ELEC546 MIMO Channel Capacity

Transcription

1 ELEC546 MIMO Channel Capacity Vincent Lau Simplified Version.0 //2004

2 MIMO System Model Transmitter with t antennas & receiver with r antennas. X Transmitted Symbol, received symbol Channel Matrix (Flat Fading) h ht r t H = mn h r h tr t Y // r 2 ( σ ) Y= HX+ Z Z~ CN 0, z Ir 2 mn = h ~ i.i.d. complex Gaussian with zero mean and E h

3 MIMO Channel Capacity Similar to the scalar case, we have Capacity with CSIT and CSIR Capacity with CSIR only Capacity with CSIT only Capacity with no CSIT and no CSIR We will consider the first two cases in MIMO //2004 3

4 MIMO Channel Capacity (CSIT & CSIR) Based on the SVD on channel matrix H=UDV where U,V are NrxNr and NtxNt unitary matrices, D= diagonal matrix (with N=min(Nr,Nt) non-zero singular values. = + = + Y HX Z UDV X Z Y = U Y= DX + Z where X = V X and Z = U Z y dx z 2 Y = and E =E = + ZZ U ZZ U = σ zin y N dnx N z N //2004 4

5 Capacity of Vector Channel The channel capacity of the vector channel is given by: N N 2 dne C = E max I( X,..., X N; Y,..., Y N d,.., d N) = E max I( Xn; Yn d n) = max log2 + p( x,..., x N d,.., dn ) p( x ) n dn E n= 2 n= n σ z The capacity achieving input distributions for ( x,..., x N ) are i.i.d. Gaussian with variance E n satisfying the N constraint: n= E n = E The optimization problem is given by: 0 N 2 N dne n Select ( E,..., EN) such that L( µ, E,..., EN) = log2 + E is maximized 2 n n= σ z µ n= n //2004 5

6 Water-Filling Solution The solution of the optimization problem: L/ E n = 0 E n µ σ z / dn if µ σn = 0 otherwise µ σ / d 2 2 z n σ / d 2 2 z n Transmitted Power Levels //2004 6

7 MIMO Channel Capacity (CSIT,CSIR) Structure of MIMO encoder with perfect CSIR & CSIT X = V X X= VX //2004 7

8 MIMO Channel Capacity (CSIT,CSIR) Structure of MIMO receiver with perfect CSIR & CSIT. Y= U Y= DX+ Z //2004 8

9 MIMO Channel Capacity with CSIR only Structure of MIMO Encoder Isolated encoding (t-branches) could achieve capacity of the MIMO channel Evenly distribute transmit power across the t-branches. Structure of MIMO Decoder No special structure could be derived for the case of perfect CSIR only. The optimal MIMO receiver is in general an ML decoder. Later on (in BLAST), we will see that MMSE-SIC receiver is also optimal. received codeword and n T N tx codeword // X ( yx ) y x R mˆ = arg max p ( m) where and (m) are the n N m

10 Evaluation of MIMO Capacity with CSIR only Nt = t (# of tx antennas) Nr = r (# of rx antennas) { } log INr ( / σ H z t ) C E P N HH 2 = + = P + 2 µ 0 0 σ zn t E { } log U 0 0 U H Two factors acting against each other 0 0 P a) Increase in the number of spatial 2 µ + m σ zn channels contribute to capacity t increase. m P b) Power splitting across the spatial = E{ } log + µ where U is eigenvector matrix of HH µ n 2 n channels capacity per spatial n= Ntσ z channel decrease At the end, factor a) wins //2004 0

11 Evaluation of MIMO Capacity with CSIR only Ergodic Channel Let t=r; Split power across t antennas capacity per spatial channel decreases # of spatial channel increases as t increases Overall capacity is increased as t increases. CMIMO m constant when t and r are very large //2004

12 Evaluation of MIMO capacity with CSIR only r C = E log ( P ) + hh = E log + P hi i= 2 //2004 2

13 Evaluation of MIMO Capacity with CSIR only P P C = E + = E + h r 2 log I hh log i nt n T i= //2004 3

14 MIMO Outage Capacity for slow fading channels For non-ergodic channels, the capacity itself is a random variable Outage probability: P C( ) = log r + hh h I ( ) ( ) 2 2 ntσ z Pout R = Pr{ C h R} //2004 4

15 MIMO Outage Capacity for slow fading channels //2004 5

16 MIMO Outage Capacity for slow fading channels What nt is very large, at a fixed nr, we have hh In n R T in probability by law of large number. Hence, the MIMO capacity as nt gets large is: T 2 ( h) log ( / ) Cn = nr + P σ z 0 Pout = Pr C( h) < R C C n n T T > R < R //2004 6

17 Example Questions Q) What happen to the MIMO capacity if elements of H is not i.i.d.? Can we still obtain linear increase in capacity w.r.t. # of antennas? Q2) Given a fixed power and fixed bandwidth, it seems that we can increase the capacity to an arbitrarily large value by installing more and more antennas on both sides. What do you think? Is there any limitation? //2004 7

18 Summary of Main Points MIMO capacity in fading channels - Enjoy linear capacity increases at the same power and bandwidth budget -Rely on the full rank of channel matrix scattering channels. -E.g. in free space, the capacity will not be increased linearly by using more tx and rx antennas Fast Fading Channels Slow Fading Channels CSIT and CSIR -Based on SVD of channel matrix, the MIMO channel is decomposed into m parallel spatial channels -Transmitter adapts the power allocation across the m spatial channels -Achieves higher ergodic capacity CSIR only -Capacity is a random variable cannot guarantee 00% reliability -Yet, as nt large, channel hardening the capacity the average capacity - packet outage 0 CSIR only // Optimal structure isolated encoding per transmit antenna -Uniform power allocation per transmit antenna -Large nt and nr capacity increases linearly with m=min(nt,nr)

19 Appendix Advanced Topics //2004 9

20 Appendix A: Background on Circularly Symmetric Complex Gaussian Complex Gaussian random vector x or xˆ To specify the distribution of, we need to specify the following: A complex Gaussian vector is circularly symmetric if x where Q is a Hermitian non-negative definite matrix The probability density of circular symmetric complex Gaussian vector x n with mean vector, µ = E[ x] and covariance matrix Q ( )( ) n n = E x µ x µ, is given by: n //

21 Background //2004 2

22 Appendix B: MIMO Channel Capacity with CSIT & CSIR (Alternative Way) MIMO Capacity with perfect CSIT & perfect CSIR C = E I( XY H= h) Capacity is given by: CSIR, CSIT h max ; p( xh) The mutual information is given by: I XY ; h = H Yh H YXh, = H Yh H Z ( ) ( ) ( ) ( ) ( ) Hence, maximizing the mutual information is equivalent to maximizing H(Y h). This is achieved by choosing the input distribution to be circularly symmetric complex Gaussian with covariance matrix: E xx h = Q ( h) The mutual information is thus given by: ( Xh ) ( XY h) = I + hqh max I ; log r p 2 2 σ z Hence, the optimization problem is equivalent to: Select Q to optimize log 2 Ir + hqh with the transmit energy constraint tr Q P 2 σ z // ( )

23 MIMO Channel Capacity Solution (Spatial & Temporal Water-filling) //

24 MIMO Channel Capacity Solution (Spatial & Temporal Water-filling) For i.i.d. channel matrix, the rank of channel matrix H is min(r,t) with probability. r r t t U ~eigenvector matrix of hh V ~eigenvector matrix of The mutual information could be expressed as: log I + hqh = log I + UDV QVDU = log I + D V QV D 2 r 2 2 r 2 2 r 2 σ z σ z σ z ( ) h h Since the input covariance matrix is Hermitian, it could be expressed by SVD as: Q = WQW where tr ( Q) = tr ( Q ) Since D is diagonal, observe that the determinant is maximized if (VQV) is diagonal. W=V. The mutual information is thus given by: min( tr, ) min( tr, ) λnq nn log2 Ir + DQD = log ~ eigenvalue of, 2 λn h h E [ q nn] P σ z n= σ z h n= //

25 MIMO Channel Capacity Solution (Spatial & Temporal Water-filling) The optimal solution is given by: q nn = n [,min( t, r)] n + 2 σ z µ λ tr + 2 ( log 2 ( µλn / σ z )) where µ is chosen to satisfy [ h nn] min(, ) C = E q P n= For the general case of block-fading channel (where an encoding frame spans over ergodic fading process), the solution has the general form of spatial and temporal water-filling. Energy is allocated in the temporal domain according to the strength of eigen-values. Within each fading block, energy is further distributed according to the individual eigen-values. // n

26 MIMO Channel Capacity For the special case of quasi-static fading channel, an encoding frame spans over a fading state only the solution consists of spatial water-filling. This is equivalent to the vector channel case {energy distribution across min(t,r) vector channels}. Structure of MIMO Transmitter with perfect CSIT & perfect CSIR. The capacity achieving input distribution X is circularly symmetric complex Gaussian with covariance matrix given by = E = This could be achieved by: Q xx VQV V QΨ r t t r r r r X = where Ψ ~ CN(0, It) or ~ CN(0, Ir) t V QΨ r > t t t t t t V ~ eigen-beam forming matrix, Q ~ power water-filling matrix //

27 Appendix C: Proof of MIMO Capacity with CSIR only Without CSIT, the input co-variance matrix could not be a function of H. The channel capacity with CSIR only is given by: hqh CCSIR = max Eh I( ; ) max E log2 r p( ) XY H= h = h I + : tr 2 X Q ( Q) P σ Since Q is non-negative definite, we could write Q=UDU where D is non-negative (real) and diagonal and U is unitary. ( hu) D( hu) hqh hdh f ( Q) = E log2 r E log h I + 2 = h 2 Ir + = E log 2 2 r + = f( ) h I 2 D σ z σ z σ z Hence, without loss of generality, we could consider non-negative diagonal Q. z //

28 MIMO Channel Capacity with CSIR only On the other hand, given any non-negative diagonal matrix Q and a permutation matrix T, consider = Since HT has the same distribution as H, we have f Q = f Q ( ) ( ) T Since log(det) is a concave function on the set of positive definite matrix, hence f(q) is also concave function of Q. Let Q = QT, we have f ( Q ) f ( Q ) = f ( ) T Q. t! T Observe that Q= αi optimal input distribution is t given by circularly complex Gaussian with covariance matrix E xx = ( P / t) I t t QT TQT //