Transcript Moratuwa Campus Final Year BSC Engineering December

University Of Moratuwa
Lecture
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
(b) 9
(c) 16
(d)33
(e) 67
(f) 120
(g) 520
(h) 1028
(i) 2050
(j) 4100
(k) 8200
(l) 16401
•
•
•
•
Answer to Exercise 1
(a) 5=101
(c) 16=10000
(e) 67=1000011
(g) 520=1000001000
(i) 2050=100000000010
• (k) 8200=10000000001000
•
•
(b) 9=1001
(d)33=100001
(f) 120=1111000
(h) 1028=10000000100
(j) 4100=1000000000100
(l) 16401=100000000010001
Exercise 2
Convert 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
(b) 110
(c) 1001
(d) 11101
(e) 100000
(f) 1011010
(g) 111000111
5
6
9
29
32
90
455
Exercise 3
Convert 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
(a) 9
(b) 20
(c) 36
(d) 129
(e) 518
(f) 1030
(g) 4095
(h) 8200
Hexa
9
14
24
81
206
406
FFF
2008
Binary
1001
10100
100100
10000001
1000000110
10000000110
111111111111
10000000001000
Convert the following samples into encoded
format and calculate the signal /noise ratio
• 700mV
-400mV
300mV
• 100mV
1515mV
-95mV
Answers
• 700mV
• 11011101
175
• 100mV
10110001
25
-400mV
300mV
01010001
50
1515mV
11001001
∞
-95mV
11110000
72
0011000
295
Pcm equipment
Pcm equipment(2) contd
PART 3
• HIGHER ORDER PCM
Technological Evolution
(Fill the blanks)
Multiplex Level
STM1
STM4
STM16
STM64
STM256
Speed
Period of the Pulse
No: of voice
channels
Technological Evolution at a glance
Multiplex Level
Speed
Period of the Pulse
No: of voice
channels
STM1
155.52Mbps
1890
STM4
622.08Mbps
6.4ns
1.6ns
STM16
2.5Gbps
400ps
30240
STM64
10Gbps
100ps
120960
STM256
40Gbps
25ps
483840
7560
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.
Air
Glass
Water
Air
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
n2
Sin Ө 1
=
Sin Ө 2
n2
=
n1
Sin Ө 2
n1
n 1 Sin Ө 1 = n 2 Sin Ө 2
PO - Ray of Incidence
medium 1
OQ - Ray of Refraction
medium 2
Ө 1 - Angle of Incidence
Ө 2 - Angle of Refraction
n 1 - RI for
n 2 - RI for
TOTAL INTERNAL REFLECTION
n 1 Sin Ө 1 = n 2 Sin Ө 2
With the increase of the angle of
incidence, the angle of
refraction 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 Fibre
optical 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
6.4ns
620Mbps
2.5Gbps
10Gbps
40Gbps
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
Future Scenarios
Theoretical Maximum of an Optical Fibre Cable
488ns
100 ps
Transponders
λ1
1
10Gbps
2
λ2
2399
λ2399
TDM
2Mbps
2400
Optical Fibre
Only 1 core is needed
λ2400
Number of wavelengths = ( 24 * 103 Gb ) / 10 Gb
= 2400 wavelengths
PART 5
COMMON CHANNEL SIGNALLING
MESSAGE TYPES
• BASIC MESSAGE
• HOMOGENIOUS MESSAGE
• NON HOMOGENEOUS MESSAGE
Basic Message
Instruction
Data
Data
Fixed
Variable
=
K1
Message for Homogenous Network
Label
Instruction
Data
= K2
OPC
14bits
DPC
CIC
14bits
12bits
OPC – Originating Point Cord
DPC – Destination Point Cord
CIC – Circuit Identification Cord
Message for Non-Homogenous Network
SIO
K2
Label
National or
International
Message
User
4bits
4bits
Instruction
Data
SIO - Service Information Octel
K2 - Message for Homogenous Network
Part
Now we are ready with the complete message, can we
transmit it just as it is?
NO
WHY NOT?
IAM
ACM
ANC
CBK
H1
H0
0000
0000
0001
0010
0011
0100
0101
0110
0111
IAM
SAM
0010
OSM
COT
CCF
0011
ORQ
0100
ACM
CHO
0101
SEC
COC
NNC
ADI
CFL
SSB
UNN
0110
ANC
ANN
CBK
CLF
RAN
FOT
CCL
0111
RLG
BLO
BLA
UBL
UBA
CCR
RSC
1000
MGB
MBA
MGU
MU
A
HOA
HBA
HGU
1001
CFM
CPM
CPA
CSV
CVM
CHM
CLI
1100
1110
1111
1010
1011
1100
LOS
SST
ACB
DPN
MPR
1101
1110
1111
Spare reserved for national use
EUM
EAM
HUA
GRS
GRA
SGB
Spare reserved for international and basic national use
1011
1101
1001
Spare reserved for national use
0001
1010
1000
SBA
SGU
SUA
Basic concept of message transmission to establish a call
IAM
B
ACM
ANC
A
Ringing current to
subscriber “B” n
ringback tone to
subscriber “A”
Speech
CBK
Node X
IAM
(Initial address
message)
ACM
(Answer complete
mesaage )
ANC
CBK
Variable
Node Y
H0
H1
0001
0001
H0
H1
0100
0001
H0
H1
0110
0001
H0
H1
0110
0011
20 Bits
4 Bits
Fixed (8 Bits)
No Data
No Data
Dial Number
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
Need to transfer message between A to B
P0f
P1f
P15f
X
Y
exchange
exchange
P0b
P1b
P15b
Customer A
Customer B
Helicopter View
Exchange X
Exchange X
IAM ( P0f TS16 )
( P0b TS16 )
ACM
RBT
( P9 TS28)
( P0b TS16 ) ANS
( P9f TS28)
speaking
( P0f TS16 ) CBR
( P9b TS28)
Name
Standards
Purpose
IAM
Initial Address
Message
Dialing Information
ACM
Address Complete
Message
B customer free or not
RBT
Ring Back Tone
Tone herd by A
ANS
Answer Signal Charge
B customer answer or not
Call Back Tone
Release the circuit
CBR
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
= (31 * 14) + (30 *2)
494
Channel number that we use
=
305
If we numbered voice channel from 1 to 494 : Select related TS
30 + 30 =60
305 – 60 = 245
245 / 31 = 7 mod 28
7+2=9
P9 TS28
(PCM no = 9 , TS no = 28)
Number that we dial = 15904607
IAM
0001 0001
4 bits 4 bits
0011
8
1000,1010,1001,0000,1000,1100,0000,1110
4 4 bits
4 * 8 bits
K=56 bits
Message
8 bits
CRC
SCF
16
16
bits
bits
SIO
1
2
305
K
8
12
12
14
56
bits
bits
bits
bits
bits
Total bits = 150
8 bits
ACM
0001
0110
4 bits
4 bits
K=16 bits
Message
8
bits
8 bits
CRC
SCF
SIO
1
2
305
K
16
bits
16
bits
8
bits
12
bits
12
bits
14
bits
16
bits
Total bits = 110
8
bits
ANC
0001
0110
4 bits
4 bits
K=8 bits
Message
8
bits
CRC
SCF
SIO
1
2
305
K
16
bits
16
bits
8
bits
12
bits
12
bits
14
bits
8
bits
Total bits = 102
8
bits
CBR
0001
0110
4 bits
4 bits
K=8 bits
Message
8
bits
CRC
SCF
SIO
1
2
305
K
16
bits
16
bits
8
bits
12
bits
12
bits
14
bits
8
bits
Total bits = 102
8
bits
Conclusion
• time for forward message = 2.34 ms
• time for forward message = 4.906 ms
Phases of a call
Dial Tone
Dialing
Signaling
Answer
Ring back Tone
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
A
•
•
•
•
•
+
Timing equations
An + In = Dn+1
Dn = Cn+1
Cn = Bn+1
An + Bn = An+1
B
C
Input
Data
X0
D
+
I
An + Bn = An+1
Step
Cn = Bn+1
Cn+1= Dn+1
An + In = Dn+1
0
A
0
B
0
C
0
D
0
Input
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 + Bn = An+1
Cn = Bn+1
Cn+1= Dn+1
An + In = Dn+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
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 MESSAGE
E(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
0
1
1
1
0
0
1
1
1
0
0
0
0
1
1
1
0
1
1
1
0
0
0
1
1
1
0
1
0
0
1
1
0
1
1
0
0
0
1
1
1
1
0
0
0
0
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
1
0
1
1
0
1
0
1
0
1
1
0
0
0
0
0
0
1
1
0
0
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
How CCITT No:7 works- Study about the layered structure
Layer 4
Instructions DATA
Layer 3
SIO
Label
k1
k2
k1
Layer 2
W0
W5
Layer 1
Layer 3
Layer 2
Layer 4
CRC=0
SCF
K2
SCF
K2
K2
K1
W127
SCF=Sequence control field
DPC=st B
LABEL CONTENTS
OPC|DPC|CIC
User Part
Actual message
Signaling Link
Message Signal
handline control
Station A
SCF
Clear W5
Link Control
Message Error
detection
type
and
correction
Station B
How reroutine is done?
Layer 4
DATA
Layer 3
SIO
Label
k1
k2
k1
Layer 1
Layer 2
W0
W5
Layer 2
Layer 3
CRC=0
SCF
K2
SCF
K2
K2
DPC=st C
W127
SCF=Sequence control field
Station A
SCF
Clear W5
SCF
FSN10 ,BIB0
K1
K2
SCF
K2
Station C
K2
W10
Station B
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 MESSAGE
E(X)=X6 + X11
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
0
0
0
0
0
1
0
0
1
0
1
1
0
1
0
0
0
0
0
0
0
1
1
0
0
1
1
0
1
1
1
0
0
0
0
0
0
1
1
0
0
1
1
0
1
1
1
0
0
0
0
0
0
0
1
0
0
1
1
1
0
0
0
0
1
0
0
0
0
0
1
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
1
1
0
0
0
1
1
1
0
0
0
0
0
0
0
0
1
1
0
0
0
0
1
1
1
0
1
0
0
1
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
Output
C
Analogy
D
C
A
B
A
A
B
B
Input
No of
points =4
Full available Switch
Non blocking switch
All the voltages generated in the phone can be seen in the
points
D
C
D
4
4 switching
16
No: of * points =
Is it fully available = Yes
Is it non-blocking = Yes
When the number of inputs and no:
of ouptuts increases , we have to
think about a alternative solution
QUALITY FACTORS OF A SWITCHING
NETWORK
A
B
E
F
C
D
G
H
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:
A
B
C
D
How to make this non-blocking?
E
F
G
H
A
B
C
D
When A is connected to E, can B be connected to F?
No, therefore this is a blocking network
E
F
2
2
2
2
G
H
2 stage Link switching networks
full available but blocking. How to make non blocking
2 stage Link switching networks
full 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
2
1
Non-blocking 3 stage switching
network
Near minimum cross points
Working of T switch
Assume a master switch is connected with 2
RSU’s
A
B
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
B
A
Now assume that A is speaking with B
A speaks in p1 TS5 and B in p16 TS10
What will happen at the master switch C ?
Switching Equation
A’s hello !!
P1f TS5-------  P16b TS10
B will hear.
B’s hello !!
P16f TS10---  P1b TS5
A will hear.
Detailed working of T-switch
Timing chart
A’s hello !!
P1f TS5 ------  P16b TS10
B will hear.
B’s hello !!
P16fTS10 ----  P1b TS5
A will hear.
P1f
TS5
P16b
TS10
P16f
TS10
P1b
TS5
T=0
T=125µs
What really happens at the T-switch
A’s hello !!
B’s hello !!
P1f TS5 ------  P16b TS10
P16fTS10 ----  P1b TS5
B will hear.
A will hear.
4
4
2
W0
P1f TS5
W5
2
1
1
P16f TS10
3
W522
Go to W522
W0
W5
Go to W5
W522
3
W1023
8 Bits
Buffer Memory
Microprocessor-2
actions for each
channel in 125µs
W1023
10 Bits
Control Memory
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
HEART
BRAIN
MIND
TRUTH
FREE
WILL/GOOD
TELEPHOE NODE
SWITCHING
NETWORK
CONTROL
NETWORK
SOFTWARE
PROGRAMMING
ANIMAL
HEART
BRAIN
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 SIGNALLING
SIGNALLING & VOICE IN ONE TUNNEL-WASTAGE IN
SIGNALLING, AND LESS THAN 50% EFFICIENT IN VOICE
MEDIA DUE TO THE CHARACTERISTICS OF VOICE.
COMMON CHANNEL SIGNALLING
SIGNALLING 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
Packet 1
l
MESSAGE
Packet 2
Packet 3
l
l
.
•
Shown above is a message
•
The message will be divided in to equal length packets
•
Each Packet will have a Header
The header will consist the following details:
– Originating Point Code
– Terminating Point Code
– Packet Number
Packet 1
A
B
C
D
Packet 1
Packet 2
Packet 1
Packet 3
Packet 2
Packet 3
T1
Packet 1
Packet 2
Time
Slots
How Many
Transactio
ns?
T2
1
T3
2
3
Packet 3
T4
2
T5
TOTAL DELAY α S + H - 1
Where
S - Number of Packets
H - Number of Hops
1
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 FOLLOWING
OVERLAP 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
Datagram
Each packet contains
addressing information
and is routed separately
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
Computer
A
1 2 3
Need to transmit ‘123’ from
computer A to computer B
First data is
broken to small
pieces
(PACKETS)
Computer B
Packets contain header
information that includes
a destination address.
Computer A
3
Routers in the
network read this
address and forward
packets along the
most appropriate
path to that
destination.
1 2
Computer B
Computer
A
2
1
3
Computer B
Computer
A
2
1
3
Computer B
Computer
A
2
1
3
Computer B
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
Advantages
• 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
Difference between channel associated common
channel and packet signaling networks
Channel associated signalling(CAS)
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.
Common Channel signalling – All supervisory
Signals of voice channels are in one time slot
And the voice channels have similar inefficiency as CAS
Voice channel
Supervisor signal
Signalling and voice are going on
packets whenever it is needed
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 overcom