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The single example of measurement filtering is the case where a position sensor provides a sequence of measurements X = Xa + d; +

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The single example of measurement filtering is the case where a position sensor provides a sequence of measurements X = Xa + d; + {{i (3.1) where Edi is the actual position of the stationary vehicle is a deterministic error in the ith measurement Ei is a random error in the ith measurement The deterministic error can be caused, for example, by a known sensitivity to temperature and determined a priori by calibration over the temperature range of operation. The random error is usually due to unknown causes or causes not worth the trouble to investigate. The mean value of the random error & is zero; otherwise, it would be compensated. For ease of analysis, we assume that these random errors have a fixed standard deviation over time and are independent from one measurement to the next. The error in each position measurement x; is seen to be the random error wwwww = X d Xa (3.2) The best estimate of position in this case is the average of the measurements N N N [X; - Ed] i=1 where the standard deviation of the error in the estimated position SN - XN - Xa (3.3) (3.4) 08N N (3.5) Hence the sequence of measurements can be processed to obtain an ever- improving estimate of the vehicle's position. To avoid storing all past mea- surements, the Nth estimate can be written in the recursive form XN = N 1 N - XN XN - 1 N (3.6) where only the (N 1)th estimate and the Nth measurement are required. 3.9.2 Multiple Sensors, Stationary Vehicle A more complex algorithm is employed if the position of a stationary vehicle is measured using sequences from multiple position sensors. In this case, assume that the optimal estimate from the Kth sensor using N measurements is denoted NK and the standard deviation of its zero mean error is SNK: For the case of two position sensors, the optimal minimum error estimate of position of the vehicle after the Nth measurement from each of the two independent sensors is - XN = \ N + (1 1) N2 (3.7)

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