5.1 Sampling Theorem 267 | cea aentbass spectrum. GCA) is obtained using convolution of a train of er with lower and higher cut-off frequencies 96 MHz. and 98 MHz will produce the fs, the shaded part of the spectrum.ble. This can be ea the ideal reconsin: yer), we need to impler to generate. For exam “must first use the noni denote the new signal5.2 Pulse Code Modulation 283 rr ges can be extremely reliable without deterioration. Analog messages: films, for example, lose quality at each successive stage of reproduction and eally from one distant place to another, often at relatively high cost. continues to halve every two or three time period. And there is no end in si digital technology. As a result, digital 1 of storage technologies. years, while performance or capacity ight yet to this breathtaking and relentless technologies today dominate in any given ‘applied what tums out to be advanced mathematical concepts for describing ted the first known description of a binary numeral system, possibly as early Others, like R. Hall in Mathematics of Poetry place him later, circa 200 BCE. 1646-1716) was the first mathematician in the West to work out systematically Is and Os) for any number. He felt a spiritual significance in this discovery, unity, was clearly a symbol for God, white 0 represented nothingness. He ‘can be represented merely by the use of 1 and 0, this surely proves that God of nothing! a ‘signals come from a variety of sources. Some sources suc as computers are are analog, but are converted into digital form by a variety of techniques h (DM), which will now be analyzed. The rest of this section provides CM and its various aspects, such as quantizing, encoding, synchronizing, the dwidth, and SNR. ve limit the amplitude of the message signal m(?) to the range (—mp, my), as shown not necessarily the peak amplitude of m(#). The amplitudes of m(r) beyond +m, s,m, is not a parameter of the signal m/(1); rather, it is the limit of the quantizer. ‘mp) is divided into L uniformly spaced intervals, each of width Av = 2imp/L. ed by the midpoint of the interval in which it lies (Fig. 5.14). The quantized mitted as binary pulses. At the receiver some pulses may be detected incorrectly. ces of error in this scheme: quantization error and pulse detection error. Im almost : pulse detection error is quite small compared to the quantization error and can be therefore, we shall assume that the error in the received signal is caused mple of the signal m(:), and if (kT) is the corresponding quantized sample, then la in Eq. (5.10), _ m=) m(KT,) sine 2x Bt— kr) a _ inl) = J) WKT;) sine 2x Br ~ kre) k ted from quantized samples. The distortion component q(’) in the = tin(t) — m(1). Thus,5.2 Pulse Code Modulation 285 ‘equally likely to lie anywhere in the range (—Av/2, Av/2), the mean square i ib Av Jap 2 = on (5.32) mB = ; 6.33) square value or power of the quantization noise, we shall denote it by Nq, ee ia uae Te, Ng= P= 375 detection error at the receiver is negligible, the reconstructed signal 7(t) at the BO = =m), tae So/Nov the SNR, is a linear function of the message signal power m?(j) (see Fig. ( +f Progressive Taxation: Nonuniform Quantization is anindienton of he ialty ofthe recived signal. Ideally we would like to same quality) for all ‘values of the message signal power mr mP(). Unfortunately, the e theory of probability can an derive this result directly by noting that the probability density ofthe Co ao: ‘Hence,5.2 Pulse Code Modulation 287 the output signal y. The compressor maps input signal increments Am into larger “emall input signals, and vice versa for large input signals. Hence, a given interval Am “number of steps (or smaller step size) when m is small, The quantization noise is lower power. An approximately logarithmic compression characteristic yields a quanti- ‘proportional to the signal power m?(t), thus making the SNR practically independent il power over a large dynamic range® (see later Fig. 5.18). This ‘approach of equaliz- similar to the use of progressive income tax to equalize incomes. The loud talkers ‘are penalized with higher noise steps Av to compensate the soft talkers and weaker choices, two compression laws have been accepted as desirable standards by the 1Tu-T®: ‘America and Japan, and the A-law used in Europe and the rest of the world and on and the A-law curves have odd symmetry about the vertical axis. The (6.35a) amplitudes) is ania T+InA my y (5.35b) 1 Am\ 12m rat (1+) Ae b j are shown in Fig. 5.16. ‘parameter j1 (or A) determines the degree of compression. To obtain a nearly constant s range of input signal power 40 dB, 1 should be greater than 100. Barly North American other digital terminals used a value of s1 = 100, which yielded the best results for 7B ‘An optimum value of =255 has been used for all North American 8-bit (256-level) pt Be Figure 5.16.0] Low charac (b]Alaw choosernd-limited BH te 3 2nB pieces of Pieces of informa infor tion Persea 2a higher than this mining a er af5.2 Pulse Code Modulation 291 24 x 64,000 = 1.536 Mbit/s, which requires a minimum of of transmission bandwidth. of the Output SNR , and the output SNR in Eq. (5.34) or Eq. (5.36) can be expressed as (6.39) 3 mn fu J eae incompressed case, in Eq. (5.34)) eau [compressed case, in Eq. (5.36)] nto Eq. (5.39) yields So _ (gy eri (ac wid PHASE da Sale che ce upp aca Hee WB ene benefit in terms of SNR. This relationship is clearly seen by using the decibel sca no So Fla” Pv8() = 1010810 [e)™* = (a + 6n) dB 6 = 10logig¢ + 2nlogyo2 9¢. This shows that increasing m by | (increasing one bitin the codeword) quadruples crease). Thus, if we increase n from 8 to 9, the SNR quadruples, but the transmis. 32 kHz to 36 kHz (an increase of only 12.5%), This shows that in PCM, $ ission bandwidth. We shall see later that frequency and phase modulation a doubling of the bandwidth to quadruple the SNR. In this respect, PCM is strikiHANA lsitihii.,) lon WONTWOIY 4, Mi NY OF 4, He friniiy, OH HOMINCEN NI, 4), 108, AIO show sn5.4 Digital Multiplexing 299 8928 js bs bs b bs a 54 ——_—_—_—> 4 in four successive frames and repeat thereafter, bits in the frame for S, is not unique. $a typical format, that of the DM1/2 multiplexer, We have here bit-by-bit interleaving ata rate of 1.544 Mbit/s. The main frame (multiframe) consists of four subframes. x overhead bits: for example the subframe 1 (first line in Fig. 5.23) has overhead bits and Fj. In between these overhead bits are 48 interleaved data bits from the four ts from each channel). We begin with overhead bit Mo, followed by 48 multiplexed data ‘overhead bit Cy followed by the next 48 multiplexed bits, and so on. Thus, there are = 1152 data bits and 6 x 4 = 24 overhead bits making a total 1176 bits/frame. The ith subscript 0 are always 0 and those with subscript us, Mo, Fo are all Os and M, and F) are all 1s. The F digits are periodic 010101 .... and Pattern, which the multiplexer uses to synchronize on the frame. After locking demultiplexer searches for the O111 pattern formed by overhead bits MoMiM;M). s the four subframes, each corresponding to a line in Fig. 5.23. It is possible, although ts will also have a pattern 101010. .. The receiver could lock onto this wrong sequence MiMi) provides verification of the genuine FoF)FoF sequence. The C bits are used jon about bit stuffing, as discussed later, cases, not all incoming channels are active all the time: some transmit data, and the system is underutilized. We can, therefore, accept more input channels to take ctivity, at any given time, of at least one channel. This obviously involves much more ‘operations, and also rather careful system planning. In any random traffic situation hat the number of transmission channels demanded will not exceed the number avail- ecount of the statistics of the signal sources, it is possible to ensure an acceptably low [48] Cy [48] Fy [48] Cy (48) Cy [48], 48) M; (481 Cy [48] Fy [48] Cp [48] Cy [48] FB), 148] Co [48] Fo (48) Co 48) Co 48) Fy LB) (48) Cp 148) Fy [48] Cy (48) Cy (48), a8] Figure 5.23 0M1/2 multiplexer former.51-2 514 517 S18 Problems 333 Determine the Nyquist sampling rate and the Nyquist sampling interval forthe signals: (a) sine (10071) (b) sinc? (10071) (©) sine (10078) + sine (5071) (@ sine (100#) + 3 sinc? (6071) (©) sinc (504) sine (10071) eo 810 = cos(S0n1) and g2(1) = cos(250z1) are ideally sampled using an impulse train with = 100Hz. (@) Show that the sampled signals 8; (1) and (1) are indistinguishable () Verify the above by computing spectra Gj (f) and G(f), A signal g(®) band-limited to B Hz is sampled by YY 4 periodic pulse train pz, (f) made up of a rectangular pulse of width 1/8B second (centered at the origin) repeating at the Nyquist rare (2b a GarTy hae iyquist rate (2B pulses per second). Show that the A) ws 2 not BO = je + x = sin(™*) g(t) cos nm Bt Show that the signal @(¢) can be recovered by passing Z(t) through an ideal lowpass filter of bandwidth B Fiz anda gain of 4, A signal g(® with maximum frequency 5 kHz is sampled using a pulse train (see Fig. 5.10), The pulse duration is 20 ps and the pulse repetition rate is 100 js. At the receiver, the sampled signal is multiplied with cos(40n 10°1) and then passed through a lowpass filter with cut-off 5 kHz. Show that the filter. ‘output is a scaled copy of g(r). A signal g(#) = sinc* (S7rt) is sampled (using uniformly ‘spaced impulses) at a rate of (i) 5 Hz; (ii) 10 Hz; (iii) 20 Hz. For each of the three cases: (a) Sketch the sampled signal. (b) Sketch the spectrum of the sampled signal. (© Explain whether you can recover the signal g(t) from the sampled signal. (@) Ifthe sampled signal is passed through an ideal lowpass filter of bandwidth 5 Hz, sketch the spectrum of the output signal Signals gi(®) = 104T(10#) and go(t) = 8(¢) are applied at the inputs of ideal lowpass filters Hy (f) = T1(f/20,000) and H3(f) = T1(f/10,000) (Fig. PS.1-7). The outputs y; (1) and y>(#) of these filters are multiplied to obtain the signal y(t) = y, (#)y2(t). Find the Nyquist rate of y; (0), y2(#), and y(t). Use the convolution property and the width property of convolution to determine the bandwidth of y} (N)y2(2). See also Prob. 5.1-1 8) y(1)=y,(1)9, (1) ure P.5.1-7 A zero-order hold circuit (Fig. P5.1-8) is often used to reconstruct a signal g(#) from its samples, ©(0 and shove that th is cir samy ple tops connected hy wm wa age of 1 voll. Fit in part (w)Problems 335 Forpractical reasons discussed inthe tex, signals are sampled ata rate well above the Nyquist rate, Practical (© Eps use 44,100 samples Per second. If L = 65,536, determine the number of bits per second required to encode the signal, and the minimum bandwidth required to transmit the encoded signal, ot sesso i ed ae hhas a bandwidth of 4.5 MHz. This signal is sampled, quantized, and binary (a) Determine the sampling rate if the signal isto be sampled at arate 20% above the Nyquist rate. (@) Ifthe samples are quantized into 1024 levels, determine the number of binary pulses required to encode ‘each sample. {@) Determine the binary pulse rate (bits per second) of the binary-coded signal, and the minimum bandwidth required to transmit this signal 24 Five telemetry signals, each of bandwidth 240 Hz, are to be transtnitted simultaneously by binary PCM. The 7 signals must be sampled at least 20% above the Nyquist rate, Framing and synichronizing requires an additional (0.5% extra bits. A PCM encoder is used to convert these signals before they are time-multiplexed into a single data stream. Determine the minimum possible data rate (bits per second) that must be transmitted, and the ‘minimam bandwidth required to transmit the multiplex signal, 535 Iris desired to set up a central station for simultaneous monitoring of the electrocardiograms (ECGs) of 10 7 pogpital patients, The data from the 10 patients are brought to a processing center over wires and are sampled, bbinary-coded, and time-division-multiplexed. The multiplexed data are now transmitted to the mon ‘toring station (Fig. P5.2-5). The ECG signal bandwidth is 100 Hz. The maximum acceptable error im sample amplitudes is 0.25% of the peak signal amplitude,’The sampling rate must be at least twice the Nyquist rate. Determine the minimum cable bandwidth needed to transmit these data. Monitoring station Processing, ite Receiver Figure P.5.2-5 5246 A message signal m(t) is transmitted by binary PCM without compression, Ifthe SQNR is requited to be at le: 47.48, determine the minimum value of L. = 2" required, assuming that m(2) is sinusoidal. Determine the ac | SQNR obtained with this minimum L. 527 Repeat Prob, 5.2-6 for m(t) shown in Fig, PS.2-7, ie- 336 SAMPLING AND ANALOG-TODIGITAL CONVERSION Hint: The power of a periodic signal is its energy averaged over one cycle. In this h signal amplitude takes on the same values every quarter cycle, the power can also jy. signal energy over a quarter cycle. m(0) Figure P.5.2-7 52-8 ForaPCM signal, determine L if the compression parameter 1 = 100 and the minirn yin Determine the output SONR with this value of &, Remember that L must be a power. > binary PCM. 52.9 A signal band-limited to 1 MHz is sampled ata rate 50% higher than the Nyquist rat levels by using a j-law quantizer with = 255: (a) Determine the SQNR. (@) The SQNR (the received signal quality) found in part (a) was unsatisfactory. 1) my.) by 1048. Would you be able to obtain the desired SQNR without increasing the tr it was found that a sampling rate 20% above the Nyquist rate is adequate? If so, exp), ‘maximum SQNR that can be realized in this way? $.2-10 The output SQNR of a 10-bit PCM was found to be insufficient at 30 dB. To achieve the de it was decided to increase the number of quantization levels L. Find the fractional inctease bandwidth required for this increase in L. 52-11 Amplitude of a signal m(t) ranges from —1.5V to +1.5V and its mean square value 1°) transmitted using PCM. The required minimum SQNR is 38 dB. Maximum frequency in mr (@) Calculate the required minimum symbol rate when coding is ternary (—1, 0, 1), ls transmission bandwidth. Assume that transmission of 1 symbol/s requires | Hz of transmiss (b) Repeat the above for binary coding (0, 1). Assume that transmission of 1 bivs requires | bandwidth. © 5,41 Inacertain telemetry system, there are four analog signals m, (¢), mp(),m3(#), and m4(). The is 3.6 kHz, but for each of the remaining signals it is 1.4 kHz. These signals are to be sampled their respective Nyquist rates and are to be word-by-word multiplexed. This can be achieves » PAM samples of the four signals and then binary coding the multiplexed samples (as in the ‘carrier in Fig. 5.20a), Suggest a suitable multiplexing scheme for this purpose. What is the com (in rotations per second)? Note: In this case you may have to sample some signal(s) at sts Nyguist rate(s), 54-2 Repeat Prob. 5.4-1 if there are four signals my (1), mp(t), m3(t), and mg(t) with banciwisits 200 Hz, respectively. Hint: First multiplex my, m3, and mg and then multiplex this composite signal with 1) (