Secure transmission and power allocation in multiuser distributed massive MIMO systems

Abstract

In this paper, we investigate the secure downlink transmission in a multi-user massive multiple-input multiple-output system assuming there is a passive multi-antenna eavesdropper (Eve) that intends to intercept information of a target user. Specifically, we employ the distributed massive antenna sets at the base station, which are known as remote radio heads (RRHs). With cooperative maximum-ratio transmission beamforming, artificial noise (AN) generation and autonomous power allocation at each RRH, the closed-form deterministic lower bound of the ergodic secrecy rate is derived by random matrix theory. Based on a simplified channel model, the impacts of various parameters on secrecy performance, such as uplink training energy, eavesdropper’s antennas number, power allocation factor, have been analyzed in detail, which provide intuitive insights for optimization and performance evaluation. Moreover, by exploiting complementary geometric programming, an power optimization algorithm over signal and AN power is derived subject to total power and Eve’s signal-to-interference-plus-noise ratio constraints. Numerical results have been presented to reveal the system’s secrecy performance and confirm all analysis results in this paper.

Appendices

Appendix 1

According to the statistic characteristics of the MMSE channel estimate and estimation error, exploiting random matrix theory, the intended signal, multi-user interference, and AN leakage can be rewritten as

$$\begin{aligned}{\left| {{{\mathbb {E}}}\left[ {{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1/2}}{{\mathbf{w}}_k}} \right] } \right| ^2}&= {\left| {{{\mathbb {E}}}\left[ {\left( {\hat{\mathbf{g}}_k^T + \tilde{\mathbf{g}}_k^T} \right) {\mathbf{P}}_d^{{1/2}}{{\mathbf{w}}_k}} \right] } \right| ^2} \\& = {\left| {{{\mathbb {E}}}\left[ {\hat{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1 /2}}{{\mathbf{w}}_k}} \right] } \right| ^2}, \end{aligned}$$

(37)

$$\begin{aligned}&{{\mathbb {E}}}\left[ {{{\left| {{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1/2}}{{\mathbf{w}}_l}} \right| }^2}} \right] \\&\quad = {{\mathbb {E}}}\left[ {{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1/2}}{{\mathbf{w}}_l} {\mathbf{w}}_l^T{\mathbf{P}}_d^{{1 /2}}{{\mathbf{g}}_k}} \right] ,\quad \forall l \ne k, \end{aligned}$$

(38)

$$\begin{aligned}&{{\mathbb {E}}}\left[ {{{\left| {{\mathbf{g}}_k^T{\mathbf{P}}_a^{{1/2}} \mathbf{{Az}}} \right| }^2}} \right] \\&\quad = {{\mathbb {E}}}\left[ {\left( {\hat{\mathbf{g}}_k^T + \tilde{\mathbf{g}}_k^T} \right) {\mathbf{P}}_a^{{1/2}} \mathbf{Az}{\mathbf{z}^H}{\mathbf{A}^H} {\mathbf{P}}_a^{{1 /2}}\left( {\hat{\mathbf{g}}_k^* + \tilde{\mathbf{g}}_k^*} \right) } \right] \\&\quad = {{\mathbb {E}}}\left[ {\tilde{\mathbf{g}}_k^T{\mathbf{P}}_a^{{1/2}} \mathbf{Az}{\mathbf{z}^H}{\mathbf{A}^H} {\mathbf{P}}_a^{{1 /2}}\tilde{\mathbf{g}}_k^*} \right] . \end{aligned}$$

(39)

The above statistics can be easily obtained by the law of large numbers and central limit theorem with infinite BS antenna number ${N_T} \rightarrow \infty$. Furthermore, we can easily get that

$$\begin{aligned}&\frac{1}{{N_T^2}}{\left| {{{\mathbb {E}}}\left[ {{\mathbf{g}}_k^T\mathbf{P }_d^{{1/2}}{{\mathbf{w}}_k}} \right] } \right| ^2} \\&\quad -\frac{1}{{N_T^2}}{\left( {\sum \limits _{m = 1}^M {p_{d,m}^{{1 /2}}{\alpha _m}{\mathrm{tr}}\left( {{{{\varvec{\Psi }}} _{mk}}} \right) } } \right) ^2} \mathop \rightarrow \limits ^{a.s.} 0, \end{aligned}$$

(40)

$$\begin{aligned}&\frac{1}{{{N_T}}}{{\mathbb {E}}}\left[ {{{\left| {{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1/2}}{{\mathbf{w}}_l}} \right| }^2}} \right] \\&\quad -\frac{1}{{{N_T}}}\sum \limits _{m = 1}^M {{p_{d,m}}\alpha _m^2{\mathrm{tr}}\left( {{{\mathbf{R}}_{mk}}{{{\varvec{\Psi }}} _{ml}}} \right) } \mathop \rightarrow \limits ^{a.s.} 0, l \ne k, \end{aligned}$$

(41)

$$\begin{aligned}&\frac{1}{{{N_T}}}{{\mathbb {E}}}\left[ {{{\left| {{\mathbf{g}}_k^T{\mathbf{P}}_a^{{1/2}} \mathbf{Az}} \right| }^2}} \right] \\&\quad -\frac{1}{{{N_T}\left( {{N_T} - K} \right) }}\sum \limits _{m = 1}^M {{p_{a,m}}{\mathrm{tr}}\left( {{\mathbf{A}}{{\mathbf{A}}^H}{{\mathbf{E}}_{mk}}} \right) } \mathop \rightarrow \limits ^{a.s.} 0. \end{aligned}$$

(42)

Then we focus on the variance term, i.e.

$$\begin{aligned} {\mathrm{var}} \left[ {{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1 /2}}{{\mathbf{w}}_k}} \right]&= {{\mathbb {E}}}\left[ {{{\left| {{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1/2}}{{\mathbf{w}}_k}} \right| }^2}} \right] - {\left| {{{\mathbb {E}}}\left[ {{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1/2}}{{\mathbf{w}}_k}} \right] } \right| ^2}\\&= {{\mathbb {E}}}\left[ {{{\left| {\hat{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1/2}} {{\mathbf{w}}_k}} \right| }^2}} \right] + {{\mathbb {E}}}\left[ {{{\left| {\tilde{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1/2}} {{\mathbf{w}}_k}} \right| }^2}} \right] \\&\quad - {\left| {{{\mathbb {E}}}\left[ {{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1/2}} {{\mathbf{w}}_k}} \right] } \right| ^2}. \end{aligned} $$

(43)

Obviously, the last two terms at far right of the above equation can be obtained easily by (40) and (41). Meanwhile, the first component can be unfolded as

$$\begin{aligned} {{\mathbb {E}}}\left[ {{{\left| {\hat{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1/2}} {{\mathbf{w}}_k}} \right| }^2}} \right] = {{\mathbb {E}}}\left[ {{{\left( {\sum \limits _{m = 1}^M {p_{d,m}^{{1/2}}\hat{\mathbf{g}}_{mk}^T{{\mathbf{w}}_{mk}}} } \right) }^2}} \right] . \end{aligned} $$

(44)

It is recalled that channels from different RRHs are mutually independent. Also it is noted that ${\left\| {{{\hat{\mathbf{g}}}_{mk}}} \right\| ^2} \sim {{\mathcal{W}}_1}\left( {{N_T},{\phi _{mk}}} \right)$ follows Wishart distribution. Based on the expectation properties of Wishart matrix [35], we can get that ${{\mathbb {E}}}\left[ {{{\left\| {{{\hat{\mathbf{g}}}_{mk}}} \right\| }^4}} \right] = {\left( {{\mathrm{tr}}\left( {{{{\varvec{\Psi }}} _{mk}}} \right) } \right) ^2} + {\mathrm{tr}}\left( {{{\varvec{\Psi }}} _{mk}^2} \right)$, thus

$$\begin{aligned} &\frac{1}{{{N_T}}}{{\mathbb {E}}}\left[ {{{\left| {\hat{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1/2}}{{\mathbf{w}}_k}} \right| }^2}} \right] - \frac{1}{{{N_T}}}\sum \limits _{m = 1}^M {\left\{ {{p_{d,m}}\alpha _m^2{\mathrm{tr}}\left( {{{\varvec{\Psi }}} _{mk}^2} \right) } \right. } \\&\quad \left. { + p_{d,m}^{{1/2}} {\alpha _m}{\mathrm{tr}}\left( {{{{\varvec{\Psi }}} _{mk}}} \right) \sum \limits _{n = 1}^M {p_{d,n}^{{1/2}}{\alpha _n}{\mathrm{tr}}\left( {{{{\varvec{\Psi }}} _{nk}}} \right) } } \right\} \mathop \rightarrow \limits ^{a.s.} 0 \end{aligned} $$

(45)

Hence, we can finally obtain the variance term as

$$\begin{aligned} &\frac{1}{{{N_T}}}{\mathrm{var}} \left[ {{\mathbf{g}}_k^T{\mathbf{P}}_d^{{1/2}}{{\mathbf{w}}_k}} \right] - \frac{1}{{{N_T}}}\sum \limits _{m = 1}^M {{p_{d,m}}\alpha _m^2{\mathrm{tr}}\left( {{{\mathbf{R}}_{mk}}{{{\varvec{\Psi }}} _{mk}}} \right) } \mathop \rightarrow \limits ^{a.s.} 0. \end{aligned} $$

(46)

Inserting these results into (11), we can yield the expression as (13), which concludes the proof.

Appendix 2

According to the assumption, all UTs are located uniformly in the region. So we can easily consider that there are $\frac{K}{M}$ UTs in the major service region of each RRH. Based on these conditions and assumptions, the intended signal, multi-user interference, and AN leakage in (13) can be rewritten as

$$\begin{aligned}&{\left( {\sum \limits _{m = 1}^M {p_{d,m}^{{1/2}}{\alpha _m}{\mathrm{tr}}\left( {{{{\varvec{\Psi }}} _{mk}}} \right) } } \right) ^2} \\&\quad = {p_{d,m}}\alpha _m^2{\left( {\sum \limits _{m = 1}^M {{\mathrm{tr}}\left( {{{{\varvec{\Psi }}} _{mk}}} \right) } } \right) ^2} = \frac{{\varphi {P_T}a{N_T}}}{K}, \end{aligned}$$

(47)

$$\begin{aligned}&\sum \limits _{m = 1}^M {\sum \limits _{l = 1}^K {{p_{d,m}}\alpha _m^2{\mathrm{tr}}\left( {{{\mathbf{R}}_{mk}}{{{\varvec{\Psi }}} _{ml}}} \right) } } \\&\quad = {p_{d,m}}\alpha _m^2\sum \limits _{m = 1}^M {\sum \limits _{l = 1}^K {{\lambda _{mk}}{\mathrm{tr}}\left( {{{{\varvec{\Psi }}} _{ml}}} \right) } } = \frac{{\varphi {P_T}b}}{K}, \end{aligned}$$

(48)

$$\begin{aligned}&\frac{1}{{{N_T} - K}}{p_{a,n}}{\mathrm{tr}}\left( {{\mathbf{A}}{{\mathbf{A}}^H}{{\mathbf{E}}_{nk}}} \right) \\&\quad = \frac{1}{{{N_T} - K}}{\varepsilon _{nk}}{p_{a,n}}{\mathrm{tr}}\left( {{\mathbf{A}}{{\mathbf{A}}^H}} \right) = {\varepsilon _{nk}}\left( {1 - \varphi } \right) {P_T}. \end{aligned}$$

(49)

Recalling (13), we can simply SINR of kth UT as

$$\begin{aligned} {\lambda _k} = \frac{{\varphi {P_T}a{N_T}}}{{\varphi {P_T}b + K{\varepsilon _{nk}}\left( {1 - \varphi } \right) {P_T} + K}}. \end{aligned} $$

(50)

Similarly, we can also obtain the SINR of Eve as

$$\begin{aligned} {\bar{\lambda }_E} = \frac{{\varphi c{N_E}\left( {{N_T} - K} \right) }}{{\left( {1 - \varphi } \right) aK\left( {{N_T} - K - {N_E}} \right) }}. \end{aligned} $$

(51)

Plugging these SINR expressions of potential UT and Eve into (25) and (26), we finally achieve ergodic secrecy rate of kth UT as (27). This completes the proof.

Appendix 3

According to definition in [36], we know that one set with $\left\{ {\left. x \right| {g_1}\left( x \right) \le 0, \ldots ,} \right. \left. {{g_K}\left( x \right) \le 0} \right\}$ is regular when: (1) this set is nonempty and compact; (2) For any ${x^*}$ with a nonempty index set ${\mathcal{K} }= \left\{ {\left. k \right| {g_k}\left( {x'} \right) = 0} \right\}$, the convex hull of $\nabla {g_k}\left( {{x^*}} \right) ,k \in {\mathcal{K}}$ doesn’t contain the origin. For problem ${J_2}$, we similarly define an index set ${\mathcal{K}} = {{\mathcal{K}}_P} \cup {{\mathcal{K}}_{UT}} \cup {{\mathcal{K}}_E}$, where ${{\mathcal{K}}_P} = \left\{ {\left. p \right| {g_{P,p}} = 0} \right\} , {{\mathcal{K}}_U} = \left\{ {\left. u \right| {g_{U,u}} = 0} \right\}$ and ${{\mathcal{K}}_E} = \left\{ {\left. e \right| {g_{E,e}} = 0} \right\}$ with ${g_{P,p}}, {g_{U,u}}, {g_{E,e}}$ representing the constraints of total power, UT’s and Eve’s SINR. As there are $M + 2$ variables in these equations, so we know that $\left| {\mathcal{K}} \right| \le M + 2$ always holds true. As the feasible set of ${J_2}$ is not empty and compact, we only need prove that the origin does not lie in the convex hull of $\nabla {g_{P,p}} , p \in {{\mathcal{K}}_P},\nabla {g_{U,u}} , u \in {{\mathcal{K}}_U}$ and $\nabla {g_{E,e}} , e \in {{\mathcal{K}}_E}$. Note that $\left| {{{\mathcal{K}}_P}} \right| = 1$. We need to prove that there is not non-negative vector to satisfing the equation ${\mathbf{A}}\theta = {\mathbf{b}}$, where

$$\begin{aligned} {\mathbf{A}}= & \left[ {\begin{array}{cccccc} 1&\quad \cdots&\quad {\frac{{\partial {g_{U,u}}}}{{\partial {{\mathbf{p}}_d}}}}&\quad \cdots &\quad {\frac{{\partial {g_{E,e}}}}{{\partial {{\mathbf{p}}_d}}}}&\quad \cdots \\ 1&\quad \cdots &\quad {\frac{{\partial {g_{U,u}}}}{{\partial {p_{a,n}}}}}&\quad \cdots &\quad {\frac{{\partial {g_{E,e}}}}{{\partial {p_{a,n}}}}}&\quad \cdots\\ 0&\quad \cdots &\quad {\frac{{\partial {g_{U,u}}}}{{\partial \omega }}}&\quad \cdots &\quad {\frac{{\partial {g_{E,e}}}}{{\partial \omega }}}&\quad \cdots\\ 1& \quad \cdots &\quad 1&\quad \cdots &\quad 1&\quad \cdots \end{array}} \right] ,\\ {\mathbf{b}}= & \left[ {\begin{array}{c} {{\mathbf{{0}}_{\left( {M + 2} \right) \times 1}}}\\ 1 \end{array}} \right] . \end{aligned}$$

Obviously, the dimensions of A and b are $\left( {M + 3} \right) \times \left| {\mathcal{K}} \right|$ and $\left( {M + 3} \right) \times 1$. The last cow of A represents the convex combination of paraments. We definite an augmented matrix $\bar{\mathbf{A}} = \left[ {{\mathbf{A}}\left| {\mathbf{b}} \right. } \right]$. Recalling that the last elements of the first $\left( {M + 2} \right)$ rows are all zero and $\left| {\mathcal{K}} \right| \le M + 2$, the last row of matrix ${\bar{\mathbf{A}}}$ can always be transformed into $\left[ {{{\mathbf{0 }}_{1 \times \left| {\mathcal{K}} \right| }}\left| b \right. } \right]$, where $b \ne 0$. Hence, there is no ${\varvec{\uptheta }}$ existing to satisfy equations ${A{\varvec{\uptheta } }} = {\mathbf{b}}$. Now we finish the proof.