Optimal Data and Training Symbol Ratio for Communication over Uncertain Channels

Optimal Data and Training Symbol Ratio for Communication over Uncertain Channels Ather Gattami Ericsson Research Stockholm, Sweden Email: athergattami@ericssoncom arxiv:50502997v [csit] 2 May 205 Abstract We consider the problem of determining the power ratio between the training symbols and data symbols in order to maximize the channel capacity for transmission over uncertain channels with a channel estimate available at both the transmitter and receiver The receiver makes an estimate of the channel by using a known sequence of training symbols This channel estimate is then transmitted back to the transmitter The capacity that the transceiver maximizes is the worst case capacity, in the sense that given a noise covariance, the transceiver maximizes the minimal capacity over all distributions of the measurement noise under a fixed covariance matrix known at both the transmitter and receiver We give an exact expression of the channel capacity as a function of the channel covariance matrix, and the number of training symbols used during a coherence time interval This expression determines the number of training symbols that need to be used by finding the optimal integer number of training symbols that maximize the channel capacity As a bi-product, we show that linear filters are optimal at both the transmitter and receiver NOTATION deta) deta) = i λ i, where λ i } are the eigenvalues of the square matrix A A B denotes the kronecker product between the matrices A and B E } Ex} denotes the expected value of the E } stochastic variable x Ex y} denotes the expected value of the stochastic variable x given y cov covx, y} = E xy } hx) Denotes the entropy of x hx y) Denotes the entropy of x given y Ix; y) N m, V ) A Background Denotes the mutual information between x and y Denotes the set of Gaussian variables with mean m and covariance V I INTRODUCTION This work considers the problem of determining the power ratio between the training symbols and data symbols in order to maximize the channel capacity for transmission over uncertain channels with channel state information at the transmitter and receiver While the problem of MIMO communication over a channel known at the transmitter and receiver is well understood, the problem of uncertain channels still needs a deeper understanding of how much power we should spend on estimating the channel in order to transmit as much data as possible This problem poses a trade-off between the power ratio allocated for training and data transmission On one hand, if we spend more power on training and less on data transmission, the data throughput will be small of course On the other hand, if we spend most of the power on data transmission and much less on training, the channel estimate will be bad and therefore, the channel estimation error noise will be large, causing a rather low data rate Given per symbol power constraints, one might need a number of training symbols in order to achieve a certain quality of channel state information, which leaves a smaller number of symbols for data transmission This is an important constraint that we take into account in this work B Previous Work There has been a lot of work on MIMO communication in the context of uncertain channel state information The seminal paper of Telatar [7] studied the problem of communication over uncertain Gaussian channels under the assumption that the channel realization is available at the receiver but not the transmitter However, in practice, the transmitter and receiver don t have full knowledge of the channel The work was extended in [6] for the case of slowly varying channels The channel estimation extension of [7] was presented in [3], where the problem of power ratio between the training and data symbols was studied under average and per symbol power constraints over the channel coherence timethe time where the channel is roughly constant) The crucial assumption of average power constraint allows for spending only one symbol on training, since the power is only limited by the total power resource available during the coherence time For the case of per symbol power constraints, the power ratio is harder to compute The case where a channel estimate is available at the transmitter and reciever was studied in [] under the specific signaling strategy of zero-forcing linear beamforming and given Gaussian measurement noise Calculation of the channel capacity with respect to a channel estimate available at the receiver only was given in [2] C Contribution We consider communication over uncertain channels with channel state feedback at the transmitter The receiver makes

an estimate of the channel by using a known sequence of training symbols This channel estimate is then transmitted back to the transmitter The capacity that the transceiver maximizes is the worst case capacity, in the sense that given a noise covariance, the transceiver maximizes the minimal capacity over all distributions of the measurement noise under a fixed covariance matrix known at both the transmitter and receiver The channel estimation error implies that the covariance of the transmitted symbols over time affects both the covariance of the transmitted information symbols and the total noise covariance, which makes the signal structure more complicated, where it s not clear if the symbols should be uncorrelated over time For the single input multiple output SIMO) channel, we give an exact expression of the channel capacity as a function of the channel covariance matrix, the noise covariance matrix, and the number of training symbols used during a coherence time interval This expression determines the number of training symbols that need to be used by finding the optimal integer number of training symbols that maximize the channel capacity Numerical examples illustrate the trade-off between the number of training and data symbols The results indicate that when the transmission power or equivalently the signal to noise ratio) is high, a smaller number of training symbols is required to maximize the capacity compared to the low transmission power case We confirm these observations theoretically considering the asymptotic behavior of the power as it grows large or decreases to very small values Proof: Consult [5] Proposition 3 AM-GM Inequality): Let a, a 2,, a n be n nonnegative real numbers Then n a i n n a i n with equality if and only if a = a 2 = = a n Proof: The proof may be found in most standard textbooks on Calculus III PROBLEM FORMULATION Let H be a random channel such that vec H) CN 0, C) The realization of the channel is assumed to be constant over a coherence time interval corresponding to a block of T symbols Let xt) CN 0, X) be the transmitted symbol over the Gaussian channel at time t The received signal at time t is given by yt) = Hxt) + wt) where wt)} is random white noise process, independent of H and x, with zero mean and covariance given by E wt)w t)} = I m known at both the transmitter and receiver Without loss of generality, we assume that W is invertible Let x),, xt )) be the block symbols transmitted within the channel coherence time T The average power constraint imposed on the block is given by T T E xt) 2} P t= The power constraint above could allow for some symbols xt) to have a larger power than P However, in the real world, we have hard constraints on the peak average power per symbol, so we will impose power constraints per symbol given by E xt) 2} P, for t =,, T II PRELIMINARIES Definition Kronecker Product): For two matrices A Suppose that we allocate training symbols for the channel estimation part and T d = T symbols for data transmission If a performance criterion is measured over the total R m n and B R p q, the Kronecker product is defined as channel coherence time T, then clearly the optimal strategy a B a 2 B a n B is to transmit the training symbol sequence first followed a 2 B a 22 B a 2n B by the data symbols Also, when no energy constraints are A B = present, the optimal power allocation is for each symbol to be transmitted with full power P However, it s not clear what a m B a m2 B a mn B the optimal ratio between the training and data symbols That is, what is the optimal choice of T Proposition Mutliplication Property of the Kronecker Product): τ to maximize the channel capacity? To this end, we will derive the exact expression of For any set of matrices A, B, C, D, we have that the channel capacity SIMO channel A B)C D) = AC BD A SIMO Channel Estimation Proof: Consult [4] Consider a random Gaussian channel H taking values in Proposition 2 Determinant Property): For any two matrices R m Let x τ be a deterministic training symbol known at the A R m n and B R n m, we have that detab + I m ) = detba + I n ) transmitter and receiver with x τ 2 training sequence is given by = P The transmitted xt) = x τ, for t =,, At the receiver, we obtain the measurement symbols yt) = Hx τ + wt), Introduce the vectors y) y2) y τ =, w τ = y ) for t =,, w) w2) w )

and let the covariance matrix of w τ be E w τ wτ } = I m I Tτ = I mtτ It s well known that the optimal estimator Ĥ of H given y τ that minimizes the MSE is given by where Ĥ = E H y τ } ) = cov H, y τ ) cov y τ, y τ ) y τ 2) = E Hyτ } E y τ yτ } y τ 3) = x τ C P C + ) I m yt) 4) t= = x τ C P C + ) I m Hx τ + wt)) 5) wt) = t= wt) The covariance of wt) is easily obtained from the expression above, and it s given by E wt) w t)} = I m The channel estimation error is given by H = H Ĥ 6) = H x τ C P C + ) I m Hx τ + wt)) 7) = I m P C P C + ) I m )H x τ C P C + ) I m wt) 8) = P C + ) I m H x τ C P C + ) I m wt) 9) Now we have that } Ĉ = E Ĉ 0) = P C P C + ) I m C ) It s well known that Ĥ and H are independent since H and y τ are jointly Gaussian Thus, B Channel Capacity C } = E H H 2) = E H Ĥ)H Ĥ) } 3) = C Ĉ 4) = C P C P C + ) I m C 5) = I m P C P C + ) I m )C 6) = P C + I m ) C 7) In this section, we will derive a formula for the channel capacity under the assumption that the transmitted and receiver have a common estimate of the channel The estimate is the optimal estimate that minimizes the expected value of the variance of the estimation error Suppose that we transmit training symbols during the time interval t =,, and T d = T data symbols during the time interval t = +,, T The received noisy measurements of the data symbols are given by yt) = Hx d t) + wt) 8) = Ĥx dt) + Hx d t) + wt) 9) = Ĥx dt) + vt), 20) for t = +,, T Note that vt) = Hx d t) + wt) is uncorrelated with x d t) and Ĥx dt) jointly Introduce the vectors y + ) x d + ) y + 2) y d =, x x d + 2) d = yt ) x d T ) w + ) v + ) w + 2) v + 2) w d = wt ), v d = vt ) and let the resepective covariance matrices of w d and v d be and W d = E w d wd} = I m I Td = I mtd V 0 0 V d = E v d vd} 0 V 0 = V I Td = 0 0 V The worst case capacity of the channel with repsect to the measurement noise I mtd is given by C ) = x d E x d t) 2 }=P inf Ix v d ; y d ) 2) d Ev d v d }=I mt d

Thus, the problem that we want to solve is as follows Problem : Find the optimal integer [, T ] such that C ) = is maximized x d E x d t) 2 }=P inf Ix v d ; y d ) d Ev d v d }=I mt d for i, j =,, T Thus, Ĥ d x d x dĥ d = x dx d ) ĤĤ ) and given Ĥ, we get E Hd x d x H } d d = X d Ĉ Similarly, we get Hd x d x H d d = x dx d ) H H ) and E Hd x d x H } d d = X d C IV MAIN RESULTS It s well known that for a deterministic channel Ĥ d signal measurement y d = Ĥdx d + v d and This gives in turn V d = E v d v d} = I mtd + X d C In particular, the block diagonal elements of V d are equal, with the block elements given by with measurement noise v d uncorrelated with the transmitted signal x d, the optimal transmitting strategy is for x d to be Gaussian in order to minimize the worst case noise v d which is also shown to be Gaussian In other words, the Gaussian input and noise form a Nash equilibrium The case when the noise v d has an arbitrary covariance matrix V d has been solved in [3] In our case, the situation is different The covariance matrix of v d has a structure that also depends on the choice of the covariance matrix of the transmitted signal x d More precisely, recall that we have assumed that wt)} is a temporally uncorrelated noise process with arbitrary distribution Introduce Ĥ 0 0 Ĥ d = Ĥ I 0 Ĥ 0 T d = 0 0 Ĥ and H 0 0 H d = H 0 H 0 I Td = 0 0 H We may write w d = v d H d x d, with v d uncorrelated with Ĥ d and x d Furthermore, E v d vd} E Hd x d x H } d d = E w d wd} = I mtd Let X d = E x d x d } Then we have Ĥx d + ) Ĥx d + 2) Ĥ d x d = Ĥx d T ) and the blocks of Ĥdx d x dĥ d at position i, j) will be given bysince x d + i) is a scalar for all i =,, T ) [Ĥdx d x dĥ d] ij = Ĥx d + i)x d + j)ĥ = x d + i)x d + j)ĥĥ, V = I m + E HP H } = I m + P C Now the cost given by 2) may be written in terms of v d instead That is, C ) = E x d t) 2 }=P inf v d Ix d ; y d ) 22) Ev d v d }=I mt d +X d C Theorem : Consider a communciation channel given by yt) = Hxt) + wt) with H taking values in R m and H CN 0, C), t =,, T Let Ĥ be the channel estimate that is available at both the transmitter and receiver, based on training symbols Under the power constraint E x 2 t) } P, the worst case capacity C ) is given by C ) = T )log 2 detp C + I m ) log 2 detp C + I m )) with C = P C + ) I m C Furthermore, the optimal } receiver is given by the linear estimator E x d Ĥx d + v d with v d CN 0, P C ) Proof: The proof is deferred to the appendix Theorem can be checked for the extreme cases of = 0 and = T For the case = 0, clearly no channel estimation is obtained and the expression of C reveals that this error becomes infinite Thus, the capacity becomes and no information may be transmitted The other extreme, = T, means that all power is spent on channel estimation and no data may be transmitted We see that the expression of the channel capacity formula gives zero capacity, agreeing with the physical model V NUMERICAL EXAMPLES Example : Consider a communication SIMO channel H with perfect feedback and a randomly generated covariance matrix ) 07426 07222 C = 07222 64075 The length of the block is assumed to be T = 00 Figure shows that the optimum number of training symbols is = 4 when the power cosntraint is given by P = 00, whereas for P = 00, Figure 2 shows that the capacity is maximized for = 27, which is 27% of the available transmission power

capacity bits/s) capacity bits/s) 000 3 900 800 25 700 2 600 500 5 400 300 200 05 00 0 0 0 20 30 40 50 60 70 80 90 00 number of training symbols 0 0 0 20 30 40 50 60 70 80 90 00 number of training symbols Fig A plot of the 2 channel capacity as a function of the number of training symbols with tranmission power given by P = 00 Fig 2 A plot of the 2 channel capacity as a function of the number of training symbols with tranmission power given by P = 00 Example 2: Consider a communication SIMO channel H with perfect feedback and a randomly generated covariance matrix C given in the appendix The length of the block is assumed to be T = 00 Figure 3 shows that the optimum number of training symbols is = 2 when the power cosntraint is given by P = 00, whereas for P = 00, Figure 4 shows that the capacity is maximized for = 9 The example clearly shows that the number of pilots needed is smaller when the number of receiving antennas increases This is because of the increasement of the received signal power which makes the system less sensitive to channel estimation errors VI ASYMPTOTIC RESULTS The previous numerical examples clearly show the the dependence of the number of training symbols on the the symbol transmit power We may understand this relation by looking at the asymptotic results when the power P tends to 0 or infinity low respectively high signal to noise ratio) Recall that the covariance of the channel estimation error is given by C = P C + I m ) C Cleary, if P is very small and is small, then I m P C, and so P C + I m will be dominated by I m Thus C ) I m C = C as P 0 This implies that the capacity would tend to zero too Therefore, for small P, the number of training symbols must be large in order for I m to be of the same order as P C and so making C in some sense Now consider the other extreme, that is when P is large and recall the capacity formula C ) = T )log 2 detp C + I m ) log 2 detp C + I m )) In this case, we get I m P C Inspecting the formula for C again, we see that P C + I m will be dominated by P C and thus C P C) C = P Hence, ) log 2 detp C + I m ) log 2 det I m + I m T τ ) = m log 2 + Note also that for large P, we have that log 2 detp C + I m ) log 2 detp C)

capacity bits/s) capacity bits/s) 6000 60 5000 50 4000 40 3000 30 2000 20 000 0 0 0 0 20 30 40 50 60 70 80 90 00 number of training symbols 0 0 0 20 30 40 50 60 70 80 90 00 number of training symbols Fig 3 A plot of the 0 channel capacity as a function of the number of training symbols with tranmission power given by P = 00 Fig 4 A plot of the 0 channel capacity as a function of the number of training symbols with tranmission power given by P = 00 So for large P, the capacity may be well approximated by ) C ) T )log 2 detp C) m log 2 + ) We see that the capacity increases linearly with decreasing while it increases logarithmically with increasing Therefore, the optimal choice of the number of training symbols as P VII CONCLUSION We considered the problem of deciding the power ratio between the training symbols and data symbols in order to maximize the channel capacity for transmission over uncertain channels with channel state information at the transmitter and receiver We considered the worst case capacity as a performance measure, where the transceiver maximizes the minimal capacity over all distributions of the measurement noise with a fixed covariance known at both the transmitter and receiver We presented an exact expression of the channel capacity as a function of the channel covariance matrix, the noise covariance matrix, and the number of training symbols used during a coherence time interval This expression determines the number of training symbols that need to be used by finding the optimal integer number of training symbols that maximize the channel capacity We also showed by means of numerical examples the trade-off between the number of training and data symbols The results indicate that when the transmission power or equivalently the signal to noise ratio) is high, a smaller number of training symbols is required to maximize the capacity compared to the low transmission power case We confirm these observations theoretically considering the asymptotic behavior of the power as it grows large or decreases to very small values Future work considers the general MIMO case, which is more involved due to combinatorial issues arising in choosing the number of training symbols for different transmitting antennas REFERENCES [] G Caire, N Jindal, M Kobayashi, and N Ravindran Multiuser MIMO achievable rates with downlink training and channel state feedback Information Theory, IEEE Transactions on, 566):2845 2866, June 200 [2] G Fodor, P Di Marco, and M Telek Performance analysis of block and comb type channel estimation for massive mimo systems In 5G for Ubiquitous Connectivity 5GU), 204 st International Conference on, pages 62 69, Nov 204 [3] B Hassibi and B M Hochwald How much training is needed in multiple-antenna wireless links? Information Theory, IEEE Transactions on, 494):95 963, April 2003 [4] R A Horn and C R Johnson Topics in Matrix Analysis Cambridge University Press, 99 [5] R A Horn and C R Johnson Matrix Analysis Cambridge University Press, 999 [6] T L Marzetta and BM Hochwald Capacity of a mobile multipleantenna communication link in Rayleigh flat fading Information Theory, IEEE Transactions on, 45):39 57, Jan 999

[7] I Emre Telatar Capacity of multi-antenna Gaussian channels European Transactions on Telecommunications, 06):585 595, 999 Proof of Theorem and Define C = C = APPENDIX inf v d CN 0,V d ) y d CN 0,X d C+W d ) V d =I mtd +X d C inf v d CN 0,V d ) Ex d t)x d t)}=x d V d =I mtd +X d C Ix d ; y d ) 23) Ix d ; y d ), 24) Let the estimator at the receiver be a linear function of the received signal y d, that is the estimate of x d is given by ˆx d = L d Ĥdx d + v d ) for some matrix L d R T d mt d Since the receiver could be chosen nonlinear, we will get a lower bound on the capacity C ) Introduce x d = x d ˆx d Then Thus, C = Ix d ; y d ) 25) = hx d ) hx d y d ) 26) = hx d ) hˆx d + x d y d ) 27) hx d ) h x d ) 28) inf v d Ix d ; y d ) = hx d ) h x d ) 29) V d =I mtd +X d C which is attained for v d such that y d = Ĥdx d +v d is Gaussian, which implies in turn that ˆx d = L d Ĥdx d + v d ) and x d are Gaussian, and x d is independent of y d, so equality holds in 28) Also, hx d ) is maximized for x d when it s Gaussian under a fixed covaraince Hence, C ) C Now consider an arbitrary receiver, that is not necessarily linear and pose that v d is Gaussian and independent of x d This gives the capacity upperbound C C ) We have that Ix d ; y d ) = hy d ) hy d x d ) 30) = hy d ) hv d x d ) 3) = hy d ) hv d ) 32) where the inequality 32) holds since v d is assumed to be independent of x d Since hy d ) is maximized when y d is Gaussian, we get C ) C = C Since we already have the inequality C ) C, we conclude that C ) = C, and clearly x d and y d Gaussian give the worst case capacity C ) = C Now let the eigenvalue decompositions of X d, C, and C be given by X d = UΣU, Σ = diagσ,, σ Td ), C = Ū ΣŪ, Σ = diag Σ,, Σ Td ), and C = Ũ ΣŨ, Σ = diag σ,, σ Td ) Then, the mutual information between the Gaussian input x d and Gaussian output y d satisfies Ix d ; y d ) = hy d ) hy d x d ) 33) = log 2 detx d C + I m I Td ) log 2 detx d C + I m I Td ) 34) = log 2 detx d C + I mtd ) log 2 detx d C + I mtd ) 35) = log 2 detu Ū)Σ Σ)U Ū) + I mtd ) log 2 detu Ũ)Σ Σ)U Ũ) + I mtd ) 36) = log 2 detu Ū) U Ū)Σ Σ) + I mtd ) log 2 detu Ũ) U Ũ)Σ Σ) + I mtd ) 37) = log 2 detσ Σ + I mtd ) log 2 detσ Σ + I mtd 38) Td ) = log 2 detσ i Σ + Im ) Td ) log 2 detσ i Σ + Im ) Td = T d log 2 T d detσ i Σ + Im ) detσ i Σ + Im ) ) T d detσ i Σ + Im ) T d log 2 detσ i Σ + Im ) 39) ) T d 40) 4) where 36) follows from Proposition, 37) follows from Proposition 2, 38) follows from Proposition, and 4) follows from Proposition 3, with equality if and only if σ = σ 2 = = σ Td Since σ + σ 2 + + σ Td = Tr Σ) = Tr X) = np, we must have σ = σ 2 = = σ Td = P This implies that the capacity maximizing input covariance is X = P I Td Thus, the maximum capacity is given by C ) = log 2 detp I Td ) Ĉ + C) + I m I Td ) log 2 detp I Td ) C + I m I Td ) 42) = T d log 2 detp C + I m ) log 2 detp C + I m )) 43) = T )log 2 detp C + I m ) and the proof is complete log 2 detp C + I m )) 44)

Supplement for Example 2 The random matrix C used in Example 2 is given by 268 2535 22424 965 5896 2535 87639 20577 43889 207 22424 20577 60997 02384 894 965 43889 02384 7327 523 C = 5896 207 894 523 02643 25086 00 6082 03583 3222 45906 0499 45764 0705 077 398 373 239 5362 236 9345 2956 06284 22747 06889 40798 02636 05846 0775 27 25086 45906 398 9345 40798 00 0499 373 2956 02636 6082 45764 239 06284 05846 03583 0705 5362 22747 0775 3222 077 236 06889 27 2366 02868 08628 2528 3 02868 84323 09609 23883 20394 08628 09609 204969 05705 327 2528 23883 05705 97425 0307 3 20394 327 0307 335