Answers of Exercise 6

1. Answers of Exercise 6 1. For an analogy TV signal, its frequency range is in [0, 6MHz]. To transmit the TV signal across a digital network, it is necessary to convert the analogy signal to a digital TV signal. What is the minimum sampling frequency in such conversion? Suppose that every sample will be encoded into 16 bits binary value (this is called TV PCM coding). Calculate bit rate of the digital TV signal after PCM coding. Answer: According to Nyquist sampling theorem, the minimum sampling frequency Fs = 2B = 2 × 6 = 12MHz Bit rate of TV PCM coding = 12 × 16 = 192 Mbps 2.Summarize the features and performance of typical long-distance connection technologies including T and OC series services, conventional modem, ISDN, xDSL and Cable modem. Answer: a) T and OC services provide digital connections for digital telephones. They can be also used for computer communications and building WANs. T lines use coax cables and usually offer digital services of 64Kbps, 1.5Mbps (T1), 6.3Mbps (T2), and 44.7Mbps (T4). OC lines use optical fibers and offer services of 51.8Mbps (OC-1), 155.5Mbps (OC-3), 622Mbps (OC-12), etc. b) Conventional modem, ISDN and xDSL are local loop technologies for transmissions of digital data between subscribers and local central offices. All of them use same copper pairs which already exist. Modem is used for data transmissions over the analogy telephone line in low frequency range (300~3300Hz), while xDSL use high frequency range (>25KHz) to transmit data. ISDN provides integrated services of voice, low-speed video and data transmissions. c) Cable modem is used for data transmissions over CATV.

2. Answers of Exercise 6 3. Explain why bit rates of the upstream and the downstream in ADSL are not fixed? Answer:Characteristic of each line is different from others. According to Shannon channel capacity, the maximum data amount transmitted through a line is depended upon its signal-to-noise ratio. If a line has a good quality, more data can be sent. Otherwise it can only send less data. On the other hand, the quality of a given line is not fixed but dynamically changed. Therefore, bits rates in ADSL are changed and not fixed. 4. In a packet switch network, the address of each computer consists two parts: one identifies a switch and other identifies a computer attached to that switch. Why? Answer: In a packet switch network, after a computer sends out a packet to its switch, the packet will be forwarded from switch to switch, and finally arrives a destination switch and is delivered to the destination computer. When a switch receive a packet, it checks packet’s destination switch. If the destination switch is same as the current switch, the packet will be delivered to a computer connecting the switch. If not, the switch must forward the packet to next hop (switch) only based on its destination switch. Thus, such two-part addressing makes switch forwarding easier and efficient.

3. Answers of Exercise 6 5. Suppose that a packet switch network with a five nodes is given below. Give a routing table for each of the five nodes. 2 3 5 1 4 Answer (one solution): node 1 node 2 node 3 node 4 node 5 dest next dest next dest next dest next dest next 1 - 1 (2,1) 1 (3,4) 1 (4,1) 1 (5,4) 2 (1,2) 2 - 2 (3,2) 2 (4,3) 2 (5,4) 3 (1,4) 3 (2,3) 3 - 3 (4,3) 3 (5,4) 4 (1,4) 4 (2,3) 4 (3,4) 4 - 4 (5,4) 5 (1,4) 5 (2,3) 5 (3,4) 5 (4,5) 5 - 1 - 1 (2,1) 2 (3,2) 1 (4,1) 5 - 2 (1,2) 2 - 3 - 4 - * (5,4) * (1,4) * (2,3) * (3,4) 5 (4,5) * (4,3) Routing tables without default routes Routing tables with default routes