同济大学通信系统原理(英文班)练习题附答案

Chapter 1Homework:Fill-in Questions:log M) bits 1. In a M-ary communication system, each symbol contains ( 2information content.2. The purpose of communication is to ( transfer information ). Efficiency of the digital communication system can be measured by the specifications such as (R B) , (R b) , ( η). And reliability can be measured by ( P e) , (P b).3. The basic factors for measuring the merit of a communication system are (efficiency ), and ( reliability ).4. The main influence of constant parameter channel on signal transmission are usually described by their ( Amplitude-Frequency ) and ( Phase-Frequency ) characteristics.5. Common characteristics of random parameter channels are (transmission attenuation of the signal is varying with time ), (transmission delay of the signal varies with time ), and (signal arrives at the receiver over several paths ).Multiple-Choice Questions:1.For analog and digital communications, which is (are) true? ( A B C D ).A. digital communication typically uses more bandwidth.B. analog communication cares more about fidelityC. digital communication cares more about probability for correct decisionD. digital communication typically uses analog carrier to carry baseband signalsSingle-Choice Questions:1. The symbol rate of 16-ary digital signal is 1200Bd, then the corresponding information rate is ( D ); If with the same information rate, the symbol rate of 8-ary digital signal is ( D ).A. 1600b/s, 1200BdB. 1600b/s, 3200BdC. 4800b/s, 2400BD. 4800b/s, 1600B2. Assume the delay difference of two paths in a random parameter channel is τ, the frequencies of maximum transmission loss are ( B ), the frequencies =1msof minimum transmission loss are ( B ).A. n kHz, (n+0.5)kHzB. (n+0.5)kHz, n kHzC. n kHz, (n+0.2)kHz C. (n+0.2)kHz, n kHzTrue or FalseWhen there is no input signal, the additive interference doesn’t exist, but the multiplicative interference still exists. ( F )Homework: 2.1, 2.4, 2.5, 2.6, 2.8, 2.11Fill-in Questions:1. AWGN noise is short term for ( Additive White Gaussian Noise ).2. The power spectral density and ( the autocorrelation function ) of a stationary randomprocess are a pair of Fourier transform.3. If bandwidth is 10 MHz, signal-to-noise ratio is 20dB, the Shannon capacity is C=( 66.6Mbps ).Single-Choice Questions:1. Assume X(t) is a generalized stationary random process with zero mathematical expectation, then the average power of X(t) is ( D ).A. E[X (t )]B. E 2[X (t )]C. R(∞)D. D[X (t )]2. For a narrow band random process 00()()cos ()sin C S X t X t t X t t ωω=-, if ()X t is a Gaussian process, then ( A ).(A) ()C X t and ()S X t are also Gaussian processes; (B) Only ()C X t is also Gaussian process;(C) ()C X t and ()S X t are not Gaussian processes; (D) Only ()S X t is also Gaussian processes.parison of AM, DSB, SSB and FM these four communication system,which has the best efficiency? Which has the best reliability? Which have the same efficiency?SSB has best efficiency. FM has the best reliability. AM and DSB have the same efficiency.2.What is the requirement on the characteristics of the filter for producing theVSB signal?[H(f+f0)+H(f−f0)]=C3.What kind of baseband signal is suitable for VSB modulation?The analog baseband signal with D.C. component and low frequency component is suitable for VSB modulation.4.Let the expression of a FM modulated signal be10cos(2π×106t+5sin103πt)Find: (1) the maximum frequency deviation of the modulated signal.(2) the frequency of the carrier.(3) the bandwidth and the average power of the modulated signal.(4) if the FM circuit constant is k f=5kHz/V, then the expression of the baseband signal can be written as:=5,∴∆f=m f∙f m=5×500=2500Hz(1) ∵m f=∆ff m(2) f0=106Hz(3) B=2(m f+1)f m=2×(5+1)×500=6000HzP =A 22=1022=50 W(4) ∵∆f =k f ∙A m ∴A m =∆fk f =25005000=0.5 V∴the baseband signal can be written as:m (t )=0.5cos 103πt5. Assume the amplitude of a FM modulated signal is 10V , the instantaneousfrequency of the FM signal is:f (t )=106+104cos (2π×103t ) (Hz)Find: (1) the expression of the FM modulated signal.(2) the maximum frequency deviation , the modulation index, and the bandwidth of the modulated signal.(1) s (t )=10cos [2π×106t +10sin (2π×103t )](2) ∆f =104, m f =10, B =2(m f +1)f m =2×(10+1)×1000=22 kHzFill-in Questions:1. Assume the expression of a FM signal is )102sin 5102cos(536t t ⨯+⨯ππ, the carrierfrequency of this FM signal is (106)Hz, the maximum frequency deviation is (5000) Hz, the bandwidth of this FM signal is (12×103 ) Hz ，the average power is (12.5 )W.Single-Choice Questions:1.Which analog modulation has the highest spectrum efficiency? ( C )(A) AM (B) DSB(C) SSB (D) VSBCalculation Questions:1. Let a baseband modulating signal be a sinusoidal wave with the frequency 10kHz, and theamplitude 1V . It modulates the phase of a carrier with frequency 10MHz, and the maximum phase deviation of modulation is 10rad. (1) Calculate the approximate bandwidth of the phase modulated signal. (2)If the frequency of the modulating signal is changed to 5kHz, calculate the bandwidth of the phase modulated signal.Answer: (1) We has known that: m 10 kHz, A 1 V m f == , and maximum phase deviation max 10 rad ϕ=.The instantaneous phase of the carrier can be represented as:()()p t k m t ϕ=, where 10p k =.Then, the instantaneous frequency deviation of the carrier can be represented as: d ()sin p m m d t k t dtϕωω=, and maximum frequency deviation is p m k ωω∆=. Hence, the frequency modulation index is:10p m f p m m k m k ωωωω∆====Thus, the approximate bandwidth of the phase modulated signal is2(1)2(110)10220 kHz f m B m f =+=+⨯=(2) If the frequency of the modulating signal is changed to 5kHz, then the approximate bandwidth of the phase modulated signal is:2(1)2(110)5110 kHz f m B m f =+=+⨯=.Chapter4Fill-in Questions:1. The basic process, from( Sampling the analog signal ), ( the quantization ), and( converting to binary symbols ) , is usually called as pulse code modulation (PCM).Single-Choice Questions:1. The purpose of nonuniform quantization is ( B ).A. Protect big signalB. Protect small signalC. remove quantization noiseD. Increase quantization levels 2. In Delta modulation, the best way to avoid overload quantization noise is ( A ).A. Increase sampling frequencyB. Increase quantization stepC. Reduce sampling frequencyD. Reduce quantization step3. The sampling theorem points out that if the highest frequency of a low-pass analog signal is H f , then the signal can be represented by its samples when the sampling frequency no less than ( B ).(A) H f (B) 2H f (C) 3H f (D) 5H f4. If the samples obey uniform distribution over [,]a a -, then the quantization noise power is determinedby( C )(A) a (B) a and quantization interval ∆ (C) quantization interval ∆ (D) other parameters. Multiple-Choice Questions:1. Digitization of analog signals includes ( B 、D 、E )(A). Modulation (B). Quantization (C). Multiplexing (D). Source coding (E). SamplingShort-Answer Questions:1. In Delta modulation system, why we increase sampling frequency s f rather thanquantization step ∆ to avoid overload quantization noise?Answer:Given the sampling frequency and the quantization step ∆, then the slope of a step is:s k f =∆⋅It is the maximum possible slope of a step-shaped wave, or it is called the maximum tracking slope of the decoder. When the slope of the input signal of the delta modulator exceeds this maximum, overload quantization noise will be generated. Therefore, in order to avoid overload quantization noise, it is necessary to make the product of ∆ and s f large enough, so that the slope of the signal can not exceed this product. One the other hand, the value of ∆ is directly related to the magnitude of the basic quantization noise. If the value of ∆ is large, then the basic quantization noise must also be large. Therefore, only the method of increasing s f to increase the product s k f =∆⋅ can ensure that the basic quantization noise and the overload quantization noise do not exceed the limit.Fill-in Questions:1.( 2 )Baud/Hz is the highest possible unit bandwidth rate, and is also called as the Nyquistrate.2.The time-domain equalizer is used to overcome (intersymbol interference (ISI)).3.The basic process, from ( Sampling the analog signal), ( the quantization ), and( converting to binary symbols ) , is usually called as pulse code modulation (PCM). Single-Choice Questions:1. If the samples obey uniform distribution over [,]a a-, then the quantization noise power is determined by ( C)(A) a(B) a and quantization interval ∆(C) quantization interval ∆(D) other parameters.Multiple-Choice Questions:1. Digitization of analog signals includes ( B, E )(A). Modulation (B). Quantization (C). Multiplexing (D). Source coding (E). Sampling2. Which are symbol code type for baseband digital signals? (A, B, C, D, E )(A). HDB3 (B). AMI (C). CMI (D). Biphasic code (E). 5B6B (F). BPSK(G).MSKShort-Answer Questions:1.Which are the design principles of code for digital baseband signal?As we know, the performance of a practical baseband transmission system can be found in its eye pattern (refer to the figure below). Please find out the characteristics in the figure 1, and point out their impacts on transmission performance.Answer:(1) The location of the central perpendicular line is the optimum sampling instant.(2) The middle horizontal line represents the optimum decision threshold level.(3) The perpendicular height of the shadow region represents the distortion range of thereceived signal.(4) The slope of the bevel edge of the “eye” represents sensi tivity of the sampling instantto the timing error.(5) Under no noise situation, the degree of the opening of the “eye is the noise tolerance;if noise at the sampling instant exceeds this tolerance, then error decision may happen.Calculation Questions:1.Assume there are 4 overall transfer characteristics of baseband transmission systemsshown in Figure 2. If symbol rate is 2000Bd, please illustrate whether they can transmit information without ISI? Which system has better transmission characteristic?Solution: ∵symbol rate is 2000Bd,∴ISI exists in system (a) and (c), ISI does not exist in system (b) and (d).Since system (b) can be realized physically, and its η=1.33 Bd/Hzsystem (d) can’t be realized physically, and its η=1 Bd/Hzso system (b) has better transmission characteristic.1000-1000 (Hz) (Hz)1500-1500 500-500(Hz)750-750 (Hz)2000-2000Chapter61. Assume there is a space-ground communication system, the symbols rate is 0.5MB, the bandwidth of receiver is 1MHz. Antenna gains of ground station and space station are respectively 40dB and 6dB ，the path loss is (60+10lg d )dB, where d is the distance (km). Given the transmitting power is 10W, the double-side power spectral density of white noise is 2×10−12W/Hz. Requiring the symbol error probability of the system is P e =10−5, try to find the maximum communication distance under these conditions as below.(1) Adopting 2FSK modulation and coherent demodulation. (2) Adopting 2DPSK modulation and coherent demodulation. 解：（1）采用2FSK 方式传输，进行相干解调，其误码率为 510221-==rerfcP e 可计算出所需要的信噪比为r=18.3，噪声功率为 ()W B n n 661202104101104--⨯=⨯⨯⨯==σ 接收端信号幅度平方为()W r a n 462210464.11043.1822--⨯=⨯⨯⨯==σ 传输中信号功率衰减为()dB a A 35.5110464.1102lg 1022lg 10422=⨯⨯=-根据给定条件，51.35+40+6=60+10lgd ，可以算出传输距离为 ()km d 543210735.3==(2) 采用2DPSK 方式传输，选择差分相干方式进行解调，其误码率5e 1e 102r P --==，可计算出所需要的信噪比r =10.82，接收端信号幅度平方为2265n 2210.824108.65610 W a r σ--==⨯⨯⨯=⨯传输中信号功率衰减为 225221010lg 10lg 53.63 dB 28.65610A a -⨯==⨯根据给定条件，53.63+40+6=60+10lgd ，可以算出传输距离3.963109183 km d ==。