计算机网络(第四版)课后习题(英文)+习题答案(中英文) - 图文 下载本文

如果网络容易丢失分组,那么对每一个分组逐一进行确认较好,此时仅重传因为 丢失 许多无线设备需要移动,电池使用寿命不长也是其缺点之一。

的分组。

Chapter 2 The Physical Problems 如果网络高度可靠,那么在不发差错的情况下,仅在整个文件传送的结尾发送一

次确认,从而减少了确认的次数,节省了带宽;不过,即使有单个分组丢失,也需 要重传整个文件。 26. Why does ATM use small, fixed-length cells?(E) 因为这样可以迅速地经由交换机转发,并且这是在硬件上完成的。这样的设计使 2. A noiseless 4-kHz channel is sampled every 1 msec. What is the maximum data rate?(E) 得制造可以同时并行处理多个 CELLS 的硬件设备更加容易。另外,它们不会阻碍 传输线路很久,更加容易保证提供出高质量的服务。 28. An image is 1024 x 768 pixels with 3 bytes/pixel. Assume the image is uncompressed. How long does it take to transmit it over a 56-kbps modem channel? Over a 1-Mbps cable modem? Over a 10-Mbps Ethernet? Over 100-Mbps Ethernet?(E) 该图像大小为 1024 * 768 * 3 * 8 = 18,874,368 bits. 传输速率为 56Kbits/sec,需要 18,874,368 / 56,000 = 337.042 sec. 传输速率为 1Mbits/sec, 需要 18,874,368 / 106 = 18.874 sec. 传输速率为 10Mbits/sec,需要 18,874,368 / 107 = 1.887 sec. 传输速率为 100Mbits/sec,需要 18,874,368 / 108 = 0.189 sec. 29. Ethernet and wireless networks have some similarities and some differences. One property of Ethernet is that only one frame at a time can be transmitted on an Ethernet. Does 802.11 share this property with Ethernet? Discuss your answer.(E) 想象一下隐藏终端的问题。假设一个无线网络里有五台终端,从 A 至 E,使它们 每一台都只可以联系到与其相邻的两个邻居之一,那么 A 在与 B 通讯的同时 D 可 以与 E 进行通讯。因此无线网络有潜在的并行性,这与以太网上不同的。 30. Wireless networks are easy to install, which makes them inexpensive since installation costs usually far overshadow equipment costs. Nevertheless, they also have some disadvantages. Name two of them.(E)

无线网络的缺点:一是安全性,偶然出现在无线网络内的人都能监听到网络上传

递的消息;再有就是可靠性,无线网络在传输过程中会出现很多错误;另外,

由尼奎斯特定理,无噪声信道最大数据传输率=2Hlog2V b/s。依题有带宽 H =

4kHz,因此最大数据传输率决定于每次采样所产生的比特数(log2V)。 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 8.2Mbps。

3. Television channels are 6 MHz wide. How many bits/sec can be sent if four-level digital signals are used? Assume a noiseless channel.(E) 依题有带宽 H = 6MHz,每次采样 log2V = 2bit

由尼奎斯特定理,可发送的最大数据传输率为 2Hlog2V = 24Mbps。 4. If a binary signal is sent over a 3-kHz channel whose signal-to-noise ratio

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dB, what is the maximum achievable data rate?(M)

由香农定理信道比为 S/N 的有噪声信道的最大数据传输率 = Hlog2(1+S/N)。

依题知带宽 H = 3kHz,信噪比为 10lgS/N = 20 dB,知 S/N =100 由于 log2101≈6.658,该信道的信道容量为 3log2(1+100)=19.98kbps 再根据尼奎斯特定理,发送二进制信号的 3kHz 信道的最大数据传

输速率为

2Hlog2V = 2*3 log22 = 6kbps 综上,可以取得的最大数据传输速率为 6kbps。

5. What signal-to-noise ratio is needed to put a T1 carrier on a 50-kHz line?(M)

T1 信号的带宽 = 1.544 * 106Hz,为发送 T1 信号,由香农定理,最大数据传输率

= Hlog62(1+S/N) = 1.544 * 10Hz ,依题知带宽 H = 50 kHz,解得 S/N = 231–1 再由尼奎斯特定理 2Hlog2V = 2Hlog2S/N = 93 dB 所以,在 50kHz 线路上使用 T1 载波需要 93dB 的信

噪比。

7. How much bandwidth is there in 0.1 micron of spectrum at a wavelength of 1

micron?(M)

依题知频段为 0.1 ,波长为 1

因此,在 0.1 的频段中可以有 30THz。

12. Multipath fading is maximized when the two beams arrive 180 degrees out of

The screen is 480 x 640 pixels, each pixel being 24 bits. There are 60 screen images phase. How much of a path difference is required to maximize the fading for a per second. How much bandwidth is needed, and how many microns of wavelength 50-km-long 1-GHz microwave link?(E) are needed for this band at 1.30 microns?(M) 由公式 ,这里光速 c =300000km/s,依题 f=1GHz,所以微波的波长是 传输数据的速率为 480×\慇2X640×24×\慇2X60bps,即 442Mbps。

8. It is desired to send a sequence of computer screen images over an optical fiber.

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。 18. A simple telephone system consists of two end offices and a single toll office to which each end office is connected by a 1-MHz full-duplex trunk. The average telephone is used to make four calls per 8-hour workday. The mean call duration is 6 min. Ten percent of the calls are long-distance (i.e., pass through the toll office). What is the maximum number of telephones an end office can support? (Assume 4 kHz per circuit.)(E) 需要 442Mbps 的带宽,对应的波长范围是 。 9. Is the Nyquist theorem true for optical fiber or only for copper wire?(D) 尼奎斯特定理是一个数学性质,不涉及技术处理。该定理说,如果你有一个函每部电话每小时做 0.5 次通话,每次通话 6 分钟。因此一部数,

电话每小时占用一条 它的傅立叶频谱不包含高于 f 的正弦和余弦,那么以 2 f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 的频率采样该函数,那么 途, 你就可以获取该函数所包含的全部信息。因此尼奎斯特定理适用于所有介质。 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 10. In Fig. 2-6 the lefthand band is narrower than the others. Why?(E) 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。. 21. The cost of a fast microprocessor has dropped to the point where it is now possible to put one in each modem. How does that affect the handling of telephone line errors?(E) 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. A modem constellation diagram similar to Fig. 2-25 has data points at the following coordinates: (1, 1), (1, -1), (-1, 1), and (-1, -1). How many bps can a modem 得小,才能保持⊿f 大约相等。

with these parameters achieve at 1200 baud?(E)

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. Radio antennas often work best when the diameter of the antenna is equal to

the wavelength of the radio wave. Reasonable antennas range from 1 cm to 5 meters

in diameter. What frequency range does this cover?(E) 由于这 3 个波段的频率范围大体上相等,根据公式 , 小的波段⊿ 也

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每个波特有 4 个合法值,因此比特率是波特率的两倍。