Chapter 13 Multiple Access.

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Presentation transcript:

Chapter 13 Multiple Access

Figure 13.1 Multiple-access protocols

13.1 Random Access MA CSMA CSMA/CD CSMA/CA

Figure 13.2 Evolution of random-access methods

Figure 13.3 ALOHA network

Figure 13.4 Procedure for ALOHA protocol

Figure 13.5 Collision in CSMA

Figure 13.6 Persistence strategies

13.7 CSMA/CD procedure

Figure 13.8 CSMA/CA procedure

13.2 Control Access Reservation Polling Token Passing

Figure 13.9 Reservation access method

Figure 13.10 Select

Figure 13.11 Poll

Figure 13.12 Token-passing network

Figure 13.13 Token-passing procedure

13.3 Channelization FDMA TDMA CDMA

In FDMA, the bandwidth is divided into channels. Note: In FDMA, the bandwidth is divided into channels.

In TDMA, the bandwidth is just one channel that is timeshared. Note: In TDMA, the bandwidth is just one channel that is timeshared.

In CDMA, one channel carries all transmissions simultaneously. Note: In CDMA, one channel carries all transmissions simultaneously.

Figure 13.14 Chip sequences

Figure 13.15 Encoding rules

Figure 13.16 CDMA multiplexer

Figure 13.17 CDMA demultiplexer

Figure 13.18 W1 and W2N

Figure 13.19 Sequence generation

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 = [+1, +1, -1, -1] . [+1, +1, -1, -1] = 1 + 1 + 1 + 1 = 4 If two sequences are different, the inner product is 0. B . C = [+1, -1, +1, -1] . [+1, +1, -1, -1] = 1 - 1 - 1 + 1 = 0

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 ) = [+1, +1, -1, -1] . [-1, -1, +1, +1] = - 1 - 1 - 1 - 1 = -4 The inner product of a code with the complement of another code is 0. B . (-C ) = [+1, -1, +1, -1] . [-1, -1, +1, +1] = -1 + 1 + 1 - 1 = 0