Presenters: Puneet Gupta Sol Lederer
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Mobile Assisted Localization in Wireless Sensor Networks
N.B. Priyantha, H. Balakrishnan, E.D. Demaine, S. TellerMIT Computer Science
Presenters:
Puneet Gupta
Sol Lederer
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Case for Mobile Assisted Localization
Obstructions, especially in indoor environments
Sparse node deployments Geometric dilution of precision (GDOP)
Hence, finding 4 reference points for each node for localization is difficult
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Overview of scheme
Initially no nodes know their location Mobile node finds cluster of nearby
nodes Explores “visibility region” and
measures distance # of measurements required is linear
in the # of nodes Virtual nodes are discarded
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Theorem 1
A graph is globally rigid if it is formed by starting from a clique of 4 non-coplanar nodes and repeatedly adding a node connected to at least 4 nodes.
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MAL: Distance Measurement
First case: Two nodes, n0 and n1 , single unknown ||n0 - n1||
Adding mobile node, m, introduces 3 unknowns (mx, my, mz), making problem more difficult
Necessary condition: # deg of freedom (unknowns – knowns) ≤ 0.
Solution: Use three mobile locations along the same line in a plane containing n0 and n1
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Case of 2 nodes solved 6 constraints from
measurements of ||ni – mj|| for I = 0,1 and j = 0,1,2
Extra constraint obtained from colinearity of mobile points
unknowns – knowns = 0 Solve system of
polynomial equations
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Case of 3 nodes
Three nodes, n0 n1 n2, three unknowns, ||n0 - n1|| ||n1 - n2|| ||n0 - n2||
Each mobile position gives #unknowns (mx, my, mz) = 3 #constraints (||m – ni||, i = 0,1,2) = 3
Three additional constraints needed
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Case of 3 nodes Solution
Restriction: All mobile positions lie in a common plane k mobile locations k-3 additional co-
planarity constraints Solution: k = 6, geometry of n0, n1, n2
above the plane containing 6 coplanar points m0, m1, m2, m3, m4, m5 no three of which are collinear, determined by the distances ||mi - nj||, i = 0…5 & j = 0...2
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Case of 4 or More
Number of nodes = j ≥ 4 Initially: Number of unknowns = (3j – 5)
3 coordinates per node Minus 3 deg of translational motion Minus 2 deg of rotational motion
Each mobile node adds (j – 3) deg of freedom (j distances – 3 coordinates of mobile position)
j – 3 >= 1
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Case of 4 or more Solution
Require at least (3j – 5)/(j – 3) mobile positions
E.g. for j = 4, required mobile positions to uniquely determine the geometry = 7
But, no 4 of the 11 nodes (4 + 7) may be coplanar
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MAL: Movement Strategy
Initialize: Find 4 nodes that can all be seen from a common
location Move the mobile to 7 nearby locations & measure
distances Compute pair-wise distances
Loop: Pick a localized stationary node (not yet considered by
this loop) Move mobile in perimeter of this node, searching for
positions to hear a non-localized node Localize this node
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AFL: Anchor-free localization
Elect five nodes as shown
Get crude coordinates based on hop count to anchors
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AFL
Use non-linear optimization algorithm to minimize sum-squared energy E
Coordinate assignments satisfy all 1-hop node distances when E = 0
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Graph from running AFL—using RF connectivity information
Graph obtained by MAL
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Performance
Layout of nodes in test scenario
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Estimate error
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Critique
Pros: Innovative stategy
Cons: In a cumbersome terrain (e.g. forest) it
may not be feasible to deploy a roving node.
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The End