【文章內(nèi)容簡(jiǎn)介】 ts = Mbps. It can be used to acmodate DS2 service. 39. Message switching sends data units that can be arbitrarily long. Packet switching has a maximum packet size. Any message longer than that is split up into multiple packets. 40. 答： 當(dāng)一條線路（例如 OC3）沒(méi)有被多路復(fù)用，而僅從一個(gè)源輸入數(shù)據(jù)時(shí)，字母 c（表示 conactenation，即串聯(lián)）被加到名字標(biāo)識(shí)的后面，因此， OC3 表示由 3 條單獨(dú)的OC1 線路復(fù)用成 ，而 OC3c 表示來(lái)自單個(gè)源的 的數(shù)據(jù)流。 OC3c 流中所包含的 3 個(gè) OC1 流按列 交織編排，首先是流 1 的第 1 列，流 2 的第 1 列，流 3 的第1 列，隨后是流 1 的第 2 列，流 2 的第 2 列， ??以此類推，最后形成 270 列寬 9 行高的幀。 OC3c 流中的用戶實(shí)際數(shù)據(jù)傳輸速率比 OC3 流的速率略高（ 和），因?yàn)橥烽_(kāi)銷僅在 SPE 中出現(xiàn)一次，而不是當(dāng)使用 3 條單獨(dú) OC1 流時(shí)出現(xiàn)的 3 次。換句話說(shuō)， OC3c 中 270 列中的 260 列可用于用戶數(shù)據(jù)，而在 OC3 中僅能使用258列。更高層次的串聯(lián)幀（如 OC12c）也存在。 OC12c 幀有 12*90=1080 列和 9 行。其中段開(kāi)銷和線路開(kāi)銷占 12*3=36 列，這樣同步載荷信封就有 108036=1044 列。 SPE 中僅 1 列用于通路開(kāi)銷，結(jié)果就是 1043 列用于用戶數(shù)據(jù)。 由于每列 9 個(gè)字節(jié)，因此一個(gè) OC12c 幀中用戶數(shù)據(jù)比特?cái)?shù)是 8 ＝ 75096。每秒 8000 幀，得到用戶數(shù)據(jù)速率 750968000 =600768000bps，即 。 所以，在一條 OC12c 連接中可提供的用戶帶寬是 。 41. 答： The three works have the following properties: 星型：最好為 2，最差為 2，平均為 2； 環(huán)型：最好為 1，最差為 n/2，平均為 n/4 如果考慮 n 為奇偶數(shù)， 則 n 為奇數(shù)時(shí)，最壞為（ n1） /2，平均為（ n+1） /4 n 為偶數(shù)時(shí)，最壞為 n/2，平均為 n2/4(n 全連接：最好為 1，最差為 1，平均為 1。 42. 對(duì)于電路交換， t= s時(shí)電路建立起來(lái)； t ＝ s+ x /d 時(shí)報(bào)文的最后一位發(fā)送完畢； t = s+ x/b+kd時(shí)報(bào)文到達(dá)目的地 。而對(duì)于分組交換，最后一位在 t=x/b 時(shí)發(fā)送完畢。 為到達(dá)最終目的地，最后一個(gè)分組必須被中間的路由器重發(fā) k 次，每次重發(fā)花時(shí)間p/ b，所以總的延遲為 為了使分組交換比電路交換快，必須： 所以： 43. 答： 所需要的分組總數(shù)是 x /p ，因此總的數(shù)據(jù)加上頭信息交通量為 (p+h)x/p位。 源端發(fā)送這些位需要時(shí)間為 (p+ /pb 中間的路由器重傳最后一個(gè)分組所花的總時(shí)間為 (k1) + / b 因此我們得到的總的延遲為 對(duì)該函數(shù)求 p 的導(dǎo)數(shù)，得到 令 得到 因?yàn)?p＞ 0，所以 故 時(shí)能使總的延遲最小。 44. Each cell has six neighbors. If the central cell uses frequency group A, its six neighbors can use B, C, B, C, B, and C respectively. In other words, only 3 unique cells are needed. Consequently, each cell can have 280 frequencies. 45. First, initial deployment simply placed cells in regions where there was high density of human or vehicle population. Once they were there, the operator often did not want to go to the trouble of moving them. Second, antennas are typically placed on tall buildings or mountains. Depending on the exact location of such structures, the area covered by a cell may be irregular due to obstacles near the transmitter. Third, some munities or property owners do not allow building a tower at a location where the center of a cell falls. In such cases, directional antennas are placed at a location not at the cell center. 46. If we assume that each microcell is a circle 100 m in diameter, then each cell divide it by the area of 1 microcell, we get 15,279 microcells. Of course, it is impossible to tile the plane with circles (and San Francisco is decidedly threedimensional), but with 20,000 microcells we could probably do the job. 47. Frequencies cannot be reused in adjacent cells, so when a user moves from one cell to another, a new frequency must be allocated for the call. If a user moves into a cell, all of whose frequencies are currently in use, the user’s call must be terminated. 48. It is not caused directly by the need for backward patibility. The 30 kHz channel was indeed a requirement, but the designers of DAMPS did not have to stuff three users into it. They could have put two users in each channel, increasing the payload before error correction from 260 50= 260 =kbps. Thus, the quality loss was an intentional tradeoff to put more users per cell and thus get away with bigger cells. 49. DAMPS uses 832 channels (in each direction) with three users sharing a single channel. This allows DAMPS 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). DAMPS uses 30 KHz = MHz. The difference can be mainly attributed to the better speech quality provided by GSM (13 Kbps per user) over DAMPS (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. 51. By definition If T sends a 0 bit instead of 1 bit, its chip sequence is negated, with the ith Ti . Thus, 52. When two elements match, their product is +1. When they do not match, their mismatches. Thus, two chip sequences are orthogonal if exactly half of the corresponding elements match and exactly half do not match. 53. Just pute the four normalized inner products: = 0 The result is that A and D sent 1 bits, B sent a 0 bit, and C was silent. 54. 答： 可以，每部電話都能夠有自己到達(dá)端局的線路，但每路光纖都可以連接許多部電話。忽略語(yǔ)音壓縮，一部數(shù)字 PCM電話需要 64kbps 的帶寬。如果以 64kbps 為單元來(lái)分割10Gbps，我們得到每路光纜串行 156250 家?，F(xiàn)今的有 線電視系統(tǒng)每根電纜串行數(shù)百家。 55. 答： 它既像 TDM，也像 FDM。 100 個(gè)頻道中的每一個(gè)都分配有自己的頻帶（ FDM），在每個(gè)頻道上又都有兩個(gè)邏輯流通過(guò) TDM 交織播放（節(jié)目和廣告交替使用頻道）。 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 2Mbps downstream bandwidth guarantee to each house implies at most 50 houses per coaxial cable. Thus, the cable pany 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 QAM64, this is 1200 Mbps. Using QAM256, this is 1600 Mbps. 58. Even if the downstream channel works at 27 Mbps, the user interface is nearly always 10Mbps Ether. There is no way to get bits to the puter any faster than 10Mbps under these circumstances. If the connection between the PC and cable modem is fast Ether, then the full 27 Mbps may be available. Usually, cable operators specify 10 Mbps Ether because they do not want one user sucking up the entire bandwidth. 第 3 章 數(shù)據(jù)鏈路層 1. 答： 由于每一幀有 的概率正確到達(dá)，整個(gè)信息正確到達(dá)的概率為 p==。 為使信息完整的到達(dá)接收方，