Power allocation policies with full and partial inter-system channel state information for cognitive radio networks

Abstract

This paper investigates several power allocation policies in orthogonal frequency division multiplexing -based cognitive radio networks under the different availability of inter-system channel state information (CSI) and the different capability of licensed primary users (PUs). Specifically, we deal with two types of PUs having different capabilities: a dumb (peak interference-power tolerable) PU and a more sophisticated (average interference-power tolerable) PU. For such PU models, we first formulate two optimization problems that maximize the capacity of unlicensed secondary user (SU) while maintaining the quality of service of PU under the assumption that both intra- and inter-system CSI are fully available. However, due to loose cooperation between SU and PU, it may be difficult or even infeasible for SU to obtain the full inter-system CSI. Thus, under the partial inter-system CSI setting, we also formulate another two optimization problems by introducing interference-power outage constraints. We propose optimal and efficient suboptimal power allocation policies for these four problems. Extensive numerical results demonstrate that the spectral efficiency achieved by SU with partial inter-system CSI is less than half of what is achieved with full inter-system CSI within a reasonable range of outage probability (e.g., less than 10 %). Further, it is shown that the average interference-power tolerable PU can help to increase the saturated spectral efficiency of SU by about 20 and 50 % in both cases of full and partial inter-system CSI, respectively.

Appendix

1.1 A Proof of Lyapunov condition

$\underline{Lyapunov\,condition}:$

$$ \lim_{N \rightarrow \infty} \frac{\left(\sum_{n=1}^{N} r_{n}^{3} \right)^{\frac{1}{3}}}{\left(\sum_{n=1}^{N} {\sigma_{n}^{2}} \right)^{\frac{1}{2}}} = 0, $$

(38)

where r _n is defined as the third central moment of random variable X _n , i.e., $E \left[(X_{n}-m_{n})^3\right];\, m_{n}$ and σ ² _n are the finite mean and variance of the X _n , respectively.

Proof

The third central moment of the random variable X _n can be written as

$$ r_{n}^{3} = E\left[(X_{n} - m_{n})^3\right] $$

(39)

$$ = E\left[X_{n}^{3}\right]-3m_{n} E\left[X_{n}^{2}\right]+ 2m_{n}^{3}. $$

(40)

The random variable X _n  = p ⁿ g ⁿ₁ is independently exponential distributed with mean p ⁿλ₂₁. By plugging the following statistics (41)-(41) for X _n into (40), we can readily obtain the third central moment, r ³ _n  = 2(p ⁿλ₂₁)³.

$$ \begin{aligned} \hbox{mean: }\quad\quad E[X_{n}] &= m_{n} = p^{n} \lambda_{21},\\ \hbox{variance:}\quad\,Var[X_n] &= \sigma_{n}^{2} = (p^{n} \lambda_{21})^{2},\\ \hbox{2th moment: }\quad\quad E[X_{n}^{2}] &= 2(p^{n} \lambda_{21})^{2},\\ \hbox{3rd moment: }\quad\quad E[X_{n}^{3}] &= 6(p^{n} \lambda_{21})^{3}. \end{aligned} $$

The power allocation p ⁿ for the subchannel $n\in\mathcal{N}=\{1,2,\ldots,N\}$ is a nonnegative value, whereas the average channel gain λ₂₁ is a positive value. To exclude meaningless summations in (38), we define the set of subchannels with positive power as $\mathcal{N}^{\prime} \doteq \left\{ \,n\,|\, p^{n} > 0, \forall n \in \mathcal{N} \right\}$. Since λ₂₁ and p ⁿ are finite, there exists positive maximum and minimum values of p ⁿλ₂₁ over the set $\mathcal{N}^{\prime}$. Let us define the maximum and minimum values as $M=\max_{n \in \mathcal{N}^{\prime}}\,{p^{n} \lambda_{21}}>0$ and $m=\min_{n \in \mathcal{N}^{\prime}}{p^{n} \lambda_{21}}>0$, respectively. Thus, in the (38), the numerator can be upper-bounded and the denominator can be lower-bounded as follows:

$$ \left(\sum_{n\in{\mathcal{N}}}{r_{n}^{3}}\right)^{\frac{1}{3}} = \left(\sum_{n\in{\mathcal{N}}^{\prime}}2(p^{n}\lambda_{21})^{3}\right)^{\frac{1}{3}} \leq \left(|{\mathcal{N}}^{\prime}| \cdot 2M^{3}\right)^{\frac{1}{3}}, $$

(41)

$$ \left(|{\mathcal{N}}^{\prime}| \cdot m^{2}\right)^{\frac{1}{2}} \leq \left(\sum_{n\in{\mathcal{N}}^{\prime}}(p^{n}\lambda_{21})^{2}\right)^{\frac{1}{2}} =\left(\sum_{n\in{\mathcal{N}}}{\sigma_{n}^{2}}\right)^{\frac{1}{2}}, $$

(42)

where $|\mathcal{N}^{\prime}|$ denotes the cardinality of set $\mathcal{N}^{\prime}$. Note that as N tends to infinity, $|\mathcal{N}^{\prime}|$ goes to infinity as well. Using the upper-bound of the numerator (41) and the lower-bound of the denominator (42), we can obtain the Lyapunov condition in the (38).

$$ \lim_{N \rightarrow \infty} \frac{\left(\sum_{n=1}^{N} r_{n}^{3} \right)^{\frac{1}{3}}}{\left( \sum_{n=1}^{N} \sigma_{n}^{2} \right)^{\frac{1}{2}}} \leq \lim_{N \rightarrow \infty} \frac{\left(|{\mathcal{N}}^{\prime}| \cdot 2M^{3}\right)^{\frac{1}{3}}}{\left(|{\mathcal{N}}^{\prime}| \cdot m^{2}\right)^{\frac{1}{2}}} $$

(43)

$$ = \lim_{|{\mathcal{N}}^{\prime}| \rightarrow \infty} \frac{2^{\frac{1}{3}}M}{m} \cdot \frac{1}{|{\mathcal{N}}^{\prime}|^{\frac{1}{6}}} = 0. $$

(44)

This completes the proof. $\square$