Cooperative Prediction for Cognitive Radio Networks

Abstract

Combining spectrum sensing (SS) and primary user (PU) traffic forecasting provides a cognitive radio network with a platform from which informed and proactive operational decisions can be made. The success of these decisions is largely dependent on prediction accuracy. Allowing secondary users (SU) to perform these predictions in a collaborative manner allows for an improvement in the accuracy of this process, since individual SUs may suffer from SS and prediction inaccuracies due to poor channel conditions. To overcome these problems a collaborative approach to forecasting PU traffic behaviour, that combines SS and forecasting through SU cooperation, has been proposed in this article. Both pre-fusion and post-fusion scenarios for cooperative prediction were investigated and a number of binary prediction methods were considered (including the authors’ own simple technique). Cooperative prediction performance was investigated, under various PU traffic conditions, for a group of ten SUs experiencing different channel conditions and a sub-optimal cooperative forecasting algorithm was proposed to minimise cooperative prediction error. Simulation results indicated that the accuracy of the prediction methods was influenced by the PU traffic pattern and that cooperative prediction lead to a significant improvement in prediction accuracy under most of the traffic conditions considered. However, this came at the cost of increased computational complexity. The pre-fusion scenario was found to be the most accurate scenario (up to 25 % improvement), but was also eleven times more complex than when no fusion was employed. The cooperative forecasting algorithm was found to further improve these results.

Appendix

In this appendix the expression for $\lambda _{i}$ is derived.

Lemma 1

The signal detection threshold $\lambda _{i}$ is given by the following expression,

$$\begin{aligned} \lambda _{i} = \frac{ \mu _{s}\sigma _{n}^2 - \sigma _{n}\sigma _{s}\sqrt{ 2\ln {(\frac{\sigma _{n}}{\sigma _{s}})}(\sigma _{n}^2-\sigma _{s}^2)+\mu _{s}^2} }{ \sigma _{n}^2 - \sigma _{s}^2 } \end{aligned}$$

(33)

where $\mu _{s}$ and $\sigma ^2_{s}$ are the mean and variance of the information carrying component of a received signal respectively and $\sigma ^2_{n}$ is the variance of the noise component.

Proof

Let the probability density functions of the information and noise components be,

$$\begin{aligned} \rho _{n}(x) = \frac{1}{\sigma _{n}\sqrt{2\pi }}\exp { \frac{(x-\mu _{n})^2}{2\sigma _{n}^2} } \end{aligned}$$

(34)

and

$$\begin{aligned} \rho _{s}(x) = \frac{1}{\sigma _{s}\sqrt{2\pi }}\exp { \frac{(x-\mu _{s})^2}{2\sigma _{s}^2} } \end{aligned}$$

(35)

respectively.

To solve for x, let $\rho _{n}(x)=\rho _{s}(x)$ such that,

$$\begin{aligned} \frac{1}{\sigma _{n}\sqrt{2\pi }}\exp { \frac{(x-\mu _{n})^2}{2\sigma _{n}^2} } = \frac{1}{\sigma _{s}\sqrt{2\pi }}\exp { \frac{(x-\mu _{s})^2}{2\sigma _{s}^2} } \end{aligned}$$

(36)

This expression can be simplified and written in the form of a second order polynomial, $ax^2 + bx + c = 0$, where the coefficient are given as,

$$\begin{aligned}&a = x^2(\sigma _{n}^2-\sigma _{s}^2) \end{aligned}$$

(37)

$$\begin{aligned}&b = 2x(\sigma _{s}^2\mu _{n}-\sigma _{n}^2\mu _{s}) \end{aligned}$$

(38)

$$\begin{aligned}&c = \sigma _{n}^2\mu _{s}^2 - \sigma _{s}^2\mu _{n}^2 - 2\sigma _{n}^2\sigma _{s}^2\ln {(\frac{\sigma _{n}}{\sigma _{s}})} \end{aligned}$$

(39)

If $\lambda _{i}=x$, is the solution for Eq. (36), then,

$$\begin{aligned} \lambda _{i} = \frac{ \mu _{s}\sigma _{n}^2-\mu _{n}\sigma _{s}^2}{ \sigma _{n}^2 - \sigma _{s}^2 } \pm \frac{ \sigma _{n}\sigma _{s} \sqrt{ 2\ln {(\frac{\sigma _{n}}{\sigma _{s}})}(\sigma _{n}^2-\sigma _{s}^2) + \mu _{n}^2 + \mu _{s}^2 - 2\mu _{n}\mu _{s}} }{ \sigma _{n}^2 - \sigma _{s}^2 } \end{aligned}$$

(40)

But, under the assumption that $\rho _{n} = \mathcal {N}\left( 0,\sigma ^2_{n} \right)$, $\mu _{n}=0$ must be substituted into Eq. (40) to give the following expression,

$$\begin{aligned} \lambda _{i} = \frac{ \mu _{s}\sigma _{n}^2 \pm \sigma _{n}\sigma _{s}\sqrt{ 2\ln {(\frac{\sigma _{n}}{\sigma _{s}})}(\sigma _{n}^2-\sigma _{s}^2)+\mu _{s}^2} }{ \sigma _{n}^2 - \sigma _{s}^2 } \end{aligned}$$

(41)

If it is assumed that $\mu _{s} > 0$, then it follows that,

$$\begin{aligned} \mu _{s}\sigma _{n}^2 > \sqrt{ 2\ln {(\frac{\sigma _{n}}{\sigma _{s}})}(\sigma _{n}^2-\sigma _{s}^2)+\mu _{s}^2} \end{aligned}$$

Therefore, since $\lambda _{i} \le \mu _{s}$, Eq. (41) becomes,

$$\begin{aligned} \lambda _{i} = \frac{ \mu _{s}\sigma _{n}^2 - \sigma _{n}\sigma _{s}\sqrt{ 2\ln {(\frac{\sigma _{n}}{\sigma _{s}})}(\sigma _{n}^2-\sigma _{s}^2)+\mu _{s}^2} }{ \sigma _{n}^2 - \sigma _{s}^2 } \end{aligned}$$

(42)

$\square$