ARAIM based on fault detector reuse for reducing computational load

Abstract

Since Advanced Receiver Autonomous Integrity Monitoring (ARAIM) evaluates monitored subsets of visible satellites to validate the integrity of position solutions, the increased number and conservative fault probabilities of satellites have resulted in more fault hypotheses, thus incurring a challenge for the computing power of receivers. Based on the fact that computational cost is caused by the fault detector that computes subset solutions and error statistics, we develop a technique of fault detector reuse in which multiple subsets share one detector to reduce the computational complexity significantly. The effect of the technique on ARAIM performance is explored to derive two factors, the number of detectors and geometry of subset satellites, in designing feasible methods. According to the framework of detector reuse and its performance analysis, this study proposes a specific approach, Detector Reuse with Preserving Geometry (DRPG), based on satellites’ spatial distribution to guarantee subsets’ geometry. Compared with three existing algorithms in aspects of the number of detectors, PLs, and availability, DRPG stably enhances ARAIM performance while significantly decreasing the number of fault detectors, as indicated by simulations. Our work provides another feasible perspective for lowering the computational through fault detector reuse. The specific approach, DRPG, could be a helpful extension of multi-constellation ARAIM over its superior simplicity and good integrity performance.

Appendix: Foundations of analyzing the impact on ARAIM performance and a numerical example for DRPG

The appendix gives foundations for analyzing how detector reuse affects ARAIM performance. Section A simplifies the detection threshold to further evaluate the EMT, Section B explores the influence of difficult point $\sigma_{q}^{(k)}$ on the PL, and Section C provides a numerical example of DRPG for replication.

• A. Simplified expression of detection thresholds

It is necessary to specify that ${\mathbf{C}}_{int}$ is used in (9) for conservativeness and simplifying the calculation.

$$\begin{gathered} \sigma_{ss,q}^{(k)2} = (({\mathbf{S}}^{(k)} - {\mathbf{S}}^{(0)} ){\mathbf{C}}_{int} ({\mathbf{S}}^{(k)} - {\mathbf{S}}^{(0)} )^{T} )_{q} \hfill \\ = ({\mathbf{S}}^{(k)} {\mathbf{W}}^{{(0)}{^{ - 1} }} {\mathbf{S}}^{{(k)}{^{T} }} + {\mathbf{S}}^{(0)} {\mathbf{W}}^{{(0)}{^{ - 1} }} {\mathbf{S}}^{{(0)}{^{T} }} - {\mathbf{S}}^{(k)} {\mathbf{W}}^{{(0)}{^{ - 1} }} {\mathbf{S}}^{{(0)}{^{T} }} - {\mathbf{S}}^{(0)} {\mathbf{W}}^{{(0)}{^{ - 1} }} {\mathbf{S}}^{{(k)}{^{T} }} )_{q} \hfill \\ \end{gathered}$$

(A-1)

In the baseline algorithm, we have:

$${\mathbf{W}}^{(0)} = {\mathbf{C}}_{int}^{ - 1} ;{\mathbf{W}}^{(k)} = {\mathbf{W}}^{(0)} {\text{diag}} ({\mathbf{F}}_{k} )$$

(A-2)

where diag (${\mathbf{F}}_{k}$) returns a square diagonal matrix with the elements of vector ${\mathbf{F}}_{k}$ on the main diagonal. ${\mathbf{F}}_{k}$ only consists of 0 or 1, so:

$$\begin{gathered} {\mathbf{S}}^{(k)} {\mathbf{W}}^{{(0)}{^{ - 1} }} {\mathbf{S}}^{{(k)}{^{T} }} = ({\mathbf{G}}^{T} {\mathbf{W}}^{(k)} {\mathbf{G}})^{ - 1} {\mathbf{G}}^{T} {\mathbf{W}}^{(k)} {\mathbf{W}}^{{(0)}{^{ - 1} }} {\mathbf{W}}^{{(k)^{T} }} {\mathbf{G}}(({\mathbf{G}}^{T} {\mathbf{W}}^{(k)} {\mathbf{G}})^{ - 1} )^{T} \hfill \\ = ({\mathbf{G}}^{T} {\mathbf{W}}^{(k)} {\mathbf{G}})^{ - 1} {\mathbf{G}}^{T} {\mathbf{W}}^{(k)} {\mathbf{G}}(({\mathbf{G}}^{T} {\mathbf{W}}^{(k)} {\mathbf{G}})^{ - 1} )^{T} \hfill \\ = (({\mathbf{G}}^{T} {\mathbf{W}}^{(k)} {\mathbf{G}})^{ - 1} )^{T} \hfill \\ = ({\mathbf{G}}^{T} {\mathbf{W}}^{(k)} {\mathbf{G}})^{ - 1} \hfill \\ = {\mathbf{H}}^{(k)} \hfill \\ \end{gathered}$$

(A-3)

Similarly:

$${\mathbf{S}}^{(0)} {\mathbf{W}}^{{(0)}{^{ - 1} }} {\mathbf{S}}^{{(0)}{^{T} }} = {\mathbf{S}}^{(k)} {\mathbf{W}}^{{(0)}{^{ - 1} }} {\mathbf{S}}^{{(0)}{^{T} }} { = }{\mathbf{S}}^{(0)} {\mathbf{W}}^{{(0)}{^{ - 1} }} {\mathbf{S}}^{{(k)}{^{T} }} { = }({\mathbf{G}}^{T} {\mathbf{W}}^{(0)} {\mathbf{G}})^{ - 1} = {\mathbf{H}}^{(0)}$$

(A-4)

Substitute (A-3) and (A-4) into (A-1) to obtain (A-5):

$$\sigma_{ss,q}^{(k)2} = {\mathbf{H}}_{q}^{(k)} - {\mathbf{H}}_{q}^{(0)} = \sigma_{q}^{(k)2} - \sigma_{q}^{(0)2}$$

(A-5)

• 2 Impact of $\sigma_{q}^{(k)}$ on PL performance

(B-1) is the PL equation:

$$2Q\left( {\frac{{PL_{q} - b_{q}^{(0)} }}{{\sigma_{q}^{(0)} }}} \right) + \sum\limits_{k = 1}^{{N_{fault} }} {p_{fault,k} } Q\left( {t_{k} } \right) = PHMI_{q}$$

(B-1)

where

$$t_{k} = \frac{{PL_{q} - T_{q}^{(k)} - b_{q}^{(k)} }}{{\sigma_{q}^{(k)} }} = \frac{{PL_{q} - Q^{ - 1} \left( {\frac{{PFA_{q} }}{{2N_{det} }}} \right)\left( {\sigma_{q}^{(k)2} - \sigma_{q}^{(0)2} } \right)^{\frac{1}{2}} - b_{q}^{(k)} }}{{\sigma_{q}^{(k)} }};t_{k} > 0$$

(B-2)

and

$$T_{q}^{(k)} = Q^{ - 1} \left( {\frac{{PFA_{q} }}{{2N_{det} }}} \right)\left( {\sigma_{q}^{(k)2} - \sigma_{q}^{(0)2} } \right)^{\frac{1}{2}}$$

(B-3)

We first examine the functions $Q$ and $Q^{ - 1}$:

$$\frac{{dQ(t_{k} )}}{{dt_{k} }} = \frac{1}{{\sqrt {2\pi } }}\frac{d}{{dt_{k} }}\left(\int_{{t_{k} }}^{\infty } {e^{{ - \frac{{u^{2} }}{2}}} } \text{d}u\right) = - \frac{1}{{\sqrt {2\pi } }}e^{{ - \frac{{t_{k}^{2} }}{2}}}$$

(B-4)

$$\frac{{dQ^{ - 1} (t_{k} )}}{{dt_{k} }} = 1/\frac{{dQ(t_{k} )}}{{dt_{k} }} = - \sqrt {2\pi } e^{{\frac{{t_{k}^{2} }}{2}}}$$

(B-5)

(B-4) and (B-5) are always negative. As mentioned previously, an increase of $t_{k}$ lowers PLs, because $PL_{q}$ shall fall to hold (B-1) again. Then, we prove that $t_{k}$ rises with reducing $\sigma_{q}^{(k)}$.

$$\frac{{dt_{k} }}{{d\sigma_{q}^{(k)} }} = = - \frac{{PL_{q} - T_{q}^{(k)} - b_{q}^{(k)} }}{{\sigma_{q}^{(k)2} }} - \frac{1}{{\sigma_{q}^{(k)} }}\frac{{dT_{q}^{(k)} }}{{d\sigma_{q}^{(k)} }}$$

(B-6)

where

$$\frac{{dT_{q}^{(k)} }}{{d\sigma_{q}^{(k)} }} = \sigma_{q}^{(k)} \left( {\sigma_{q}^{(k)2} - \sigma_{q}^{(0)2} } \right)^{{ - \frac{1}{2}}} Q^{ - 1} \left( {\frac{{PFA_{q} }}{{2N_{det} }}} \right)$$

(B-7)

Substitute (B-7) into (B-6) to obtain (B-8):

$$\frac{{dt_{k} }}{{d\sigma_{q}^{(k)} }}{ = } - \frac{{t_{k} }}{{\sigma_{q}^{(k)} }} - \frac{1}{{\sqrt {(\sigma_{q}^{(k)2} - \sigma_{q}^{(0)2} )} }}Q^{ - 1} \left( {\frac{{PFA_{q} }}{{2N_{det} }}} \right)$$

(B-8)

(B-8) is always negative because it is easily known that $t_{k} > 0$ and $\frac{{PFA_{q} }}{{2N_{det} }} < \frac{1}{2}$.

• 3 Numerical example for the DRPG flow

We consider two constellations GPS and Galileo in Tsinghua University at 12:00. To simplify the representation, the 8 GPS visible satellites and the 9 Galileo satellites are noted as follows:

• GPS = {1,2,3,4,5,6,7,8}; Galileo = {9,10,11,12,13,14,15,16,17}.

Then, as an example with GPS, their elevation and azimuth angles are:

• Elevation = [0.8975, 0.5661, 0.5114, 0.5947, 0.2163, 1.4013, 0.0919,0.3735];

• Azimuth = [0.9819, 2.4253, 5.1940, 4.0427, 1.5882, 4.4486, 6.2018, 5.6676]

  $$N_{SO,det} = 2$$

This means that the GPS requires two single-order fault detectors, i.e., two unit faults. Next, four groups of satellites need to be obtained.

• Group_el = {1,6}.

• Size (Group_az1) = 2; Size (Group_az2) = 2; Size (Group_az3) = 2;

• Group_az1 = {5,2}; Group_az2 = {4,3}; Group_az3 = {8,7}.

Taking out one satellite from each group sequentially, we have:

• Unit fault 1 = {1,5,4,8} = {1,4,5,8}; Unit fault 2 = {6,2,3,7} = {2,3,6,7}.

Similarly, we get three unit faults for Galileo:

• Unit fault 3 = {10,11,12,17}; Unit fault 4 = {9,13,14,15}; Unit fault 5 = {16}.

Unit fault 1 means satellites 1, 4, 5, and 8 share one single-order detector, and all single-order detectors can be known like this. Then, we acquire all detectors corresponding to all the original subsets using iterations of single-order detectors.