Joint Relay Selection and Destination Assisted Cooperative Jamming with Power Allocation for Secure DF Relay Network

Abstract

To enhance the physical-layer security of a decode-and-forward (DF) cooperative relay network in the presence of a potential eavesdropper, a joint relay selection and destination assisted cooperative jamming (CJ) scheme is proposed. In phase I, the source transmits its confidential signal, and concurrently, it cooperates with the friendly jammers and the destination to send jamming signals to confuse the eavesdropper without affecting the forwarding relay which is preselected. In phase II, the forwarding relay retransmits the decoded signal, and simultaneously, the particular relay cooperates with the friendly jammers to send jamming signal to confuse the eavesdropper without affecting the destination. The achievable secrecy rate maximization (SRM) problem is investigated by optimally allocating the power between the data signal and jamming signals. Although SRM problem is non-convex, the power allocation problem can be solved by a bisection method together with a one-dimensional search. Optimal relay selection and suboptimal relay selection schemes are also proposed to further improve the secrecy rate. Simulation results are presented to show that the joint scheme can attain higher achievable secrecy rates than some existing works.

Appendices

Appendix 1: Derivation of Lemma 1

Denote $f_{1} (P^{\prime}_{S} ,{\mathbf{w^{\prime}}}_{1} ) = \log_{2} \left( {\frac{{1 + P^{\prime}_{S} \left| {h_{SR} } \right|^{2} }}{{1 + \frac{{P^{\prime}_{S} \left| {h_{SE} } \right|^{2} }}{{1 + \left| {{\mathbf{w^{\prime}}}_{1}^{\dag } {\mathbf{h}}_{1} } \right|^{2} }}}}} \right),$ and $f_{2} (P^{\prime}_{S} ,{\mathbf{w^{\prime}}}_{1} ,P^{\prime}_{R} ,{\mathbf{w^{\prime}}}_{2} ) = \log_{2} \left( {\frac{{1 + P^{\prime}_{R} \left| {h_{RD} } \right|^{2} }}{{1 + \frac{{P^{\prime}_{S} \left| {h_{SE} } \right|^{2} }}{{1 + \left| {{\mathbf{w^{\prime}}}_{1}^{\dag } {\mathbf{h}}_{1} } \right|^{2} }} + \frac{{P^{\prime}_{R} \left| {h_{RE} } \right|^{2} }}{{1 + \left| {{\mathbf{w^{\prime}}}_{2}^{\dag } {\mathbf{h}}_{2} } \right|^{2} }}}}} \right).$ Since $P_{S}$ can be fixed, $f_{1} (P_{S} ,{\mathbf{w}}_{1} )$ is a strictly increasing function of $\left\| {{\mathbf{w}}_{1} } \right\|^{2} .$ Consequently, if $P^{\prime}_{S} + \left\| {{\mathbf{w^{\prime}}}_{1} } \right\|^{2} < P_{1} ,$ there exists $(P^{\prime}_{S} ,{\mathbf{w^{\prime}}}_{3} ),$$\left\| {{\mathbf{w^{\prime}}}_{3} } \right\|^{2} > \left\| {{\mathbf{w^{\prime}}}_{1} } \right\|^{2}$ such that $f_{1} (P^{\prime}_{S} ,{\mathbf{w^{\prime}}}_{3} ) > f_{1} (P^{\prime}_{S} ,{\mathbf{w^{\prime}}}_{1} ).$ This contradicts the assumption that $(P^{\prime}_{S} ,{\mathbf{w^{\prime}}}_{1} )$ is optimum. Thus, it holds that $P^{\prime}_{S} + \left\| {{\mathbf{w^{\prime}}}_{1} } \right\|^{2} = P_{1} .$ Similarly, we may show $P^{\prime}_{R} + \left\| {{\mathbf{w^{\prime}}}_{2} } \right\|^{2} < P_{0} - P_{1}$ is impossible. This completes the proof.

Appendix 2: Derivation of Lemma 2

Before proceeding, we denote the objective function of (15) as $H(P_{R} )$. It is easy to see that the problem of (15) is an optimization problem of a single $P_{R}$ variable, and the maximum is achieved at one of the following points: 0, $P_{0} - P_{1}$ and the points with zero derivative of $H(P_{R} )$$\left( {H^{\prime}(P_{R} ) = 0} \right).$ And we have the following property w.r.t. $H^{\prime}(P_{R} ).$

Property 1

Whenever function $H^{\prime}(P_{R} )$ crosses the value 0 at any point belongs to $[0,P_{0} - P_{1} ],$ we must have $H^{\prime\prime}(P_{R} ) < 0$ at this point.

Proof

The first- and the second-derivatives of $H(P_{R} )$ are respectively given by

$$ H^{\prime}(P_{R} ) = \frac{d}{{dP_{R} + 1}} - \frac{es}{{\left( {\left( {e - mf - f} \right)P_{R} + sm + s} \right)\left( {s - fP_{R} } \right)}} $$

(22)

$$ H^{\prime\prime}(P_{R} ) = - \frac{{d^{2} }}{{(dP_{R} + 1)^{2} }} + {\kern 1pt} {\kern 1pt} \frac{{es(2mf^{2} P_{R} - 2efP_{R} + 2f^{2} P_{R} - 2mfs + es - 2fs)}}{{\left( {\left( {e - mf - f} \right)P_{R} + sm + s} \right)^{2} \left( {s - fP_{R} } \right)^{2} }} $$

(23)

and hence

$$ H^{\prime}(0) = d - \frac{e}{(sm + s)} $$

(24)

$$ H^{\prime}(P_{0} - P_{1} ) = \frac{d}{{d(P_{0} - P) + 1}} - \frac{es}{{\left( {\left( {e - mf - f} \right)\left( {P_{0} - P} \right) + sm + s} \right)\left( {s - f\left( {P_{0} - P} \right)} \right)}} $$

(25)

If $H^{\prime}(P_{R}^{*} ) = 0$ with $P_{R}^{*} \in [0,P_{0} - P_{1} ],$ then we have

$$ H^{\prime\prime}(P_{R}^{*} ) = \frac{2esf}{{\left( {\left( {m + 1} \right)\left( {fP_{R}^{*} - s} \right) - eP_{R}^{*} } \right)\left( {fP_{R}^{*} - s} \right)^{2} }}. $$

(26)

Note that $fP_{R}^{*} - s \le f(P_{0} - P_{1} ) - ((P_{0} - P_{1} )f + 1) < 0,$ then we must have $H^{\prime\prime}(P_{R}^{*} ) < 0.$ This completes the proof. Based on the above property, we could also have the following result stated in Property 2.

Property 2

$H^{\prime}(P_{R} )$ could only have the following behaviors over $P_{R} \in [0,P_{0} - P_{1} ]$

• i. If $H^{\prime}(P_{R} )$ cross the value 0 at point $P_{R}^{*} \in [0,P_{0} - P_{1} ],$ $P_{R}^{*}$ must be unique. And for $P_{R} \in [0,P_{R}^{*} ],$ we have $H^{\prime}(P_{R} ) > 0;$ for $P_{R} \in [P_{R}^{*} ,P_{0} - P_{1} ],$ we have $H^{\prime}(P_{R} ) < 0.$

• ii. If $H^{\prime}(P_{R} )$ does not cross the value 0 at all, $H^{\prime}(P_{R} ) > 0$ or $H^{\prime}(P_{R} ) < 0$ in $[0,P_{0} - P_{1} ].$

Proof

First, we claim that if $H^{\prime}(P_{R} )$ crosses the value 0, it can cross the value 0 at most once. To see why this is the case, we should note that $H^{\prime}(P_{R} )$ is quadratic w.r.t $P_{R} ,$ it cannot cross value 0 more than twice. Thus, we assume that $H^{\prime}(P_{R} )$ crosses the value 0 twice at the two points $0 \le P_{R}^{*} < \tilde{P}_{R}^{*} \le P_{0} - P_{1} .$ From Property 1, we know that $H^{\prime\prime}(P_{R}^{*} ) < 0$ and $H^{\prime\prime}(\tilde{P}_{R}^{*} ) < 0$. According to the continuity of $H^{\prime}(P_{R} )$, there exists a very small $\sigma > 0$ such that $H^{\prime}(P_{R}^{*} + \sigma ) < 0$ and $H^{\prime}(\tilde{P}_{R}^{*} - \sigma ) > 0$. From the intermediate value theorem, we know that there is at least another one point $\bar{P}_{R}^{*} \in \left[ {P_{R}^{*} ,\tilde{P}_{R}^{*} } \right]$ that makes $H^{\prime}(\bar{P}_{R}^{*} ) = 0$ and $H^{\prime\prime}(\bar{P}_{R}^{*} ) > 0$ (otherwise $H^{\prime}(P_{R} )$ cannot cross value 0), which contradicts the assumption that $H^{\prime}(P_{R} )$ crosses value 0 twice and Property 1. Therefore it should cross 0 exactly once. If $H^{\prime}(P_{R} )$ crosses the value 0 at point $P_{R}^{*} \in [0,P_{0} - P_{1} ]$, because $H^{\prime\prime}(P_{R}^{*} ) < 0$, we know that $H^{\prime}(P_{R} ) > 0$ for $P_{R} \in [0,P_{R}^{*} ]$ and $H^{\prime}(P_{R} ) < 0$ for $P_{R} \in [P_{R}^{*} ,P_{0} - P_{1} ]$.

If $H^{\prime}(P_{R} )$ does not cross the value 0 at all, it is obviously that $H^{\prime}(P_{R} ) > 0$ or $H^{\prime}(P_{R} ) < 0$ in $[0,P_{0} - P_{1} ]$. This completes the proof of Property 2.

If $H^{\prime}(0) > 0$ and $H^{\prime}(P_{0} - P_{1} ) < 0$, according to Property 2 (i) $H^{\prime}(P_{R} )$ crosses the value 0 at the unique point $P_{R}^{*} \in [0,P_{0} - P_{1} ]$, $H^{\prime}(P_{R} ) > 0$, i.e., $H(P_{R} )$ is increasing for $P_{R} \in [0,P_{R}^{*} ]$, and $H^{\prime}(P_{R} ) < 0$, i.e., $H(P_{R} )$ is decreasing for $P_{R} \in [P_{R}^{*} ,P_{0} - P_{1} ]$, so $H(P_{R} )$ attains its maximum at $P_{R} = P_{R}^{*}$. To obtain the exact value of $P_{R}^{*}$, let $H^{\prime}(P_{R} ) = 0$, and we get an equation

$$ A_{1} (P_{R} )^{2} + B_{1} P_{R} + C_{1} = 0. $$

(27)

The solution to (27) is $\frac{{ - B_{1} \pm \sqrt {B_{1}^{2} - 4A_{1} C_{1} } }}{{2A_{1} }}$, while $\frac{{ - B_{1} }}{{2A_{1} }} = \frac{{2\left[ {1 + f\left( {P_{0} - P_{1} } \right)} \right]}}{f - g} \ge 2\left( {\frac{1}{f} + P_{0} - P_{1} } \right),$ thus $P_{R}^{*} = \frac{{ - B_{1} - \sqrt {B_{1}^{2} - 4A_{1} C_{1} } }}{{2A_{1} }}$ should be the only choice. This means if $H^{\prime}(0) > 0$ and $H^{\prime}(P_{0} - P_{1} ) < 0$, the solution of (15) is $G(P_{S} ) = \frac{{ - B_{1} - \sqrt {B_{1}^{2} - 4A_{1} C_{1} } }}{{2A_{1} }}$.

If $H^{\prime}(0) < 0$ and $H^{\prime}(P_{0} - P_{1} ) > 0$, there must be a point $P_{R}^{*} \in [0,P_{0} - P_{1} ]$ with $H^{\prime}(P_{R}^{*} ) = 0$ that makes $H^{\prime\prime}(P_{R}^{*} ) > 0$, which contradicts with Property 1, thus this case is impossible.

If $H^{\prime}(P_{0} - P_{1} ) > 0$ and $H^{\prime}(0) > 0$, it is obviously $H^{\prime}(P_{R} )$ cannot crosses value 0 exactly once, according to Property 2, we can obtain that $H^{\prime}(P_{R} ) > 0$, i.e., $H(P_{R} )$ is strictly increasing on $[0,P_{0} - P_{1} ]$, and its maximum happens to be $H(P_{0} - P_{1} )$, i.e., $G(P_{S} ) = P_{0} - P_{1}$.

Similarly, if $H^{\prime}(P_{0} - P_{1} ) < 0$ and $H^{\prime}(0) < 0$, we can obtain that $H^{\prime}(P_{R} ) < 0$, i.e., $H(P_{R} )$ is strictly decreasing on $[0,P_{0} - P_{1} ]$. The maximum value is $H(0) = \frac{1}{1 + m} < 1$, which implies that the secrecy rate is negative. In this case, the security of the system cannot be guaranteed.

This completes the proof.

Appendix 3: Derivation of Lemma 3

Now we first show $J_{2} (P_{S} )$ is a strictly decreasing function of $P_{S}$.

If $G(P_{S} ) = P_{0} - P_{1} ,$ then $J_{2} (P_{S} ) = \frac{{1 + d(P_{0} - P_{1} )}}{{1 + m + e(P_{0} - P_{1} )}}.$ It is easy to see that $J_{2} (P_{S} )$ decreases with $m$. Since $m$ increases with $P_{S}$, then $J_{2} (P_{S} )$ is decreasing function of $P_{S}$.

If $G(P_{S} ) = 0$, then $J_{2} (P_{S} ) = \frac{1}{1 + m}$. Thus in this case $J_{2} (P_{S} )$ is still a decreasing function of $P_{S}$.

Finally, if $G(P_{S} ) = \frac{{ - B_{1} - \sqrt {B_{1}^{2} - 4A_{1} C_{1} } }}{{2A_{1} }}$, substituting (18b)–(18d) into (17), we can transform $g$ into a function of $G(P_{S} )$.

$$ g = \frac{{d(1 + fP_{2} - fG\text{(}P_{S} \text{)})^{2} }}{{1 + fP_{2} + dfG\text{(}P_{S} \text{)}^{2} }} $$

(28)

where $P_{2} = P_{0} - P_{1}$. We now have

$$ J_{2} (P_{S} ) = \frac{{1 + dG\text{(}P_{S} \text{)}}}{{\frac{e}{g} + \frac{{eG\text{(}P_{S} \text{)}}}{{1 + (P_{2} - G\text{(}P_{S} \text{)})f}}}} $$

(29)

Substitue (28) into (29), we can further obtain

$$ \begin{aligned} & J_{2} (P_{S} ) \\ & {\kern 1pt} {\kern 1pt} = \frac{{1 + dG\text{(}P_{S} \text{)}}}{{e\left( {\frac{{1 + fP_{2} + dfG\text{(}P_{S} \text{)}^{2} }}{{d\left( {1 + fP_{2} - fG\text{(}P_{S} \text{)}} \right)}} + \frac{G}{{1 + fP_{2} - fG\text{(}P_{S} \text{)}}}} \right)}} \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} & = \frac{{d(1 + fP_{2} - fG\text{(}P_{S} \text{)})^{2} }}{{e(1 + fP_{2} )}}. \\ \end{aligned} $$

(30)

Since $G\text{(}P_{S} \text{)} \in [0,P_{2} ]$, $J_{2} (P_{S} )$ monotonically increases with $G(P_{S} )$, and we know that $G(P_{S} )$ decreases with $P_{S}$. As a result, $J_{2} (P_{S} )$ decreases with $P_{S}$.

From the above three cases, we conclude that $J_{2} (P_{S} )$ is strictly decreasing and thus quasi-concave. Due to the convexity of $J_{1} (P_{S} )$ with fixed $P_{1}$, $J(P_{S} )$ is a point-wise minimum of a family of affine functions, as a result quasi-concave [30]. And its maximum value can be obtained via a bisection method. This completes the proof.