Part 3 : Answer the following questions briefly. Q.21 In Ch.7 we talked about pi
ID: 2082209 • Letter: P
Question
Part 3 : Answer the following questions briefly.
Q.21 In Ch.7 we talked about piggybacking. What does it refer to ?
Q.22 In AMI alternating 1's have di erent polarities. What is the objective of this alternating the polarity (i.e. what is achieved by it) ?
Q.23 (a) Name a media access control technique that is used for the transmission of short messages over long distances (such as satellite communications).
(b) Name a media access control technique that used in Ethernet.
Q.24 (a) List 2 major characteristics of the rst generation North American cellular system (AMPS).
(b) List at least 2 major characteristics of the 2G (second generation) cellular systems that is di ererent from 1G.
Q.25 Suppose a cellular service provider is given suffcient bandwidth that would permit 280 simultaneous links without frequency reuse. Now let the system use 7-cell clusters for frequency reuse. How many simultaneous links can be served if the coverage area is made up of 56 cells ?
Explanation / Answer
21) In general communication system the sender/transmitter sends messages as frame and it is received at the receiving end. After that, an ack/ acknowledgement is sent by the receiver that refers to the successful reception of the message. But in piggybacking technique, the receiver does not send the ack immediately after the reception. It waits untill whole message packet is received and the next packet passes the network layer. Then the receiver sends the delayed ack attached with dataframe. This process is used in two way communication of data transmission.
22) AMI/Alternate Mark Inversion code is digital telecommunicationstechnique to maintain system synchronization. This is a bipolar data communication technique. Here binary '1's are taken as +5V and for '0's it is -5V, instead of 0V. Now, what is the benifit? The answer is when a large data containing too many '1's and '0's are transmitted , due to noise in channel the voltage levels are distorted and error occurs. As for example the +5V distorted to +2V and the 0V distorted to +2V then receiver can not detects the actual data. Now in AMI the alternating polarity is introduced to detect exacty the data. Thus, the main advantage is error deduction.
23. a) Both CDMA(Code Division Multiple Access) and FDMA(Frequency Division Multiple Access) technique can be used. But for FDMA, a guard band should be used.
b) The most common media access control used in ethernet is Carrier-sense multiple access with collision detection (CSMA/CD) .
24. a)AMPS was originally standardized by American National Standards Institute (ANSI). Two major characteristics are,
1.AMPS operated in 850MHz band.For each market area, the United States Federal Communications Commission (FCC) allowed two licensees (networks) known as "A" and "B" carriers. Each carrier within a market used a specified "block" of frequencies consisting of 21 control channels and 395 voice channels. Originally, the B (wireline) side license was usually owned by the local phone company, and the A (non-wireline) license was given to wireless telephone providers.
2.Each duplex channel was composed of 2 frequencies. 416 of these were in the 824–849 MHz range for transmissions from mobile stations to the base stations, paired with 416 frequencies in the 869–894 MHz range for transmissions from base stations to the mobile stations. Each cell site used a different subset of these channels than its neighbors to avoid interference. This significantly reduced the number of channels available at each site in real-world systems. Each AMPS channel had a one way bandwidth of 30 kHz, for a total of 60 kHz for each duplex channel.
b)Two major characteristics of 2g are,
1. The communication system is digitally encrypted. Thus it has higher data security than 1g and the it is more efficient for communication.
2. In 2g, mobile data use facility is introduced. Thus it gave the opertunity of mobile data and MMS.
25. Service provider provides 280 simultaneous links without frequency reuse.
newly formed clusters have 7 cells each. thus, q=7
total number of cluster=56/7=8
Thus, frequency reuse factor N can be found as, 3N=q^2=49
=> N=49/3=16.33
again, N=i^2+j^2+ij
Taking i=3, j=2 , N=19>16.33
thus reuse factor is 19.
Thus for single tier approximation, channel/cell= 280/19*8=1.84 taken as 1 channel /cell