Question

Wu Wulis. For each case, compare the puits Ubtained in Prob. 6.2-3 and Prob. 6.2-4. 6.2-6 A duobinary line coding proposed by Lender is also ternary like bipolar. In this code, a 0 1 transmitted by no pulse, and a lis transmitted by a pulse p(t) or (t) according to the following, 15 Cncoded by the same pulse as that used for the previous 1 if there are an even (including zero) number of Os between them. It is encoded by a pulse of opposite polarity if there, are an odd number of Os between them. Like bipolar, this code also has single-error-detection capability, because correct reception implies that between successive pulses of the same polarity, an even number of Os must occur, and between successive pulses of opposite polarity, an odd number of Os must occur. there are an ev them. Like hinale Oded by a pulse of on use correct recan

Answer 1

Aug-612-6-(d) In this, we obseave that on the average, half the pulses have ako and the remaining half have ak=1 on-1. Hence, ce Ro=Lim W (+1) + N (6)] = 1/2 Four R2, we need to determine akakti. There are four possible equally likely sequences of two bits: 11, 10,01,00. Since bit o es encoded by no pulse (ako), the product of akakti=0 foor the last three of these sequences. This means on the average 3N combinations have akart=0 and only N combinations have nonzero akakti. Because of the duotinary rule, the bit Sequence 11 can only te encoded by two consecutive pulses of the same Polarity (both positive or both negative). This means ak and anti are 1 and 1 ou -1 and -1 respectively, an either care au akti = 1. Thus, there N combinations have akakti = 1. Therefore, R = Lim [N (2) + 3N (6)] = N₃0 N For R2, We need to otseave the product akautz,

Foor this we need to observe all possible comtinations of three bits in Sequence. There are eight equally likely Comtinations: 111, 101, 1o, loo, bll, olo, ool,000. The last six combinations have either the first and our the last bito. Hence, akak+2=0 for all these Six Combinations. The first two combinations are the only ones which Yield nonzero akakt2. Using the duotinary rule, the frost Combination is encoded by three pulses of the Same polarity Call! Positive or negative), Theus ak and auta are 1 and 1 an 1 and -1, gespectively, yielding akakt2=1. Similarly, because of the ductinary rule, the biorst and the thiad pulses in the Second bit combination lol are of opposite Palacity yielding av akt2 =-1. Thus on the average, akakta =1 foor N terims, -1 for N teams, and o for 3 teams. Hence, R2 - Lim ho po cestop 62) + 3N0] = 0 In a similar way we can show that, Rmco, n>1. Sy (W) = 1P (W) 1² re do cs ить Sy(w) = 1 PCB)12 (1+ cos w To) 2 To No So, sy (W) = 1 Rotae Rn N=1 Roto to

[Sycw) = 1 PCW/ 2 (08? (WTO) Foor half-width Pulse, Plat) - orect ( 24/76) sy (W) - To Sine? (WTO) (os? (WTO) A Sy(w) Тъ гурту -> R6/2 Rb. f(Hz) Sin From PSD, we observe that the bandwidth is approximately R6/2 Hz.