2020年杭州电子科技大学考研试题通信原理
电子科技大学《通信原理》20春期末考试参考答案
B.DSB
C.VSB
D.FM
答案:D
15.对于SSB调制而言，下列叙述不正确的是（）
A.传输带宽比AM调制少一半
B.信道利用率少一半
C.同等条件下比AM省功率
D.在军用短波中由广泛应用
答案:B
16.由香农公式可知连续信道的容量受“三要素”的影响，“三要素”是指（）
A.带宽、信号功率、信息量
D.计算机输入输出信号
答案:D
19.下列是数字通信系统的有效性指标的是（）
A.传码率
B.容错率
C.误信率
D.误码率
答案:A
20.某黑白图片每个象素有16个亮度等级，设所有亮度等级统计独立，每个象素所含的平均信息量为（）
A.1bit
B.2bit
C.3bit
D.4bit
答案:D
21.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（）。
B.带宽、信号功率、噪声功率谱密度
C.带宽、信号功率、传信率
D.信息量、带宽、噪声功率谱密度
答案:B
17.在数字通信系统中，传输速率属于通信系统性能指标中的（）
A.有效性
B.可靠性
C.适应性
D.标准型
答案:A
18.下列是数字信号的是（）
A.电话送出的语音信号
B.电视摄像机输出的图像信号
C.温度传感器的输出信号
B.实现信号的多路复用
C.提高系统抗干扰能力
D.降低系统功耗
答案:D
26.已知某四进制数字传输系统的传信率为2400b/s，则系统的传码率为（）。
A.1200B
B.2400B
杭州电子科技大学--移动通信与个人通信期末A试卷
共2页第2页
3、有哪4种不同的切换信道分配方案？在什么情况下，预留信道方案（RCS）比其他方案更需要？（5分）
4、在切换过程中，导致切换失败有哪4种原因？（4分）
5、号码便携性可分哪3类？固定网络的号码便携性有哪4种基本的解决方案？（7分）
6、SSD是什么？（2分）
7、什么是TMN？为什么要采用？（4分）
共2页第1页
5、移动性管理的所有消息都是由实现的。
二、名词解释：（每个1分，共计10分）
PIN：
CCS：
SRSN：
DRSN：
WAP:
SGSN：
GGSN：
WiMAX：
OFDM：
OFDMA：
三、简答题：（共计40分）
1、NCHO和MCHO比较,谁所需的切换时间更长?（5分）
2、SS7网络由哪3个不同部分组成？画出PCS网络与PSTN之间互联结构（6分）。
8、什么是MHLR，借助MHLR，DRDN有哪2种注册策略？（4分）
9、被我国认定的3G的标准目前有哪3种？（3分）
四、综合题：（每题10分，共计40分）
1、位置管理分为两个部分？为什么说这两部分设计存在优化问题？
2、降低国际呼叫发送成本有哪4种方法？简述之。
3、什么是IS-41系统间切换？存在哪4种系统间切换？
杭州电子科技大学研究生考试卷
考试课程移动通信与个人Fra bibliotek信考试日期
年月日
成绩
学院
学号
姓名
一、填空题：（每空1分，共计10分）
1、MS一般有和两部分组成。
2、GSM安全性体现在两个方面：和。
3、切换检测是基于对的测量。可用来决定切换三种测量分别是1）；2）；3）。
电子科技大学大三电子信息专业通信原理英文考试卷及答案 (4)
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 二0x 至二0x 学年 第 学期<< 通信原理 >> 课程考试题（A 卷） (120分钟) 考试日期 20xx 年 1 月 14 日1. (16 Points) Blank Filling:1) The PSD of a baseband signal s(t) is shown in the following figure, the absolute bandwidth,3dB bandwidth and first zero-crossingbandwidth is____________, _______________ and ________________ respectively，________________ and ________________.4) If the analog signal is )40002cos()42sin()(t t t s ⨯+⨯=ππ, the minimum sampling rate is ________________.5) If the data rate is 50kbps, the first null bandwidth for the unipolar NRZ signal, the polar NRZ signal and the Manchester NRZsignal are______________, ______________, ______________ separately.6) For no ISI, and the symbol rate of communication system is 50 kbaud , the 6 dB bandwidth of the raised cosine-rolloff filter, f 0,should designed to be _________ .7) If the modulated signal )100002cos()]42cos(5.01[5)(t t t s ⨯⨯+=ππ, its complex envelope is ____________.8) Assume that the input signals is T)]-U(t -][U(t)cos [)(0t e t s at ω-=. When the input noise is white, the impulse response of thematched filter h(t)=_________________________________and the optimum sampling time =0t ___________________________. 1 5MHz, 3.5MHz ， 4MHz2 Sampling, quantizing, encoding3 Channel receiver4 8kHz5 50kHz, 50kHz, 100kHz 625kHz7 )]42cos(5.01[5)(t t g ⨯+=π8 U(-t)]-t)-][U(T cos [)(0)(t e t h t T a ω--= t0=TIMPORTANT NOTICE: Among the following 7 questions, you are required to answer any 6 of them. Make sure you CROSS the one you will NOT select.(重要提醒：下面7题中只需选择6题，并记住在你不做的题号上划叉) 2． (9 Points) The transmission system is shown as following. The bandwidth of the baseband analog signal m(t) is 15kHz. Assumethat the peak SNR of receiver output is 50 dB. Find (a) the minimum sampling rate fs;(b) the minimum number of bits in the PCM word; (c) the minimum data rate of the digital signal s(t)Solution:(a) 332215103010S f B Hz ==⨯⨯=⨯(b) 6.02 4.77pk SNR n =+( 4.77)/6.02(50 4.77)/6.027.5pk n SNR =-=-≈ Because n is an integer, we have n=8. (c) 338301024010S R nf bps ==⨯⨯=⨯3. (9 Points)Consider a random data pattern consisting of binary 1’s and 0’s, where the probability o f obtaining either a binary 1 orbinary 0 is 1/2. Assume that these data are encoded into a polar-type waveform such that the pulse shape of each bit is given by/2102()10/2b b b t T t f t t T T ≤⎛⎫⎧=∏=⎨⎪>⎩⎝⎭where b T is the time needed to send one bit. Find the expression for the PSD of this waveform. Solution:()()22()5/21 k=0()[]0 k 0|()|()(0)25/2b b n n k b b b F f T Sa fT R k E a a F f f R T Sa fT T πψπ+=⎧==⎨≠⎩==4. (9 Points)Assume that an AM modulated signal ist t t t s a c c a c )sin(100cos 500)sin(100)(ωωωωω--++= Where the un-modulated carrier is t c ωcos 500(a) Find the complex envelope for the modulated signal. (b) Write the expression for the modulating signal m(t).(c) What are the percentages of positive and negative modulation (d) Evaluate the average power of the modulated signalSolution:(a)tt t t t tt t t s c a a c c a c c a c ωωωωωωωωωωcos )sin 200500(sin cos 200cos 500)sin(100cos 500)sin(100)(+=+=--++=)sin 4.01(500sin 200500)(t t t g a a ωω+=+= (b) t t m a ωsin 4.0)(=(c) percentages of positive modulation :%405005004.1500=-⨯;percentages of negative modulation :%405006.0500500=⨯- (d) w t g P av 1350)24.01(50021)(21222=+⨯==5. (9 Points)An angle-modulation system is indicated in Fig.Where D p is the phase sensitivity of the phase modulator, the carrier frequency Hz f c 410=, if the modulated waveform s ( t ) is given by:)2000sin 21102cos(100)(4t t t s ππ+⨯=(a) For modulating signal m ( t ), determine whether s ( t ) is a phase modulating wave or frequency modulating wave;(b) Find the expression of m ( t );(c) Find the maximum phase deviation and maximum frequency deviation; (d) Find the transmission bandwidth by using Carson’s rule.Solution:(a) ])(cos[)](cos[)(⎰∞-+=+=tp c c p p c c d m D t A t m D t A t s ττωω∴ s ( t ) is a frequency modulated wave.(b) t d m D t tp πττθ2000sin 21)()(==⎰∞- → t t m ππ2000cos 200)(= (c) 21)](max[==∆t θθ ; Hz t dt d F 500]2000sin 21[21max =⎭⎬⎫⎩⎨⎧=∆ππ (d) Hz f F B B m f FM 3000)1000500(2)(2)1(2=+=+∆=+=β6．(9 Points)Assume that the channel bandwidth could be used is 1000~4000Hz, the carrier frequency is 2500Hz. (a) if DPSK modulation and 0.2r = raised cosine-rolloff filter is used, find the maximum data rate could be supported.(b) if QPSK modulation and 5.0=r raised cosine-rolloff filter is used, find the maximum data rate could be supported.Solution:400010003000T B Hz =-=2(1)(1)1T T basebandlB RB B D r r R l r==+=+⇒=+(a) for DPSK130002500/110.2T lB R b s r ⨯===++(b) for QPSK 230004000/110.5T lB R b sr ⨯===++7．(9 Points)Assume that 9600/R bit s =.For rectangular data pulses, (a) （1）calculate the first null-to-null bandwidth of OOK.(b) （2）calculate the first null-to-null bandwidth of 16QAM. Solution:(a)()22219.2T basebandOOKRB B D kHz l====(b) ()16222 4.8T baseband QAM RB B D KHz l ==⋅==8. (9 Points)Suppose that a 2FSK system with non-coherent detector has a BER of 5210-⨯ at 0/b E N of 13dB. (a) With the same 0/b E N , what is BER for coherent detection?(b) To obtain the same BER, what is the 0/b E N when non-coherent OOK system is used? Solution: （9分）(1)For coherent-2FSK,e P Q Q ==(2) While OOK and 2FSK signals have the same BER-formula, so 0/b E N is the same, 13dB.。
《通信原理》自测题1答案
1电子科技大学网络教育考卷（A1卷）答案(20 年至20 学年度第 学期)考试时间 年 月 日(120分钟) 课程 通信原理 教师签名__ ___一、选择题：（单选，每小题3.5分，共35分）1. 数字通信系统模型中，包括调制器和解调器的广义信道称（ B ）。
A 、调制信道；B 、编码信道；C 、通讯信道；D 、传输信道；E 、都不对。
2. 对通信系统而言，下面哪个目的不能通过采用调制手段来达到（ D ）。
A 、提高频率便于辐射；B 、实现信道复用；C 、改变信号占据的带宽；D 、提高信息传输速率。
3. 用单音f(t)=cosωm t 对载波C (t)=cosω0t 进行调频，若J 0（βFM ）=0，则边带功率是载波功率的（ D ）。
A 、0％；B 、50％；C 、80％；D 、100％；E 、都不对。
4. PCM 通信系统中，收端重建信号与原始信号肯定不同，主要原因是存在（ A ）。
A 、量化误差；B 、抽样误差；C 、编码误差；D 、传输误差；E 、都不对。
5. 若载波频率为ω0，基带调制信号s(t)是周期为T 的方波，则二进制ASK 信号的传输带宽为（ E ）Hz 。
A 、1/4T ；B 、1/2T ；C 、1/T ；D 、2/T ；E 、4/T ；F 、都不对。
6. 一相位不连续的二进制FSK 信号，发“1”码时的波形为Acos(3000πt+θ1)，发“0”码时的波形为Acos(9000πt+θ2)。
若码元速率为200波特，则系统传输带宽最小应为（ B ）Hz 。
A 、3000；B 、3400；C 、4200；D 、9000；E 、都不对。
7. 假定调制信号是由ωm ，2ωm 和3ωm 分量组成，他们的振幅相等，即：f(t)＝cos ωm t+cos2ωm t+cos3ωm t ，设K f ＝2π×105rad/v ，则其FM 波的最大频偏为（ C ）KHz 。
A 、3；B 、30；C 、300；D 、3000；E 、都不对。
电子科技大学《通信原理》课程试题B卷含答案
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 --⎡⎤===-⎣⎦⎰⎰。