Understanding Robot Localization Using Kalman Filters

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Robot localization in a hallway is achieved through Kalman-like filters that use sensor data to estimate the robot's position based on a map of the environment. This process involves incorporating measurements, updating state estimates, and relying on Gaussian assumptions for accuracy. The robot's uncertainty in location evolves as it moves and encounters identifiable landmarks like doors. However, challenges arise when the robot is unsure of the specific door it is sensing, highlighting the limitations of Gaussian assumptions in certain scenarios.


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  1. Histogram filter Non-parametric Filters 1 10/1/2024

  2. Localizing a Robot in a Hallway consider a robot moving down a hall equipped with a sensor that measures the presence of a door beside the robot the pose of the robot is simply its location on a line down the middle of the hall the robot starts out knowing how far down the hallway it is located Kalman-like filters require an initial estimate of the location robot has a map of the hallway showing it where the doors are 2 10/1/2024

  3. Kalman Localization robot starts out knowing how far down the hallway it is located 3 10/1/2024

  4. Kalman Localization as the robot moves forward, its uncertainty in its location shifts and grows according to its motion model 4 10/1/2024

  5. Grid Localization when it reaches a door that can be uniquely identified, it can incorporate this measurement into its state estimate measurement liklihood updated state estimate 5 10/1/2024

  6. Grid Localization as the robot moves forward, its uncertainty in its location shifts and grows according to its motion model 6 10/1/2024

  7. Gaussian Assumption Kalman-like filters assume that quantities can be represented accurately as a mean + covariance e.g., the state is a random variable with Gaussian distribution e.g., measurements are random variables with Gaussian distribution 7 10/1/2024

  8. Gaussian Assumption assumption is ok here EKF UKF 8 10/1/2024

  9. Gaussian Assumption assumption is (possibly) not ok here EKF UKF 9 10/1/2024

  10. Gaussian Assumption assumption is not ok here (robot does not know which door it is measuring) p(robot is sensing a door | state) 10 10/1/2024

  11. Non-parametric Filters non-parametric filters do not rely on a fixed functional form of the state posterior instead, they represent the posterior using a finite number of values each roughly corresponding to a region (or point) in state space two variations partition state space into a finite number of regions e.g., histogram filter represent the posterior using a finite number of samples e.g., particle filter 1. 2. 11 10/1/2024

  12. Histogram table of frequencies bins 12 10/1/2024

  13. Histogram Filter histogram filter uses a histogram to represent probability densities in its simplest form, the domain of the densities is divided into subdomains of equal size with each subdomain being a bin of the histogram the value stored in the bin is proportional to the density 13 10/1/2024

  14. Histogram Filter suppose the domain of the state x is [-5, 5] and that x is a random variable with Gaussian density (mean 0, variance 1) using bins of width w = 0.1 we can represent the density using the following histogram histogram p height of bar Gaussian PDF n = pdf ( / ) x w , i c i center of bin i i = 1 in x 14 10/1/2024

  15. Histogram Filter suppose we want to pass the density through some non-linear function = 3 ( ) 1 f x x reminder: this is the solution obtained by passing 500,000 random samples through f (x), not the result of using a histogram filter = = , 0 1 15 10/1/2024

  16. Histogram Filter create an empty histogram h with bins xc,i for each i yi = f (xc,i) ni = p(xc,i) find the bin bk that yi belongs in h (bk) = h (bk) + ni 1. 2. 1. 2. 3. 4. 16 10/1/2024

  17. A Simple Implementation dx = 0.05; % width of x bins xc = -5:dx:5; % bin centers x y = nthroot(xc 1, 3); % y = f(xc) n = normpdf(xc, 0, 1); % n = p(xc) dy = 0.1; % width of y bins yc = -2:dy:2; % bin centers y h = zeros(size(yc)); % histogram for i = 1:length(y) bk = find(y(i) > yc (dy / 2) & y(i) < yc + (dy / 2)); h(bk) = h(bk) + n(i); end bar(yc, h, 1); 17 10/1/2024

  18. Histogram Filter alternatively create an empty histogram h with bins xc,i for each bin bk in_bin ) ( ) ( , i 1. 2. = ( , ) h b p x y b 1. k c i i k y = ( ) b f x , i c i in_bin ( , ) y probability that yi is in bin bk i k 18 10/1/2024

  19. Histogram Filter = 3 ( ) 1 f x x = = , 0 1 19 10/1/2024

  20. Grid Localization grid localization uses a histogram filter over a grid decomposition of pose space consider a robot moving down a hall equipped with a sensor that measures the presence of a door beside the robot the pose of the robot is simply its location on a line down the middle of the hall the robot starts out having no idea how far down the hallway it is located robot has a map of the hallway showing it where the doors are grid decomposes the hallway into a finite set of non-overlapping intervals e.g., every 50cm would yield intervals [0, 0.5], (0.5, 1], (1, 1.5], 20 10/1/2024

  21. Grid Localization the robot starts out having no idea how far down the hallway it is located the histogram of its state density is uniform 21 10/1/2024

  22. Grid Localization because the robot is beside a door, it has a measurement it can incorporate this measurement into its state estimate measurement liklihood updated state estimate 22 10/1/2024

  23. Grid Localization as the robot moves forward, its uncertainty in its location shifts and grows according to its motion model 23 10/1/2024

  24. Grid Localization when it reaches a door, it can incorporate this measurement into its state estimate it now has a pretty good idea where it is in the hallway measurement liklihood updated state estimate 24 10/1/2024

  25. Grid Localization as the robot moves forward, its uncertainty in its location shifts and grows according to its motion model 25 10/1/2024

  26. Grid Localization Algorithm algorithm_grid_localization( ) for all k do motion_model( ) measurement_model( ) endfor return } { ,t { }, , , p u z m 1. , 1 k t t t 2. i = = p p p x x mean ( ), , mean ( ) u 3. , , 1 k t i t k t i p x , mean ( ), z m 4. , , k t k t t k 5. k p 6. { } p histogram , k t u control input t z measurement t m map x mean( ) center of mass of grid cell xi i 26 10/1/2024

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