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...The MSE matrix is then given by (3.31) and the MSEs by (3.32): MS[E.sub.i] = [[(vI + [P.sup.[dagger]][H.sup.[dagger]]HP).sup.-1]].sub.ii] , where v = for the ZF receiver and v = 1 for the MMSE receiver. Therefore, the problem of minimizing the transmit power subject to individual MSE QoS constraints as a function of the linear precoder P at the transmitter can be formulated as (recall that QoS constraints in terms of the SINRs and BERs can always be reformulated as MSE QoS constraints):

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (3.94)

which is a nonconvex problem for the same reasons as (3.44).

We will start by obtaining a suboptimal solution based on imposing a diagonal structure on the transmission. As shown in Section 3.3.2, this is accomplished by imposing the following form on the transmit matrix:

P = [V.sub.H][SIGMA], (3.95)

where [V.sub.H] is a (semi-)unitary matrix with columns equal to the right singular vectors of the channel matrix H corresponding to the L largest singular values and [SIGMA] = diag ([square root of p]) is a diagonal matrix containing the square-root of the power allocation p over the channel eigenmodes. Under such a diagonal structure, the expression for the MSEs simplifies to the scalar and convex expression:

MS[E.sub.i] = 1/v + [p.sub.i][[lambda].sub.H,i] 1 [less than or equal to] i [less than or equal to] L, (3.96)

where the [[lambda].sub.H,i]'s denote the L largest eigenvalues of matrix [H.sup.[dagger]]H. Problem (3.94) becomes then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (3.97)

which is a convex problem.

However, the simple reformulation in (3.97) need not be optimal in the sense that its solution need not be an optimal solution to the original problem formulation in (3.94). In the following, we provide a truly equivalent simple reformulation of the original complicated nonconvex problem (3.94) based on majorization theory (cf. Chapter 2 and [97]) as was originally derived in [114]. (15)

We will start with the simple case where the MSE QoS constraints are equal, [[rho].sub.i] = [rho] for all i, and then...

NOTE: All illustrations and photos have been removed from this article.



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