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.。