Hdlc

CS254 Network Technologies

Lecture 4: HDLC & Medium Access Control Dr Nikos Antonopoulos Department of Computing University of Surrey Autumn 2006

11.5 HDLC Configurations and Transfer Modes Frames Frame Format Examples Data Transparency

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11.15

NRM

11.16

ABM

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11.17

HDLC frame

11.18

HDLC frame types

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11.19

I-frame

11.20 S-frame control field in HDLC

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11.21

U-frame control field in HDLC

Table 11.1 U-frame control command and response Command/response

Meaning

SNRM

Set normal response mode

SNRME

Set normal response mode (extended)

SABM

Set asynchronous balanced mode

SABME

Set asynchronous balanced mode (extended)

UP

Unnumbered poll

UI

Unnumbered information

UA

Unnumbered acknowledgment

RD

Request disconnect

DISC

Disconnect

DM

Disconnect mode

RIM

Request information mode

SIM

Set initialization mode

RSET

Reset

XID

Exchange ID

FRMR

Frame reject

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Example 3 Figure 11.22 shows an exchange using piggybacking where is no error. Station A begins the exchange of information with an I-frame numbered 0 followed by another I-frame numbered 1. Station B piggybacks its acknowledgment of both frames onto an I-frame of its own. Station B’s first I-frame is also numbered 0 [N(S) field] and contains a 2 in its N(R) field, acknowledging the receipt of A’s frames 1 and 0 and indicating that it expects frame 2 to arrive next. Station B transmits its second and third I-frames (numbered 1 and 2) before accepting further frames from station A. Its N(R) information, therefore, has not changed: B frames 1 and 2 indicate that station B is still expecting A frame 2 to arrive next.

11.22

Example 3

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Example 4 In Example 3, suppose frame 1 sent from station B to station A has an error. Station A informs station B to resend frames 1 and 2 (the system is using the Go-BackN mechanism). Station A sends a reject supervisory frame to announce the error in frame 1. Figure 11.23 shows the exchange.

11.23

Example 4

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Note: Bit stuffing is the process of adding one extra 0 whenever there are five consecutive 1s in the data so that the receiver does not mistake the data for a flag.

11.24 Bit stuffing and removal

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11.25 Bit stuffing in HDLC

Chapter 13

Multiple Access

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Figure 13.1 Multiple-access protocols

13.1 Random Access MA CSMA CSMA/CD CSMA/CA

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Figure 13.2

Evolution of random-access methods

Figure 13.3

ALOHA network

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Figure 13.4 Procedure for ALOHA protocol

Figure 13.5

Collision in CSMA

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Figure 13.6 Persistence strategies

13.7

CSMA/CD procedure

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Figure 13.8

CSMA/CA procedure

13.2 Control Access Reservation

Polling Token Passing

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Figure 13.9

Reservation access method

Figure 13.10 Select

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Figure 13.11

Poll

Figure 13.12

Token-passing network

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Figure 13.13

Token-passing procedure

13.3 Channelization FDMA

TDMA

CDMA

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

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

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

Figure 13.14

Chip sequences

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Figure 13.15

Encoding rules

Figure 13.16

CDMA multiplexer

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0

Figure 13.17

CDMA demultiplexer

Figure 13.18

W1 and W2N

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1

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

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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

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