Summary of PCM systems Communication By

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 Province deport District deport Local deport

Telecom Network in Summary LE Domestic Transport LE Local EX 03 LE LE Access 02 LE LE International Transport IEX 01 Land line 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 ü Equate the sample to a quantize level. ü Then transmit verification will be easy at the receiver ü Quantizing noise is inevitable § Encoding ü Convert this quantized level in to binary level ü 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=10000010 • (k) 8200=1000001000 • • • (b) 9=1001 (d)33=100001 (f) 120=1111000 (h) 1028=10000000100 (j) 4100=100000100 (l) 16401=10000010001

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 1000000110 10000000110 111111 1000001000

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 R 1 R 2 125 For a given signal 125µs period the samples to be send R 1 is idling too long. To make it efficient 32 value signals are sampled and send with in 125µs TS 0 TS 16 TS 31 Each TS = 3. 9µs TS- Time Slot Practically TS 0, TS 16 not used for normal voice signal. But for Synchronizing + Signaling respectively R 1 R 2 the speed is 2. 048 mb/s

Convert the following samples into encoded format and calculate the signal /noise ratio • 700 m. V -400 m. V 300 m. V • 100 m. V 1515 m. V -95 m. V

Answers • 700 m. V • 1101 175 • 100 m. V 10110001 25 -400 m. V 300 m. V 01010001 50 1515 m. V 11001001 ∞ -95 m. V 11110000 72 0011000 295

Cont… Supervisory Signaling Analog Register Characteristics üSupervisory is always present with voice. üRegister is always prior to voice hence analogue channel exchange will be as follows. üExchange to another exchange will be as follows V = Voice R = Register Sup. Signals are on M, E, Wires

Cont… ü Multiframe in a PCM SYSTEM for supervisory signals only TS 16 is available. CCJTT has allocated 4 bits for each channel. ü To send 30 channels supervisory signals on TS 16, You need 15 frames. ü To align SIG TR module to SIG RX module one TS 16 is used. Hence Multiframe consist 16 Frames. f 0 MF Sys f 1 CH 1 f 2 CH 1 f 15 CH 1 2 ms

Structure of Multiframe One Multiframe= 16 Frames TS 1 -15 TS 0 TS 17 -31 Practical Channels TS 16 TS 0 1 2 15 17 31 There are two kinds of synchronization words odd and even Odd actually synchronization Even alarm signaling F 0 TS 16 is used for Multiframe alignment all other TS 16 are used for Channel Associated signaling

Pleslouronus Digital Multiplexing 2/8 First Order or primary Second Order order 400 110 8/34 Third Order 25 34/140 Fourth Order 7 140/620 Fifth Order 1. 7

Channel Associated Signaling At a Glance F 0 TSI 6 TS 31 F 1 TS 0 CH 17 TS 31 F 14 TS 0 CH 4 CH 30 TS 31 F 15 TS 0 CH 15 CH 31 TS 31

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 9 2161 125µs 1 s 2430 270 125µs 2. Structure for 2 Mb/s and 34 Mb/s 1 2. 048 Mb/s 1 2 3 4 34. 368 Mb/s 1 1 2 3 4 125 µs = 36 × 8 1 s = 2304 kb Path over head + Justification =0. 256 (12. 5%) For 34 mb structure 21 Nos 2. 048 Mb/s can be placed 1 84 756 36 9 2430 × 8 Bits 155. 52 Mbits 1 84 125 µs = 36 × 8 1 s = 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. 368 mb/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 140 Mb/s Similarly for 140 Mb/s (actual 139. 264 Mb/s) 1 2 1465 1 258 For , 125µs=2322 × 8 bits 1 s=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 34 Mb/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 63 Nos, or b. 34 3 Nos, or c. 140 1 Nos, 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 STM 1 STM 4 STM 16 STM 64 STM 256 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 STM 1 155. 52 Mbps 1890 STM 4 622. 08 Mbps 6. 4 ns 1. 6 ns STM 16 2. 5 Gbps 400 ps 30240 STM 64 10 Gbps 100 ps 120960 STM 256 40 Gbps 25 ps 483840 7560