,,布，其概率分布函数。 exp,,，，px2,,2,2,,,,
(3) ，，，，，，,,,,，，
Rt,t,EXtXt,E(xcos2,t,xsin2,t)xcos2,t,xsin2,t121211211222X
2 ,,,,cos2,tcos2,t，sin2,tsin2,t1212
2 ，， ,,cos2,t,t21
习题2.5 试判断下列函数中哪些满足功率谱密度的条件:
22(1)，，; (2)，，; (3)，， a，,f,a,f，cos2,fexpa,f
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《通信原理》习题第一章
解:根据功率谱密度P(f)的性质:?P(f)，非负性;?P(-f)=P(f) ，偶函数。,0可以判断(1)和(3)满足功率谱密度的条件，(2)不满足。
习题2.6 试求X(t)=A的自相关函数，并根据其自相关函数求出其功率。cos,t
解:R(t，t+)=E[X(t)X(t+)] = EAtAtcos*cos(),,,，,,,,
21A2 ,，，,,,,,,,,,AEtRcoscos(2)cos(),,22
2A功率P=R(0)= 2
习题2.7 设和是两个统计独立的平稳随机过程，其自相关函数分别，，，，XtXt12
为。试求其乘积X(t)=的自相关函数。，，，，R,和R,XtXt()()XX1212
解:(t,t+)=E[X(t)X(t+)]=E[] XtXtXtXt()()()()，，,,1212
== EXtXtEXtXt()()()()，，,,RR,,,,XX()(),,112212
习题2.8 设随机过程X(t)=m(t)，其中m(t)是广义平稳随机过程，且其自cos,t
相关函数为
,42,10,10 kHZ10 kHZff,,,Pf(),,X 0,其它,
(1)试画出自相关函数的曲线;(2)试求出X(t)的功率谱密度和功率P。
R(),Pf()XX
1, 10，,,,,,,
,R,,,,101解:(1) ，，,,,,x
,0,其它,
其波形如图2-1所示。
，，R,x
12
,0 ,1 1
图2-1信号波形图
(2)因为广义平稳，所以其功率谱密度，，，，。由图2-8可见，，，
X(t)P,,R,R,XXX
4
《通信原理》习题第一章的波形可视为一个余弦函数与一个三角波的乘积，因此
,11,,2,,,,,,,,，，,，,,，，，，，，,,Sa，
1P,,x00,222,, ，，,，,,,,1,,,,2200,Sa，Sa,,,,,,422,,,,，，
,111,d,,,0, ，，，，PP,,或SRxx,,,222,
sin,f习题2.9设信号x(t)的傅立叶变换为X(f) =。试求此信号的自相关函数f,
。
2sin,f2解:x(t)的能量谱密度为G(f)== Xf()f,
1, 10，,,,,,,
,，,jf2,,RGfedf,,,,,()101其自相关函数，，,,,,X,,,,其它0,,
k-k,习题2.10 已知噪声的自相关函数，k为常数。，，，，ntR,,en2
(1)试求其功率谱密度函数和功率P;(2)画出和的曲线。，，，，，，
R,PfPfnnn
2kk,k,，,,，,,jj,,,,解:(1) ()(),,,,,,PfRedeednn,,,,,,222(2)，,kf
，， P,R0,k2n
(2)和的曲线如图2-2所示。，，R(),Pfnn
，，，，R,Pfnn 1
k2
0 f0 ,图2-2
习题2.11 已知一平稳随机过程X(t)的自相关函数是以2为周期的周期性函数: R()1, 11,,,,,,,,
试求X(t)的功率谱密度并画出其曲线。 Pf()X
解:详见例2-12
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《通信原理》习题第一章
习题2.12 已知一信号x(t)的双边功率谱密度为
,42,10,10 kHZ10 kHZff,,, Pf(),,X0,其它,
试求其平均功率。
334f2，,,10*10424108解: ,,,,PPfdffdf()2102*10**10X,,00,,33
,t/,,et,0,习题2.13 设输入信号，将它加到由电阻R和电容C组成的高
xt(),,0,0t,,
通滤波器(见图2-3)上，RC,。试求其输出信号y(t)的能量谱密度。
解:高通滤波器的系统函数为
H(f)= Xttt()2cos(2), ,，,,,,,,,
输入信号的傅里叶变换为
1,X(f)= ,C 1，12jf,,，jf2,,R ,
输出信号y(t)的能量谱密度为
R,22图2-3RC 高通滤波器 ()()()(),,,GfYfXfHfy11()(1)，，R22jfCjf,,, 习题2.14 设有一周期信号x(t)加于一个线性系统的输入端，得到的输出信号为y(t)=式中，为常数。试求该线性系统的传输函数H(f). ,dxtdt()/,,, 解:输出信号的傅里叶变换为Y(f)=,所以H(f)=Y(f)/X(f)=j ,,*2*()jfXf2,,f 习题2.15 设有一个RC低通滤波器如图2-7所示。当输入一个均值为0、双边n0功率谱密度为的白噪声时，试求输出功率谱密度和自相关函数。 2
解:参考例2-10
习题2.16 设有一个LC低通滤波器如图2-4所示。若输入信号是一个均值为0、
n0双边功率谱密度为的高斯白噪声时，试求 2
(1) 输出噪声的自相关函数。(2)输出噪声的方差。
解:(1)LC低通滤波器的系统函数为 L
C
图2-4LC低通滤波器
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《通信原理》习题第一章
2
1jfC2,,H(f)= 22214,fLC,jfL2，,jfC2,
n120 输出过程的功率谱密度为,,,,,PPH()()()i02,,21LC
CnC0对功率谱密度做傅立叶反变换，可得自相关函数为 R()exp(),,,,04LL(2) 输出亦是高斯过程，因此
Cn20 ,,,,,,RRR(0)()(0)0004L
习题2.17若通过图2-7中的滤波器的是高斯白噪声，当输入一个均值为0、双边
n0功率谱密度为的白噪声时，试求输出噪声的概率密度。 2
解:高斯白噪声通过低通滤波器，输出信号仍然是高斯过程。由2.15题可知n20E(y(t))=0 , ,,,(0)Ry04RC
所以输出噪声的概率密度函数
212xRC ,,px()exp()yn,n00
2RC
第三章习题习题3.1 设一个载波的表达式为，基带调制信号的表达式
为:ctt()5cos1000,,
m(t)=1+。试求出振幅调制时已调信号的频谱，并画出此频谱图。 cos200,t 解: ，，，，，，，，，，st,mtct,1，cos200,t5cos1000,t
,,,,5cos1000t，5cos200tcos1000t
5，，,5cos1000,t，cos1200,t，cos800,t2
由傅里叶变换得
55,,,,Sf,f，500，f,500，f，600，f,600，，，，，，，，，，，,,,,24 5,,，，，，,f，400，,f,4004
已调信号的频谱如图3-1所示。
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《通信原理》习题第一章
S(f)
52
54
0 400500600 ,600,500,400
图3-1 习题3.1图
习题3.2 在上题中，已调信号的载波分量和各边带分量的振幅分别等于多少, 解:由上题知，已调信号的载波分量的振幅为5/2，上、下边带的振幅均为5/4。
习题3.3 设一个频率调制信号的载频等于10kHZ，基带调制信号是频率为2 kHZ
的单一正弦波，调制频移等于5kHZ。试求其调制指数和已调信号带宽。解:由题意，已知=2kHZ，=5kHZ，则调制指数为 f,fm
,f5 m,,,2.5ff2m
已调信号带宽为 Bff,,，,，,2()2(52)14 kHZm
习题3.4 试证明:若用一基带余弦波去调幅，则调幅信号的两个边带的功率之和最大等于载波频率的一半。
'证明:设基带调制信号为，载波为c(t)=A，则经调幅后，有 cos,tmt()0 '，， stmtAt()1()cos,，,AM0，，
22'22，，已调信号的频率 PstmtAt,,，()1()cos,AMAM0，，
22'222'22 AtmtAtmtAtcos()cos2()cos,,,，，000
Bmf,，2(1)fm因为调制信号为余弦波，设，故
,,,f1000 kHZ100
2m1''2 mtmt()0, (),,,22
2A22则:载波频率为 ,,cos,PAtc02
'222mtAA()'222边带频率为 ,,,,PmtAt()coss024
P1s因此。即调幅信号的两个边带的功率之和最大等于载波频率的一半。 ,P2c 习题3.5 试证明;若两个时间函数为相乘关系，即z(t)=x(t)y(t)，其傅立叶
变换
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《通信原理》习题第一章
为卷积关系:Z()=X()*Y()。
证明:根据傅立叶变换关系，有
，,，,11，，1j,t,
FX,Y,XuY,uu,,，，，，，，，，,,,ded,,,,,,,,,22,,，，
，,，,11-1j,t，，变换积分顺序:,,,,,，，，，，，，，
FX,Y,XuY,ud,eu,,,,,,,,，，22,,
，,，,11jutj,t，， ,，，，，XueY,ed,du,,,,,,,,，，22,,
，,1jut,Xuytu，，，，ed,,, ,2
，，，，,xtyt
-1又因为，，，，，，，，,,zt,xtyt,FZ,
,1-1则 ,,，，,,，，，，FZ,,FX,,Y,
即，，，，，，Z,,X,,Y,
习题3.6 设一基带调制信号为正弦波，其频率等于10kHZ，振幅等于1V。它对频率为10mHZ的载波进行相位调制，最大调制相移为10rad。试计算次相位调制信号的近似带宽。若现在调制信号的频率变为5kHZ，试求其带宽。解:由题意，最大相移为 f,,10 kHZ , A1 V,,10 radmmmax
瞬时相位偏移为，则。 ,()()tkmt,k,10pp
dt,()瞬时角频率偏移为d则最大角频偏。 ,,,,k,ktsin,,pmpmmdt
因为相位调制和频率调制的本质是一致的，根据对频率调制的分析，可得调制指
k,,,pm数 10,,,,mkfp,,mm
因此，此相位调制信号的近似带宽为
Bmf,，,，,2(1)2(110)*10220 kHZfm
若=5kHZ，则带宽为 fm
Bmf,，,，,2(1)2(110)*5110 kHZfm
习题3.7 若用上题中的调制信号对该载波进行频率调制，并且最大调制频移为1mHZ。试求此频率调制信号的近似带宽。
,f解:由题意，最大调制频移，则调制指数 ,,f1000 kHZ1000/10100m,,,ffm 故此频率调制信号的近似带宽为
63 sttt()10cos(2*1010cos2*10),，,,
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《通信原理》习题第一章
63习题3.8设角度调制信号的表达式为。试求:
sttt()10cos(2*1010cos2*10),，,,
(1)已调信号的最大频移;(2)已调信号的最大相移;(3)已调信号的带
宽。
解:(1)该角波的瞬时角频率为
6 ,,,,()2*102000sin2000tt,，
2000, 故最大频偏 ,,,f10*10 kHZ2,
3,f10(2)调频指数 m,,,10*10f3f10m
故已调信号的最大相移。 ,,,10 rad
(3)因为FM波与PM波的带宽形式相同，即，所以已调信号Bmf,，2(1)FMfm
的带宽为
3 B=2(10+1)* 1022 kHZ,
第四章习题
,1习题4.1 试证明式，，，，。 ,f,,f,nf,s,Tn,,,
,证明:因为周期性单位冲激脉冲信号，周期为，其傅里
叶,,()()ttnT,,T,Tssn,,,
,
变换 ,,,()2(),,,,Ftn,s,nn,,,
T211s,,jnts而 Ftdt,,,()n,,T2sTTsS
,2,所以 ,,,n()(),,,,,s,Tn,,,s
,1即 ,,,fnf,,()(),s,Tn,,,s
习题4.2 若语音信号的带宽在300,400Hz之间，试按照奈奎斯特准则计算理论上信号不失真的最小抽样频率。
HzHz解:由题意，=3400,=,故语音信号的带宽为 ff300HL
BHz =3400-300= 3100
3Hz =3400=+= f，13100，3100nB，kBH31
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《通信原理》习题第一章
即=1,=。 331nk
根据带通信号的抽样定理，理论上信号不失真的最小抽样频率为
k3 ==2(+)= ，，Hz1f2B(1，)31006800s31n
习题4.3 若信号。试问: sttt()sin(314)314,
(1) 最小抽样频率为多少才能保证其无失真地恢复,
(2) 在用最小抽样频率对其抽样时，为保存3min的抽样，需要保
存多少个抽样值,
解:，其对应的傅里叶变换为 sttt()sin(314)314,
,314, ,314,, S(),,,0, 其他,
信号和对应的频谱如图4-1所示。所以 st()S(),f,,2,,3142,,50 HzHH
根据低通信号的抽样定理，最小频率为，即每秒采100f,2f,2，50,100 HzsH 个抽样点，所以3min共有:100，，360=18000个抽样值。
习题4.4 设被抽样的语音信号的带宽限制在300,3400Hz，抽样频率等于8000。试画出已抽样语音信号的频谱，并在图上注明各频率点的坐标值。 Hz 解:已抽样语音信号的频谱如图4-2所示。
st() S(,)
314,,314,,3140,314t
(a) (b)
图4-1习题4.3图
S(f)
1648,12,1612,4,8
,19.4 ,16.3 -15.7 -12.6 -11.4 -8.3 -7.7 -4.6 -3.4 -0.3 0 0.3 3.4 4.6 7.7 8.3 11.4 12.6 15.7 16.3 19.4f(kHz)
图4-2 习题4.4图
习题4.5 设有一个均匀量化器，它具有256个量化电平，试问其输出信号量噪比等于多少分贝,
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《通信原理》习题第一章
解:由题意M=256，根据均匀量化量噪比公式得
，， SN,20lgM,20lg256,48.16dBqqdB
习题4.6 试比较非均匀量化的A律和律的优缺点。 ,
答:对非均匀量化:A律中，A=87.6;律中，A=94.18。一般地，当A越大时，, 在大电压段曲线的斜率越小，信号量噪比越差。即对大信号而言，非均匀量化的律,的信号量噪比比A律稍差;而对小信号而言，非均匀量化的律的信号量噪比比A,律稍好。
习题4.7 在A律PCM语音通信系统中，试写出当归一化输入信号抽样值等于0.3时，输出的二进制码组。
解:信号抽样值等于0.3，所以极性码=1。 c1
查表可得0.3,(，)，所以0.3的段号为7，段落码为110，故=110。
13.9311.98ccc234
(11.9813.93),1第7段内的动态范围为:,，该段内量化码为,则n6416
11+=0.3，可求得,3.2，所以量化值取3。故=0011。 ccccnn，56783.9364 所以输出的二进制码组为11100011。
第五章习题
习题5.1 若消息码序列为1101001000001，试求出AMI和码的相应序列。HDB3
解: AMI 码为，1,10，100,100000，1
码为 HDB3，1,10，100,1000,10，1
习题5.2 试画出AMI码接收机的原理方框图。
解:如图5-20所示。
r(t)全波整流采样判决 ak
T
图5-1 习题5.2图
P习题5.3 设和是随机二进制序列的码元波形。它们的出现概率分别是和
g(t)g(t)12
1。试证明:若，式中，为常数，且，则此序列中将
(1,P)P,,kk0,k,1[1,g(t)/g(t)]12
无离散谱。
12
《通信原理》习题第一章
1证明:若，与t无关，且，则有 P,,k0,k,11,g(t)/g(t)12
[g(t),g(t)]21 P,1g(t)2
即 Pg(t),Pg(t),g(t),(P,1)g(t)1222
Pg(t)，(1,P)g(t),012
所以稳态波为 v(t),Pg(t,nT)，(1,P)g(t,nT),,1s2s
,[Pg(t,nT)，(1,P)g(t,nT)],0,1s2s即。所以无离散谱。得证～ P(w),0v W1习题5.4 试证明式。，，，，，，，，ht,,4sin2,WtHf，Wsin2,ftdf11,0 ,j2,ft证明:由于，由欧拉公式可得 h(t),H(f)edf11,,,
,,,h(t),H(f)(cos2ft，jsin2ft)df11,,, ,,,H(f)cos2,ftdf，
jH(f)sin2,ftdf11,,,,,,
由于为实偶函数，因此上式第二项为0，且 H(f)1
, h(t),2H(f)cos(2,ft)df11,,,
'令，，代入上式得 f,f，W,df,df'
,,h(t),2H(f'，W)cos[2(f'，W)t]df'11,W, ,,,2H(f，W)cos,2ftcos,2Wtdf，2H(f，W)sin2,ftsin2,Wtdf11,,WW,,
由于单边为奇对称，故上式第一项为0，因此 H(f)1
,,,h(t),2sin2WH(f，W)sin2fttdf11,,W W,4sin2,WH(f，W)sin2,fttdf1,0 习题5.5 设一个二进制单极性基带信号序列中的“1”和“0”分别用脉冲[见图5-2的g(t)
T有无表示，并且它们出现的概率相等，码元持续时间等于。试求:
(1) 该序列的功率谱密度的表达式，并画出其曲线;
(2) 该序列中有没有概率的离散分量,若有，试计算其功率。 f,1T
g(t)解:
A
tTOT 13
《通信原理》习题第一章
图5-2 习题5.5图1 (1)由图5-21得
,2T,,A1,t,t,,,,g(t),T2 ,,,
,0 其他,
ATwT,,2的频谱函数为: (),g(t)GwSa,,24,,由题意，，且有=,=0，所以，，，，P0,P1,P,1/2g(t)g(t)g(t)12
。将其代入二进制数字基带信号的双边功率谱密度函数的表达式中，
G(t),G(f),G(f),012
可得
2,，，
11mmm2,,,,,,,P(f)P(1P)G(f)G(f)PG(1P)Gf,,,，，,,,,,,,,,1212s,,TTTTT,,,,,,，，,,
2,11mm2,,,,,P(1P)G(f)(1P)Gf,,，,,,,,,,TTTT,,,,,,
222,1ATwT1mm,,,,,,4,,Sa，Gf,,,,,,,,4T442TTT,,,,,,,,
22,,ATwTAmm,,,,,,44SaSa,f,，,,,,,,,,164162T,,,,,,,,
曲线如图5-3所示。
P(f)s2 Av16
2 AT
16
51324 fOTTTTT图5.3 习题5.5 图2 (2)二进制数字基带信号的离散谱分量为2,Amm,,,,,4P(w)Saf ,,,,,,,,v162T,,,,,,
当m=?1时，f=?1/T，代入上式得
22,,AA11,,,,,,,,44 Pw,Sa,f，，Sa,f,(),,,,,,,,vTT162162,,,,,,,,因为该二进制数字基带信号中存在f=1/T的离散谱分量，所以能从该数字基带信号中提取码
14
《通信原理》习题第一章
元同步需要的f=1/T的频率分量。该频率分量的功率为
22222,,AAAA2A,,,,44 S,Sa，Sa,，,,,,,444,,,162162,,,,
习题5.6 设一个二进制双极性基带信号序列的码元波形为矩形脉冲，如图5-4所示，g(t)
3其高度等于1，持续时间，为码元宽度;且正极性脉冲出现的概率为，负极性脉冲Tτ =T/34
1。出现的概率为4
(1) 试写出该信号序列功率谱密度的表达式，并画出其曲线;
1(2) 该序列中是否存在的离散分量,若有，试计算其功率。 f,Tg(t)
1
,T/2,,/2,/2tT/20图5-4 习题5.6图
解:(1)基带脉冲波形可表示为: g(t)
,1 t,/2,(), gt,0 其他,
TTf,,,的傅里叶变化为: G(f),Sa(f),Sag(t),,,,,33,,
该二进制信号序列的功率谱密度为:
2,，，11mmm2,,,,,,,P(f),P(1,P)G(f),G(f)，PG，
(1,P)Gf,,,,,,,,1212,,TTTTT,,,,,,m,,,，，
,,31mm2,,,,2,,G(f)，Saf,,,,,,4T363T,,,,m,,,
曲线如图5-5所示。
P(f)
1/36
T/12
2/T3/T4/T5/T8/T9/T7/T01/Tf6/T
15
《通信原理》习题第一章
图5-5 习题5.6图 (2) 二进制数字基带信号的离散谱分量为
,1mm,,,,,2P(f)Saf ,,,,,,,,v363T,,,,m,,,
1当, 时，代入上式得 f,,m,,1T
,,1111,,,,,,,,22 ,,Pf,Saf,，Saf，(),,,,,,,,vTT363363,,,,,,,,因此，该序列中存在的离散分量。其功率为: f,1/T
22,,1sin/31sin/33,,,, P,，,,,,,v236,/336,/38,,,,,习题5.7 设一个基带传输系统接收滤波器的输出码元波形如图5-13所示。 h(t)
(1) 试求该基带传输系统的传输函数; H(f)
(2) 若其信道传输函数，且发送滤波器和接收滤波器的传输函C(f),1
数相同，即，试求此时和的表达式。 G(f),G(f)G(f)G(f)TRTR
,2T,,1-t t,,,T,,,g(t), T2解:(1)令，由图5-6可得=，因为的
gt,h(t)g(t),,,,,2,,,0 其他,
2TT,f,,2频谱函数，所以，系统的传输函数为 (),GfSa,,24,,
22,fT,fT,,jj2TTf,,,222()= H(f)GfeSae,,,24,,(2)系统的传输函数由发送滤波器、信道和接收滤波器三部分H(f)G(f)C(f)G(f)TR
组成，即=。因为，，则 H(f)C(f)G(f)G(f)C(f),1G(f),G(f)TRTR
22== H(f)G(f)G(f)TR
2,fT,j2TTf,,,4()所以 == G(f)G(f)HfSae,,,TR24,,
h(t) 1
tT/2OT图5-6 习题5.7图
习题5.8 设一个基带传输系统的传输函数如图5-7所示。 H(f)
(1) 试求该系统接收滤波器输出码元波形的表达式:
16
《通信原理》习题第一章
(2) 若其中基带信号的码元传输速率，试用奈奎斯特准则衡量R,2fB0
该系统能否保证无码间串扰传输。
H(f) 1
Offf00图5-7 习题5.8图
,1,/ f,fff00解:(1)由图5-25可得=。 H(f),0 其他 ,
,1,t/T, t,T2因为g(t),，所以。 G(f),TSa(,fT),0 其他,
2根据对称性:所以。 G(,f),g(jt),G(f),g(t),f,t,T,f,h(t),fSa(,ft)000
(2)当时，需要以为间隔对进行分段叠加，即分析在区间R,2fH(f)f,R,2fB0B0
叠加函数的特性。由于在区间，不是一个常数，所以有码间干扰。
[,ff][,ff]H(f)0,00,0
习题5.9 设一个二进制基带传输系统的传输函数为
,(1，cos2),,1/2,,f,f,000(), Hf,0 ,其他,
试确定该系统最高的码元传输速率及相应的码元持续时间T。 RB
解:的波形如图5-8所示。由图可知，为升余弦传输特性，根据奈奎斯特第一H(f)H(f)
准则，可等效为理想低通(矩形)特性(如图虚线所示)。等效矩形带宽为
111 W,，,122,4,00
1最高码元传输速率 2R,W,1B2,0
相应的码元间隔 T,1/R,2,SB0
H(f)2,0
,0
1/2,0,1/2,1/4,000图5-8 习题5.9图
17
《通信原理》习题第一章
习题5.10 若一个基带传输系统的传输函数和式(5.6-7)所示，式中。
H(f)W,W1
(1) 试证明其单位冲激响应，即接收滤波器输出码元波形为
1sint/Tcost/T,, h(t),22Tt/T1,4t/T,
1(2) 若用波特率的码元在此系统中传输，在抽样时刻上是否存在码间串T
视听说教程2网络答案
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新视野大学英语第二版第2册视听说教程答案
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新世纪视听说教程2答案听力原文
Directions: In this section, you will hear several conversations. At the end of each conversation, one or more questions will be asked about what was said. Both the conversations and the questions will be spoken only once. After each question there will be a pause. During the pause, you must read the four choices marked A), B), C) and D), and decide which is the best answer. A: How are you getting along, Julie B: After the vacation Much refreshed. A．It is interesting. B．It is tiring. C．It is exciting. D．It is refreshing. A: Why are you so cheerful while I'm so stressed all day B: Well, you should work off your stress. A．work not so hard B．work on easier problems C．eat healthy D．get rid of her stress I see you laughing all the time. B: Oh Eric. I'm ... reading a comic book. A: Comic book I didn't know you are interested in it. B: Well, it's a recent interest. When I was low in spirit three months ago, a friend gave me a comic book to read. And I became instantly addicted to such books. You know ... the pictures are so funny ... now I feel much better. A: I didn't know that. No wonder you are always in good spirits these days. B: Yes, and maybe you should read them, too, Eric. A: I ... I don't know. I'm just too busy with my thesis. It's for my master's degree and time is running out ... Ah, pretty much stress on me. B: Now you see Eric That's why you need to give yourself a break to read something amusing. It helps, I promise. A: Really Can reading comic books reduce your stress B: Of course. I'm a living example. Actually I'm having stress too ... you know, the tests, the job, things like that. But when reading the fun stuff, I just forgot all my worries. Later I find myself more powerful to deal with the issues in my life. A: That sounds magical. I'd like to try. Er ... what are you reading now B: Garfields, the cat. It's good. There's another one and you can borrow it. A: Oh, thank you, Penny. You're very helpful. Questions 3 to 5 are based on the conversation you have just heard. months ago, a friend gave Penny _____ when she was in low spirits. （岭师分享群4发布） A．a flower B．a comic book C．a comic DVD
通信原理教程(第二版)徐家恺 第三部分
3-1、设一恒参信道的幅频特性和相频特性分别为0)(K w H =，d wt w ?=)(?，其中，K 都是常数。试确定信号通过该信道后输出信号的时域表示式，并讨论之。 d t ,0)(t s 解： d jwt e K w H ?=0)()()()()()()(00d o jw O t t s K t s w S e K w S w H w S d t ?=?==? 确定信号通过该信道后，没有失真，只是信号发生了延时。 )(t s 3-2、设某恒参信道的幅频特性为，其中，t 都是常数。试确定 信号s 通过该信道后输出信号的时域表示式，并讨论之。 d jwt e T w H ?+=]cos 1[)(0d )(t 解： d jwt e T w H ?+=]cos 1[)()(]2 121[)(]cos 1[)()()()()(000w S e e e w S e T w S w H w S T t jw t T jw jwt jw O d d d d t ??+???++=+== )(2 1)(21)(00T t t s T t t s t t s d d d +?+??+?? 信号经过三条延时不同的路径传播，同时会产生频率选择性衰落。见教材第50页。 3-3、设某恒参信道可用下图所示的线形二端对网络来等效。试求它的传递函数，并说明信 号通过该信道时会产生哪些失真？ 解：jwRc jwRc jwc R R w H +=+=11 )( )()(1)(w j e w H jwRc jwRc w H ?=+= 其中 =)(w H 1)(11 2+wRc )(2)(wRc arctg w ?=π? 则群迟延2) (1)()(wRc Rc dw w d w +==?τ 可见，信号通过该信道时会频率失真和群迟延畸变。
新标准大学英语2视听说教程答案
Unit 1 Outside view Activity 1 3.5 Activity 2 one of the best universities most talented students well-known around the world have open doors good social life you want it to be on another campus it's a fun place go to concerts during the week Activity 3 library system online three / 3 four / 4 Brian leisure purposes the libraries listening in Passage 1 1.2 2. 6 3. 1 4. 5 5. 3
6. 4 2.Listen to Passage 1 again and rearrange the answers in the right column to match those questions in the left column. Ceahbgdf Passage 2 OC/O/C/O/C/C/O/C/O/OC 2.Listen to Passage 2 again and choose the best way to complete the sentences BDDBA Test CCBBD CDBCB C Unie 2 Outside view Activity 1 1.Watch the video clip and match the speakers with the statements. K/K/T/S/T/K/S/K Activity 2 2.3.5 Activity 3 regarded along communicating But If during how as plays less LISTENING IN
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Unit1 Listening skills BBDCA Listening in Task1 FTFTF & Listening in Task2 BBDAC Listening in Task3 " classical peaceful relaxing Jazz sadness 】 heavy metal energy sporting events
physical labor road accidents / Let's talk > Task 1 Good Morning to All success musical talents without % second part replaced legal action real owners · Further listening and speaking > Further listening > Task 1 special joke talent proud loud < joy
dancer talk wondered [ capture fan Further listening and speaking > Further listening > Task2 BAADC Further listening and speaking > Further listening > Task 3 TFTTT ！ Unit 1 test . Golden oldies . She is not familiar with the music they are listening to . Fast music can give people energy according to the man . They both think the tickets are expensive . The woman probably moved in not long ago band anymore - longer
通信原理教程+樊昌信+习题答案第四章
第四章习题 习题4.1 试证明式()()∑∞ -∞ =Ω-=?n nf f T f s 1δ。 证明：因为周期性单位冲激脉冲信号()()T s n t t nT δδ∞ =-∞ = -∑，周期为s T ，其傅里叶 变换 ()2() n s n F t n ωπδω∞ Ω=-∞ ?=-∑ 而 2 2 1 1()s s s T jn t n T s S F t dt T T ωδ--= = ? 所以 2()()s n s n T π ωδωω∞ Ω=-∞ ?= -∑ 即 1 ()() s n s f n f T δ ω∞ Ω=-∞ ?= -∑ 习题4.2 若语音信号的带宽在300～400Hz 之间，试按照奈奎斯特准则计算理论上信号不失真的最小抽样频率。 解：由题意，H f =3400Hz ,L f =300Hz ,故语音信号的带宽为 B =3400-300=3100Hz H f =3400Hz =13100?+3 31?3100=kB nB + 即n =1,k =3。 根据带通信号的抽样定理，理论上信号不失真的最小抽样频率为 s f =)1(2n k B +=2?3100?（1+331 ）=6800Hz 习题4.3 若信号()sin(314)314s t t t =。试问： （1） 最小抽样频率为多少才能保证其无失真地恢复？ （2） 在用最小抽样频率对其抽样时，为保存3min 的抽样，需要保 存多少个抽样值？ 解：()sin(314)s t t t =，其对应的傅里叶变换为 ()S ω=???≤其他 ,0314 ,314ωπ 信号()s t 和对应的频谱()S ω如图4-1所示。所以Hz 5023142H H ===ππωf 根据低通信号的抽样定理，最小频率为Hz 1005022H s =?==f f ，即每秒采100个抽样点，所以3min 共有：100?3?60=18000个抽样值。
新视野大学英语2视听说教程答案
新视野大学英语2视听说教程答案 【篇一：新视野大学英语视听说教程2(第二版)答案(带 unit test)】 st) unit1 listening skills bbdcalistening in task1 ftftf listening in task2 bbdac listening in task3classical，peaceful，relaxing，jazz，sadness，heavy metal，energy，sporting events，physical labor，road accidents lets talk task 1 good morning to all，success，musical talents，without，second part，replaced，legal action，real owners further listening and speaking further listening task 1 special，joke，talent，proud，loud，joy，honesty，dancer，talk，wondered，capture，fan task2baadc task 3 tfttt unit 1 test 1-5 ccdcb 1.favorite band ， 2.not anymore ， 3.no longer ， 4.a big fan ， 5.collected ， 6.the ones ， 7.going crazy ， 8.maybe to you ， 9.pressure ，10.fall in love ，11.get it ，12.from time to time ， 13.go on ，14.music video 1-5bcdda 1-5cdaba 6-10 cdcbc unit2 listening skills: making inferences adbcb listening in task 1 director，sound effects， good story，think， sad，violent， light-hearted movies， mystery movie， in the futuretask 2 dcdab task3film,directors,actors,important,the united kingdom,viewers,40 million,olympic,world cup,time,artistic value,depth,best picture,favor,science fiction lets talk task 1 abcdd further listening and speaking further listening task 1 memorizing lines，remember one line，i hear the guns roar，a loud boom，forgot his linetask 2 accadtask 3 ftftf unit 2 test 1- 5 cbabd 1.a very good 2.starring 3.best-seller 4.a new classic 5.violent 6.somewhat 7.before 8.the calmness 9.death 10.thought 1-5 dbdca 1-5 ccbab 6-10 ccbac
通信原理教程+樊昌信+习题答案第二章
第二章习题 习题2.1 设随机过程X (t )可以表示成： ()2cos(2), X t t t πθ=+-∞<<∞ 式中，θ是一个离散随机变量，它具有如下概率分布：P (θ=0)=0.5，P (θ=π/2)=0.5 试求E [X (t )]和X R (0,1)。 解：E [X (t )]=P (θ=0)2cos(2)t π+P (θ=/2)2cos(2)=cos(2)sin 22 t t t π πππ+ - cos t ω 习题2.2 设一个随机过程X (t )可以表示成： ()2cos(2), X t t t πθ=+-∞<<∞ 判断它是功率信号还是能量信号？并求出其功率谱密度或能量谱密度。 解：为功率信号。 []/2/2 /2/2 1()lim ()()1 lim 2cos(2)*2cos 2()T X T T T T T R X t X t dt T t t dt T ττπθπτθ→∞-→∞ -=+=+++? ? 222cos(2)j t j t e e πππτ-==+ 2222()()()(1)(1) j f j t j t j f X P f R e d e e e d f f πτπππττττδδ∞-∞---∞-∞==+=-++?? 习题2.3 设有一信号可表示为： 4exp() ,t 0 (){ 0, t<0 t X t -≥= 试问它是功率信号还是能量信号？并求出其功率谱密度或能量谱密度。 解：它是能量信号。X (t )的傅立叶变换为： (1)004 ()()441j t t j t j t X x t e dt e e dt e dt j ωωωωω +∞-+∞--+∞-+-∞====+??? 则能量谱密度 G(f)=2 ()X f =2 22 416 114j f ωπ=++ 习题2.4 X (t )=12cos2sin 2x t x t ππ-，它是一个随机过程，其中1x 和2x 是相互统计独立的高斯随机变量，数学期望均为0，方差均为2σ。试求： (1)E [X (t )]，E [2()X t ]；(2)X (t ) 的概率分布密度；(3)12(,)X R t t 解：(1)()[][]()[]02sin 2cos 2sin 2cos 2121=?-?=-=x E t x E t t x t x E t X E ππππ ()X P f 因为21x x 和相互独立，所以[][][]2121x E x E x x E ?=。
新视野大学英语视听说教程第二版第2册答案(全新完整版)
Unit1 II. Listening Skills 1. M: Why don’t we go to the concert today? W: I’ll go get the keys. Q: What does the woman imply? 2. W: I can’t find my purse anywhere. The opera tickets are in it. M: Have you checked in the car? Q: What does the man imply? 3. M: Are you going to buy that pirated CD? W: Do I look like a thief? Q: What does the woman imply? 4. M: Do you think the singer is pretty? W: Let’s just say that I wouldn’t/t vote for her in the local beauty contest. Q: What does the woman imply about the singer? 5. M: Have you seen Tom? I can’t find him anywhere. W: The light in his dorm was on just a few minutes ago. Q: What does the woman mean? 1.B 2.B 3.D 4.C 5.A III. Listening In Task 1: Encore! As soon as the singer completed the song, the audience cr ied, “Encore! Encore!” The singer was delighted and sang the song again. She couldn’t believe it when the audience shouted for her to sing it again. The cycle of shouts and songs was repeated ten more times. The singer was overjoyed with the response from the audience. She talked them and asked them why they were so much audience in hearing the same song again and again. One of the people in the audience replied, “We wanted you to improve it; now it is much better.” 1.F 2.T 3.F 4.T 5.F Task 2 What are your favorite songs? W: Hi, Bob! Do you want to go to a concert with me? M: A concert? What's it about?
通信原理教程 樊昌信 习题答案第三章
第三章习题 习题3.1 设一个载波的表达式为()5cos1000c t t π=，基带调制信号的表达式为：m(t)=1+cos 200t π。试求出振幅调制时已调信号的频谱，并画出此频谱图。 解： ()()()()()t t t c t m t s ππ1000cos 5200cos 1+== () t t t t t t ππππππ800c o s 1200c o s 2 5 1000c o s 51000c o s 200c o s 51000c o s 5++=+= 由傅里叶变换得 ()()()[]()()[]()()[]4004004 5 6006004550050025 -+++-+++-++= f f f f f f f S δδδδδδ 已调信号的频谱如图3-1所示。 图3-1 习题3.1图 习题3.2 在上题中，已调信号的载波分量和各边带分量的振幅分别等于多少？ 解：由上题知，已调信号的载波分量的振幅为5/2，上、下边带的振幅均为5/4。 习题3.3 设一个频率调制信号的载频等于10kHZ ，基带调制信号是频率为2 kHZ 的单一正弦波，调制频移等于5kHZ 。试求其调制指数和已调信号带宽。 解：由题意，已知m f =2kHZ ，f ?=5kHZ ，则调制指数为 5 2.52 f m f m f ?= == 已调信号带宽为 2()2(52)14 k m B f f =?+=+= 习题3.4 试证明：若用一基带余弦波去调幅，则调幅信号的两个边带的功率之和最大等于载波频率的一半。 证明：设基带调制信号为'()m t ，载波为c (t )=A 0cos t ω，则经调幅后，有 ' 0()1()cos AM s t m t A t ω??=+?? 已调信号的频率 2 2'22 0()1()cos AM AM P s t m t A t ω??==+??
新视野大学英语视听说教程2_听力答案
Unit 1 Roll over, Beethoven! II Listening skills 1-5: B B D C A III Listening in Task 1: 1-5: F T F T F Task 2: 1-5: B B D A C Task 3: 1-5: classical, peaceful, relaxing, Jazz, sadness 6-10: heavy metal, energy, sporting events, physical labor, road accidents V Let’s talk 1-8: Good Morning to All, success, musical talents, without, second part, replaced, legal action, real owners VI Further listening and speaking Task 1: 1-6: special, joke, talent, proud, loud, joy 7-12: honesty, dancer, talk, wondered, capture, fan Task 2: 1-5: B A A D C Task 3: 1-5: F F T T T Unit 2 What’s on at the cinema? II Listening skills 1-5: A D B C B III Listening in Task 1: 1-5: director, sound effects, good story, think, sad
新标准大学英语视听说教程第二版答案
UNIT1 Outside view Activity 1 Keys: 3-5-4-1-2 Activity 2 BCCAA Activity3 Activity4 1-H 2-C 3-B 4-F 5-D 6-A 7-E 8-G Activity5 不用写 News report Activity 1 AD Activity 2 DB Passage 1 Activity 1 CDAB Activity 2 1. (1) tolerant (2) rounded
2 (1) close to (2) stand up 3 (1) beauty (2) generosity. 4. (1) an opportunity (2) beyond the one 5 (1) behaved (2) selfish 6. (1) pretend (2)see through 7. (1) helping victims (2) impact 8. (1) difficult time (2) a huge amount of Passage2 Activity 1 1-D 2-B 3-E 4-A 5-C Activity 2 1. (1) strengths (2) weaknesses (3) personality(4) like (5) subjects (6) organized (7) confident (8) outgoing (9) a team (10) working alone 2. (1) future career (2) into a particular area of work (3) chosen field (4) related industries (5) leading 3.
新世纪大学英语系列教材视听说教程第二版听力答案.pdf
新世纪大学英语系列教材视听说教程4(第二版)听 力答案 新世纪大学英语视听说教程4 （第二版）答案unit 1 P2 a.8 b.2 c.1 d.4 e.1,4,5 f.6,7,8,9 g.1,3,6,7 h.1,4,5 P3 A. b B. F T F T T F T P8 OP1: A. 2 B. 1,2,5,7 OP2: A. the picture on the right P9 1.very famous buildings 2.made of glass, steel and concrete 3.designed 4.style
5.1998 6.452 meters high 7.both the modern and traditional side of my country P10 A: 1.world-famous museum, Paris, 500 years, six million 2.ancient capital, big enough, millions of, shooping center B: Louvre Museum 1989 an ugly, modern mistake Kyoto Japan Hiroshi Hara to cope with millions of visitors center P11 B: 3 C: 2,4,6,7 P13 A: TRUE:1,3,5 FALSE:2 halfway, not all the way 4 from far away, not close up B: thousands by car unimpressed i.6,7 brings new life into the city 百度搜索“就爱阅读”,专业资料,生活学习,尽在就爱阅读网https://www.360docs.net/doc/a19928569.html,,您的在线图书馆 P14