Eavesdropping-decoding compromise in spectrum sharing paradigm with ES-capable AF relay

Abstract

In this paper, an unlicensed relay is deployed to maintain wireless connection between an unlicensed source and an unlicensed destination in spectrum sharing paradigm when source-destination direct channel is unavailable. The relay is capable of energy scavenging (ES) from source signal and uses the scavenged energy to amplify-and-forward it to the destination. The relay’s information transmission is eavesdropped by a wire-tapper. For prompt system performance evaluation, the current paper suggests two closed-form formulas of eavesdropping outage probability at the wire-tapper and decoding outage probability at the destination under interference power limitation and peak transmit power limitation, from which eavesdropping-decoding compromise in the spectrum sharing paradigm with the ES-capable relay is recognized. Numerous results validate the analysis and expose that relay location, times of signal relaying and energy scavenging, received power distribution for signal processing and energy scavenging can be optimized to achieve the best eavesdropping-decoding compromise.

Appendices

Appendix 1: Proof of Lemma 3

Rewriting ${{{\mathcal {J}}}_3}\left( {a,b,c,g} \right)$ as $\int \limits _a^\infty {{e^{ - \frac{b}{z}}}\frac{{{e^{ - cz}}}}{{{{\left( {z + g} \right) }^2}}}dz}$ and then applying the series expansion for ${{e^{ - \frac{b}{z}}}}$, one can simplify ${{{\mathcal {J}}}_3}\left( {a,b,c,g} \right)$ as

$$\begin{aligned} \begin{aligned} {{{\mathcal {J}}}_3}\left( {a,b,c,g} \right)&= \int \limits _a^\infty {\left[ {\sum \limits _{v = 0}^\infty {\frac{1}{{v!}}{{\left( { - \frac{b}{z}} \right) }^v}} } \right] \frac{{{e^{ - cz}}}}{{{{\left( {z + g} \right) }^2}}}dz} \\&= \int \limits _a^\infty {\frac{{{e^{ - cz}}}}{{{{\left( {z + g} \right) }^2}}}dz} + \sum \limits _{v = 1}^\infty {\frac{{{{\left( { - b} \right) }^v}}}{{v!}}} \int \limits _a^\infty {\frac{{{e^{ - cz}}}}{{{z^v}{{\left( {z + g} \right) }^2}}}dz}. \end{aligned} \end{aligned}$$

(37)

Applying the partial fraction decomposition to $\frac{1}{{{z^v}{{\left( {z + g} \right) }^2}}}$, one can simplify (37) as

$$\begin{aligned} {{{\mathcal {J}}}_3}\left( {a,b,c,g} \right) = \int \limits _a^\infty {\frac{{{e^{ - cz}}}}{{{{\left( {z + g} \right) }^2}}}dz} + \sum \limits _{v = 1}^\infty {\frac{1}{{v!}}{{\left( {\frac{b}{g}} \right) }^v}} \int \limits _a^\infty {\left( {\frac{{{e^{ - cz}}}}{{{{\left( {z + g} \right) }^2}}} + \frac{v}{g}\frac{{{e^{ - cz}}}}{{z + g}} + \sum \limits _{u = 1}^v {\frac{{v - u + 1}}{{{{\left( { - g} \right) }^{2 - u}}}}\frac{{{e^{ - cz}}}}{{{z^u}}}} } \right) dz}. \end{aligned}$$

(38)

By denoting ${{\mathcal {G}}}\left( {a,c,g} \right) = \int \limits _a^\infty {\frac{{{e^{ - cz}}}}{{{{\left( {z + g} \right) }^2}}}dz}$ and using the definitions of ${{\mathcal {L}}}\left( {a,c,g} \right) = \int \limits _a^\infty {\frac{{{e^{ - cz}}}}{{z + g}}dz}$ and ${{\mathcal {U}}}\left( {u,p,n} \right) = \int \limits _u^\infty {\frac{{{e^{ - pz}}}}{{{z^{n + 1}}}}dz}$, it is observed that (38) accurately matches (28). As such, in order to finish the proof of Lemma 3, one must prove that ${{\mathcal {G}}}\left( {a,c,g} \right)$ can be represented as (29). Performing the integral by parts and then using the definition of ${{\mathcal {L}}}\left( {a,c,g} \right) = \int \limits _a^\infty {\frac{{{e^{ - cz}}}}{{z + g}}dz}$, it is straightforward to reduce ${{\mathcal {G}}}\left( {a,c,g} \right)$ to

$$\begin{aligned} \begin{aligned} {{\mathcal {G}}}\left( {a,c,g} \right)&= - \left. {\frac{{{e^{ - cy}}}}{{y + g}}} \right| _a^\infty - c\int \limits _a^\infty {\frac{{{e^{ - cy}}}}{{y + g}}dy} \\&= \frac{{{e^{ - ac}}}}{{a + g}} - c\int \limits _a^\infty {\frac{{{e^{ - cy}}}}{{y + g}}dy}, \end{aligned} \end{aligned}$$

(39)

which accurately matches (29), finishing the proof.

Appendix 2: Proof of Theorem 1

Rewrite (23) as

$$\begin{aligned} \text {DOP} = 1 - \Pr \left\{ {{\Phi _d} \ge x} \right\} . \end{aligned}$$

(40)

According to [31], the SNR at D, ${\Phi _d} = \frac{{{\gamma _{sr}}{\gamma _{rd}}}}{{{\gamma _{sr}} + {\gamma _{rd}} + 1}}$, in (11), can be tightly approximated as

$$\begin{aligned} {\Phi _d} \approx \min \left( {{\gamma _{sr}},{\gamma _{rd}}} \right) . \end{aligned}$$

(41)

Using (41), one can tightly approximate the DOP in (40) as

$$\begin{aligned} \begin{aligned} \text {DOP}&\approx 1 - \Pr \left\{ {\min \left( {{\gamma _{sr}},{\gamma _{rd}}} \right) \ge x} \right\} \\&= 1 - \underbrace{\Pr \left\{ {\min \left( {{\gamma _{sr}},\frac{{{{\left| {{h_{rd}}} \right| }^2}{{{\tilde{P}}}_r}}}{{\sigma _d^2}}} \right) \ge x} \right\} }_{{{\overline{\text {DOP}} }}}. \end{aligned} \end{aligned}$$

(42)

Because $\gamma _{sr}$ and ${{\tilde{P}}}_r$ contain ${P_s}{\left| {{h_{sr}}} \right| ^2}$, they are correlated. Therefore, the ${{{\overline{\text {DOP}} }}}$ can be computed through two steps: i) Step 1 computes the conditional ${{{\overline{\text {DOP}} }}}$ given ${P_s}{\left| {{h_{sr}}} \right| ^2}$; ii) Step 2 averages the conditional ${{{\overline{\text {DOP}} }}}$ over ${P_s}{\left| {{h_{sr}}} \right| ^2}$. In other words, the ${{{\overline{\text {DOP}} }}}$ must be computed as

$$\begin{aligned} \begin{aligned} {{{\overline{\text {DOP}} }}}&= {\Xi _{{\gamma _{sr}},{{{\tilde{P}}}_r}}}\left\{ {\left. {\Pr \left\{ {\min \left( {{\gamma _{sr}},\frac{{{{\left| {{h_{rd}}} \right| }^2}{{{\tilde{P}}}_r}}}{{\sigma _d^2}}} \right) \ge x} \right\} } \right| {\gamma _{sr}},{{{\tilde{P}}}_r}} \right\} \\&= {\Xi _{{\gamma _{sr}},{{{\tilde{P}}}_r}}}\left\{ {\left. {\Pr \left\{ {{\gamma _{sr}} \ge x,\frac{{{{\left| {{h_{rd}}} \right| }^2}{{{\tilde{P}}}_r}}}{{\sigma _d^2}} \ge x} \right\} } \right| {\gamma _{sr}},{{{\tilde{P}}}_r}} \right\} \\&= {\Xi _{{\gamma _{sr}},{{{\tilde{P}}}_r}}}\left\{ {\left. {\Pr \left\{ {\frac{{{{\left| {{h_{rd}}} \right| }^2}{{{\tilde{P}}}_r}}}{{\sigma _d^2}} \ge x} \right\} } \right| {\gamma _{sr}} \ge x,{{{\tilde{P}}}_r}} \right\} \\&= {\Xi _{{\gamma _{sr}},{{{\tilde{P}}}_r}}}\left\{ {\left. {1 - {F_{{{\left| {{h_{rd}}} \right| }^2}}}\left( {\frac{{\sigma _d^2x}}{{{{{\tilde{P}}}_r}}}} \right) } \right| {\gamma _{sr}} \ge x,{{{\tilde{P}}}_r}} \right\} \\&= {\Xi _{{\gamma _{sr}},{{{\tilde{P}}}_r}}}\left\{ {\left. {{e^{ - \frac{{\sigma _d^2x}}{{{{{\tilde{P}}}_r}{\lambda _{rd}}}}}}} \right| {\gamma _{sr}} \ge x,{{{\tilde{P}}}_r}} \right\} . \end{aligned} \end{aligned}$$

(43)

Inserting ${{\tilde{P}}_r} = \min \left( {\frac{{{I_m}}}{{{{\left| {{h_{rl}}} \right| }^2}}},{P_r}} \right)$ in (22) into the above, one can rewrite the ${{{\overline{\text {DOP}} }}}$ as

$$\begin{aligned} \begin{aligned} {{{\overline{\text {DOP}} }}}&= {\Xi _{{\gamma _{sr}},{{{\tilde{P}}}_r}}}\left\{ {\left. {{e^{ - \frac{{\sigma _d^2x}}{{\min \left( {\frac{{{I_m}}}{{{{\left| {{h_{rl}}} \right| }^2}}},{P_r}} \right) {\lambda _{rd}}}}}}} \right| {\gamma _{sr}} \ge x,{{{\tilde{P}}}_r}} \right\} \\&= {\Xi _{{\gamma _{sr}},{P_r}}}\left\{ {\left. {\int \limits _0^\infty {{e^{ - \frac{{\sigma _d^2x}}{{\min \left( {\frac{{{I_m}}}{y},{P_r}} \right) {\lambda _{rd}}}}}}} {f_{{{\left| {{h_{rl}}} \right| }^2}}}\left( y \right) dy} \right| {\gamma _{sr}} \ge x,{P_r}} \right\} \\&= {\Xi _{{\gamma _{sr}},{P_r}}}\left\{ {\left. {\int \limits _{\frac{{{I_m}}}{{{P_r}}}}^\infty {{e^{ - \frac{{\sigma _d^2x}}{{\frac{{{I_m}}}{y}{\lambda _{rd}}}}}}} \frac{1}{{{\lambda _{rl}}}}{e^{ - \frac{y}{{{\lambda _{rl}}}}}}dy + \int \limits _0^{\frac{{{I_m}}}{{{P_r}}}} {{e^{ - \frac{{\sigma _d^2x}}{{{P_r}{\lambda _{rd}}}}}}} \frac{1}{{{\lambda _{rl}}}}{e^{ - \frac{y}{{{\lambda _{rl}}}}}}dy} \right| {\gamma _{sr}} \ge x,{P_r}} \right\} \\&= {\Xi _{{\gamma _{sr}},{P_r}}}\left\{ {\left. {{e^{ - \frac{{\sigma _d^2x}}{{{P_r}{\lambda _{rd}}}}}}} \right| {\gamma _{sr}} \ge x,{P_r}} \right\} - {\left( {\frac{{{I_m}{\lambda _{rd}}}}{{\sigma _d^2{\lambda _{rl}}x}} + 1} \right) ^{ - 1}}{\Xi _{{\gamma _{sr}},{P_r}}}\left\{ {\left. {{e^{ - \left( {\frac{{\sigma _d^2x}}{{{\lambda _{rd}}}} + \frac{{{I_m}}}{{{\lambda _{rl}}}}} \right) \frac{1}{{{P_r}}}}}} \right| {\gamma _{sr}} \ge x,{P_r}} \right\} . \end{aligned} \end{aligned}$$

(44)

By denoting

$$\begin{aligned} \Psi \left( B \right) = {\Xi _{{\gamma _{sr}},{P_r}}}\left\{ {\left. {{e^{ - \frac{B}{{{P_r}}}}}} \right| {\gamma _{sr}} \ge x,{P_r}} \right\} , \end{aligned}$$

(45)

one can reduce (44) to

$$\begin{aligned} {\overline{\text {DOP}}} = \Psi \left( {\frac{{\sigma _d^2x}}{{{\lambda _{rd}}}}} \right) - {\left( {\frac{{{I_m}{\lambda _{rd}}}}{{\sigma _d^2{\lambda _{rl}}x}} + 1} \right) ^{ - 1}}\Psi \left( {\frac{{\sigma _d^2x}}{{{\lambda _{rd}}}} + \frac{{{I_m}}}{{{\lambda _{rl}}}}} \right) . \end{aligned}$$

(46)

By plugging (46) into (42), it is seen that (42) accurately matches (30). As such, the proof of Theorem 1 is finished if (45) is proved to match (31). Toward this end, one substitutes (3) and (12) into (45):

$$\begin{aligned} \Psi \left( B \right) = {\Xi _{{\gamma _{sr}},{P_r}}}\left\{ {\left. {{e^{ - \frac{B}{{\frac{{\alpha \eta \mu }}{{1 - \alpha }}\left( {{P_s}{{\left| {{h_{sr}}} \right| }^2} + \sigma _r^2} \right) }}}}} \right| \frac{{{P_s}{{\left| {{h_{sr}}} \right| }^2}}}{{{\hat{\sigma }} _r^2}} \ge x,{P_r}} \right\} . \end{aligned}$$

(47)

By denoting $Y = {P_s}{\left| {{h_{sr}}} \right| ^2}$ and letting M as (35), one simplifies (47) as

$$\begin{aligned} \begin{aligned} \Psi \left( B \right)&= {\Xi _Y}\left\{ {\left. {{e^{ - \frac{MB}{{Y + \sigma _r^2}}}}} \right| Y \ge {\hat{\sigma }} _r^2x} \right\} \\&= \int \limits _{{\hat{\sigma }} _r^2x}^\infty {{e^{ - \frac{MB}{{y + \sigma _r^2}}}}{f_Y}\left( y \right) dy}. \end{aligned} \end{aligned}$$

(48)

The PDF of Y can be derived through its CDF as ${f_Y}\left( y \right) = \frac{{d{F_Y}\left( y \right) }}{{dy}}$. Therefore, the CDF of Y must be derived first. Using the definition of the CDF, one obtains

$$\begin{aligned} \begin{aligned} {F_Y}\left( y \right)&= \Pr \left\{ {{P_s}{{\left| {{h_{sr}}} \right| }^2} \le y} \right\} \\&= \Pr \left\{ {\min \left( {\frac{{{I_m}}}{{{{\left| {{h_{sl}}} \right| }^2}}},{P_m}} \right) {{\left| {{h_{sr}}} \right| }^2} \le y} \right\} \\&= \Pr \left\{ {\frac{{{I_m}}}{{{{\left| {{h_{sl}}} \right| }^2}}}{{\left| {{h_{sr}}} \right| }^2} \le y,\frac{{{I_m}}}{{{{\left| {{h_{sl}}} \right| }^2}}} \le {P_m}} \right\} + \Pr \left\{ {{P_m}{{\left| {{h_{sr}}} \right| }^2} \le y,\frac{{{I_m}}}{{{{\left| {{h_{sl}}} \right| }^2}}}> {P_m}} \right\} \\&= \Pr \left\{ {{{\left| {{h_{sr}}} \right| }^2} \le \frac{{{{\left| {{h_{sl}}} \right| }^2}}}{{{I_m}}}y,\frac{{{I_m}}}{{{P_m}}} \le {{\left| {{h_{sl}}} \right| }^2}} \right\} + \Pr \left\{ {{{\left| {{h_{sr}}} \right| }^2} \le \frac{y}{{{P_m}}},\frac{{{I_m}}}{{{P_m}}} > {{\left| {{h_{sl}}} \right| }^2}} \right\} \\&= \int \limits _{\frac{{{I_m}}}{{{P_m}}}}^\infty {\left[ {\int \limits _0^{\frac{y}{{{I_m}}}q} {{f_{{{\left| {{h_{sr}}} \right| }^2}}}\left( z \right) dz} } \right] {f_{{{\left| {{h_{sl}}} \right| }^2}}}\left( q \right) dq} + \left[ {\int \limits _0^{\frac{y}{{{P_m}}}} {{f_{{{\left| {{h_{sr}}} \right| }^2}}}\left( z \right) dz} } \right] \left[ {\int \limits _0^{\frac{{{I_m}}}{{{P_m}}}} {{f_{{{\left| {{h_{sl}}} \right| }^2}}}\left( q \right) dq} } \right] \\&= \int \limits _{\frac{{{I_m}}}{{{P_m}}}}^\infty {{F_{{{\left| {{h_{sr}}} \right| }^2}}}\left( {\frac{y}{{{I_m}}}q} \right) {f_{{{\left| {{h_{sl}}} \right| }^2}}}\left( q \right) dq} + {F_{{{\left| {{h_{sr}}} \right| }^2}}}\left( {\frac{y}{{{P_m}}}} \right) {F_{{{\left| {{h_{sl}}} \right| }^2}}}\left( {\frac{{{I_m}}}{{{P_m}}}} \right) . \end{aligned} \end{aligned}$$

(49)

Plugging (19) and (20) into (49), ${F_Y}\left( y \right)$ is further simplified as

$$\begin{aligned} \begin{aligned} {F_Y}\left( y \right)&= \int \limits _{\frac{{{I_m}}}{{{P_m}}}}^\infty {\left[ {1 - {e^{ - \frac{{yq}}{{{I_m}{\lambda _{sr}}}}}}} \right] \frac{1}{{{\lambda _{sl}}}}{e^{ - \frac{q}{{{\lambda _{sl}}}}}}dq} + \left( {1 - {e^{ - \frac{y}{{{P_m}{\lambda _{sr}}}}}}} \right) \left( {1 - {e^{ - \frac{{{I_m}}}{{{P_m}{\lambda _{sl}}}}}}} \right) \\&= 1 - \left( {1 - {e^{ - \frac{{{I_m}}}{{{P_m}{\lambda _{sl}}}}}}} \right) {e^{ - \frac{y}{{{P_m}{\lambda _{sr}}}}}} - \frac{{{I_m}{\lambda _{sr}}}}{{{\lambda _{sl}}}}{e^{ - \frac{{{I_m}}}{{{P_m}{\lambda _{sl}}}}}}\frac{{{e^{ - \frac{y}{{{P_m}{\lambda _{sr}}}}}}}}{{y + \frac{{{I_m}{\lambda _{sr}}}}{{{\lambda _{sl}}}}}} \\&= 1 - \left( {1 - V} \right) {e^{ - Uy}} - VT\frac{{{e^{ - Uy}}}}{{y + T}}. \end{aligned} \end{aligned}$$

(50)

where V, T, and U are defined in (32), (33), and (34), respectively.

Taking the derivative of (50), one obtains the PDF of Y as

$$\begin{aligned} {f_Y}\left( y \right) = \left( {1 - V} \right) U{e^{ - Uy}} + VTU\frac{{{e^{ - Uy}}}}{{y + T}} + VT\frac{{{e^{ - Uy}}}}{{{{\left( {y + T} \right) }^2}}}. \end{aligned}$$

(51)

Plugging (51) into (48) and after performing the variable change ${z = y + \sigma _r^2}$, (48) is simplified as

$$\begin{aligned} \begin{aligned} \Psi \left( B \right)&= \int \limits _{{\hat{\sigma }} _r^2x}^\infty {{e^{ - \frac{MB}{{y + \sigma _r^2}}}}\left[ {\left( {1 - V} \right) U{e^{ - Uy}} + VTU\frac{{{e^{ - Uy}}}}{{y + T}} + VT\frac{{{e^{ - Uy}}}}{{{{\left( {y + T} \right) }^2}}}} \right] dy} \\&= \left( {1 - V} \right) U{e^{U\sigma _r^2}}\int \limits _{{\hat{\sigma }} _r^2x + \sigma _r^2}^\infty {{e^{ - \frac{MB}{z} - Uz}}dz} + VTU{e^{U\sigma _r^2}}\int \limits _{{\hat{\sigma }} _r^2x + \sigma _r^2}^\infty {\frac{{{e^{ - \frac{MB}{z} - Uz}}}}{{z + T - \sigma _r^2}}dz} \\&\quad + VT{e^{U\sigma _r^2}}\int \limits _{{\hat{\sigma }} _r^2x + \sigma _r^2}^\infty {\frac{{{e^{ - \frac{MB}{z} - Uz}}}}{{{{\left( {z + T - \sigma _r^2} \right) }^2}}}dz}. \end{aligned} \end{aligned}$$

(52)

By rewriting (52) in terms of ${{{\mathcal {J}}}_1}\left( {a,b,c} \right) = \int \limits _a^\infty {{e^{ - \frac{b}{z} - cz}}dz}$, ${{{\mathcal {J}}}_2}\left( {a,b,c,g} \right) = \int \limits _a^\infty {\frac{{{e^{ - \frac{b}{z} - cz}}}}{{z + g}}dz}$, and ${{{\mathcal {J}}}_3}\left( {a,b,c,g} \right) = \int \limits _a^\infty {\frac{{{e^{ - \frac{b}{z} - cz}}}}{{{{\left( {z + g} \right) }^2}}}dz}$, it is seen that (52) accurately matches (31), finishing the proof.