Dynamic estimation of local mean power in GSM-R networks

Abstract

The dynamic estimation algorithm for Rician fading channels in GSM-R networks is proposed, which is an expansion of local mean power estimation of Rayleigh fading channels. The proper length of statistical interval and required number of averaging samples are determined which are adaptive to different propagation environments. It takes advantage of signal samples and Rician fading parameters of last estimation to reduce measurement overhead. The performance of this method was evaluated by measurement experiments along Beijing–Shanghai high-speed railway. When it is NLOS propagation, the required sampling intervals can be increased from $1.1{\lambda}$ in Lee’s method to $3.7{\lambda}$ of the dynamic algorithm. The sampling intervals can be set up to $12{\lambda}$ although the length of statistical intervals decrease when there is LOS signal, which can reduce the measurement overhead significantly. The algorithm can be applied in coverage assessment with lower measurement overhead, and in dynamic and adaptive allocation of wireless resource.

Appendix Proof of Theorem 1 and 2

1.1 Proof of Theorem 1

The normlving the integral formualized estimation error P _e can be determined by $\hat{s}$ and $\sigma_{\hat{s}}$ according to Definition 1, and $\sigma_{\hat{s}}$ can be calculated by

$$\sigma_{\hat{s}}^{2}=\frac{1}{L}\int\limits_{0}^{2L}\left(1-\frac{\tau}{2L}\right)R_{p_{r}^2}(\tau)d\tau, $$

(22)

where R ² _{p_r} (τ) = E[p ² _r (x)p ² _r (x + τ)] − E[p ² _r (x)]E[p ² _r (x + τ)] is the autocovariance function of the squared envelope of p _r (x). R ² _{p_r} (τ) can be derived from Rician distribution (Eqs. 6, 7 in Sect. 2) by approximation [4] as follows:

$$R_{p_{r}^2}(\tau)=4\sigma^2\left[J_0^2\left(\frac{2\pi}{\lambda}\tau\right)+2KJ_0\left(\frac{2\pi}{\lambda}\tau\right)\cos\left(\frac{2\pi}{\lambda}\eta\tau\right)\right], $$

(23)

where $J_0(\cdot)$ is the zero-order Bessel function, and $\eta=\cos\theta_0$ is the intermediate valuable. Then $\sigma_{\hat{s}}^2$ can be calculated by substituting (23) into (22), i.e.,

$$\begin{aligned} \sigma_{\hat{s}}^{2}=\frac{4\sigma^2}{L}\int\limits_{0}^{2L}\frac{2L-\tau}{2L}[J_0^2(\frac{2\pi}{\lambda}\tau)+2KJ_0(\frac{2\pi}{\lambda}\tau)\cos(\frac{2\pi}{\lambda}\eta\tau)]d\tau\\ =\frac{\hat{s}^2(2L-\lambda)\lambda}{2(1+K)^{2}L^2}\int\limits_0^{\frac{2L}{\lambda}}[J_0^2(2\pi \rho)+2KJ_0(2\pi \rho)\cos(2\pi \eta)]\rho d\rho, \end{aligned} $$

(24)

where ρ = τ/λ is the intermediate valuable and $\sigma_{\hat{s}}^2\rightarrow0$ as $2L/\lambda\rightarrow\infty$. $\hat{s}$ can be considered as Gaussian distributed when 2L is large enough. Then $\sigma_{\hat{s}}^2$ can be represented by the simple form as follows:

$$\sigma_{\hat{s}}^2=\frac{2(n-1)}{n^2(1+K)^2}\int\limits_0^n g(K;\rho) d\rho, $$

(25)

where n: = 2L/λ represents the relationship between statistical intervals 2L and wireless prorogation wavelength $\lambda, g(K;\rho):=[J_0^2(2\pi \rho)+2KJ_0(2\pi \rho)\cos(2\pi \eta)]\rho$ is the intermediate function.

Given the definition of normalized estimation error P _e in (12), it can be calculated by substituting (25) into (12) and solving the integral formula. Then P _e can be determined by

$$\begin{aligned} P_e &:=10 \log_{10}\left(\frac{\hat{s}+\sigma_{\hat{s}}}{\hat{s}-\sigma_{\hat{s}}}\right)\\ &=10 \log_{10}\left(\frac{n(1+K)+\sqrt{2(1+n)\int\nolimits_0^n g(K;\rho) d\rho}}{n(1+K)-\sqrt{2(1+n)\int\nolimits_0^n g(K;\rho) d\rho}}\right)\\ &= 10 \log_{10}\left(\frac{\frac{2\sigma^2+\nu^2}{2\sigma^2}n+\sqrt{2(1+n)\int\nolimits_0^n g\left(\frac{\nu^2}{2\sigma^2};\rho\right) d\rho}}{\frac{2\sigma^2+\nu^2}{2\sigma^2}n-\sqrt{2(1+n)\int\nolimits_0^n g\left(\frac{\nu^2}{2\sigma^2};\rho\right) d\rho}}\right).\end{aligned} $$

(26)

1.2 Proof of Theorem 2

According to the characteristics of Rician distribution, it can be expressed that z ² _i  = x ² _i  + y ² _i where $x_i \sim N(\nu\cos \eta,\sigma^2)$ and $y_i \sim N(\nu\sin \eta,\sigma^2)$ are statistically independent normal random variables and η is any real number. Let x _0i  = x _i /σ, then $x_{0i} \sim N(\nu \sin \eta,1)$ and its sum subject to the non-central χ² distribution, that is $\sum\nolimits_{i=1}^{N}x_{0i}^2 \sim \chi_N^2(\nu^2\cos^2\eta)$. For E[χ ² _n (λ)] = n + λ and D[χ ² _n (λ)] = 2n + 4λ, the mean value and variance of $\sum\nolimits_{i=1}^{N}x_{i}^2$ can be calculated by:

$$\begin{aligned} E\left[\sum_{i=1}^{N}x_i^2\right] &= \sigma^2E\left[\sum_{i=1}^{N}x_{0i}^2\right]\\ &=\sigma^2E\left[\chi_N^2(\nu^2\cos^2\eta)\right]\\ &=\sigma^2\left(N+\nu^2\cos^2\eta\right), \end{aligned} $$

(27)

$$\begin{aligned} D\left[\sum_{i=1}^{N}x_i^2\right] &= \sigma^4D\left[\sum_{i=1}^{N}x_{0i}^2\right]\\ &= \sigma^4D\left[\chi_N^2(\nu^2\cos^2\eta)\right]\\ &= \sigma^4\left(2N+4\nu^2\cos^2\eta\right), \end{aligned} $$

(28)

and $E\left[\sum_{i=1}^{N}y_i^2\right]=\sigma^2(N+\nu^2\sin^2\eta), D\left[\sum_{i=1}^{N}y_i^2\right]=\sigma^4(2N+4\nu^2\sin^2\eta)$ can also be calculated in the same way. Then the expectation of r ² and its variance can be calculated by:

$$\begin{aligned} \bar{r^2}&=E\left[\frac{1}{N}\sum_{i=1}^{N}z_i^2\right]=\frac{1}{N}E\left[\sum_{i=1}^{N}(x_i^2+y_i^2)\right]\\ &=\frac{\sigma^2}{N}\left(N+\nu^2\cos^2\eta+N+\nu^2\sin^2\eta\right)\\ &=\frac{\sigma^2}{N}\left(2N+\nu^2\right), \end{aligned} $$

(29)

$$\begin{aligned} \sigma_{\bar{r^2}}^2 &=D\left[\frac{1}{N}\sum_{i=1}^{N}z_i^2\right]=\frac{1}{N^2}D\left[\sum_{i=1}^{N}\left(x_i^2+y_i^2\right)\right]\\ &=\frac{\sigma^4}{N^2}\left(2N+4\nu^2\cos^2\eta+2N+4\nu^2\sin^2\eta\right)\\ &=\frac{\sigma^4}{N^2}\left(4N+4\nu^2\right). \end{aligned} $$

(30)

Then the estimation error can be calculated according to (29) and (30) as follows:

$$\begin{aligned} Q_e &=10 \log_{10}\left(\frac{\bar{r^2}+\sigma_{\bar{r^2}}}{\bar{r^2}}\right)\\ &=10 \log_{10}\left(\frac{\frac{\sigma^2}{N}\left(2N+\nu^2\right)+\frac{2\sigma^2}{N}\sqrt{N+\nu^2}}{\frac{\sigma^2}{N}(2N+\nu^2)}\right)\\ &=10 \log_{10}\left(\frac{2N+\nu^2+2\sqrt{N+\nu^2}}{2N+\nu^2}\right).\end{aligned} $$

(31)