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1、数字通信2006年试卷（1）A multi-path fading channel has a multi-path spreading of Tm=5us and a Doppler spread D=100Hz. The total channel bandwidth at band-pass available for signal transmission is W=1.25MHz. The bit rate of the digital speech signal to be transmitted is R=9.6 kb/s and the modulation is BPSK.a
2、. Determine the coherence bandwidth and the coherence time.b. Is the channel frequency selective?c. Is the channel fading slowly or rapidly?d. Suppose that the channel is used to transmit the binary speech via coherently detected PSK in a frequency diversity mode. Determine how much diversity can be
3、 used.e. Suppose that a wide-band signal is used for transmission and a Rake-type receiver is used for demodulation. How many taps would you use in the Rake receiver?Solution: 【中文书习题14-4】a. 【中文书P587】信道相干带宽，信道相干时间b. 信号带宽信道相干带宽，则为频率非选择性；信号带宽信道相干带宽，则为频率选择性；，频率选择性衰落c. 信号周期信道相干时间，则为快衰落；信号周期信道相干时间，则为慢衰落；，
4、慢衰落d. For frequency diversity, Ch. Independent, frequency spacing , e. Using Rake:(1) A DS-CDMA system: the spectrum spreading sequence has a chip rate of 1.2288 Mc/s. A rate 1/3 convolutional code with dfree=10 is used. The data sequence is at a rate of 10kbits/sec. The modulation is binary PSK.a.
5、Determine the coding gain.b. Determine the processing gain.c. Determine the jamming margin assuming an Eb/J0=8dB.Solution: 【中文书P532】a. b.c.(2) A digital communication system, using coherent QPSK with Gray mapping as its modem, requires the bit error rate 10-3.a. If the channel is AWGN channel, what
6、is the required SNR per bit () for decision?b. If the channel is frequency non-selective and slow Rayleigh fading channel and with AWGN, what is the required average SNR per bit () for decision?c. In order to reduce the required, the L=4 diversity is used. If the maximum ratio combination is adopted
7、, what is the required average SNR per bit () for decision?d. For the case c, if the modem used is coherent DPSK, what is the required average SNR per bit () for decision?e. Again for QPSK in the case c, in order to reduce the required further, the linear block coding and soft decision decoding are
8、used in addition to L=4 diversity, what is the required average SNR per bit () for decision?Solution: a. 【中文书P199-200】图5-2-13示出了二、四相DPSK和相干PSK信号传输的二进制数字差错概率。通过图解法可以求解此问，已知比特误码率为10-3.根据图5-2-13可以得知比特SNR6.8dBb. 【中文书P598-599】图14-4-4所示为L=1、2、4时，两相、四相和八相DPSK信号传输的比特差错率。通过图解法可以求解此问，已知比特误码率为10-3.则符号误码率为10-3/
9、log24=510-4.根据图14-4-3用M=4，PSK可以得知比特SNR29dB。 c. 通过图解法可以求解此问，通过结论 “瑞利衰落信道上的相干PSK比DPSK好3dB”直接把读图的结果减去3dB得PSK的结果。已知比特误码率为10-3. 根据图14-4-4，用M=4，L=4得知比特SNR13dB，所以PSK的比特SNR13-3=10dB。d. 已知比特误码率为10-3. 根据图14-4-4，用M=4，L=4得知DPSK的比特SNR13dB。e. , M=4 DPSK等效于L=4的分集，根据图14-4-4，用M=4，L=4得知改善的比特SNR13dB。又根据c问的结论单纯L=4分集需要的
10、比特SNR10dB，因此新的比特SNR10-13dB=-3dB。(3) A convolutional code (2, 1, 3) with g1=111, g2=101a. Draw the block diagram of the code.b. Draw the state diagram of the code.c. Find the transfer function T(D) of the code.d. Find the minimum free distance (dfree) of the code.e. Show the corresponding path (at di
11、stance dfree from the all-zero code word) in the trellis.f. The received sequence is (010000100000). Use the Viterbi decoding algorithm to find the most likely data sequence.g. Assume that the crossover probability of the channel is 10-6 and the hard-decision Viterbi decoding is used, please use the
12、 hard-decision bound to find an upper bound on the average BER.Solution:【中文书P340-353】a. 本题卷积码是(2, 1, 3)，n=2,k=1,K=3可以表示为图2.1所示。b. 状态就是移位寄存器中最右边的K-1个寄存器的内容。(2, 1, 3)卷积码的状态就是最右边2位的内容。共有四种状态：a: 00、b: 01、c: 10、d: 11，所以状态转移图为图2.2所示。状态a: 00输入0，输出00，下一状态仍是a: 00；状态a: 00输入1，输出11，下一状态变为c: 10；状态b: 01输入0，输出11，下
13、一状态变为a: 00；状态b: 01输入1，输出00，下一状态变为c: 10；状态c: 10输入0，输出10，下一状态变为b: 01；状态c: 10输入1，输出01，下一状态变为b: 01；状态d: 11输入0，输出01，下一状态变为b: 01；状态d: 11输入1，输出10，下一状态仍是d: 11；c. 首先，把该状态图的每个分支标上D0=1, D1, D2, , DK-1，其中D的指数表示对应于每个分支的输出比特序列与全零分支的输出比特序列之间的汉明距离。在a节点处的自环可以删除，因为它在计算码字序列和全零码序列的距离特性上不起作用。另外，节点a被分成两个节点，其中之一表示状态图的输入，另
14、一个代表状态图的输出。这样得到新的状态图，根据新的状态图可以写出状态方程（只写输入的）：Xc=D2XaXb=DXc+DXdXd=DXc+DXdXe=D2Xb码的转移函数定义为T(D)=Xe/Xa。解上面的状态方程就可以得到转移函数d. 卷积码的最小距离叫做最小自由距离，用dfree来表示。从新的转移图中可以看出，从状态a出发再回到状态a（即状态e）的最小距离。显然是ac汉明距离D2，cb汉明距离D，be汉明距离D2，所以dfree=5。e. 先画出Trellis网格图，根据d问的结果，acbe，画出路线图（红线所示）。f. 先画出Trellis网格图，根据接收端的序列01, 00, 00, 1
15、0, 00, 00逐段对照把接收序列与网格图中的输出序列的汉明距离标注在Trellis图上，然后从t=0出发找到一条最短路径，如图红线所示。可知最大概率的输入序列是000000。译码时始终从状态a: 0开始，一个回朔深度一个回朔深度地判断（回朔深度=K）。g. 硬判决译码差错概率的上界公式为(8-2-34)(8-2-34)假定在由输入比特位1而引发的所有分支上引入一个因子N，那么当横越每个分支时，只有输入比特为1引发的转移才能使N指数的累积值加1。(2, 1, 3)卷积码，引入N后的状态图如图2.6所示，列出状态方程Xc=ND2XaXb=DXc+DXdXd=NDXc+NDXdXe=D2Xb码的转移函数定义为T(D, N)=Xe/Xa。解上面的状态方程就可以得到转移函数5
