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Chapter 14

Chapter 14 - Chapter 14 I l4.l-l The following signals...

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Unformatted text preview: Chapter 14 I l4.l-l The following signals represent 2 sets of 5 mutually orthogonal signals. Fig. 514.14 14.1-2 ~12, 21" f," 372- 14.1.3 1)(1,1,0)is—J-17[1+J2'sinwoz] 00:9; 0 0 1} 2) (2, l, l) is %[2-J§sinwot+ﬁcoswot] =-‘/—LT— 0 0 [2 + 2cos(a)ot + 55)] 4)2 (-1, -l. l) is 7%['%- 25in001+J§cosmot]=—j7z[—-;-+2co{wot+%)] Fig. 514.1-3 112 s, (t) 14.1.4 (3.13'41912) b) The energy of each signal is: 4 ‘ 4 1=H +:+1+16T0=31 £2=4+1+l6+16+ 1.0:“ c To Ea=9+4+9+16+17b=39 E4=4+16+4+4+ofb=28 7;, To c) F3 - F4 = (-6-8 +6+8 +0) = 0. Hence, f3(t) amt/4(1) are onhogonal. 14.2-1 Let x(t) = x" x(1+ l) = x2 x(t+2)= X3 We wish to determine leX2X3(xl9x2’x3) Since the process x(t) is Gaussian, xh x2, X3 are jointly Gaussian with identical variance (0'2 =02 =ch = x(0)=l). The covariance matrix is: X] X2 X3 2 0'” =0,” =x|x2=x(t)x(t+l) = x(l)=l an cxm or”1x3 I 2 2 1 e —— --——- 1 K = ‘7sz 0:2 0,sz Also OX2“ = cxm = X2X3 = x(t + l) x(t + 2) = Rx(l) = —e— 2 __ .— 6x3,” OX3XZ OX3 aux; = ”X3“ = xlx3 = X0) X0 +2) = x(2) = if e 50 113 ‘ l e e 2 1 I l K: — l "' and “q: ‘-—i- e e e l l ._ — 1 22 e A A llundA -A A =A ‘i-l 11—33 e2 12 21 23 32 e3 e I Ana‘A31=°: A22= "e74— And {ZAP-’1 p XIX2X3 ‘ '1 X1XZX3( ) (2 )3/2 K‘ 14.3-1 . ' .—-—+ r ' 5 S. l <- a Fig. \$14.34 w «m Wm M) P(q’"2) = Pmms) ‘ ------ ‘ Adm") = ”“0““ < 3 Hence p(qml) = p(qu) = 7,17 afar-MN 4111 = 1—47-293] and 4/2 P<qm2)= (qms>= ----- PM” I) 7‘ KW ""‘ " ”(737] Hence PM =1— no) —1..([p(m.)p(c1m,)+P<m2)P(c1m2)+ ----- +P<mM)P(c1w)} =1-_E[M+2(M-1)47;j]]“—‘ 2-(M ___1_M)Q(m) 2 2 -1 The signal energies are [:59 , (1%) , ......(:t—ME-a]2 114 Hence the average pulse energy E is Also 15 = b logzM lZlogzM Hence The average pulse energy E is ﬁller +(;)2]+;[e):+eﬂ]=2:—z This performance is considerably better than MASK in Prob. 14.3-1, which yields PM = L75Q[ 0235” for M=8 14.3-3 In this case, constants ah 's are same for k = l, 2, ..... M. Hence, the optimum receiver is the same as that in Fig. 14.8 with terms ak's omitted; We now compare r-sl, r-sz, ..... r-sM . Since r - sh = ff r cosak is the angle between I and 5,, , it is clear that we are to pick that signal 3,, with which I has the smallest angle. In short, the detector is a phase comparator. it chooses that signal which is at the smallest angle with r. 14.3-4 Because of symmetry, P(Clm|) = P(Clm2) =.....= P(C|mM) where M=2N d -d 39' =3 01' —2-’ 2 311d E|=Ez= ..... =EM8-A%= Let S1=(;d-, :1, ..... ,1) 2 2 2 Then ll6 and PM)" P(q"’|) N Pm =l-P(C)=1—[1-2Q[J—_2;§w:]] Here, M = 2N . Hence, each symbol carries the information log; M = N bits. Hence and 14.3-5 Also E " 05(0) + 02st + 025d2 s .2— Hence _ _ , 35-1112 in“: PM =1- HC) =3 231%“ + 2:75AM 14.3-6 Pm = 1- PM 14.3-7 118 Fig. 514.341) (c) P(Clm5) = Prob(noise originating from :5 remains within the square of side 5%) Mani :iJrJ meim-J and mannequin] _ 2 We also observe that E, the average energy is E = —[£——d ——---] 0.4:!2 s 2 _E__ _ 04412 02:12 and d =,_5_Eand _5__E J“ W 03, 2°"- d2320,,= 55 Th r fore P -:-4 _— ° ° (4%) 4 M) The decision region R2 for m; is shown in Fig. 3 and again in Fig. C-l. quadrant (horizontally hatched area in Fig. C-l) -A, . Thus P(Cfmz) = noise originating ﬂour :2 lie in R2 = P(noise lie in 1st quadrant)- P(noise lie in A.) R2 can be expressed as the ﬁrst 2 = [l — in - P(noise originating m :2 lie in A,) 20,, But P(noise lie in A1) = %[P(noise lie within outer square) — P(noise lie within inner square)] (See C-2) -ﬁHM in2|<§ )‘POML M‘z d '75)] a-[in-zaizzJ’i J-i 24%)“ H9 5%[442:,]+4Q[24;°'7H P(C|'"2)=“29(21]+Q[2:,,)'Q[37‘2i;] ”421.) 42rd,] =' gig-5) iii-3‘] 4mm and Fig. \$14.3-7c Moreover, by symmetry P(‘l'"2) = Winn) = P(4m:) = PHm) Hence PeM = %[§l P(£|m,-)] SE SE = W5] M 14.3-8 / >9" Fig. 5143-8 r<c>=-‘—24P(c1m.-) )=-[P(01"'x)+)++P(C|"'2)+P(C1M3)+P(C|'"4)] 165:, The decision region R, for m (see Figure) can be expressed as R, = outer square of side dJi «:- (outer square - inner square of side d) = —1- outer square of side d5 4»:- inner square of side d 120 Now P(C|m,) = Prob(noise originating from ml lies in R1) P(n lie in outer square) + 2- 4P(n lie in inner square) 1 4 =lr(I-4<—]+z”('“' g) 2 =i[“29[7dﬁ]] ii” 7%)] Similarly R2 , the decision region for I»; (see ﬁgure above) can be expressed as R; = outer square of side dJ-Z- - é-(outer square -— inner square of side d) = gouter square of side dJl7 —-1iinner square of side (1 and P(C|m2) = noise originating from m; lie in R2 1 )9 ~ 1 ' “PT i 'm, 114 n, ad '33 1: I995, f. K R3 RA a 8 ‘ The decision region R3 for M3 can be expressed as R3 = R A + R3 - RC and P=(C‘m;;) Prob( (noise originating from m lie In R3) = P(noise m RA)+ P(noise in R3)- P(noise in RC) “(M Ma) -;-.r[|n.i Inzi<7";]- %[P[|nnH"2|< 7";i“’(‘“""“2"§')] =—[1- Wu; J] [. 42m] 7i]- wen --i 24mm =;[1-20[J%d)]+%[1-29[ﬁ1] Whirl] ...g‘....‘ e- {A The decision region R4 for ":4 can be expressed as R4= RA" and P(C1’"4) = pm > —d, :12 > -d)-%{P(|nx:81n2‘< d)- PQmI’ lnzl < 57)} {new-%[1—2Q[d&ﬂz*%L-ze(fj)f [1- 1:90]2 s1-2kQ(-) For any practical scheme Q( Using this approximation, we have ““1“” 47]“‘3’f‘4 ’33) Hence P(C)— - —[P( P(C|m‘)+ P(C‘m2) )+ P(C1M3)+)+)]P(C1m4 W. --e(—- "4 F 2w Now E1==,d2 £2— - -,2d2 53 = 4:12, and £4 = 8:12. Therefore E = :(dl2 +2d2 +4d2 +8d2)=14 —d2 And 5, = J... = 5. 1032 16 4 — 2 so that EJ’. = .1.5— : 1.5.5— 16 J0 Therefore p(c)=1-3 3.55.. -3 [1.9.51 -3 2.11 2 15 J 4 15 J] 4 15 J! Moreover 3.5; » {1221 » an 15 J! 15 J! 15 J! Hence 3 8 £5 PC ;I—- —-—— ( ) g[ M] And Pm =1-P(C)=%Q["%%) Comparison of this result with that in Example 14.3[Eq.(l4.57)] shows that this conﬁguration requires approximately 1.5 times the power of the system in Example 14.3 to achieve the same performance. 14.3-9 1f :1 is transmitted, we have bl=E+a+JEn1 , L, =-E+a—JEn‘ b2=a= £112 , b.280— £112 bk=a+fink , b_k=a— Enk and P(qml)= Problbi >12.» b2. 12-2. bi. b4) Note that b1>b_1impliesE+a+JEn1>-E+a—anl or n1 >—J-E bl>b2 impliesE+a+JEnl>a+JEn2 orn2<JE+n1 bl>b-2 impliesE+a+JEnl>a- En; orm>—(Jf+nl) Hence b, >b2 and [1.2 implies -(nl+J—E_)<n2 <(n, +JE) Similarly b, >bk and b_k implies -(n, +JE)<nk <(n1+f§) Hence P(C‘ml)= Prob.(b1> b_1, 172. 5.2, b3, b_3,"' bk, b.k) =Prob.[n‘ >-JE, |n2l<(n1+JE), [n3[<(n1+‘/-£—), ...’ Inkl <(nl +JE)] Since nl , n2,--- nk are all independent gaussian random variables each with variance JU/Z , P(C]ml) = [P(n1 > —J—E_) Pﬂnz| < nl +J—E_)P(|n3| < n] +JE) Pan“ < n1 + JEN N-l I m .. 44— .. 2 =m— ‘1’: e "Wbmmf/E)‘ Man] an, Let y = I'll ﬁrs/E 41/2 25 2 . ~ {r a] /2 P,” = 1-P(qm,)= 1- e [1 290)] dy 2" 3135/.» E A150 Eb 3 log22N 14.4-1 The on-oﬁ‘ signal set and its minimum energy equivalent set are shown in Figs. (a) and (b), respectively. The minimum energy equivalent set of orthogonal signal set in Fig. (c) is also given by the set in Fig. (b). Hence, on-off (Fig. a) and orthogonal (Fig. e) have identical error probability. 1he set in Fig. (b) is polar with half the energy of on-off or orthogonal signals. , 123 9’ ____,__ \ o d “‘1 ° 9" o ) 2— 2,. La. CA3 Lb) Fig514.4-1 2 “+2 Here ¢i(t)- Ecoswor=JEcosmor wo=-;l ¢2(t) = M sin wot M Therefore 51(0‘564‘20) 51 [4’2 52(')=J§¢1(') Sit-[01 33(’)="J§¢l(’) 3=—f¢ The vectora=%25,-=%[Jﬁ¢l—‘ﬁ6¢l+J§6¢2]=J_§—é¢z , Hence the minimum energy signal set is given by 21(2) = si(r)-—“2—_°¢z(r)= J56 ¢2(:)——@¢2( )= ﬂsisinwor 32(‘)=52(‘)—:@¢2(t)= J5¢1(l)-1/—:¢2(t) =10J§coswot—-2—(g£sincoot 3 33(’)=53(’)‘£¢2 (’)=-J§5)¢i(1)-£¢2(l )e—IOﬁcoswot-zoﬁ The optimum receiver - a 3suitable form — in this case would be that shown in Fig. 14.83 or b. sin (00! Fig. 814.4-2 14.4-3 To ﬁnd the minimum energy set, we have a = th-(sl +52 + 53 + :4) = :4, 'h Hence the new minimum energy set is =s --¢. m Jam 32=’¢1+J§¢2v33='\/§¢I‘¢2,34=ﬁ'\5¢2 124 Note that all the four signals form vertices of a square because (5, 52), (£2 £3), (£3 54). and (54 31) are orthogonal. The distance between these signal pairs is always 2J5 . This set is shown in Fig. Sl4.4-3a. The signal set is now rotated so as to yield a new set shown in Fig. Sl4.4—3b. 43¢) Q." oD'I-lé'lN‘ IL: 1'; 4;: \$3 = " D ' 60 Fig, SI‘A‘s Observing symmetry we obtain PM = P(ClmI) = P(CIM2) = P(C1"'3) = “51"“) = P(n1>-Jfandn2 >—‘/-2—) l 14.“ = = , = d 3 3‘ 10 10¢l 5:710‘”l ‘2 s 10’2 an '3 lalovﬁh] 5% lg 91!? 1.» Lb) Tﬁe signal sd‘ @m Manamm «await I I I 4 ’ SI (-6 __ 3 5Jé\ 514-) aye - 3 o ‘9" ° b» 0 mo «ea '4 (d) We. minimum energy waveﬁasfm s Fig. SM.“ 125 “(611113)”le n1<-2-o£7=)=1- 47:01:: P(Clm2) = p014 < 515] = 1 —2Q(7.07) P(C . —[2(P(qm|)+P(C1m2)] - %[2 — 2117.07) + 1’— 29007)] = 1— €90.07) and P,“ = 1- P(C) = 5390.07) 5 1.03 x10"2 2 Also EI=E3=[ l ]=4x10”3 510 2 1 2 — 1 1 )+[ ) =2X10-3 33(E‘+EZ+E_3)=3X10-2 l "(wt/i3 10710 Mean energy of the minimum energy set: Emm=3(2X10-3+0+2X10 )= g: «10'3 inirnum energy set in this case is identical to 14.4-5 The use of Eq. (14.76) and signal rotation shows that the m own in Fig. Sl4. 4-4c. this situation is identical that in Prob. 14.4-4. Hence the minimum energy set is as sh to that in Prob. 14.3-5 with d = B7? From the results 1n the solution of Prob. 14.3- 5, we have =10‘3 Also, we are given 5,,(w) = (a) From the solution of Prob 14.3- 5 P,” =—-Q(702)+Q(7.12)=l.09xl (b) and (e) identical to those in Prob.2 14.4-4 0-12 14.4-6 (a) The center of gravity of the signal set is (s1 + s; )/2 Hence, the minimum energy signal set is 5 +8 — 5 +5 .- X|=Sl-—-———(122)=.1232 &12=52-———-('22)=—2——s S] The minimum energy signals are 1111(1) = 05-0.7o7s111”’—2°'- 410 = 2000;: 112(1) = 0.707 sinL‘iz‘i - 05 0.00! (b) Ex! I (05— --.0707sin-2—)2 dt=0.4984x10'5 O x2=15,,|=049114><10'5. Wearegivent.M=5><10’6 {25 :==Q[ “]= Q(4.=465) ==.'041><105 (c) We use Gram-Schmidt onhogonaliution procedure in appendix C to obtain MOP-‘10) .001 [ﬁsmmom 0 2J— y2(t)= J—sinwot- .001 = Zsmwot—-—— . =_.;_r,_ y1 LYII val=JJn2dr= JEFF 91= 0‘0 =3169. y; 2(1) =1- -cosZwo! --:-sinw0t +£— ly2|= "I ygdt =1’.[001l--8-2')=T_-—_4‘-722y2y2 1 . SKIP—161’: 1(=1) 03‘6”” 92m= 72!;5’2()+—)'1(')=°133Y2(')+°235y1( ) 351 127 -001 ...
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