In this section, we relate each of the scheduling criterion with the bounds on V-BLAST capacity. The best criterion technique is the one that is able to provide tighter bounds on
where is the minimum i th diagonal value of of the n th layer of the k th user. In this section, we will find bounds that are independent of n and i for each scheduling technique.
MaxMinSV
Choosing a user with maximum minimum singular value of H
k
is the same as choosing a user with maximum minimum eigenvalue of . We have
This means that we are looking for the minimum i th diagonal value of . We can write it in the form of the following inequality:
(13)
Since H
n
HH
n
is a squared Hermetian matrix, it can be written as
(14)
where U is a unitary matrix with orthonormal eigenvectors and Λk,n is the diagonal matrix of eigenvalues . Another approach is to use QR decomposition as done in [18].
Using simple matrix manipulations, the inverse is written as
(15)
The diagonal elements of are . Therefore,
(16)
(17)
For the sake of simplicity, we will henceforth refer to and as and , respectively. We now have
(18)
(19)
According to the inclusion principle for matrices, the minimum value of λmin occurs at n=1 and the minimum vale of λmax occurs at n=MT.
Now we have an upper bound given by
(20)
or by taking the minimum over K users
(21)
We will use this inequality to establish lower bounds on MaxMinSV.
For analytical tractability, we will focus on the case when MT=MR. The probability density function (pdf) of the smallest eigenvalue for the case of MT=MR is given by [19]
(22)
In this case, and , where μ and σ2 represent the mean and variance. From chapter 10 of [20], we see that the maximum of λmin will scale as . Therefore, we have a new bound on the capacity of VBLAST as
(23)
MinES
In this case, we start with the inequality
(24)
Here κ(k,n) is the condition number of the matrix defined as λmax/λmin. Now
(25)
The condition number κ is maximum when n=1, a lemma which follows from the inclusion principle of matrices. Therefore, we have
(26)
(27)
The pdf of s, the condition number of H
k
for the case of MT=MR, is given by [19]
(28)
As κ=s2, we can show using a variable transformation argument that
(29)
Now we find the upper bound on mink κ(k). Gordon et al. [21] provide bounds on inequalities for the minimum of random variables. These inequalities exist if a random variable ξ satisfies the (α, β) condition defined as
(31)
where α>0 and β>0.
Then, the random variable ξ is bounded as
(32)
Using the pdf of κ, we find that . Now the upper bound on the minimum is given by
(33)
Now we have
(34)
MaxSNR
We start with the established inequality
(35)
Again from the inclusion principle, the minimum of the trace occurs at n=MT.
Since we know the last layer, the pdf of for the last n becomes conditional. If we solve for the unordered VBLAST algorithm, we will get
(36)
where h
k
is a row vector containing MR i.i.d complex Gaussian entries. In this case, the trace will be a chi-squared random variable with 2MR degrees of freedom. The pdf in this case is given as
(37)
Now we use an upper bound on the maximum of the trace. Bertsimas et al. [22] provide us with a tight upper bound:
(38)
where μ and σ2 are mean and variances, respectively. In this case, μ=2MR and σ2=4MR. Now we have
(39)
Here we define an upper bound on capacity: