Dynamic power allocation based on second-order control system in two-tier femtocell networks

Abstract

In this paper, we investigate the joint uplink power allocation problem for the macrocell and femtocells in the new view of Control Theory. Since the wireless channels in the macrocell and the femtocells fluctuate with time, we aim at an energy-saving and adaptive power allocation scheme that is robust to channel fluctuations. An optimal power allocation problem, which can satisfy the requirement of signal-to-interference-plus-noise ratio and subject to the maximum power constraint, is formulated. To obtain the robust power allocation of each user (both macrocell and femtocell users), a second-order control system is created for each cell (macrocell and femtocell). According to the second-order control system and considering the maximum power constraint, we derive a distributed robust power control algorithm. The dynamic performance and convergence of the formulated system are analyzed based on control theory. The advantage of the scheme is that some dynamic control indexes can be set in advance and the dynamic process can be controlled by setting appropriate parameters. Since femtocells share the same spectrum with the macrocell, the interference from femtocells to the macrocell base station has to be controlled to protect the normal communication of the macrocell. Then an admission control algorithm is proposed to remove the user that causes the serious interference to the macrocell, if the signal-to-interference-plus-noise ratio requirement of the macrocell user can not be satisfied. Two algorithms are particularly attractive, especially in view of practical implementation issues. Numerical simulation results are presented to illustrate the effectiveness and advantages of the proposed algorithms.

Appendices

Appendix A

From (8), it can be known that

$$\begin{array}{@{}rcl@{}} C(s)=\frac{\omega_n^2}{s^2 + 2\zeta\omega_n s+\omega_n^2}\cdot{R(s)} \end{array} $$

(26)

Owing to r(t) is a step signal (r(t) = A, A is a positive constant), R(s) can be expressed as $\frac {A}{s}$. Formula (26) can be rewritten as

$$\begin{array}{@{}rcl@{}} C(s)&=&\frac{\omega_n^2}{s^2 + 2\zeta\omega_n s+\omega_n^2}\cdot\frac{A}{s}\\ &=&\frac{A}{s}-A\frac{s + 2\zeta \omega_n}{(s+\zeta \omega_n+j\omega_d)(s+\zeta \omega_n-j\omega_d)}\\ & =&A\left[\frac{1}{s}-\frac{s+\zeta \omega_n}{(s+\zeta \omega_n)^2+\omega_d^2}-\frac{\zeta \omega_n}{\omega_d}\cdot\frac{\omega_d}{(s+\zeta \omega_n)^2+\omega_d^2}\right] \end{array} $$

(27)

where $\omega _d=\omega _n\sqrt {1-\zeta ^2}$. Applying inverse Laplace transform into formula (27), we get

$$\begin{array}{@{}rcl@{}} c(t)=A&-&A\cdot \exp(-\zeta \omega_n t)\cos\omega_d t\\ &-&A\frac{\zeta}{\sqrt{1-\zeta^2}}\cdot \exp(-\zeta \omega_n t)\sin\omega_d t \end{array} $$

(28)

where $\exp (-\zeta \omega _n t)\cos \omega _d t=L^{-1}\left (\frac {s+\zeta \omega _n}{(s+\zeta \omega _n)^2+\omega _d^2}\right )$ and $\exp (-\zeta \omega _n t)\sin \omega _d t=L^{-1}\left (\frac {\omega _d}{(s+\zeta \omega _n)^2+\omega _d^2}\right )$. Organizing formula (28), we obtain

$$\begin{array}{@{}rcl@{}} c(t)&\,=\,&A\,-\,A\cdot \exp(-\zeta \omega_n t)\left( \!\cos\omega_d t\!-\frac{\zeta}{\!\sqrt{1-\!\zeta^2}}\cdot \sin\omega_d t\!\right)\\ &=&A-A\frac{\exp(-\zeta \omega_n t)}{\sqrt{1-\zeta^2}}\sin (\omega_d t+\varphi) \end{array} $$

(29)

where $\varphi =\arctan \frac {\sqrt {1-\zeta ^2}}{\zeta }$. Taking the time derivative on (29), we get

$$\begin{array}{@{}rcl@{}} && \frac{dc(t)}{dt}=A\zeta \omega_n \frac{\exp(-\zeta \omega_n t)}{\sqrt{1-\zeta^2}}\sin (\omega_d t+\varphi)\\ &&\hspace{0.5cm}-A\omega_d \frac{\exp(-\zeta \omega_n t)}{\sqrt{1-\zeta^2}}\cos(\omega_d t+\varphi) \end{array} $$

(30)

Let $\frac {dc(t)}{dt}= 0$, and the following formula holds

$$\begin{array}{@{}rcl@{}} \sin (\omega_d t+\varphi)=\frac{{\sqrt{1-\zeta^2}}}{\zeta}\cos(\omega_d t+\varphi) \end{array} $$

(31)

namely

$$\begin{array}{@{}rcl@{}} \tan (\omega_d t+\varphi)=\tan (\varphi) \end{array} $$

(32)

then, we know that ω _d t = 0,π, 2π,⋯ Since the peak time is the time that c(t) reaches the first peak, then ω _d t _p = π, i.e.,

$$\begin{array}{@{}rcl@{}} t_p=\frac{\pi}{\omega_d}=\frac{\pi}{\omega_n \sqrt{1-\zeta^2}} \end{array} $$

From formula (29), we can get $A\pm A\frac {\exp (-\zeta \omega _n t)}{\sqrt {1-\zeta ^2}}$ is the envelope equation. Then we learn that

$$\begin{array}{@{}rcl@{}} 1+\frac{\exp(-\zeta \omega_n t_s)}{\sqrt{1-\zeta^2}}= 1+{\Delta} \end{array} $$

then

$$\begin{array}{@{}rcl@{}} \exp(-\zeta \omega_n t_s)={\Delta}{\sqrt{1-\zeta^2}} \end{array} $$

(33)

According to (33), it is clear that the following formula holds

$$\begin{array}{@{}rcl@{}} -\zeta \omega_n t_s=\ln\left( {\Delta}{\sqrt{1-\zeta^2}}\right) \end{array} $$

(34)

Owing to $\sqrt {1-\zeta ^2}\approx 1$, t _s can be expressed as

$$\begin{array}{@{}rcl@{}} t_s=\frac{-\ln {\Delta}}{\zeta \omega_n} \end{array} $$

Appendix B

On the basis of differential properties of Laplace transform, we can obtain the following formula

$$\begin{array}{@{}rcl@{}} \mathscr{L}\left[\frac{d}{dt}e(t)\right]={\int}_0^{\infty}\frac{d}{dt}e(t)\cdot e^{-st}dt=sE(s)-e(0) \end{array} $$

(35)

Taking limit of the two sides of (35), the following formula sets up

$$\begin{array}{@{}rcl@{}} \lim\limits_{s\rightarrow 0}{\int}_0^{\infty}\frac{d}{dt}e(t)\cdot e^{-st}dt=\lim\limits_{s\rightarrow 0}(sE(s)-e(0)) \end{array} $$

(36)

Since

$$\begin{array}{@{}rcl@{}} \lim\limits_{s\rightarrow 0}{\int}_0^{\infty}\frac{d}{dt}e(t)\cdot e^{-st}dt&=&{\int}_0^{\infty}\frac{d}{dt}e(t)\lim_{s\rightarrow 0}e^{-st}dt\\ &=&{\int}_0^{\infty}de(t)=\lim\limits_{t\rightarrow \infty}{\int}_0^{t}de(t)\\ &=&\lim\limits_{t\rightarrow \infty}\left[e(t)-e(0)\right] \end{array} $$

(37)

we get

$$\begin{array}{@{}rcl@{}} \lim\limits_{t\rightarrow \infty}\left[e(t)-e(0)\right]=\lim\limits_{s\rightarrow 0}(sE(s)-e(0)) \end{array} $$

(38)

It is clear that

$$\begin{array}{@{}rcl@{}} \lim\limits_{t\rightarrow \infty}e(t)=\lim_{s\rightarrow 0}sE(s) \end{array} $$