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Chapter 13

# Chapter 13 - 13-1'2 Chapter 13 auf?llf 9;” 2 ﬁlter...

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Unformatted text preview: 13-1'2 Chapter 13 auf?llf 9;”! 2 ﬁlter numbed A Tb +1: Ptt) G D Fig. 513.1-1 For the integrate and dump ﬁlter (1&D), the output is the integral of p(t) Hence, at! = Tb, pa(Tb) = ATb. If we apply 6(1) at the input of this ﬁlter, the output h(t) = 1(1) — u(t - 73,) . Hence, H(w)=7},sinc(-a%n4 ””7“: and ' 2 l «N 2- 2 22;}, =— —T n00) ”'42 bsmc [2 (0 Arm-— and 2 T 127-2 This is exactly the value of p2 for the matched ﬁlter. The output p00) of this R-C ﬁlter is p,(:)=A(1—e“/‘C) OStSZ}, 3 A(]_e-T./RC)‘-(l-n)/RC I > n . The maximum value of p00) is AP, which occurs at Tb: Ap '3 Pa(Tb) = A(1"e 2 l cN' dm J an ——o——- - 2:: 2 ¢1+mznzc1 =ARC 45m?) and 105 13.2-1 13.2-2 We now maximize p2 with respect to RC. Lettingx = 13/.RC, we have LN x and 2 402 2xe"(l —e”‘)-(1 -e”) t _— = .-. 0 dx x2 This gives 2xe" = 1-9" or 14-21: = e" and I 1.26 x a 1.26 _. = ._ or RC 73 Hence, 2 Przmx =(01316)2—'L;,i Observe that for the matched ﬁlter, . xpz g 25,, = 2421; max 9“ w The energy of p(r) is 7}, times the power of p(t) . Hence, ”2(1‘m2) Azmz A27“ E = —— T T = —9- = E p 2 b + 2 b 2 b 2 Similarly, q = ”—27!- = 5,, EN = orbp(r)q(t)dr a [:3 [-A2(l —mz)cos2 me! + AZ»:2 sin2 wc1]dt 2 = "Ag-7,, + Azszb Hence, “11,0 -m2) 85,,(1—m2) w " an ' *7.— and _ 2 P. =Q(£m)= ’25,,(1 m) 2 J! Let C, be the cost of error when l is transmitted, and Co be the cost of error when 0 is transmitted. Let the optimum threshold bea, in Fig. 8132-2. Then: A ’ o C1=C10P(elm'l)'CIoQ[ ”6 a] AP +00 all Co=C01P(EIm=°)=Col Q[ 106 The average cost of an error is C= Pm(l) C, +Pm(0) Co (1) C _= %(C‘ +Co) = -;-[CIOQ[A;:% p‘ao )+cmg(ﬂ;n_ + 00 J] For optimum threshold dC/dao = 0. Hence, to compute dC/dao , we observe that PCP ‘0) k (V H.) X = _ l x e_y2/2 : 2 a : Q( ) 1 EL” dy ac A? {a and ’A? 52:-j21.a 322/ Fig.813 2-2 Hence, 2 2 jaw) (we) dC ‘ 201 — 20 -——= C e n —C e " =0 dao 720” 2” 10 on Hence, (Awe)2 (Arno) gige 20', 2 a C10 and In [91} 2°°AP and a0 =-—“—" [91] CI0 ,3, 2A,, Cw But J15 2 P .. on = —2—— and AP - 5,, Hence, at: C s—ln J—I— ” 4 [cm] 13.2-3 We follow the procedure in the solution of Prob. 132-2. The only diﬂ‘erence is Pm(l) and Pm(0) are not 0.5. Hence, c- pm(1) c. + Pm(0) c0 = 3,0) Cm Q[Apa-ao )+ Pm(0)C01Q[Apa:ao] and _(‘P"'v)2 _=(‘r+‘o!2 dC l 252 2oz ——= P 1C e " - 0 C e da 20’” 2” m() 10 Pm() 01 Hence, 1,, [\$9221.]:mmo: a: ,n [moan] PM(1)C10 0,2, 2A}: P..(l) C10 107 If; "‘ But Hence, 13.5-1 Fig. 513.5-1 2 ”(rim—n) T 4. 5,) ﬁg: = -r’/2o% 2 = ﬂ P(’l"’0) anme an 2 P(rlml) = an‘l/ﬁeJ'JP) be" The thresholds are t E P /2 and P<elmol=2 45/2] ]=2:[J-f:] < 42%] "JEN—HUME 74¢?) 13.5-2 Here, p(!) and q(t) are identiﬁed with 3 p(t) and p(t), respectively. Hence, ”(2) =[3r(-22>- manev‘wn =2p(-m)¢'-2~n 2(2) = 2:42.. - r) 1 l = 5[153p — 5p] = EPEP - 5,] = 45,, But multiplication 001(1) by a constant does not affect the performance. Hence we shall choose h(t) to be pm, - I) rather than 2pm, —t). This will also halve the threshold too, = 25],. This is shown in Fig. 513.5-2. Also, ' and f[3p(i)]p(:)dz = 35,, 108 “4W" WNW] iii] 95 + E The energy/bit is Eb = 45—1 = 55p Hence, _ ’OSEI, P: ‘4 J ] 13.5-3 ForM= 2 1J1=2xlO'3 ‘ For 256, 000 bps the baseband transmission requires a minimum bandwidth 128 kHz. But amplitude modulation doubles the bandwidth. Hence 8, = 256 kHz 10'7 = Qua]: 5,, :27 )(10”7 011 s,- = 5,12,, = 2.7 x10"7 x 256,000 = 0.069W For M = 16 256,000 B = T iog216 2(15 ” "' peM = P0108216=4x10'7_ “(l—{)4 2:51,] This yields 5,, = 5.43 x10-6 5,- = 5,,R,, = 5.43 x10'6 7: 256,000 = 139W =64kHz ForM=32 256,000 B =512kHz 7:10;, 32 Pm =Pblosz 32=5x10‘7= 2—(3—32‘) "7—3025” ) This yields 5,, =1.719 x10.s s,- = 5,55 = 1.719 x10"5 x 256,000 -= 4.417 13.54 For M = 2 andau = 2 x104 This case is identical to MASK for M = 2 10'7 =QU3§1)= 5,, = 2.7 >110-7 s, = 5,11,, = 2.7 x 10‘7 x 256,000 = 0.069W a, = 256900 x 2 = 256 kHz 2 109 For M = 16 P,” =(log216)l’l, =43, =4x10‘7 2 4x10“7 52 "21—14451 :9 5,, =1.67x10‘° 25M 3,. = Em, = 1.67 x 10" x 256,000 = 0.4275W In MPSK. the minimum bandwidth is equal to the number of M-ary pulses/second. Hence, 256,000 3 64 kHz log; 16 81-: ForM=32 PM: (103232) )P,,=sx10 7 2 2 5x10'722Q[ ”(55")2] 5,,=524x10“° 1024Ji s,. = EbRb = 524 x 10" x 256,000 = 134W 256 000 B = ’ = 512 RE 7 10g2 32 z 110 ...
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