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 0.001
m. The angle is one whose tangent is thus 0.00001. This angle is about 0.00057 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0.5 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 ,每根双绞线的两条导线在10km 长的情况下体积
是 ,即约为15708cm。由于铜的密度等于9.0g/cm
3
,每个本地
回路的质量为59×15708 =141372 g,约为141kg。这样,电话公司拥有的本地回路的总质量等于
141×1000×10
4
=1.41×10
9
kg,由于每千克铜的价格是3 美元,所以总的价值等于3×1.4×10
9
=4.2×10
9
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 4.464 Mbps downstream.
27. A 5-KB Web page has 40,000 bits. The download time over a 36 Mbps channel is 1.1 msec. If the queueing delay is also
1.1 msec, the total time is 2.2 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.
29. 答:125 的采样时间对应于每秒8000 次采样。一个典型的电话通道为4kHz。根据奈奎斯特定理,为获取一个