南海凝心计算机网络第四版习题答案(中文版)
2-28 Ten signals, each requiring 4000 Hz, are multiplexed on to a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel? Assume that the guard bands are 400 Hz wide.
有10个信号，每个都要求4000Hz，现在用FDM将它们复用在一条信道上。对于被复用的信道，最小要求多少带宽？假设防护频段为400Hz宽。
有10个4000Hz信号。我们需要9个防护频段来避免干扰。最小带宽需求是4000×10+400×9 =43,600 Hz.
2-29 Why has the PCM sampling time been set at 125 µsec?
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4kHz 的通道中的全部信息需要每秒8000 次的采样频率。
(实际上额定带宽稍有些少，截止点并不清晰)
2-30 What is the percent overhead on a T1 carrier; that is, what percent of the 1.544 Mbps are not delivered to the end
user?
每一帧中，端点用户使用193 位中的168（7*24）位，开销占25（=193-168）位，因此开销比例等于25/193=13%。
2-33 What is the difference, if any, between the demodulator part of a modem and the coder part of a codec? (After
all, both convert analog signals to digital ones.)
答：有。编码器接受任意的模拟信号，并从它产生数字信号。而解调器仅仅接受调制了的正弦（或余弦）波，产生数字信号。
