Figure 6.1 Modes of transmission: (a) baseband ...apachepersonal.miun.se/~mageri/kurser/mks/halsall/Ch06.pdf · ... baseband transmission; (b) modulated transmission. ... DLE STX

Figure 6.1 Modes of transmission: (a) baseband transmission; (b) modulated transmission.

© Pearson Education Limited 2001

0

+V

–V

0 1 0 0 1 1 0 1

Time

Binary data(a)

Transmitterline interface

Receiverline interface

0 Frequencyf

Signalpower

Bandwidth of transmission medium, f,determines maximum bit rate that can be used

0

Time

Binary data(b)

Amplitudemodulated

signal

1 0 0 1 1 0

Transmitter(modulator)

Receiver(demodulator)

0 Frequencyf2

Signalpower

fc = carrier signal (single-frequency audio tone)Bandwidth (f2–f1) determines maximum bit rate that can be used

fcf1

Figure 6.2 Effect of attenuation, distortion, and noise ontransmitted signal.


Transmitted data

Transmitted signal

Typical received signal with noise

Sampling instants

Received data 0 1 0 0 0 1 0 1

0 1 0 0 1 1 0 1

Bit error

Time, t

+V

–V

t

t

1 bit cell period

Figure 6.3 Copper wire transmission media: (a) two-wire andmultiwire open lines; (b) unshielded twisted pair; (c) shieldedtwisted pair; (d) coaxial cable.


Single pair Flat ribbon

Terminating connectors

Single pair

Insulating outer cover

Multicore

Insulating outer coverProtective screen (shield)

Insulating outer cover

Dielectric insulating material Braided outer conductor

Center conductor

(c)

(d)

(b)

(a)

each wire insulated

Figure 6.4 Optical fiber transmission media: (a) cablestructures; (b) transmission modes.


Plastic coatingOptical cladding Optical core

Single core

Multicore

Opticaltransmitter

Opticalreceiver

(i) Multimode stepped index

(ii) Multimode graded index

(iii) Monomode

Electricalinput signal

Electricaloutput signal

(b)

(a)

Figure 6.5 Satellite systems: (a) broadcast television; (b) datacommunications.


Satellite

Antenna

Down linkUp link

Earth ground stations

Earth

(a)

Hub station

VSATsVSATs

(b)

VSAT = very small aperture terminal

Figure 6.6 Ground-based radio transmission: (a) single cell; (b) multiple cells.


F1 F2 F3 F1F3

F3 F1 F2F2

F3 F1 F2F2

F1, F2, F3 = frequencies used in cell

(b)

Fixed network/computerBS

Radio field of coverageof base station

(a)

BS = base station

= user computer/terminal

BSBSBSBS

BSBSBSBS

BSBSBSBS BS

Figure 6.7 Sources of signal impairment.


0 1 0 0 1 0

+V

–V

Transmitted signal

Transmitted data

Received signal

Sampling signal

Received data 0 1 0 1 1 0

Time

Delay distortion

Line noise

Combined effect

Bit error

Attenuation

Limited bandwidth

Attenuation anddistortion effects

Figure 6.8 Effect of limited bandwidth: (a) alternative binarysignals; (b) frequency components of a periodic binary sequence;(c) examples of received signals; (d) bandwidth representations.


–T/40

–T/2–3T/4–T T/4 T/2 3T/4 TTime, t

01 0 1

–T/40

–T/2–3T/4–T T/4 T/2 3T/4 Tt

–V

+V

+V

1

Bit period, Tb

Signal period, T

v(t )

(a)

T/40

7T/4t

01 01

t

t

t

t

t

3T/4 5T/4

+V

–V

Transmitted signal, v (t)

0

+V '

–V '

0+V '/3

–V '/3

0+V '/5

–V '/5

ω0 – 3ω0

ω0 – 3ω0 + 5ω0

Examplereceivedsignals

+5ω0

ω0

–3ω0

Frequencycomponents

5f03f0f00

Signalpower

Frequency

Bandwidthalternatives

(b)

(c)

(d)

Unipolar signal

Bipolar signal

Binary signal

Figure 6.9 Examples of (binary) eye diagrams resulting fromintersymbol interference: A, ideal; B, typical.


Some examplesignal

transitions

1 1 0 1 0

0 1 0 1 1

0 0 1 1 0

1 0 1 0 1

Sampling pulses

A

B

Figure 6.10 Adaptive NEXT cancelers: (a) circuit schematic; (b) example waveforms.


AdaptiveNEXT

canceler

Transmitcircuit

Receivecircuit

DTE

–

+

(D)

(E)

Near-endcrosstalk(NEXT)

(C)

(A)

(B)

Transmitted signal

Received signal

(a)

Time, tTransmitted signal, A

Withoutcrosstalk, B

Withcrosstalk, C

Receivedsignal

(b)

t

t

t

t

Adaptively attenuatedtransmitted signal, D

Received signal, E

Figure 6.11 Asynchronous transmission: (a) principle ofoperation; (b) timing principles.


011

Transmitter

PISO

Serial-out

msb

lsb

Parallel-in

msb

lsb Parallel-out

SIPO

Serial-in ReceiverTxD RxD

PISO =TxD =

l/msb =

parallel in, serial outtransmit data outleast/most significant bit

SIPO =RxD =

serial in, parallel outreceive data in

lsb msb

0 1 1

Time

Stop bit(s)7/8-bit character/byteStart bit

TxD

TxC

Mark(ing)

Space

(a)

(b)

lsb msb

0 1 1

Time

RxD

RxC

Mark(ing)

Space

Actual edge withinone clock cycle

φ

φ

φ φ φ φ φ φ φ

φ φ φ φ φ φ φ

φ = 0 or 1 being transmitted

Figure 6.12 Examples of three different receiver clock rateratios: (a) ×1; (b) ×4; (c) ×16.


RxD

RxC(×1)

Shift(sampling)

pulse

Bit rate counterpreset to 1

Actual bit cellcenters

Time

1st data bit 2nd data bitStart bit

RxD

RxC(×4)

Shift(sampling)

pulse


Actual bit cell centers

Time

(b)


4 RxC periods 4 RxC periods2 RxCperiods

RxD

RxC(×4)

Shift(sampling)

pulse


Actual bit cell centers

Time(c)


16 RxC periods 16 RxC periods8 RxCperiods

(a)

Figure 6.13 Frame synchronization with different framecontents: (a) printable characters; (b) string of bytes.


STX "F" "R" "L" ETXMarking

Startbit

Stopbit(s)

Frame contents(printable characters)

(a)

DLE STX DLE ETXMarking

Frame contents(byte string)

(b)

Inserted DLE

Figure 6.14 Alternative bit/clock synchronization methods withsynchronuous transmission: (a) clock encoding; (b) digital phase-lock-loop (DPLL).


Transmitter(a)

ReceiverTxD RxD

Transmitter(b)

ReceiverTxD RxD

Figure 6.15 Synchronous transmission clock encoding methods:(a) Manchester; (b) differential Manchester.


Bitstream

TxC

Phase (Manchester) encodedsignal, TxD/RxD

Extracted and delayedclock, RxC

Received data

(a)

Bitstream

TxC

Differential Manchester-encoded signal, TxD

Decoded (received) data

(b)

Either

Or

1 0 0 1 1 1 0 1

1 0 0 1 1 1 0 1

Extracted clock, RxC

+–+–

+–

Figure 6.16 DPLL operation: (a) bit encoding; (b) circuitschematic; (c) in phase; (d) clock adjustment rules.


1 0 0 1 1 1 0 1Bitstream

Non-return-to-zero (NRZ) signal

NRZ inverted (NRZI) signal

Either

Or

RxC

Received bitstream, RxD

32 clocks 32 clocks

Actual transitions

Received bitstream, RxD

32 × CLK

Generated sampling(clock) pulses, RxC

32 clocks 30 clocks

Assumed transitions

32 × CLK

Generated sampling(clock) pulses, RxC

31 clocks

32 clocks

33 clocks

34 clocks

Actual transitionpossibilities

32 – 2

32 – 1

32

32 + 1

32 + 2Segment/phaseClock adjustment

A B C D E–2–1 0 +1+2

±

(b)

(a)

(c)

(d)

–+–+–+

Figure 6.17 Character-oriented synchronous transmission: (a) frame format; (b) character synchronization; (c) datatransparency (character stuffing).


STXSYNSYN ETX

Direction of transmission Time(a)

Charactersynchronization

Start-of-framecharacter

Frame contents(printable characters)

End-of-framecharacter

(b) Direction of transmissionTime

11011010000110100001101000010000000110

SYN SYN SYN

STX Frame contents

Receiver in character synchronization

Receiver entershunt mode

Receiver detectsSYN character

DLESYNSYN –––

Direction of transmission Time(c)

Start-of-framesequence

Frame contents(binary data)

End-of-framesequence

DLE ETXSTX ––– DLE DLE

Additional DLE inserted

Figure 6.18 Bit-oriented synchronous transmission: (a) framingstructure; (b) zero bit insertion circuit location; (c) exampletransmitted frame contents.


011111110111111101111110 0111111001111110

Closing flagFrame contentsOpening flag

Direction of transmission

Line idle

(a)

TxC

Transmitter

RxC

Receiver

Enable/disableEnable/disable


0111111011011001111101101111100– – 1101111110

Openingflag

Frame contents

Closingflag

Additional zero bits inserted

(b)

(c)

Figure 6.19 Parity bit method: (a) position in character; (b) XOR gate truth table and symbol; (c) parity bit generationcircuit; (d) two examples.


Transmitted character/byteTime

lsb msbStop bit(s)

Parity bit

Start bit

(a)

(b) XORBit 2Bit 1

0110

0101

0011

Output+Bit 2

Bit 1

+

B1

+

+

+

+

+

B0B2B3B4B5B6

Inverter

Even paritybit

Odd paritybit

(c)

(d) 10010011001001

1 0

(Even parity)(Odd parity)

Figure 6.20 Block sum check method: (a) row and column paritybits; (b) 1s complement sum.


B0

0 0 0 0 0 0 1 0

PR B6 B5 B4 B3 B2 B1

1 0 1 0 1 0 0 0

0 1 0 0 0 1 1 0

0 0 1 0 0 0 0 0

1 0 1 0 1 1 0 1

0 1 0 0 0 0 0 0

1 1 1 0 0 0 1 1

1 0 0 0 0 0 1 1

1 1 0 0 0 0 0 1

= STX

Framecontents

Direction oftransmission

Transverse (row)parity bits

(odd)

Longitudinal (column)parity bits (even)

= ETX

= BCC

PR = row parity bit= example of undetected error combination BCC = block check character

(a)

(b)

At sending side: At receiving side:

Example contents

0 0 0 0 0 1 01 0 1 1 0 1 11 1 0 1 1 0 00 0 0 0 0 1 1

1 0 0 1 1 0 0[1]

1 0 0 1 1 0 11

= 1s-complement sum

0 1 1 0 0 1 0 = BCCInvert

0 0 0 0 0 1 01 0 1 1 0 1 11 1 0 1 1 0 00 0 0 0 0 1 10 1 1 0 0 1 0

1 1 1 1 1 1 0[1]

1 1 1 1 1 1 11

= Zero in 1s-complement

BCC

Contents

Figure 6.21 Error burst examples.


1 1 0 1 1 0 1 1 1 1 0 1 1 1 1 0 0 0 1 1 1 1 1 0 1 1

1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1

Minimum of 6 error-free bits

6-bit error burst

Minimum of 4 error-free bits

4-bit error burst

Transmitted message =

Received message =


• • •

• • •

Figure 6.22 CRC error detection example: (a) FCS generation;(b) two error detection examples.


1 1 0 0 1 1 1 1 0 0 1 1 0 0 0 0 01 0 1 1 0 1 1 0 = Quotient (ignored)

1 1 0 0 10 0 1 0 1 1

0 0 0 0 00 1 0 1 1 1

1 1 0 0 10 1 1 1 0 0

1 1 0 0 10 0 1 0 1

0 0 0 00 1 0 1 0

1 1 0 00 1 1 0 1

1 1 0 00 0 0 1 1

0 0 0 00 1 1 0

00

01

01

00

= Remainder (FCS/CRC)

1 1 0 0 1 1 1 1 0 0 1 1 0 1 1 1 11 0 1 1 0 1 1 0

Error burst1 1 0 0 10 0 1 0 1 1

0 0 0 0 00 1 0 1 1 1

1 1 0 0 10 1 1 1 0 0

1 1 0 0 10 0 1 0 1

0 0 0 00 1 0 1 1

1 1 0 00 1 1 1 0

1 1 0 00 0 1 0 0

0 0 0 01 0 0 1

10

11

11

10

1 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 01 0 1 1 0 1 1 0(b)

1 1 0 0 10 0 1 0 1 1

0 0 0 0 00 1 0 1 1 1

1 1 0 0 10 1 1 1 0 0

1 1 0 0 10 0 1 0 1

0 0 0 00 1 0 1 0

1 1 0 00 1 1 0 0

1 1 0 00 0 0 0 0

0 0 0 00 0 0 0

00

11

11

00

(a)

Remainder ≠ 0: error detectedRemainder = 0: no errors

Frame contents: 11100110With appended zeros: 11100110 0000Generator polynomial: 11001

Transmitted frame: 11100110 0110

Figure 6.23 Idle RQ error control scheme: (a) error free; (b) corrupted I-frame; (c) corrupted ACK-frame.


I(N + 1)

Timer stoppedTimer startedTimer stopped

Timer started

Primary, P

I(N + 1)I(N)Secondary, S

Time

I(N + 1)I(N)I(N)

Timer stoppedTimer restartedTimer started

Primary, P

I(N)Secondary, S

Time

I(N + 1)I(N)I(N)

Timer stoppedTimer expires/restartedTimer started

Primary, P

I(N)Secondary, S

Time

I(N)

ACK(

N +

1)

ACK(

N)

I(N + 1)

I(N)

ACK(

N)I(N

)

I(N)

ACK(

N)

ACK(

N) I(N

)

I(N)

NAK

(N)

(a)

= Frame corrupted

Duplicate detected

I(N)

(b)

(c)

Figure 6.24 Idle RQ link utilization.


I(N + 1)I(N)

Timer stoppedTimer startedTimer stopped

Timer started

Primary, P

I(N + 1)I(N)Secondary, S

TimeACK

(N +

1)

ACK(

N) I(N

+ 1)

I(N)

= Frame propagation delay (P → S)= Frame transmission time (P → S)= Frame processing time in S= ACK propagation delay (S → P)= ACK transmission time (S → P)= ACK processing time in P

TpTixTipTpTaxTap

TpTixTipTpTaxTap

Figure 6.25 Effect of propagation delay as a function of data transmission rate; parts correspond to Example 6.8.


1 km, Tp = 5 µs

(ii)

(i)

1 Mbps

1 kbps

(a)

200 km, Tp = 1 ms

(ii)

(i)

1 Mbps

1 kbps

(b)

250 000 bits

50 000 km, Tp = 167 ms

(ii)

(i)

1 Mbps

1 kbps

(c)

1000 bits

Figure 6.26 Continuous RQ frame sequence without transmissionerrors.


I(N + 4)I(N + 3)I(N + 2)I(N + 1)I(N)Primary, P

N N

N + 1

N

N + 1

N + 2

N + 1

N + 2

N + 3

N + 2

N + 3

N + 4

N + 3

N + 4

N N + 1 N + 2 N + 3 N + 4 N + 5 V(S)

Contentsof link

retransmissionlist

Contentsof linkreceive

list

I(N + 4)I(N + 3)I(N + 2)I(N + 1)I(N)Secondary, S

N N + 1 N + 2 N + 3 N + 4

N N + 1 N + 2 N + 3 N + 4

V(R)

I(N)

I(N + 1)

I(N + 2)

I(N + 3)

I(N + 4)

ACK(

N +

3)

ACK(

N +

2)

ACK(

N +

1)

ACK(

N)

Time

V(S)V(R)

==

send sequence variablereceive sequence variable

Figure 6.27 Selective repeat: (a) effect of corrupted I-frame;(b) effect of corrupted ACK-frame.


I(N)I(N + 3)I(N + 2)I(N + 1)I(N)Primary, P

N N

N + 1

N

N + 1

N + 2

N

N + 1

N + 2

N + 2

N + 3

N

N + 3

N

N N + 1 N + 2 N + 3 N + 4 N + 4 V(S)

Contents of linkretransmission list

Contents of linkreceive list

I(N)I(N + 3)I(N + 2)I(N)Secondary, S

N N + 2 N + 3

N N + 1 N + 2 N + 3 N + 4V(R)

I(N)

I(N + 1)

I(N + 2)

I(N + 3)

I(N)

ACK(

N +

3)

ACK(

N +

2)

ACK(

N +

1)

ACK(

N)

N + 4

ACK(

N)

N + 3

I(N + 1)

N + 1

corrupted frame

(b) P entersretransmissionstate

P leavesretransmissionstate

Time


N N

N + 1

N

N + 1

N + 2

N + 1

N + 2

N + 3

N + 1

N + 3

N + 4

N + 2

N + 3

N N + 1 N + 2 N + 3 N + 4 N + 5 V(S)

I(N + 4)I(N + 3)I(N + 2)I(N)Secondary, S

N N + 2 N + 2 N + 2

N + 1 N + 1 N + 1 N + 1V(R)

I(N)

I(N + 1)

I(N + 2)

I(N + 3)

I(N + 4)

NAK

(N +

1)

ACK(

N)

I(N + 1)

N + 2

N + 3

N + 5

I(N + 1)

N + 2

N + 1

I(N + 1)

ACK(

N +

1)

(a)

N + 2

N + 4

N + 1

N + 4

N + 1

N + 5N

S enters retransmission state S leaves retransmission state

N + 3

N + 4

N + 1

N + 3

N + 4

N + 3

N + 2

P entersretransmissionstate

P leavesretransmissionstate

N + 5

N + 3

N + 4

N + 1

N + 5

Time



N + 5

N

N

N + 4 N + 4

Figure 6.28 Go-back-N retransmission strategy: (a) corrupted I-frame; (b) corrupted ACK-frame.



N N

N + 1

N

N + 1

N + 2

N + 1

N + 2

N + 3

N + 1

N + 2

N + 3

N + 1

N + 2

N N + 1 N + 2 N + 3 N + 4 N + 5 V(S)

Contents of linkretransmission

list



N N + 2 N + 3 N + 4

N N + 1 N + 1 N + 1 N + 1

I(N)

I(N + 1)

I(N + 2)

I(N + 3)

I(N + 4)

NAK

(N +

1)

ACK(

N)

Time

I(N + 1)

I(N + 1)

N + 1

N + 1

I(N + 1)

P enters retransmissionstate

N + 4

N + 3


N N

N + 1

N

N + 1

N + 2

N

N + 1

N + 2

N

N + 1

N + 2

N + 3

N + 4

N N + 1 N + 2 N + 3 N + 4 N + 5



N N + 2 N + 3

N N + 1 N + 2 N + 3 N + 4 V(R)

I(N)

I(N + 1)

I(N + 2)

I(N + 3)

I(N + 4)

ACK(

N +

3)

ACK(

N +

2)

ACK(

N +

1)

ACK(

N)

N + 5

N + 3

I(N + 1)

N + 1

corrupted frame

N + 2

(a)

(b)

N + 4

I(N + 2) I(N + 4)

P leaves retransmission state

Frames discarded


N + 4

N + 3

N + 4

V(R)

V(S)

Time

ACK(

N +

1)

Figure 6.29 Flow control principle: (a) sliding window example;(b) send and receive window limits.


Frames waiting to beacknowledged Frames waiting to

be sentFrames alreadyacknowledged

Flow stopped

Upper window edge(UWE)

Lower window edge(LWE)

Send window, K = 3

Order oftransmission

(a)

(b) Send window Receive window

1KK

1K1

Protocol

Idle RQSelective repeatGo-back-N

Figure 6.30 Sequence numbers: (a) maximum number for eachprotocol; (b) example assuming eight sequence numbers.


Maximum number of frame identifiers

22K + 1K + 1

Protocol

Idle RQSelective repeatGo-back-N

(a)

(b)

Upper window edge (UWE)

Sequencenumbers

Go-back-N, K = 7Sequence numbers incrementedmodulo 8

Lower window edge (LWE)

Figure 6.31 Example layered architecture showing the layer andsublayer interfaces associated with the idle RQ protocol.


Figure 6.32 Abbreviated names used in the specification of theidle RQ primary.


Incoming events

Name Interface Meaning

LDATAreq LS_user L_DATA.request service primitive receivedACKRCVD MAC_provider ACK-frame received from STEXP TIM_provider Wait-ACK timer expiresNAKRCVD MAC_provider NAK-frame received from S

States

Name Meaning

IDLE Idle, no message transfer in progressWTACK Waiting an acknowledgment

Outgoing events


TxFrame MAC_user Format and transmit an I-frameRetxFrame MAC_user Retransmit I-frame waiting acknowledgmentLERRORind LS_provider Error message: frame discarded for reason specified

Predicates

Name Meaning

P0 N(S) in waiting I-frame = N(R) in ACK-frameP1 CRC in ACK/NAK-frame correct

Specific actions State variables

[1] = Start_timer using TIM_user queue Vs = Send sequence variable[2] = Increment Vs PresentState = Present state of protocol entity [3] = Stop_timer using TIM_user queue ErrorCount = Number of erroneous frames [4] = Increment RetxCount received[5] = Increment ErrorCount RetxCount = Number of retransmissions for[6] = Reset RetxCount to zero this frame

Figure 6.33 Specification of idle RQ primary in the form of: (a)a state transition diagram; (b) an extended event–state table; (c) pseudocode.


ACKRCVD/NAKRCVD; [5]

LDATAreq; TxFrame, [1] [2]ACKRCVD; [3] [6]

TEXP; RetxFrame, [1] [4]NAKRCVD; RetxFrame, [1] [4]

Incomingevent

Presentstate

LDATAreq ACKRCVD TEXP NAKRCVD

IDLE

WTACK

1

4

0

2

0

3

0

3

012

34

= [5], IDLE (error condition)= TxFrame, [1] [2], WTACK= P0 and P1: [3] [6], IDLE= P0 and NOT P1: RetxFrame, [1] [4], WTACK= NOT P0 and NOT P1: [5], IDLE= RetxFrame, [1] [4], WTACK= NoAction, WTACK

(b)

(a)

Figure 6.33 Continued.


Incomingevent

Presentstate

LDATAreq ACKRCVD TEXP NAKRCVD

IDLE

WTACK

1

4

0

2

0

3

0

3

012

34

= [5], IDLE (error condition)= TxFrame, [1] [2], WTACK= P0 and P1: [3] [6], IDLE= P0 and NOT P1: RetxFrame, [1] [4], WTACK= NOT P0 and NOT P1: [5], IDLE= RetxFrame, [1] [4], WTACK= NoAction, WTACK

(b)

(c) program IdleRQ_Primary;const MaxErrCount;

MaxRetxCount;type Events = (LDATAreq, ACKRCVD, TEXP, NAKRCVD);

States = (IDLE, WTACK);var EventStateTable = array [Events, States] of 0..4;

PresentState : States;Vs, ErrorCount, RetxCount : integer;EventType : Events;

procedure Initialize; } Initializes state variables and contents of EventStateTableprocedure TxFrame;procedure RetxFrame; } Outgoing event proceduresprocedure LERRORind;procedure Start_timer; } Specific action proceduresprocedure Stop_timer;function P0 : boolean; } Predicate functionsfunction P1 : boolean;

begin Initialize;repeat Wait receipt of an incoming event

EventType := type of eventcase EventStateTable [EventType, PresentState] of

0 : beginErrorCount := ErrorCount + 1; PresentState = IDLE;if (ErrorCount = MaxErrCount) thenLERRORind end;

1 : beginTxFrame; Start_timer; Vs := Vs + 1; PresentState := WTACK end;2 : beginif (P0 and P1) then begin Stop_timer; RetxCount := 0; PresentState := IDLE end;

else if (P0 and NOTP1) then begin RetxFrame; Start_timer;RetxCount := RetxCount + 1;PresentState := WTACK end;

else if (NOTP0 and NOTP1) then begin PresentState := IDLE; ErrorCount := ErrorCount + 1if (ErrorCount = MaxErrorCount) then begin LERRORind; Initialize; end;

end;3 : begin RetxFrame; Start_timer; RetxCount := RetxCount + 1; PresentState := WTACK;

if (RetxCount = MaxRetxCount) then begin LERRORind; Initialize; end;end;

4: begin NoAction end;until Forever;

end.

Figure 6.34 Specification of idle RQ secondary: (a) abbreviatednames; (b) state transition diagram; (c) extended event–statetable; (d) pseudocode.


IRCVD+; LDATAind, TxACK, [1] [2]IRCVD–; TxNAK

Incomingevent

Presentstate

IRCVD

WTIFM 1

1 = NOT P1: TxNAK, [2]= P1 and P2: TxACK= P0 and P1: LDATAind, TxACK, [1]

(c)

(b)

(a) Incoming events


IRCVD MAC_provider I-frame received from P

States

Name Meaning

WTIFM Waiting a new I-frame from P

Outgoing events


LDATAind LS_provider Pass contents of received I-frame to user AP with an L_DATA.indication primitive

TxACK(X) MAC_user Format and transmit an ACK-frame with N(R) = XTxNAK(X) MAC_user Format and transmit a NAK-frame with N(R) = XLERRORind LS_provider Issue error message for reason specified

Predicates

Name Meaning

P0 N(S) in I-frame = VrP1 CRC in I-frame correctP2 N(S) in I-frame = Vr – 1

Specific actions State variables

[1] = Increment Vr Vr = Receive sequence variable[2] = Increment ErrorCount ErrorCount = Number of erroneous frames received

Figure 6.34 Continued.


(d) program IdleRQ_Secondary;const. MaxErrorCount;type Events = IRCVD;

States = WTIFM;var EventStateTable = array [Events, States] of 1;

EventType : Events;PresentState : States;Vr, X, ErrorCount : integer;

procedure Initialize; } Initializes state variables and contents of EventStateTableprocedure LDATAind;procedure TxACK(X); } Outgoing event proceduresprocedure TxNAK(X); procedure LERRORind;function P0 : boolean;function P1 : boolean; } Predicate functionsfunction P2 : boolean;

begin Initialize;repeat Wait receipt of incoming event; EventType := type of event;

case EventStateTable[EventType, PresentState] of1 : X := N(S) from I-frame;

if (NOTP1) then TxNAK(X);else if (P1 and P2) then TxACK(X);else if (P0 and P1) then begin LDATAind; TxACK(X); Vr := Vr + 1; end;elsebeginErrorCount := ErrorCount + 1; if (ErrorCount = MaxErrorCount) then

begin LERRORind; Initialize; end;end;

until Forever;end.

Figure 6.35 Time sequence diagram showing the link layerservice primitives: (a) connection-oriented (reliable) mode; (b)connectionless (best-effort) mode.


Sourcelink layer

V(S) := 0etc.

Source LS_user

L_CONNECT.request

L_CONNECT.confirmL_DATA.request

L_DISCONNECT.request

L_DISCONNECT.confirm

Destinationlink layer Correspondent LS_user

L_CONNECT.indication

L_DATA.indication

L_DISCONNECT.indication

SETUP-frame

UA-

frame

I-frame

ACK-

frame

DISC-frame

UA-

frame

Time

Source Destination(a)

V(R) := 0etc.

L_UNITDATA.request

L_UNITDATA.indication

Frames

Event control blocks (ECBs)

(b)

I-frame

Figure 6.36 HDLC frame format and types: (a)standard/extended format; (b) standard control field bitdefinitions; (c) extended control field bit definitions.


Note: With the indicated direction of transmission, all control field types are transmitted bit 8/16 first.

Figure 6.37 HDLC normal response mode: example framesequence diagram with single primary and secondary (i.e. nopiggyback acknowledgments).


1

01

12

23

0

0

03

N(R) = 3 ∴ I(2)acknowledged


Retransmit from I(1)

0

0 0V(S) V(R)

Contents ofretransmission list

0 0V(S) V(R)

0 1

0 1

0 1

0 2

N(S) = V(R) ∴ frame accepted

N(S) ≠ V(R) ∴ frame rejected


frame corrupted

I(0, 0/P = 1)

RR(1/F = 1)

I(1, 0)

I(2, 0/P = 1)

I(1, 0)

RR(3/F = 1)

Sender (P)

REJ (1/F = 1)

03

0 3N(S) = V(R) ∴ frame accepted

0 3

I(2, 0/P = 1)

Time


RR(2/F = 1)2

Receiver (S)

Figure 6.38 HDLC asynchronous balanced mode: piggybackacknowledgment procedure.


743

54

56

0

01 3

V(S) V(R)


3 0

V(S) V(R)


I(0, 3)

Combined P/S Combined P/S

Time

12 3

12

3 5

23

4 6

34

5 6

45

6 7

12

23

5


N(R) = 1 ∴ I(0) acknowledged

N(S) = V(R) ∴ frame acceptedN(R) = 2 ∴ I(1) acknowledged

N(R) = 5 ∴ I(2, 3, 4)acknowledged


5 7

6 0

6 0

6 0

3 6

2 5

2 4

0 3

34 0

34

5 0

45

6 0

56

7 1

67

0 2

7 2

6 1

0 3

0 4

0 5

0 6

67

67

7


N(S) = V(R) ∴ frame acceptedN(R) = 5 ∴ I(3) I(4) acknowledged




N(R) = 0 ∴ I(7) acknowledged

0 6

I(3, 0)

I(4, 0)

I(5, 0

)

I(6, 1

)

I(7, 2

)

RR(5)

RR(6)

I(1, 3)

I(2, 5)I(4, 6)

I(3, 6)

I(5, 7)

RR(0)





321

210

10

0

3

43

543

Figure 6.39 HDLC window flow control procedure.


0

V(S) V(R)


00

V(S) V(R)

I(0, 0)

Combined P/S Combined P/S

Time

0 0

0I(0, 0

)


RetxCount

0

RetxCount

0

1 01

1 11

1 21

1 31

101

202

303

313

313

213

314

114

1

2

3

1

3

RR(1)

RR(3)

I(1, 0)

I(2, 0)

RR(1)

I(3, 1)1 30

1 40

0

10

21

2

K = 3

A

A

A = window closed

Figure 6.40 HDLC summary: (a) service primitives; (b) statetransition diagram (ABM).


Summary

Figure 6.41 Summary of topics discussed relating to digital communications.


Digital communication basics

Digital transmission

Signal impairmentsTransmission media

Transmission control modes

SynchronousAsynchronous

Bit/clock synchronization Block/frame synchronizationCharacter/byte synchronization

Error detection methods

Cyclic redundancy checkParity Block sum check

Protocol basics

Link managementError control Flow control

Protocol specification methods

HDLC

Example 6.1


A 1000-bit block of data is to be transmitted between two computers.Determine the ratio of the propagation delay to the transmission delay,a, for the following types of data link:

(i) 100m of twisted-pair wire and a transmission rate of 10kbps,(ii) 10 km of coaxial cable and a transmission rate of 1Mbps,(iii) 50 000 km of free space (satellite link) and a transmission rate of

10 Mbps.

Assume that the velocity of propagation of an electrical signal withineach type of cable is 2 × 108 ms–1, and that of free space 3 × 108 ms–1.

Answer:

S 100(i) Tp = = = 5 × 10–7 s

V 2 × 108

N 1000Tx = = = 0.1 s

R 10 × 103

Tp 5 × 10–7

a = = = 5 × 10–6

Tx 0.1

6.1 Continued


S 10 × 103

(ii) Tp = = = 5 × 10–5 sV 2 × 108

N 1000Tx = = = 1 × 10–3 s

R 1 × 106

Tp 5 × 10–5

a = = = 5 × 10–2

Tx 1 × 10–3

S 5 × 10–7

(iii) Tp = = = 1.67 × 10–1 sV 3 × 108

N 1000Tx = = = 1 × 10–4 s

R 10 × 106

Tp 1.67 × 10–1

a = = = 1.67 × 103

Tx 1 × 10–4

Example 6.2


A transmission channel between two communicating DTEs is made upof three sections. The first introduces an attenuation of 16 dB, thesecond an amplification of 20dB, and the third an attenuation of 10dB.Assuming a mean transmitted power level of 400 mW, determine themean output power level of the channel.

Answer:

Either:

400For first section, 16 = 10 log10 Hence P2 = 10.0475 mW

P2

P2For second section, 20 = 10 log10 Hence P2 = 1004.75mW10.0475

1004.75For third section, 10 = 10 log10 Hence P2 = 100.475mW

P2

That is, the mean output power level = 100.475mW

Or:

Overall attenuation of channel = (16 – 20) + 10 = 6 dB

400Hence 6 = 10 log10 and P2 = 100.475mW

P2

Example 6.3


A binary signal of rate 500 bps is to be transmitted over a communica-tions channel. Derive the minimum bandwidth required assuming(i) the fundamental frequency only, (ii) the fundamental and thirdharmonic, and (iii) the fundamental, third, and fifth harmonics are tobe received.

Answer:

The worst-case sequence 101010… at 500 bps has a fundamental fre-quency component of 250 Hz. Hence the third harmonic is 750 Hz andthe fifth harmonic 1250 Hz. The bandwidth required in each case is asfollows:

(i) 0–250Hz; (ii) 0–750Hz; (iii) 0–1250Hz.

Example 6.4


Data is to be transmitted over the access line to a PSTN using a trans-mission scheme with eight levels per signaling element. If thebandwidth of the PSTN is 3000 Hz, deduce the Nyquist maximum datatransfer rate.

Answer:

C = 2W log2 M= 2 × 3000 × log2 8= 2 × 3000 × 3= 18000bps

In practice the data transfer rate will be less than this because of othereffects such as noise.

Example 6.5


Assuming that a circuit through a PSTN has a bandwidth of 3000 Hzand a typical signal-to-noise power ratio of 20 dB, determine the maxi-mum theoretical information (data) rate that can be achieved.

Answer:

SSNR = 10 log10 ( )N

STherefore: 20 = 10 log10 ( )N

SHence: = 100

N

SNow: C = W log2 (1 + )N

Therefore: C = 3000 × log2 (1 + 100)

= 19963bps

Example 6.6


A block of data is to be transmitted across a serial data link. If a clock of19.2kHz is available at the receiver, deduce the suitable clock rate ratiosand estimate the worst-case deviations from the nominal bit cell centers,expressed as a percentage of a bit period, for each of the following datatransmission rates:

(i) 1200bps

(ii) 2400bps

(iii) 9600bps

Answer:

It can readily be deduced from Figure 6.12 that the worst-case deviationfrom the nominal bit cell centers is approximately plus or minus onehalf of one cycle of the receiver clock.

Hence:

(i) At 1200 bps, the maximum RxC ratio can be × 16. The maximumdeviation is thus ± 3.125%.

(ii) At 2400 bps, the maximum RxC ratio can be × 8. The maximumdeviation is thus ± 6.25%.

(iii) At 9600 bps, the maximum RxC ratio can be × 2. The maximumdeviation is thus ± 25%.

Clearly, the last case is unacceptable. With a low-quality line, especiallyone with excessive delay distortion, even the second may be unreliable.It is for this reason that a ×16 clock rate ratio is used whenever possible.

Example 6.7


A series of 8-bit message blocks (frames) is to be transmitted across adata link using a CRC for error detection. A generator polynomial of11001 is to be used. Use an example to illustrate the following:

(a) the FCS generation process,

(b) the FCS checking process.

Answer:

Generation of the FCS for the message 11100110 is shown in Figure6.22(a). Firstly, four zeros are appended to the message, which is equiv-alent to multiplying the message by 24, since the FCS will be four bits.This is then divided (modulo 2) by the generator polynomial (binarynumber). The modulo-2 division operation is equivalent to performingthe exclusive-OR operation bit by bit in parallel as each bit in the divi-dend is processed. Also, with modulo-2 arithmetic, we can perform adivision into each partial remainder, providing the two numbers are ofthe same length, that is, the most significant bits are both 1s. We do notconsider the relative magnitude of both numbers. The resulting 4-bitremainder (0110) is the FCS, which is then appended at the tail of theoriginal message when it is transmitted. The quotient is not used.

At the receiver, the complete received bit sequence is divided by thesame generator polynomial as used at the transmitter. Two examples areshown in Figure 6.22(b). In the first, no errors are assumed to be pre-sent, so that the remainder is zero – the quotient is again not used. Inthe second, however, an error burst of four bits at the tail of the transmit-ted bit sequence is assumed. Consequently, the resulting remainder isnonzero, indicating that a transmission error has occurred.

Example 6.8


A series of 1000-bit frames is to be transmitted using an idle RQ proto-col. Determine the link utilization for the following types of data linkassuming a transmission bit rate of (a) 1 kbps and (b) 1 Mbps. Assumethat the velocity of propagation of the first two links is 2 × 108 ms–1 andthat of the third link 3 × 108 ms–1. Also the bit error rate is negligible.

(i) a twisted-pair cable 1km in length,

(ii) a leased line 200km in length,

(iii) a satellite link of 50000km.

Answer:

The time taken to transmit a frame Tix is given by:

Number of bits in frame, NTix =

Bit rate, R, in bps

At 1kbps:

1000Tix = = 1 s

103

At 1Mbps:

1000Tix = = 10–3 s

106

S 1Tp = and U =

V 1 + 2a

103(i) Tp = = 5 × 10–6 s

2 × 108

5 × 10–6(a) a = = 5 × 10–6 and hence (1 + 2a) � 1 and U = 1

1

5 × 10–6(b) a = = 5 × 10–3 and hence (1 + 2a) � 1 and U = 1

10–3

200 × 103(ii) Tp = = 1 × 10–3 s

2 × 108

1 × 10–3(a) a = = 1 × 10–3 and hence (1 + 2a) � 1 and U = 1

1

1 × 10–3 1(b) a = = 1 and hence (1 + 2a) > 1 and U = = 0.33

10–3 1 + 2

6.8 Continued


50 × 106

(iii)Tp = = 0.167s3 × 108

0.167 1(a) a = = 0.167 and hence (1 + 2a) > 1 and U = = 0.75

1 1 + 0.334

0.167 1(b) a = = 167 and hence (1 + 2a) > 1 and U = = 0.003

10–3 1 + 334

Example 6.9


Use the frame sequence diagram shown earlier in Figure 6.23 and thelist of abbreviated names given in Figure 6.34(a) to specify the opera-tion of the idle RQ secondary using (i) a state transition diagram, (ii)an extended event–state table, (iii) pseudocode.

Answer:

The specification of the idle RQ secondary in each form is given inFigure 6.34(b), (c), and (d) respectively. Note that just two state vari-ables are needed for the secondary: the receive sequence variable –shown as Vr in the specification – which holds the sequence number ofthe last correctly received I-frame, and ErrorCount which keeps arecord of the number of erroneous I-frames received. Again, ifErrorCount reaches a defined maximum limit an error message –LERRORind – is output to the network layer in an ECB.

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Figure 6.1 Modes of transmission: (a) baseband ...apachepersonal.miun.se/~mageri/kurser/mks/halsall/Ch06.pdf · ... baseband transmission; (b) modulated transmission. ... DLE STX