Secrecy performance analysis of cell-free massive MIMO in the presence of active eavesdropper with low resolution ADCs

Abstract

This paper investigates the secrecy performance of the Cell-Free massive multiple-input multiple-output network with finite resolution analog-to-digital converters at the access points (APs) and users in presence of an active eavesdropper. Using the additive quantization noise model, the uplink minimum mean squared error channel estimation and downlink data precoding are respectively operated. Specifically, the lower bound on the achievable ergodic rate and upper bound for the information leakage to the eavesdropper are theoretically derived in details. Thereby, the closed-form expression for the achievable ergodic secrecy rate is accordingly obtained with respect to the number of APs, number of each APs antenna, number of users, pilot and data transmission power and quantization bits, etc. In addition, the asymptotic approximation for the ergodic secrecy rate has been presented. Moreover, the path-following power control algorithm has been proposed aiming at maximizing the secrecy rate subject to both power and achievable rate constraints. Finally, extensive simulations are provided to corroborate the theoretical analytical results and the validity of the proposed power allocation scheme.

Appendices

Appendix A

To derive the closed-form expression of (24), we should analyze the different components seperately. Firstly, we focus on the term ${\mathbb E}\left\{ {\sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} } \right\}$ as

$$\begin{aligned} \begin{aligned}&{\mathbb E}\left\{ {\sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} } \right\} \\&= \sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\mathbb E}\left\{ {{\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} \right\} } \\&= \sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} \left( {{\mathbb E}\left\{ {{\hat{\varvec{g}}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} \right\} + {\mathbb E}\left\{ {{\varvec{\tilde{g}}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} \right\} } \right) }. \end{aligned} \end{aligned}$$

(53)

Due to the independence between the channel estimation ${{\hat{\varvec{g}}}_{mk}}$ and estimation error ${{\varvec{\tilde{g}}}_{mk}}$, it is easy to obtain that

$$\begin{aligned} \begin{aligned} {\mathbb E}\left\{ {\sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} } \right\} = \sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\lambda _{mk}}} . \end{aligned} \end{aligned}$$

(54)

Then, we calculate

$$\begin{aligned} \begin{aligned}&\mathrm{Var}\left( {\sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} } \right) = \\&{\mathbb E}\left\{ {{{\left| {\sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} - {\mathbb E}\left\{ {\sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} } \right\} } \right| }^2}} \right\} \\&= \sum \limits _{m = 1}^M {{\eta _{mk}}{\mathbb E}\left\{ {{\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*{\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} \right\} } \\&- {\left( {{\mathbb E}\left\{ {\sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} } \right\} } \right) ^2} \\& \quad + \sum \limits _{m = 1}^M {\sum \limits _{m' \ne m}^M {\sqrt{{\eta _{mk}}{\eta _{m'k}}} {\mathbb E}\left\{ {{\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*{\varvec{g}}_{m'k}^T{\hat{\varvec{g}}}_{m'k}^*} \right\} } } . \end{aligned} \end{aligned}$$

(55)

According to some existing literature[35]-[36], we can get that

$$\begin{aligned} \begin{aligned} {\mathbb E}\left\{ {{\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*{\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} \right\}&= {\mathbb E}\left\{ {{{\left\| {{{\varvec{g}}_{mk}}} \right\| }^4}} \right\} + {\mathbb E}\left\{ {\left| {{\varvec{\tilde{g}}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} \right| } \right\} \\&= {N^2}\lambda _{mk}^2 + N{\lambda _{mk}}{\beta _{mk}}, \end{aligned} \end{aligned}$$

(56)

$$\begin{aligned} \begin{aligned} {\mathbb E}\left\{ {{\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*{\varvec{g}}_{m'k}^T{\hat{\varvec{g}}}_{m'k}^*} \right\} = {N^2}{\lambda _{mk}}{\lambda _{m'k}}, m \ne m'. \end{aligned} \end{aligned}$$

(57)

Substituting (56) and (57) into (55), we can get that

$$\begin{aligned} \begin{aligned} \mathrm{Var}\left( {\sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} } \right) = N\sum \limits _{m = 1}^M {{\eta _{mk}}{\lambda _{mk}}{\beta _{mk}}}. \end{aligned} \end{aligned}$$

(58)

Then, we compute

$$\begin{aligned} \begin{aligned} \mathrm{U}{\mathrm{I}_k}&= \gamma _k^2{\rho _d}\sum \limits _{i \ne k}^K {\sum \limits _{m = 1}^M {{\eta _{mi}}E\left\{ {{\varvec{g}}_{mk}^T{\hat{\varvec{g}}}_{mi}^*{\hat{\varvec{g}}}_{mi}^T{\varvec{g}}_{mk}^*} \right\} } }\\&= N\gamma _k^2{\rho _d}\sum \limits _{m = 1}^M {{\eta _{mi}}{\lambda _{mi}}{\beta _{mk}}} . \end{aligned} \end{aligned}$$

(59)

Now we can focus on the last term as

$$\begin{aligned} \begin{aligned} \mathrm{Q}{\mathrm{N}_k}&= {\gamma _k}\left( {1 - {\gamma _k}} \right) {\mathbb E}\left\{ {{{\left| {{r_k}} \right| }^2}} \right\} = {\gamma _k}\left( {1 - {\gamma _k}} \right) \left( {1 + {\rho _d}{\zeta _k}} \right) . \end{aligned} \end{aligned}$$

(60)

where

$$\begin{aligned} \begin{aligned} {\zeta _k} =&N\sum \limits _{i = 1}^K {\sum \limits _{m = 1}^M {{\eta _{mi}}{\lambda _{mi}}{\beta _{mk}}} } +\\&{N^2}\sum \limits _{m = 1}^M {\sum \limits _{m' = 1}^M {\sqrt{{\eta _{mk}}{\eta _{m'k}}} {\lambda _{mk}}{\lambda _{m'k}}} }. \end{aligned} \end{aligned}$$

(61)

Substituting (54), (58), (59) and (60) into (24) yields the results given by (27), which completes the proof.

Appendix B

For analytical tractability, we can analyze the numerator (desired signal) and denominator (the effective noise) of the Eve’s SINR. Hence, we start with the desired signal component that

$$\begin{aligned} \begin{aligned}&{\mathbb E}\left\{ {{{\left| {\sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\varvec{g}}_{mE}^T{\hat{\varvec{g}}}_{mk}^*} } \right| }^2}} \right\} = \sum \limits _{m = 1}^M {{\eta _{mk}}{\mathbb E}\left\{ {{{\left| {{\varvec{g}}_{mE}^T{\hat{\varvec{g}}}_{mk}^*} \right| }^2}} \right\} }\\&= \sum \limits _{m = 1}^M {{\eta _{mk}}{\mathbb E}\left\{ {{{\left| {\left( {{\hat{\varvec{g}}}_{mE}^T + {\varvec{\tilde{g}}}_{mE}^T} \right) {\hat{\varvec{g}}}_{mk}^*} \right| }^2}} \right\} } \\&= \sum \limits _{m = 1}^M {{\eta _{mk}}\left( {{\mathbb E}\left\{ {{{\left| {{\hat{\varvec{g}}}_{mE}^T{\hat{\varvec{g}}}_{mk}^*} \right| }^2}} \right\} + {\mathbb E}\left\{ {{{\left| {{\varvec{\tilde{g}}}_{mE}^T{\hat{\varvec{g}}}_{mk}^*} \right| }^2}} \right\} } \right) } \end{aligned} \end{aligned}$$

(62)

According to the property of the MMSE algorithm used in this paper, we know that

$$\begin{aligned} \begin{aligned} {\mathbb E}\left\{ {{{\left| {{\hat{\varvec{g}}}_{mE}^T{\hat{\varvec{g}}}_{mk}^*} \right| }^2}} \right\}&= \frac{{{\rho _E}\beta _{mE}^2}}{{{\rho _u}\beta _{mk}^2}}{\mathbb E}\left\{ {{{\left| {{\hat{\varvec{g}}}_{mk}^T{\hat{\varvec{g}}}_{mk}^*} \right| }^2}} \right\} \\&= \frac{{N\left( {N + 1} \right) {\rho _E}\beta _{mE}^2}}{{{\rho _u}\beta _{mk}^2}}\lambda _{mk}^2. \end{aligned} \end{aligned}$$

(63)

Due to mutual independence between the channel estimation and estimation error, it is not hard to achieve that

$$\begin{aligned} \begin{aligned} {\mathbb E}\left\{ {{{\left| {{\varvec{\tilde{g}}}_{mE}^T{\hat{\varvec{g}}}_{mk}^*} \right| }^2}} \right\}&= N{\lambda _{mk}}\left( {{\beta _{mE}} - {\lambda _{mE}}} \right) \\&= N{\lambda _{mk}}\left( {{\beta _{mE}} - \frac{{{\rho _E}\beta _{mE}^2{\lambda _{mk}}}}{{{\rho _u}\beta _{mk}^2}}} \right) . \end{aligned} \end{aligned}$$

(64)

Next, we calculate the multi-user interference component using the independence of different terms as

$$\begin{aligned} \begin{aligned} \sum \limits _{i \ne k}^K {{\mathbb E}\left\{ {{{\left| {\sum \limits _{m = 1}^M {\sqrt{{\eta _{mi}}} {\varvec{g}}_{mE}^T{\hat{\varvec{g}}}_{mi}^*} } \right| }^2}} \right\} } = \sum \limits _{i \ne k}^K {N{\eta _{mi}}{\lambda _{mi}}{\beta _{mE}}}. \end{aligned} \end{aligned}$$

(65)

Plugging (62) and (65) into (28), we finally achieve the result as (29). This finishes the proof.

Appendix C

From (27) and (29), it is easy to see that ${R_k}$ and ${R_E}$ are increasing with M. Then, by relaxing M to be a continuous real number, after some simple algebraic manipulations, we know that $\frac{{\partial R_{\sec }^k}}{{\partial M}} > 0$ which indicates that system can obtain more secrecy capacity by equipping more APs.

Note that the expression (27) can be rewritten as

$$\begin{aligned} \begin{aligned} {R_k} = {\log _2}\left( {1 + \frac{1}{{\frac{{1 - {\gamma _k}}}{{{\gamma _k}}} + L_k^{\left( 1 \right) } + L_k^{\left( 2 \right) }}}} \right) , \end{aligned} \end{aligned}$$

(66)

where

$$\begin{aligned} \begin{aligned} L_k^{\left( 1 \right) } = \frac{{\sum \limits _{i = 1}^K {\sum \limits _{m = 1}^M {{\eta _{mi}}{\lambda _{mi}}{\beta _{mk}}} } }}{{N{\gamma _k}{{\left( {\sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\lambda _{mk}}} } \right) }^2}}}, \end{aligned} \end{aligned}$$

(67)

$$\begin{aligned} \begin{aligned} L_k^{\left( 2 \right) } = \frac{1}{{{N^2}{\gamma _k}{\rho _d}{{\left( {\sum \limits _{m = 1}^M {\sqrt{{\eta _{mk}}} {\lambda _{mk}}} } \right) }^2}}}. \end{aligned} \end{aligned}$$

(68)

Then, defining that ${\mathrm{\Gamma } _k} = \min \left\{ {\sqrt{{\eta _{mk}}} {\lambda _{mk}}} \right\} , 1 \le m \le M$ and ${\mathrm{\Theta } _k} = \max \left\{ {\sum \limits _{i = 1}^K {{\eta _{mi}}{\lambda _{mi}}{\beta _{mk}}} } \right\} , 1 \le m \le M$, we can get that

$$\begin{aligned} \begin{aligned} L_k^{\left( 1 \right) } \le \frac{{M{\mathrm{\Theta } _k}}}{{N{\mathrm{\gamma } _k}{M^2}\mathrm{\Gamma } _k^2}} = \frac{{{\mathrm{\Theta } _k}}}{{MN{\mathrm{\gamma } _k}\mathrm{\Gamma } _k^2}}, \end{aligned} \end{aligned}$$

(69)

$$\begin{aligned} \begin{aligned} L_k^{\left( 2 \right) } \le \frac{1}{{{M^2}{N^2}{\mathrm{\gamma } _k}{\rho _d}\mathrm{\Theta } _k^2}}, \end{aligned} \end{aligned}$$

(70)

Obviously, we can derive that $\mathop {\lim }\limits _{M \rightarrow + \infty } L_k^{\left( 1 \right) } = 0$ and $\mathop {\lim }\limits _{M \rightarrow + \infty } L_k^{\left( 2 \right) } = 0$.

Consequently, it can be derived that

$$\begin{aligned} \begin{aligned} \mathop {\lim }\limits _{M \rightarrow + \infty } {R_k} = {\log _2}\left( {1 + \frac{{{\gamma _k}}}{{1 - {\gamma _k}}}} \right) . \end{aligned} \end{aligned}$$

(71)

Similarly, with the definition of ${A_k}$, ${B_k}$, ${C_k}$ and ${D_k}$, we can get that

$$\begin{aligned} \begin{aligned} {\log _2}\left( {1 + \frac{{M{A_k}}}{{M{B_k} + 1}}} \right) \le {R_E} \le {\log _2}\left( {1 + \frac{{M{C_k}}}{{M{D_k} + 1}}} \right) . \end{aligned} \end{aligned}$$

(72)

Evidently, we note that ${A_k}$, ${B_k}$, ${C_k}$ and ${D_k}$ are all positive terms, and don’t scale with M. Hence, we can further get that

Plugging (71) and (72) into (36) can complete the proof.