Chapter 10 - Kỹ thuật thông tin vô tuyến

Kỹ thuật thông tin vô tuyến

Chapter 10 (a) (AAH )T = (AH )T AT T = (AT ) AT = AAH ∴ (AAH )H For AAH , = AAH λ = λ, i.e eigen-values are real AAH = QΛQH (b) X H AAH X = (X H A)(X H A)H = X H A ≥ ∴ AAH is positive semidefinite (c) IM + AAH = IM + QΛQH = Q(I + Λ)QH AH positive semidefinite ⇒ λi ≥ 0∀i ∴ + λi > 0∀i ∴ IM + AAH positive definite (d) det[IM + AAH ] = det[IM + QΛQH ] = det[Q(IM + ΛM )QH ] = det[IM + ΛM ] Rank(A) = Πi=1 (1 + λi ) det[IN + AH A] = det[IN + QΛQH ] = det[Q(IN + ΛN )QH ] = det[IN + ΛN ] Rank(A) = Πi=1 (1 + λi ) AAH and AH A have the same eigen-value ∴ det[IM + AAH ] = det[IN + AH A] H = U ΣV T  −0.4793 0.8685 −0.1298 U =  −0.5896 −0.4272 −0.6855  −0.6508 −0.2513 0.7164    1.7034 0 0.7152  Σ= 0 0.1302   −0.3458 0.6849 0.4263  −0.5708 0.2191 0.0708   V =  −0.7116 −0.6109 0.0145  −0.2198 0.3311 −0.9017 H = U ΣV T Let   U =  0   V =  0 Σ= 0   0 ∴H=  0 Check the rank of each matrix rank(HI ) = ∴ multiplexing gain = rank(H2 ) = ∴ multiplexing gain = RH C= log2 + i=1 Constraint Vi = ρ λi ρ Mt λi = constant ∴ ∂C ρ 1 ρ = − =0 λi ρ ∂λi Mt ln (1 + ) Mt ln (1 + λi ρ ) Mt Mt ⇒ λi = λj ∴ when all RH singular values are equal, this capacity is maximized (a) Any method to  13 08 .05 09 D= 23 13 show H ≈ U ΛV is acceptable For example:  11 H −H 14 where : dij = ij ij × 100 H 10 (b) precoding filter M = V −1 shaping filter F = U −1   −.5195 −.3460 −.7813 F = −.0251 −.9078 4188  −.8540 2373 4629   −.2407 −.8894 3887 M = −.4727 −.2423 −.8473 −.8478 3876 3622 Thus Y = F(H)MX + FN = U ∗ U ΛVV ∗ X + U ∗ N = ΛX + U ∗ N (c) Pi P 1 1 γo − γi for γi > γo , P γi = λio B = 94.5 for i = 1, N Assume γ2 > γ0 > γ3 since = else 6.86 for i = 2, 68 for i = γ3 = 68 is clearly too small for data transmission Pi P 1 = ⇒ γ0 − γ1 − γ2 = ⇒ γ0 = 1.73 P1 P2 P = 5676 P = 4324 C = B log2 + γ1 P1 + log2 + γ2 P2 P P = 775.9 kbps (d) With equal weight beamforming, the beamforming vector is given by c = √1 [1 1] The SNR (3) is then given by: cH H H Hc = (.78)(100) = 78 N0 B SN R = (1) This gives a capacity of 630.35 kbps The SNR achieved with beamforming is smaller than the best channel in part (c) If we had chosen c to equal the eigenvector corresponding to the best eigenvalue, then the SNR with beamforming would be equal to the largest SNR in part(c) The beamforming SNR for the given c is greater than the two smallest eigenvalues in part(c) because the channel matrix has one large eigenvalue and two very small eigenvalues C = max B log2 det[IM γ + HRX H H ] RX : Tγ (RX ) = ρ If the channel is known to the transmitter, it will perform an SVD decomposition of H as H = U ΣV HRX H H = (U ΣV )RX (U ΣV )H n×n By Hadamard’s inequality we have that for A ∈ det(A) ≤ Πn Aii i=1 with equality iff A is diagonal We choose RX to be diagonal, say = Ω then det(IM R + HRX H H ) = det(I + ΩΣ2 ) ∴ C = max Bσi log2 (1 + λi ρi ) i √ where λi are the singular values ρi ≤ρ The capacity of the channel is found by the decomposition of the channel into RH parallel channels, where RH is the rank of the channel matric H C= ρi : max i ρi ≤ρ B log2 (1 + λi ρi ) i √ where λi are the RH non-zero singular values of the channel matrix H and ρ is the SNR constraint γi = λi ρ Then the optimal power allocation is given as Pi = P γ0 − γi γi ≥ γ0 γi < γ for some cut-off value γ0 The resulting capacity is given as C= B log2 (γi /γ0 ) i:γi ≥γ0 (2) For   H=  1  −1 1 −1 −1   1  1 −1 RH = 3, γ1 = 80, γ2 = 40, γ3 = 40 We first assume that γ0 is less than the minimum γi which is 40 γ0 = 1+ i=1 γi which gives γ0 = 2.8236 < mini γi , hence the assumption was correct C = 12.4732 bits/sec/Hz B For  1 −1  1 −1   H=  −1 1  −1 −1 −1  RH = 4, γ1 = 40, γ2 = 40, γ3 = 40, γ4 = 40 We first assume that γ0 is less than the minimum γi which is 40 γ0 = 1 + γi i=1 which gives γ0 = 3.6780 < mini γi , hence the assumption was correct C = 13.7720 bits/sec/Hz B  h11   H =    hMr  h1Mt      hMr Mt M r ×Mt Denote G = HH T  Gii = lim Mt →∞ Mt =   lim [hi1 hiMt ]   Mt →∞ Mt  Mt →∞ Mt       Mt lim = Ej hij hi1 hiMt hij j=1 = σ2 = ∀i lim Mt →∞,i=j Gij Mt    = lim [hi1 hiMt ]   Mt →∞ Mt  = lim Mt →∞ Mt hj1 hjMt       Mt hik hjk k=1 = Ek hik hjk = Ek (hik )Ek (hjk ) = ∀i, j, i = j HH T M ρ HH T + M ∴ lim M →∞ ∴ lim B log2 det IM M →∞ = IM = B log2 det [IM + ρIM ] = B log2 [1 + ρ] det IM = M B log2 [1 + ρ] 10 We find the capacity by randomly generating 103 channel instantiations and then averaging over it We assume that distribution is uniform over the instantiations MATLAB CODE clear; clc; Mt = 1; Mr = 1; rho_dB = [0:25]; rho = 10.^(rho_dB/10); for k = 1:length(rho) for i = 1:100 H = wgn(Mr,Mt,0,’dBW’,’complex’); [F, L, M] = svd(H); for j = 1:min(Mt,Mr) sigma(j) = L(j,j); end sigma_used = sigma(1:rank(H)); gamma = rho(k)*sigma_used; %% Now we water filling\ gammatemp = gamma; gammatemp1 = gammatemp; gamma0 = 1e3; while gamma0 > gammatemp1(length(gammatemp1)); gammatemp1 = gammatemp; gamma0 = length(gammatemp1)/(1+sum(1./gammatemp1)); gammatemp = gammatemp(1:length(gammatemp)-1); end C(i) = sum(log2(gammatemp1./gamma0)); end Cergodic(k) = mean(C); end 25 Mt = Mr = Mt = Mr =3 Mt = Mr = Mt = Mr =1 Mt = Mr = 20 Cergodic 15 10 0 10 15 20 25 ρ (dB) Figure 1: Problem 10 11 We find the capacity by randomly generating 104 channel instantiations and then averaging over it We assume that distribution is uniform over the instantiations MATLAB CODE clear; clc; Mt = 1; Mr = 1; rho_dB = [0:30]; rho = 10.^(rho_dB/10); for k = 1:length(rho) for i = 1:1000 H = wgn(Mr,Mt,0,’dBW’,’complex’); [F, L, M] = svd(H); for j = 1:min(Mt,Mr) sigma(j) = L(j,j); end sigma_used = sigma(1:rank(H)); gamma = rho(k)*sigma_used; C(i) = sum(log2(1+gamma/Mt)); end Cout(k) = mean(C); pout = sum(C 01 Cout(k) = Cout(k)-.1; pout = sum(C