Chpt_7_solutions - Chapter 7 W 7.2-1 For full width rect...

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Unformatted text preview: Chapter 7 W 7.2-1 For full width rect pulse p(t) = rect[TL] [7 0T P(w) = TI, sine [7”) For polar signaling [see Eq. (7.12)] Sy(w) = lP(w)i2 = Tb sinc2 2 For on-off case [see Eq. (7.18b)] Sy(w)=|1’(w)|2 [HE E 5[w_2_zm]] 4Tb Tb nz-C) =Z£sinc2 1+2—7r Z 6 (0—2—70:— 4 2 Tb "=40 7}; . 2 (1)7}, 2701 But smc 7 =0 for co=—T— foralln¢0,and=lforn=0. Hence, I) T Sy(w)= ismcz (225—)+£6(w) =12,sinc2 w—n’ sin2 2 2 The PSDs of the three cases are shown in Fig. 872-]. From these spectra, we find the bandwidths for all three cases to be Rb Hz. The bandwidths for the three cases, when half—width pulses are used, are as follows: Polar and on-off: 2Rb Hz; bipolar: Rb Hz. Clearly, for polar and on-off cases the bandwidth is halved when full-width pulses are used. However, for the bipolar case, the bandwidth remains unchanged. The pulse shape has only a minor influence in the T bipolar case because the term sin2 in Sy (0)) determines its bandwidth. 7.2-2 7.2-3 7.2-4 SyUOD 9—H. (a U." 7; '° 7;, ( lo ) Fig. S7.2—2 t + g:- t _ % P0) = “CI E -—rect i 2 2 and = £311“: eijb/4 + £51)“: e_ja’7;I/4 2 4 2 4 =ij sinc sin 4 4 IP(w)|2 2 w. 2 0T. Sy(”’) = = 7}, Sim: sin 7}, 4 4 From Fig. 87.2-2, it is clear that the bandwidth is :1 rad / s or 2Rb Hz. b For differential code (Fig. 7.17) R0 = lim N—Nn 1 N 2 N 2 —- -——- 1 + -——— -l = l N i 2 ( ) 2 ( ) :I To compute R], we observe that there are four possible 2-bit sequences 11, 00, 01, and 10, which are equally likely. The product akak +1 for the first two combinations is l and is —1 for the last two combinations. Hence, R1: lim iv (i)+£(—i)]= 0 N—->cn N 3 2 Similarly, we can show that R" = 0 n > 1 Hence, 2 P a) T T . Sy(w) = l ( =. [—b) sinc2 [—60 b) 7}, 4 4 (a) Fig. 87.2-4 shows the duobinary pulse train y(t) for the sequence 1110001101001010. (b) To compute R0, we observe that on the average; half the pulses have a], = 0 and the remaining half have ak = l or -1. Hence, . l N 2 N 1 = lim ————'_l:1 +—0 :— Ro N—mN[2() 2 To determine RI , we need to compute akak+]. There are four possible equally likely sequences of two bits: 11, 10, 01, 00. Since bit 0 is encoded by no pulse (ak = 0), the product of akak+l = 0 for the last three of . 3N . . N . . these sequences. This means on the average —— combinations have akakH = 0 and only —4— combinations 4 57. 7.3-1 have nonzero akakH. Because of the duobinary rule, the bit sequence 11 can only be encoded by two consecutive pulses of the same polarity (both positive or both negative). This means at and oh, are l and l or —-1 and -1 respectively. In either case aka“, = 1. Thus, N these? combinations have akak.” =1. Therefore, 1 N 3N l R— lim ——l+—'O =— lN—no N[4() 4 U] 4 To compute R2 in a similar way, we need to observe the product akak+2. For this we need to observe all possible combinations of three bits in sequence. There are eight equally likely combinations: 111, 101, 110, 100, 011, 010, 001, and 000. The last six combinations have either the first and/or the last bit 0. Hence, akahz = 0 for all these six combinations. The first two combinations are the only ones which yield nonaero akahz. Using the duobinary rule, the first combination is encoded by three pulses of the same polarity (all positive or negative). Thus ak and ak+2 are l and 1 or -1 and - l,respectively, yielding akak+2 = 1. Similarly, becausc of the duobinary rule, the first and the third pulses in the second bit combination 101 are of opposite polarity yielding akakn; -1. Thus on the average, aka/{+2 = l for E terms,—1 for—[Y— terrns, and 0 for g— terms. Hence, . l N N 3N R = lim ———l+——l+—0 =0 2 M N[8<) 8() 4 u] in a similar way we can show that R,l = 0 n > 1 , and from Eq. (7.10c), we obtain 2 2 Sy(“’) = Psi—:30 +cosco75) = cosz(w_72,) Ref 7— Rb Fig. 57.2-4 For half-width pulse P(t) = rect(2t / 7b). (co) = Esmcz w—n)cosz 4 4 2 From Fig. 87.2-4 we observe that the bandwidth is approximately Rb /2 Hz Sy From Eq. (7.32) (l+r)6000 1 —-—-————:> r = — 4000 = 2 3 58 7# Dam/5 WE 7359 JED/L )4 [409m] JIM/H, U936; DWDE (DITH % (mm-1+ (2&7: Wages (Sea H6 7. [7) 7H9 ?SDJ§/((u>. '“ SPB/j We“ 7769 Ofl r/t 97000”) fl/A/flfl/ flew/w r; Gm» 5/ @{139 " 2 CU 9(va L47? (K0 71 2 2,2,, cmm’) /7:/ ’ £0: Z/m MUST DHWINb fin ) r) 20 _ W $1; - LDC/6’53)” [gain-m fooCEVLf'NTm B/CoD/X/e ; I A i (3 TWJml/ii‘u‘b f} flag; @971th 73 beh—) l§¢7flffpflmolz§ gr?) ' AN) 4 0 u ,c a k _/r MEéflnwofi rt _,/< a ‘ SEE J Sf;va Kim/(6 iW/Dflf PULSE: Tim 1’: 5 0/; 17% 7W5 biTLSv‘ftam arc 453m”: I l O | O i o o I O 6 O i "‘ ‘" i \ 1 ‘ i l - I (‘3‘ Hue Edna rover~ RFC)!“ .1, (1K ([1 ilkeiy "ha be .I,‘ or. v, Far LIME "0 ' q (i 0;. HR ' Uzi] I, f) + _’ 7 an 771‘: 0 fun? ' y has +0 Q I or ~ bin J. A] i 2 z ' _ “a I 210 + 4mlai Al~>0o N werfr’ 40%,“4/83 flv:/5:I:”w;é 5emLm'7/u; 01W GL qk ConfresPaménio, $0 a i ‘3. ed 0/4.qu‘ gm We parka;er “Si/3:2,? V" “a /0 n“ 1) m (11109 (fiwQKH ‘0 “my 4») :30” y j —H o- .— ,( ‘ 0m QKH “ I -v. 7cm ‘ ,0 ,LIJ—L ...
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Chpt_7_solutions - Chapter 7 W 7.2-1 For full width rect...

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