计算机网络(第4版) 清华大学出版社 习题答案(中文版) 下载本文

第 2 章 物理层 1.

答;本题是求周期性函数的傅立叶系数。而题面中所给出的为信号在一个周期内的解析式。

即;

2. 答:无噪声信道最大数据传输率公式:最大数据传输率=2Hlog2V b/s。因此最大数据传输率决定于每次采样所产生的比特数,如果每次采样产生16bits,那么数据传输率可达128kbps;如果每次采样产生1024bits,那么可达。注意这是对无噪声信道而言的,实际信道总是有噪声的,其最大数据传输率由香农定律给出。

3. 答:采样频率12MHz,每次采样2bit,总的数据率为24Mbps。

4. 答:信噪比为20 dB 即 S/N = 100.由于 log2101≈,由香农定理,该信道的信道容量为3log2(1++100) =。

又根据乃奎斯特定理,发送二进制信号的3kHz 信道的最大数据传输速率为 2*3 log22=6 kbps。

所以可以取得的最大数据传输速率为6kbps。

5. 答:为发送T1 信号,我们需要

所以,在50kHz 线路上使用T1 载波需要93dB 的信噪比。

6. 答:无源星没有电子器件,来自一条光纤的光照亮若干其他光纤。有源中继器把光信号转换成电信号以作进一步的处理。

7. 答:

因此,在的频段中可以有30THz。

8. 答:数据速率为480× 640×24× 60bps,即442Mbps。

需要442Mbps 的带宽,对应的波长范围是

9. 答:奈奎斯特定理是一个数学性质,不涉及技术处理。该定理说,如果你有一个函数,它的傅立叶频谱不包含高于f 的正弦和余弦,那么以2 f 的频率采样该函数,那么你就可以获取该函数所包含的全部信息。因此奈奎斯特定理适用于所有介质。

10. 答:3 个波段的频率范围大约相等,根据公式

小的波段⊿ 也小,才能保持⊿f 大约相等。

顺便指出,3 个带宽大致相同的事实是所使用的硅的种类的一个碰巧的特性反映。

11. 答:

12. 答:1GHz 微波的波长是30cm。如果一个波比另一个波多行进15cm,那么它们到达时将180异相。显然,答案与链路长度是50km 的事实无关。

13. 答:

If the beam is off by 1 mm at the end, it misses the detector. This amounts to a triangle with base 100 m and height m. The angle is one whose tangent is thus . This angle is about degrees.

14. With 66/6 or 11 satellites per necklace, every 90 minutes 11 satellites pass overhead. This means there is a transit every 491 seconds. Thus, there will be a handoff about every 8 minutes and 11 seconds.

15. The satellite moves from being directly overhead toward the southern horizon, with a maximum excursion from the vertical of 2. It takes 24 hours to go from directly overhead to maximum excursion and then back.

16. The number of area codes was 8× 2× 10, which is 160. The number of prefixes was 8× 8 ×10, or 640. Thus, the number of end offices was limited to 102,400. This limit is not a problem.

17. With a 10-digit telephone number, there could be 1010 numbers, although many of the area codes are illegal, such as 000. However, a much tighter limit is given by the number of end offices. There are 22,000 end offices, each with a maximum of 10,000 lines. This gives a maximum of 220 million telephones. There is simply no place to connect more of them. This could never be achieved in practice because some end offices are not full. An end office in a small town in Wyoming may not have 10,000 customers near it, so those lines are wasted.

18. 答:每部电话每小时做 次通话,每次通话6 分钟。因此一部电话每小时占用一条电路3 分钟,60/3=20,即20 部电话可共享一条线路。由于只有10%的呼叫是长途,所以200 部电话占用一条完全时间的长途线路。局间干线复用了1000000/4000=250 条线路,每条线路支持200 部电话,因此,一个端局可以支持的电话部数为200*250=50000。

19. 答:双绞线的每一条导线的截面积是在10km 长的情况下体积是

,每根双绞线的两条导线

,即约为15708cm。

由于铜的密度等于cm3,每个本地回路的质量为9×15708 =141372 g,约为141kg。这样,电话公司拥有的本地回路的总质量等于141×1000×104= × 10 9kg,由于每千克铜的价格是3 美元,所以总的价值等于3× ×10 9= × 109美元。

20. Like a single railroad track, it is half duplex. Oil can flow in either direction, but not both ways at once.

21. 通常在物理层对于在线路上发送的比特不采取任何差错纠正措施。在每个调制解调器中都包括一个CPU 使得有可能在第一层中包含错误纠正码,从而大大减少第二层所看到的错误率。由调制解调器做的错误处理可以对第二层完全透明。现在许多调制解调器都有内建的错误处理功能。

22. 每个波特有4 个合法值,因此比特率是波特率的两倍。对应于1200 波特,数据速率是2400bps。

23. 相位总是0,但使用两个振幅,因此这是直接的幅度调制。

24. If all the points are equidistant from the origin, they all have the same amplitude, so amplitude modulation is not being used. Frequency modulation is never used in constellation diagrams, so the encoding is pure phase shift keying.

25. Two, one for upstream and one for downstream. The modulation scheme itself just uses amplitude and phase. The frequency is not modulated.

26. There are 256 channels in all, minus 6 for POTS and 2 for control, leaving 248 for data. If 3/4 of these are for downstream, that gives 186 channels for downstream. ADSL modulation is at 4000 baud, so with QAM-64 (6 bits/baud) we have 24,000 bps in each of the 186 channels. The total bandwidth is then Mbps downstream.

27. A 5-KB Web page has 40,000 bits. The download time over a 36 Mbps channel is msec. If the queueing delay is also msec, the total time is msec. Over ADSL there is no queueing delay, so the download time at 1 Mbps is 40 msec. At 56 kbps it is 714 msec.

28. There are ten 4000 Hz signals. We need nine guard bands to avoid any interference. The minimum bandwidth required is 4000× 10 + 400×9 =43,600 Hz.