A Cooperative Transmission Scheme for the Secure Wireless Multicasting

Abstract

In this paper, a wireless multicast scenario with secrecy constraints is considered, where the source wishes to send a common message to two intended destinations in the presence of a passive eavesdropper. One destination is equipped with multiple antennas, and all of the other three nodes are equipped with a single antenna. Different to the conventional direct transmission, we propose a cooperative transmission scheme based on the cooperation between the two destinations. The basic idea is to divide the multicast scenario into two cooperative unicast transmissions at two phases and the two destinations help each other to jam the eavesdropper in turns. Such a cooperative transmission does not require the knowledge of the eavesdropper’s channel state information. Both analytic and numerical results demonstrate that the proposed cooperative scheme can achieve zero-approaching outage probability.

Appendix

Proof of Theorem 1

For $N=2$, from (12), we have

$$\begin{aligned} P_{out}^{(2)}&\mathop {\le }\limits ^{(a)} Pr \left( \frac{\min \{\gamma _1,\gamma _2\}}{1+\frac{|h_{SE}|^2}{|h_{E2}|^2}+\frac{|h_{SE}|^2}{|\mathbf {h}_{ez}|^2}} <2^{2R}\right) \nonumber \\&\le Pr \left( \frac{\min \{\gamma _1,\gamma _2\}}{1+\frac{2|h_{SE}|^2}{\min \{|h_{E2}|^2,|\mathbf {h}_{ez}|^2\}}} <2^{2R}\right) \nonumber \\&\mathop {=}\limits ^{(b)} Pr \left( \frac{x_1}{1+2x_2} <2^{2R}\right) , \end{aligned}$$

(15)

where $(a)$ is obtained by fixing $\rho =0$ and $(b)$ is obtained by defining $x_1=\min \{\gamma _1,\gamma _2\}$ and $x_2=\frac{z_1}{z_2}$ with $z_2=\min \{|h_{E2}|^2,|\mathbf {h}_{ez}|^2\}$. When $N=2$, according to (5), the PDF of $\gamma _1$ is exponential distributed. So $x_1$ is also exponential distributed with PDF

$$\begin{aligned} f_{x_1}(x_1)=\frac{2}{P}e^{-\frac{2x_1}{P}}\hbox { for }x_1>0. \end{aligned}$$

Thus, continuing from (15), we have

$$\begin{aligned} P_{out}^{(2)}&\le \mathbf {E}_{x_2} \left[ F_{x_1}\left( \left( 1+2x_2\right) 2^{2R}\right) \right] \nonumber \\&= \mathbf {E}_{x_2} \left[ 1-\exp \left\{ -\frac{2^{2R+1}(1+2x_2)}{P}\right\} \right] \nonumber \\&= 1-e^{-\frac{\epsilon }{2}}\mathbf {E}_{x_2} [e^{-\epsilon x_2}] , \end{aligned}$$

(16)

where $\epsilon =2^{2(R+1)}/P$ and $\mathbf {E}_{x_2}$ denotes the expectation with respect to the variable $x_2$.

Since the PDFs of $z_1$ and $z_2$ are $f_{z_1}(z_1)=e^{-z_1}$ $f_{z_2}(z_2)=2e^{-2z_2}$ when $N=2$, the cumulative density function (CDF) of $x_2$ can be obtained as

$$\begin{aligned} F_{x_2}(x_2)&=\int \limits _{0}^{\infty }\left( \int \limits _{0}^{x_2\cdot z_2} f_{z_1} d z_1\right) f_{z_2} d z_2\nonumber \\&=1-\frac{2}{x_2+2}. \end{aligned}$$

(17)

So the PDF of $x_2$ can be calculated as

$$\begin{aligned} f_{x_2}(x_2)=\frac{2}{(2+x_2)^2} \hbox { for } x_2>0. \end{aligned}$$

Then the outage probability can be upper bounded as

$$\begin{aligned} P_{out}^{(2)}&\le 1-e^{-\frac{\epsilon }{2}} \int \limits _0^{\infty } \frac{2e^{-\epsilon x}}{(x_2+2)^2} d x_2 \nonumber \\&= 1-e^{-\frac{\epsilon }{2}}\left( -2e^{2\epsilon }\int \limits _2^{\infty } e^{-\epsilon t}d\frac{1}{t}\right) \nonumber \\&= 1-e^{-\frac{\epsilon }{2}}\left[ 2e^{2\epsilon }\left( \frac{e^{-2\epsilon }}{2}-2\epsilon \int \limits _{2\epsilon }^{\infty } \frac{e^{-z}}{z}dz\right) \right] \nonumber \\&= 1-e^{-\frac{\epsilon }{2}}\left( 1+2\epsilon \cdot e^{2\epsilon }E_i(-2\epsilon )\right) . \end{aligned}$$

(18)