Chapter 11

Chapter 11 - .1-1 This is clearly a non-stationary process For example i amplitudes of all sample functions are zero at same instants(one is shown

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Unformatted text preview: Chapter 11 11.1-1 This is clearly a non-stationary process. For example, i amplitudes of all sample functions are zero at same : instants (one is shown with a dotted line). Hence, the I statistics clearly depend onr. C 7 C 7 ti, t -’ Fig. S! 1.1-! 11.1-2 Ensemble statistics varies with t. This can be seen by W La finding :(73=Acos(ax+0)=AIgoocos(M+9)p(w)da> a a D A 100 to = 1-06 0 cos(ax + 9)da). This is a function of 1. Hence, the process is non-stationary. ' % L; Fig. Sll.l-2 11.1-3 This is clearly a non-stationary process since its statistics depend on t. For example, at! = 0, the amplitudes of all sample fimctions is b. This is not the case at other values of t. i: a >E—<‘o . O t"? Fig. 511 1-3 lid-4 x(r) = acoiax + 9) . m = acos(ax + 0) = 3cos(ax + 0) = cos(ax + 0) Ianp.(a)da K 2 FaCK.) A -- cos(ox+0/2A ada=0 [ ) 1L4 - - A A 1-9 Rx(rl,12) a a2 eos(axl +0)cos(mtz +0) s eos(ax, +9)eos(at2 +0):2 = cos(ax, +0)codarz +9) I14 $110 Fig. Sll.l-4 2 sis—com, +0)cos(axz +9) 83 11.1.5 703 = acos(ax +0) = jg°°cos(ax + 0)p(a)) do) a too 0 10° 8—- co an 9 d =--—' a 100 0 s( + ) a) lo()ts111(¢211+)‘0 = 73355111000: + 0)-sin0] Using this result, we obtain 2 Rx(tl ,tz) = a2 cos(aztl + 6) cos(ax2 + 9) a %eos[w(t, + 12) + 20] + cosa)(tl - 12) 2 ‘ a =2006'+12)[sin[100(ti+t2)+20]-sin291+2W:_12)[sin100(rl42)] [1.1-6 R73=az+b=ix+u But§=o Hence, x(t)=b _ — 3 2 _ 2_ 2 2 la 4 Also, 8—0, a —J_2a p(a)da=:-? —28-3- ______———-——— Rx(11,12)=(8!1 +b)(8lz +b) = 821112 +a(tlb+12b)+b2 - 4 = 321112 + ab(t1+12)+b2 = —t,tz +b2 3 ' PCK) 11.1.7 (11) x_(t')=E=° "’ (c) '* ° ‘ -— ‘7 l 2 l ‘ 2 1 1 Rx(n.rz)=KK=K =I_.K “WWI—F “‘3 K2. t” s - — 1 (d) The process IS w.s.s. Since x(t)=08nd Rx(’h'2)‘§ ° t" (e) The process is not ergodic since the time mean of each k sample function is different from that of the other and it is -3 O t _’ not equal to the ensemble mean (i = 0) (f) x2=Rx0=-;: ‘4' I O b... Fig.511.1-7 0.5- pm _._ P607 21! _. 111-8 x(t)=acos(wct+9) ‘E... oi 1 9 3W _ 1 - a9 ; "-' 0 82 = 3' i I C 2 Q f (b) méacoiwcwflsacofiwcnoko t"? (c) Rx(t,,tz) a: a? cos(wct; +0)cos(wctz +6) (d) The process is w.s.s. Fig. 5111-8 84 (e) The process is not ergodic. Time means of each sample function is different and is not equal to the ensemble mean. (0 F‘ Rx(°)=% 11.2-1 (a), (d), and (e) are valid PSDs. Others are not valid PSDs. PSD is always a real, non-negative and even function of w. Processes in (b), (c), (i), and (g) violate these conditions. 11.2-2 (a) Let x(t) = x1 and x(t+ r) = X; Then, (X1 iX2)2 = X12 +X22 +2XIX2 2 0, X12 ‘l‘XZ2 Zt 2X1X2 But, m; = R,(r) and x,2 = x22 = R,(o) Hence, Rx(0)z|Rx(r)| (b) Rx(r) = x(t)x(1+ r), lim Rx(r) = x(r)x(r + 1') 1-D” As r —-> oo, x(1) and x(t + r) become independent, so lim Rx(r) = x(t)x(t + r) = mm = *2 f—ND 11.2-3 R,‘(r) = 0 for r = 15% and its Fourier transform S,(a)) is bandlimited to B Hz. Hence, R,(1) is a waveform bandlimited to B Hz and according to Eq. 6.l0b Q Rx(r) = 2 124-215) sinc (27rBr -n). Since Rx = 0 for all» exceptn = 0. "=4 R40) 28 Rx(r) = Rx(0) sine (2mm and S,(w) 5 Rafi-1’5) . Hence, x(t) is a white process bandlimited to B Hz. 11.2.4 Rx(r) = Pm (1, 1)+ Pm: (-1,-1)- me(-l, 1)— Pm: (1,-1) But because of symmetry of 1 and 0, Pqu (11 l) = Pqu ('19-!) “d lex; ("'1’ l) = Px,x2(l»-1) and Rx(r) = 2pm (1, 1)- PW (1-1)] 1-6—- "I: = ZPXI (])[Px2|x, - szlxl (- W = 2px, (1)[Px2[x‘ (1") -(| - Px2|x1' 2PX2IX‘ 0“)— I Consider the case :17}, < < (n+ l)1}, . in this case, there are at least anodes and a possibility of (n +1) —- T r ‘ Dd pm 1 0d =L.E_L=._.- n es b[(n+ )n es] 13 TD n Prob(n nodes) - l- Prob[(n + l)nodes] (n + l) - é ‘ The event (x2 = llxl = 1) can occur if there are N nodes and no state change at any node or state change at only 2 nodes or state change at only 4 nodes, etc. Hence, szlxl (ill) a Prob{(n+ l)nodes] Prob(state change at even number of nodes) 4- Prob(n nodes) Prob(State changes at eveen number of nodes) The number of ways in which changes at K nodes out of N nodes occur is ( ,1: Hence, Pm,“ (111) s 3*')(o.6)°(o.4)"*‘ +(5+')(o.6)2(o.4)""+.....] (Lg- a} 85 [(3)(o.6)°(o.4)" + (g)(o.s)2(o.4)”'2 [H 1- and M?) = 2mm (“0—1 This yields Rana—12% M<n (no) = --0.44+0.24-‘7-‘ti 7}, <|r| < 27}, (as I) b =0.l36-0.048§ 273<Ir|<3Tb (n=2) b and so on. Rx 0") T 37 9.1 Fig. 511.24 The PSD can be found by differentiating Rx(t) twice. The second derivative a'2 R, /dt2 is a sequence of impulses as shown in Fig. SI 1.2-4. From the time-differemiation property, 86 .— 2 : R,‘(r) H (jm)2 S,(a)) = ~0st (w) . Hence, recalling that 6(1— T) c9 e’jn’r , we have T —wzsx(w) = Ti[—2.4+1.44(a~n +e’1"Tb)—ozss(efl“”b +e-I2~T~)+....] b = -T1—[—2.4 +2.88cosw7}, -0576eoszw1;, +0.1 152 cos3w7}, +....] b and l l l l S w =——-—— 2.4-2.8 eosmT -— 52 T — os3wT -— 4 7+...) A) nm2[ 8( 5 sec nip-25¢ I, nscoswl, ] 11.2-5 Because Sm(a)) is a white process bandlimited to B , Rm(r) = R",(0) sine (28:) and fl R—-—=, = , , "(28) o n :11213 Thisshowsthatxt r+—"— =R 3—)=0 ( )4 23] “(23 Thus, all Nyquist sample are uncorrelated. Now, from Eq. 11.29, 2 Q Sy (w) = mm)! [R0 + z RIn common] Tb "eel n 2 l and where a k is the kth Nyquist sample. Rn = 3m". = 0 R0 = 3% = x2 = Rm(0). Hence, we)? 71, 3mm) = ZBRm(0)|P(w)2l since 2', = 5% Sy(w) = “.245 For duobinary P .,(1)= f.k(-i)=025 and 5*(o)=05 -— l l l ‘* =“>z*(")z*°[3)‘° l l “0 "i =(‘)2i+(“)2i*°’(5)‘5 RI = Mm = Z zflkfiml’apm (Mam) ‘k ak+l Because ak and a“! each can take 3 values (0, l, -l), the double sum on the right-hand side of the above equation has 9 terms out of which only 4 are nonzero. Thus, R] a (1X1)Plklt+| (1’1)+(—lX—lfliglk.‘(-IX-l)+(‘X-I)Plglkq (lX-1)"(-1)(1)Plklk+1 ('00) Because of duobinary rule, the neighboring pulses must have the same polarities. Hence, I P‘k'h-l (1’1) 3 PM (1)P.k+ll'k a = E Similarly, P.k.M(-l,—l) =% Hence, R, a; A150 R2 3 aka“; In this case, we have the same four terms as before, but a; and a“; are the pulse strengths separated by one time slot. Hence, by duobinuy rule, 1 l 1 Pulse: (1'1) = PM(1)},11‘441'1'0“)IB = 1—5 . . l Sunilarly, Flu“: (—1,—1) = E 87 1 In a similar way, we can show that Pun”: (1,—1) = an,“ (-1,!) -- 16 Hence, R2 = 0 Using a similar procedure, we can show that R,I = 0 for n z 2. Thus, from Eq. (1 1.29) am noting that R" is 2 2 . i w P a) T 1 an even function of n, we obtain Sy (w) = (l + cosafl}, = L(7J-cos2 [22-h] « b b Hiram—73 For half-width rectangular pulse P(¢u) = gisinc (Earl) and Sy (w) = 75inc2( 4 )cos’( 2 ) 11.2-7 if = 99+(—1)(1- Q) = 29-1 Rout =<1)29+(—1)’<1—2)=1 Because all digits are independent, 2 Q sy(m) = We)» [1+2(2Q-i)2[zcosnwn]] Tb n=l 11.2-8 Approximate impulses by rectangular pulses each of heighth and width 5 such that he -- 1 and e —> 0 (Fig. 51 1.2-83) Rx(7)=zlex2lexz(xlx2) 7! ’l ’2 ‘ Since xl and x; can take only two values h and 0, there will only be 4 terms in the summation, out of which only one is nonzero (corresponding to X] = h, x; = h ). Hence, Rx“) = hz PM!) g hz PX] (h)szlxl Since there are a pulses/second, pulses occupy a: fraction of time. Hence, 2 PX] = as and RX(’) s h ad’lexl : ahpxzbq Now, consider the rangelri < a. I’m,“ is the _“ e at. Prob(x2) = h, given that x, = h. This means x, lies on one f..- of the impulses. Mark off an interval oft from the edge of g- , — this impulse (see fig. Sl 1.2'8b). if x1 lies in the hatched interval, at; falls on the same pulse. Hence, PlexlUIIh) = Pmb(x. lie in the hatched region) a 5—? = 1—2 E-T: p . and Rx(r)=a.h(l——;-) I Since Rx(t) is an even function of r, Rx(r) == ah[ - t- In the limit as s -+ 0, R,(r) becomes an impulse of strength (2. R803 © R,<r)=aa(r) M=o. ‘ When I > a, x, and x2 become independent. Hence, 0 T9 Px2|xl W') " Px, (h) = as I R,‘(r)=«tzzh.‘::=ar2 ' |r|>0 Hence, R,(r) s a6(r)+a2 Fig. 511.243 11.2-9 In this case the autocorrelation function at r = Oremain same as in Prob “2-8. But for t > 0 whenever x(t), x(t + r) are both nonzero, the product x(t)x(t + r) is equally likely to be h2 and — II2 . Hence, Rx(r) = 0, r at 0 and R,(r) == a6(r) 11.2-10 The process in this problem represents the model for the thermal noise in conductors. A typical sample function of this process is shown in Fig. 8112-10. The signal x(t) changes abruptly in amplitude at random instants. The average number of changes or shifts in amplitudes are pper second, and the number of changes are Poisson-distributed. The amplitude after a shit! is independent of the amplitude prior to the shin. The first-order probability density of the process is p(x;t). It can be shown that this process is stationary of order 2. Hence, p(x;t) can be expressed as We have Rx“) ' Ejlxlxszm (thz)d*1d*2 ‘ mm, (xi)sz (le31 = 3|)d'idxz (1) To calculate p,‘2 (x2|x‘ = x1), we observe that in 1' seconds (interval betwun x, and x2 ), there are two mutually exclusive possibilities; either there may be no amplitude shift (it; = x1), or there may be an amplitude shift (X; at x1). We can therefore express p,‘2 (lex. = it!) as px2 (1:21)“ = n) = px1(x2|xl = x,, no amplitude shifl)P(no amplitude shin) + p,2 (lex1 = x,, amplitude shifi)P(amplitude shifi) 1: 13+“: " ‘1 Fig. 511.240 The number of amplitude shifis are given to have Poisson distribution. The probability oflt shifis in r seconds is given by * , mm (5;!) 6" where there are on the avenge fl shins per second. The probability of no shifls is obviously po(f) , where Pom = e"! The probability of amplitude shifi- l- p0(f) = l—e'p' . Hence p,‘2 (lex, = xi) 2: e'p’pxz(x21x1 = x], no amplitude shifi)+ (1- (‘0');7x2 (x2|x, = x1 , amplitude shift) _ ' (2) when there is no shin, x2 = x1 and the probability density of x; is concentrated at the single value xl. This is obviously an impulse located at x2 = x,. Thus, p,‘2 (lex, = x]. no amplitude shifi) = 60:; - x1) (3) whenever there are one or more shifis involved, in general, it; =2 x1. Moreover, we are given that the amplitudes before and after a shift are independent. Hence, p,2 (x21x, = x1, amplitude mm) = pxl (x2) = p(x) (4) 89 where p,‘2 (x2) is the first-order probability density of the process. This is obviously p(x). Substituting Eqs. (3) and (4) in Eq. (2), we get Px1(x2|xl = x1) ‘ {#502 " xx) 41-9-“)sz (x2) x e’p’lflxz - x1)+(epf - 1)p,‘2 (x2 Substituting this equation in Eq. (1), we get RAT) = e‘mfilzxtxzm. (xr)[5(x2 '1‘1)+(¢’3 ' '1)Px,(x2)}btdx2 s {pillfioxlxszl (31»(32 ‘ xlfi‘ld’n +£I:"IZ(JT '1)’,"‘(x')P‘1(x2)dr'dx2] = e-prblxipxr (MM! +(efl' '1)I:"PXI(")¢‘J:’ZP‘2 (x2)dx2] = {#1: + (efl’ - l)i2] where i and X2 are the mean and the mean-square value of the process. Eq. (5) becomes For a thermal noise i = 0 and Rx(r) = xze'p' r > 0 Since autocorrelation is an even function of r, we have m) a x26“ and 11.3-1 For any real number a, (ax — y)2 2 0 azx2 +y2 -2ax-y 2 0 Therefore the discriminant of the quadratic in a must be non-positive. .Hence, (23f «$2.720: (a)2 < x2 y2 Now, identify x with 2:0,) and y with y(12) , and the result follows. 1m mmumsm a Rx(r)+ Ry(1’)+ ny(f) + 1- Rx(r)+ Ry(t) since x(t) and y(t) are independent. 4 =4Rx(f)+9Ry(f) since ny(t)= Ry,(r) =0 Rw(t) = [x(t)+y(t)12x(t + r)+3y(r + 1)] = znx(r)+ 3R,(r) Rw(t) = RW(-t) = 2Rx(t)+ 3Ry(r) —_____—.——-—--— 11.3-3 ny(r) = ABcos(wot + e) cos[nwo(t + t) + n¢] (5) coiwot + nwo(t + r) + (n + 1”] = 3‘; :K c0409! + nwo(t + t) + OH W1!“ = 0 ____._._______ Similarly, cos[muo(t + r) - 001+(n—l)¢]= o and ny(r) = 0 11.34 x(t) = co + 2C" cosmoo - b)+ 0,, n=l co 8 Co + ZCnOuvot — "mob + 0,.) ":1 Since I: is a r.v. uniformly distributed in the range(0, 73), mob = 3%! is 3 1w. uniformly distributed in the 6 range (0, 27:). Using the argument in problem “3-3, we observe that all harmonics are incoherent. Hence the autocorrelation function of Rx(r) is the sum of autocorrelation function of each term. Hence follows the result. “.44 (a) 51(0)) 2: 2KTR. and 52(0) = 2KTR2 Since the two sources are incoherent, the principle of superposition applies to the PSD. If 5",1 (a2) and S02 (w) are the PSD's at the output terminals due to 51(w) and S2(a2) respectively, then Salim)=ln.(w)l’sl(w)andsoz(w)=1Hz(m)|’Sz(w) where R2 / ij R2 H‘(w) ___ = ijZC H = R3 = R2 Rl(ja)RzC+ '0' R2 ch+ R] + R2 RHJLJ’ELC. “42.. R2+l/jcaC jatR2C+l (b) "2010’ R' " R R2(ijIC+ 1)+R1 ij1R2C+ R1 + R2 2 2 _ 21mm; ‘ 2mm 5400’ 2R2R2C2 2 “"1 S°2(°’) 2 2 2 2 2 (D i 2 +(R1+R2) a) R} ch +(R‘+R2) 2er R; R +112 530(0) = So. (w)+5oz(0) = 73%;“)? .8255. m R|R2C +(R1+R2) I Risa; C 1/ jazC Rpm; A (b) H w = = . a. T ( ) _L_+_RJ.£2_ Ja’CRlR2+(RI+R2) 45-333} ij Rpm, RM“; 2 ZKTR R 3» =|”(‘”) lid? 2 (RI +R2) 'ZKTleg _ 2KTR1R2(RI +R2) = ——_.._————I-—'—- 022C2R12R§ +(Rl + 122)2 R: + R2 wznfigc’ +(R, + sz which is the same as that found in part (a). 9i 11.4.2 y(t) = j: h(a)x(t — a)da ny(t) = x(t)y(t + r) = x0”: h(a)x(t + t— a)da = h(a)x(t)x(t + f - a)da = h(a)Rx(r — a)da = h(r)t Rx(1) and Sxy(rn) = H(w).S'X (w) l . jazC 1 Inan. 11.13, H a) a- = _ ( ) R+ .1 1wRC+1 10C and 5mm») = 2KTR/(ijC +1)md Rm°(r) = 2107: e"’RCu(r) 11.4-3 (a) We have found Rx(t) of impulse noise in Prob. 112-8 R,‘(r) = (16(1) 4» a2,and Sx(m) = a +2xa26(m) Hence, Sy(w) - ll-I(m)|2[a +21ra26(a))] = 21ra2|H(0)|2 6(m)+ a|H(a))|2 and RM = 9"[sy(w)]= «Flaw»2 +ah<r)-h(-r> ' i (b) h(z) =%.e-'/'u(t). "(0) =%. 1 1 ' , jut-+— r . 2 2 2 ' 2 q - q a: — r v |H(w)l =1+w2t2,and Rx(!) = azq2 +¢£¥ {141,272}: azq2 +—2-r—.e W ’_ I I n I X“) m y“) W 1—» I” Fig. 3114-3 ' 11.5-1 n(t) = nc(t)cosmct + n,(t)sinwct The PSD oan!) and n,(t) are identical. They are shown in Fig. 811.5-1. Also, :17 is the area under - 4 l» 5,,(w) ,and isgivenbynz -2 flxlo‘ +'—°—[:”--l) =125x10‘au ‘ ‘ 2 - 2 2 2 v n§(or n3) is the area unclerSnc (w), and is given by n: = n} =2[5000JU+%-%x5000]= 125x 104w 11.54 We follow a procedure similar to that of the solution of Prob. 11.5-1 except that the center frequencies are different. For the 3 center fi'equencies Sufi (m)[or S“, (m) ]are shown in Fig. SI 1.5—2. In all the three cases, the area under Sn: (m) is the same, viz., 1.25 x mm. Thus in all 3 cases n2 = n§=125x104w ‘N' “N— -IOK 5K 10K ‘IOK 5K 10K IDA K ~F H3 -’ 5? H5 9 .p #3 a Center (at) o [05 k can'lev (“‘1‘ 45 k cam-er 40 K Fig. $11.52 y» Fig.511.5-3 6 S (w) 2 6 1 11.54 H =___£————=_L+_L—= = (a) "pm Sm(w)+Sn(w) __.5_+5 6mz+60 w2+10 9+0)2 1 - (b) hap(;)=m.e Mill (c) The time constant is 7:5 . Hence, a reasonable value of time-delay required to mnke this filter realizable is 7:? = 0.949 sec. (6) Noise power at the output of the filter is ‘ '° Sm(a’)sn(a’) I 5 -l 0’ a 3 6 N = — —-—-—-—-dw = —- do) = tan = ° 2” -°° Sm(w)+ Sn(w) 2x -° m2 +10 2x710 710 .., 710 The signal power at the output and the input are identical 1 «o 6 SI‘So=§-,;L-—idw=1 9+¢u 514R: .19; 3 fl: 1.054 No 3 93 4 _ sm(a’) ___.__a_’2_‘—"—4——- 11.55 (s) Hap(m)'————_—Sm(w)+sn(m)— 4 + 32 03244 (Dz-+64 m2+64 1[ 53.33] l+ 902 +96 5 m2 + 10.67 (b) kg,(:)= -;-J(t)+8.163e’3'266m I (e) The time constant of the filter is 0.306 sec. o - A reasonable value of time-delay required to make this filter realizable is 3 x 0306 = 0.918 sec. ((1) Noise power at the output of the filter is No =._l_ MM =_‘_J'°° ___..2__—da, =0544 2n '°° Sm(w)+3n(w) 2” “'° 9(m2 +1067 The signal power is 1 no 4 _ hop“) S, —S,=E; «mm—l £0. = __‘._ = 1338 Na 0544 Fig. $11.55 94 ...
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This note was uploaded on 05/07/2008 for the course ECE S-306 taught by Professor Dandakar during the Spring '08 term at Drexel.

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Chapter 11 - .1-1 This is clearly a non-stationary process For example i amplitudes of all sample functions are zero at same instants(one is shown

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