計算機網(wǎng)絡答案

1、第 1 章 概述1. 答：狗能攜帶 21千兆字節(jié)或者 168 千兆位的數(shù)據(jù)。 18 公里 /小時的速度等于 0.005 公里 / 秒，走過 x 公里的時間為 x / 0.005 = 200x 秒， 產(chǎn)生的數(shù)據(jù)傳輸速度為 168/200x Gbps 或者 840 /x Mbps 。因此，與通信線路相比較，若 x24 X答：數(shù)據(jù)速率為 480 X需要442Mbps的帶寬，對應的波長范圍是。9. 答：奈奎斯特定理是一個數(shù)學性質(zhì)，不涉及技術處理。該定理說，如果你有一個函數(shù)，它的傅立葉頻譜不包含高于f的正弦和余弦，那么以2f的頻率采樣該函數(shù)，那么你就可以獲取該函數(shù)所包含的全部信息。因此奈奎斯特定理適用于

2、所有介質(zhì)。10. 答：3個波段的頻率范圍大約相等，根據(jù)公式也小，才能保持f大約相等。小的波段順便指出，3個帶寬大致相同的事實是所使用的硅的種類的一個碰巧的特性反映。11. 答：12. 答：1GHz微波的波長是30cm。如果一個波比另一個波多行進15cm，那么它們到達時將180異相。顯然，答案與鏈路長度是50km的事實無關。13. 答：If the beam is off by 1 mm at the end, it misses the detector. This amounts to a triangle with base100 m and height 0.001 m. The an

3、gle is one whose tangent is thus 0.00001. This an gle 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 tran sit every 491 sec on ds. Thus, there will be a handoff about every 8 minu tes and 11 sec on ds.15. The

4、satellite moves from being directly overhead toward the southern horizon, with a maximum . It takes 24 hours to go from directly excursion from the vertical of 2 overhead to maximum excurs ion and the n back.16. The number of area codes 810, or 640. Thus, 10, which is 160. The number of prefixes was

5、8 X2 Xwas 8 Xhe 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

6、 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

7、near it, so those lines are wasted.18. 答：每部電話每小時做 0.5 次通話， 每次通話 6 分鐘。 因此一部電話每小時占用一條電 路3 分鐘，60/3=20，即20 部電話可共享一條線路。 由于只有 10%的呼叫是長途， 所以 200 部電話占用一條完全時間的長途線路。局間干線復用了 1000000/4000=250 條線路，每條線 路支持 200 部電話，因此，一個端局可以支持的電話部數(shù)為 200*250=50000 。19. 答：雙絞線的每一條導線的截面積是 ，每根雙絞線的兩條導線在 10km 長的情況下體 積是，即約為15708cm。由于銅的密度等

8、于9.0g/cm3，每個本地回路的質(zhì)量為9X15708=141372 9kg，由于每千克銅的價格是3 10 1.41 X,約為141kg。這樣，電話公司擁有的本地回路的總質(zhì)量等于141X1000X104= 109美元。X 9=4.2 1.4 X0美元，所以總的價值等于 3X20. Like a single railroad track, it is half duplex. Oil can flow in either direction, but not both ways at once.21. 通常在物理層對于在線路上發(fā)送的比特不采取任何差錯糾正措施。在每個調(diào)制解調(diào)器中都包括一個 CP

9、U 使得有可能在第一層中包含錯誤糾正碼,從而大大減少第二層所看到的錯誤率。由調(diào)制解調(diào)器做的錯誤處理可以對第二層完全透明。現(xiàn)在許多調(diào)制解調(diào)器都有內(nèi)建的錯誤處理功能。22. 每個波特有 4 個合法值,因此比特率是波特率的兩倍。對應于 1200 波特,數(shù)據(jù)速率是 2400bps。23. 相位總是 0,但使用兩個振幅,因此這是直接的幅度調(diào)制。24. If all the points are equidistant from the origin, they all have the same amplitude, so amplitude modulation is not being used.

10、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 an

11、d 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.the queueing delay is als

12、o 1.1 msec, the total time is 2.2 msec. Over ADSL there is no queueing delay, so the dow nl oad 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 interferenee. The minimum bandwidth required is 400 9 =43,600 Hz. -+10 4000 x2

13、9. 答：125的采樣時間對應于每秒8000次采樣。一個典型的電話通道為4kHz。根據(jù)奈奎斯特定理，為獲取一個4kHz的通道中的全部信息需要每秒8000次的采樣頻率。(Actually the nominal ban dwidth is somewhat less, but the cutoff is not sharp.)30. 每一幀中，端點用戶使用 193位中的168(7*24)位，開銷占25 (=193-168)位，因此 開銷比例等于 25/193=13%。31. 答：比較使用如下方案的無噪聲4kHz信道的最大數(shù)據(jù)傳輸率：(a) 每次采樣2比特的模擬編碼16kbps(b) T1 PCM

14、 系統(tǒng)56kbpsIn both cases 8000 samples/sec are possible. With dibit en cod ing, two bits are sent per sample.With T1, 7 bits are sent per period. The respective data rates are 16 kbps and 56 kbps.32. 答：10個幀。在數(shù)字通道上某些隨機比特是0101010101模式的概率是1/1024。察看10個幀，若每一幀中的第一位形成比特串0101010101，則判斷同步成功，而誤判的概率為1/1024，小于0.00

15、1。33. 答：有。編碼器接受任意的模擬信號，并從它產(chǎn)生數(shù)字信號。而解調(diào)器僅僅接受調(diào)制了 的正弦(或余弦)波，產(chǎn)生數(shù)字信號。34. 答：a. CCITT 2.048Mbps標準用32個8位數(shù)據(jù)樣本組成一個125的基本幀，30個信道用于傳信息，2個信道用于傳控制信號。在每一個4kHz信道上發(fā)送的數(shù)據(jù)率就是8*8000=64kbps 。b. 差分脈碼調(diào)制(DPCM )是一種壓縮傳輸信息量的方法，它發(fā)送的不是每一次抽樣的二進制編碼值，而是兩次抽樣的差值的二進制編碼?，F(xiàn)在相對差值是4位，所以對應每個4kHz信道實際發(fā)送的比特速率為4*8000=32bps。c. 增量調(diào)制的基本思想是：當抽樣時間間隔s

16、t很短時，模擬數(shù)據(jù)在兩次抽樣之間的變化很小，可以選擇一個合適的量化值？作為階距。把兩次抽樣的差別近似為不是增加一個？就是減少一個？。這樣只需用1bit二進制信息就可以表示一次抽樣結果，而不會引入很大誤差。因此，此時對應每個4kHz信道實際發(fā)送的數(shù)據(jù)速率為 1*8000=8kHz。35. 答：在波的1/4周期內(nèi)信號必須從 0上升到A。為了能夠跟蹤信號，在T/4的時間內(nèi)(假定波的周期是 T)必須采樣8次，即每一個全波采樣32次，采樣的時間間隔是1/x ,因此波的全周期必須足夠的長，使得能包含32次采樣，即T 32/x，或f max =x/32。36. 的漂移意味著109秒中的1秒，或1秒中的10-

17、9秒。對于 OC-1答：10-9速率，即 51.840Mbps，取近似值50Mbps，大約一位持續(xù) 20ns。這就說明每隔 20秒，時鐘就要偏離 1位。這就說明，時鐘必須每隔 10秒或更頻繁地進行同步，才能保持不會偏離太大。37. 答：基本的 SONET 幀是美 125 產(chǎn)生 810 字節(jié)。由于 SONET 是同步的，因此不論是 否有數(shù)據(jù)，幀都被發(fā)送出去。每秒 8000 幀與數(shù)字電話系統(tǒng)中使用的 PCM 信道的采樣頻率 完全一樣。810 字節(jié)的 SONET 幀通常用 90 列乘以 9 行的矩形來描述，每秒傳送 51.84Mbps ，即 8 810疋000= 51840000bps。這就是基本的

18、 SONET信道，它被稱作同步傳輸信號 STS-1, 所有的 SONET 干線都是由多條 STS-1 構成。每一幀的前 3 列被留作系統(tǒng)管理信息使用，前 3 行包含段開銷，后 6 行包含線路開銷。 剩下的87列包含879X8X8000= 50112000bps。被稱作同步載荷信封的數(shù)據(jù)可以在任何位置 開始。線路開銷的第一行包含指向第一字節(jié)的指針。同步載荷信封（SPE ）的第一列是通路開銷。通路開銷不是嚴格的 SONET 結構，它在嵌入在載荷信封中。通路開銷端到端的流過網(wǎng)絡， 因此把它與端到端的運載用戶信息的 SPE 相關聯(lián)是有意義的。然而，它確實從可提供給端 點用戶的 50.112Mbps 中

19、又減去 1X9X8X8000= 576000bps， 即 0.576Mbps， 使之變成 49.536Mbps 148.608 Mbps。 3= 。 OC-3 相當于 3 個 OC-1 復用在一起，因此其用戶數(shù)據(jù)傳 輸速率是 49.546 X38. VT1.5 can 8 bits =1.728 Mbps. It can be 9 rows Xaccommodate 8000 frames/sec 3 coluXmns X 9 rows used to accommodate DS-1. VT2 can accommodate 8000 frames/sec 4 columns X8bits

20、= 2.304 Mbps. It can be used to accommodate European CEPT-1 service. X 8 bits = 6.912 Mbps. It 9 rows X12 columns XVT6 can accommodate 8000 frames/sec can bXe used to accommodate DS-2 service.39. Message switching sends data units that can be arbitrarily long. Packet switching has a maximum packet s

21、ize. Any message longer than that is split up into multiple packets.40. 答：當一條線路（例如 0C-3 ）沒有被多路復用，而僅從一個源輸入數(shù)據(jù)時，字母c （表示conactenation,即串聯(lián)）被加到名字標識的后面， 因此，0C-3表示由3條單獨的0C-1線 路復用成155.52Mbps，而OC-3C表示來自單個源的 155.52Mbps的數(shù)據(jù)流。OC-3c流中所 包含的 3 個 0C-1 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

22、 2 列, ?以此類推,最后形成 270 列寬 9 行高的幀。OC-3c 流 中 的 用 戶 實 際 數(shù) 據(jù) 傳 輸 速 率 比 OC-3 流 的 速 率 略 高 （ 149.760Mbps 和 148.608Mbps ）,因為通路開銷僅在 SPE中出現(xiàn)一次，而不是當使用 3條單獨OC-1流時出 現(xiàn)的 3 次。 換句話說， OC-3c 中 270 列中的 260 列可用于用戶數(shù)據(jù)，而在 OC-3 中僅能使 用258列。更高層次的串聯(lián)幀（如 OC-12C）也存在。OC-12c 幀有 12*90=1080 列和 9 行。其中段開銷和線路開銷占 12*3=36 列，這樣同步載荷 信封就有 1080-

23、36=1044 列。 SPE 中僅 1 列用于通路開銷， 結果就是 1043 列用于用戶數(shù)據(jù)。 由于每列 9 9 X1043= 75096。每秒 8000 幀，得到用戶數(shù)據(jù)速率 75096X8000 個字節(jié)，因此一 個OC-12C幀中用戶數(shù)據(jù)比特數(shù)是 8 0600768000bps，即600.768Mbps。所以，在一條 OC-12c 連接中可提供的用戶帶寬是 600.768Mbps。41. 答： The three networks 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

24、數(shù)，則n為奇數(shù)時，最壞為（n-1） 12，平均為（n +1） /41） -n為偶數(shù)時，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。x /d - = s+ 42.對于電路交換，t= s時電路建立起來；t壬 時報文的最后一位發(fā)送完畢；t時 發(fā)送完畢。x/b+kd時報文到達目的地。而對于分組交換，最后一位在t=x/b s+1次，每次重發(fā)花時間p/-為到達最終目的地，最后一個分組必須被中間的路由器重發(fā)k 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。/pb ）x源端發(fā)送這

25、些位需要時間為（p+h）/ +h中間的路由器重傳最后一個分組所花的總時間為（k-1）（p b因此我們得到的總的延遲為對該函數(shù)求p的導數(shù)，得到令得到p 0，所以因為故時能使總的延遲最小。44. Each cell has six n eighbors. If the cen tral cell uses freque ncy group A, its six n eighbors canuse B, C, B, C, B, and C respectively. In other words, only 3 unique cells are needed. Con seque ntly, eac

26、h cell can have 280 freque ncies.45. First, i nitial deployme nt simply placed cells in regi ons where there was high den sity of huma n or vehicle populatio n. 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 bu

27、ildi ngs or mountains. Depe nding on the exact locati on of such structures, the area covered by a cell may be irregular due to obstacles near the transmitter. Third, some communities or property owners do not allow building a tower ata location where the center of a cell falls. In such cases, direc

28、tional antennas are placed at a locati on not at the cell cen ter.46. If we assume that each microcell is a circle 100 .If we take the area of San dm in diameter, then each cell has an area of 2500 108 m2 and divide it by the area of 1 microcell, we get 15,279 Fran cisco, 1.2 microcells. Of course,

29、it is impossible to tile the pla ne with circles (and San Fran cisco is decidedly three-dime nsion al), but with 20,000 microcells we could probably dothe job.47. Freque ncies cannot be reused in adjace nt cells, so whe n a user moves from one cell to ano ther,a new freque ncy must be allocated for

30、the call. If a user moves into a cell, all of whose freque ncies are curre ntly in use, the user s call must be term in ated.48. It is not caused directly by the need for backward compatibility. The 30 kHz channel wasin deed a requireme nt, but the desig ners of D-AMPS did not have to stuff three us

31、ers into it. They could have put two users in each cha nn el, i ncreas ing the 19.5 kbps. 75 =13 kbps to 260 payload before error correct ion from 26050= Thus,the quality loss was an inten ti onal trade-off to putmore users per cell and thus get away with bigger cells.49. D-AMPS uses 832 cha nn els

32、(in each direct ion) with three users shari ng a sin gle cha nn el. This allows D-AMPS to support up to 2496 users simulta neously per cell. GSM uses 124 cha nn els with eight users shari ng a sin gle cha nn el. This allows GSM to support up to 992 users simulta neously. Both systems use about the s

33、ame amount of spectrum (25 MHz in each direct ion).124 892 = 26.76 MHz. GSM uses 200 KHz XD-AMPS uses 30 KHzX =24.80 MHz. The difference can be mainly attributed to the better speech quality provided by GSM (13 Kbps per user) over D-AMPS (8 Kbps per user).50. The result is obtained by negating each

34、of A, B, and C and then adding the three chip seque nces. Alter natively the three can be added and the n n egated.The 1 +1). 1 3 1 result is 什3 +1 +151. By defi nitionIf T sends a 0 bit in stead of 1 bit, its chip seque nce is n egated, with the i-th eleme nt Ti . Thus, becoming52. When two eleme n

35、ts match, their product is 1. To make the sum 0, there must +1. When they do not match, their product is be as many matches as mismatches. Thus, two chip seque nces are orthogo nal if exactly half of the corresp onding eleme nts match and exactly half do not match.53. Just compute the four no rmaliz

36、ed inner products:3 1 +1 ( 1 +1 +1)/8 = 1 1 +1 +1 1 1 3 +1 +1) d ( 1 +13 +1 +1) d 1 3 +11 +1( 1 1)/8 = 1 +1 +1 +1 1 +1 1(1 +1 +1 1 +1 3 +1+1) d( 1 3 +1 1 +1 ( 1)/8 = 0 1 +11)/8 = 1 +1 1 1 11 +13 +1 +1) d ( 1 3 +1 1 +1 ( 1The result is that Aand D sent 1 bits, B sent a 0 bit, andC was sile nt.54. 答：可

37、以， 每部電話都能夠有自己到達端局的線路， 但每路光纖都可以連接許多部電話。 忽略語音壓縮，一部數(shù)字 PCM 電話需要 64kbps 的帶寬。如果以 64kbps 為單元來分割 lOGbps,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個頻道中的每一個都分配有自己的頻帶( FDM)， 在每個頻道上又都有兩個邏輯流通過 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 neithe

38、r is a fantastic example of TDM because the alternation is irregular.56. A 2-Mbps downstream bandwidth guarantee to each house implies at most 50 houses per coaxial cable. Thus, the cable company will need to split up the existing cable into 100 coaxial cables and connect each of them directly to a

39、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 QAM-64, this is 1200 Mbps. Using QAM-256, this is 1600 Mbps.58. Even if the downstream channel works at 27 Mbps, the user interface is nearly always 10-Mbps Ethernet. There is no way to get bits to the computer any faster than 10-Mbps under these circumstances. If the connection between the PC and cable modem is fast Ethernet, then the full 27 Mbps may be available. Usually