An efficient adaptive method for estimating the distance between mobile sensors

Abstract

The received signal strength (RSS) is a common source of information used for estimating the distance between two wireless nodes, whether these nodes are stationary or mobile. Minimum mean squared error distance estimation methods that use the RSS require prior knowledge of both the variance of the noise and, in the case of mobile sensors, the dynamics of the nodes’ mobility. In mobile applications, where low computational complexity is important, pseudo-optimal estimations are preferred, as they do not require such information. In this case, the maximum likelihood estimator (MLE) is often used. In this paper, we propose an efficient pseudo-optimal log-power based distance estimation method using RSS under lognormal shadowing, that improves the MLE. It does not require a priori knowledge either of the movement dynamics or of the variance of the noise. The method is based on adaptively minimizing the variance of the prediction error, using a random walk model with correlated increments. It is analytically demonstrated that the distance estimation error variance of the proposed method improves the MLE in both the static and mobile cases. We use a simulated velocity model example to compare its performance with other algorithms in this group, such as the linear mean square filter and the Gauss–Newton search.

Appendix

We first show that there is a unique minimizer of $\sigma ^{2}_\epsilon$ in the interval $\theta (k)\in [0,1]$ for the family of movement dynamics whose autocorrelation can be represented by a Fourier series expansion. Various motion dynamic cases can be represented using this representation, ([14]). By using (26), the prediction error at step $k+1$ can be represented by the following recursion:

$$\begin{aligned} \epsilon (k+1)= & {} \, p_r(k+1)-p_r(k)+\theta \epsilon (k)\end{aligned}$$

(40)

$$\begin{aligned}= & {} \, r(k+1)+\eta (k+1)-\eta (k)+\theta \epsilon (k). \end{aligned}$$

(41)

where $r(k)$ is a zero mean, incremental log-power sequence given by

$$\begin{aligned} r(k)=p(k)-p(k-1). \end{aligned}$$

(42)

By taking the expected value of $\epsilon ^2(k)$ and considering that $\eta (k)$ is an $i.i.d.$ sequence, independent of $r(k)$, we obtain the following relationship for the prediction error variance:

$$\begin{aligned} \sigma _\epsilon ^2=2\sigma _\eta ^2+\theta ^2\sigma _\epsilon ^2+\sigma _r^2- 2\theta {\mathcal {E}}[\eta (k)\epsilon (k)]+2\theta {\mathcal {E}}[r(k+1)\epsilon (k)], \end{aligned}$$

(43)

where ${\mathcal {E}}[\eta (k+1)\epsilon (k)]=0$. The expectations of the last two terms are given by

$$\begin{aligned}&{\mathcal {E}}[\eta (k)\epsilon (k)]=\sigma _\eta ^2, \end{aligned}$$

(44)

$$\begin{aligned}&{\mathcal {E}}[r(k+1)\epsilon (k)]= {\mathcal {E}}[r(k+1)r(k)+\theta r(k+1)\epsilon (k-1)]\nonumber \\&= {\mathcal {E}}[r(k+1)r(k)+\theta r(k+1)r(k-1) +\theta ^2r(k+1)\epsilon (k-2)]\nonumber \\&\vdots \nonumber \\&=\sum _{i=1}^{\infty }\theta ^{i-1}R(i), \end{aligned}$$

(45)

where $R(i)={\mathcal {E}}[r(k)r(k-i)]$. By using it in (43) and taking into account that $\sigma _r^2=R(0)$, the final expression of $\sigma _\epsilon ^2$ is obtained as

$$\begin{aligned} \sigma _\epsilon ^2=\frac{2(1-\theta )\sigma _\eta ^2-\sigma _r^2+ 2\sum _{i=0}^{\infty }\theta ^{i}R(i)}{1-\theta ^2}. \end{aligned}$$

(46)

In order to analyze the possible minimizers of cost function $\sigma _\epsilon ^2$ with respect to $\theta$, we need to write the correlation function $R(i)/\sigma _r^2$. To this end, let us consider the following complex series expansion representing the autocorrelation:

$$\begin{aligned} \frac{R(i)}{\sigma _r^2}=\frac{1}{2N}\sum _{n=1}^N(\alpha _n^i+\alpha _n^{*i}), \end{aligned}$$

(47)

where $\alpha _n$ is complex with $|\alpha _n|\le 1$, $(*)$ means compex conjugate, and $R(i)$ is a positive definite function, [25]. Thus, the last term in the numerator of (46) gives

$$\begin{aligned} \frac{\sigma _r^2}{N}\sum _{i=0}^{\infty }\theta ^{i}\sum _{n=1}^N(\alpha _n^i +\alpha _n^{*i})&=\frac{\sigma _r^2}{N}\sum _{n=1}^N\sum _{i=0}^{\infty }( \theta \alpha _n)^i+(\theta \alpha _n)^{*i}\nonumber \\&=\frac{\sigma _r^2}{N}\sum _{n=1}^N\frac{1}{1-\theta \alpha _n}+\frac{1}{1-\theta \alpha _n^*}. \end{aligned}$$

(48)

By denoting $\alpha _n=\rho _n e^{j\varphi _n}$ and taking into account that both $\theta$ and $\rho _n$ are positive and less than one, a family of possible values for each element of the sum are given by the following parametrization:

$$\begin{aligned} \frac{1}{1-\theta \alpha _n}+\frac{1}{1-\theta \alpha _n^*}=2\frac{1-\theta \rho _n \cos (\varphi _n)}{(1-\theta \rho _n e^{j\varphi _n})(1-\theta \rho _n e^{-j\varphi _n})}, \end{aligned}$$

(49)

where $\rho _n\in [0, 1]$ and $\varphi _n\in [0, 2\pi ]$. Now, we need the following lemma:

Lemma

[26]: Let $g$ and $f$ be defined on a convex set $\varTheta$, such that $f(\theta ) \ne 0$ for all $\theta \in \varTheta$. Then, $g/f$ has only one minimum on $\varTheta$ (quasi-convex) if both $f(\theta )> 0$ is concave and $g(\theta )\ge 0$ is convex for all $\theta \in \varTheta$.

First, from (46) and (49), we can write

$$\begin{aligned} g(\theta )&=2(1-\theta )\sigma _\eta ^2-\sigma _r^2+\frac{2\sigma _r^2}{N} \sum _{n=1}^{\infty }\frac{1-\theta \rho _n \cos (\varphi _n)}{(1-\theta \rho _n e^{j\varphi _n})(1-\theta \rho _n e^{-j\varphi _n})}\end{aligned}$$

(50)

$$\begin{aligned} f(\theta )&=1-\theta ^2. \end{aligned}$$

(51)

From (51), $f(\theta )>0$ concave, for $\theta \in (0, 1)$, and from (50), $g(\theta )$ is always positive definite. In order to study its convexity we obtain the second derivative with respect to $\theta$ which gives

$$\begin{aligned} \frac{4\cos ^2(\varphi _n)-\cos (\varphi _n)[2(\theta \rho _n)^3+6\theta \rho _n] +6(\theta \rho _n)^2 -2}{[(\theta \rho _n)^2-2\theta \rho _n\cos (\varphi _n)+1]^3}. \end{aligned}$$

(52)

By calculating derivatives we find that the minimum of the above is given at $\theta \rho _n=0$ for any possible value of $\cos (\varphi _n)$. Then, for $\cos (\varphi _n)\ge \sqrt{1/2}$, the second derivative is always positive, which restricts the value of $\varphi _n$ in the interval [$-\pi /4, \pi /4$]. This means that the higher-frequency sinusoidal component of the Fourier Series Expansion of $R(i)$, (47), must contain at least eight samples per period, which is a weak restriction. Thus, we can conclude that, under mild assuptions, there exists only one minimizer for the minimum variance prediction error for $\theta$, within the interval ($0, 1$).

In order to prove (iii), let us compute the upper bound of $\sigma _\epsilon ^2$. By taking into account that $R(i)/\sigma _r^2$ is an autocorrelation function, from (46) the following inequality holds:

$$\begin{aligned} \sigma _\epsilon ^2\le & {} \, \frac{2(1-\theta )\sigma _\eta ^2+\sigma _r^2\left( 2\sum _{i=0}^{\infty } \theta ^{i}-1\right) }{1-\theta ^2}\end{aligned}$$

(53)

$$\begin{aligned}= & {} \, \frac{2(1-\theta )\sigma _\eta ^2+\sigma _r^2\left( \frac{1+\theta }{1-\theta } \right) }{1-\theta ^2}=h(\theta ). \end{aligned}$$

(54)

Now, let us define the minimizer of $h(\theta )$ as

$$\begin{aligned} \bar{\theta }=\arg \min _{\theta }\{h(\theta )\}. \end{aligned}$$

(55)

By multiplying both sides of (54) by $1-\theta ^2$, within the admissible interval of $\theta$, and performing the derivative, with respect to $\theta$, the following upper bound for the minimum of $\sigma _{\epsilon }^2$ is obtained:

$$\begin{aligned} \sigma _{\epsilon min}^2\le \frac{\sigma _\eta ^2}{\bar{\theta }}-\frac{\sigma _r^2}{2 \bar{\theta }(1-\bar{\theta })^2}\le \frac{\sigma _\eta ^2}{\bar{\theta }}. \end{aligned}$$

(56)

In order to find the relationship between the variance of $e(k)$ and the variance of the prediction error, we use (22) as follows:

$$\begin{aligned} \sigma _e^2&={\mathcal {E}}[(p(k)-\hat{p}(k))^2]\end{aligned}$$

(57)

$$\begin{aligned}&={\mathcal {E}}[\eta ^2(k)]+\theta ^2{\mathcal {E}}[\epsilon ^2(k)]-2\theta {\mathcal {E}}[\eta (k)\epsilon (k)],\end{aligned}$$

(58)

$$\begin{aligned}&=(1-2\theta )\sigma _\eta ^2+\theta ^2\sigma _\epsilon ^2, \end{aligned}$$

(59)

where the last term of (58) is ${\mathcal {E}}[\eta (k)\epsilon (k)]=\sigma _\eta ^2$. Finally, by replacing the bound (56) in (59) we obtain an upper bound on the pseudo-optimal minimum error attainable for any correlation function as follows:

$$\begin{aligned} \sigma _e^2\le & {} \, (1-2\bar{\theta })\sigma _\eta ^2+\bar{\theta }^2\sigma _{\epsilon min}^2\end{aligned}$$

(60)

$$\begin{aligned}\le & {} \, (1-2\bar{\theta })\sigma _\eta ^2+\bar{\theta }\sigma _\eta ^2\end{aligned}$$

(61)

$$\begin{aligned}\le & {} \, \sigma _\eta ^2. \end{aligned}$$

(62)

In particular, in the static case, $\sigma _r^2=0$, the minimizer is $\bar{\theta }=1$, which leads to $\sigma _{\epsilon min}^2=\sigma _\eta ^2$ and $\sigma _e^2=0$.