University Of Moratuwa Lecture 2012
Feb 02, 2016
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