发布时间 : 星期二 文章计算机网络(第4版) 清华大学出版社 习题答案(中文版)更新完毕开始阅读
49. D-AMPS uses 832 channels (in each direction) with three users sharing a single channel. This allows D-AMPS to support up to 2496 users simultaneously per cell. GSM uses 124 channels with eight users sharing a single channel. This allows GSM to support up to 992 users simultaneously. Both systems use about the same amount of spectrum (25 MHz in each direction).
D-AMPS uses 30 KHz× 892 = MHz. GSM uses 200 KHz × 124 = MHz. The difference can be mainly attributed to the better speech quality provided by GSM (13 Kbps per user) over D-AMPS (8 Kbps per user).
50. The result is obtained by negating each of A, B, and C and then adding the three chip sequences. Alternatively the three can be added and then negated.
The result is (+3 +1 +1 .1 .3 .1 .1 +1).
51. By definition
If T sends a 0 bit instead of 1 bit, its chip sequence is negated, with the i-th element becoming .Ti . Thus,
52. When two elements match, their product is +1. When they do not match, their product is .1. To make the sum 0, there must be as many matches as mismatches. Thus, two chip
sequences are orthogonal if exactly half of the corresponding elements match and exactly half do not match.
53. Just compute the four normalized inner products: (.1 +1 .3 +1 .1 .3 +1 +1) d (.1 .1 .1 +1 +1 .1 +1 +1)/8 = 1 (.1 +1 .3 +1 .1 .3 +1 +1) d (.1 .1 +1 .1 +1 +1 +1 .1)/8 = .1 (.1 +1 .3 +1 .1 .3 +1 +1) d (.1 +1 .1 +1 +1 +1 .1 .1)/8 = 0 (.1 +1 .3 +1 .1 .3 +1 +1) d (.1 +1 .1 .1 .1 .1 +1 .1)/8 = 1
The result is that A and D sent 1 bits, B sent a 0 bit, and C was silent.
54. 答：可以，每部电话都能够有自己到达端局的线路，但每路光纤都可以连接许多部电话。忽略语音压缩，一部数字PCM电话需要64kbps 的带宽。如果以64kbps 为单元来分割10Gbps，我们得到每路光缆串行156250 家。现今的有线电视系统每根电缆串行数百家。
55. 答：它既像TDM，也像FDM。100 个频道中的每一个都分配有自己的频带（FDM），在每个频道上又都有两个逻辑流通过TDM 交织播放（节目和广告交替使用频道）。 This example is the same as the AM radio example given in the text, but neither is a fantastic example of TDM because the alternation is irregular.
56. A 2-Mbps downstream bandwidth guarantee to each house implies at most 50 houses per coaxial cable. Thus, the cable company will need to split up the existing cable into 100 coaxial cables and connect each of them directly to a fiber node.
57. The upstream bandwidth is 37 MHz. Using QPSK with 2 bits/Hz, we get 74 Mbps
upstream. Downstream we have 200 MHz. Using QAM-64, this is 1200 Mbps. Using QAM-256, this is 1600 Mbps.
58. Even if the downstream channel works at 27 Mbps, the user interface is nearly always 10-Mbps Ethernet. There is no way to get bits to the computer any faster than 10-Mbps under these circumstances. If the connection between the PC and cable modem is fast Ethernet, then the full 27 Mbps may be available. Usually, cable operators specify 10 Mbps Ethernet because they do not want one user sucking up the entire bandwidth.
第 3 章 数据链路层
1. 答：由于每一帧有 的概率正确到达，整个信息正确到达的概率为 p==。
为使信息完整的到达接收方，发送一次成功的概率是p ，二次成功的概率是(1-p)p，三次成功的概率为(1-p )2 p，i 次成功的概率为(1-p)i-1 p，因此平均的发送次数等于：
2. The solution is
(a) 00000100 01000111 01111110 (b) 01111110 01000111 01111110 01111110
(c) 01111110 01000111 1 0 0 01111110
3. After stuffing, we get A B ESC ESC C ESC ESC ESC FLAG ESC FLAG D.
4. If you could always count on an endless stream of frames, one flag byte might be enough. But what if a frame ends (with a flag byte) and there are no new frames for 15 minutes. How will the receiver know that the next byte is actually the start of a new frame and not just noise on the line The protocol is much simpler with starting and ending flag bytes.
5. The output is 01111010.
6. 答：可能。假定原来的正文包含位序列01111110 作为数据。位填充之后，这个序列将变成01111010。如果由于传输错误第二个0 丢失了，收到的位串又变成01111110，被接收方看成是帧尾。然后接收方在该串的前面寻找检验和，并对它进行验证。如果检验和是16 位，那么被错误的看成是检验和的16 位的内容碰巧经验证后仍然正确的概率是1/216。如果这种概率的条件成立了，就会导致不正确的帧被接收。显然，检验和段越长，传输错误不被发现的概率会越低，但该概率永远不等于零。
7. 答：如果传播延迟很长，例如在探测火星或金星的情况下，需要采用前向纠错的方法。还有在某些军事环境中，接收方不想暴露自己的地理位置，所以不宜发送反馈信号。如果错误率足够的低，纠错码的冗余位串不是很长，又能够纠正所有的错误，前向纠错协议也可能是比较合理和简单的。
8. Making one change to any valid character cannot generate another valid character due to the nature of parity bits. Making two changes to even bits or two changes to odd bits will give another valid character, 所以Hamming 距离为2
9. Parity bits are needed at positions 1, 2, 4, 8, and 16, so messages that do not extend beyond bit 31 (including the parity bits) fit. Thus, five parity bits are sufficient. The bit pattern transmitted is 01110101
10. The encoded value is .
11. If we number the bits from left to right starting at bit 1, in this example, bit 2 (a parity bit) is incorrect. The 12-bit value transmitted (after Hamming encoding) was 0xA4F. The original 8-bit data value was 0xAF.
12. 答：单个错误将引起水平和垂直奇偶检查都出错。两个错误，无论是否同行或者同列，也容易被检测到。对于有三位错误的情况，就有可能无法检测了。for example, if some bit is inverted along with its row and column parity bits. Even the corner bit will not catch this.
13. 答：用n 行k 列的矩阵来描述错误图案，在该矩阵中，正确的位用0 表示，不正确的位用1 表示。由于总共有4 位传输错误，每个可能的错误矩阵中都恰有4 个1。则错误矩阵的个数总共有C nk4个。而在错误矩阵中，当4 个1 正好构成一个矩形的4 个顶点的时候，这样的错误是检测不出来的。则检测不出来的错误矩阵的个数为C2n × C2k
所以，错误不能检测的概率是：
即；
14. 答：如所列的除式，所得的余数为x2+x+1。
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