Joint sensing time and power allocation in cognitive networks with amplify-and-forward cooperation

Abstract

Cognitive radio is a novel approach to cope with spectrum scarcity, in which either a network or a wireless node changes its transmission or reception parameters to communicate efficiently. However, it is difficult to avoid the interference between licensed and unlicensed users in various scenarios. This paper analyzes the jointly optimized allocation of sensing time and power for a two-user, amplify-and-forward (AF) cognitive network developed by maximizing the average aggregate throughput of its secondary network. In particular, this paper discusses diverse cooperation ratios for different scenarios and a unique cooperation ratio in spite of scenario changes. The observations of experiment results indicate that the sensing duration is within a strict interval. The results show that the optimized sensing time is 14.111 ms and the aggregate throughput equals to 1.1451 bps/Hz which are tractable by sequential optimization. This result indicates that by adopting the fixed cooperation ratios, the achievable throughput of the system is decreased. The system innovatively creates multiple independent fading channels to achieve technological diversity among partners.

Appendix 1: Proof of R(β1, l, β2, l) as a concave function

$$ R\left({\beta}_{i,l}\right)=\mu {R}_1\left({\beta}_{1,l},{\beta}_{2,l}\right)+\left(1-\mu \right){R}_2\left({\beta}_{1,l},{\beta}_{2,l}\right)\kern10.5em =\frac{1}{2}\mu \underset{2}{\log}\left(1+{\beta}_1{\gamma}_1+\frac{\gamma_2{\gamma}_3{\beta}_1\left(1-{\beta}_2\right)}{1+{\beta}_1{\gamma}_3+\left(1-{\beta}_2\right){\gamma}_2}\right)+\frac{1}{2}\left(1-\mu \right)\underset{2}{\log}\left(1+{\beta}_2{\gamma}_2+\frac{\gamma_1{\gamma}_4{\beta}_2\left(1-{\beta}_1\right)}{1+{\beta}_2{\gamma}_4+\left(1-{\beta}_1\right){\gamma}_1}\right) $$

The proof starts with the characteristics of a concave function. If the function R(β_{1, l}, β_{2, l}) is a second order derivative in a certain interval, the necessary and sufficient condition for this function to be a concave function is that $\frac{\partial^2R\left({\beta}_i\right)}{\partial^2{\beta}_1}<0$ and $\frac{\partial^2R\left({\beta}_i\right)}{\partial^2{\beta}_2}<0$.Therefore, it is necessary to prove that the second derivative of $\frac{\partial^2\mathrm{R}\left({\upbeta}_{\mathrm{i}}\right)}{\partial^2{\upbeta}_1}<0$ and $\frac{\partial^2\mathrm{R}\left({\upbeta}_{\mathrm{i}}\right)}{\partial^2{\upbeta}_2}<0$. The following steps are to prove $\frac{\partial^2R\left({\beta}_i\right)}{\partial^2{\beta}_1}<0$. The same explanation is applicable to the proof of $\frac{\partial^2\mathrm{R}\left({\upbeta}_{\mathrm{i}}\right)}{\partial^2{\upbeta}_2}<0$.

• (1) Calculate the first derivative:

$$ \frac{\partial R\left({\beta}_i\right)}{\partial {\beta}_1}=\frac{1}{2}\mu /\mathit{\ln}2\mathit{\ln}\left(1+{\beta}_1{\gamma}_1+\frac{\gamma_2{\gamma}_3{\beta}_1\left(1-{\beta}_2\right)}{1+{\beta}_1{\gamma}_3+\left(1-{\beta}_2\right){\gamma}_2}\right)+\frac{1}{2}\left(1-\mu \right)/\mathit{\ln}2\mathit{\ln}\left(1+{\beta}_2{\gamma}_2+\frac{\gamma_1{\gamma}_4{\beta}_2\left(1-{\beta}_1\right)}{1+{\beta}_2{\gamma}_4+\left(1-{\beta}_1\right){\gamma}_{1.}}\right)=\frac{1}{2}\mu /\mathit{\ln}2\frac{\gamma_1+\frac{\gamma_2{\gamma}_3\left(1-{\beta}_2\right)\left(1+\left(1-{\beta}_2\right){\gamma}_2\right)}{{\left(1+{\beta}_1{\gamma}_3+\left(1-{\beta}_2\right){\gamma}_2\right)}^2}}{1+{\beta}_1{\gamma}_1+\frac{\gamma_2{\gamma}_3{\beta}_1\left(1-{\beta}_2\right)}{1+{\beta}_1{\gamma}_3+\left(1-{\beta}_2\right){\gamma}_2}}+\frac{1}{2}\left(1-\mu \right)/\mathit{\ln}2\frac{\frac{-{\gamma}_1{\gamma}_4{\beta}_2\left(1+{\beta}_2{\gamma}_4\right)}{{\left(1+{\beta}_2{\gamma}_4+\left(1-{\beta}_1\right){\gamma}_1\right)}^2}}{1+{\beta}_2{\gamma}_2+\frac{\gamma_1{\gamma}_4{\beta}_2\left(1-{\beta}_1\right)}{1+{\beta}_2{\gamma}_4+\left(1-{\beta}_1\right){\gamma}_1}}=a\frac{\gamma_1+\frac{bc}{{\left({\beta}_1{\gamma}_3+b\right)}^2}}{1+{\beta}_1{\gamma}_1+\frac{c{\beta}_1}{\beta_1{\gamma}_3+b}}+d\frac{\frac{- ef}{{\left(e+\left(1-{\beta}_1\right){\gamma}_1\right)}^2}}{e+\frac{f\left(1-{\beta}_1\right)}{e+\left(1-{\beta}_1\right){\gamma}_1}}=a\frac{\gamma_1{\left({\beta}_1{\gamma}_3+b\right)}^2+ bc}{\left(1+{\beta}_1{\gamma}_1\right){\left({\beta}_1{\gamma}_3+b\right)}^2+c{\beta}_1\left({\beta}_1{\gamma}_3+b\right)}+\frac{- def}{e{\left(e+\left(1-{\beta}_1\right){\gamma}_1\right)}^2+f\left(1-{\beta}_1\right)\left(e+\left(1-{\beta}_1\right){\gamma}_1\right)} $$

where:

$$ {\displaystyle \begin{array}{ll}a=\frac{1}{2}\upmu /\ln 2;& \mathrm{b}=1+\left(1-{\upbeta}_2\right){\mathrm{r}}_2;\\ {}\mathrm{c}={\mathrm{r}}_2{\mathrm{r}}_3\left(1-{\upbeta}_2\right);& \mathrm{d}=\frac{1}{2}\left(1-\upmu \right)/\ln 2;\\ {}\mathrm{e}=1+{\upbeta}_2{\mathrm{r}}_4;& \mathrm{f}={\mathrm{r}}_1{\mathrm{r}}_4{\upbeta}_2\end{array}} $$

• (2) Calculate the second derivative:

$$ \frac{\partial^2R\left({\beta}_i\right)}{\partial^2{\beta}_1}=a\bullet \frac{-{\left[{r}_1{\left({\beta}_1{r}_3+b\right)}^2+ bc\right]}^2-2 bc{r}_3\left[\left(1+{\beta}_1{r}_1\right)\left({\beta}_1{r}_3+b\right)+c{\beta}_1\right]}{{\left[\left(1+{\beta}_1{r}_1\right){\left({\beta}_1{r}_3+b\right)}^2+c{\beta}_1\left({\beta}_1{r}_3+b\right)\right]}^2}+ def\frac{-\left[e+\left(1-{\beta}_1\right){r}_1\right]\left[2{r}_1\left(1+{\beta}_2{r}_2\right)+f\right]-{r}_1f\left(1-{\beta}_1\right)}{\left[\left(1+{\beta}_2{r}_2\right){\left(e+\left(1-{\beta}_1\right){r}_1\right)}^2+f\right(1-{\beta}_1\left(e+\left(1-{\beta}_1\right){r}_1\right)\Big]{}^2} $$

where:

$$ {\displaystyle \begin{array}{c}{\gamma}_1={Ph}_10/{\sigma}_0^2;\\ {}{\gamma}_4={Ph}_21/{\sigma}_1^2;\end{array}}\kern0.5em {\displaystyle \begin{array}{c}{\gamma}_2={Ph}_20/{\sigma}_0^2;\\ {}{\beta}_{i,l}\in \left[0,1\right],\kern1.36em i\in \left\{1,2\right\};\end{array}}\kern0.5em {\displaystyle \begin{array}{c}{\gamma}_3={Ph}_12/{\sigma}_2^2;\\ {}\end{array}} $$

So we know that,

$$ \left\{\begin{array}{c}\begin{array}{c}a=\frac{1}{2}\upmu /\ln 2>0\\ {} def=\left(\frac{1}{2}\left(1-\upmu \right)/\ln 2\right)\left(1+{\upbeta}_2{\mathrm{r}}_4\right)\left({\mathrm{r}}_1{\mathrm{r}}_4{\upbeta}_2\right)>0\\ {}{\left[{r}_1{\left({\beta}_1{r}_3+b\right)}^2+ bc\right]}^2>0\end{array}\\ {}2 bc{r}_3\left[\left(1+{\beta}_1{r}_1\right)\left({\beta}_1{r}_3+b\right)+c{\beta}_1\right]>0\\ {}{\left[\left(1+{\beta}_1{r}_1\right){\left({\beta}_1{r}_3+b\right)}^2+c{\beta}_1\left({\beta}_1{r}_3+b\right)\right]}^2>0\\ {}-\left[e+\left(1-{\beta}_1\right){r}_1\right]\left[2{r}_1\left(1+{\beta}_2{r}_2\right)+f\right]-{r}_1f\left(1-{\beta}_1\right)<0\\ {}\left[\left(1+{\beta}_2{r}_2\right){\left(e+\left(1-{\beta}_1\right){r}_1\right)}^2+f\right(1-{\beta}_1\left(e+\left(1-{\beta}_1\right){r}_1\right)\Big]{}^2>0\end{array}\right. $$

And we can say that,

$$ \left\{\begin{array}{c}a\bullet \frac{-{\left[{r}_1{\left({\beta}_1{r}_3+b\right)}^2+ bc\right]}^2-2 bc{r}_3\left[\left(1+{\beta}_1{r}_1\right)\left({\beta}_1{r}_3+b\right)+c{\beta}_1\right]}{{\left[\left(1+{\beta}_1{r}_1\right){\left({\beta}_1{r}_3+b\right)}^2+c{\beta}_1\left({\beta}_1{r}_3+b\right)\right]}^2}<0\\ {} def\frac{-\left[e+\left(1-{\beta}_1\right){r}_1\right]\left[2{r}_1\left(1+{\beta}_2{r}_2\right)+f\right]-{r}_1f\left(1-{\beta}_1\right)}{\left[\left(1+{\beta}_2{r}_2\right){\left(e+\left(1-{\beta}_1\right){r}_1\right)}^2+f\right(1-{\beta}_1\left(e+\left(1-{\beta}_1\right){r}_1\right)\Big]{}^2}<0\end{array}\right. $$

From the process we know that $\frac{\partial^2R\left({\beta}_i\right)}{\partial^2{\beta}_1}$<0. Similarly, it is easy to prove that $\frac{\partial^2R\left({\beta}_i\right)}{\partial^2{\beta}_2}$<0. In conclusion, we knew that R(β_{1, l}, β_{2, l}) is a concave function.