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Transmission: Issues in Dio; Issues in digital transmission : Frequenc, (_ Coding and their Power Spectral density, ' Division Multiplexing, Time Division Moltiplering, Line 2% Chapter Outline 1 Introduction North American Hierarchy = T Lines = A PCM-TDM System : Tt Carrier. System mE Lines ™ A Baseband Digital Communication System ™ Introduction to Discrete PAM Signals (Digital Data Formats) i= Line Coding and its Properties ® Various PAM Formats or Line Codes = Unipolar RZ and NRZ = Polar RZ and NRZ ® Bipolar NRZ § hase Manchester Format = Polar Quaternary NR2Z Format m= High Density Bipolar Signalling B&ZS Line Code = Power Spectra of Discrete PAM Signals (Various Line Codes) | Power Spectral Density (psd) of - NRZ Unipolar Format ~ NRZ Polar Format NRZ Bipolar Format Manchester Format —— aa itiplexing is the process of transmitting multiple signals over a single commu tal System TDM Hierarchy, In last two chapters, we have discussed the Pulse analog modulation and pulse digital modulation ‘Methods. In this chapter, we shall diseuss an important process in communication system known as multiplexing. In the articles tis multiplexing and what are the Jypes of multiplexing and then proceed for a detailed aspepts of multiplexing. MULTIPLEXING* (Mtultipiexing may be defined as a technique which allows communication channel simultaneously, Phere-are two major types of multiplexing techniques. They are as under: 4) Frequency division multiplexing (FD), «(97 Time division multiplexing (TDM). © 7.2.1. Frequency Division Multiplexing (FDM) () Definition i This technique permits a fixed frequenc: hand a user in the complete channel bandwidth. Such freqiensy slot is allotted continuously to that user. As an i fre consider that the channel bandwidth is 1 ae ith, Then be ten users, each requiring upto 100 KHz bandwittr st the complete channel bandwidth of 1 MHz ia aadeyiry into ten frequency bands, i.e., each of 100 band. This user can be allotted one independent reauency AA technique is known as Frequency Division (FDM). cation channel 312 8 Issues in Digital Transmission: Multiplexing a —— Ind Line Coding mi fo It is mainly used for modulated signal. This is due to» », POXOUKNOW? the fact that a modulated si r . and by just ‘changing thi The number of simultaneous conversations that can be transmitted using FDM depends on the total bandwidth available which in turn, varies with the medium. are independent tuned circuits and demodulators. 7.2.2.Time Division Multiplexing (TDM) i ition ; - en earlier, in PAM, PPM and PDM, the pulses present for a short duration and for most if the time between the two pulses, no signal is present.(This free space between the pulses can be ‘ re ed by pulses from other channels. Thisis known as Time Division Muliplesing TD ‘Thus, time division multiplexing (TDM) makes maximum utilization of the transmission channel ‘i ision with FDM wee ee rareyve ean oaj thatin FDM, alle signals are Gxruanitted simultaneously over the “Communication medium, and the signals occupy frequency slots. However, in TDM, the cies Soe multiplexed are transmitted sequentially one after the other. Each signal occupies Geary He aecantaramerm in sina are slated foeach oer in th total Jot in the frequency spectrum re, in'TDM, die i ee ene snmmeatgn shannel is availabe to each signal bing the complet transmitted), = ear iii) Conceptual Diagram — —— ‘ oe 7.1 explains the concept of TDM. Message 1|Message 2|Message 3] Message i,__—— One frame ————? s: Fig. 7.1. Illustration of TOM concept or ree oer acd nae that in context of TDM, we define one important term ic. t his stage, i i i jasmit all the signals once on the frame(One frame corresponds to the time period required toTra esigals eco i total pan ‘is has been shown in figure 7.1. Here, we have Seat Al tiene ones one a ed to multiplex analog or digital to be transmitt SS per fe four signals once on the channel: ay signalsyhowever it is more suitable for the A PAMITDM SYSTEM Elie Manic Comnuhatts i ibines the concepts of PAM Dae /M system. Infact, this system com! and TDM both as shown in figure 7.2. eee " ing Principle ing switch or commutator. b 4 fev multiplexer is a single pole Te aE rotations per second. As the switch 7 "and it ro! ‘4 mechanical switch or an electronic SLE its vesition 1 2, 30r NTor as ort time. There are (UP. Tech., Sem. Exam, 2005-08) it is goi Ne cts. Hence, the switch arm rau Core ae axe connected to these conf N analog signals, to be mu ‘ ASASSAAA earn if | qo - M Principles a of Comm ™ will connect these N oo of a TDM signal which ig , ———— : : is ) ‘otatory switch samples ae t j © one frame, therefore, one inication channel, The waveform message during each of its n figure 7.3. It shows that the Stes rotations. Since, each rotation corresponds in T, seconds wh: ae ere T= If, toknow that this switch must tthe same speed as that of the commutator at the tra Reconstruction and its position must be synchronized with commutator inorder to ensure proper Seat aig iieness The Tow pass filters (LPF) on the receiver side are used forthe reconstruction of eri inputs } message signals. —— Higinal Message. agi i ination i ct) egaie 7.3.1.Signalling Rate and its Determination ina PAM/TDM System eyes? (i) Definition ‘As a matter of fact,{the signalling rate of a TDM system is defined as the number of pulses transmitted per secort@)[t is represented by r — (ji) Derivation of Expression [Commutatog. | ; Let us now derive an expression for the signalling rate of the PAM/TDM system in the form Tees <= [Pecan of following few points: Timing pulses G) Let, = maximum frequency of all the input signals x, to xy. (ii) Therefdve, as per Nyquist criterion, the sampling frequency f,2 2f,,. Hence, the speed of igh 7.2. Block diagram of a PAM/TDM system, rotation of commutators is f, rotations per second with f,>? f,, : Gi) AAs shown in figure 7.4, one revolution of commutators corresponding to one frame contains Hence, the functi : ecu action of be ommutator is two fold as under: sample from each input signal. sam) i Gi) T FLOW. SARE! of each input message at a rate f, which is higher than 2f,,. Hence, 1 revolution = 1 frame=N pulses ATA ‘o sequentially inter] cj nee i i leave the N samples inside the interval T.= Uf. er Now, W, i multiplexed signal* at the output of the commutator is applied to.a pulse amplitude modulatgr(]g converts the PAM pul: i Ises into a form suitable fc i iF a le for transmission over the com: fh Flusnnel ihe innut ae signals are passed through low pass filters before rian arte | r. These filters gre actually the anti aliasing filters which avoi iasi cut off frequency of each low pass filter (LPF) is f, Hz, shi avid ths assis jay ; x T - 1 = One revolution of commutator _| Le Nopulses in (1,) =, seconds : - Fig. 7.4. Evaluation af number of pulses per second for PANITDM system. J+ one revolution ne (iv) One frame period is (If) eT, ceconds. Therefore, in T, seconds, N number of pulses te a t bare transmitted. Hence, the pulse to pulse spacing within the frame is given by, : ae 4 1 aaa Pulse to pulse spacing = 4 = Nf (7.2) period of one pulse (ON + OF) is (UNE) seconds, the number of pulses per second (v) As th % % Multiplexed x ML % a Astl PAM wave . y - at “in givenby,— : [Si | i a 1] | Number of pulses per second = Nf, ate Ree ‘Therefore, signalling rate of a TDM system = += Ni, pulsesisecond. rene Saleen But, as f,2 2fqy therefore, : gnalling rate of a TDM system Fig, 7.3 (a) Sampling of the first input (b) Multiplexed PAM Signalling rate of a ys Signal transmitted on the transmission channel, vn ithportage Potat ey sated Soft Timilipioning Is sronowte sce mulligle communications cn lake paee as high ax posible. Te io evident rom fhe for the cost of a single link plus the multiplexing Syuitinant tiie ues aioe F oe reused by inereasing the sampling Fle fy ida di a Sk => 2 Nf, pulses per second. hat TDM system is supposed to have its signalling rate the expression above that the signalling aie can be ‘and or the number of input signals N- Ethel of Communication : 1 {ss.0s in Digital Transmission: Multiplexing and Line Coding w nt Features of Toys Utilized for each channel. Gi). Lae srannel bandwidth o nb ation distorti 2 istortion is absent. EXAMPLE 7.3. Twenty-four voice signals are sampled uniformly and then time division multiplexed, Gi) TDM circu Civ) The eeUlttY is not very complex Problem of crosstalk is not sex ve i ere ‘The sampling operation uses Mat top samples with 1 ps duration. The multiplexing operat Inoludes provision for synchronization by adding an extra pulse of appropriate amplitude and 1 a. 3.7. Drawbacks of tom ts duration. The highest frequency component of each voice signal is 5.4 ktiz. (i) Assuming a sampling rate of 8 kilz, calculate the spacing between successive pulses of (v) The channel bandwidth will be 120002 = 6000 Hz Ans, is essential for Proper operati operation, the multiplexed signal. (i) Repeat (i) assuming the use of Nyquist rate sampling. (GATE Examination 1997) Gi) Due to si % low narrowb: EXAMPLE 7.1. To analog si and fading, all the TDM channels may get wiped out. means of tine oe analog signals x,(¢) and y out. : S(O 1s 4.5 LH aeision multiplexing (TDN) age t8 be transmitted aver a evinmon channel by Solution) Chen iy Ssh . What will be the minimum valve cyenest freawency of x(t) is 4 kHz and that of Sampling rate = 8 kH2-= 8000 samplesisec ion: The highest frequency component e of permissible sampling rate? ‘There are 24 voice signals + 1 synchronizing pulse. Paleo width of each voice channel and synchronizing pulse is Ls. 1 Therefore, the mi tof th i + the minimum value of permi 1¢ composite signal consisting of permissible sampling rate will be, g of x(t) and x,(t) is 4.5 kElz, Now, time taken by the commutator for 1 rotation = agg = 125 H ses = Eoin) = 24.5 kH2=9kH2 Any, AMPLE 7.2. A si = fh a Jt Signal x(t) is bandlimi 29441225 x,(t) which mat! andlimited to $ kHz. Number of pulses produced in 1 rotation = 24 +1 a ch are bandlimited to 1 kHz each. These siznale ecto he onsen a and e : 4 are to be transmitted by a TDM system, spherefore, the lading odgus of the-palses are at “g—- =5 ft seeands distances ax hown in gute 7.9 G) Design a TDM seh. a eme where each signal i i i AN ee must he the speed of the couniitater? ot es aie ae alculate the minimum transmission bandwidth of the cham f ual lution: . () Table 7.1 shows different message signals with corresponding Nyquist rates. AL ! ae TABLE 7.1 pea - @ Synchronization j 4 rotation Fig.7.9 Hence, spacing between successive pulsea=5— 1 4s “nt SKE: kHz sy Nyquist rate of sampling = 2¥ 3.4 kHa = 68 kHz : oon ves axe produced per second. One rotation of commutator takes at) “Lk 2kHe t : me He ae ‘This means that 6800 samo) < a ane +6800 = 147 ps time 1147 see corresponds to 25 pulses to 5.88 psec. the spacing b djacent pulses will be 5.88. set- Aivision multiplexed and ‘pandwidth of the PAM! [S.No. | Message signal | Bandwidth =] Nyquist rate Therefore, ‘Therefore, 1 pulse corresponds ae ‘As the pulse width of each pulse is 1} s2¢, ‘we assume t= 0 then the spacing between the a If the sampling commutator rotates at the rate of 2000 rotations per second then the signals x,(0), x,(0 d See etween adjacent pulses willbe 4.88 us and x0 will be sampled at their Nyquist rate, Bat, we have to sample x,(t) also at its Nyquist rate which ~———J is three times higher than that of the other three. In_,(t ges Le ce signals each of b EXAMPLE 74, Six messge signal sa ein ta achieve this, we should sample x,(t) three = xf aa ssnH = |channel Senin \e 81 yn: ‘The number of channels N= 6 andwidth 5 kHz are time imum channel order times in one rotation of the commutator. Therefore, the commutator must have atleast 6 poles connected to the signal i . 2 0 the signals as shown in figure 7.8. re (ii) The speed of rotation of the commutator is Bandwidth of each channel, fq= 5 kHe 2000 rotations/sec, Therefore, minimum sampling te =25 \Hiz = 10 kHe (iii) Number of samples produced per second is Signaling rate = Number of bits per second calculated as under: = 6x10 kHz = 60 K bitslsec. Ans- x,(t) produces 3 x 2000 = 6000 samples/sec. Minimum channel bandvicth to avoid cross talk in PAM/TDM is, inimut = =80kHz Ans. BW NE, = 6% 5 kia 80 KH xy(t), Xq(t) and x, (t) produce 2000 samples/see. each, ‘Therefore, number of samples per second = 6000 + (3 x-2000) = 12000 samples/sec. Signaling rate = 12000 samples/sec. : TIERS at 5 Temas he aes requency com oe aes Benste vlog signal This is called the sampling theorem. saa “FV Boih digital and analog Tenals ay be tansmitied by TDM ae ie El IN Principles of Communication i EXAMPLE 7.5. SI following PAM pulse repeti required.* leaving scheme for the time tiplexing of the lephone channels and one 20 jc channel. Find the xed signal and estimate the minimum system bandwidth (i) The 20 kHz music cl ‘red (Gif) The sampled signals are apy (iv) These signals are finally m 5 Telephone channels het itt TOM iuk commutator 40K samplesisec. 4htiz ‘4bit ‘Sampler AD converter 180 K samples/sec 160 K samples/see Fig. 7.40. PAM/TDM system for Example 7.5. (+) The TDM commutator output has a pulse repetition rate of 49 k samples/see as there are 5 channel ing rate is § kHz v x, the output of the separate sampler has a pulse repetition rate of 40 K samples cee. The outputs of A-D converters hes pulse repetition rates SE ii f 40 x 4 = 160 K samplesisec. erefore, pulse repetition rate at the output ofa multiplexer sre “are 160 + 160 = 320 K samplesfsec. different standard: i for TDM, One Hence, pulse repetition rate of the system = 320 kHz ised art i ‘Therefore, bandwidth required = bit rate = 820 kHz. Ans, ~4, INTRODUCTION TO DIGITAL MULTIPLEXERS (@) Basic Idea I : : i j inant @ we have discussed the idea of TDM, In this section, et us eansider the multiplexing ce ea at different bit rates. The digital multiplexing will enable us to combine many NY : signals such as computer outputs, digital voiee, digitized facsimile and TV signals. (ii) Block Diagram ; Boe 7a illustrates the concept of digital multiplexing and demultiplexing. e digital data can be multiplexed by using a bit-by-bit interleaving procedure, This can be achieved by ising a selector switch which sequentially selects a bit Input and places it # Pulse-amplitudé modulation signals may be multiplexed | ‘llowin, s¢-aniplittide modulation signals may be mull Fy allowing samaples ol several signals to be interleaved into adjacent time slots. BEAD 2 49 2 wal high speed tran are separated out and di Principle multiplexing is based on the principle of ‘ymbols from two oF more digital signals. This and waveform preservation. The signals which are to be multiph ital data sources or rom analog Sources that have been digitally coded. Gh digital multiplexer is used to merge the input bits from different sources to form one signal from transmission ¥ ‘The multiplexed signal consists Of source digits 2. Important Functions ‘The important functions that must be performed by 2 multiplexer are asunder: ‘To establish a frame, A frame i i As the number of multiplexed ‘yoice signals inereases, 30 does the to have ‘a wider. frequency “ esponse and variations of time- _ delay with frequency tobe held to alow level... ii) "To insert ¢ (iy) To make allow: 7.5. CLASSIFICATION OF DIGITAL MULTIPLEXERS. ‘The various digital sources that are to be multiplexed will have different bit rates. In practice, the bit rate variation!possess the most serious design problem and leads to the three categories of multiplexers, as under: (@) Synchronous multiplexers ‘Asynchronous multiplexers Quasi-synchronous multiplexers. 3) Synchronous Multiplexers: a} 2 the synchronous digital ‘multiplexers, a master clock governs all the sources. Therefore, all the sources will operate at the same pit rate. As the bitrate variations arecomplets cae the synchronous multiplexing systems attain a very high throughput efficiency. But, they nee elaborate provisions to distribute the master ‘lock signal, to all the sourees. ee A ae ——- Fist TORT sates in we fade use me poiiaan RODE vodddddd PUTT TTT TTT Principles of Communication im BW Issues in Digital +; ete. are the names of the servi t es use the Ti tay implement i canes Ee as services. The T lines have Gi , vais My ieee pi ‘the bitrates of the correspondit ais le 7.3. Thus, T-1 ineimplements DS-iservice -1 service, ‘T-2 implements D; 8-2 servi at vane aries ervice and so on. DSO is defined as Spo vouKnoweaeg The DS-1 signal and TI carrie! represent the lowest level in hierarchy of TDM signals wit higher bit rates, All of thes signals contain PCM audi signals, each sampled 8000 time: per second. : i TABLE 7.3. Relation between DS and T lines Rate (Mbps) 1.544 6.312 44.736 274.176 % Important Point: 7 lines are digital lines w aucio or vide Sea SNe : si a But, the T lines can also be used or analo; xample, T1 line can be use 1g communication. For example, be used 194 PCM-TDM SYSTEM: T1 CARRIER SYSTEM (U.P. Tech, Sem. Exam, 2006-07, 2007-08) 1. Block Diagram When a large numberof PCM signals are tobe transmitted overa commot channel, multiplexing of these PCM signals is required. Figure 7.17 shows the basic time division multiplexing scheme, called as the T1-digital system or T1 earrier system. This system is used to eanvey multiple signals over telephone lines using wideband coaxial cable. 2. Working Operation of the T1 Carrier System* The working operation of the PCM-TDM system shown in figure 7.17 form of few points as under: ( () This system has been designed to accommodate 24 voice channels marked S, to S,,. Each signal is bandlimited to 3.3 kHz, and the sampling is done at a standard rate of 8 kHz. This is higher than the Nyquist'rate. The sampling is done by the ‘commutator switch SW). ji) These voice signals are selected one by commutator switch SW,. z iii) Fach sampled signal is then applied to the PCM transmitter which converis it into a digital signal ess of A to D conversion and companding, as explained earlier. (iv) The resulting digital waveform is transmitted over a can be explained in the co-axial cable. ~ ans if cominon POM sy. one and connected to a PCM transmitter by the Samples £2000 per second Voice signals Sas (s) Periodically, after every : D Repeaters. ‘They eliminate the distortion introduced. suberimpost that the received sit (vi) At the destination, the signal is compan nd demultiplexed, using a PCM receiver. The PCM receiver output is connected to different low pass filters via the decommutator switch SW. (vii) Synchronization between the transmitter and receiver commutators SW, and SW, is essential in order to ensure proper communication. 3. Important Terms Related to a T1 Carrier System* Now, let us discuss few important terms related toa TI carrier system as under: (i) Bits/Frame ‘The commutators sweep continuously fromS, toS,, andback to, at therate of 3000 revolutions per second. This will generate 8000 samples persecontt of each signal (G, to S.,). Bach sample is iferencoded (converted) into an eight bit digital word, ; "Thus, the digital signal generated during one complete sweep (revolution) of the commutator is given by 1 Frame = 1 revolution = 24 channels = 24 * 8 bits = 192 figure 7.18. Bach voice signal from S, to Sa 38 frame corresponds to the time ‘corresponding transmission of esc «ds to one-revelution of the commutator, “ise Tapecal TST iret called a cde Thandle conversion and the ADC, DAC, serial —to-parallel conversion circuits, companders, One frame of T1 carrier system is shown in encoded into eight bits. One signal once. Hence 1-frame correspon te i Principles of Communication m transmission over the channel, This Process is Known as line coding or ing: There are several igning waveforms (ie, Pulses) to the digital data, In the binary case (two symbols), for example, conceptually the simplest line code is on. off where a 1 is transmitted by a pulee P() and a 0 is transmitted by no pulse 0 is transmitted by a pulse p(t) as shown in figure 7.200). The polar scheme is the most Power efficient code, since for a given ) On-off (RZ). (b) Polar (RZ), (e) off (NRZ), (e) Polar (NZ). a pulse p(t) or the previous 1 is encode + Pulses representing consecutive 1's alternate in sign, as shown in figure 7.20(c). This code has the advantage that if an error is made in the detection of pulses, the received pulse sequence will violate the bipolar rule and the error is immediately detected (although net corrected), Another line code that in the past appeared Promising is the duobinary (and modified duobinary) proposed by Lender. Although this code is better than the bipolar in terms of bandwidth efficiency, it has lost ite appeal due to some Practical problems and will not be discussed here. In our discussion so far, we have used half-width pulses just for the sake of j lustration. We can select oth full-width pulses are often used in some applications. Whenever full-width pulses are used, the pulse amplitude is held to a constant value throughout the pulse interval (it does not have a chance to 80 to zero before the pext pulse begins). For this reason these schemes are known as non-re In contrast to return-to-zero (RZ) schemes as shown in figure 7.20(a), (b), (c). Figure 7.20(d) shows an on-off NRZ signal, whereas figure 7.20(e) illustrates a polar NRZ signal, 4. Regenerative Repeater Regenerative repeaters are used at regularly spaced intervals along a digital Ap transmission line to detect the incoming iT, wvgg + digital signal and regenerate new clean pulses for further transmission along Ae: the line. This process periodically JnUaun! O eliminates, and ‘thereby combats, the Ae: accumulation of noise and signal ial —! DU O U0 ake oe he eeninion ath “f OOoOoonon noon If the pulses are trai zy of R, pulses per second, we require the Fig. 7.21. An on-off signal is the sum of a polar signal and a periodic timing information—the clock Glock frequency periodic signal, 1g and Line Coding i ‘i i tion can be _ This timing informal ulses at a repeater, properly. The polar signal in {the Hine code is chose Pra of click frequency Ry He ‘ re ncy R, Ha. When this signal is applied si 8 Oe id of reueney ty Hs, ing, the output, ‘d.as the sum of a periodic signal (of clock be errr Because of the presence ofthe pes " i sonant circt bur octal Hee component, pecan ra when ir an on-fl sigal : : ths the same way as that for 8 ‘ing bit pattern. In to the clock frequen ‘Thus, 4 DIP the same W {ut Sssenitveotheinsoming erat Hing information can syura! nt cuit OUP) sere if there are ine he reer namitted by "no ules. Hens cenit and the sin 4 eee seer g information. We ae a Tene code in which the Ditpatiorn dvs be cent line code. The pol id oe parent, whereas, the on-off and the on-off or bipolar ac 4 sequence {no js no tarts scheme (Where each iy jg transmitte Dipolar ahenes ae tamer M SIGNALS (DIGITAL DATA FORMATS)" RETE PA sat the other 7.11. INTRODUCTION to DISC! ignal to digital signal. But, ‘ d to convert the analog euters. The information frott ae ‘There are various techniques rom the source such nae mitted aver abandlimite discrete si Legace aan arte ef oath ee eee ie iapersed. This causes the pulse a ore chin, we should Subanon Sf a mee ea Such a distortion is called anes on the transmission ned core aici utaaaias rant a PAM cue melee namie over a ‘waveform for baseband represe a PAM suitable form: fe eal ion. This means that digit are alsocalledasline codes. These ofan appropriate ‘Stan ons nh ata i its transmission from oA ie itt cs ee ‘used for transmitting the outpu ane Se Lt : see Reels modulate some carrier to gel signal symbol resented a8, train, Therefore, such signal may be rep “ «(7.18) xy =D 4 Meek) ee : ; ste f codulated by a, i tude of symbol in the message sence) Heep Se ca: signal, ic. is abasic pulse nae te pul ae — HS wation (0, Seer eae oe he ae greed Fie wp istheretangular pulse anditcan take variable duty eve. i ae pt)=1 for t=0 ue 9 0 for t=4T, 42 2 toms eb sm iginal digital sign te checking is the det form. ‘To recover the orginal digital signal from x0 ne x(t) is oe ae sia check the aan Cee arin re, then x(t) at some fixe . From equation (7.15), we obser riaiored is tine A; the transmitted ade vo sample 30) when pO) i rT Thee aero TSI timation is preeentitranemitied tae Pl : {p(t) = 0) no dig ally at t= KC where k=0, #1 el ” Dane, the rectangular pulse and may be Also, p way of obtaining digital data is > source is inherently discrete in nal gnal 
