Problems and answers on princiles of communications0522-2012.doc

Answers to Problems(Edition 6, Fan Changxin, Chinese Text-book)Chapter 1 Introduction1Problems and Answers1Chapter 2 Deterministic Signals2Problems and Answers2Chapter 3 Random Process4Problems and Answers4Chapter 4 Channels5Problems and Answers5Chapter 6 Digital Base-band Transmission System7Problems and Answers7Chapter 7 Digital Band-pass Transmission System10Problems and Answers10Chapter 8 New-type Digital Band-pass Modulation Technology11Problems and Answers11Chapter 9 Digital Transmission on Analogue Signals14Problems and Answers14Chapter 11 Error Control Coding16Problems and Answers16Chapter 12 Orthoganal Coding and Pesudorandom Sequence18Problems and Answers18Chapter 13 Syncronization Principle20Chapter 14 Communication Network20Problems and Answers20Chapter 1 Introduction Problems and Answers1-1. Suppose the appearing probability of English letter is 0.105, is 0.002. Please try to calculate the amount of information of and .Solution: The appearing probability of English letter is , and from the formula for the amount of information, we can know its amount of information is： (b)The appearing probability of English letter is , also from the formula for the amount of information, we can know its amount of information is： (b)1-6. Assume the output of a source of information is made up by 128 different symbols, the appearing probability of 16 of these is 1/32, and the rest 112 is 1/224. The information source sends 1000 symbols every second, and every symbol is independent of each other. Please try to calculate the average information rate of this information source.Solution: (b/s)Chapter 2 Deterministic Signals Problems and Answers2-2. Assume a signal can be expressed as . Please try to judge that it is power signal or energy signal.Solution:For and so isnt energy signal because is an integer.so is a power signal.Because s selfcorrelation function and Power spectral density are Fourier transform, namely so 2-5 Please try to work out the autocorrelation function of , and its power from its autocorrelation function.Solution:for , even because = so also because so 2-9 If we know the bilateral power spectral density of a signal is now try your best to calculate its average power.Solution: for so its average power is (HZ)Chapter 3 Random Process Problems and Answers3-3. Assume a random process is , if and are the gaussian random variables which are independent of each other and their average is 0、variance is . Now calculate:(1) 、；(2) One-dimensional distribution density function of ;(3) and .Solution: (1) (2) Because the linear combination of gaussian variables still be gaussian variables, so we can get from (1)its one-dimensional distribution density function (3) Chapter 4 ChannelsProblems and Answers4-5. A signal source is A, B, C and D are four symbols composed. Set up each character are independent, the probability of their occurrence 1/4, 1/4, 3/16, 5/16. After channel transmission, each character the right to accept the probability is 1021/1024, wrong for these characters in the conditional probability is 1/1024. Try to draw the model of the channel and find the channel capacity C is equal to b / character.Solution:Channel model diagram：The information entropy:Known by the subject mean：P(A/A)=P(B/B)=P(C/C)=P(D/D)=P(A/B)=P(A/C)=P(A/D)=P(B/A)=P(B/C)=P(B/D)=P(C/A)=P(C/B)=P(C/D)=P(D/A)=P(D/B)=P(D/C)=So receiver receives A、B、C and D probability:Chapter 6 Digital Base-band Transmission System Problems and Answers6-13. Digital baseband signal for transmission symbol rate is .Which kind of transmission characteristics should be taken in the following figure ? And please give a brief description of the reasons.Solution:Three transmission characteristics satisfy the Nyquist criterion, derived from Figure baseband system can be no inter symbol interference transmission . For the actual transmission system needs to assess the transmission characteristics of the pros and cons.(1). For (a) , its bandwidth is , and its symbol rate is , so bandwidth efficiency is Transfer function of the frequency domain expression:So shows that the tail of the impulse response of the transfer function has speed attenuation.(a) physically practical applications can be achieved（2）For (b) ,its bandwidth is , and its symbol rate is , so bandwidth efficiency is Transfer function of the frequency domain expression:So shows that the tail of the impulse response of the transfer function has speed attenuation.（b） physically difficult to achieve.(3) For (c) ,its bandwidth is , and its symbol rate is , so bandwidth efficiency is Transfer function of the frequency domain expression:So shows that the tail of the impulse response of the transfer function has speed attenuation.(c) physically practical applications can be achieved.In summary, more appropriate to choose (c).6-25. Please design a Zero Forcing equalizer with three taps. The value of input signal at every sampling points are , others are 0.(1) Work out the best coefficient of these three taps;(2) Compare the peak distortion from start to finish.Solution: (1) List matrix equation: =substitute sample value into the equation above, we can get the equations below: solve the equations the solution of three taps best coefficient : (2) , , , , Inputing x(t) peak distortion： Outputting x(t) the peak distortion： So the finishing peak distortion reduce 7.5 times to （than）starting.Chapter 7 Digital Band-pass Transmission System Problems and Answers7-13. If we know the input SNR of a modem in a binary phase shift keying(PSK) system is , now try to work out the bit error rate(BRE) when its coherent demodulation 2PSK、coherent demodulation - yards inverse transform 2DPSK and differential coherent demodulation respectively.Solution:a. BRE when its 2PSK coherent demodulation: According to we can get , so b. BRE when its yards inverse transform 2DPSK coherent demodulation: c. BRE when its 2DPSK differential coherent demodulation: 7-20. If we transmit 2400b/s date by 4PSK modulation:(1) How much is the minimum theoretical bandwidth?(2) While the transmission bandwidth doesnt change, but bitrate double, how the modulation method should change?Solution:(1) when we transmit 2400b/s date by 4PSK, Set the minimum theoretical bandwidth is B , then: So that the minimum theoretical bandwidth is 1200Hz(2) to keep the transmission bandwidth same when bit rate double, we can set the modulation method is MPSK , then Get , so the modulation method should be chosen is 16PSKChapter 8 New-type Digital Band-pass Modulation Technology Problems and Answers8-1.If there is one digital signal sequence +1 -1 -1 -1 -1 -1 +1, try to draw use it modulated FSK signal phase changes in Fig. Symbol rate of 1000B and the carrier frequency is 3000Hz, try to draw MSK signal waveform Fig.Solution :When digital signal sequence is +1 then waveforms phase increases and digital signal sequence is -1 then waveforms phase decreases that according to the characteristics of MSK.So signal phase changes Fig is Symbol rate of 1000B and the carrier frequency is 3000Hz that get by the subject mean, so one MSK symbol period there are three periodic signals.So signal waveform Fig is 8-2. Suppose one MSK signal whose code element rate is , express code element “1”, while express “0”, if , try to work out , and draw the wave shape when the code element is “101”.Solution: the waveform coefficients of MSK orthogonal modulation signal is 0, we know the two formulas on right are “0”, the first formula (k=1,2,3, )while k=1, so (Hz)the waveform of “101”Chapter 9 Digital Transmission on Analogue Signals Problems and Answers9-9 . Adopt 13 broken-line A law to code and image the smallest quantized interval is 1 unit, if we also know the value of sampling pulse is +635 units:(1) Try to get the code-block outputted by the coder, and calculate the quantization error;(2) Write out the uniform quantization 11 yards to the 7 bit code.Solution: (1) Polarity yards: Because +635 0 , so =1 Because 635 128 , so =1 Because 635512 , so =1 Because 6351024 , so =0 From these comparisons, we can know +635 is in the seventh condition.Period of the code: the seventh segments started level is 512, uniform quantization =32 units. Because 512+32=768 635768, so =0 Because 512+324=640 635576, so =1 Because 512+323=608 635608, so =1The coders outputed code-block is 11100011, quantitative output is +608 quantization units, quantization error is 635-608=27 quantization units. (2) Excepting the polarity yards, 7 is non-linear code-block 1100011, corresponding to which the average quantization of 11 units code is 01001100000.9-13. Transmite a 10-way analog signal which bandwidth is 300Hz3400Hz by PCM time-division multiplexing. The ratio of sampleing is 8000Hz. Then it was encoded to natural binary code after getting through one 8-classe quantization. The wave shape of code element is a rectangular pulse whose bandwidth isand its duty ratio is 1. Please try to work out the Nyquist basic bandwidth at the time to transport the PCM time-division multiplexing signal.Solution:The time slot width that every signal needs is =ms After sampling, it goes one 8-class quantization codingFrom N=, we can get N=3, according to which we know it goes 3 bit coding. Every bit width is =ms, and because duty ratio is 1, we can get the pulse width is =. So system bandwidth is: B=240kHz.Chapter 11 Error Control Coding Problems and Answers11-6. If we know a linear-codes supervision matrix is: Please try to get all the possible code-blocks.Solution: Because is supervision matrix and n=7, k=4, r=n-k=3 in the problem, and the model generator matrix is G=, =Then G= The code-block that can be used is =G. So the code-block that can be used is: 0000000 1000111 0001011 1001100 0010101 1010010 0011110 1011001 0100110 1100001 0101101 1101010 0110011 1110100 0111000 111111111-17. if we know a convolutional code that ,its fundamental matrix is .please try to work out its generator matrix G and its parity-check matrix H.Solution:Because its basic generator matrix is and .so that is a square matrix and ,.So its generator matrix is:Because so .So that its parity-check matrix is:Chapter 12 Orthoganal Coding and Pesudorandom SequenceProblems and Answers 12-2. Now we know a original state of three linear feedback shift registers is 111, try to get two kinds of output sequence of sequence .Solution: From the question we know =3, then =7 and +1=(+1)(+1+)(17+) We can get from the concept of primitive polynomial ()=1+ ()=1+ They are both primitive polynomials, if get anyone between them as feedback shift registers property polynomial, both can get sequence From ()=1+,we can get sequence ()=()+()+ =+ =(1+)+(1+)+ So sequence is 11110101.From ()=1+,we can get sequence ()=()+()+ =+ =(1+)+(1+)+ So sequence is 11101001.12-7. Try to verify the quadratic residue sequence is sequence when and .Solution: (1) while: =1 (mod3) =1 (mod3) So 1 is mod3s quadratic residue,2 is m