Outage probability optimization of UAV relay system based on elliptical trajectory

Abstract

Unmanned aerial vehicles (UAVs), which enable both high flexibility and low cost, have been widely used in both military and civilian application scenarios. In this paper, we investigate a fixed-wing UAV relay system, where the fixed-wing UAV acts as a mobile relay with an elliptical trajectory between base station and ground users with broken communication links because of obstructions. Different time and power allocations are utilized to compute the outage probability among the Rician fading channel model. Due to the complexity of the formulated issue, we approximate the asymptotic solution using the first-order Maclaurin series convex optimization. Finally, we contrast the elliptical trajectory relay’s outage probability performance with that of the circular trajectory relay and the fixed relay. In the meanwhile, we examine the impact of the elliptical trajectory’s various long and short axis ratios on the outage probability of a relay system. The efficiency of the suggested elliptical trajectory relay method has been demonstrated in simulations.

Appendix

As for formula (22), the objective function is a function of the variables $\lambda$ and $\beta$.We define

$${f_1}(\lambda ,\beta )=\frac{{\lambda \left( {{2^{R/\lambda }} - 1} \right)}}{\beta }$$

(28)

$${f_2}(\lambda ,\beta )=\frac{{\left( {1 - \lambda } \right)\left( {{2^{R/\left( {1 - \lambda } \right)}} - 1} \right)}}{{\left( {1 - \beta } \right)}}$$

(29)

The objective function can be demonstrated to be a convex function using the convexity criterion if ${f_1}(\lambda ,\beta )$ and ${f_2}(\lambda ,\beta )$ are convex functions. By obtaining the function’s Hessian matrix, we may determine if it is a convex function. The second order partial derivatives of the function ${f_1}(\lambda ,\beta )$ can be expressed as

$$\frac{{{\partial ^2}{f_1}(\lambda ,\beta )}}{{\partial {\lambda ^2}}}=\frac{{{{(\ln (2)R)}^2}{2^{R/\lambda }}}}{{{\lambda ^3}\beta }}>0$$

(30)

$$\frac{{{\partial ^2}{f_1}(\lambda ,\eta )}}{{\partial {\beta ^2}}}=\frac{{2\lambda \left( {{2^{R/\lambda }} - 1} \right)}}{{{\beta ^3}}}>0$$

(31)

$$\frac{{{\partial ^2}{f_1}(\lambda ,\beta )}}{{\partial \beta \partial \lambda }}=\frac{{{\partial ^2}{f_1}(\lambda ,\beta )}}{{\partial \lambda \partial \beta }}=\frac{{\ln (2)R{2^{R/\lambda }}}}{{\lambda {\beta ^2}}} - \frac{{\left( {{2^{R/\lambda }} - 1} \right)}}{{{\beta ^2}}}$$

(32)

According to the above partial derivatives, the determinant of the Hessian matrix can be obtained as

$$\begin{aligned} & H_{1} (\lambda ,\beta ) \triangleq \frac{{\partial ^{2} f_{1} (\lambda ,\beta )}}{{\partial \lambda ^{2} }} \cdot \frac{{\partial ^{2} f_{1} (\lambda ,\beta )}}{{\partial \beta ^{2} }} \\ & - \frac{{\partial ^{2} f_{1} (\lambda ,\beta )}}{{\partial \beta \partial \lambda }} \cdot \frac{{\partial ^{2} f_{1} (\lambda ,\beta )}}{{\partial \lambda \partial \beta }} \\ & = \frac{{(\ln (2)R)^{2} 2^{{R/\lambda }} }}{{\lambda ^{3} \beta }} \cdot \frac{{2\lambda \left( {2^{{R/\lambda }} - 1} \right)}}{{\beta ^{3} }} \\ & \quad - \left( {\frac{{\ln (2)R2^{{R/\lambda }} }}{{\lambda \beta ^{2} }} - \frac{{\left( {2^{{R/\lambda }} - 1} \right)}}{{\beta ^{2} }}} \right)^{2} \\ \end{aligned}$$

(33)

Similarly, the second order partial derivatives of the function ${f_2}(\lambda ,\beta )$ can be expressed as

$$\frac{{{\partial ^2}{f_2}(\lambda ,\beta )}}{{\partial {\lambda ^2}}}=\frac{{{{(\ln (2)R)}^2}{2^{R/\left( {1 - \lambda } \right)}}}}{{{{\left( {1 - \lambda } \right)}^3}\left( {1 - \beta } \right)}}>0$$

(34)

$$\frac{{{\partial ^2}{f_2}(\lambda ,\eta )}}{{\partial {\beta ^2}}}=\frac{{2\lambda \left( {{2^{R/\left( {1 - \lambda } \right)}} - 1} \right)}}{{{{\left( {1 - \beta } \right)}^3}}}>0$$

(35)

$$\frac{{{\partial ^2}{f_2}(\lambda ,\eta )}}{{\partial \beta \partial \lambda }}=\frac{{{\partial ^2}{f_2}(\lambda ,\beta )}}{{\partial \lambda \partial \beta }}=\frac{{\ln (2)R{2^{R/\left( {1 - \lambda } \right)}}}}{{\left( {1 - \lambda } \right){{\left( {1 - \beta } \right)}^2}}} - \frac{{\left( {{2^{R/\left( {1 - \lambda } \right)}} - 1} \right)}}{{{{\left( {1 - \beta } \right)}^2}}}$$

(36)

The determinant of the Hessian matrix can be obtained as

$$\begin{aligned} & H_{2} (\lambda ,\beta ) \triangleq \frac{{\partial ^{2} f_{2} (\lambda ,\beta )}}{{\partial \lambda ^{2} }} \cdot \frac{{\partial ^{2} f_{2} (\lambda ,\beta )}}{{\partial \beta ^{2} }} \\ & - \frac{{\partial ^{2} f_{2} (\lambda ,\beta )}}{{\partial \beta \partial \lambda }} \cdot \frac{{\partial ^{2} f_{2} (\lambda ,\beta )}}{{\partial \lambda \partial \beta }} \\ & = \frac{{(\ln (2)R)^{2} 2^{{R/\left( {1 - \lambda } \right)}} }}{{\left( {1 - \lambda } \right)^{3} \left( {1 - \beta } \right)}} \cdot \frac{{2\lambda \left( {2^{{R/\left( {1 - \lambda } \right)}} - 1} \right)}}{{\left( {1 - \beta } \right)^{3} }} \\ & \quad - \left( {\frac{{\ln (2)R2^{{R/\left( {1 - \lambda } \right)}} }}{{\left( {1 - \lambda } \right)\left( {1 - \beta } \right)^{2} }} - \frac{{\left( {2^{{R/\left( {1 - \lambda } \right)}} - 1} \right)}}{{\left( {1 - \beta } \right)^{2} }}} \right)^{2} \\ \end{aligned}$$

(37)

The simplification and derivation of ${H_1}(\lambda ,\beta )$ and ${H_2}(\lambda ,\beta )$ show that both of them are bigger than zero. As a result, the sum of them is a convex function, while the objective function is also a convex function. The optimization problem (formula) of the above interruption probability is an inequality-constrained convex optimization problem. The interior point method in the convex programming method finds the numerical solution.