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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Chapter 13 Multiple Access
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.1 Multiple-access protocols
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 13.1 Random Access MACSMA CSMA/CD CSMA/CA Each Station has the right to the medium without being controlled by any other station
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.3 ALOHA network The earliest random-access method,developed in 1970s Designed to be used on wireless Local area Network at 9600bps Based on Following rules Multiple Access Any station can send a frame when it has to send a frame Acknowledgment After sending a frame the station waits for an acknowledgment.If it does not receive the Ack within 2times maximum propagation delay,It tries sending the frame again after a random amount of time
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.4 Procedure for ALOHA protocol
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.5 Collision in CSMA Sense before transmit The possibility of collision still exist because of the propagation delay
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.6 Persistence strategies It defines the procedure for a station that senses a busy medium Station senses the line,If the line is Idle station sends the frame immediately If the line is not Idle station waits for a random period of time and then senses the line again This method has two variations I-PERSISTANT Station senses the line,If the line is Idle station sends the frame immediately(with a probability of 1),this method increases the chance of collision p-PERSISTANT If the line is Idle station sends the frame with probability p,and refrain from sending with a probability 1-p Station runs a Random number between 1-100
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 13.7 CSMA/CD procedure In the exponential back of method the station waits an amount of time between 0 and 2 N x Maximum propagation time N is the number of attempted transmission
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.8 CSMA/CA procedure IFG Time =Inter frame Gap Used in Wireless LAN
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 13.2 Control Access Reservation Polling Token Passing
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.9 Reservation access method
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.10 Select
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.11 Poll
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.12 Token-passing network
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.13 Token-passing procedure
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 13.3 Channelization FDMA TDMA CDMA
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 In FDMA, the bandwidth is divided into channels. Note:
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 In TDMA, the bandwidth is just one channel that is timeshared. Note:
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 In CDMA, one channel carries all transmissions simultaneously. Note:
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.14 Chip sequences Each station is assigned a code,which is a sequence of numbers called CHIPS Chip period is always less than a bit period Chip rate is always an integral multiple of bit rate
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.15 Encoding rules If a station needs to send 0 bit,it sends a –1 If a station needs to send 1 bit,it sends a +1 If a station needs to send no signal,it sends a 0
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.16 CDMA multiplexer In the following example Four stations sharing the link during the 1-bit interval is shown below
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.17 CDMA demultiplexer There is only one sequence flowing through the channel,the sum of the sequences But each receiver can detect its data from the sum
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 ORTHOGONAL SEQUENCES The sequences are not chosen randomly They should be orthogonal to each other Orthogonal sequences satisfies following properties If we multiply a sequence by –1,every element in the sequence is complemented If we multiply two sequences element by element and add the results, we get a number called inner product If two sequences are same we get N (N=Number of sequences),if they are different we get 0 (A.A=N A.B=0) A.(-A) = -N For the generation of the sequences,WALSH TABLE method is used
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 WALSH TABLE It is a two dimensional table with equal number of rows and columns Each row is a sequence of chip For W 1 ( one Row one Column Walsh Table),we can choose –1 or +1 If we know the table for N sequences W N, we can create the table for 2N sequences W 2N
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.18 W1 and W2N
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Figure 13.19 Sequence generation Now each row can be used for each station as its chip code
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Example 1 Check to see if the second property about orthogonal codes holds for our CDMA example. Solution The inner product of each code by itself is N. This is shown for code C; you can prove for yourself that it holds true for the other codes. C. C = If two sequences are different, the inner product is 0. B. C =
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Example 2 Check to see if the third property about orthogonal codes holds for our CDMA example. Solution The inner product of each code by its complement is N. This is shown for code C; you can prove for yourself that it holds true for the other codes. C. ( C ) = The inner product of a code with the complement of another code is 0. B. ( C ) =
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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 Orthogonal Codes of CDMA In Our previous example At the Multiplexer the sequence S that comes out is S=-A-B+D Station 1 is sending –A (-1 was multiplied by A) Station 2 is sending –B Station 3 is sending an empty sequence Station 4 is sending D At the De multiplexer the sequence S is received Station 1 finds the inner product of S and A S.A=(-A-B+D).A=-4+0+0=-4 The result is divided by 4 (–4/ 4 = -1), And –1 represents bit 0 Same is true for stations 2,3 and 4
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