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2.12.1 Inter Symbol Interference (ISI) f . ce (ISI) is e Definition of ISI: In telecommunication, inter symbol Seine ia sea of | distortian of a signal in which one symbol interferes with su a ecantae hue ‘ tee | . d phenomenon as the previous symbols have similar effec ) aking | unwanted pheno st | the communication less reliable. 4 icati is di iscrete pulse modulation ISI arises when the communication channel is dispersive. In discr p the amplitude position or duration of the transmitted pulses is varied according to the digita| information to be transmitted. ¢ These pulse modulation schemes are referred to as pulse amplitude modulation (PAM), pulse position modulation (PPM) and pulse duration modulation (PDM) schemes. In this methods, PAM system are more efficient in terms of power and bandwidth utilization. ¢ The element of a baseband binary PAM system, are shown in Fig. 2.37. Clock pulses Pulse generator data {bp} Clock recovery network Transmitting filter A7(f) pulse Output A/D Gaussian data ( (bp} Fig. 2.37 Baseband binary data transmission system ¢ The pulse Benerator output is a pulse waveform oo SO= > a,P(- KT,) (2.39) k=-co Principle of Digital Data Transmission sii 97 where Q,, depends on the k,, input bit. The sequence {6,} into a new sequence of normalized such that P(O) = ] pulse amplitude modulator modifies this binary short pulses for convenience we will assume that P(t) is aif the ky, input bit is 1 and aq = 4 ~aif the ky, input bit is 0 e The signal S(t) is modified ag a result of transmission through the channel of impulse response h(t). The noisy signal then passes through receiving filter H(f), and the output y(t) of the receiving filter is sampled by analog to digital converter. The transmitted bit is regenerated by the A/D converter based on the sampled values of y(t). A set of typical waveform that occurs at various stage in the system is shown in Fig. 2.38. P(t) | Ty t Input bits a 0 0 1 1 0 "oT n r] fle. | 2T, 57, , [0 [ [ 37% 47%, | | agg awiaesew — Transmitting output t Receiving filter output (d) Error due to noise and ISI in binary PAM system (a) pulse generator output (b) transmitting filter output Fig. 2. ical waveform ; ig. 2.38 Typical w (c) Receiving filter output (d) Output bits Principle of Digital Data Transmission eee a a ——————— 99 eigieee ee EEE en — — a A a tm eS a ey ) r(t) r(t) (t _ sin 2nf,,t 1 2 fm H(f) | | 1/2 fan ~J Sus Fig. 2.40 Pulse shaping by the channel 2.12.2 Nyquist Criterion For Zero |S| 1. One method of controlling ISI is to shape the transmitted pulses properly. we know that “sinc function” produces zero ISI. So 1 sin(nt/T,) 0 T, merT, fo NS s h(t) = 42) This is the impulse response of an ideal low pass filter whose frequency response shown in Fig. 2.41(b). hit) H(w) Ts E =F in O x =9T, "-Ts 0 Ts ~— 2Ts Ts Ts Fig. 2.41 oN Note that impulse response A(t) goes through zero at equally spaced intervals that are | multiple of 7, except at the center. Fig. 2.42 Pulse with Zero ISI “YrUTE avy = a y we i 1d 6 ry . If the channel passes frequency upto /,,,, the zeros of the pulse are spaced by 1/2 Tai hig e cha _l it is clear that pulse of the same shape that are Spaced 2fm |’ T, or an integer multiple of 7, will not interfere. is the nyquist interval. If |T, 2.12.3 Raised-cosine Filter ee Raised-Cosine filter: The raised-cosine filter is a filter frequently used for pulse shaping in digital modulation due to its ability to minimise intersymbol interference (I SD. Its name shen from the fact that the non-zero portion of the frequency spectrum, of its simplest form (a = 1) is a cosine function, ‘raised’ up to sit above the f (horizontal) axis. —~—J The ideal band limited pulse shape is difficult to achieve due to sharp corners on the frequency spectrum. A desirable compromise is the raised cosine characteristics. The fourier transform of this pulse is similar to the Square transform of the ideal low-pass filter except that the tran. sition from maximum to minimum follows a sinusoidal curve for this purpose. We use raised. cosine filter in which the frequency response decreases towards zero gradually. The frequency response of the Raised-cosine filter is given by r 1 O(G+oa)W Here H(w) is Impulse response of raised cosine filter. where W = of. The corresponding impulse response is hii) = (el cosaWt | ...(2.44) § T,\ We Jiq_ (20Wet/ 2)? The parameter & is called roll-off factor. 1. When o = 9 it will coincides with t 2. When a = 1, j function is he ideal low-pass filter. tis known as the full-cosine roll-off characteristics, its frequency response (2.45) _—E———_ Digital Communicati, Ng 02 OS 4H @) Fig. 2.44 le 2.1 eight message signals having aan tae: Exampre 2 7” The error in sampling ampHtude cannot each are time-division multiplexed using a a nero the minimum transmission bandwidth be greater than 1 percent of the peak amplitude. ling rate m required if raised cosine pulses with roll off factor « = 0.2 are used. The samphng ust be at least 25 percent above the Nyquist rate. Solution: We know that the maximum quantizing error must satisfy 5 Inacertain telemetry system, A (Q) max = ry m = £.¢ 0.01m, Hence L > 100 and we choose L = 128, L = 2? = 2” n= 7. The number of bits per sample required is 7. Since the Nyquist sampling rate is 2f,, = 4000 samples/s. The sampling rate for each signal is f, = 1.25 (4000) = 5000 samples/s. There are eight time-division multiplexed signals requiring a total of 8(5000) = 40,000 samples/s Since each sample is encoded by 7 bits, the resultant bit rate is “3 = 7(40,000) = 280 Kb/s T, Then minimum transmission bandwidth required is fy = LO _ 140.2 27, > 9 (280) = 168 kHz, Ans: Principle of Digital Data Transmission 103 aes A ROR NDE AIT. FLARE RAB He - Example 2.16 A certain telephone line bandwidth is 3.5 kHz. Calculate the data rate (in b/s) that can be transmitted if we use binary signaling with the raised-cosine pulses and a roll-off factor o = 0.25. Z Solution: We know that pulse rate or data rate is = — fp T, 1+©4 1 _ 2(3500) T, 1+0.25 = 5600b/s Example 2.17 A communication channel of bandwidth 75 kHz is required to transmit binary data at a rate of 0.1 Mb/s using raised-cosine pulses. Determine the roll-off factor &. Solution: We know that 1 + « = 2f,T, Here T,, —§—— =10°s fp = 75 kHz = 75(10°) Hz Now, 1+ 0a = 2f,T, = 2(75) (10°) (10°) 1+a=1.5 Hence we obtain a = 0.5. mipivas Vu ‘UCALIONS a __ a 106 .< ghown in Fig. 2.45 satisfies Nyquist, 18 Pf) ate at which bing se spectrum _ maximum r ry Example 2.21 4 aver he 1.2 MHz, determine 6 on What is the roll-off factor» ' = 0.8 Za 2 gs C . Criterion. If /; data can be transmitted by this pulse using Nyquist ! ! i —F208 1 fy=i.2 finMHz~ 0 ss Fig. 2.45 ullon: n S case — 7. ence we can ransmit data at a rat R, i Mg I 1 9 A Also, fg = 1.2 MHz 1+@ mma) we know that fy = = (R,) after substituting the value of f, = 1.2 MHz, R, = 2 MHz in eqn (I) we get 1.2 x 106 = 14% (2x 108) = lta=1.2 = a=0.2 Example 2.22 Binary data of a rate of 1 Mbit/s is to be transmitted by using Nyquist pulse with PU/) shown in Fig. 2.45. The frequencies f, and f, of the spectrum are adjustable. The channel available for the transmission of this data has a bandwidth of 700 kHz. Determine f and /, and the roll-off factor a. Solution: Given f, = 700 kHz, R, = 1M bit/s Also Ry = = 500 kHz and f, = 700 - 500 R f, = 200 en i- Hence, a=. excess bandwidth theoretical minimum bandwidth qa=. te — 2h, = 200 | 05h, R, 50074 and f, = Kh - 12 “fx = 500 - 200 = goo kHz 114 0 = MANES 2.14 EYE DIAGRAM The performance of baseband PAM system depends on the amount of ISI and channel noise. The distribution of ISI an channel noise in the system can be easily observed by displaying the received waveform Y(t) on an oscilloscope. We need only a basic oscilloscope to generate the eye diagram. Given a baseband signal at the channel output. Yt) = 2a, Pt - kT,) ...(2.50) This Y(t) can be applied to the vertical input of the oscilloscope. The time base of the scope is triggered at the same rate - as that of the incoming pulses. b The oscilloscope show th ae output 20. p 8 the superposition of many traces of length T, from the channel The resulting pattern on th e oscilloscope lo to ur as the eye pattern of the system, pe looks like a human eye. This is widely know? Consider the transmis; The eye diagram of Fig. 2 51(a) is shown in Fj pulses as shown in Fig. 2.51(@) respectively, . © Shown in Fig. 2.51 (b) for the time base of T. and 2! > 116 rrors is shown by the rate of closing of the eye (iv) The sensitivity of the system to timing e as sampling time is varied. (v) The sampling time is midway bet from zero crossings, then the amount 0 of “Jitter” or variations in clock rate and phase. . the channel since 1n a Strictly (vi) Asymmetries in the eye pattern indicate non linearities in 1 linear system with truly random data all the eye openings will be identical. e Fig. 2.53(b) shows a distorted v eform and the corresponding eye pattern The eye pattern appears “close ow distributed about +A gs; if the clock information is deriveg ween zero crossin : —_ f zero crossings indicates the amount f distortion 0 ersion of the wav d@” and the sampled values are n Waveforms ae Sampling times Fig. 2.53 E i H ye diagrams of a binary PAM system. (a) Ideal (b) Distorted (c) Distortion + noi oise e Anosily di ; e Eye even ne version of the received waveform and the correspondi are a in generally used for monitoring the performance of Rasdlena PAM eae systems 2. 15 DIGITAL RECEIVERS AND REGENERATIVE REPEATERS (iii) Extracti ’ ng the timing j ; instants. ing infor Principle of Digital Data Transmission 117 e A regenerative repeater is shown in Fig. 2.54. e A regenerative repeater consists of a receiver plus a “regenerator”. e A complete regenerative re peater may also include provision for separation of dc power from ac signals. e This separation is generally accomplished by transformer-Coupling the signals and by passing the dc around the transformers to the power supply circuit. Sampler and Decision Timing extraction Fig. 2.54 Regenerative Receivers 
