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Unformatted text preview: Chapter 11 11.11 This is clearly a nonstationary 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!
11.12 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 = 106 0 cos(ax + 9)da). This is a function of 1.
Hence, the process is nonstationary. ' %
L; Fig. Sll.l2 11.13 This is clearly a nonstationary 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 13 lid4 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 19
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.l4 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)] [1.16 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=:? —283 ______—————— 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.511.17
0.5 pm _._ P607
21! _.
1118 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. 51118
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(°)=% 11.21 (a), (d), and (e) are valid PSDs. Others are not valid PSDs. PSD is always a real, nonnegative and even
function of w. Processes in (b), (c), (i), and (g) violate these conditions. 11.22 (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)zRx(r)
(b) Rx(r) = x(t)x(1+ r), lim Rx(r) = x(r)x(r + 1') 1D” As r —> oo, x(1) and x(t + r) become independent, so lim Rx(r) = x(t)x(t + r) = mm = *2 f—ND 11.23 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 124215) sinc (27rBr n). Since Rx = 0 for all» exceptn = 0. "=4
R40)
28 Rx(r) = Rx(0) sine (2mm and S,(w) 5 Rafi1’5) . 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 (11)] 16— "I: = ZPXI (])[Px2x,  szlxl ( W
= 2px, (1)[Px2[x‘ (1") (  Px2x1' 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.l360.048§ 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 1.24. From the timedifferemiation 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 +eI2~T~)+....]
b = T1—[—2.4 +2.88cosw7}, 0576eoszw1;, +0.1 152 cos3w7}, +....]
b and
l l l l
S w =———— 2.42.8 eosmT — 52 T — os3wT — 4 7+...)
A) nm2[ 8( 5 sec nip25¢ I, nscoswl, ] 11.25 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 righthand side of the above equation has 9 terms out of which only 4 are nonzero. Thus,
R] a (1X1)Plklt+ (1’1)+(—lX—lﬂiglk.‘(IXl)+(‘XI)Plglkq (lX1)"(1)(1)Plklk+1 ('00) Because of duobinary rule, the neighboring pulses must have the same polarities. Hence,
I P‘k'hl (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(7Jcos2 [22h] «
b b Hiram—73 For halfwidth rectangular pulse P(¢u) = gisinc (Earl) and Sy (w) = 75inc2( 4 )cos’( 2 ) 11.27 if = 99+(—1)(1 Q) = 291
Rout =<1)29+(—1)’<1—2)=1 Because all digits are independent, 2 Q
sy(m) = We)» [1+2(2Qi)2[zcosnwn]] Tb n=l 11.28 Approximate impulses by rectangular pulses each of heighth and width 5 such that he  1 and e —> 0 (Fig. 51 1.283)
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 ET: 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 Px2xl W') " Px, (h) = as I
R,‘(r)=«tzzh.‘::=ar2 ' r>0
Hence, R,(r) s a6(r)+a2 Fig. 511.243 11.29 In this case the autocorrelation function at r = Oremain same as in Prob “28. 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) 11.210 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. 811210. 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 Poissondistributed. The amplitude after a shit! is independent of the amplitude prior to the shin. The ﬁrstorder 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 (x2x‘ = 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(x2xl = 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 (x2x, = 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 ﬁrstorder probability density of the process. This is obviously p(x). Substituting Eqs. (3) and (4) in Eq. (2), we get
Px1(x2xl = x1) ‘ {#502 " xx) 419“)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]
= eprblxipxr (MM! +(eﬂ' '1)I:"PXI(")¢‘J:’ZP‘2 (x2)dx2] = {#1: + (eﬂ’  l)i2] where i and X2 are the mean and the meansquare 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 11.31 For any real number a, (ax — y)2 2 0
azx2 +y2 2axy 2 0
Therefore the discriminant of the quadratic in a must be nonpositive. .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) —_____—.———— 11.33 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 “33, 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 RR2C +(R1+R2) I Risa; C
1/ jazC Rpm; A
(b) H w = = . a. T
( ) _L_+_RJ.£2_ Ja’CRlR2+(RI+R2) 45333}
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) 11.43 (a) We have found Rx(t) of impulse noise in Prob. 1128
R,‘(r) = (16(1) 4» a2,and Sx(m) = a +2xa26(m) Hence,
Sy(w)  llI(m)2[a +21ra26(a))] = 21ra2H(0)2 6(m)+ aH(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 +—2r—.e W ’_
I I n I X“) m y“) W
1—» I”
Fig. 31143 ' 11.51 n(t) = nc(t)cosmct + n,(t)sinwct The PSD oan!) and n,(t) are identical. They are shown in Fig. 811.51. 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.
11.51 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 camer 40 K Fig. $11.52 y» Fig.511.53
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 timedelay 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 timedelay 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 S306 taught by Professor Dandakar during the Spring '08 term at Drexel.
 Spring '08
 Dandakar

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