Improved Least Squares Approaches for Differential Received Signal Strength-Based Localization with Unknown Transmit Power

Abstract

In this paper we consider the problem of improving unknown node localization by using differential received signal strength (DRSS). Many existing localization approaches, especially those using the least squares methods, either ignore nonlinear constraint among model parameters or utilize them inefficiently. In this paper, we develop four DRSS-based localization methods by utilizing different combinations of covariance and weight matrices. Each method constructs a two-stage procedure. During the first stage, an initial coarse position estimate is obtained. The second stage results the refined localization by accounting for nonlinear dependency among estimator variables. The proper choice among these proposed methods may be offered, depending on a particular signal to noise range. We implement these methods and compare them with some of the state-of-art methods in this particular problem domain and verify their performances by simulations.

Appendix: Derivation of Covariance Matrix \({{{\mathbf {C}}}}_{\varvec{ \beta }}\) Appendix

Lemma 1

If a normaln-dimensional multivariate random variable${\varvec{x}}=\left[ x_1, \ldots ,x_n\right]$has the distribution${{\mathcal {N}}}\left( \varvec{\mu },\varvec{\Sigma }\right)$, then its moment generating function (MGF) is given by [45]

$$\begin{aligned} {\mathbf {M}}_{{\mathbf {x}}}({\varvec{a}})&=\,{{\mathbb {E}}}_{{{\mathbf {x}}}}\left\{ \text {exp}\left( {{{\varvec{a}}}}^{\mathrm{T}}\cdot {\varvec{x}}\right) \right\} =\text {exp}\left( {{{\varvec{a}}}}^{\mathrm{T}}\cdot {\varvec{\mu }}{\mathbf +}\frac{1}{2}{{{\varvec{a}}}}^{\mathrm{T}}\cdot {\mathbf \Sigma }\cdot {{\varvec{a}}}\right) \end{aligned}$$

(35)

where${\varvec{a}}=\left[ a_1, \ldots ,a_n\right]$is a known vector withnreal elements.

we may substitute the normal n-dimensional multivariate random vector $\left[ R_d\left( 2,1\right) ,\ldots ,R_d\left( L,1\right) \right]$ for ${{\varvec{x}}}$. According to (4), the joint multivariate distribution of random variables $R_d\left( l\mathrm{,1}\right) \mathrm{,\ }l\mathrm{=2,}\ldots \mathrm{,}L$ is

$$\begin{aligned} \begin{aligned} f_{R_d\left( 2,1\right) ,\ldots ,R_d\left( L,1\right) }&\left( {\left[ r_1, \ldots ,r_{L-1}\right] }^\mathrm{T}\right) \\&=\frac{1}{{\left( 2\pi \right) }^{\left( L-1\right) /2}{\left| {\mathrm{C}}_R\right| }^{1/2}}exp\left[ -\frac{1}{2}\left( {\widetilde{{\varvec{r}}}}^\mathrm{T}{\mathrm{C}}^{-1}_R\widetilde{{\varvec{r}}}\right) \right] , \\&\widetilde{{\varvec{r}}}=\left[ r_1+10\alpha {\text {Log}}_{10}\left( \frac{d_2}{d_1}\right) ,\ldots ,r_{L-1}+10\alpha {\text {Log}}_{10}\left( \frac{d_L}{d_1}\right) \right] \end{aligned} \end{aligned}$$

(36)

and covariance matrix ${\mathrm{C}}_R$ is a $\left( L-1\right) \times \left( L-1\right)$ matrix whose all off-diagonal elements are $\sigma ^2$ and its diagonal elements are ${2\sigma }^2$, where

$$\begin{aligned} \mathbf {\mathrm{C}}_R=&\begin{bmatrix} {2\sigma }^2&\sigma ^2&\cdots&\sigma ^2 \\ \sigma ^2&{2\sigma }^2&\cdots&\sigma ^2 \\ \vdots&\ddots&\ddots&\vdots \\ \sigma ^2&\cdots&2\sigma ^2&\sigma ^2 \\ \sigma ^2&\cdots&\sigma ^2&{2\sigma }^2 \end{bmatrix}. \end{aligned}$$

(37)

Then ${{\varvec{x}}}$ has the distribution defined in (36) where $\left[ -10\alpha {\text {Log}}_{10}\left( {d_2}/{d_1}\right) ,\right.$$\left. \cdots ,-10\alpha {\text {Log}}_{10}\left( {d_L}/{d_1}\right) \right]$ and ${\mathrm{C}}_R$ are substituted for ${\varvec{\mu }}$ and ${\mathbf \Sigma }$ respectively in (35).

By using Lemma 1, the ${\left( i,j\right) }^{\mathrm{th}}$ entry of covariance matrix ${{{\mathbf {C}}}}_{{\beta }}$ is given by

$$\begin{aligned} \begin{aligned} {{{{\mathbf {C}}}}_{\beta }}_{i,j}=&{{{\mathbb {E}}}}_{ \begin{array}{c} R_d\left( l,1\right) \\ l=2, \ldots ,5 \end{array} }\left\{ \beta ^{-1}_{i+1\mathrm{,1}}\beta ^{-1}_{j+1\mathrm{,1}}\right\} \\&-{{{\mathbb {E}}}}_{ \begin{array}{c} R_d\left( l,1\right) \\ l=2, \ldots ,5 \end{array} }\left\{ \beta ^{-1}_{i+1\mathrm{,1}}\right\} \cdot {{{\mathbb {E}}}}_{ \begin{array}{c} R_d\left( l,1\right) \\ l=2, \ldots ,5 \end{array} }\left\{ \beta ^{-1}_{j+1\mathrm{,1}}\right\} ,\\&\ i,j\in \left\{ 1, \ldots ,L-1\right\} ,i\ne j. \end{aligned} \end{aligned}$$

(38)

For example, by setting $i=1$ and $j=2$, then $\beta ^{-1}_{2\mathrm{,1}}\beta ^{-1}_{3\mathrm{,1}}$ and vector ${{\varvec{a}}}$ equal to, respectively, $\text {exp}({{{\varvec{a}}}}^\mathrm{T} \cdot {\left[ R_d\left( 2,1\right) , \ldots ,R_d\left( L,1\right) \right] }^\mathrm{T})$ and ${-}\left( {2\text {Ln}\left( \mathrm{10}\right) }/{\mathrm{10}\alpha }\right) \cdot {\left[ -1,-1,0,0\right] }^T$.

According to (35), we have

$$\begin{aligned} \begin{aligned}&{{{\mathbb {E}}}}_{ \begin{array}{c} R_d\left( l,1\right) \\ l=2, \ldots ,5 \end{array} }\left\{ \beta ^{-1}_{2\mathrm{,1}}\beta ^{-1}_{3\mathrm{,1}}\right\} \\&\quad = \text {exp}\left\{ -2\text {Ln}\left( 10\right) \cdot {\text {Log}}_{10}\left( \frac{d_2{\cdot d}_3}{d^2_1}\right) +\frac{1}{2}\cdot {\left( \frac{2\text {Ln}\left( \mathrm{10}\right) }{\mathrm{10}\alpha }\right) }^2\cdot \left( {6\sigma }^2\right) \right\} \end{aligned} \end{aligned}$$

(39)

In order to write (39), we have used the relation

$$\begin{aligned} \begin{bmatrix} 1&1&0&0 \end{bmatrix} \cdot \begin{bmatrix} 2\sigma ^2&\sigma ^2&\sigma ^2&\sigma ^2 \\\sigma ^2&2\sigma ^2&\sigma ^2&\sigma ^2 \\ \sigma ^2&\sigma ^2&2\sigma ^2&\sigma ^2 \\\sigma ^2&\sigma ^2&\sigma ^2&2\sigma ^2 \end{bmatrix}\cdot \begin{bmatrix}1 \\ 1 \\ 0 \\ 0 \end{bmatrix}=\left( {6\sigma }^2\right) \end{aligned}$$

(40)

Similarly, by taking $\beta ^{-1}_{2\mathrm{,1}}$ as $\text {exp}({{{\varvec{a}}}}^T\cdot {\left[ R_d\left( 2,1\right) , \ldots ,R_d\left( L,1\right) \right] }^T)$, then ${{\varvec{a}}}$ equals ${-}\left( {2\text {Ln}\left( \mathrm{10}\right) }/{\mathrm{10}\alpha }\right) \cdot {\left[ -1,0,0,0\right] }^T$ and according to (35), we have

$$\begin{aligned} \begin{aligned} \mathbb {E}_{\underset{l=2,\ldots ,5}{R_d\left( l,1\right) }}\left\{ \beta ^{-1}_{2,1}\right\}&= \text {exp}\left\{ -2\text {Ln}\left( 10\right) \cdot \text {Log}_{10}(\frac{d_2}{d_1}) + \frac{1}{2}\cdot {\left( \frac{2\text {Ln}(10)}{10\alpha }\right) }^2\cdot \left( {2\sigma }^2\right) \right\} . \end{aligned} \end{aligned}$$

(41)

In the same way, ${{{\mathbb {E}}}}_{ \begin{array}{c} R_d\left( l,1\right) \\ l=2,\ldots ,5 \end{array} }\left\{ \beta ^{-1}_{3\mathrm{,1}}\right\}$ is given by

$$\begin{aligned} \begin{aligned} \mathbb {E}_{\underset{l=2,\ldots ,5}{R_d\left( l,1\right) }}\left\{ \beta ^{-1}_{3,1}\right\}&= \text {exp}\left\{ -2\text {Ln}\left( 10\right) \cdot \text {Log}_{10}\left( \frac{d_3}{d_1}\right) +\frac{1}{2}\cdot {\left( \frac{2\text {Ln}\left( 10\right) }{10\alpha }\right) }^2\cdot \left( {2\sigma }^2\right) \right\} . \end{aligned} \end{aligned}$$

(42)

As a result, ${{{\mathbb {E}}}}_{ \begin{array}{c} R_d\left( l,1\right) \\ l=2,\ldots ,5 \end{array} }\left\{ \beta ^{-1}_{3\mathrm{,1}}\right\} \cdot {{{\mathbb {E}}}}_{ \begin{array}{c} R_d\left( l,1\right) \\ l=2,\ldots ,5 \end{array} }\left\{ \beta ^{-1}_{2,1}\right\}$ is achieved by

$$\begin{aligned} \begin{aligned}&\mathbb {E}_{\underset{l=2,\ldots ,5}{R_d\left( l,1\right) }}\left\{ \beta ^{-1}_{3,1}\right\} \cdot \mathbb {E}_{\underset{l=2,\ldots ,5}{R_d\left( l,1\right) }}\left\{ \beta ^{-1}_{2,1}\right\} \\&\quad = \text {exp} \left\{ -2\text {Ln}\left( 10\right) \cdot \text {Log}_{10}\left( \frac{d_2\cdot d_3}{d_1 ^2}\right) + {\left( \frac{-2\text {Ln}\left( \mathrm{10}\right) }{10\alpha }\right) }^2\cdot \left( {2\sigma }^2\right) \right\} . \end{aligned} \end{aligned}$$

(43)

Substituting equation from (39) to (43) in (38), ${{{{\mathbf {C}}}}_{{\beta }}}_{2,3}\$ is given by

$$\begin{aligned} \begin{aligned} {{{\mathbf C}}_{{\beta }}}_{2,3}&=\text {exp}\left\{ -2\text {Ln}\left( 10\right) \cdot {\text {Log}}_{10}\left( \frac{d_2{\cdot d}_3}{d^2_1}\right) +\frac{1}{2}\cdot {\left( \frac{2\text {Ln}\left( \mathrm{10}\right) }{\mathrm{10}\alpha }\right) }^2\cdot \left( {6\sigma }^2\right) \right\} \\&-\text {exp}\left\{ -2\text {Ln}\left( 10\right) \cdot {\text {Log}}_{10}\left( \frac{d_2{\cdot d}_3}{d^2_1}\right) +{\left( \frac{2\text {Ln}\left( \mathrm{10}\right) }{\mathrm{10}\alpha }\right) }^2\cdot \left( {2\sigma }^2\right) \right\} \end{aligned} \end{aligned}$$

(44)

and the diagonal elements of the covariance matrix ${{{\mathbf C}}_{{\beta }}}$ are given by

$$\begin{aligned} \begin{aligned} {\mathbf {C}}_{\beta _{i,i}}={{{\mathbb {E}}}}_{ \begin{array}{c} R_d\left( l,1\right) \\ l=2, \ldots ,5 \end{array} }\left\{ \beta ^{-2}_{i+1,1}\right\} {-}{{{\mathbb {E}}}}^2_{ \begin{array}{c} R_d\left( l,1\right) \\ l=2, \ldots ,5 \end{array} }\left\{ \beta ^{-1}_{i+1\mathrm{,1}}\right\} ,\ \ i\in \left\{ 1, \ldots ,L-1\right\} \end{aligned} \end{aligned}$$

(45)

By setting $i=1$, and taking $\beta ^{-2}_{i+1\mathrm{,1}}$ as $\text {exp}({{{\varvec{a}}}}^\mathrm{T}\cdot {\left[ R_d\left( 2,1\right) , \ldots ,R_d\left( L,1\right) \right] }^\mathrm{T})$, then ${{\varvec{a}}}$ equals ${-}({2\text {Ln}\left( \mathrm{10}\right) }/{\mathrm{10}\alpha })\cdot {\left[ -2,0,0,0\right] }^\mathrm{T}$. According to (35), we have

$$\begin{aligned} \begin{aligned} {{{\mathbb {E}}}}_{ \begin{array}{c} R_d\left( l,1\right) \\ l=2, \ldots ,5 \end{array} }\left\{ \beta ^{-2}_{2\mathrm{,1}}\right\} =\text {exp}\left\{ -4\text {Ln}\left( 10\right) \cdot {\text {Log}}_{10}\left( \frac{d_2}{d_1}\right) +\frac{1}{2}\cdot {\left( \frac{4\text {Ln}\left( \mathrm{10}\right) }{\mathrm{10}\alpha }\right) }^2\cdot \left( {2\sigma }^2\right) \right\} \end{aligned} \end{aligned}$$

(46)

${{{\mathbb {E}}}}_{ \begin{array}{c} R_d\left( l,1\right) \\ l=2, \ldots ,5 \end{array} }\left\{ \beta ^{-1}_{2\mathrm{,1}}\right\}$ is obtained in (41). Substituting (46) and (41) in (45), ${\mathbf {C}}_{\beta _{2,2}}$ is given by

$$\begin{aligned} \begin{aligned} \mathbf {C}_{\beta _{2,2}}&=\text {exp}\left\{ -4\text {Ln}\left( 10\right) \cdot {\text {Log}}_{10}\left( \frac{d_2}{d_1}\right) +\frac{1}{2}\cdot {\left( \frac{4\text {Ln}\left( \mathrm{10}\right) }{\mathrm{10}\alpha }\right) }^2\cdot \left( {2\sigma }^2\right) \right\} \\&-\text {exp}\left\{ -4\text {Ln}\left( 10\right) \cdot {\text {Log}}_{10}\left( \frac{d_2}{d_1}\right) +\cdot {\left( \frac{2\text {Ln}\left( \mathrm{10}\right) }{\mathrm{10}\alpha }\right) }^2\left( {2\sigma }^2\right) \right\} \end{aligned} \end{aligned}$$

(47)