University Of MoratuwaLecture

2012

PART 1

• TELEPHONE NET WORK

PART 2

• PULSE CODE MODULATION

Exercise 1:Convert the following denary numbers to

binary(Don’t use the method of dividing by 2, use the finger method)

• (a) 5 (g) 520• (b) 9 (h) 1028• (c) 16 (i) 2050• (d)33 (j) 4100• (e) 67 (k) 8200• (f) 120 (l) 16401

Answer to Exercise 1

• (a) 5=101 (b) 9=1001• (c) 16=10000 (d)33=100001• (e) 67=1000011 (f) 120=1111000 • (g) 520=1000001000 (h) 1028=10000000100 (i) 2050=100000000010 (j) 4100=1000000000100• (k) 8200=10000000001000 (l)

16401=100000000010001 •

•

Exercise 2Convert the following from binary

to Denary(Using fingers only)• (a) 101• (b) 110• (c) 1001• (d) 11101• (e) 100000• (f) 1011010• (g) 111000111

Answers to Exercise 2

• (a) 101 5

• (b) 110 6• (c) 1001 9• (d) 11101 29• (e) 100000 32• (f) 1011010 90• (g) 111000111 455

Exercise 3Convert the following denary numbers

to hexa and then to binary• (a) 9• (b) 20• (c) 36• (d) 129• (e) 518• (f) 1030• (g) 4095• (h) 8200

Answers to Exercise 3• Denary Hexa Binary• (a) 9 9 1001• (b) 20 14 10100• (c) 36 24 100100• (d) 129 81 10000001• (e) 518 206 1000000110• (f) 1030 406 10000000110• (g) 4095 FFF 111111111111• (h) 8200 2008 10000000001000

Convert the following samples into encoded format and calculate the signal /noise ratio

• 700mV -400mV 300mV

• 100mV 1515mV -95mV

Answers• 700mV -400mV 300mV

• 11011101 01010001 11001001 175 50 ∞• 100mV 1515mV -95mV

10110001 11110000 0011000 25 72 295

Pcm equipment

Pcm equipment(2) contd

PART 3

• HIGHER ORDER PCM

Technological Evolution(Fill the blanks)

Multiplex Level Speed Period of the Pulse No: of voice channels

STM1

STM4

STM16

STM64

STM256

Technological Evolution at a glance

Multiplex Level Speed Period of the Pulse No: of voice channels

STM1 155.52Mbps 6.4ns 1890

STM4 622.08Mbps 1.6ns 7560

STM16 2.5Gbps 400ps 30240

STM64 10Gbps 100ps 120960

STM256 40Gbps 25ps 483840

PART 4

• BASICS OF OPTICAL FIBRE

What is Snell’s Law?• This describes the bending of light rays when it

travels from one medium to another.

Glass

Air

Air

Water

Snell's law states that the ratio of the sines (Sin) of the angles of incidence and refraction is equivalent to the ratio of velocities in the two media, or equivalent to the opposite ratio of the indices of refraction.

Sin Ө 1 n 2 =Sin Ө 2 n 1

Sin Ө 1 n 2 =Sin Ө 2 n 1

n 1 Sin Ө 1 = n 2 Sin Ө 2

PO - Ray of Incidence n 1 - RI for medium 1 OQ - Ray of Refraction n 2 - RI for medium 2Ө 1 - Angle of IncidenceӨ 2 - Angle of Refraction

TOTAL INTERNAL REFLECTIONn 1 Sin Ө 1 = n 2 Sin Ө 2

With the increase of the angle of incidence, the angle ofrefraction increases accordingly.

When reaches φ2 90°, there is no refraction and φ1 reaches a critical angle (φc )

Beyond the critical angle, light ray becomes totally internally reflected

Attenuation in Fibreoptical fibre behaves differently for different wavelength of light. The following diagram shows that. The three windows of wavelengths where the attenuation is lower is given

below. Hence these 3 windows are mostly used for practical purposes.

1. General Observation on Attenuation and the Present Day Technology

• Attenuation is low between 1500nm-1700nm in wavelength.• This gives rise to operate 24Tbps speed • How?

C=fλ where C=3*108

• And f1-f2=[c/(1500nm)]-[c/1700nm]=24Tbps• The present day technology goes up to 10Gbps or 40Gbps.

• STM1 STM4 STM16 STM64…… STM256 155.52Mbps 620Mbps 2.5Gbps 10Gbps 40Gbps

6.4ns 1.6ns 400ps 100ps 25ps

SEA-ME-WE 4 Cable System Configuration Diagram

Present day technology adapting to the optical fibre

The following 2 major factors play a vital role in designing the maximum capacity of an optical fibre• How far the digital multiplexing can be achieved

• As at present , 488ns micro information of a bit pertaining to 2Mbps PCM stream will be reduced to 25ps when it goes through STM64 (10Gbps). If the technology improves to shrink less than 25ps , then the number of bits in the higher order PCM will be more than 10Gbps.

•To transmit 10Gbps, the optical fibre requires a bandwidth of around 0.078ns = 78ps ( for 1 wavelength)•If the available bandwidth in the optical fibre is 200ns , the number of wavelengths that can be produced is around 2400 , which will result in producing a total of 24Tbps.•Hence both Time Division Multiplexing and Dense Wave Division Multiplexing can further improve the traffic carrying capacity of an optical fibre up to a total of 24Tbps.

Optical Fibre

Optical Fibre

Number of wavelengths = ( 24 * 103 Gb ) / 10 Gb = 2400 wavelengths

Future ScenariosTheoretical Maximum of an Optical Fibre Cable

Only 1 core is needed

Transponders

Optical Fibre

1 λ1

2 λ2

2399 λ2399

2400λ2400

TDM

10Gbps2Mbps

488ns 100 ps

PART 5

COMMON CHANNEL SIGNALLING

MESSAGE TYPES

• BASIC MESSAGE

• HOMOGENIOUS MESSAGE

• NON HOMOGENEOUS MESSAGE

Basic Message

Message for Homogenous Network

= K1

= K2Instruction

DataLabelLabel

OPC DPC CIC

14bits 14bits 12bits

Instruction

Data

Fixed Variable

Instruction

Data

OPC – Originating Point Cord

DPC – Destination Point Cord

CIC – Circuit Identification Cord

WHY NOT?

SIO - Service Information OctelK2 - Message for Homogenous Network

Message for Non-Homogenous Network

SIO K2SIO K2Instruction

DataLabel

4bits4bits

National or

International Message

User

Part

Now we are ready with the complete message, can we transmit it just as it is?

NO

IAM

ACM

ANCCBK

0000

0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

0000

0001 IAM SAM

0010 OSM COT CCF

0011 ORQ

0100 ACM CHO

0101 SEC COC NNC ADI CFL SSB UNN LOS SST ACB DPN MPR EUM

0110 ANC ANN CBK CLF RAN FOT CCL EAM

0111 RLG BLO BLA UBL UBA CCR RSC

1000 MGB MBA MGU MUA HOA HBA HGU HUA GRS GRA SGB SBA SGU SUA

1001 CFM CPM CPA CSV CVM CHM CLI

1010

1011

1100

1101

1110

1111

H0H1

Spare reserved for international and basic national use

Spare reserved for national use

Spare reserved for national use

Basic concept of message transmission to establish a call

A

B

Node X Node Y

IAM

ACM

ANC

CBK

Speech

Ringing current to subscriber “B” n ringback tone to subscriber “A”

IAM(Initial address

message)

ACM(Answer complete

mesaage )

ANC

CBK

0001 0001 Dial Number

0100 0001

0110 0001

0110 0011

Fixed (8 Bits)

H0 H1

H0 H1

H0 H1

H0 H1

20 Bits 4 Bits

Variable

No Data

No Data

HOW THE COMMON CHANNEL SIGNALLING WORKS

• ASSUME A CALL IS ESTABLISHED IN A NETWORK WHERE THERE ARE TWO EXCHANGES(EX X & EX Y) ARE CONNECTED WITH 16 PCM SYSTEMS.

• THE CALL IS CONNECTED VIA CIRCUIT NUMBER 305. ASSUME P(0) TS16 & P1(1) IS USED FOR COMMON CHANNEL SIGNALLING.

• DRAW HOW THE SIGNALS ARE ESTABLISHED BETWEEN THE EXCHANGES(assume the call is establised, and after the call, A keeps the receiver first)

• Calculate the total times taken for forward & backward signalling

X

exchange

Y

exchange

P0f

P1f

P15f

P0b

P1b

P15b

Need to transfer message between A to B

Customer A Customer B

Helicopter ViewExchange X Exchange X

IAM

ACM

ANS

CBR

( P0f TS16 )

( P0b TS16 )

RBT( P9 TS28)

( P0b TS16 )

( P9b TS28)speaking( P9f TS28)

( P0f TS16 )

Name Standards Purpose

IAMInitial Address Message

Dialing Information

ACMAddress Complete Message

B customer free or not

RBT Ring Back Tone Tone herd by A

ANS Answer Signal Charge B customer answer or not

CBR Call Back Tone Release the circuit

ERROR CONTROL

• FORWARD ERROR CORRECTION• Detect and correct the error• In unidirectional transmission

• BACKWARD ERROR CORRECTION• Detect the error and request for

retransmission• In bydirectional transmission

Understanding cyclic redundancy code of error correction

(Question)

CALCULATION !

Number of voice channel for voice communication between X and Y

Channel number that we use

If we numbered voice channel from 1 to 494 : - Select related TS 30 + 30 =60 305 – 60 = 245 245 / 31 = 7 mod 28

P9 TS28 (PCM no = 9 , TS no = 28)

= (31 * 14) + (30 *2) 494

= 305

7 + 2 = 9

Number that we dial = 15904607

0001 0001 0011 1000,1010,1001,0000,1000,1100,0000,1110

IAM

4 bits 4 bits 8 4 4 bits 4 * 8 bits

K=56 bits

CRC SCF SIO 1 2 305 K

Message

16 bits

16 bits

8 bits

12bits

12bits

14bits

56bits8 bits

8 bits

Total bits = 150

ACM

0001 0110

4 bits 4 bits 8 bits

Message

CRC SCF SIO 1 2 305 K

16

bits

16

bits

8

bits

12

bits

12

bits

14

bits

16

bits

8bits

8bits

Total bits = 110

K=16 bits

ANC0001 0110

4 bits 4 bits

Message

CRC SCF SIO 1 2 305 K

16

bits

16

bits

8

bits

12

bits

12

bits

14

bits

8

bits

8bits

8bits

Total bits = 102

K=8 bits

CBR0001 0110

4 bits 4 bits

Message

CRC SCF SIO 1 2 305 K

16

bits

16

bits

8

bits

12

bits

12

bits

14

bits

8

bits

8bits

8bits

Total bits = 102

K=8 bits

Conclusion

• time for forward message = 2.34 ms

• time for forward message = 4.906 ms

Phases of a callDial Tone

Dialing Signaling

Ring back Tone

Answer Speak Release

ERROR CONTROL

• FORWARD ERROR CORRECTION• Detect and correct the error• In unidirectional transmission

• BACKWARD ERROR CORRECTION• Detect the error and request for

retransmission• In bydirectional transmission

CYCLIC REDUNCY CODE OR FRAME CHECH SEQUANCE

• DESIGNED TO DETECT NOISE BURST• ACCORDING TO THE NOISE CHARACTERISTICS A

POLYNOMIAL IS IDENTIFIED(N+1 BITS)• SHIFT THE MESSAGE BY N BITS• THEN DIVIDE BY MOULO 2 THE SHIFTED MESSAGE BY

THE POLYNOMIAL• GET THE RESIDUAL OF N BITS & SHIFT THE MESSAGE

BY THESE BITS AS CRC• AT THE RECEIVER IF THERE ARE NO ERRORS, YOU WILL

NOT GET ANY RESIDUAL WHEN YOU DIVIDE THE RECIEVED MESSAGE BY THE SAME POLYNOMIAL

EXAMPLE ON CRC

Understanding cyclic redundancy code of error correction

(Question)

Hence there are no errors

Hint to answer• Write the polynomial in x• Draw the 1 bit shift registers and the circuit

diagram• Write the timing equations for n+1 th step for

each output• Sketh the output map– no of columns=no of

outputs+steps+input(pl add to the message the no of zeros or crc depending upon the situation, no of rows has to be input+2

• Carryout the timing equation for each step, the last step will give you the output

CRC• Polynomial:P=11001,P(x)=x4+x3+x0

• X4 X3 X2 X1 X0

• Timing equations

• An + In = Dn+1

• Dn = Cn+1

• Cn = Bn+1

• An + Bn = An+1

+ +A B C D

Input Data

I

Step A B C D Input

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

0 0 0 0 Reset

0 0 0 0 1

0 0 0 1 0

0 0 1 0 0

0 1 0 0 1

1 0 0 1 0

1 0 1 1 1

1 1 1 0 0

0 1 0 1 1

1 0 1 1 0

1 1 1 1 1

0 1 1 0 0

1 1 0 0 0

0 0 0 1 0

0 0 1 0 0

0 1 0 0 Out put

An + In = Dn+1Cn+1= Dn+1Cn = Bn+1An + Bn = An+1

Step A B C D Input

0 0 0 0 0 Reset

1 0 0 0 0 1

2 0 0 0 1 0

3 0 0 1 0 0

4 0 1 0 0 1

5 1 0 0 1 0

6 1 0 1 1 1

7 1 1 1 0 0

8 0 1 0 1 1

9 1 0 1 1 0

10 1 1 1 1 1

11 0 1 1 0 0

12 1 1 0 0 0

13 0 0 0 1 0

14 0 0 1 0 0

15 0 1 0 0 Out put

An + In = Dn+1Cn+1= Dn+1Cn = Bn+1An + Bn = An+1

QUESTION

• SHOW THE FOLLOWING RECEIVED MESSAGE IS IN ERROR,FOR THE SAME TRASMITTED MESSAGE ie 10010101010100

• Received message:10110100010100• WRITE THE ERROR MESSAGE EQUATION

The remainder 00010 implies that there is an error

ERROR EQUATION

• TRANSMITTED MESSAGE + RECIEVED= ERROR MESSAGE

• 10010101010100• 10110100010100 00100001000000 = ERROR MESSAGEE(X)=X6 + X11

INSTANCES WHERE THE CRC IS FAILED TO ANSWER?

• THER ARE INSTANCES WHERE THE CRC WILL FAILED TO ANSWER, ONE SUCH INSTANCES WILL BE WHEN THERE ARE ERRORS INTRODUCED EQUAL TO THE POLYNOMIAL

WHEN ERROR MESSAGE IS EQUAL TO THE POLYNOMIAL (EXAMPLE)

• ASSUME THE FOLLOWING• TRANSMITTED MESSAGE

• 100101010100• RECEIEVED MESSAGE

• 100101001101• POLYNOMIAL

• 1101• SHOW THAT CRC IS FAILED TO IDENTIFY THE ERROR IN

THE MESSAGE?

1 1 1 1 0 0 1 0 01 1 0 1 1 0 0 1 0 1 0 1 0 1 0 0

1 1 0 10 1 0 0 0

1 1 0 10 1 0 1 1

1 1 0 10 1 1 0 0

1 1 0 10 0 0 1 1

0 0 0 00 0 1 1 0

0 0 0 00 1 1 0 1

1 1 0 10 0 0 0

THOUGH THE RESIDUAL IS 0 THERE IS AN ERROR IN THE RECEIEVED MESSAGE

• Hint divide the received message by mod 2• Then observe that no residuals• Write the error message & compare with the

polynomial

TRY A CRC SUM

• TRANSMIT MESSAGE• 11001011101• POLYNOMIAL• 101101• FIND OUT THE CRC• DRAW THE CIRCUIT DIAGRAM AND SHOW

CLEARLY HOW YOU PRODUCE CRC?

How a message is transmitted

Preventive cycle retransmission method of error correction

Question on basic method

4 layers of CCITT no:7

Layer 4

Instructions DATA

User Part

Layer 3

LabelSIO

Signaling Link

Layer 2

W0

W127

Link Control

FSN=5,FIB=1CRC=0

Layer 1

Station A Station B

k1

k1

k2

Actual message

SCF K2

Message handline

Signal control

Message type

Error detection and correction

SCF=Sequence control field

Layer 2

SCF K2

Layer 3

K2

Layer 4

K1

SCF

BSN=5,BIB=1Clear W5

W5

OPC|DPC|CICLABEL CONTENTS

DPC=st B

How CCITT No:7 works- Study about the layered structure

How reroutine is done?

Layer 4

DATA

Layer 3

LabelSIO

Layer 2

W0

W127

FSN=5,FIB=1CRC=0

Layer 1

Station A

k1

k1

k2

SCF K2

SCF=Sequence control field

Layer 2

SCF K2 K2

SCF

BSN=5,BIB=1Clear W5

W5

Station B

Layer 3

DPC=st C

Station C

SCF K2

SCF K2K2

W10

FSN10 ,BIB0

K1

QUESTION

• SHOW THE FOLLOWING RECEIVED MESSAGE IS IN ERROR,FOR THE SAME TRASMITTED MESSAGE ie 10010101010100

• Received message:10110100010100• WRITE THE ERROR MESSAGE EQUATION

The remainder 00010 implies that there is an error

ERROR EQUATION

• TRANSMITTED MESSAGE + RECIEVED= ERROR MESSAGE

• 10010101010100• 10110100010100 00100001000000 = ERROR MESSAGEE(X)=X6 + X11

1 0 0 0 0 0 0 12 0 0 0 0 0 1 13 0 0 0 0 1 1 04 0 0 1 1 0 0 05 0 1 1 0 0 0 06 1 1 0 0 0 0 07 0 0 0 1 1 1 18 0 0 1 1 1 1 19 1 1 1 1 1 1 1

10 0 1 0 0 1 0 011 1 0 1 0 0 0 112 1 1 1 0 0 0 013 0 1 1 0 0 0 014 1 1 0 1 0 0 115 0 0 0 0 0 0 016 0 0 0 0 0 0 017 0 0 0 0 0 0

How a message is transmitted

Basic method of error correction

Preventive cycle retransmission method of error correction

Question on basic method

4 layers of CCITT no:7

Link switching networks

3 stage link switching networks

Part 6

• Switching network

Basic analogue switch

Input

Output

A

B

C D

No of points =4Full available SwitchNon blocking switchAll the voltages generated in the phone can be seen in the points

A

C D

B

AB

CD

Analogy

4 4 switching

No: of * points =Is it fully available =Is it non-blocking =

16YesYes

When the number of inputs and no: of ouptuts increases , we have to think about a alternative solution

QUALITY FACTORS OF A SWITCHING NETWORK

To have full availability, we should have at least 1 link from the input small switch to a small output switch as shown above

Blocking:

When A is connected to E, can B be connected to F?

AB

CD

EF

GH

No, therefore this is a blocking network

ABCD

EFGH

How to make this non-blocking?AB

CD

EF

GH

2

2

22

2 stage Link switching networksfull available but blocking. How to make non blocking

2 stage Link switching networksfull available and non blocking.

3

3

3

3

3

3

3

3

3

ANALYSIS OF3 stage link switching networks for non blocking

ANALYSIS OF3 stage link switching networks for non blocking

1

2

1

2

Non-blocking 3 stage switching network

Near minimum cross points

Working of T switch

A B

Assume a master switch is connected with 2 RSU’s

Working of T switch

Assume a master switch is connected with 2 RSU’s as shown in the figure below

Assume 16 PCm systems are connetcted to from RSU to MSU

Now assume that A is speaking with BA speaks in p1 TS5 and B in p16 TS10

A B

What will happen at the master switch C ?

A’s hello !! P1f TS5------- P16b TS10 B will hear.

Switching Equation

B’s hello !! P16f TS10--- P1b TS5 A will hear.

Detailed working of T-switchTiming chart

A’s hello !! P1f TS5

T=0 T=125µs

TS5

TS10

P1f

P16b

B’s hello !! P16fTS10

TS10

TS5

P16f

P1b

------ P16b TS10 B will hear.

---- P1b TS5 A will hear.

What really happens at the T-switchA’s hello !! P1f TS5 ------ P16b TS10 B will hear.

8 Bits

W0

W1023

P1f TS5 W5

Buffer Memory

W0

W1023

Go to W5 W522

Control Memory 10 Bits

1

2

3

4

B’s hello !! P16fTS10 ---- P1b TS5 A will hear.

P16f TS10 W5221

Go to W522 W5

2

3

4

Microprocessor-2 actions for each channel in 125µs

TYPES OF T SWITCHES• WHAT WE HAHE STUDIED NOW IS OUTPUT CONTROLLED T

SWITCH• THE INPUT PCM TSS ARE CYLICLY STORE IN THE BUFFER

MEMORY• THE OUT PUT PCM ARE CYCLICLY ADDRESSING THE

CONTROLLED MEMORY, THE CONTENTS WILL TELL YOU WHERE TO READ AND TRANSPORT IT TO THE DESTINATION.

• THE OUTPUT PCMS ARE RIGIDLY CONNECTED TO THE CONTROLED MEM,ORY THATS WHY IT IS CALLED OUT PUT CONTROLLED T SWITCH

• SIMILARLY CAN YOU TRY A INPUT CONTROLLED T SWICH AND WRITE ITS CHARACTERISTICS?

CHARACTERISTICS OF INPUT CONTROLLED T SWITCH

• INPUT PCMS ARE RIGIDLY CONNECTED TO THE CONNTROLLED MEMORY

• THE INPUT PCMS ARE CYCLICLY ADDRESSING THE CONTROLLED MEMORY. THE CONTENTS WILL TELL YOU WHER TO STORE IN THE BUFFER MEMORY.

• THE OUTPUT PCMS ARE CYCLICLY READING THE BUFFER MEMORY.

BASIC COMPONANTS OF A SWITCHING SYSTEM

• SWITCHING UNIT• CONTROLLED UNIT• PERIPARALS• SOFTWARE

• IS IT ANALOGUES TO HUMAN BODY?• YES,EXCEPT FREE WILL OF THE HUMAN,THIS

SYSTEM WILL HAVE HEART, BRAIN & MIND

Comparison of Human with Animal & SWITCHING NODE

HUMAN

TELEPHOE NODE

HEART BRAIN MIND FREE WILL/GOOD

SWITCHING NETWORK CONTROL

NETWORK SOFTWARE

TRUTHTRUTH

ANIMAL

HEARTBRAIN

PROGRAMMING

INSTINC

SWITCHING UNIT• MAIN FUNCTION—connecting to an input to a

output• In the case of local node, input will be the

customer, and the output will be a route, where the call is destined to

• Limiting factor: no of connections that can be established simultaneously is the limiting factor

• MEASURED IN ERLANG• Present day technology : analogue & digital

switches are now obsolete, now packet switching routers are deployed.

CONTROLLED UNIT• FUNCTION: ALL THE MANAGEMENT FUNCTIONS THAT NEED TO

CARRYOUT IN ESTABLISHING ACONNECTION WILL BE DONE BY THE CONTROLLED UNIT.

• THE MANAGEMENT FUNCTIONS ARE /CALL ESTABLISHMENT, SENDING INFORMATION TO THE OTHER NODES, CALL BILLING FUNCTION, CUSTOMER FACCILITY MANAGEMENT ETC.....

• LIMITATION WILL BE THE OCCUPANCY OF THE PROCESSOR. NORMALLY MORE THAN 80% WILL NOT BE ADVISABLE FOR ANY PROCESSOR. ANOTHER MEASUREMENT WILL BE TLME TAKEN TO ESTABLISH A CONNECTION

• TECHNOLOGY: CENTRALISED CONTROL FUNCTION HAS BEEN SHIFTED TO DITRIBUTED PROCESSER FUNCTION. MODERN NGN SWITCH WILL HAVE MOST OF THE MANAGEMENT FUNTIONS CENTRALISED TO THE CONTROL PART OF SOFT SWITCH, WHILE ROUTING PART IS DISTRIBUTED TO THE ROUTERS. ANOLGUE CONTROL(WIRED LGIC),IS OBSELETE.

PERIPARALS

• THEY ARE THE ANCILIARY EQUIPMENT TO CARRY OUT THE MAJOR FUNTIONALITIES OF THE SYSTEM. THEY BARE REGISTERS, TONE GENERETORS, TIMING DEVICES, ETC...

SOFTWARE

• SOFTWARE WILL PROVIDE ALL DETAILED ACTION PLANS IS BECOMING HIGLY COMPLEX.

• MODULAR KIND OF SOFTWARE IS NOW ENCOURAGED.

• WITH NGN TECHNOLOGY THE SOFTWARE HAS BECOME A VITAL ELEMENT FOR THE PROPER FUNCTIONING OF MUTIPLE SERVICE FACILIETIES.

Understanding traffic concepts

v

Packet Switching

THE IP WORLD

• TODAY WE ARE IN THE IP WORLD

• ALL THE NETWORKS ARE PUSHING TO THE IP APPLICATIONS

• MODERN CODING METHODS PUSH IP NETWORKS TO GO FOR REAL TIME APPLICATIONS

• TDM & ATM NETWORKS ARE REPLACED BY IP

• LETS STUDY THE PACKET CONCEPTS

Major switching types

Circuit ,Message and Packet switching

CHRACTERISTICS OF SIGNALLING & VOICE

• CHANNEL ASSOCIATED SIGNALLINGSIGNALLING & VOICE IN ONE TUNNEL-WASTAGE IN

SIGNALLING, AND LESS THAN 50% EFFICIENT IN VOICE MEDIA DUE TO THE CHARACTERISTICS OF VOICE.

COMMON CHANNEL SIGNALLINGSIGNALLING IS COMMONLY USED FOR MANY VOICE

CHANNELS, HENCE SIGNALLING CHANNEL IS EFFICIENT. VOIE CHANNELS CARRIES THE SAME INEFFICIENCY AS IN THE CASE OF CHANNEL ASSOCIATED SIGNALLING

HENCE THE CONCEPTS OF MESSAGE AND PACKET SWITCHING NETWORK TO BE CONSIDERED.

DELAY ANALYSIS OF A MESSAGE SWITCH

• TAKE THE PREVIOUS EXAMPLE, THE FOLLOWING ARE KNOWN

• MESSAGE SIZE=30, OVERHEAD=3, NO OF HOPS=3,THE DELAY FOR ONE HOP/OCTET=1 MSEC.

• THE TOTAL DELAY FOR 3 HOPS=33*3 MSEC• TOTAL DELAY=NO OF HOPS*NO OF OCTET IN

MESSAGE+NO OF OVERHEAD OCTETS IN THE MESSAGE(THIS RESULT IS TALLYING)

Deciding Optimum Packet Size

• Shown above is a message

.

Packet 1 Packet 3Packet 2MESSAGE

• The message will be divided in to equal length packets

l l l

Packet 1

• Each Packet will have a HeaderThe header will consist the following details:

– Originating Point Code– Terminating Point Code– Packet Number

Packet 3

Packet 3

Packet 1

Packet 2

A B C D

Packet 1

Packet 2

Packet 3

Time Slots

How Many Transactio

ns?

1

2

3

2

1

Packet 1T1T1

T2T2

T3T3

T4T4

T5T5

Packet 1Packet 2

Packet 2Packet 3

TOTAL DELAY α S + H - 1

Where S - Number of PacketsH - Number of Hops

Therefore in this case the total delay time is α 3 + 3 – 1 = 5

DELAY ANALYSIS FOR PACKET SWITCHING NETWORK

• TAKE TWO EXAMPLES THE 30 OCTET MESSAGE IS (1) MADE TO TWO PACKETS OF 15 OCTET EACH, (2) MADE TO 3 PACKETS OF 10 OCTETS EACH

• YOU WILL OBSERVE THE FOLLOWINGOVERLAP SENDING OF PACKETS, FOR EXAMPLE,

WHEN 1 ST PACKET RCEIVED AT B, THIS PACKET IS SEND FROM B TO C,DURING THIS TIME THE 2ND PACKET IS SEND FROM A TO B. NOTE THAT IN AGIVEN TIME 2 ACTIONS HAVE BEEN MADE

THE TOTAL DELAY EXPERIENCED IN (1)72 MSEC (2) 65 MSEC

Message vs Packet delay

• It is easy to calculate message delay rather than packet delay ,why ?

• When sending packets from one node to another the following process can be adopted as against message transmission

• (a)Packets can be sent to another node through different paths simultaneously

• (b)Packetizing at the node and sending packets over the hop can be made in different times to maximize the sending of packets over a hop .

• (C)Hence , overlap sending & receiving of packets can be achieved in a node . Hence the delay introduced in packet mode is rather complicated (than message) although it is efficient .

Contd….

• In message mode the delay introduced in (H-1) hops is rather simple and is equal to

• Message delay = K * (H-1) • In packet mode the delay is proportional to addition

delay to message length and it is assumed to be • D = K*(H-1) + term proportional to message length • Delay={ S+ (H-1) } * ZT

Summary

Calculation of the optimum packet size

• Deciding the optimum packet size will depend upon 2 factors

• Overall delay• Overall overhead bits compared to the

message• Calculate the optimum packet size of 30 octet

message, where the overhead for aq packet is 3 octet?

What is the packet network?

• A packet is a unit of data that is transmitted across a packet-switched network.

• A packet-switched network is an interconnected set of networks that are joined by routers or switching routers.

Example

• Packet Switching technology TCP/IP

• largest packet-switched network Internet

Why ?

• Data traffic is bursty– Logging in to remote machines– Exchanging e-mail messages

• Don’t want to waste reserved bandwidth– No traffic exchanged during idle periods

• Better to allow multiplexing– Different transfers share access to same links

Goals

• To optimize utilization of available link capacity

• To increase the robustness of communication

Concept

• A method of transmitting messages through a communication network, in which long messages are subdivided into short packets.

• The packets are then sent through the network to the destination node.

Packet-Switching Techniques

Each packet contains addressing information and is routed separately

Datagram Virtual Circuits

A logical connection is established before any packets are sent; packets follow the same route.

Datagram

• Each packet treated independently• Packets can take any practical route• Packets may arrive out of order• Packets may go missing• Up to receiver to re-order packets and recover

from missing packets

1ComputerA

Computer B

2 3Need to transmit ‘123’ from computer A to computer B

First data is broken to small

pieces (PACKETS)

1

Computer A

Computer B

23

Packets contain header information that includes

a destination address.

Routers in the network read this

address and forward packets along the most appropriate

path to that destination.

3

ComputerA

Computer B

1

2

3

ComputerA

Computer B

1

2

3

ComputerA

Computer B

1

2

Virtual Circuits v Datagram• Virtual circuits

– Network can provide sequencing and error control– Packets are forwarded more quickly

• No routing decisions to make– Less reliable

• Loss of a node loses all circuits through that node

• Datagram– No call setup phase

• Better if few packets– More flexible

• Routing can be used to avoid congested parts of the network

• Line efficiency– Single node to node link can be shared by many packets over time– Packets queued and transmitted as fast as possible

• Data rate conversion– Each station connects to the local node at its own speed– Nodes buffer data if required to equalize rates

• Packets are accepted even when network is busy– Delivery may slow down

• Priorities can be used

Advantages

Difference between channel associated common channel and packet signaling networks

Supervisor signal

Voice channel

Each voice channel will have a supervisory channel(either direct or associate).Highly inefficient for the signalling channel and less than 50% efficient for the voice channel.

Channel associated signalling(CAS)

Common Channel signalling – All supervisory Signals of voice channels are in one time slotAnd the voice channels have similar inefficiency as CAS

Packet network=Signalling and voice are sent in packets ,highly efficient for voice as well as signalling.The deficiecy experienced for voice channels, in CAS & CCS has overcome

Signalling and voice are going on packets whenever it is needed