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Unformatted text preview: Chapter 7 W 7.21 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 onoff case [see Eq. (7.18b)] Sy(w)=1’(w)2 [HE E 5[w_2_zm]] 4Tb Tb nzC) =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 onoff: 2Rb Hz; bipolar: Rb Hz. Clearly, for polar and onoff cases the bandwidth is halved when fullwidth 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. 7.22 7.23 7.24 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. 87.22, 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 2bit 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. 87.24 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. 7.31 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. 57.24 For halfwidth pulse P(t) = rect(2t / 7b). (co) = Esmcz w—n)cosz 4 4 2 From Fig. 87.24 we observe that the bandwidth is approximately Rb /2 Hz Sy From Eq. (7.32) (l+r)6000 1
——————:> r = — 4000 =
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