ece 153 hw4

ece 153 hw4 - ECE I53 Homework #4— 1' sinfl 2 7.1 _ a)...

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Unformatted text preview: ECE I53 Homework #4— 1' sinfl 2 7.1 _ a) SX(f)=f[g =AT( £2) Table in Appendix B. 2 b) Sm) = g - 1LT Rx<r)=AW ) 2 7,3 Sy(t) = f[Rx(T)cos21rfor] j2fifoT+ -j2‘ll'foT = [+]] = ér-[Rmpfi'fw] + %T[Rx(r)e'fl”f°7] = 55,4 f — m + ism + fo) Note that Sy(f) = TG(f)G(f) where . T 2 am #61:}; ) and 1‘ 9(7) —-T - T T -'-RY(T) = T9(T)*9(T) = T /_ wg(A)g(r-—A)dA and ‘ EIY2(t)] Mo) = T /_:g<A)g(—A)dA TA: 9200(1) since g()\) = g(——A) 1 2T]: (1 — %)2d)\ 315? 3 va(T) = €[X(t + T)[X(t) — X (t - d)]] 7'7 a) = Rm) — Rx(‘r + d) _ mm = f[Rx(T)] - f[Rx(T + d)] = Sx(f) — Sx(f)e”""' b mm = £[(X(t + 'r) — X(t + 7' — d))(X(t) — X(t — d))] ) ‘ = RX(T)—Rx(T+d)—Rx(T—d)+RX(7-) = 2Rx(’T) — Rx(T + d) - Rx(T -- d) Sy(f) = 25X( f) — Sx(f)ej2”fd — Sx(f)e‘j2”fd = 2Sx(f)[1— cos 27rfd] 7.8 5W” = EIX (t)18[Y(t)] = Tnme 32“, t + T) = EIX (t)Y(t)X (t + T)Y(t + 7)] = SIX (M (t + T)]8[Y(t)Y(t + 7)] = Rx(T)Ry(T) &U>= flfififlmww=5An*&u) 7.19 a) The impulse response is 1‘,,1f,,1t-T,, h(t) = T£_T6(t)dt=T/_m6(t)dt—T m 6(t)dt 1 é.Twm—uu-Tn 0 T t 1 T . 11_e-j2rft lejzwf§1_e—jzarfg1 ' I -— _ ‘J2Wft =_ =__ —J21|’f2 Hm " T/oe dt T j27rf T j27rf e = lsin7rfT6_J-,rfl T 7rf 19) 5Y(f) = |H(f)l25X(f) . 2 ;T 531111127r72;2 5X) b) Sy(f) = |H(f)l25x(f) = N0 /2 1 + 47r2f2 No Rm) = P115an = —4—e-"' c) Ry(0) = :1? Noise Power SNR 7.25 Z(t) = X(t) — Y(t) a) wave + 7)] 52(f) b) 5 [Z (02] Also —fc+W fc+W /_ C_W SN(f)df+ [My SN(f)df f fc+W 2 ff KW SN(f)df a2/2 2&3va SN(f)df Y(t) = h(t) *X(t) 5[(X(t) — Y(t))(X(t + T) — Y(t + TD] Rx(7') + Ry(7') — Ryx(-T) — ny(-T) Rx(T) + Ry(T) —- RXy(T) — ny(—T) 5X(f) + SYU) — SXY(f) — 33w”) Sx(f) + |H(f)|25x(f) * H"(f)5x(f) — H(f)5x(f) |{1 - H(f)}|25x(f) Rxm) + Ry(0) — 2ny(0) £[X’(t)] + / /_°Oh(s)h(r)RX(s — r)dsdr —2 f: h(r)RX(—r)dr H II II II II 1 + |H(f)|2 - (H"(f) + H(f)) 5x(f) 2Re[H(f)l €[Z’(t)] = [:11 — H(f)|"7Sx(f)df 7.27 X(t) From Problem 7.25: r l -W 0 Sz(f) = 11- H(f)|25x(f) _ 0 if] < W ~ m m > W 00 4a a: = 21rf 2 W 4a2 + 47r2f2 df d3 = 27rdf 0° 1 dz 80/2 «w 4a2 + x2 5; 4d -1—— tan 7r 2a 20: 21W 2 [1r _1 1rW ‘ 5[Z2(t)] oo 4: ll _. I I 2' l 7.28 Y(t) = a cos(wc(t) + 6) + N(t) We will assume that G and N (t) are statistically independent: Ry = a cos wc'r + RN(T) , 2 2 Sm) = 7760 — 19+ 3—60‘ + is) + Sm) SN(f) —- -— _fc fc Signal Power = /_—ff:;w (“1260‘ — fc) + 01250 + fcl) df c 4 2 + (“1w — 12) + “Z60 + m) df a2 a2 a2 ‘ = I + I = ‘2’ Noise Power -.fc+W fc+W d j a W S~(f)df+ M) f __f _. f¢+w 2 f” Sn(f)df a2/2 2mm? SN(f)df lI SNR W 1 7.29 a) HQ) = g (5) e'jm" = 1 _ %e-j2«f 0° 1 n —j2« n 1 = e f = 1 _ lefjh'f N syu) = |H(f)l’-1:-° = _ 2 _ No/2 32m - |G(f)l Sm) — G _ cos 21 f) (% _ NW) N b) 5m (f) = Hum (f) = I: => Rwy(lc) = E29- "(’0 - . No/2 Swz(f) = H(f)G(f)Sx(f) = (1 _ %e-,-m_)(1 _ %e-j2,,) 1v0 No/2 ' = 1 _ %e—j2«f — 1 _ Ilie—flrf I: N I I: szw) = (No (i) ‘ ?° (i) )“("°) _ _______N°_/2______. _ ___.§j_v.°___ _ %N° c) Sz(f) - _cos21rf) (% _ _1icos2,rf) _ g— cos21rf -}§ — i-cos21rf From Problem 7.9 we know that " fllalk]% f: lave—W ; 1 ‘ “2 k=_°° 1+a2—2acos21rf 8 4 - 122(k) _ —No—.7-“1[ 7 3 2 1+6) —2( -5NoEf-l — _ 32 1|"I 64 1|"l -' #45) WW) €[ZZ] = Rz(0) . 32 64 - (E'TGE)N° = 16.1“, 7.38 For the random telegraph signal: 4 .S'x(f) = m I sym = $5M + fc) + 35x0 — fa) _ 2a 2a - m+w4a2+wmm ' 2c: . 2a 4a2('w + 21rf,_.)2 + 46!2 + (w — 2‘Irfc)2 -2061 ' —2a0 2a 402 20a 7.49 X(t) = aX(t1) + bX(t2) a) e(t) X(t) — X(t) aX(t1) + bX(t2) — Orthogonality condition implies that 5[(aX(t1) + bX(t2) - X (t))X (t1)] = 8[(aX(t1)+bX(t2)-X(t))X(t2)] = 0 => aRx(0) + bRx(t2 - t1) GRX(t1 — t2) + [I ll 5" A F.- l 5“ v g [ R3921) Right” l m = [ $23: :3 [ a 1 _ 1 [ Rx(0)Rx(t — t1) — Rx(t‘, — t1)RX(t'— t,) b _ R340) — Rut, .— t1) Rx(0)Rx(t — t2) — Rx(t2 — t1)Rx(t — t1) b) £[e’(t)] e[e(t)[ax(t1) + bX(t2) —X(t)n a£[e(t)X(t1)]+68[e(t)X(t2)l-€[e(t)X(t)l —a£ [X (t‘1)X (t)] — b£[X(t2)X + 8 [X (t)X (t)] stew = Rx(0) , ~ Rx(0)Rx(t — t2) — Rx(t2 — t1)Rx(t _ t1) ‘ " R}(0)—R§(ta—tl) R"("”) Rx(t - t1) Check: If t = t1 then a = 1, b = 0 and E[e’(t)] = 0. ...
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This note was uploaded on 11/11/2009 for the course ECE NA taught by Professor Najmabadi during the Spring '08 term at UCSD.

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ece 153 hw4 - ECE I53 Homework #4— 1' sinfl 2 7.1 _ a)...

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