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学院_______________________ 系别____________ 班次_____________ 学号__________ 姓名________________………….……密…..……….封……..……线………..…以………..…内………....答…………...题…………..无…….….效…..………………..电子科技大学《通信原理》期末考试（A 卷）1. 填空:1) PCM 编码的三个基本步骤 是__抽样_, ___量化__, __编码___. 2) 通信系统包含哪三个子系统: __发射机__ , 信道 and _接收机__. 3) 设模拟信号为 )1000042sin()(t t t s ππ+⨯=, 其最小抽样频率为 _20kHz_.4) 若数据速率是 100kbps,单极性 NRZ 信号, 双极性 NRZ 信号及Manchester NRZ 信号 的第一零点带宽分别为__100kHz___, __100kHz _, ___200kHz __.5) 对于无 ISI 得情形, 若升余弦滚降滤波器的6dB 带宽, f 0, 为 50kHz, 则通信系统的符号速率是 _100k _ baud.6) 若 AM 信号表达式为)100002cos()]42cos(5.01[5)(t t t s ⨯⨯+=ππ, 则该信号的复包络是_)]42cos(5.01[5)(t t g ⨯+=π_.7) 对于非均匀量化两个主要的压扩律 是: __u __律, ___A__律. 8) 设输入信号为 0()[cos][()-(-)] s t t t U t U t T . 当输入噪声为白噪声, 匹配滤波器的冲击响应为h(t)=___)]()())[(cos()(00000T t t U t t U t t t t +-----ω ___，最佳采样时刻为 =0t __T __.9) 若调制信号的带宽为 50kHz, 则AM 、DSB-SC 、SSB 信号的传输带宽分别为 _100kHz__, __100kHz__, __200kHz__.10) 若 s(t)的PSD 为 []22222)4(12cos 16)(f T fT T A f b b b c g -=ππp ,其第一零点带宽是 2. 设()m t 是 单极性 NRZ 信号，数据速率为 1000 bits/s (其峰值为 1V ,二进制 1 码或 0码 发生的概率均为 1/2). OOK 信号表达式为 ()()cos c c s t A m t t ω=, 其中1c A V =. (a) 计算 OOK 信号的PSD. 设载波频率为 10 KHz. (b) 画出OOK 信号的 PSD 并给出第一零点带宽的值. (c) 计算 OOK 信号的频谱效率.Solution:(a) ()()cos c c s t A m t t ω= , m (t )为单极性 ()()c g t A m t =()sin ()()b b b b fT t f t F f T T fT ππ⎛⎫=↔= ⎪⎝⎭∏22:01,2210,2c n n kn For k A a a a and I +=⎧⎪⎪===⎨⎪⎪⎩依概率依概率222111(0)()0222cn n i i c i AR a a P A ===⨯+⨯=∑:0For k ≠,()[][][]24cn n k n n k AR k E a a E a E a ++===()22,02,04c c A k R k A k ⎧=⎪⎪=⎨⎪≠⎪⎩ ()()()222221()1()4111()()444sin sin sin c b g n bb c b bbA T n bP f f T T bA T T b b f f T bbfT fT fT fT fT fT δδδππππππ∞=-∞⎡⎤=+-⎢⎥⎣⎦⎡⎤=+=+⎢⎥⎣⎦⎛⎫∑⎪⎝⎭⎛⎫⎛⎫⎪ ⎪⎝⎭⎝⎭3311/1010b T R ms === ()()221()[()()]41()()16sin ()sin ()()()s g g c c b bc cP f P f f P f f c b c b T T f f f f c b c b f f T f f T f f T f f T δδππππ=-+--⎡⎤⎢⎥=++-++⎢⎥⎢⎥⎣⎦-+⎛⎫⎛⎫ ⎪ ⎪-+⎝⎭⎝⎭(b)2null B KHz =(c) 1/1//22null R Kb s b s Hz B KHz η===3. 某模拟信号，其带宽为3400Hz ，要在一个 PCM 系统中传输。
2019电⼦科技⼤学研究⽣试卷答案电⼦科技⼤学研究⽣试卷（考试时间：⾄，共 2 ⼩时）课程名称图论及应⽤教师学时 60 学分 3 教学⽅式堂上授课考核⽇期 2019 年 5 ⽉⽇成绩考核⽅式：（学⽣填写）⼀．填空题(每空3分，共15分) 1. 图G 的邻接矩阵为0111101111001100?? ? ? ? ? ???, 则G 的⽣成树的棵数为 8 . 2. 设1G 是11(,)n m 简单图,2G 是22(,)n m 简单图,则1G 和2G 的(Cartesian)积图12G G ?的边数()m G =1221n m n m +. 3. 图1中最⼩⽣成树T 的权值()W T = 23 .4. 图2中S 到T 的最短路的长度为 8 .5. 设G 是n 阶简单图,且不包含三⾓形,则其边数⼀定不超过24n . ⼆．单项选择题(每题3分，共15分) 学号姓名学院……………………密……………封……………线……………以……………内……………答……………题……………⽆……………效……………………座位号图1 图21. 关于彼得森(Petersen)图, 下⾯说法正确的是 ( B )A. 彼得森图是哈密尔顿图;B. 彼得森图是超哈密尔顿图;C. 彼得森图可1-因⼦分解;D. 彼得森图是可平⾯图.2. 下⾯说法正确的是 ( C )A. 有割点的三正则图⼀定没有完美匹配;B. 有割边的三正则图⼀定没有完美匹配;C. 存在哈密尔顿圈的三正则图必能1因⼦分解;D. 正则的哈密尔顿图必能2因⼦分解.3. 关于图的度序列, 下⾯说法正确的是 ( B )A. 任意两个有相同度序列的图都同构;B. 若图G 度弱于图H ，则图G 的边数⼩于等于图H 的边数;C. 若⾮负整数序列12(,,,)n d d d π=满⾜1ni i d =∑为偶数，则它⼀定是图序列;D. 如果图G 所有顶点的度和⼤于或等于图H 所有顶点的度和，则图G 度优于图H.4. 关于图的补图, 下⾯说法错误的是 ( A )A. 若图G 连通，则其补图必连通;B. 若图G 不连通，则其补图必连通;C. 图G 中的⼀个点独⽴集，在其补图中的点导出⼦图必为⼀个团;D. 存在5阶的⾃补图.5. 关于欧拉图, 下⾯说法正确的是 ( D )A. 每个欧拉图有唯⼀的欧拉环游;B. 每个顶点的度均为偶数的图是欧拉图;C. 欧拉图中⼀定没有割点;D. 欧拉图中⼀定没有割边.(三).(10分)若阶为25且边数为62的图G 的每个顶点的度只可能为3，4，5或6，且有两个度为4的顶点，11个度为6的顶点，求G 中5度顶点的个数。
习题1. 已知某2ASK 系统的码元速率为1000波特，所用载波信号为()6cos 410A t π⨯。
（1） 假定比特序列为{0110010}，试画出相应的2ASK 信号波形示意图； （2） 求2ASK 信号第一零点带宽。
解：由1000b s R R bps ==，6210cf Hz =⨯， 621020001000b c c b T f T R ⨯=== （1）一个码元周期内有2000个正弦周期：111{}n a ()2ASK s t 0（2）022000b B R Hz ==2.(mooc)某2ASK 系统的速率为2b R =Mbps ，接收机输入信号的振幅40μV A =，AWGN 信道的单边功率谱密度为180510N -=⨯W/Hz ，试求传输信号的带宽与系统的接收误码率。
解：传输信号的带宽24T b B R MHz ==，平均码元能量：24bb A T E =。
()()222100102cos 21cos 422,0,2bb T T b bc c b b b b A T A E A f t dt f t dt E E E E ππ==+=+==⎰⎰()2622618000401040444210510b b b E A T A N N R N --⨯====⨯⨯⨯⨯ 系统的接收误码率：（1） 若是非相干解调，非相干解调误码率公式，222200222BPFn b A A A R N B N σ===γ//24092( 5.11221 5.11.0306102)b E N e P e e e ----=≈==⨯γ4表 （2） 若是相干解调：由相干解调误码率公式得（最佳），1001.26981040b e E P Q Q N -⎛⎫===⎪⨯ ⎪ ⎝⎭也是MF 接收机的结果。
3. 某2FSK 发送“1”码时，信号为()()111sin s t A t ωθ=+，0s t T ≤≤；发送“0”码时，信号为()()000sin s t A t ωθ=+，0s t T ≤≤。
电子科技大学22春“电子信息工程”《通信原理》期末考试高频考点版（带答案）一.综合考核(共50题)1.编码信道包括()。
A.调制信道B.调制器C.编码器和译码器D.解调器参考答案：ABD2.已知两码组为(0000)，(1111)，若用于纠错，能纠正()位错。
A、1B、2C、3D、4参考答案：A3.下列用来实现载波同步的方法有()。
A.插入导频法B.非线性变换滤波法C.同相正交法D.起止式同步法参考答案：ABC4.模拟信号是利用()来实现其数字传输的。
A.A/DB.数字传输系统C.D/AD.检错5.当被抽样信号f(t)的最高频率为fmax时，则f(t)的全部信息都包含在其抽样间隔不大于1/2fmax秒的均匀抽样里。
()A.错误B.正确参考答案：B6.某分组码的最小码距是6，若用于检错，可保证检出()位错。
A、3B、4C、5D、6参考答案：C7.某m序列由n级移存器组成。
将此m序列延迟一位后同原序列模2相加，所得序列的周期是()。
A、2^n-1B、2^nC、2^n+1D、2^(n+1)参考答案：A8.若信息速率为10Mbps，则半占空的NZ码的主瓣带宽是()MHz。
A、6.25B、10MC、12.5MD、20M参考答案：D将N路频率范围为0.1~4KHz的语音信号用FDM方式进行传输，则采用AM调制方式时N路语音信号的最小传输带宽将分别为()。
A.4KHzB.4NKHzC.8KHzD.8NKHz参考答案：D10.若16进制ASK系统的信息传输速率为6400bps，试求该系统的码元传输速率。
()A.6400BB.800BC.3200BD.1600B参考答案：D11.某信源的速率是1kbps，用2PSK调制，其主瓣宽度是()。
A、1kHzB、2kHzC、4kHzD、8kHz参考答案：B12.在数字通信中，同步分为()。
A.载波同步B.位同步C.帧同步D.调频信号的同步参考答案：ABCD13.A.AMB.FMBD.LSB参考答案：C14.随参信道对所传信号的影响有()。
学院_______________________ 系别____________ 班次_____________ 学号__________ 姓名________________………….……密…..……….封……..……线………..…以………..…内………....答…………...题…………..无…….….效…..………………..电子科技大学《通信原理》课程试题（B 卷）(120分钟)一 二三四五六七八九十总分评卷教师1. Blank Filling:(1) A analog waveform can be converted to digital signal (for a PCM system) by three basic operations:sampling 、 quantizing 、 coding .(2) A law and µ law companding is the two main nonuniform quantizing used in PCM system.(3) The input amplitude of a linear circuit is 5V .The output amplitude is 10V . Then the gain of the circuit is 6dB.(4) BANDPASS SAMPLING THEOREM: If a (real) bandpass waveform has a nonzero spectrum only over thefrequency interval f 1 < | f | < f 2, where the transmission bandwidth B T is taken to be the absolute bandwidth B T = f 2 – f 1, then the waveform may be reproduced from sample values if the sampling rate is fs=2B T .(5) All communications systems involve three main subsystems : transmitter 、 channel 、 receiver .2. Multiple Choice:(1) If the bandwidth of an modulation signal m(t) is f 0, the bandwidth of AM, DSB-SC and SSB signal are_____(A)________ respectively(A) 2f 0, 2f 0, f 0 (B) f 0, 2f 0, f 0 (C) 2f 0, f 0, 2f 0 (D) 2f 0, 2f 0, 2f 0(2) In the following definition, the right is __(D)______(A) A bandpass waveform has a spectral magnitude that is zero for the frequency in the vicinity of the origin andnegligible elsewhere(B) A baseband waveform has a spectral magnitude that is zero for the frequency in some band concentratedabout a frequencyc f f ±=, where 0f c >>, and negligible elsewhere(C) Modulation is the process of imparting the source information onto a bandpass signal with a carrierfrequency by introduction of amplitude or phase perturbations or both. The baseband source signal is called the modulated signal(D) Modulation is the process of imparting the source information onto a bandpass signal with a carrierfrequency by introduction of amplitude or phase perturbations or both. The baseband source signal is called the modulating signal.(3) In a communication system, the random data pattern consist of binary 1’s and 0’s, and its PSD is22sin(/2)()||4/2b b s b A T fT f fT ππ=Pwhere RT b 1= is the time needed to send 1 bit of data. The first-null bandwidth of the signal and thebandwidth efficiency are__(C)____respectly(A) R, 1 (B)4R, 0.25 (C) 2R, 0.5 (D) R, 1(4) If the PSD(Power Spectral Density) of a baseband signal s(t) is shown in the following figure, the absolutebandwidth 、3dB bandwidth and first zero-crossing bandwidth is___(D)_______ respectively.(A) f3 f2, (f1+f3)/2(B) f3, f2, (f1+f2)/2(C) f2, f3, (f1+f2)/2 (D) f3, (f1+f2)/2, f2(5)The amplitude spectra of a Raised cosine-rolloff signal is shown in the following Figure, its 6-dB bandwidth is-__(B) ____(A) (B T+f1)/4(B) (B T+f1)/2(C) (B T-f1)(D) f1(6)For distortionless transmission of bandpass signal, the channel transfer function need to satisfy__(D)______(A) The amplitude response or phase response is constant(B) The product of the amplitude response and phase response are constant(C) The amplitude response or the derivative of the phase response is constant(D)Both the amplitude response and the derivative of the phase response are constant(7)In the following description about MSK, the wrong is __(B)___(E)_____(A) MSK is a continuous FSK(B) MSK’s modulation index is minimum, and equal to 1(C) Two signals representing ‘1’ and ‘0’ are ortho gonal over the bit interval(D) MSK is identical with OQPSK when the pulse shape is sinusoidal type(E) MSK bandwidth is 3R/2 when the transmission rate of baseband signal is R(8)In the following description about QPSK, π/4 QPSK and OQPSK signal, the wrong is ___(D)______(A) They are 4-ary modulated bandpass signaling(B) For rectangular-shaped data pulses, the envelope of the QPSK signal is constant and there is no AM on thesignal(C) For nonrectangular-shaped data pulses, the phase shift and AM of QPSK are the maximum ,while the phaseshift and AM of OQPSK are the minimum(D) For rectangular-shape data pulses, the phase shift and AM of QPSK are the maximum while the phase shiftand AM of OQPSK are the minimum(9)In the following description about the matched filter, the right is ____(B)___(A) MF is a Gaussian filter(B) MF maximize the instantaneous output signal power to the average output noise power for a given inputsignal waveshape(C)For the case of white noise, the impulse response of MF is h(t)=Cs(-t) when the input signal is s(t) and C is aconstant(D)Realizing forms have the integrate-and-dump method, the correlator method and the LPF configurations(10)N onlinear amplifier output-to-input characteristic, hard (ideal) limiter characteristic and envelope detector’soutput-to-input characteristic are__(A)_____ respectivelyFig.1 Fig.2 Fig.3(A) Fig.2, Fig.3 and Fig.1 (B) Fig.1, Fig.2 and Fig.3(C) Fig.3, Fig.2 and Fig.1 (D) Fig.2, Fig.2 and Fig.13. A binary base band signal (polar NRZ) is passed through a transmitted system, the data rate is 26Mb/s, and the transponder has a bandwidth of 100MHz.(a)If the system is a base band communication system, what is the null-to-null bandwidth of transmitted signal,what is the minimum transmission bandwidth for it?(b)If the base band signal is modulated by a carrier (using BPSK, and the carrier frequency is 10GHz) and thenbe transmitted through band pass communication system, what is the null-to-null bandwidth of transmitted signal, what is the minimum transmission bandwidth for it?(c)In question b), if the base band signal is passed through a raised cosine roll off filter with a 50% roll offfactor and is then modulated onto the carrier, what is the transmission bandwidth for it?(d)If QPSK and the raised cosine roll off filter with a 50% roll off factor are used, what is the maximum data rate that the band pass system can support?Solution: (a)min26132null RB R MHz B MHz ==== (b)min2522262null RB R MHz B MHz ===⋅=(c)(1)26 1.539T B R r MHz =+=⨯=(d)max (1)(1)2200100 1.5133.3/2 1.5T RB D r r R R Mb s=+=+=⨯==4. A modulated RF waveform is given by[]t cos 50t cos 200i c ω+ω, where ,kHz 2f i = and.MHz 90f c =(a)If the phase deviation constant is 200 rad/V , find the mathematical expression for the corresponding PMvoltage m(t). What is the peak phase deviation? (b) If the frequency deviation constant is 106 rad/V-s, find the mathematical expression for the correspondingFM voltage m(t). What is the peak frequency deviation? (c) If the RF waveform appears across a 50Ω load, determine the average power and the PEP. Solution: (a)()()[]()()cos 200cos 50cos 50cos 50cos 400050200PM c c P c i P i S A t D m t t t D m t t m t tωωωωπθ=+=+∴=∴=∆=(b)()()[]()()()()65cos 200cos 50cos 50cos 50cos 4000504000sin 40000.2sin 40001011504000sin 40002210tFM c c f c i tf i d S A t D m d t t D m d t tm t t td t f t t dt F Hzωσσωωσσωπππππθππππ-∞-∞=+=+∴==∴=-⋅⋅=-⎡⎤==-⨯⨯⋅⎢⎥⎣⎦∴∆=⎰⎰5. The equivalent transfer function of a binary baseband system is as:()⎪⎩⎪⎨⎧τ≤τπ+τ=other ,021f ,f 2cos 1)f (H 000(a) Find the maxmum baud rate that is no ISI. (b) Find the spectral efficiency of this system.Solution: (a)000111(1)22(11)2D r D τττ+=→==+(b) 1/TD Bd Hz B η==6. Eleven analog signal, each with a bandwidth of 3400Hz, are sampled at an 8kHz rate and multiplexed together witha synchronization channel(8kHz) into a TDM PAM signal, Assume that the analog samples will be encoded into 4-bit PCM words. Then the TDM PCM signal is pass though a channel with an overall raised cosine-rolloff Nyquistfilter characteristic of 75.0=γ . (a) Indicating the multiplexing rate f s , and the overall bit rate of the TDM PCM signal. (b)Evaluate the absolute bandwidth required for the channel.Solution: (a) 8s f kHz =121248384/s R nf kb s ==⨯⨯=(b)()384(1) 1.75336()22T R B kHz γ=+==7. The receiver is shown in Fig 7. Assume that the Bandpass input signaling is)t (n )t f 2cos(200)t f 2cos()t 800sin(100)t (r c c +π+ππ= [V],where n(t) is white noise with a PSD of20102/N -= [W/Hz]. Find the out )N /S (Figure 7Solution:()100sin(800)cos(2)200cos(2)1200cos(2)1sin(800)2AM c c c s t t f t f t f t t πππππ=+⎡⎤=+⎢⎥⎣⎦()()()2222230111200sin(800)2281200831.252221041010lg 14.9C m c outoutA m t t A m t A m t S N NB S dBN π-====⨯⎛⎫=== ⎪⨯⨯⨯⨯⎝⎭⎛⎫= ⎪⎝⎭8.Assume that the input signal ⎩⎨⎧≤<=-otherwise ,0T t 0 ,e )t (s t is a known signal. Finda) The impulse response of the matched filter for the case of white noise.b) The maximum signal level occurs at the matched filter output.Solution:()(), 0t T ()0, otherwise T t e h t s T t --⎧<≤⎪=-=⎨⎪⎩()()2220112T TtTo s T s t dt e dt e --⎡⎤===-⎣⎦⎰⎰。
习题1. 已知某2ASK 系统的码元速率为1000波特，所用载波信号为()6cos 410A t π⨯。
（1） 假定比特序列为{0110010}，试画出相应的2ASK 信号波形示意图； （2） 求2ASK 信号第一零点带宽。
解：由1000b s R R bps ==，6210cf Hz =⨯， 621020001000b c c b T f T R ⨯=== （1）一个码元周期内有2000个正弦周期：111{}n a ()2ASK s t 0（2）022000b B R Hz ==2.(mooc)某2ASK 系统的速率为2b R =Mbps ，接收机输入信号的振幅40μV A =，AWGN 信道的单边功率谱密度为180510N -=⨯W/Hz ，试求传输信号的带宽与系统的接收误码率。
解：传输信号的带宽24T b B R MHz ==，平均码元能量：24bb A T E =。
()()222100102cos 21cos 422,0,2bb T T b bc c b b b b A T A E A f t dt f t dt E E E E ππ==+=+==⎰⎰()2622618000401040444210510b b b E A T A N N R N --⨯====⨯⨯⨯⨯ 系统的接收误码率：（1） 若是非相干解调，非相干解调误码率公式，222200222BPF n b A A A R N B N σ===γ//24092( 5.11221 5.11.0306102)b E N e P e e e ----=≈==⨯γ4表 （2） 若是相干解调：由相干解调误码率公式得（最佳），1001.26981040b e E P Q Q N -⎛===⨯ ⎝也是MF 接收机的结果。
3. 某2FSK 发送“1”码时，信号为()()111sin s t A t ωθ=+，0s t T ≤≤；发送“0”码时，信号为()()000sin s t A t ωθ=+，0s t T ≤≤。
