Evaluating the Performance of Full-Duplex Energy Harvesting Vehicle-to-Vehicle Communication System over Double Rayleigh Fading Channels

Abstract

In this paper, we analyze the performance of vehicle-to-vehicle (V2V) communication system, which employs full-duplex (FD) and energy harvesting (EH) techniques at source and relay nodes from power beacon (PB) through radio frequency. Unlike previous systemswhere all nodes located at fixed locations, we investigate the case that three nodes (source, relay,and destination) are moving vehicles. Therefore, the channels between them follow double (cascade) Rayleigh fading distributions. Furthermore, the source and relay nodes can harvest the energy from PB for data transmission when they move on the road. We derive the exact expressions of the outage probability (OP) and symbol error probability (SEP) of the proposed system and then intensively study the impacts of various parameters such as the number transmission antennas of PB, the time duration for EH, the distances between nodes, and the residual self-interference (RSI) at the FD relay node on the system performance. Monte-Carlo simulations validate all theory analysis. Numerical results show that system performance is strongly impacted by the number of transmission antennas of the power beacon, the EH duration, the RSI, and the distances between nodes. Moreover, for a given transmission of power beacon and the SIC capability of the FD relay node, there exist optimal EH duration and optimal distance from the source to relay, which provide the best system performance.

Appendices

Appendix A

This appendix presents step-by-step derivations of the mathematical expression of the outage probability in Theorem 1. Firstly, the probability Pr{γ_R < x} is calculated as Eq. A1.

$$ \begin{array}{@{}rcl@{}} \text{Pr}\{\gamma_{\mathrm{R}} < x\} &=& \text{Pr}\left\{\frac{\eta \alpha P |h_{\text{SR}}|^{2} }{(1-\alpha)d_{\text{SR}}^{\beta}(\gamma_{\text{RSI}} +\sigma^{2})} \sum\limits_{i = 1}^{\mathrm{K}} |h_{i\mathrm{S}}|^{2} < x\right\}\\ &=& \text{Pr}\left\{|h_{\text{SR}}|^{2} < \frac{(1-\alpha)d_{\text{SR}}^{\beta}(\gamma_{\text{RSI}} +\sigma^{2})x }{\eta \alpha P \sum\limits_{i = 1}^{\mathrm{K}} |h_{i\mathrm{S}}|^{2} } \right\} \end{array} $$

(A1)

Due to the fact that $|h_{\text {SR}}|^{2}=|h_{\text {SR}_{1}}|^{2}|h_{\text {SR}_{2}}|^{2}$, the CDF of |h_SR|² is computed as

$$ \begin{array}{@{}rcl@{}} F_{|h_{\text{SR}}|^{2}}(x)&=& \Pr(|h_{\text{SR}_{1}}|^{2}|h_{\text{SR}_{2}}|^{2} < x)\\ &=&{\int}_{0}^{\infty}\Pr\Big(|h_{\text{SR}_{2}}|^{2} < \frac{x}{y}\Big)f_{|h_{\text{SR}_{1}}|^{2}}(y)dy. \end{array} $$

(A2)

From Eqs. 17 and 18, we can rewrite (A2) as

$$ \begin{array}{@{}rcl@{}} F_{|h_{\text{SR}}|^{2}}(x)&=&1-\frac{1}{{{\varOmega}}_{3}}{\int}_{0}^{\infty}\exp\left( -\frac{y}{{{\varOmega}}_{3}}-\frac{x}{y{{\varOmega}}_{4}}\right)dy \\ &=&1-\sqrt{\frac{4x}{{{\varOmega}}_{3}{{\varOmega}}_{4}}} K_{1}\Bigg(\sqrt{\frac{4x}{{{\varOmega}}_{3}{{\varOmega}}_{4}}}\Bigg). \end{array} $$

(A3)

We can see from Eq. A3 that, the CDF of |h_SR|² implicitly takes into account the effects of scatterers around the transmitter and the receiver which include the fluctuating amplitude and phase of signal, the Doppler shifts caused by the movement of vehicles [26].

Using Eqs. A3, 20, and A1 becomes (A4). It is also noted that, we set ${{\varOmega }}_{1} =\mathbb {E}\{|h_{i\mathrm {S}}|^{2}\}$, ${{\varOmega }}_{3} =\mathbb {E}\{|h_{\text {SR}_{1}}|^{2}\}$ and ${{\varOmega }}_{4} =\mathbb {E}\{|h_{\text {SR}_{2}}|^{2}\}$ in Eq. A4.

$$ \begin{array}{@{}rcl@{}} \text{Pr}\{\gamma_{\mathrm{R}} < x\} &=& \int\limits_{0}^{\infty} \Bigg[1-\sqrt{\frac{4(1-\alpha)d_{\text{SR}}^{\beta}(\gamma_{\text{RSI}} +\sigma^{2})x}{{{\varOmega}}_{3}{{\varOmega}}_{4}\eta \alpha P y}} K_{1}\Bigg(\sqrt{\frac{4(1-\alpha)d_{\text{SR}}^{\beta}(\gamma_{\text{RSI}} +\sigma^{2})x}{{{\varOmega}}_{3}{{\varOmega}}_{4}\eta \alpha P y}}\Bigg) \Bigg] \frac{y^{\mathrm{K}-1}}{{{\varOmega}}_{1}^{\mathrm{K}} {\Gamma}(\mathrm{K})} \exp\Big(-\frac{y}{{{\varOmega}}_{1}}\Big) \\ &=& \int\limits_{0}^{\infty} \Bigg[1-\sqrt{\frac{Ax}{y}} K_{1}\Bigg(\sqrt{\frac{Ax}{y}}\Bigg) \Bigg] \frac{y^{\mathrm{K}-1}}{{{\varOmega}}_{1}^{\mathrm{K}} {\Gamma}(\mathrm{K})} \exp\Big(-\frac{y}{{{\varOmega}}_{1}}\Big)\\ &=& \frac{1}{{{\varOmega}}_{1}^{\mathrm{K}} {\Gamma}(\mathrm{K})} \Bigg[\int\limits_{0}^{\infty} y^{\mathrm{K}-1} \exp\Big(-\frac{y}{{{\varOmega}}_{1}}\Big)dy - \sqrt{Ax} \int\limits_{0}^{\infty} K_{1}\Bigg(\sqrt{\frac{Ax}{y}}\Bigg) y^{\mathrm{K}-\frac{3}{2}} \exp\Big(-\frac{y}{{{\varOmega}}_{1}}\Big) dy\Bigg] . \end{array} $$

(A4)

For the first integral in Eq. A4, we apply [42, 3.351.3] to have

$$ \begin{array}{@{}rcl@{}} \int\limits_{0}^{\infty} y^{\mathrm{K}-1} \exp\Big(-\frac{y}{{{\varOmega}}_{1}}\Big)dy ={{\varOmega}}_{1}^{\mathrm{K}} {\Gamma}(\mathrm{K}). \end{array} $$

(A5)

For the second integral in Eq. A4, we set $t = \exp \Big (-\frac {y}{{{\varOmega }}_{1}}\Big )$. After some algebra calculations, we obtain the following integral

$$ \begin{array}{@{}rcl@{}} &&{}\int\limits_{0}^{\infty} K_{1}\Bigg(\sqrt{\frac{Ax}{y}}\Bigg) y^{\mathrm{K}-\frac{3}{2}} \exp\Big(-\frac{y}{{{\varOmega}}_{1}}\Big) dy \\ &=& {{\varOmega}}_{1} {\int\limits_{0}^{1}} K_{1}\Bigg(\sqrt{\frac{Ax}{-{{\varOmega}}_{1} \ln t}}\Bigg) (-{{\varOmega}}_{1} \ln t)^{\mathrm{K}-\frac{3}{2}} dt. \end{array} $$

(A6)

Applying the Gaussian-Chebyshev quadrature method in [45], Eq. A6 becomes

$$ \begin{array}{@{}rcl@{}} &&{}{{\varOmega}}_{1} {\int\limits_{0}^{1}} K_{1}\Bigg(\sqrt{\frac{Ax}{-{{\varOmega}}_{1} \ln t}}\Bigg) (-{{\varOmega}}_{1} \ln t)^{\mathrm{K}-\frac{3}{2}} dt \\ &=& \frac{{{\varOmega}}_{1} \pi}{2M} \sum\limits_{m=1}^{M} \sqrt{1-{\phi_{m}^{2}}} K_{1}\Bigg(\sqrt{\frac{Ax}{-{{\varOmega}}_{1}\ln u}}\Bigg) (-{{\varOmega}}_{1}\ln u)^{{\mathrm{K}-\frac{3}{2}}}. \end{array} $$

(A7)

Substituting Eqs. A5 and A7 into Eq. A4, we get (A8).

$$ \begin{array}{@{}rcl@{}} \text{Pr}\{\gamma_{\mathrm{R}} < x\} &=& \frac{1}{{{\varOmega}}_{1}^{\mathrm{K}} {\Gamma}(\mathrm{K})} \left[{{\varOmega}}_{1}^{\mathrm{K}} {\Gamma}(\mathrm{K}) - \sqrt{Ax} \frac{{{\varOmega}}_{1} \pi}{2M} \sum\limits_{m=1}^{M} \sqrt{1-{\phi_{m}^{2}}}\right.\\ &&{\kern3.4pc}\left.\times K_{1}\Bigg(\sqrt{\frac{Ax}{-{{\varOmega}}_{1}\ln u}}\Bigg) (-{{\varOmega}}_{1}\ln u)^{{\mathrm{K}-\frac{3}{2}}} \right]\\ &=& 1 - \frac{\pi\sqrt{Ax}}{2M({{\varOmega}}_{1})^{\mathrm{K}-1} {\Gamma}(\mathrm{K})} \sum\limits_{m=1}^{M} \sqrt{1-{\phi_{m}^{2}}}\\ &&\times K_{1}\Bigg(\sqrt{\frac{Ax}{-{{\varOmega}}_{1}\ln u}}\Bigg) (-{{\varOmega}}_{1}\ln u)^{{\mathrm{K}-\frac{3}{2}}} \end{array} $$

(A8)

By doing the same calculations as for Pr{γ_R < x} for Pr{γ_D < x}, we have (A9).

$$ \begin{array}{@{}rcl@{}} \text{Pr}\{\gamma_{\mathrm{D}} < x\} &=& \text{Pr}\left\{\frac{\eta \alpha P |h_{\text{RD}}|^{2} }{(1-\alpha)d_{\text{RD}}^{\beta} \sigma^{2}} \sum\limits_{i = 1}^{\mathrm{K}} |h_{i\mathrm{R}}|^{2} < x\right\}\\ &=& \text{Pr}\left\{|h_{\text{RD}}|^{2} < \frac{(1-\alpha)d_{\text{RD}}^{\beta}\sigma^{2}x }{\eta \alpha P \sum\limits_{i = 1}^{\mathrm{K}} |h_{i\mathrm{R}}|^{2} } \right\}\\ &=& 1 - \frac{\pi\sqrt{Bx}}{2N({{\varOmega}}_{2})^{\mathrm{K}-1} {\Gamma}(\mathrm{K})} \sum\limits_{n=1}^{N} \sqrt{1-{\phi_{n}^{2}}}\\ &&\times K_{1}\Bigg(\sqrt{\frac{Bx}{-{{\varOmega}}_{2}\ln v}}\Bigg) (-{{\varOmega}}_{2}\ln v)^{{\mathrm{K}-\frac{3}{2}}}. \end{array} $$

(A9)

Then, plugging Eqs. A8 and A9 into Eq. 15, we obtain (16) in Theorem 1 which is the closed-form expression of the outage probability of the proposed system. The proof is complete.

Appendix B

In this section, we provide the detailed derivations of Eq. 22 in Theorem 2.

Replacing F(x) in Eq. 23 with P_out in Eq. 16, we obtain (B1).

$$ \begin{array}{@{}rcl@{}} \text{SEP} &=& \frac{a \sqrt b}{2\sqrt {2\pi }}\int\limits_{0}^{\infty} \frac{e^{-b x/2}}{\sqrt x} \Bigg[ 1-\frac{\pi^{2} \sqrt{AB}x}{4MN({{\varOmega}}_{1}{{\varOmega}}_{2})^{\mathrm{K}-1} {\Gamma}^{2}(\mathrm{K})}\\ &&\times\sum\limits_{m=1}^{M} \sum\limits_{n=1}^{N} \sqrt{(1-{\phi_{m}^{2}})(1-{\phi_{n}^{2}})} K_{1}\Bigg(\sqrt{\frac{Ax}{-{{\varOmega}}_{1}\ln u}}\Bigg) \\ &&\times K_{1}\Bigg(\sqrt{\frac{Bx}{-{{\varOmega}}_{2}\ln v}}\Bigg) ({{\varOmega}}_{1}{{\varOmega}}_{2}\ln u\ln v)^{\mathrm{K}-\frac{3}{2}}\Bigg]dx \\ &=& \frac{a \sqrt b}{2\sqrt {2\pi }} \Bigg[\int\limits_{0}^{\infty} \frac{e^{-b x/2}}{\sqrt x} dx -\frac{\pi^{2} \sqrt{AB} ({{\varOmega}}_{1}{{\varOmega}}_{2}\ln u\ln v)^{\mathrm{K}-\frac{3}{2}}}{4MN({{\varOmega}}_{1}{{\varOmega}}_{2})^{\mathrm{K}-1} {\Gamma}^{2}(\mathrm{K})}\\ &&\times\sum\limits_{m=1}^{M} \sum\limits_{n=1}^{N} \sqrt{(1-{\phi_{m}^{2}})(1-{\phi_{n}^{2}})}\\ &\times& \int\limits_{0}^{\infty} x^{\frac{1}{2}} e^{-b x/2} K_{1}\Bigg(\sqrt{\frac{Ax}{-{{\varOmega}}_{1}\ln u}}\Bigg) K_{1}\Bigg(\sqrt{\frac{Bx}{-{{\varOmega}}_{2}\ln v}}\Bigg)dx \Bigg]. \end{array} $$

(B1)

For the first integral in Eq. B1, we use [42, 3.361.2] to solve it, i.e.,

$$ \begin{array}{@{}rcl@{}} \int\limits_{0}^{\infty} \frac{e^{-b x/2}}{\sqrt x}dx = \sqrt{\frac{2\pi}{b}}. \end{array} $$

(B2)

For the second integral in Eq. B1, we set z = e^−bx/2. After some mathematical calculations, we get (B3).

$$ \begin{array}{@{}rcl@{}} &&{}\int\limits_{0}^{\infty} x^{\frac{1}{2}} e^{-b x/2} K_{1}\Bigg(\sqrt{\frac{Ax}{-{{\varOmega}}_{1}\ln u}}\Bigg) K_{1}\Bigg(\sqrt{\frac{Bx}{-{{\varOmega}}_{2}\ln v}}\Bigg)\\ {\kern12pt}dx &=& \frac{2}{b} {\int\limits_{0}^{1}} \sqrt{\frac{-2 \ln z}{b}} K_{1}\Bigg(\sqrt{\frac{2A\ln z}{b{{\varOmega}}_{1}\ln u}}\Bigg) K_{1}\Bigg(\sqrt{\frac{2B \ln z}{b{{\varOmega}}_{2}\ln v}}\Bigg)dz. \end{array} $$

(B3)

Due to the complexity of the integral in Eq. B3, it is hard to find the closed-form expression for this integral. Therefore, we use the Gaussian-Chebyshev quadrature method in [45] to solve this integral. After that, we have (B4).

$$ \begin{array}{@{}rcl@{}} &&{}\frac{2}{b} {\int\limits_{0}^{1}} \sqrt{\frac{-2 \ln z}{b}} K_{1}\Bigg(\sqrt{\frac{2A\ln z}{b{{\varOmega}}_{1}\ln u}}\Bigg) K_{1}\Bigg(\sqrt{\frac{2B \ln z}{b{{\varOmega}}_{2}\ln v}}\Bigg) dz = \frac{\pi}{Jb} \\ &&{\kern3pt}\times\sum\limits_{j=1}^{J} \sqrt{1-{\phi_{j}^{2}}} \sqrt{\frac{-2 \ln w}{b}} K_{1}\Bigg(\sqrt{\frac{2A\ln w}{b{{\varOmega}}_{1}\ln u}}\Bigg) K_{1}\Bigg(\sqrt{\frac{2B \ln w}{b{{\varOmega}}_{2}\ln v}}\Bigg). \end{array} $$

(B4)

Substituting Eqs. B2 and B4 into Eq. B1, we obtain (22) in Theorem 2 which is the closed-form expression of the SER of the proposed system. The proof is complete.