Answers of Exercise 6

munchdrabNetworking and Communications

Oct 30, 2013 (3 years and 1 month ago)

75 views

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.

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.

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

1

4

3

5

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