ADuenclix A Solutions to the Exercises 550 Solution 17.12 + + a) E{(.x(t) g ( l ) ) ’ } = E{:x2(t) + r ( l ) y ( t ) y ” t ) } = + = + E(s2(t)} 2E{r(t)y(t)t } E{&)} = E { ( ) } E{yZ(t)} To sitt’isfy the condition: E{x(t)y(t)} = + b) ‘ ’ ~spccial case of r = i n (17.48),y ~ ~ , = , ; h’{.?;.(t z)w(t)} (17.46) a i d /is = E ( : r ( t ) }yield: E{&)Y(~))= W4t)) E{y(tj) c) Eitbcr E { z ( l ) )or b ; { g ( t ) } ! or both expected expected values must be zero (at, least onc of Ihe randoiii signal inust have zero rrieaiz) Solution 17.13 LVith dctcriniriislic signal, t h e quantities in qurstioii must IF calculated with tiiueaverages, because the enserriblr means rorrcspond to the signal ilself Power: lirri - Power of t’he ac coiiiponenl: (d(L) - $o)z Solution 17.14 Forinirig the time-average according t,o ( 17.18) is linear With p = cl(t),tlierefore: ~~ ( d ( t ) p)’i = & ( t )- 2pd(1) + p2 = @ ( t )- 2/rd(t) ~ + /? = q i j - j62 ser derivat’ioii of ( 17.8) Solution 17.15 ,vL:!,( r) = E{(X(t) ~ = E{z(t)~ p z ) (y(t ~ ~ p y ) >= (- tC} - p,n-E{y(t - T)} - /i, F:{z(t)}+ /~~,!i.~ = vXg( Z) b:xpy Solution 17.16 For t 4cm a,riy c:liosen random processes are gcnern.lly iincorrcla.tcd: ~