Minimum BER Linear Transceivers for BlockCommunication SystemsLecturer: Tom LuoStanford University EE392o Z.Q. LuoOutline• Block-by-block communication– Abstract model– Applications– Current design techniques• Minimum BER precoders for zero-forcing equalization– Average BER andconvexity– Analytic solution–The Message: Minimum BER precoders are MMSE precoders with a special choiceof the unitary matrix degree of freedom– Performance Analysis: SNR gains of several dB• Extensions1Stanford University EE392o Z.Q. LuoBlock-by-Block CommunicationPSfrag replacementsModDetects(n)Hx(n)v(n)r(n)ˆs(n)+• a vector memoryless system• r(n) = Hx(n) + v(n)•Applications– block transmission over ISI channels (OFDM, DMT)– multiple antenna flat fading channels– multiple antenna frequency-selective channels2Stanford University EE392o Z.Q. LuoBlock transmission over ISI channelsPSfrag replacementsModCEDetects(n)Hv(n)ˆs(n)+S/P1:MP/SP:1h(n)S/P1:PP/SM:1• H is circulant if E and C are based on cyclic-prefix extensions• H is Toeplitz (and full rank) if E and C based on “zero-padding”• Relative advantages:– CP: channel indep. diagonalization via DFT; simple equalization– ZP: achieves maximum diversity without coding; no prefix3Stanford University EE392o Z.Q. LuoMultiple antenna flat fading channelsPSfrag replacementsModDetects(1), . . . , s(N)H (n)x(1), . . . , x(N)v(1), . . . , v(N)r(1), . . . , r(N)ˆs(1), . . . ,ˆs(N)+• Flat channels: r(n) = H (n)x(n) + v(n)•Space-time coding based on accumulating s(n) and r(n) as columns• If H is constant, R = H X + V•Take vec’s vec(R) = (I ⊗ H ) vec(X) + vec(V )•Hence H = I ⊗ H4Stanford University EE392o Z.Q. LuoHow should we communicate?PSfrag replacementsModDetects(n)Hx(n)v(n)r(n)ˆs(n)+Depends on• Structure of the transmitter and receiver– in single-user and multiple access transmitter is usually linear,x(n) = F s(n)– structure of the receiver sometimes fixed• Accuracy of channel knowledge at the transmitter and receiver• Design objective– rate maximization; performance maximization; power minimization5Stanford University EE392o Z.Q. LuoToday’s focusPSfrag replacementsFGHv1:MS/PP/SP/SS/PP:11:PM:1Detectoryrxs˜sˆs+DetectorDetectorDetector• Linear transmission• Receiver: linear equalizer with elementwise detection• Accurate channel knowledge at both ends• Performance orientated design• Uniform constellation assignment in s6Stanford University EE392o Z.Q. LuoCurrent performance orientated designsPSfrag replacementsFGHv1:MS/PP/SP/SS/PP:11:PM:1Detectoryrxs˜sˆs+DetectorFGHv1:MS/PP/SP/SS/PP:11:PM:1Detectoryrxs˜sˆs+DetectorDetectorDetectorTypically, maximize performance subject to power bound for uniform constellation• Minimize MSE (arithmetic mean of elementwise MSEs)• Minimize MSE subject to zero forcing equalization• Maximize total signal power to total noise power subject to zero forcing• Maximize arithmetic mean of elementwise SINRs• Maximize geometric mean of SINRs7Stanford University EE392o Z.Q. LuoOutline• Block-by-block communication– Abstract model– Applications– Current design techniques• Minimum BER precoders for zero-forcing equalization– Average BER andconvexity– Analytic solution–The Message: Minimum BER precoders are MMSE precoders with a specialchoice of the unitary matrix degree of freedom– Simulation results: SNR gains of several dB• Extensions8Stanford University EE392o Z.Q. LuoAnalysisPSfrag replacementsFGHv1:MS/PP/SP/SS/PP:11:PM:1Detectoryrxs˜sˆs+DetectorFGHv1:MS/PP/SP/SS/PP:11:PM:1Detectoryrxs˜sˆs+DetectorDetectorDetector•˜s = GHF s + Gv•Choose length(s) ≤ rank(H)•Choose G to be a zero-forcing equalizer: G = (HF )†.• Hence,˜s = s + (HF )†v•Design problem: Find F that minimizes BER9Stanford University EE392o Z.Q. LuoAverage bit error rate• Pe: average bit error rate over all possible transmitted vectors.Pe= E{Pe|s}•For BPSK/QPSK signals, Pecan be expressed asPe=1M2MXj=1PsjMXm=1mPm|sj,•Zero-forcing equalizer =⇒˜s = s + Gv. HencePe=12MXmerfcÃ1p2σ2[GGH]mm!where E{vvH} = σ2I, [GGH]iiis the ith diagonal entry of GGH, M is the blocksize.10Stanford University EE392o Z.Q. LuoKey Observation: Convexity• If φ(x) = erfc¡1√2σ2x¢, for x > 0, thend2φdx2=1√π(2σ2)−12exp³−12σ2x´³−32+12σ2x´x−52.• Hence, if x <13σ2, then∂2f(x)∂x2> 0.• Therefore, if1σ2[GGH]mm> 3then Peis a convex function of [GGH]mm11Stanford University EE392o Z.Q. LuoDesign of the Minimum BER Precoder• Our goal:minimizeFPesubject to trace(F FH) ≤ p0• In the region that Peis convex, applying Jensen’s inequality, we havePe=12MXmerfcÃ1p2σ2[GGH]mm!≥12erfcÃ1q2σ2MPMm=1[GGH]mm!=12erfcÃ1q2σ2(trace(GGH)M)!, Pe,LBequality holds if [GGH]mmare equal, ∀m ∈ [1, M].12Stanford University EE392o Z.Q. LuoDesign of the Minimum BER Precoder II• Pe,LBdefines a lower bound on Pe; minimum BER precoders can be designed in twostages.– Stage 1: MinimizePe,LBsubject to the power constraint and the convex condition.minimizeFPe,LBsubject to trace(F FH) ≤ p0[GGH]mm<13σ2, ∀m ∈ [1, M]– Stage 2: Show that a particular solution for Stage 1 achieves the minimized lowerbound.13Stanford University EE392o Z.Q. LuoSolving Stage 1• Parameterize via SVD: F = U [Φ0] V ; H = Q [Σ0] W•If number columns of F ≤ rank(H), then zero forcing equalizers exist• In those cases GGH= VHΦ−1Z(U)Φ−1V ,where Z(U) = [IM0] UHW Σ−2WHU£IM0¤• Recall that Pe,LB=12erfc³¡2σ2trace(GGH)/M¢−1/2´•Exploit: monotonicity of erfc(·);• Hence, minimizing Pe,LBis equivalent to minimizing trace¡Φ−2Z(U)¢14Stanford University EE392o Z.Q. LuoReformulation• Minimizing Pe,LBsubject to power and validity constraints is equivalent tominimizeU ,Φ,Vtrace¡Φ−2Z(U)¢subject to trace(Φ2) ≤ p0£VHΦ−1Z(U)Φ−1V¤mm<13σ2, ∀m ∈ [1, M]• Awkward due to the last constraint• Lemma: For symmetric A ≥ 0, with A = ΨΓΨH,minV VH=Imaxm[VHAV ]mm=trace(A)Mand a minimizing V = ΨD, where D is the DFT matrix.15Stanford University EE392o Z.Q. LuoRemaining Problem• To complete minimization of lower bound we must solve:minimizeU ,Φtrace¡Φ−2Z(U)¢subject to trace(Φ2) ≤ p0(2)• Solution is the MMSE precoder for zero-forcing equalizationU = W ; Φ = ΦMMSE-ZF=rp0trace(Σ−1)Σ−1/2where, for simplicity we assumed that length(s) = rank(H).• Hence, if trace¡Φ−2MMSE-ZFZ(W )¢> M/(3σ2) thenFmin, LB= W