Energy harvesting with adaptive transmit power for cognitive radio networks

Abstract

This article derives the packet error probability (PEP) of different relay selection techniques for cognitive radio networks (CRN). The power of secondary source and relays is adaptive to not cause high interference to primary receiver $P_R$. Secondary nodes harvest energy from radio frequency signals to be able to communicate. CRN with secondary nodes transmitting with adaptive power was already studied in the literature. However, in all previous studies, secondary nodes use their own batteries to transmit. The main motivation of our paper is to study CRN with energy harvesting and adaptive transmit power (ATP). We derive new expressions of PEP and throughput for CRN with energy harvesting and ATP. Our results are valid for opportunistic relaying, partial relay selection and reactive relay selection. Our main contribution is to optimize harvesting duration to maximize the system throughput.

Appendices

Appendix A

The CDF of $E_S$ is expressed as

$$\begin{aligned} F_{E_S}(x)= & {} P(E_{S}<x)\nonumber \\= & {} P\left( \min \left( E_{available},\frac{I}{|g_{SP_{R}}|^{2}}\right) <x\right) \nonumber \\= & {} 1-P\left( \min \left( \mu |g_{AS}|^2,\frac{I}{|g_{SP_{R}}|^{2}}\right) >x\right) , \end{aligned}$$

(52)

Assuming that $g_{AS}$ and $g_{SP_{R}}$ are independent, we deduce

$$\begin{aligned} F_{E_S}(x)= & {} 1-P(\mu |g_{AS}|^2>x)P\left( \frac{I}{|g_{SP_{R}}|^{2}}>x\right) \nonumber \\= & {} 1-P(\mu |g_{AS}|^2>x)P\left( |g_{SP_{R}}|^{2}<\frac{I}{x}\right) , \end{aligned}$$

(53)

$|g_{AS}|^2$ and $|g_{SP_{R}}|^{2}$ follow an exponential distribution with mean $E(|g_{AS}|^2)=\frac{1}{\alpha _{AS}}$ and $E(|g_{SP_R}|^2)=\frac{1}{\alpha _{SP_R}}$. Therefore, we have

$$\begin{aligned} F_{E_S}(x)=1-\alpha _{AS}e^{-\alpha _{AS}\frac{x}{\mu }} \left[ 1-\alpha _{SP_R}e^{-\alpha _{SP_R}\frac{I}{x}}\right] , \end{aligned}$$

(54)

Appendix B

The SNR at $R_k$ is the product of two random variables $E_S$ and $\frac{|h_{SR_k}|^2}{N_0}$.

$$\begin{aligned} \gamma _{SR_{k}}=E_S \frac{|h_{SR_k}|^2}{N_0}, \end{aligned}$$

(55)

Let $X=\frac{|h_{SR_k}|^2}{N_0}$, the CDF of SNR at relay $R_k$ is given by

$$\begin{aligned} F_{\gamma _{SR_{k}}}(x)= & {} P(\gamma _{SR_{k}} \le x)=\int _0^{+\infty }f_X(v)P\nonumber \\&\times \left( E_S\le \frac{x}{v}\right) dv=\int _0^{+\infty }f_X(v)F_{E_S}\left( \frac{x}{v}\right) dv,\nonumber \\ \end{aligned}$$

(56)

where $f_X(v)$ is the PDF of $X=\frac{|h_{SR_k}|^2}{N_0}$:

$$\begin{aligned} f_X(v)=N_0 \alpha _{SR_k} e^{-N_0 \alpha _{SR_k}v}, \forall v \ge 0, \end{aligned}$$

(57)

where $\alpha _{SR_k}=\frac{1}{E(|g_{SR_k}|^2)}$.

Using the expression of CDF of $E_S$ provided in “Appendix A”, we can write

$$\begin{aligned} F_{\gamma _{SR_{k}}}(x)= & {} 1-\int _0^{+\infty }\alpha _{AS}e^{-\alpha _{AS}\frac{x}{v\mu }}\nonumber \\&\times \left[ 1-\alpha _{SP_R}e^{-\alpha _{SP_R}\frac{Iv}{x}}\right] f_X(v)dv, \end{aligned}$$

(58)

We deduce

$$\begin{aligned}&F_{\gamma _{SR_{k}}}(x)=1-N_0 \alpha _{SR_k}\alpha _{AS}\int _0^{+\infty }e^{-\alpha _{AS}\frac{x}{v\mu }} e^{-N_0 \alpha _{SR_k}v} dv\nonumber \\&\quad +\,N_0 \alpha _{SR_k}\alpha _{SP_R}\alpha _{AS}\nonumber \\&\quad \times \int _0^{+\infty }e^{-\alpha _{AS}\frac{x}{v\mu }}e^{-\alpha _{SP_R}\frac{Iv}{x}}e^{-N_0 \alpha _{SR_k}v} dv, \end{aligned}$$

(59)

We use the following result [39]

$$\begin{aligned} \int _0^{+\infty }e^{-ax-\frac{b}{x}}dx=2\sqrt{\frac{b}{a}}K_1(2\sqrt{ba}), \forall a>0,b>0 \end{aligned}$$

(60)

to write the CDF of SNR as

$$\begin{aligned}&F_{\gamma _{SR_{k}}}(x)=1-2\sqrt{\frac{\alpha _{AS}x}{\mu N_0\alpha _{SR_k}}}K_1\left( 2\sqrt{N_0\alpha _{SR_k}\frac{\alpha _{AS}x}{\mu }}\right) \nonumber \\&\quad +\,2N_0 \alpha _{SR_k}\alpha _{SP_R}\alpha _{AS} \sqrt{\frac{\alpha _{AS}x^2}{\mu (N_0\alpha _{SR_k}x+\alpha _{SP_R}I)}}\nonumber \\&\quad K_1\left( 2\sqrt{N_0\alpha _{SR_k}\alpha _{AS}\frac{x}{\mu }+\frac{\alpha _{AS}}{\mu }\alpha _{SP_R}I}\right) \end{aligned}$$

(61)

Appendix C

Using (9) and (10), we have

$$\begin{aligned}&F_{\gamma _{SR_k}}(x)=1-2e^{N_0\frac{\alpha _{P_TR_k}}{E_{P_T}}}\frac{\alpha _{P_TR_k}}{E_{P_T}}\sqrt{\frac{\alpha _{AS}x}{\mu \alpha _{SR_k}}}\int _{N_0}^{+\infty }\nonumber \\&\quad \sqrt{u}e^{-u\frac{\alpha _{P_TR_k}}{E_{P_T}}} K_1\left( 2\sqrt{\alpha _{SR_k}\frac{\alpha _{AS}ux}{\mu }}\right) du,\nonumber \\&\quad +\,2e^{N_0\frac{\alpha _{P_TR_k}}{E_{P_T}}}\frac{\alpha _{P_TR_k}}{E_{P_T}}\alpha _{SR_k}\alpha _{SP_R}\alpha _{AS} x\nonumber \\&\quad \int _{N_0}^{+\infty }\sqrt{\frac{\alpha _{AS}u^2}{\mu (\alpha _{SR_k}xu+\alpha _{SP_R}I)}}e^{-u\frac{\alpha _{P_TR_k}}{E_{P_T}}}\nonumber \\&\quad \times \, K_1\left( 2\sqrt{\alpha _{SR_k}\alpha _{AS}\frac{ux}{\mu }+\frac{\alpha _{AS}}{\mu }\alpha _{SP_R}I}\right) du \end{aligned}$$

(62)

We use the following result [39]

$$\begin{aligned} \int _{0}^{+\infty }\sqrt{y}K_{1}(2\beta \sqrt{y})e^{-\alpha _{3}y}dy=\frac{ e^{\frac{\beta ^{2}}{2\alpha _{3}}}}{2\beta \alpha _{3}}W_{-1,0.5}\left( \frac{ \beta ^{2}}{\alpha _{3}}\right) ,\nonumber \\ \end{aligned}$$

(63)

where $W_{\mu ,\nu }(x)$ is the Whittaker function.

$$\begin{aligned}&F_{\gamma _{SR_k}}(x)=1+2e^{N_0\frac{\alpha _{P_TR_k}}{E_{P_T}}}\frac{\alpha _{P_TR_k}}{E_{P_T}}\sqrt{\frac{\alpha _{AS}x}{\mu \alpha _{SR_k}}}\int _{0}^{N_0}\nonumber \\&\quad \sqrt{u}e^{-u\frac{\alpha _{P_TR_k}}{E_{P_T}}}K_1\left( 2\sqrt{\alpha _{SR_k}\frac{\alpha _{AS}ux}{\mu }}\right) du,\nonumber \\&\quad -\,W_{-1,0.5}\left( \frac{\alpha _{SR_k}x\alpha _{AS}E_{P_T}}{\mu \alpha _{P_TR_k}}\right) e^{N_0\frac{\alpha _{P_TR_k}}{E_{P_T}}}e^{\frac{\alpha _{SR_k}x\alpha _{AS}E_{P_T}}{2\mu \alpha _{P_TR_k}}}\nonumber \\&\quad \times \frac{1}{\alpha _{SR_k}}+\,2e^{N_0\frac{\alpha _{P_TR_k}}{E_{P_T}}}\frac{\alpha _{P_TR_k}}{E_{P_T}}\alpha _{SR_k}\alpha _{SP_R}\alpha _{AS} x\nonumber \\&\quad \int _{N_0}^{+\infty }\sqrt{\frac{\alpha _{AS}u^2}{\mu (\alpha _{SR_k}xu+\alpha _{SP_R}I)}}e^{-u\frac{\alpha _{P_TR_k}}{E_{P_T}}}\nonumber \\&\quad \times \, K_1\left( 2\sqrt{\alpha _{SR_k}\alpha _{AS}\frac{ux}{\mu }+\frac{\alpha _{AS}}{\mu }\alpha _{SP_R}I}\right) du \end{aligned}$$

(64)

where the two integrals are evaluated numerically using MATLAB.