Continuous Estimation in WLAN Positioning By Tilen Ma Clarence Fung.
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Transcript of Continuous Estimation in WLAN Positioning By Tilen Ma Clarence Fung.
Continuous Estimation in Continuous Estimation in WLAN PositioningWLAN Positioning
ByBy
Tilen MaTilen Ma
Clarence FungClarence Fung
ObjectiveObjective
Area-Based Probability (ABP)Area-Based Probability (ABP)
Continuous Space Estimation(CSE)Continuous Space Estimation(CSE)– Center of MassCenter of Mass– Time-AveragingTime-Averaging
Point MappingPoint Mapping
ConclusionConclusion
Applying Area-based ApproachApplying Area-based Approach
Area-Based Probability (ABP)Area-Based Probability (ABP)
Advantages:Advantages:
Presents the user an understanding of the Presents the user an understanding of the system in a more natural and intuitive system in a more natural and intuitive mannermanner
High accuracyHigh accuracy
More mathematical approachMore mathematical approach
Steps in using ABPSteps in using ABP
Decide the Areas
Measure Signals at Different Areas
Create a Training Set
Measure Signals at Current Position
Create a Testing Set
Find out the Probability of Being at Different Areas
Calculate Probability Density
Return the Area with Highest Probability
Area Based ProbabilityArea Based Probability
We compute P(SWe compute P(St t |A|Aii) for every area A) for every area Aii ,i=1…m, using the ,i=1…m, using the Gaussian assumptionGaussian assumption
Finding Probability DensityFinding Probability Density– the object must be at one of the 12 areas the object must be at one of the 12 areas – ΣP(Ai | St) =1 for all i ΣP(Ai | St) =1 for all i
Max{P(AMax{P(Ai i |S|Stt) } = Max{c*P(S) } = Max{c*P(St t |A|Aii) } ) } = Max{P(S= Max{P(St t |A|Aii) }) }
Return the area AReturn the area Aii with top probability with top probability
Applying Area-based ApproachApplying Area-based Approach
There are two approach to estimate There are two approach to estimate position:position:– Discrete Space EstimationDiscrete Space Estimation– Continuous Space EstimationContinuous Space Estimation
Discrete Space EstimationDiscrete Space Estimation
LimitationLimitation
Only one of the discrete locations in the traOnly one of the discrete locations in the training set is returnedining set is returned
Cannot return the intermediate locationsCannot return the intermediate locations
Low accuracyLow accuracy
Not desirable for location-based applicatioNot desirable for location-based application. Eg. Tour guiden. Eg. Tour guide
Introduction to CSEIntroduction to CSE
Continuous Space EstimationContinuous Space EstimationAdvantage:Advantage:– Return locations may or may not be in the Return locations may or may not be in the
training settraining set– Higher accuracyHigher accuracy– Suitable for mobile applicationSuitable for mobile application
Two techniques:Two techniques:– Center of MassCenter of Mass– Time-AveragingTime-Averaging
Center of MassCenter of Mass
Assume n locationsAssume n locations
Treat each location in the training set as Treat each location in the training set as an objectan object
Each object has a weight equals to its Each object has a weight equals to its probability densityprobability density
Obtain Center of Mass of n objects using Obtain Center of Mass of n objects using their weighted positionstheir weighted positions
Center of MassCenter of Mass
Let p(i) be the probability of a location xLet p(i) be the probability of a location x ii, i=1,2 …, i=1,2 …
nn
Let Y be the set of locations in 2D space and Y(i) Let Y be the set of locations in 2D space and Y(i) is the corresponding position of xis the corresponding position of x ii
The Center of Mass is given by:The Center of Mass is given by:
Center of MassCenter of Mass
ExampleExample
Time-AveragingTime-Averaging
Use a time-average window to smooth the Use a time-average window to smooth the resulting location estimatedresulting location estimated
Obtain the result location by averaging the Obtain the result location by averaging the last W locations estimated by discrete-last W locations estimated by discrete-space estimatorspace estimator
Time-AveragingTime-Averaging
Given a stream of location estimates xGiven a stream of location estimates x11,x,x22,…,x,…,xtt
The current location xThe current location xcc is estimated by is estimated by
Time-AveragingTime-Averaging
Problem with CSEProblem with CSE
Locations in training setLocations in training set
Estimated positionEstimated position
Point MappingPoint Mapping
Goal : Map the result to the closest point in Goal : Map the result to the closest point in the set of all possible locationsthe set of all possible locations
Step 1: Divide the corridor into several line Step 1: Divide the corridor into several line segments L segments L ii
Step 2: We define each line segment L Step 2: We define each line segment L ii by by
an equation:an equation:
PP = = PPi1i1 + u + uii ( (PPi2i2 – – PPi1i1) )
– PPi1i1 (x (xi1i1,y,yi1i1) is the starting point of L ) is the starting point of L ii
– PPi2i2 (x (xi2i2,y,yi2i2) is the end point of L ) is the end point of L ii
Let the point Let the point EE(x(xee,y,yee) be the estimated point) be the estimated point
Let ILet Iii be the point of intersection between L be the point of intersection between L ii(P(Pi1i1PPii
22) and the line at the tangent to L) and the line at the tangent to L ii passing throug passing throug
h h EE
Step 3: Finding the distances DStep 3: Finding the distances D ii between the est between the est
imated point E and L imated point E and L i i for all ifor all i
the dot product of the tangent and Lthe dot product of the tangent and L ii is 0, thus is 0, thus
(E - I(E - Iii) dot () dot (PPi2i2 – – PPi1i1) = 0 ) = 0
Solving this we have,Solving this we have,
Substituting uSubstituting uii into the equation of L into the equation of Lii gives gives
the point of intersection Ithe point of intersection Iii as as
x = xx = xi1i1 + u + uii (x (xi2i2 – x – xi1i1))
y = yy = yi1i1 + u + uii (y (yi2i2 – y – yi1i1) )
DDii is equal the distance between I is equal the distance between Iii and E and E
Step 4:check if IStep 4:check if Iii lies in the line segment L lies in the line segment Lii
i.e. ui.e. uii lies between 0 and 1 lies between 0 and 1
Step 5:Step 5:
Return IReturn Iii lying in L lying in Lii and with smallest D and with smallest Dii
ConclusionConclusion
Continuous Space Estimation solves the Continuous Space Estimation solves the limitation in Discrete Space Estimationlimitation in Discrete Space Estimation
Continuous Space Estimation improves Continuous Space Estimation improves the accuracy in determining position the accuracy in determining position
Point Mapping overcome out of bound Point Mapping overcome out of bound problem in CSE problem in CSE
THE ENDTHE END