Performance Analysis of Full-Duplex Vehicle-to-Vehicle Relay System over Double-Rayleigh Fading Channels

Abstract

In this paper, we study the performance of a full-duplex (FD) relay system in vehicle-to-vehicle (V2V) communication. In this relay communication system, the communication link from the source node to the relay node can be modeled by Rayleigh fading or double (cascaded) Rayleigh fading distributions while the link from the relay node to the destination node is modeled by double Rayleigh fading distribution. Through the numerical calculation, we obtain the exact analytical expressions for the outage probabilities (OPs) and symbol error rates (SERs) in two cases, i.e. case A (the first hop is the Rayleigh fading channel and the second hop is double Rayleigh fading channel) and case B (two hops are the double Rayleigh fading channels). From these obtained mathematical expressions, the impacts of the distances between the communication nodes and the residual self-interference caused by the imperfect self-interference cancellation at the FD relay are studied. In addition, the effects of path loss exponent and the transmission power at the FD relay are also investigated. Numerical results show that the system performance in terms of OP and SER in the case of double Rayleigh fading channels is significant lower than the case of Rayleigh fading channels. Monte-Carlo simulations are conducted to validate the correctness of these numerical results.

Appendices

Appendix A

This appendix gives detailed derivations of the outage probability of FD-V2V communication system in case A and case B corresponding to one and two communication links are double Rayleigh fading channels, respectively.

Let us start with the Rayleigh fading channel, the probability density function (PDF) and cumulative distribution function (CDF) of the instantaneous channel gain |g|² are given by

$$ f_{|g|^{2}}(z) = \frac{1}{\Omega}\exp\left( - \frac{z}{\Omega}\right), $$

(17)

$$ F_{|g|^{2}}(z)=1-\exp\left( -\frac{z}{\Omega}\right). $$

(18)

where ${\Omega } = \mathbb {E}\{|g|^{2}\}$.

In the case of double Rayleigh fading channels, the instantaneous channel gain |h|² is the multiplication of two independent variables |g₁|² and |g₂|², where |g₁|² and |g₂|² are the instantaneous channel gains of the Rayleigh fading channels between two vehicles [32]. Thus, we have the CDF of |h|² as

$$ \begin{array}{@{}rcl@{}} F_{|h|^{2}}(z)&=& \Pr(|g_{1}|^{2}|g_{2}|^{2} \le z)\\ &=&{\int}_{0}^{\infty}\Pr\left( |g_{2}|^{2}\le \frac{z}{|g_{1}|^{2}}\right)f_{|g_{1}|^{2}}(t)dt\\ &=&1-\frac{1}{{\Omega}_{1}}{\int}_{0}^{\infty}\exp\left( -\frac{t}{{\Omega}_{1}}-\frac{z}{t{\Omega}_{2}}\right)dt \\ &=&1-\sqrt{\frac{4z}{{\Omega}_{1}{\Omega}_{2}}} K_{1}\left( \sqrt{\frac{4z}{{\Omega}_{1}{\Omega}_{2}}}\right). \end{array} $$

(19)

From Eq. 19, we can obtain the PDF of |h|² as

$$ f_{|h|^{2}}(z) =\frac{2}{{\Omega}_{1}{\Omega}_{2}} K_{0}\left( \sqrt{\frac{4z}{{\Omega}_{1}{\Omega}_{2}}}\right), $$

(20)

where ${\Omega }_{1} = \mathbb {E}\{|g_{1}|^{2}\}, {\Omega }_{2} = \mathbb {E}\{|g_{2}|^{2}\}$.

From the PDF and CDF functions of the instantaneous channel gains given in Eqs. 18, 17, 19, and 20, we can derive the OP in the case A and B as follows.

1.1 Case A

To derive the OP of the system in this case, we firstly derive the CDF of γ_R and γ_D. Since the S →R link is Rayleigh fading channel, its instantaneous channel gain is defined as |h_SR|² = |g₁|² and the average channel gain ${\Omega }_{1}=\mathbb {E}\{|g_{1}|^{2}\}$. Then, the CDF of γ_R can be given by

$$ \begin{array}{@{}rcl@{}} \text{Pr}\{\gamma_{\mathrm{R}} <x\} &=& \text{Pr}\left\{\frac{ P_{\mathrm{S}}|h_{\text{SR}}|^{2}}{{d}_{\text{SR}}^{\alpha} (\gamma_{\text{RSI}} +\sigma^{2})} <x\right\} \\ &=& \text{Pr}\left\{\frac{ P_{\mathrm{S}}|g_{1}|^{2}}{{d}_{\text{SR}}^{\alpha} (\gamma_{\text{RSI}} +\sigma^{2})} <x\right\} \\ &=& 1 - \exp \left( -\frac{{d}_{\text{SR}}^{\alpha} (\gamma_{\text{RSI}} +\sigma^{2})x}{{\Omega}_{1} P_{\mathrm{S}}} \right)\\ &=& 1 - \exp(-X_{\mathrm{A}} x). \end{array} $$

(21)

Meanwhile, the R →D link is double Rayleigh fading channels, thus |h_RD|² = |g₃|²|g₄|² and ${\Omega }_{3}=\mathbb {E}\{|g_{3}|^{2}\}$, ${\Omega }_{4}=\mathbb {E}\{|g_{4}|^{2}\}$. Consequently, the CDF of γ_D can be calculated as

$$ \begin{array}{@{}rcl@{}} \text{Pr}\{\gamma_{\mathrm{D}} <x\} &=& \text{Pr}\left\{\frac{P_{\mathrm{R}}|h_{\text{RD}}|^{2}}{d_{\text{RD}}^{\alpha} \sigma^{2}} <x\right\} \\ &=& \text{Pr}\left\{\frac{P_{\mathrm{R}}|g_{3}|^{2}|g_{4}|^{2}}{d_{\text{RD}}^{\alpha} \sigma^{2}} <x\right\}\\ &=& 1 - \sqrt{\frac{4d_{\text{RD}}^{\alpha} \sigma^{2}x}{{\Omega}_{3}{\Omega}_{4} P_{\mathrm{R}}}} K_{1} \left( \sqrt{\frac{4d_{\text{RD}}^{\alpha} \sigma^{2}x}{{\Omega}_{3}{\Omega}_{4} P_{\mathrm{R}}}} \right)\\ &=& 1 - \sqrt{4Y_{\mathrm{A}} x} K_{1}(\sqrt{4Y_{\mathrm{A}} x}). \end{array} $$

(22)

Based on Eq. 12, we have the OP of the FD-V2V communication system in the case A as in Eq. 9.

1.2 Case B

Similar to case A, we have

$$ \begin{array}{@{}rcl@{}} \text{Pr}\{\gamma_{\mathrm{R}} <x\} &=& \text{Pr}\left\{\frac{ P_{\mathrm{S}}|h_{\text{SR}}|^{2}}{{d}_{\text{SR}}^{\alpha} (\gamma_{\text{RSI}} +\sigma^{2})} <x\right\} \\ &=& \text{Pr}\left\{\frac{ P_{\mathrm{S}}|g_{1}|^{2}|g_{2}|^{2}}{{d}_{\text{SR}}^{\alpha} (\gamma_{\text{RSI}} +\sigma^{2})} <x\right\} \\ &=& 1 - \sqrt{\frac{{4d}_{\text{SR}}^{\alpha} (\gamma_{\text{RSI}} +\sigma^{2})x}{{\Omega}_{1}{\Omega}_{2} P_{\mathrm{S}}}}\\ &&K_{1} \left( \sqrt{\frac{{4}d_{\text{SR}}^{\alpha} (\gamma_{\text{RSI}} +\sigma^{2})x}{{\Omega}_{1}{\Omega}_{2} P_{\mathrm{S}}}} \right)\\ &=& 1 - \sqrt{4X_{\mathrm{B}} x} K_{1}(\sqrt{4X_{\mathrm{B}} x}), \end{array} $$

(23)

where |h_SR|² = |g₁|²|g₂|², ${\Omega }_{1}=\mathbb {E}\{|g_{1}|^{2}\}$ and ${\Omega }_{2}=\mathbb {E}\{|g_{2}|^{2}\}$. Based on the similar calculations as in Eq. 22, we have

$$ \text{Pr}\{\gamma_{\mathrm{D}} <x\} = 1 - \sqrt{4Y_{\mathrm{B}} x} K_{1}(\sqrt{4Y_{\mathrm{B}} x}). $$

(24)

Using Eq. 12, the OP of the FD-V2V communication system in the case B is derived as in Eq. 10. The proof is completed.

Appendix B

The section gives detailed derivations of the SER of the FD-V2V communication system in case A and case B.

1.1 Case A

From Eq. 16, substituting F(x) by OP in Eq. 9, we have

$$ \begin{array}{@{}rcl@{}} {\text{SER}}_{\mathrm{A}} \!&=&\! \frac{a \sqrt b}{2\sqrt {2\pi }}\left[\int\limits_{0}^{\infty}\frac{e^{-b x/2}}{\sqrt x}dx \right.\\ &&\!- \left.2\int\limits_{0}^{\infty} \frac{\exp (-x(X_{\mathrm{A}}+\frac{b}{2})) \sqrt{Y_{\mathrm{A}} x} K_{1}(2\sqrt{Y_{\mathrm{A}} x})}{\sqrt x} dx \right]\\ \end{array} $$

(25)

For the first integral in Eq. 25, using [38, Eq.3.361.2], we have

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

(26)

For the second integral in Eq. 25, using [38, Eq.6.643.3], we have

$$ \begin{array}{@{}rcl@{}} &&\int\limits_{0}^{\infty} \exp \left.\left( -x(X_{\mathrm{A}}+\frac{b}{2}\right)\right) \sqrt{Y_{\mathrm{A}}} K_{1}(2\sqrt{Y_{\mathrm{A}} x}) dx \\ &=& \frac{{\Gamma}\left( \frac{3}{2}\right) {\Gamma}\left( \frac{1}{2}\right)}{2\sqrt{X_{\mathrm{A}}+\frac{b}{2}}} \exp\left( \frac{Y_{\mathrm{A}}}{2(X_{\mathrm{A}}+\frac{b}{2})}\right) W_{-\frac{1}{2},\frac{1}{2}}\left( \frac{Y_{\mathrm{A}}}{X_{\mathrm{A}}+\frac{b}{2}}\right)\\ \end{array} $$

(27)

Substituting (26) and (27) into Eq. 25, we obtain the SER in case A as Eq. 14.

1.2 Case B

Similar to case A, we have

$$ \begin{array}{@{}rcl@{}} {\text{SER}}_{\mathrm{B}} &=& \frac{a \sqrt b}{2\sqrt {2\pi }}\left[\int\limits_{0}^{\infty}\frac{e^{-b x/2}}{\sqrt x}dx \right.\\ &&- \left.4\int\limits_{0}^{\infty} \frac{e^{-b x/2}\sqrt{X_{\mathrm{B}}Y_{\mathrm{B}} x^{2}} K_{1}(2\sqrt{X_{\mathrm{B}} x}) K_{1}(2\sqrt{Y_{\mathrm{B}} x})}{\sqrt x} dx \right]\\ \end{array} $$

(28)

The first integral in Eq. 28 can be easily obtained as in Eq. 26. For the second integral, setting z = e^−bx/2 results in $x=-\frac {2}{b}\ln z$. Hence, we can rewrite the second integral as

$$ \begin{array}{@{}rcl@{}} &&\int\limits_{0}^{\infty} e^{-b x/2}\sqrt{X_{\mathrm{B}}Y_{\mathrm{B}} x} K_{1}(2\sqrt{X_{\mathrm{B}} x}) K_{1}(2\sqrt{Y_{\mathrm{B}} x}) dx \\ &=& {\int\limits_{0}^{1}} z \sqrt{X_{\mathrm{B}}Y_{\mathrm{B}} \left( -\frac{2}{b}\ln z\right)} K_{1}\left( 2\sqrt{X_{\mathrm{B}} \left( -\frac{2}{b}\ln z\right)}\right)\\ &&\times K_{1}\left( 2\sqrt{Y_{\mathrm{B}} \left( -\frac{2}{b}\ln z\right)}\right) \frac{2}{bz}dz =J. \end{array} $$

(29)

Using the Gaussian-Chebyshev quadrature method in [39], we can obtain J as

$$ \begin{array}{@{}rcl@{}} J&=& \frac{\pi}{Mb} \sum\limits_{n=1}^{M} \sqrt{1-{\phi_{n}^{2}}} \sqrt{-\frac{2X_{\mathrm{B}}Y_{\mathrm{B}}\ln y}{b}} \\ &&\times K_{1}\left( 2 \sqrt{-\frac{2X_{\mathrm{B}}\ln y}{b}}\right) K_{1}\left( 2 \sqrt{-\frac{2Y_{\mathrm{B}}\ln y}{b}}\right). \end{array} $$

(30)

Then, the SER in case B can be obtained as in Eq. 15. The proof is completed.