3. Range Detection: Signal is Relayed Without Loss.
We have
dab
v
+ ??tb +
dsb
v
??’
dsa
v
= t2 ??’ t1, (1)
dbc
v
+ ??tc +
dsc
v
??’
dsb
v
= t3 ??’ t2, (2)
which gives
dsa = dsb + k1, where k1 = dab ??’ v ?· (t2 ??’ t1 ??’ ??tb), (3)
dsc = dsb + k2, where k2 = ??’dbc + v ?· (t3 ??’ t2 ??’ ??tc). (4)
Equations (3)(4) show that k1, k2 can be obtained by measuring t1, t2, t3 with
S??™s local timer, learning the positions of A,B,C and time di?®erences ??tb,??tc
from the beacon signals. We are going to apply trilateration with k1, k2 to
compute coordinates (x, y) for sensor S in Step 3.
Note that TPSS can still work if some beacon signals get lost during the
transmission from the initiator or any intermediate relay nodes. As long as a
sensor S receives one signal from three di?®erent relay beacons, S??™s location
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can be computed with TPSS. For example (Figure 4), M is a beacon signal
travelling along beacons 1, 2, 3, 4 and 5. The messages relayed by beacons 1
and 4 are lost or destroyed during the transmission. S receives M only from
beacons 2, 3, 5 at time t0, t1, t2, respectively. Let dij(dsj) denote the distance
between node i(s) and j, and ??ti be the time di?®erence information conveyed
by beacon node i. We have:
d23
v
+ ??t3 +
ds3
v
??’
ds2
v
= t1 ??’ t0, (5)
d34
v
+ ??t4 +
d45
v
+ ??t5 +
ds5
v
??’
ds3
v
= t2 ??’ t1. (6)
It follows that,
ds2 = ds3 + k1, where k1 = d23 ??’ v ?· (t1 ??’ t0 ??’ ??t3), (7)
ds5 = ds3 + k2, where k2 = ??’(d34 + d45) + v ?· (t2 ??’ t1 ??’ ??t4 ??’ ??t5).
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