Secrecy Performance of Multi-RIS-Assisted Wireless Systems

Abstract

This paper presents a theoretical framework using multiple reconfigurable intelligent surfaces (RISs) for enhancing the secrecy performance of wireless systems. In particular, multiple RISs are exploited to support transmitter-legitimate user communication under the existence of an eavesdropper. Two practical scenarios are investigated, i.e., there are only transmitter-eavesdropper links (case 1) and there are both transmitter-eavesdropper and transmitter-RIS-eavesdropper links (case 2). We mathematically obtain the closed-form expressions of the average secrecy capacity (ASC) of the considered system in these two investigated cases over Nakagami-m fading channels. The impacts of the system parameters, such as the locations of the RISs, the number of REs, and the Nakagami-m channels, are deeply evaluated. Computer simulations are used to validate our mathematical analysis. Numerical results clarify the benefits of using multiple RISs for improving the secrecy performance of wireless systems. Specifically, the ASCs in cases 1 and 2 are significantly higher than that in the case without RISs. Importantly, when the locations of the RISs are fixed, we can arrange the larger RSIs near either the transmitter or legitimate user to achieve higher ASCs. In addition, when the numbers of reflecting elements (REs) in the RISs increase, the ASCs in cases 1 and 2 are greatly enhanced.

Appendix

This appendix detailedly provides the step-by-step derivations of the CDFs of $\beta _d$, $\beta _e^{c1}$, and $\beta _e^{c2}$.

Firstly, $F_{\beta _e^{c1}}(x)$ can be derived directly by using the CDF of the channel gain following Nakagami-m fading channels [37], i.e.,

$$\begin{aligned} \nonumber F_{\beta _e^{c1}}(x)&= \textrm{Pr} \left\{ h_{se}^2 \rho _e< x \right\} = \textrm{Pr} \left\{ h_{se}^2 < \frac{x}{\rho _e} \right\} \\ \nonumber&= \frac{1}{\Gamma (m_{se})} \gamma \Bigg ( m_{se}, \frac{m_{se} x}{\Omega _{se} \rho _e } \Bigg ) \\&= 1 - \frac{1}{\Gamma (m_{se})} \Gamma \Bigg ( m_{se}, \frac{m_{se} x}{\Omega _{se} \rho _e } \Bigg ). \end{aligned}$$

(37)

Secondly, $F_{\beta _d}(x)$ and $F_{\beta _e^{c2}}(x)$ are calculated as follows. Let $\mathcal {X}_{dkl}= g_{kl} h_{kl}$, $\mathcal {Y}_{dk} = \sum _{l=1}^{L_k} \mathcal {X}_{dkl}$, $\mathcal {Z}_{d} = \sum _{k=1}^{K} \mathcal {Y}_{dk}$, $\mathcal {T}_{d} = h_{sd} + \mathcal {Z}_d$, $\mathcal {X}_{ekl}= g_{kl} r_{kl}$, $\mathcal {Y}_{ek} = \sum _{l=1}^{L_k} \mathcal {X}_{ekl}$, $\mathcal {Z}_e = \sum _{k=1}^{K} \mathcal {Y}_{ek}$, and $\mathcal {T}_e = h_{se} + \mathcal {Z}_e$ be new variables, Eqs. 23 and 25 respectively become

$$\begin{aligned} F_{\beta _d}(x) = \textrm{Pr} \left\{ \mathcal {T}_{d}^2 \rho _d < x \right\} , \end{aligned}$$

(38)

$$\begin{aligned} F_{\beta _e^{c2}}(x) = \textrm{Pr} \left\{ \mathcal {T}_e^2 \rho _e < x \right\} . \end{aligned}$$

(39)

It is obvious that $F_{\beta _d}(x)$ and $F_{\beta _e^{c2}}(x)$ have similar types. Thus, in the following parts, we focus on deriving $F_{\beta _d}(x)$, $F_{\beta _e^{c2}}(x)$ can be derived similarly as $F_{\beta _d}(x)$.

Since the Nakagami-m fading channels are considered, the nth moment of $h_{sd}$ is given by [28]

$$\begin{aligned} \mu _{h_{sd}}(n) \triangleq \mathbb {E}\{h_{sd}^n\} = \frac{\Gamma (m_{sd}+n/2)}{\Gamma (m_{sd})} \Big ( \frac{m_{sd}}{\Omega _{sd}}\Big )^{-n/2}. \end{aligned}$$

(40)

From Eq. 40, we obtain the first and second moments of $h_{sd}$ as

$$\begin{aligned} \mu _{h_{sd}}(1) = \frac{\Gamma (m_{sd}+1/2)}{\Gamma (m_{sd})} \sqrt{\frac{\Omega _{sd} }{m_{sd}}}, \end{aligned}$$

(41)

$$\begin{aligned} \mu _{h_{sd}}(2) = \frac{\Gamma (m_{sd}+1)\Omega _{sd}}{\Gamma (m_{sd})m_{sd}} = \Omega _{sd}. \end{aligned}$$

(42)

Since $\mathcal {X}_{dkl}= g_{kl} h_{kl}$, the PDF of $\mathcal {X}_{dkl}$ is calculated as

$$\begin{aligned} f_{\mathcal {X}_{dkl}}(y) = \int _{0}^{\infty } \frac{1}{x} f_{h_{kl}} \Big (\frac{y}{x}\Big ) f_{g_{kl}} (x) dx. \end{aligned}$$

(43)

Replacing the PDF given in Eq. 11 into Eq. 43, we have

$$\begin{aligned} \nonumber f_{\mathcal {X}_{dkl}}(y) =&\frac{4}{\Gamma (m_{g_k}) \Gamma (m_{h_k}) } \Big (\frac{m_{g_k}}{ \Omega _{g_k}}\Big )^{m_{g_k}} \Big (\frac{m_{h_k}}{ \Omega _{h_k}}\Big )^{m_{h_k}} \\ \nonumber&\times y^{2 m_{h_k}-1} \int _{0}^{\infty } x^{2 m_{g_k} - 2 m_{h_k} -1} \\&\times \exp \Big (- \frac{m_{g_k} x^2}{\Omega _{g_k}} -\frac{y^2 m_{h_k}}{ \Omega _{h_k} x^2} \Big ) dx. \end{aligned}$$

(44)

Applying [36, Eq. (3.478.4)], Eq. 44 becomes

$$\begin{aligned} f_{\mathcal {X}_{dkl}}(y) = \frac{4 \alpha _{kl}^{m_{g_k}+m_{h_k}} }{\Gamma (m_{g_k}) \Gamma (m_{h_k})} y^{m_{g_k}+ m_{h_k}-1} \mathcal {K}_{m_{g_k} - m_{h_k}} (2 \alpha _{kl} y), \end{aligned}$$

(45)

where $\alpha _{kl} = \sqrt{\frac{m_{g_k} m_{h_k}}{\Omega _{g_k} \Omega _{h_k}}}$.

Now, the nth moment of $\mathcal {X}_{dkl}$ is computed as

$$\begin{aligned} \mu _{\mathcal {X}_{dkl}}(n) \triangleq \mathbb {E}\{\mathcal {X}_{dkl}^n\} = \int _{0}^{\infty } y^n f_{\mathcal {X}_{dkl}}(y) dy. \end{aligned}$$

(46)

Using [36, Eq. (6.561.16)], Eq. 46 becomes

$$\begin{aligned} \mu _{\mathcal {X}_{dkl}}(n) = \alpha _{kl}^{-n} \frac{\Gamma (m_{g_k}+n/2) \Gamma (m_{h_k}+n/2) }{\Gamma (m_{g_k}) \Gamma (m_{h_k}) }. \end{aligned}$$

(47)

Then, the CDF of $\mathcal {X}_{dkl}$ is given by

$$\begin{aligned} \nonumber&F_{\mathcal {X}_{dkl}}(x) \approx \frac{1}{\Gamma \Big ( \frac{[\mu _{\mathcal {X}_{dkl}}(1)]^2 }{\mu _{\mathcal {X}_{dkl}}(2) - [\mu _{\mathcal {X}_{dkl}}(1)]^2} \Big )} \\&\times \gamma \Big ( \frac{[\mu _{\mathcal {X}_{dkl}}(1)]^2 }{\mu _{\mathcal {X}_{dkl}}(2) - [\mu _{\mathcal {X}_{dkl}}(1)]^2}, \frac{\mu _{\mathcal {X}_{dkl}}(1) x}{\mu _{\mathcal {X}_{dkl}}(2) - [\mu _{\mathcal {X}_{dkl}}(1)]^2} \Big ) \cdot \end{aligned}$$

(48)

Now, we can derive the CDF of $\mathcal {Y}_{dk} = \sum _{l=1}^{L_k} \mathcal {X}_{dkl}$ as

$$\begin{aligned} \nonumber&F_{\mathcal {Y}_{dk}}(x) \approx \frac{1}{\Gamma \Big ( \frac{L_k [\mu _{\mathcal {X}_{dkl}}(1)]^2 }{\mu _{\mathcal {X}_{dkl}}(2) - [\mu _{\mathcal {X}_{dkl}}(1)]^2} \Big )} \\&\times \gamma \Big ( \frac{L_k [\mu _{\mathcal {X}_{dkl}}(1)]^2 }{\mu _{\mathcal {X}_{dkl}}(2) - [\mu _{\mathcal {X}_{dkl}}(1)]^2}, \frac{\mu _{\mathcal {X}_{dkl}}(1) x}{\mu _{\mathcal {X}_{dkl}}(2) - [\mu _{\mathcal {X}_{dkl}}(1)]^2} \Big ) \cdot \end{aligned}$$

(49)

Based on [38], we obtain the nth moment of $\mathcal {Y}_{dk}$ as

$$\begin{aligned} \nonumber \mu _{\mathcal {Y}_{dk}}(n)&\triangleq \mathbb {E}\{\mathcal {Y}_{dk}^n\} \\ \nonumber&= \sum _{n_1=0}^{n} \sum _{n_2=0}^{n_1} \cdots \sum _{n_{L_k -1}=0}^{n_{L_k -2}} {n \atopwithdelims ()n_1} {n_1 \atopwithdelims ()n_2} \cdots {n_{L_k -2} \atopwithdelims ()n_{L_k -1}} \\&\times \mu _{\mathcal {X}_{dk1}}(n -n_1) \mu _{\mathcal {X}_{dk2}}(n_1 -n_2) \cdots \mu _{\mathcal {X}_{dkL_k}}(n_{L_k -1}), \end{aligned}$$

(50)

where ${a \atopwithdelims ()b} = \frac{a!}{b!(a-b)!}$, and the nth moment of $\mathcal {Z}_d = \sum _{k=1}^{K} \mathcal {Y}_{dk}$ is

$$\begin{aligned} \nonumber \mu _{{\mathcal {Z}_d}}(n)&\triangleq \mathbb {E}\{\mathcal {Z}_d^n\} \\ \nonumber&= \sum _{n_1=0}^{n} \sum _{n_2=0}^{n_1} \cdots \sum _{n_{K -1}=0}^{n_{K -2}} {n \atopwithdelims ()n_1} {n_1 \atopwithdelims ()n_2} \cdots {n_{K -2} \atopwithdelims ()n_{K -1}} \\&\times \mu _{\mathcal {Y}_{d1}}(n -n_1) \mu _{\mathcal {Y}_{d2}}(n_1 -n_2) \cdots \mu _{\mathcal {Y}_{dK}}(n_{K -1}). \end{aligned}$$

(51)

From Eqs. 47,50 and 51, we compute the first and second moments of ${\mathcal {Z}_d}$ as

$$\begin{aligned} \mu _{\mathcal {Z}_d}(1) = \sum _{k=1}^{K} \sum _{l=1}^{L_k} \mu _{\mathcal {X}_{dkl}}(1) , \end{aligned}$$

(52)

$$\begin{aligned} \nonumber \mu _{\mathcal {Z}_d}(2) =&\sum _{k=1}^{K} \Big [ \sum _{l=1}^{L_k} \mu _{\mathcal {X}_{dkl}}(2) +2 \sum _{l=1}^{L_k} \mu _{\mathcal {X}_{dkl}}(1) \sum _{l'=l+1}^{L_k} \mu _{\mathcal {X}_{dkl'}}(1) \Big ] \\&+ 2 \sum _{k=1}^{K} \Big [ \sum _{l=1}^{L_k} \mu _{\mathcal {X}_{dkl}}(1) \Big ] \sum _{k'=k+1}^{K} \Big [ \sum _{l=1}^{L_{k'}} \mu _{\mathcal {X}_{dk'l}}(1) \Big ] . \end{aligned}$$

(53)

Then, the nth moment of ${\mathcal {T}_{d}} = h_{sd} + {\mathcal {Z}_d}$ is calculated as

$$\begin{aligned} \nonumber \mu _{\mathcal {T}_d}(n)&\triangleq \mathbb {E}\{(h_{sd} + \mathcal {Z}_d)^n\} = \mathbb {E}\left\{ \sum _{i=0}^{n} {n \atopwithdelims ()i} h_{sd}^i \mathcal {Z}_d^{n-i} \right\} \\&= \sum _{i=0}^{n} {n \atopwithdelims ()i} \mu _{h_{sd}}(i) \mu _{\mathcal {Z}_d}(n-i). \end{aligned}$$

(54)

Consequently, the first and second moments of ${\mathcal {T}_d}$ calculated from Eq. 54 are

$$\begin{aligned} \mu _{\mathcal {T}_d}(1) = \mu _{h_{sd}}(1) + \mu _{\mathcal {Z}_d}(1), \end{aligned}$$

(55)

$$\begin{aligned} \mu _{\mathcal {T}_d}(2) = \mu _{h_{sd}}(2) + \mu _{\mathcal {Z}_d}(2) + 2 \mu _{h_{sd}}(1) \mu _{\mathcal {Z}_d}(1). \end{aligned}$$

(56)

Similarly, the first and second moments of ${\mathcal {T}_e}$ are expressed as

$$\begin{aligned} \mu _{\mathcal {T}_e}(1) = \mu _{h_{se}}(1) + \mu _{\mathcal {Z}_e}(1), \end{aligned}$$

(57)

$$\begin{aligned} \mu _{\mathcal {T}_e}(2) = \mu _{h_{se}}(2) + \mu _{\mathcal {Z}_e}(2) + 2 \mu _{h_{se}}(1) \mu _{\mathcal {Z}_e}(1). \end{aligned}$$

(58)

Therefore, the CDFs of ${\mathcal {T}_d}$ and ${\mathcal {T}_e}$ are, respectively, given by

$$\begin{aligned} \nonumber F_{\mathcal {T}_d}(x) =&\frac{1}{\Gamma \Big ( \frac{[\mu _{\mathcal {T}_d}(1)]^2 }{\mu _{\mathcal {T}_d}(2) - [\mu _{\mathcal {T}_d}(1)]^2} \Big )} \\ \nonumber&\times \gamma \Bigg ( \frac{[\mu _{\mathcal {T}_d}(1)]^2 }{\mu _{\mathcal {T}_d}(2) - [\mu _{\mathcal {T}_d}(1)]^2}, \frac{\mu _{\mathcal {T}_d}(1) x}{\mu _{\mathcal {T}_d}(2) - [\mu _{\mathcal {T}_d}(1)]^2} \Bigg ) \\ \nonumber =&\frac{1}{\Gamma (\Xi _d)} \gamma ( \Xi _d, \Psi _d x ) \\ =&1-\frac{1}{\Gamma (\Xi _d)} \Gamma ( \Xi _d, \Psi _d x ) , \end{aligned}$$

(59)

$$\begin{aligned} F_{\mathcal {T}_e}(x) = 1- \frac{1}{\Gamma (\Xi _e)} \Gamma ( \Xi _e, \Psi _e x ) , \end{aligned}$$

(60)

where

$$\begin{aligned} \Xi _d =\frac{[\mu _{\mathcal {T}_d}(1)]^2 }{\mu _{\mathcal {T}_d}(2) - [\mu _{\mathcal {T}_d}(1)]^2}, \end{aligned}$$

(61)

$$\begin{aligned} \Psi _d =\frac{\mu _{\mathcal {T}_d}(1) }{\mu _{\mathcal {T}_d}(2) - [\mu _{\mathcal {T}_d}(1)]^2}, \end{aligned}$$

(62)

$$\begin{aligned} \Xi _e = \frac{[\mu _{\mathcal {T}_e}(1)]^2 }{\mu _{\mathcal {T}_e}(2) - [\mu _{\mathcal {T}_e}(1)]^2}, \end{aligned}$$

(63)

$$\begin{aligned} \Psi _e =\frac{\mu _{\mathcal {T}_e}(1) }{\mu _{\mathcal {T}_e}(2) - [\mu _{\mathcal {T}_e}(1)]^2}. \end{aligned}$$

(64)

Next, we can calculate the CDFs of $\beta _d$ and $\beta _e^{c2}$ from Eqs.  38 and 39 as

$$\begin{aligned} F_{\beta _d}(x) = \textrm{Pr} \left\{ \mathcal {T}_{d}^2< \frac{x}{\rho _d} \right\} = \textrm{Pr} \left\{ \mathcal {T}_{d} < \sqrt{\frac{x}{\rho _d}} \right\} = F_{\mathcal {T}_{d}}\Big (\sqrt{\frac{x}{\rho _d}}\Big ), \end{aligned}$$

(65)

$$\begin{aligned} F_{\beta _e^{c2}}(x) = \textrm{Pr} \left\{ \mathcal {T}_{e}^2< \frac{x}{\rho _e} \right\} = \textrm{Pr} \left\{ \mathcal {T}_{e} < \sqrt{\frac{x}{\rho _e}} \right\} = F_{\mathcal {T}_{e}}\Big (\sqrt{\frac{x}{\rho _e}}\Big ). \end{aligned}$$

(66)

Applying Eqs. 59 and 60,65, and 66 become Eqs. 26 and 28, respectively. The proof is thus complete.