278 SAMPLING AND ANALOGTOLDIGITAL CONVERSION The PCM Encoder ‘The multiplexed PAM output supplied ceach sample into 0 group of n binary d discussherethe digitaat-ntine encosir, which ma coco be We sal vor genet fon af reference inptthe i} omer nee (age vain coi resistors K, 2, 2R. 27, hether the sam- Aepenting en wheter the smple sn the upper a he lower lf of he nae: TO step, another digit 1 or Os generated, depending on whether the saraple i i the MOET lower half ofthe subinterval in whieh it has been located. This process conti inary digit in the code has been generated. . iiss mf conn ete averae fener Tn i ase cachofthe gts ap a ecoding is the inverse of encealing.. vo ie cates iit is 3 a resistor 2°, The currents in a of different value. The kth digit is applied to Br aery ood ae aulded. The sunv is proportional tothe quantize sample ‘word 10010110 will give a current proportional t02? + 0+ 0 ‘This completes the DIA conversion, $2 40427 +2140 = 150, 6.2.4 Transmission Bandwidth and the Output SNR Forabinary PCM, weassigna distinctgroup of binary digits (bts) 1oeach ofthe F.quantization levels. Because a sequence of binary digits ean be arranged in 2" distinet pattems, L logyt cach quantized sample is, thus, encoded into bits. Because a signal m(t) band-limite Hz requites a minimum of 28 samples per second, we require a total of 2nB bits, that is, 2nB pieces of information per second, Becavsea unit bandwidth (1 Hz) can transmit maximum of two pieces of information per second (Sec. 6.1.3), we requirea minimum channel of bandwidth &; Hz, given by By =nB Hz (6.38) ‘This isthe theoretical minimura transmission bandwidth required to transmit the PCM signal, In Secs. 7.2and 7.3, we shall see that for practical reasons We may use a transmission bandwidth higher than this minimum. Example 6.2 A signal m(t) band-limited to 3kHz is sampled at a rate 334% higher than the Nyquist rate, ‘The maximum acceptable error in the sample amplitade (the maximum quantiaation error) is 0.5% of the peak amplitude ip, The quantized samples are binary coded. Find the minimum, bandwidth of a channel required to transinit the encoded binary signal, If 24 such signals are time division- multiplexed, determine the minimum transmission bandwidth required to transmit the multiplexed signal The Nyquist sampling rate is Ry = 2 x 3000 = 6000 He (samples per second). ‘The ‘actual sampling rate is Ry = 6000 x (14) = 8000 Hz. ‘The quantization step is Av, and the maximum quantization error is +kAw/2,“Therefore, fom £4. (630+ that is a oi 2 Hence, the next higher value of oe ty zee Em bee fe. We require to transmit ammereverNT Yet mer ol 2 = Bbitspersampl quire to canst crunieliic wong raismission hanavideh Br = a} ttn weeps min tannin Fane = se is i requires minim of 1536/2 — 0.768 Mle of transmission band Exponential Increase of the Output SNR sanbecxpressed a Fon Eq, (6.37), L? = 2, and the output SNR in Eq. (6.34) or Eg. (6.36) in be exp! ee cay ak, %~ (639) Ms an wh where Am juncompressed ease, in Eg (6.34)) = ™p a essed case, in Eq. (6.36) int'+ap rm % Substitution of Eq, (6.38) into Eq. (6.39) yields $2 = c(aytti8 (6.40) No From Eq. (6.40) we observe that the SNR increases exponentially with the transmission bande width 87. Tis rade of SNR for bendwidth is attractive and comes close tothe upper theoresieal frat A small increase n bandwidth yiels lrge bene in terms of SNR. This relationship is clearly seen by using the decibel scale to rewrite Eq. (6.39) as = 10logiole(2)?"] 1Ologige + 2nlogig2 =(@+6n) dB (641) where = 10 logigc. This shows that increasing n by 1 increasing one bit in the ‘quadruples the output SNR (a 6 dB incr ara vad uples, but the transmission bandwith increases only fom 32 kHiecs 36 kHz (ani of only 12.59), This shows that in PC) rennin eae ‘We shall see later that frequency and ph the bandwidth to quadruple the SNR. In this respect, PCM isoING AND en ANALOGTODIGITAL CONVERSION Evo ple 6.3 A ignal y of bandwidth & = 4 kEtz is teansmiued using a binary companded PCM with, ) eS Compare the case Of L = 64 with the case of L = 256 from the point of view of se) m bandwidth and the output SNR. For = 64, m= 6, and the transmission bandwidth is a = 24 KH, Le Se Fe = @ +39) uB 3 3 _ oe Ole ieaone = ~*5! - Hence, Se pn 27490B For L = 256, n= 8 and the transmission bandwidth is 32 kHz, Braton= 394068 The difference between the two SNRs is 12 dB, which is a ratio of 16. Thus, the SNR. for L = 256 is 16 times the SNR for L = 64, The former requices just about 33% more bandwidth compared to the latter. eee Comments on Logarithmic Units Logarithmic units and logarithmic scales are very convenient when a variable has a large dynamic range. Such is the case with frequency variables or SNRs. A logarithmic unit for the Power ratiois the decibel (dB), defined as 10 logio (power ratio). Thus, an SNR is xdB, where = 10 ogi9 5 ‘We use the same unit to express power gain or loss over a certain transmission medium. For instance, ifovera certain cable the signal powers attenuated by a factor of 15, the cable painis G = 10 logo Fy = 11.76 08 (or the cable attenuation (loss) is 11,76 dB. /Afouah the decibels a measure of power ratios, itis often ued a8 a measure of power REI For instance, “100 wat” may be considered to be a power retio of 100 with respect to |-watt power, and is expressed in units of dBW as a Pasw = 10 logig 100 20 aBwpom in TH Cortes Systems 281 expect 10 1 mN¥ passer Fs tal Telephony: measured with 1 “Thus, 100-wat poner is 20 AW. Similarly, power MEAS Bm, For instance, 100-wait power is 100. OW _ soabm Pann = 10108 Taw 5 6.3 DIGITAL TELEPHONY: PCM IN T1 CARRIER SYSTEMS A Historical Note : ' Because af the unavailability of suitable switching devices, morethan 5 transistor, were not only bulky, but they were poor switches and dissipateda lot of heat Systems having vacuum tubes as switches were large, rather unreliable, and tended to ‘overheat. PCM. was just waiting for the invention of the transistor, which happens to be a small device that consumes tittle power and is nearly ideal switch, Coincidentally, at about the time the transistor was invented, the demand for telephone service had become so heavy that the existing system was overloaded, particularly in large ics. It was not easy to install new underground cables because space available under the reets in many cities was alrendy occupied by other services (water, gas, sewer, ete,), Moreover, sing up streets and causing many dislocations was not very attractive. An attempt was made ‘on a limited scale to increase the capacity by frequency-division-multiplexing several voice channels through amplitude modulation. Unfortunately, the cables were primarily designed for the audio voice range (0-4 KH?) und suffered severely from noise. Furthermore, cross talk between pairs of channels on the same cable Was unacceptable at high frequencies. Ironically, PCM—requiring a bandwidth several times larger than thatrequired for FDM signals—offered the solution. This is because digital systems with closely spaced regenerative repeaters can work satisfactorily on noisy lines that give poor high-frequency performance.” The repeaters, spaced approximately 6000 feet apart, clean up the signal and regenerate new pulses before the pulses get too distorted and noisy. This is the history of the Bell System’s TI carrier system.> 10 Apair of wires that used to transmit one audio signal of bandwidth 4 kHz is now used to transmit 24 time-division-multiplexed PCM telephone signals with a total bandwidth of 1.544 MHz. 20 yearselapsedbetween TI Time Division Multiplexing A schematic of a Tl carrier system is shown in Fig. 6.20a. All 24 channels are sampled in a sequence. The sampler output represents a time-division-multiplexed PAM signal. The multiplexed PAM signal is now applied to the input of an encoder that quantizes each sample and encodes it into eight binary pulses—a binary codeword" (see Fig, 6.20b). The signal, now converted to digital form, is sent over the transmission medium, Regenerative repeaters spaced approximately 6000 feet apart detect the pulses and retransmit new pulses, At the receiver, the decoder converts the binary pulses into samples (decoding). The samples are then demultiplexed (ie., distributed to each of the 24 channels). The desired audio signal is reconstryeted by passing-the samples through a low-pass filter in each channel “Inaneatier version, each sample was encoded by seven bits. An additional bit was added fo signaling, 2 eeyar a AND ANALOG-TODIGITAL CONVERSION Ms Channel os | OF Digital prmany °, processor Digitat processor @ ch. 2 SS. Coder output o ‘The commutators in Fig. 620 are not mechanical but are high-speed electronic switching circuits, Several schemes are available for this purpose.!' Sampling is done by electronic gates (such as a bridge diode circuit, as shown in Fig. 4.5a) opened periodically by narrow pulses of 2 yas duration. The 1.544 Mbit/s signal of the TI system, called digital signal level { (DS1), is used further to multiplex into progressively higher level signals DS2, DS3, and DS4, as, described next, in See. 6.4 ‘After the Bell System introduced the TI carrier system in the United States, dozens of variations were proposed or adopted elsewhere before the ITU-T standardized its 30-channel PCM system with a ate of 2.048 Mbités (in contrast to TL, with 24 channels and 1.544 Mbités). ‘The 30-channel system is used all over the World, exceptin North America and Japan. Because ‘of the widespread adoption of the TI carrier system in the United States and Japan before the ITU-T standardization, the two standards continue to be used in different parts of the world, ‘with appropriate interfaces in intemational connections.Figure 6.21 Tesyster ‘ Signaling format. Frame nos. 456171811 OB 1A Me Scnmmation Information Tnformation i a its ening Framing, ign cain 1 Contin Systems 263 6.3 Digi! Telephony: F bit bit Frame no, Framing bit Al frames except ——— eo 1,7 13,19, Taforclation Information Information bits bits bits Synchronizing and Signaling Binary codewords corresponding to samples of each of the 24 channels are multiplexed in a sequence, 35 shown in Fig. 6.21. A segment containing one codeword (corresponding to {ne sample) from each of the 24 channels is called a frame. Each frame has 24 x 8 = 192 Information bits, Because the sampling rate is $000 sarmples per second, each frame takes 125 jis. To separate information bits correctly at the receiver, itis necessary to be sure where ‘each frame begins. Therefore, a framing bit is added at the beginning of euch frame, This ‘makes a total of 193 bits per frame. Framing bits are chosen so that a sequence of framing bits, ‘one ut the beginning of each frame, forms « special pattern that is unlikely to be formed in a speech signal. “The sequence formed by the first bit from each frame is examined by the logic of the receiving terminal. If this sequence does not follow the given code pattern (framing bit pattern), ‘a synchronization loss is detected, and the next position is examined to determine whether it is actually the framing bit, It takes about 0.4 t0 6 ms to detect and about $0 ms (in the worst possible case) to reframe. In addition to information and framing bits, we need to transmit signaling bits corre- sponding to dialing pulses, as well as telephone on-hoolvoff-hook signals, When channels developed by this system are used to transmit signals between telephone switching systems, the switches must be able to communicate with each other to use the channels effectively. Since all eight bits are now used for transmission instead of the seven bits used in the earlier version.” the signaling channel provided by the cighth bit is no longer available. Since only a rather low-speed signaling channel is required, rather than create extra time slots for this infor- mation, we use one information bit (the least significant bit) of every sixth sample of a signal * Inthe earlier version of T1, quantizing levels L = 128 required. information bit i Seskeew ing required only seven information bits.“The eighth bit was