chap_1-vectorist

# chap_1-vectorist - Solutions Manual to Accompany Digital...

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Unformatted text preview: Solutions Manual to Accompany Digital Communications Second Edition Bernard Sklar Prentice Hall PTR Upper Saddle River, New Jersey 07458 F“ L/ﬁ/t‘ 94"!- ‘K 7; -T7/«&)#:J: Tﬂm'éy-ffxf ....1 73"1’42. An. I, 1;_7:1_2_:df’=__2:_ 15" ()%2= Amzrﬁt M"Zz_éf£§ 0 W541. L _. Nerf-2040]” =— A.“ "'14. 411’ 1-1 at) 4555):: wt +£15.26 #442“ ﬂZe/vw ? H’Q/v .. ’5 cafe-25m gtaé'fj zrfanl 7:“:A “[3:21]- 0 W m w W (xwra- T wane-0 Em =[:9c‘({-)¢t = 17" M = T "9;, PK = £:@@)4§=[°¢§ {£31566 Mayne 4:; e i 'ch1" 1-2 _T%_ 4; I: %_ V0 '1: ”7775— ﬂ ' ~53 mo Wm + 1, , Tr #17 00 do: 2-01.: Ia- To :ﬂl—fa, A yuo(l+wa 2.0-6)4 400 C1+m¥0f)# c EST; Doot++oafjgp = 250 W 3:: Mar 2 Ial‘fécwﬁ) =fw Q6?) cal-F == 1§+2C+Ioo4~roo Fab = 2.579 W 1-3 5!, Myth} 62;”; . (at) 9:60;).— ' 5:“?! o . -’ _J::\:Li_ -1 {1 It {1. «60) : 756-8”) V’ (b in ’Xﬁc’)=3(t)+m Wﬁm _;__ 1. 45(9) \$6 9:09.345? Cc—v 04(6)= @750“) I. «(7.9 = rx(—€’) / 2. 7((0) 7% 7(0) x Ii? 4.1 a») Xécw; 569+ Mm; 1-5 ,3 @(a’) .. <4 ”(mag-m) A magma 275g 24¢ p m<>ﬁéd1Wde ’“mif’é’di W team (w Wm} "‘T’é ,5?) ' . W . £— 01) R155) "‘ @m/Df’+f3€ can 202/ .._._ 2. L W em? -= I0 1—6 ""' D f = m w £+ 113% 1" “IE/2.4 (1 5 lJ—m ‘ 3> _ XE; 11< I. Y? g (a)[:capéﬁfée-s)oéﬁ = mm (6)1: ,0 5(6) (;+~:)"Léé = m (d) I; Jﬁf+4)(6"+66 +2)ﬂf = "7 (d) J: wogt’) [55—23% a 5.0/83 Li [email protected])____'é[m_m+ Jame, y] X, G) * X166) = Xﬁ) MEé—mdéﬁ] UACM—TW: (0'4 -I = 0’25‘5’4 ﬂaZS’KHL M” 3' H Lw‘l/m] H6! Lat Me) =- Me) -II——..._.__ -I—I-—__ 666) = 366) “ {(15 ‘1‘») £66) IL. {afar} \$604293 cm . 25> '6- K .9 ﬂ[)— ,9 W w) 1 “t '6 1-13 1.20 (d) The normalized spectrum, me/ 6,1106), appears as: 0.?5 6 6 6 6 9.105 92-105 94-105 45-105 .8-10 1-10‘5 LUZ-10610410 Loo-10 1.08-10 1.1-10 f Applying numerical methods with Mathcad ®, the two-sided 99% bandwidth can be found by numerical integration as: J106+ f1 1H Ge(f)df EGAN where fl is found to be equal to 103 kHz. Thus, the two—sided 99% bandwidth is equal to 206 kHz. Since this bandwidth corresponds to the given spectrum (with signaling rate : 10,000 symbols/s), normalizing it relative to one symbol per second, yields the two-Sided 99% bandwidth as (206x103/104) = 20.6 Hz for one symbol per second, or in general the 99% bandwidth in terms of the signaling rate, R, is 20.6xR Hz. 2 0.99 1.20 (e) 35-dB Bandwidth: 35—dB attenuation :> 10'3-5 = 3.16 X 10“ 1-14 1-15 ...
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• Spring '08
• Nagaraj,SV

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