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handout1 - EE544Q: Digital Communications Handout 1...

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Unformatted text preview: EE544Q: Digital Communications Handout 1 Representation of a narrowband random process Let X (t) be a WSS random process with zero _mean and power spectral density 5X(w) such that SX(w) : U for |wl 2 2wc. It may be that S'X(w) = 0 except for ma — wb < w < wc + wb, Where my, << we, but this is not required. Under this condition, it is possible to represent X (if) as X05) = Xc(t) cos(wct) — X305) sin(wct). Furthermore, if X05) is Gaussian, then both XCU) and X36) are Gaussian. Claim: ' (1) E[Xc(t)Xc(t + a] = we = war) (2) Ema—m + 1-)] x 3mm = 73mm 7 (3) E[X(t)X(t + 7)] : RX (T) = RXJT) cos(wc7*) — BXC,_X5(T) sin(wc'r) (4) SXW) = %[5Xc(w~wcl+3xc(w+wc)l+%i5xc,xs(w—wcl~5Xc,Xs(w+wc)l (5) chfiv) : S_X5(w) = LP[SX(w m we) + S_X(1U + 106)] (6) »9Xc,.x'5(wl : j ' LP[SX(w * wc) — 5M1” + wc)l Where LPH denotes an ideal lowepass filter with unit gain passing only fre— quencies from —wc to we. Proof: By using the trigonometric equalities 1 sin(a) cos(b) = §[sin(a + b) + sin(a — 5)], cos(a) sin(b) : %[sin(a + 5) " Sinta * bl]: cos(a) cos(b) : $[cos(a — b) + cos(a + 5)], sin(a) sin(b) = %[cos(a — 5) — C0301 + blla 1 the autocorreiation function of X (t), 3X01"), can be expressed as E[X(t)X(t + a} RXC (T) cos(wct) cos[wc[t + T)] + 3X50") sin(wct) sin[wc[t + 7-)] fiRXDXa (7‘) cos(wct) sin[wc(t + 7)] — RXMXCU) sin(wct) cos[wc(t + éRxc(T){cos[wc(2t + 7)] —i— cos[wc'r)} + iRX5(1-){cos(wc7) a cos[wc(2t + MgRXC'XJTHsiMwCQt + a] + sin(wcr)} — $RXS,XC(T){sin[wC(2t + TH — 5311mm} Rx(7‘) Fl %[RXC(T) + RX5(T)] c0s(wCT) —— %[RXC’X5(T) — RXS,XC(T)]Sin(1UcT) gem) — we] cosiwcflt + 7)] —‘ alarm) + R.X5,XC(T)]SiI1[wc(2t + 1-)]. Since X (if) is WSS, RX(T) is independent of if, and from the last expression of RX[T) we have ' RXJT) : 3X47), RXCXJT) —RX5,XC(T)= and 34(7) = RXC (7') cos(wcr) — RXCIXJT) sinfingr). We have thus proven (l), (2), and Since RX(7') : RXC (T) cos(wc’r) — RXC,X5(T)Si1’l('LUCT) 1 1 : RXC(T)§(ejwcT + e-jwc'r) _ RXE,XS(T)2_j(€waT i €_jwcT)7 we have 1 . SKW) I §[5Xc(w “ W) + SKEW + 7«Mi + %[5X0,Xs (w — we) * 31cm (10 + chJ by the shift property of the Fourier transform. Claim (4) is thus proved. Also since S_X(w) = 0 for lw| 2 2106, we have SXCCLU) I 0 and chzxjw) = 0 for |w| 2 we. Hence S'Xn(w *- 2wc) : SXJw + 21%) = SXCJSCLU — 21190) = SXCIXJw + 210C) 2 0 for Jw| S we. Then for |w| S we, 5X03 7 we) and 5X(w + we) are given by and 1 . S_X(w + (LDC) I + é‘chle respectively. Claim (5) and (6) can then be proved. ...
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handout1 - EE544Q: Digital Communications Handout 1...

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