Coordinated Multicell Beamforming Based on Power Minimization in an Uplink–Downlink Configured Massive MIMO Network

Abstract

In this paper, the coordinated beamforming problem is investigated to minimize the sum transmit power of the system in a two-cell network, downlink transmission in one cell and uplink transmission in the other cell. It is assumed that the base stations (BSs) and the users are equipped with multiple input multiple output (MIMO) antennas, and the coordinated beamforming vectors are designed for the BS in the downlink cell and the users in the uplink cell under the signal-to-interference plus noise ratio (SINR) constraints. Then, the problem is extended for a scenario where the BSs and users have massive MIMO antennas, and using the random matrix theory, an algorithm is proposed to design the coordinated beamforming vectors. Simulation results show that this coordinated beamforming algorithm for the massive MIMO scenario accurately tracks the optimal sum transmit power minimization beamforming approach, while it reduces the computational complexity of the coordinated beamforming.

Appendix

1.1 The proof of Theorem 1

In order to prove Theorem 1, we use the concept of standard function defined in Yates (1995) and Lemma 1.

Definition 1

A function $f: \mathrm {R}^{n}\rightarrow \mathrm {R}^{n}$, where for ${\mathbf {x}}=[x_1,\dots ,x_n]$, $f(x_{1},\dots ,x_{n})=[f_1({\mathbf {x}}),\dots ,f_n({\mathbf {x}})]^{T}$ is said to be a standard function if it satisfies the following conditions,

• 1. Positivity: If ${\mathbf {x}}=[x_{1},\dots ,x_{n}]^{T}$ and $\forall i\ x_{i}\ge 0$ then $\forall i$ $\ f_{i}({\mathbf {x}})\ge 0$.

• 2. Monotonicity: If ${\mathbf {x}}=[x_{1},\dots ,x_{n}]^{T}$ , ${\mathbf {y}}=[y_{1},\dots ,y_{n}]^{T}$ and for all i, $x_i \ge y_i$ then $\forall i \ f_i({\mathbf {x}})\ge f_i({\mathbf {y}})$.

• 3. Scalability: For all $\rho \ge 0$, and ${\mathbf {x}}$, $\forall i$ $\rho f_i({\mathbf {x}})\ge f_i(\rho {\mathbf {x}})$.

Lemma 1

If f be a standard function, then the fixed point iteration algorithm converges to a unique solution.

Proof

The proof could be found in Yates (1995). $\square$

Now, we prove that the function defined in Theorem 1 is standard and then its fixed point could be found using the fixed point iteration algorithm. Accordingly, we investigate the conditions of the standard function,

• 1. Positivity: It must be shown that for $i=1,2$, $\forall k$, $f_{i,k}( {\varvec{\lambda }}_{i},{\varvec{\lambda }}_{2})>0$

  $$\begin{aligned}&\forall k \lambda _{1,k}, \lambda _{2,k}>0\Longrightarrow {\varvec{\Sigma }}_{1}\succ 0\Longrightarrow \big ({\varvec{\Sigma }}_{1}+\mathbf{I}_{N}\big )\succ 0 \Longrightarrow \big ({\varvec{\Sigma }}_{1}+\mathbf{I}_{N}\big )^{-1}\succ 0 \Longrightarrow \\&\quad \forall k \ f_{1,k}( {\varvec{\lambda }}_{1},{\varvec{\lambda }}_{2})>0. \\&\quad \forall k \lambda _{1,k}, \lambda _{2,k}>0\Longrightarrow {\varvec{\Sigma }}_{2,k}\succ 0\Longrightarrow \big ({\varvec{\Sigma }}_{2,k}+\mathbf{I}_{M}\big )\succ 0 \Longrightarrow \\&\quad \big ({\varvec{\Sigma }}_{2,k}+\mathbf{I}_{M}\big )^{-1}\succ 0 \Longrightarrow \forall k\ f_{2,k}( {\varvec{\lambda }}_{1},{\varvec{\lambda }}_{2})>0. \end{aligned}$$

• 2. Monotonicity: It must be confirmed that,

  $$\begin{aligned}&\forall i,k \ \lambda _{i,k}>\lambda _{i,k}^{\prime } \Longrightarrow f_{i,k}( {\varvec{\lambda }}_{1},{\varvec{\lambda }}_{2}) >f_{i,k}( {\varvec{\lambda }}_{1}^{\prime },{\varvec{\lambda }}_{2}^{\prime }). \\&\quad {\varvec{\Sigma }}_{1}={\varvec{\Sigma }}_{1}^{\prime }+\big ({\varvec{\Sigma }}_{1}-{\varvec{\Sigma }}_{1}^{\prime }\big ) \Longrightarrow {\varvec{\Sigma }}_{1}=\\&\quad \sum _{m=1}^{K}\bigg (\frac{\lambda _{\mathrm{1},m}^{\prime }}{N}{\mathbf {H}} _{\mathrm{{1,1}},m} {\mathbf {v}}_{\mathrm{{1}},m} ^{DL,H} {\mathbf {v}}_{\mathrm{{1}},m} ^{DL} {\mathbf {H}} _{\mathrm{{1,1}},m} ^{H}+\frac{\lambda _{\mathrm{2},m}^{\prime }}{M} {\mathbf {H}} _{\mathrm{{1,2}}}^{H} {\mathbf {w}}_{\mathrm{{2}},m} ^{UL,H} {\mathbf {w}}_{\mathrm{{2}},m} ^{UL} {\mathbf {H}} _{\mathrm{{1,2}}}\bigg )+\\&\quad \sum _{m=1}^{K}\bigg (\frac{(\lambda _{\mathrm{1},m}-\lambda _{\mathrm{1},m}^{\prime })}{N}{\mathbf {H}} _{\mathrm{{1,1}},m} {\mathbf {v}}_{\mathrm{{1}},m} ^{DL,H} {\mathbf {v}}_{\mathrm{{1}},m} ^{DL} {\mathbf {H}} _{\mathrm{{1,1}},m} ^{H}+\frac{(\lambda _{\mathrm{2},m}-\lambda _{\mathrm{2},m}^{\prime })}{M} {\mathbf {H}} _{\mathrm{{1,2}}}^{H} {\mathbf {w}}_{\mathrm{{2}},m} ^{UL,H} {\mathbf {w}}_{\mathrm{{2}},m} ^{UL} {\mathbf {H}} _{\mathrm{{1,2}}}\bigg ). \end{aligned}$$

  As mentioned, $\lambda _{i,k}>\lambda _{i,k}^{\prime }$ and then ${\varvec{\Sigma }}_{1}-{\varvec{\Sigma }}_{1}^{\prime }\succ 0$; on the other hand, based on the Proposition 4 in Wiesel et al. (2006), for nonnegative matrices $\mathbf{C}$ and $\mathbf{D}$ and vector $\mathbf{x}$ in the range of $\mathbf{C}$, the following equation is satisfied,

  $$\begin{aligned} \dfrac{1}{{\mathbf {x}}^{H} \big ({\mathbf {C}}+{\mathbf {D}}\big )^{-1}{\mathbf {x}}}\ge \dfrac{1}{{\mathbf {x}}^{H} {\mathbf {C}}^{-1}{\mathbf {x}}}, \end{aligned}$$

  (41)

  then with defining ${\mathbf {x}}={\mathbf {H}}_{1,1,k} {\mathbf {v}}_{1,k}^{DL,H}$, ${\mathbf {C}}=\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}^{\prime }$ and ${\mathbf {D}}={\varvec{\Sigma }}_{1}-{\varvec{\Sigma }}_{1}^{\prime }$, the result would be,

  $$\begin{aligned}&\dfrac{1}{\frac{1}{N}\big (1+\frac{1}{\gamma _{1,k}}\big ){\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}} \ge \\&\quad \dfrac{1}{\frac{1}{N}\big (1+\frac{1}{\gamma _{1,k}}\big ){\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}^{\prime }\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}}; \end{aligned}$$

  then, $f_{1,k}( {\varvec{\lambda }}_{1},{\varvec{\lambda }}_{2}) >f_{1,k}( {\varvec{\lambda }}_{1}^{\prime },{\varvec{\lambda }}_{2}^{\prime })$. This equation could be similarly proved for $f_{2,k}$.

• 3. Scalability: It should be proved that for $\rho > 1$,

  $$\begin{aligned} \forall i,k \ \rho f_{i,k}( {\varvec{\lambda }}_{1},{\varvec{\lambda }}_{2}) > f_{i,k}(\rho {\varvec{\lambda }}_{1},\rho {\varvec{\lambda }}_{2}). \end{aligned}$$

Assume that $\rho > 1$ be arbitrary, then,

$$\begin{aligned} \rho f_{1,k}( {\varvec{\lambda }}_{1},{\varvec{\lambda }}_{2})=\dfrac{1}{\frac{1}{N}\big (1+\frac{1}{\gamma _{1,k}}\big ){\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\rho \mathbf{I}_{N}+\rho {\varvec{\Sigma }}_{1}\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}}, \end{aligned}$$

where

$$\begin{aligned}&\rho \mathbf{I}_{N}+\rho {\varvec{\Sigma }}_{1}=\rho \mathbf{I}_{N}+\rho \sum _{m=1}^{K}\bigg (\frac{\lambda _{\mathrm{1},m}}{N}{\mathbf {H}} _{1,1,m} {\mathbf {v}}_{1,m} ^{DL,H} {\mathbf {v}}_{1,m} ^{DL} {\mathbf {H}} _{1,1,m} ^{H}+\\&\quad \frac{\lambda _{\mathrm{2},m}}{M} {\mathbf {H}} _{1,2}^{H} {\mathbf {w}}_{2,m} ^{UL,H} {\mathbf {w}}_{2,m} ^{UL} {\mathbf {H}} _{1,2}\bigg )=(\rho -1)\mathbf{I}_{N}+\mathbf{I}_{N} \\&\quad +\rho \sum _{m=1}^{K}\bigg (\frac{\lambda _{\mathrm{1},m}}{N}{\mathbf {H}} _{1,1,m} {\mathbf {v}}_{1,m} ^{DL,H} {\mathbf {v}}_{1,m} ^{DL} {\mathbf {H}} _{1,1,m} ^{H}\\&\quad +\frac{\lambda _{\mathrm{2},m}}{M} {\mathbf {H}} _{1,2}^{H} {\mathbf {w}}_{2,m} ^{UL,H} {\mathbf {w}}_{2,m} ^{UL} {\mathbf {H}} _{1,2}\bigg ). \end{aligned}$$

As mentioned, $\rho >1$ and then, $(\rho -1)\mathbf{I}_{N}\succeq 0$; hence, defining ${\mathbf {C}}=\mathbf{I}_{N}+\rho {\varvec{\Sigma }}_{1}$ and ${\mathbf {D}}=(\rho -1)\mathbf{I}_{N}$, based on (41), the following equation is resulted,

$$\begin{aligned}&\rho f_{1,k}( {\varvec{\lambda }}_{1},{\varvec{\lambda }}_{2}) = \dfrac{1}{\frac{1}{N}\big (1+\frac{1}{\gamma _{1,k}}\big ){\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\rho \mathbf{I}_{N}+\rho {\varvec{\Sigma }}_{1}\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}}\ge \\&\quad \dfrac{1}{\frac{1}{N}\big (1+\frac{1}{\gamma _{1,k}}\big ){\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\mathbf{I}_{N}+\rho {\varvec{\Sigma }}_{1}\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}}\\&\quad = f_{1,k}(\rho {\varvec{\lambda }}_{1},\rho {\varvec{\lambda }}_{2}). \end{aligned}$$

This equation could be similarly confirmed for $f_{2,k}$.

Accordingly, f is a standard function, and based on Lemma 1, the fixed point of this function could be found using the fixed point iteration algorithm.

1.2 The Proof of Theorem 2

In order to prove Theorem 2, we first state the following Lemmas.

Lemma 2

(Silverstein and Bai (1995)) Assume that ${\mathbf {A}} \in {\mathbb {C}}^{N\times N}$ be a hermity and invertible matrix. Then, for any vector ${\mathbf {x}} \in {\mathbb {C}}^{N\times 1}$ and $\tau \in {\mathbb {C}}$, if ${\mathbf {A}}+\tau {\mathbf {x}}{\mathbf {x}}^{H}$ is an invertible matrix, then the following equations are held

$$\begin{aligned} ({\mathbf {A}}+\tau {\mathbf {x}}{\mathbf {x}}^{H})^{-1}&={\mathbf {A}}^{-1}-\dfrac{{\mathbf {A}}^{-1}\tau {\mathbf {x}}{\mathbf {x}}^{H}{\mathbf {A}}^{-1}}{1+\tau {\mathbf {x}}^{H}{\mathbf {A}}^{-1} {\mathbf {x}}}, \end{aligned}$$

(42)

$$\begin{aligned} {\mathbf {x}}^{H}({\mathbf {A}}+\tau {\mathbf {x}}{\mathbf {x}}^{H})^{-1}&=\dfrac{{\mathbf {x}}^{H}{\mathbf {A}}^{-1}}{1+\tau {\mathbf {x}}^{H}{\mathbf {A}}^{-1} {\mathbf {x}}}. \end{aligned}$$

(43)

Lemma 3

(Bai and Silverstein (1998)) Assume that ${\mathbf {x}},{\mathbf {y}}\sim {{\mathcal {C}}}{{\mathcal {N}}}(\mathrm{0},\frac{1}{N} \mathbf{I}_{N}) \in {\mathbb {C}}^{N }$ be independent and $\mathbf{A}$ be a hermity matrix independent of ${\mathbf {x}}$ and ${\mathbf {y}}$, then ${\mathbf {x}}^{H}{\mathbf {A}}{\mathbf {x}} \mathop {\longrightarrow }\limits _{N\rightarrow \infty }^{a.s.} \frac{1}{N} tr\{{\mathbf {A}}\}$ and ${\mathbf {x}}^{H}{\mathbf {A}}{\mathbf {y}}\mathop {\longrightarrow }\limits _{N\rightarrow \infty }^{a.s.} 0.$

Lemma 4

For $k=1,\dots ,K$,

$$\begin{aligned} {\mathbf {H}}_{1,1,k}{\mathbf {v}}_{1,k}^{DL,H}&\sim {{\mathcal {C}}}{{\mathcal {N}}}(0,\Vert {\mathbf {v}}_{1,k}^{DL}\Vert ^{2}\sigma _{1,1,k}^{2}\mathbf{I}_{N}),\\ {\mathbf {H}}_{2,2,k}^{H}{\mathbf {w}}_{2,k}^{UL,H}&\sim {{\mathcal {C}}}{{\mathcal {N}}}(0,\Vert {\mathbf {w}}_{2,k}^{UL}\Vert ^{2}\sigma _{2,2,k}^{2}\mathbf{I}_{M}). \end{aligned}$$

Proof

The proof is straightforward. $\square$

Now, considering (24) we have,

$$\begin{aligned} \lambda _{1,k}=\dfrac{1}{\frac{1}{N}\big (1+\frac{1}{\gamma _{1,k}}\big ){\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}} , \end{aligned}$$

and on the other hand based on Lemma 2,

$$\begin{aligned} \begin{aligned}&{\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}\\&\quad =\dfrac{{\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}^{\prime }\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}}{1+\frac{\lambda _{1,k}}{N}{\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}^{\prime }\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}} , \end{aligned} \end{aligned}$$

then,

$$\begin{aligned} \lambda _{1,k}=\dfrac{1+\frac{\lambda _{1,k}}{N}{\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}^{\prime }\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}}{\frac{1}{N}\big (1+\frac{1}{\gamma _{1,k}}\big ){\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}^{\prime }\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}}, \end{aligned}$$

hence,

$$\begin{aligned} \lambda _{1,k}=\dfrac{\gamma _{1,k}}{\frac{1}{N}{\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}^{\prime }\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}}. \end{aligned}$$

(44)

Also, based on Lemma 4, ${\mathbf {H}}_{1,1,k}{\mathbf {v}}_{1,k}^{DL,H}\sim {{\mathcal {C}}}{{\mathcal {N}}}(0,\Vert {\mathbf {v}}_{1,k}^{DL}\Vert ^{2}\sigma _{1,1,k}^{2}\mathbf{I}_{N})$ and it is clear that the matrix ${\varvec{\Sigma }}_{1,k}^{\prime }$ and vector ${\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}$ are independent, then based on Lemma 3,

$$\begin{aligned} \begin{aligned}&\frac{1}{N}{\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}^{\prime }\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}\\&\quad \mathop {\longrightarrow }\limits _{N\rightarrow \infty }^{a.s.} \frac{1}{N} \Vert {\mathbf {v}}_{1,k}^{DL}\Vert ^{2}\sigma _{1,1,k}^{2} tr\{\big (\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}^{\prime }\big )^{-1}\}. \end{aligned} \end{aligned}$$

On the other hand, according to the assumption of the theorem, $m_{{\varvec{\Sigma }}_{1,k}^{\prime }}(-1)=\frac{1}{N}tr\{\big (\Sigma _{1,k}^{\prime }+\mathbf{I}_{N}\big )^{-1}\}$, hence,

$$\begin{aligned} \begin{aligned}&\frac{1}{N}{\mathbf {v}}_{1,k} ^{DL} {\mathbf {H}} _{1,1,k}^{H}\big (\mathbf{I}_{N}+{\varvec{\Sigma }}_{1}^{\prime }\big )^{-1} {\mathbf {H}} _{1,1,k} {\mathbf {v}}_{1,k} ^{DL,H}\\&\quad \mathop {\longrightarrow }\limits _{N\rightarrow \infty }^{a.s.} \Vert {\mathbf {v}}_{1,k}^{DL}\Vert ^{2}\sigma _{1,1,k}^{2} m_{{\varvec{\Sigma }}_{1,k}^{\prime }}(-1), \end{aligned} \end{aligned}$$

(45)

then considering (44), we result the following equation for the Lagrangian multipliers,

$$\begin{aligned} \lambda _{1,k}=\dfrac{\gamma _{1,k}}{ \Vert {\mathbf {v}}_{1,k}^{DL}\Vert ^{2}\sigma _{1,1,k}^{2} m_{{\varvec{\Sigma }}_{1,k}^{\prime }}(-1)}. \end{aligned}$$

Now, we prove the second equation of Theorem 2; considering (24), we have,

$$\begin{aligned} \lambda _{2,k}=\dfrac{1}{\frac{1}{M}\big (1+\frac{1}{ \gamma _{2,k}}\big ){\mathbf {w}}_{2,k} ^{UL}{\mathbf {H}} _{2,2,k}\big (\mathbf{I}_{M}+{\varvec{\Sigma }}_{2,k}\big )^{-1}{\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H}}, \end{aligned}$$

also based on Lemma 2,

$$\begin{aligned} \begin{aligned}&{\mathbf {w}}_{2,k} ^{UL}{\mathbf {H}} _{2,2,k}\big (\mathbf{I}_{M}+{\varvec{\Sigma }}_{2,k}\big )^{-1}{\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H}=\\ &\quad =\dfrac{{\mathbf {w}}_{2,k} ^{UL}{\mathbf {H}} _{2,2,k}\big (\mathbf{I}_{M}+{\varvec{\Sigma }}_{2,k}^{\prime }\big )^{-1}{\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H}}{1+\frac{\lambda _{2,k}}{M}{\mathbf {w}}_{2,k} ^{UL}{\mathbf {H}} _{2,2,k}\big (\mathbf{I}_{M}+{\varvec{\Sigma }}_{2,k}^{\prime }\big )^{-1}{\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H}}, \end{aligned} \end{aligned}$$

then,

$$\begin{aligned} \lambda _{2,k}=\dfrac{1+\frac{\lambda _{2,k}}{M}{\mathbf {w}}_{2,k} ^{UL}{\mathbf {H}} _{2,2,k}\big (\mathbf{I}_{M}+{\varvec{\Sigma }}_{2,k}^{\prime }\big )^{-1}{\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H}}{\frac{1}{M}\big (1+\frac{1}{ \gamma _{2,k}}\big ){\mathbf {w}}_{2,k} ^{UL}{\mathbf {H}} _{2,2,k}\big (\mathbf{I}_{M}+{\varvec{\Sigma }}_{2,k}^{\prime }\big )^{-1}{\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H}}, \end{aligned}$$

hence,

$$\begin{aligned} \lambda _{2,k}=\dfrac{\gamma _{2,k}}{\frac{1}{M}{\mathbf {w}}_{2,k} ^{UL}{\mathbf {H}} _{2,2,k}\big (\mathbf{I}_{M}+{\varvec{\Sigma }}_{2,k}^{\prime }\big )^{-1}{\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H}}. \end{aligned}$$

(46)

But based on (22), the matrix ${\varvec{\Sigma }}_{2,k}^{\prime }$ is not independent of the vector ${\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H}$ then the term $\frac{1}{M}{\mathbf {w}}_{2,k} ^{UL}{\mathbf {H}} _{2,2,k}\big (\mathbf{I}_{M}+{\varvec{\Sigma }}_{2,k}^{\prime }\big )^{-1}{\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H}$ could not be accurately approximated. However, if it be assumed that the eliminated term of matrix ${\varvec{\Sigma }}_{2,k}$, $\frac{\lambda _{2,k}}{M}{\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H} {\mathbf {w}}_{2,k} ^{UL} {\mathbf {H}} _{2,2,k}$ is the dominant term among the terms containing the random matrix ${\mathbf {H}} _{2,2,k}^{H}$,

then ${\varvec{\Sigma }}_{2,k}^{\prime }$ and ${\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H}$ could be considered independent, then based on Lemma 3 and Lemma 4,

$$\begin{aligned} \begin{aligned}&\frac{1}{M}{\mathbf {w}}_{2,k} ^{UL}{\mathbf {H}} _{2,2,k}\big (\mathbf{I}_{M}+{\varvec{\Sigma }}_{2,k}^{\prime }\big )^{-1}{\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H} \\&\quad \mathop {\longrightarrow }\limits _{M\rightarrow \infty }^{a.s.}\frac{1}{M}\Vert {\mathbf {w}}_{2,k}^{UL}\Vert ^{2}\sigma _{2,2,k}^{2}tr\{\big (\mathbf{I}_{M}+{\varvec{\Sigma }}_{2,k}^{\prime }\big )^{-1} \}, \end{aligned} \end{aligned}$$

hence, based on the assumption of the theorem we have,

$$\begin{aligned}&\frac{1}{M}{\mathbf {w}}_{2,k} ^{UL}{\mathbf {H}} _{2,2,k}\big (\mathbf{I}_{M}+{\varvec{\Sigma }}_{2,k}^{\prime }\big )^{-1}{\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H} \\&\quad \mathop {\longrightarrow }\limits _{M\rightarrow \infty }^{a.s.} \Vert {\mathbf {w}}_{2,k}^{UL}\Vert ^{2}\sigma _{2,2,k}^{2}m_{{\varvec{\Sigma }}_{2,k}^{\prime }}(-1), \end{aligned}$$

then based on (45), the result would be,

$$\begin{aligned} \lambda _{2,k}\simeq \dfrac{\gamma _{2,k}}{m_{{\varvec{\Sigma }}_{2,k}^{\prime }}(-1)\Vert {\mathbf {w}}_{2,k}^{UL}\Vert ^{2}\sigma _{2,2,k}^{2}}, \end{aligned}$$

where using $\simeq$ is because of that it was connived to consider ${\varvec{\Sigma }}_{2,k}^{\prime }$ and ${\mathbf {H}} _{2,2,k}^{H} {\mathbf {w}}_{2,k} ^{UL,H}$ independent.

1.3 The proof of Theorem 3

In order to prove Theorem 3, we first state the following theorem from the random matrix theory,

Theorem 4

(See Theorem 5 in Lakshminarayana et al. (2015)) Consider matrix $\mathbf{B}=XTX^{ H}$, where $\mathbf{X}=\frac{ 1}{\sqrt{ N}}\mathbf{Y}\in {\mathbb {C}}^{ N\times LK}$ with components $Y(p,q)\sim {{\mathcal {C}}}{{\mathcal {N}}}(0,1)$ and ${\mathbf {T}}=\mathrm {diag}\big (t_{1},\dots ,t_{LK}\big )\in {\mathbb {R}}^{LK\times LK}$ be a non-random diagonal matrix. Assume that $m_{{\mathbf {B}}}(z)=\frac{1}{N}tr\{\big ({\mathbf {B}}-z\mathbf{I}_{N}\big )^{-1}\}$, $z \in {\mathbb {R}}$ then,

$$\begin{aligned} m_{{\mathbf {B}}}(z)\mathop {\longrightarrow }\limits _{N,K\rightarrow \infty }^{a.s.}{\bar{m}}_{{\mathbf {B}}}(z), \end{aligned}$$

where ${\bar{m}}_{{\mathbf {B}}}(z)$ is the unique solution of following fixed point equation,

$$\begin{aligned} {\bar{m}}_{{\mathbf {B}}}(z)=\bigg (\frac{1}{N}\sum \limits _{i=1}^{LK}\dfrac{t_{i}}{1+t_{i}{\bar{m}}_{{\mathbf {B}}}(z)}-z\bigg )^{-1}. \end{aligned}$$

(47)

Now, we rewrite matrix ${\varvec{\Sigma }}_{1,k}^{\prime }$ as,

$$\begin{aligned} {\varvec{\Sigma }}_{1}^{\prime }&=\sum _{m\ne k}^{K}\frac{\lambda _{\mathrm{1},m}}{N}{\mathbf {H}} _{1,1,m} {\mathbf {v}}_{1,m} ^{DL,H} {\mathbf {v}}_{1,m} ^{DL} {\mathbf {H}} _{1,1,m} ^{H}\\&\qquad +\sum _{m=1}^{K}\frac{\lambda _{\mathrm{2},m}}{M} {\mathbf {H}} _{1,2}^{H} {\mathbf {w}}_{2,m} ^{UL,H} {\mathbf {w}}_{2,m} ^{UL} {\mathbf {H}} _{1,2}\\ &\quad =\bigg [\frac{{\mathbf {H}} _{1,1,1} {\mathbf {v}}_{1,1} ^{DL,H} }{\sqrt{N}\sigma _{1,1,1}},\dots ,\frac{{\mathbf {H}} _{1,1,k-1} {\mathbf {v}}_{1,k-1} ^{DL,H} }{\sqrt{N}\sigma _{1,1,k-1}},\frac{{\mathbf {H}} _{1,1,k+1} {\mathbf {v}}_{1,k+1} ^{DL,H} }{\sqrt{N}\sigma _{1,1,k+1}},\\&\qquad \dots ,\frac{{\mathbf {H}} _{1,1,K} {\mathbf {v}}_{1,K} ^{DL,H} }{\sqrt{N}\sigma _{1,1,K}},\frac{{\mathbf {H}}_{1,2}^{H} {\mathbf {w}}_{2,1} ^{UL,H}}{\sqrt{M}\sigma _{1,2}},\dots ,\frac{{\mathbf {H}} _{1,2}^{H} {\mathbf {w}}_{2,K} ^{UL,H}}{\sqrt{M}\sigma _{1,2}}\bigg ] \\&\quad \mathrm {diag}\bigg (\lambda _{1,1}\sigma _{1,1,1}^{2}\Vert {\mathbf {v}}_{1,1}^{DL}\Vert ^{2},\dots ,\lambda _{1,k-1}\sigma _{1,1,k-1}^{2}\Vert {\mathbf {v}}_{1,k-1}^{DL}\Vert ^{2},\\&\lambda _{1,k+1}\sigma _{1,1,k+1}^{2}\Vert {\mathbf {v}}_{1,k+1}^{DL}\Vert ^{2},\dots ,\lambda _{1,K}\sigma _{1,1,K}^{2}\Vert {\mathbf {v}}_{1,K}^{DL}\Vert ^{2}\bigg )\\&\bigg [\frac{{\mathbf {H}} _{1,1,1} {\mathbf {v}}_{1,1} ^{DL,H} }{\sqrt{N}\sigma _{1,1,1}},\dots ,\frac{{\mathbf {H}}_{1,1,k-1} {\mathbf {v}}_{1,k-1} ^{DL,H} }{\sqrt{N}\sigma _{1,1,k-1}},\frac{{\mathbf {H}}_{1,1,k+1} {\mathbf {v}}_{1,k+1} ^{DL,H} }{\sqrt{N}\sigma _{1,1,k+1}},\\&\dots ,\frac{{\mathbf {H}} _{1,1,K} {\mathbf {v}}_{1,K} ^{DL,H} }{\sqrt{N}\sigma _{1,1,K}},\frac{{\mathbf {H}}_{1,2}^{H} {\mathbf {w}}_{2,1} ^{UL,H}}{\sqrt{M}\sigma _{1,2}},\dots ,\frac{{\mathbf {H}}_{1,2}^{H} {\mathbf {w}}_{2,K} ^{UL,H}}{\sqrt{M}\sigma _{1,2}}\bigg ]^{H}\\&={\mathbf {X}}_{1}{\mathbf {T}}_{1}{\mathbf {X}}_{1}^{H}, \end{aligned}$$

where

$$\begin{aligned}&{\mathbf {X}}_{1}=\bigg [\frac{{\mathbf {H}} _{1,1,1} {\mathbf {v}}_{1,1} ^{DL,H} }{\sqrt{N}\sigma _{1,1,1}},\dots ,\frac{{\mathbf {H}} _{1,1,k-1} {\mathbf {v}}_{1,k-1} ^{DL,H} }{\sqrt{N}\sigma _{1,1,k-1}},\frac{{\mathbf {H}} _{1,1,k+1} {\mathbf {v}}_{1,k+1} ^{DL,H} }{\sqrt{N}\sigma _{1,1,k+1}},\\&\quad \dots ,\frac{{\mathbf {H}} _{1,1,K} {\mathbf {v}}_{1,K} ^{DL,H} }{\sqrt{N}\sigma _{1,1,K}},\frac{{\mathbf {H}}_{1,2}^{H} {\mathbf {w}}_{2,1} ^{UL,H}}{\sqrt{M}\sigma _{1,2}},\dots ,\frac{{\mathbf {H}} _{1,2}^{H} {\mathbf {w}}_{2,K} ^{UL,H}}{\sqrt{M}\sigma _{1,2}}\bigg ], \end{aligned}$$

and matrix $\mathbf{T}$ is,

$$\begin{aligned}&\mathrm {diag}\bigg (\lambda _{1,1}\sigma _{1,1,1}^{2}\Vert {\mathbf {v}}_{1,1}^{DL}\Vert ^{2},\dots ,\lambda _{1,k-1}\sigma _{1,1,k-1}^{2}\Vert {\mathbf {v}}_{1,k-1}^{DL}\Vert ^{2},\\&\quad \lambda _{1,k+1}\sigma _{1,1,k+1}^{2}\Vert {\mathbf {v}}_{1,k+1}^{DL}\Vert ^{2},\dots ,\lambda _{1,K}\sigma _{1,1,K}^{2}\Vert {\mathbf {v}}_{1,K}^{DL}\Vert ^{2}\bigg ). \end{aligned}$$

The vector ${\mathbf {Y}}_{1}=\sqrt{ N}{\mathbf {X}}_{1}$ is a zero mean normal random vector with variance one, then based on Theorem 4 when N and K limit to infinity, the following would be resulted,

$$\begin{aligned}m_{{\varvec{\Sigma }}_{1,k}^{\prime }}(-1)=\frac{1}{N}tr\{\big (\Sigma _{1,k}^{\prime }+\mathbf{I}_{N}\big )^{-1}\}\mathop {\longrightarrow }\limits _{N,K\longrightarrow \infty }^{a.s.}{\bar{m}}_{{\varvec{\Sigma }}_{1,k}^{\prime }}(-1),\end{aligned}$$

where ${\bar{m}}_{{\varvec{\Sigma }}_{1,k}^{\prime }}(-1)$ is the unique solution of Eq. (37).

Equation (38) could be similarly proved by rewriting matrix ${\varvec{\Sigma }}_{2,k}^{\prime }$.