Multihop Multibranch Spectrum Sensing with Energy Harvesting

Abstract

This paper deals with spectrum sensing using the energy detector (ED) with multiple hops multibranch relaying where the primary user (PU) and relays harvest energy from radio frequency signal transmitted from a given node A. The harvested energy is used by PU and relays to transmit signals to the fusion center where the ED is used to detect the PU. The study is valid for amplify and forward relaying, any number of hops and any number of branches. We also suggest a new lower bound of the detection probability using the cumulative distribution function of signal-to-noise ratio (SNR). When there are many available branches, only the best one is activated. The activated branch offers the highest end-to-end SNR.

Appendix

If $Y_{1}$ and $Y_{2}$ are two exponential r.v. with mean 1/$\eta _{1}$ and 1/$\eta _{2}.$

The CDF of $Y=Y_{1}Y_{2}$ is written as

$$\begin{aligned} F_{Y}(y)=P(Y_{1}Y_{2}\le y)=\int _{0}^{+\infty }P(Y_{1}\le \frac{y}{z} )\eta _{2}e^{-\eta _{2}z}dz. \end{aligned}$$

(31)

We can write

$$\begin{aligned} F_{Y}(y)= & {} \int _{0}^{+\infty }\left[ 1-e^{-\eta _{1}\frac{y}{z}}\right] \eta _{2}e^{-\eta _{2}z}dz \\= & {} 1-\int _{0}^{+\infty }e^{-\eta _{1}\frac{y}{z}}\eta _{2}e^{-\eta _{2}z}dz \end{aligned}$$

(32)

We have

$$\begin{aligned} \int _{0}^{+\infty }e^{-\frac{e}{z}}e^{-\frac{z}{f}}dz=2\sqrt{\frac{e}{f}} K_{1}(2\sqrt{\frac{e}{f}}). \end{aligned}$$

(33)

Using (32), (33), $e=\eta _{1}y$ and $f=\frac{1}{ \eta _{2}}$, we have

$$\begin{aligned} F_{Y}(y)=1-2\sqrt{\eta _{1}\eta _{2}y}K_{1}(2\sqrt{\eta _{1}\eta _{2}y}). \end{aligned}$$

(34)

The PDF of Y is expressed as

$$\begin{aligned} f_{Y}(y)=-\frac{\sqrt{\eta _{1}\eta _{2}}}{\sqrt{y}}K_{1}(2\sqrt{ \eta _{1}\eta _{2}y})-2\sqrt{\eta _{1}\eta _{2}y}K_{1}^{^{\prime }}\left(2\sqrt{\eta _{1}\eta _{2}y}\right)\frac{\sqrt{\eta _{1}\eta _{2}}}{ \sqrt{y}}. \end{aligned}$$

(35)

We have

$$\begin{aligned} K_{1}^{^{\prime }}(u)=-K_{0}(u)-\frac{K_{1}(u)}{u}, \end{aligned}$$

(36)

Therefore,

$$\begin{aligned} f_{Y}(y)=2\eta _{1}\eta _{2}K_{0}(2\sqrt{\eta _{1}\eta _{2}y}), \end{aligned}$$

(37)