Transcript Summary of PCM systems

Summary of PCM systems
Communication

By Directional Interaction

Public and Personal
Technical Characteristics
Public- Unidirectional Tx does not know how many receivers are ON
Personal – By directional interactive
Operation of Domestic Delivery
Network
District deport
Local deport
Province deport
Province deport
District deport
Local deport
Telecom Network in Summary
Local EX 03
Domestic Transport
International
Transport
IEX
Access
02
IEX
01
Land line
IEX
IEX
Why Telecom more Popular
 Electronically dist=0
 Answer only Charge
 Tell No. 15 digits (Universal)
CC
AC
CC- Country Code
AC- Area Code
DN- Directory No
 Demarcation of Telecom
Transmission
DN
Cont…
 Digital
Problem to achieve Digital Tx
Noise
Tx
Rx
Media
The Samples cannot
be Reproduced
Attenuation
Find a technique Digital Tx
(1) Tx Info
Tr Media
(1) Rx Info
(2) Verify the Rx Info
Verification Difficult
Cont…
 Quantizing
 Equate the sample to a quantize level.
 Then transmit verification will be easy at the receiver
 Quantizing noise is inevitable
 Encoding
 Convert this quantized level in to binary level
 Verification will be more easy
Quantizing
In linear quantizing S/N is good only for high valued samples and
90% of the samples are within ½ of maximum voltages
Hence the samples will be equate to 1/256 levels
Hence Quantizing Noise (∆V) is inherent in PCM transmission,
since there is a difference between actual sample to Quantized
level.
The A law Signaling Compression and
Characteristics
Segment No
Voltage Range
Voltage range
7
Vm – Vm/2
3072 – 1536
6
Vm/2 – Vm/4
1536 – 768
5
Vm/4 – Vm/8
4
Change over to
next segment
Level range
Increment per
Level
127 – 111
96
>1512
111 – 95
48
768 – 384
>756
95 – 79
24
Vm/8 – Vm/16
384 – 192
>378
79 – 63
12
3
Vm/16 – Vm/32
192 – 96
>189
63 – 47
6
2
Vm/32 – Vm/64
96 – 48
>94.5
47 – 31
3
1
Vm/64 – Vm/128
48 – 24
>47.25
31 – 15
1.5
0
Vm/128 –
24 – 0
>23.25
15 – 0
1.5
Cont…
Note : A Total of 256 quantisation steps covers
line peak to peak range of nomal speech
intensities
A law gives lower quantising dislortion …. Law
There are 16 segments shown in this graph
positive 0,1 and negative 0,1 consai one linear
segment.
hence there are 10 linear segments.
Encoded 8 bit format
S
Sign
A B
C
W
X Y Z
No of seg
No of pos in the
Segment
If S=1 it is positive sample
If S=0 it is Negative sample
Vm – Maximum voltage = 3072 mv
N – Na of quantised levels =256
Some times ’A’ low is named as Eurpean law (C.E.P.T)
Equation for logaribimic part y=n ln Ax / ln A (1/A<x<1)
Linear part y=Ax (0<x<1/A)
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
Encoding
The quantized level is then converted in to 8 bits. This 8 bits
represent,
S ABC WXYZ
S = sign + or ABC = No of segments
WXYZ = No of level in that segments
Summary of process involved,
equate
Sample
To a quantize Convert
level 1/256
8 bit
Difference Codes used in digital
Transmission
Frequency
Time Division Multiplexing
Media
250
R1
R2
125
For a given signal 125µs period the samples to be send
R1 is idling too long. To make it efficient 32 value signals are sampled
and send with in 125µs
TS0
TS16
TS31
Each TS = 3.9µs
TS- Time Slot
Practically TS0,TS16 not used for normal voice signal.
But for Synchronizing + Signaling respectively
R1
R2 the speed is 2.048 mb/s
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
Cont…
Supervisory
Signaling
Analog
Register
Characteristics
Supervisory is always present with voice.
Register is always prior to voice hence analogue channel
exchange will be as follows.
Exchange to another exchange will be as follows
V = Voice
R = Register
Sup. Signals are on M, E, Wires
Cont…
 Multiframe in a PCM SYSTEM for supervisory signals only
TS16 is available. CCJTT has allocated 4bits for each channel.
 To send 30 channels supervisory signals on TS16, You need
15 frames.
 To align SIG TR module to SIG RX module one TS16 is used.
Hence Multiframe consist 16 Frames.
f0
MF Sys
f1
CH1
CH1
f2
CH1
CH1
f15
CH1
CH1
2 ms
Structure of Multiframe
One Multiframe= 16 Frames
TS 1-15
TS 0
TS 17-31
Practical Channels
TS 16
TS 0
TS 0
TS 0
1
2
15
17
31
There are two kinds of
synchronization words odd
and even
Odd actually synchronization
Even alarm signaling
F0 TS16 is used for Multiframe
alignment all other TS16 are
used for Channel Associated
signaling
Pleslouronus Digital Multiplexing
First Order or
primary order
400
2/8
8/34
Second Order
Third Order
110
25
34/140
Fourth Order
7
140/620
Fifth Order
1.7
Channel Associated Signaling At a
Glance
F0
TS0
F1
TS0
CH 1 CH 17
TS31
F14
TS0
CH 4 CH 30
TS31
F15
TS0
CH 15 CH 31
TS31
TSI6
TS31
Block Diagram of PCM System
* = Except 16
1 – Signaling Compartment
2 – SYNC Compartment
3 – (V + R) Compartment
C – Combiner
D - Distributor
Transcoding
Code Conversion to suit for the Transmission media
Out put of a PCM System either RZ, NRZ
1 bite named as mark NRZ means, Mark will return to zero before
the period of CLK pulse, but at the period of the click pulse.
RZ, means mark will NOT come to zero before the period of the
CLK pulse, but at the period of the CLK pulse if the following is
not a MARK.
Practical Transcording wave Forms
High Density Bipolar 3.
Rules
1. Don’t allow more than 3 Consecutive Zero’s to be present in the wave
form (media). Introduce a violation bit. Violation bit has to be of the
same polarity of the previous MARK.
2. Two Consecutive violation bits has to be of opposite polarity.
3. Between two consecutive violation bits if there are even number of
last violation will be boove where B is the stuffing BIT and will be of
opposite polarity to the previous MARK.
Process Involved
Basic Structure of SDH
1. Basic structure
1
1
2
270
125µs
1s
2430 × 8 Bits
155.52 Mbits
2430 270
9 2161
125µs
2. Structure for 2Mb/s and 34Mb/s
1
2.048 Mb/s
1
2
3
4
34.368 Mb/s
1
1
2
3
4
125 µs = 36 × 8
1s = 2304 kb
Path over head + Justification =0.256 (12.5%)
For 34 mb structure 21Nos 2.048 Mb/s can be
placed
84
756
36
9
1
1
84
125 µs = 36 × 8
1s = 2304 kb
Path over head + Justification =14.02 (40%)
Cont…
3. Observations
 For 34 Mb/s in PDH 2.048 Mb/s, 16 streams can be Multiplexed
 In SDH 21 No can be Multiplexed WHY?
 For PDH, CEPT 34.368mb/s and PDH American Equipment is 44.736 Mb/s, Hence 84
columns is used for 444.736 Mb/s American Systems, SDH stream stems from
American SONET. Hence it has been designed for American 44.736 Mb/s.
 Every basic structure has to placed, it needs more two columns to accommodate PDH
+Justification.
Hence fir 34 direct to be placed, it needs more two columns to accommodate PDH +
Justification.
Hence
1 2 3
86
3.5 if we fill with 21 Nos of 2.048 Mb/s, these first 2 columns are spare
Cont…
4. Structure for 140Mb/s
Similarly for 140 Mb/s (actual 139.264Mb/s)
1
2
1465
1
258
For ,
125µs=2322 × 8bits
1s=148.605
2322
258
Spare bits for POH + Justification=9.344(6.7%)
5. Observations

For 140 Mb/s is PDH (CEPT) there are 4 Nos 34Mb/s streams. But in SDH
only 3 Nos 344 Mb/s can be accommodated.

Hence in SDH, 63 Nos 2.048 Mb/s in STM can be accommodated.

No equipment for PDH 140 Mb/s (America)
Cont…
6. Similar reasoning as for 3.4: in order to accommodate direct 140
Mb/s into SDH 3 columns are used for PDH + Justification
1 2 34
261
7. If we fill with 3 of 34 Mb/s, these first 3 columns are spare
8. Accommodation of bit rates for SDH
Maximum of
a. 2.048 63Nos, or
b. 34 3Nos, or
c. 140 1Nos, or
d. Combination of
a& b
If,
1
2
No 34 mb/s then maximum of 42 no of 2 Mb/s
No 34 mb/s then maximum of 21 no of 2 Mb/s
270
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