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Summary of PCM systems. Communication By Directional Interaction Public and Personal Technical Characteristics Public- Unidirectional Tx does not know.

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Presentation on theme: "Summary of PCM systems. Communication By Directional Interaction Public and Personal Technical Characteristics Public- Unidirectional Tx does not know."— Presentation transcript:

1 Summary of PCM systems

2 Communication By Directional Interaction Public and Personal Technical Characteristics Public- Unidirectional Tx does not know how many receivers are ON Personal – By directional interactive

3 Operation of Domestic Delivery Network Local deport District deport Province deport District deport Local deport

4 Telecom Network in Summary LE Land line Local EXDomestic Transport International Transport 01 02 03 AccessIEX

5 Why Telecom more Popular Electronically dist=0 Answer only Charge Tell No. 15 digits (Universal) Demarcation of Telecom CCACDN CC- Country Code AC- Area Code DN- Directory No Transmission

6 Cont… Digital Problem to achieve Digital Tx Find a technique Digital Tx Attenuation Tx Rx Media Noise Tr Media (1) Tx Info (1)Rx Info (2)Verify the Rx Info Verification Difficult The Samples cannot be Reproduced

7 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

8 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.

9 The A law Signaling Compression and Characteristics Segment NoVoltage RangeVoltage rangeChange over to next segment Level rangeIncrement per Level 7Vm – Vm/23072 – 1536127 – 11196 6Vm/2 – Vm/41536 – 768>1512111 – 9548 5Vm/4 – Vm/8768 – 384>75695 – 7924 4Vm/8 – Vm/16384 – 192>37879 – 6312 3Vm/16 – Vm/32192 – 96>18963 – 476 2Vm/32 – Vm/6496 – 48>94.547 – 313 1Vm/64 – Vm/12848 – 24>47.2531 – 151.5 0Vm/128 –24 – 0>23.2515 – 01.5

10 Cont… 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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/5/1475611/slides/slide_10.jpg", "name": "Cont… 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

11 Exercise 1: Convert the following denary numbers to binary(Dont 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

12 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

13 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

14 Answers to Exercise 2 (a) 101 5 (b) 1106 (c) 10019 (d) 1110129 (e) 10000032 (f) 101101090 (g) 111000111455

15 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

16 Answers to Exercise 3 DenaryHexaBinary (a) 991001 (b) 2014 10100 (c) 3624 100100 (d) 129 81 10000001 (e) 518206 1000000110 (f) 1030406 10000000110 (g) 4095 FFF 111111111111 (h) 82002008 10000000001000

17 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 To a quantize level 1/256 Convert 8 bit Sample

18 Difference Codes used in digital Transmission Frequency

19 Time Division Multiplexing 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 Practically TS0,TS16 not used for normal voice signal. But for Synchronizing + Signaling respectively R1 R2 the speed is 2.048 mb/s 125 250 R1 R2 Media TS31TS16TS0 Each TS = 3.9µs TS- Time Slot

20 Convert the following samples into encoded format and calculate the signal /noise ratio 700mV -400mV 300mV 100mV 1515mV -95mV

21 Answers 700mV -400mV 300mV 11011101 01010001 11001001 175 50 100mV 1515mV -95mV 10110001 11110000 0011000 25 72 295

22 Cont… SignalingAnalog Supervisory 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

23 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. MF Sys CH1 f0 f1 f2 f15 2 ms

24 Structure of Multiframe TS 0 TS 1-15 TS 16 TS 17-31 TS 0 One Multiframe= 16 Frames TS 0 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 Practical Channels 1 2 15 17 31

25 Pleslouronus Digital Multiplexing 400 1102571.7 First Order or primary order Second Order Third OrderFourth OrderFifth Order 2/8 8/34 34/140 140/620

26 Channel Associated Signaling At a Glance TSI6TS0TS31 TS0TS31 TS0TS31 TS0TS31 CH 1CH 17 CH 4CH 30 CH 15CH 31 F0F0 F1F1 F 14 F 15

27 Block Diagram of PCM System * = Except 16 1 – Signaling Compartment 2 – SYNC Compartment 3 – (V + R) Compartment C – Combiner D - Distributor

28 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.

29 Practical Transcording wave Forms High Density Bipolar 3. Rules 1. Dont allow more than 3 Consecutive Zeros 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

30 Basic Structure of SDH 1.Basic structure 2. Structure for 2Mb/s and 34Mb/s 1 270 1 9 2161 2 270 2430 125µs 125µs 2430 × 8 Bits 1s 155.52 Mbits 1 1234 36 4321 9 1184 756 841 34.368 Mb/s2.048 Mb/s 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 125 µs = 36 × 8 1s = 2304 kb Path over head + Justification =14.02 (40%)

31 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. 3.5 if we fill with 21 Nos of 2.048 Mb/s, these first 2 columns are spare Hence 123 86

32 Cont… 4.Structure for 140Mb/s Similarly for 140 Mb/s (actual 139.264Mb/s) 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) 12258 2322 2581 1465 For, 125µs=2322 × 8bits 1s=148.605

33 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 7.If we fill with 3 of 34 Mb/s, these first 3 columns are spare 8. Accommodation of bit rates for SDH If, 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 1232614 Maximum of a.2.048 63Nos, or b.34 3Nos, or c.140 1Nos, or d.Combination of a& b 1 2 270

34 Technological Evolution (Fill the blanks) Multiplex LevelSpeedPeriod of the PulseNo: of voice channels STM1 STM4 STM16 STM64 STM256

35 Technological Evolution at a glance Multiplex LevelSpeedPeriod of the PulseNo: of voice channels STM1155.52Mbps 6.4ns 1890 STM4622.08Mbps 1.6ns 7560 STM162.5Gbps 400ps 30240 STM6410Gbps 100ps 120960 STM25640Gbps 25ps 483840


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