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1

ECE 194C Acoustic Target Tracking in Sensor

Networks

www.ece.ucsb.edu/Faculty/Iltis/ece194c

• Methods for acoustic target tracking.

– Near Field

• Signal-strength ratios.

• Cross-correlation with broadcast acoustic signal

• Sum cross-correlations (no prior signal knowledge)

– Far-field

• Maximum-likelihood (single source)

• MUSIC (multiple sources)

• Alternating maximization (multiple sources.)

2

Cell-Based LocalizationRef: D. Li, K. Wong, Y. Hu and A. Sayeed “Detection, classification…” IEEE Sig.

Proc. Mag. 2002

3

Localization of Acoustic Targets

Time-of-arrival (line-of-bearing) obtained via beamforming in direction of

strongest acoustic signature.

Figure, Ref: Wang and Chandrakasan, IEEE Sig. Proc. Mag. July 2002.

4

Woodpecker Localization

Ref: H. Wang et. al. “Acoustic Sensor Networks for Woodpecker Localization,” SPIE

Conference Proc. 2005, also “Platform for collaborative Acoustic Signal Processing.”

5

Near-Field vs. Far Field

Plane-wave

Approximation.

Wavefront

curvature

6

Near-Field Propagation Model

Source

α|||| 1

1xx −

=s

Tsr

α|||| 2

2xx −

=s

Tsr

22 )()(|||| isisis yyxx −+−=− xx

222 ),( x=yx

x

y

111 ),( x=yx

),( sss yx=x

7

Acoustic Tracking using Energy Ratios

Problem: Source power ST is unknown

Solution: Use received energy ratios at different sensors

(Ref: Li, Wong, Hu Sayeed IEEE Sig. Proc. Magazine 2002)

α

α

ααρ

||||

||||

||||/

||||,

is

js

js

T

is

T

j

iji

ss

r

r

xx

xx

xxxx

−=

−−==

22

,

2

,

||||||||

||||||||

isjijs

isjijs

xxxx

xxxx

−=−⇒

−=−

α

αα

ρ

ρ

8

Energy Ratio – Target Localization

For each pair of sensors, candidate target position lies on a circle

[ ]

+−+−

−+−=

−=

−=

=−+−

−+−=−+−

−=−

ρ

ρ

ρ

ρρ

ρ

ρ

ρ

ρ

ρ

ρ

αα

α

α

1

)(

)1(

)()(

)1(

)(,

)1(

)(

)()(

)()()()(

||||||||

2222

2

22

2

22

2

,

22

222

,

22

22

,

2

iijjijij

ij

iijj

c

iijj

c

jicscs

isisjijsjs

isjijs

yxyxyyxxr

yyy

xxx

ryyxx

yyxxyyxx

xxxx

9

Example – 3 Sensors

Form circles 1,2 1,3 2,3

0 100 200 300 400 500 600 700 800 900 10000

100

200

300

400

500

600

700

800

900

1000

x-pos (meters)

y-p

os (

mete

rs)

x2

x1

x3

||x1 - x||2 = ρ1,2

||x2 - x||2

10

Least-Squares Solution

Model for measured ratios with additive noise

ji

is

js

ji n ,,||||

||||+

−=

xx

xxρ

Nonlinear least-squares solution (exhaustive search.)

2

1 1

,||||

||||minarg ∑∑

= +=

−−=

s sN

i

N

ij i

j

jisxx

xx

xx ρ

1=α

11

Example – 4 Sensors/Least-Square

Solution

0 100 200 300 400 500 600 700 800 900 10000

100

200

300

400

500

600

700

800

900

1000

x-pos (meters)

y-p

os (

mete

rs)

12

Coherent Correlation

Example acoustic signature (truck)

13

Autocorrelation

∑−

=

−=1

0

)()()(N

n

knxnxkρ

14

Sensor Network Coherent CorrelationRef: Q. Wang, W. Chen, R. Zheng, K. Lee and L. Sha “Acoustic target tracking

using tiny wireless sensor devices,” ISPN 2003

Use Reference Beacon Synchronization (RBS) to map sensor clocks to global time.

Clusterhead transmits beacon to sensors, sensors reply with their individual clock

readings. Clusterhead computes correction factor.

Clusterhead

Beacon

Beacon

Bea

con

Clock 2

Clock 1

Clo

ck 3

15

Coherent CorrelationAll sensors record circular buffer of sound.

Clusterhead determines when a signal-of-interest is present.

Clusterhead transmits its received waveform to sensors.

Sensors cross-correlate their recorded sound with broadcast packet.

16

Coherent Correlation

[ ]

1

1

0

1

1

0

11

1

0

11

)()(

)()()(

)()()(

vknxdnx

knxnndnx

knxnrk

N

n

N

n

N

n

+−−

=−+−

−=

=

=

=

ρ

)()()(

)()()(

2

1

0

2

1

0

22

nvknxdnx

knxnrk

N

n

N

n

+−−

−=

∑−

=

=

ρ

d1

d2

sfd /11 ≈τDelay estimate in sec.

17

Cross-Correlation Time-of-Arrival

Estimates

∑−

=

−=1

0

)()(max

argˆN

n

i nxnr ττ

τ

18

TriangulationSensors transmit arrival time estimates to clusterhead.

Clusterhead uses range equations to solve for target position and start time.

Range equations – c = speed of sound approx. 331 m/s.

0

2

3

2

33

0

2

2

2

22

0

2

1

2

11

/)()(

/)()(

/)()(

tcyyxx

tcyyxx

tcyyxx

ss

ss

ss

+−+−=

+−+−=

+−+−=

τ

τ

τ

Time t0 of source transmission is unknown, but we have three equations, three

unknowns (x,y,t).

19

Least-Squares Triangulation

( )∑=

−−+−−=

+=

+−+−=

sN

i

isisi

ss

ss

ii

isisi

tcyyxxtyx

tyx

v

tcyyxx

1

2

0

22

0

0

1

0

22

/)()(ˆ,,

minarg)ˆ,ˆ,ˆ(

ˆ

/)()(

τ

ττ

τ

Assume time-of-arrival measurement errors are i.i.d. Gaussian.

Optimal solution for position/transmission time is nonlinear least-squares

Solutions via modified grid search

20

Coherent Tracking Scenario

21

Tracking a Moving Acoustic Source

22

Noncoherent Acoustic Localization

Correlation-based methods requires knowledge of acoustic waveform.

Alternative approach based on cross-correlation between sensors –

independent of waveform structure. Ref. Chen, Hudson and Yao

“Maximum-likelihood source localization and unknown sensor location

estimation…” IEEE Trans. Sig. Proc. 2002.

Consider cross-correlation between two sensors:

)(nxp

)(nxq

qτqppq

q

N

n

p

N

n

qppq

c

nsns

nxnxc

τττ

τττ

ττ

−=

−−−=

−=

∑−

=

=

)(maxarg

)()(

)()()(

1

0

1

0

23

Sum Cross-CorrelationConsider delay-difference – transmission time t0 cancels.

( )cc

tctc

c

nsnsnxnxc

qspsspq

qspsqp

qppqpq

pqq

N

n

p

N

n

qppqpq

/||||/||||)(

/||||/||||

)(maxarg

)()()()()(

00

1

0

1

0

xxxxx

xxxx

−−−==

+−−+−=−

−=

−−−=−= ∑∑−

=

=

τ

ττ

τττ

τττττ

Localize by maximizing sum cross-correlation – Do not need to know s(n)

or start time t0!

∑∑= =

=s sN

p

N

q

spqpqs cJ1 1

))(()( xx τ

24

FFT Implementation of Sum Cross-

Correlation

Nki

qpqq

N

n

Nkni

pp

N

p

N

q

N

n

pqqp

N

p

N

q

spqpqs

pq

s s

s s

ekXnx

enxkX

nxnx

cJ

/2

1

0

/2

1 1

1

0

1 1

)()(

)()(

)()(

))(()(

τπ

π

τ

τ

τ

=

= =

=

= =

⇔−

=

−=

=

∑∑∑

∑∑ xx

Recall time-shift

corresponds to linear

phase shift in FFT

)(τpqc

pqττ =

25

FFT Implementation (Cont’d.)

( )

∑∑∑

∑∑∑ ∑

∑∑∑

=

=

−−

= =

= =

=

=

= =

=

=

=

=

−=

1

0

2

)|,(|

1

*/21

0

),(

1

/2

1 1

1

0

1

0

/2/2

1 1

1

0

|),(|1

)()(1

)(1

)(

)()()(

2

N

k

s

kB

N

p

p

NkiN

k

kB

N

q

q

Nki

N

p

N

q

N

n

N

k

NkniNki

qp

N

p

N

q

N

n

pqqps

kBN

kXekXeN

eekXN

nx

nxnxJ

s

s

p

s

s

q

s s

pq

s s

x

x

xx

444 3444 2144 344 21

τπτπ

πτπ

τ

26

Beamformer Interpretation

)(nxp

n

21

0

2

222

1

)(2

2

1

)(22

2

|)(||)(|)(

|)(||)(|

|)(||),(|

)()(

kSckSNJ

kSNekSe

kXekB

ekSkX

N

k

ss

s

kiN

p

ki

p

N

p

ki

s

ki

p

p

s

sp

s

sp

p

==

==

=

=

=

=

=

x

x

x

x

τπτπ

τπ

τπ

Nki

p

pp

pekSkX

nsnx

/2)()(

)()(

τπ

τ

−=

−=

27

Example of FFT-Based Noncoherent

Localization

28

Beamformer Magnitude

450500

550600

650700

200

300

400

5000

2

4

6

8

x 108

x-pos (meters)y-pos (meters)

J(x

s)

xs = (588,396)

)()/()()()(

|)(|)(

22

1

0

2

1

)(2

samplescfyyxx

kXeJ

spspssp

N

k

p

N

p

ki

s

s

sp

−+−=

=∑ ∑−

= =

x

xx

τ

τπ

29

Far-Field Direction-of-ArrivalIn the Wang and Chandrakasan scenario (IEEE Sig. Proc. Mag. 2002) isolated clusters of

sensors operate in the far field.

Use lines-of-bearing (direction-of-arrival) to triangulate)

Simplifies communication – just transmit LOBs to a central processor for triangulation.

30

Far-Field Uniform Linear Array Model

cd /sinθτ =

cd /sin2 θτ =

−−

=

)/sin)1((

)/sin2(

)/sin(

)(

)(

cdNns

cdns

cdns

ns

n

θ

θ

θ

M

x

)/sin)1(()( cdpnsnxp θ−−=

d

d

θ

31

ULA Signals – LOB = 0

32

ULA Signals – LOB = 45 deg

cd /)4/sin(π

cd /)4/sin(3 π

33

Cross-Correlation for LOB Estimation

)(max

argˆ

))(()(

/sin)()(max

arg

)/sin)1(()/sin)1((

)()()(

1 1

1

0

1

0

θθ

θ

θτθ

θττ

τθθ

ττ

J

cJ

cdqpc

cdqnscdpns

nxnxc

s sN

p

N

q

pqpq

pqpq

pq

pq

N

n

N

n

pqqppqpq

=

=

−=

−−−−−=

−=

∑∑

= =

=

=

34

FFT-Based BeamformingUse analysis for noncoherent near-field case

∑∑∑−

= =

=

==

−−==

1

0

2

1

)(21

1

2

)(2

|)(||),(|)(

)/)sin()1(2exp()()()(

N

k

p

N

p

kiN

k

ki

p

kXekBJ

cdpkikSekSkX

s

p

p

θτπ

θτπ

θθ

θπ

Clusterhead computes line-of-bearing and FFT (Ref. Wang and Chandrakasan)

35

ULA 2 Sensor Scenario

36

Beamformer Outputs

)()/()sin()1()(

|)(||),(|)(

,

1

0

2

,

1

)(21

1

2 ,

samplescfdp

kXekBJ

spi

N

k

pi

N

p

kiN

k

i

s

pi

θθτ

θθθτπ

−=

== ∑∑∑−

= =

=

37

Far-Field Multiple Source Localization

(RF)

cd /sinθτ =

cd /sin2 θτ =

)(

))/sin)1((2exp(

))/sin(2exp(

)2exp(

)( t

cdMtfiA

cdtfiA

tfiA

t

c

c

c

vz +

−−

−=

θπ

θπ

π

Response of an antenna array to single RF source, freq. fc.

38

Maximum-Likelihood Solution for RF

Localization

)()()(

))sin)1(exp(

))sinexp()( kk

MiA

iA

A

k vavz +=+

= θ

θπ

θπ

M

Discrete-time model after downconversion

2||)()(||min

argˆ θθ

θ az −= kML

Gaussian noise – ML solution (exhaustive search.)

39

Multiple Source Localization

∑∑==

+=+

=N

n

nn

N

n

nn

nn

n

kk

MiA

iA

A

k11

)()()(

))sin)1(exp(

))sinexp()( vavz θ

θπ

θπ

M

Array response to N sources

2

121

||)()(||,...,

minargˆ ∑

=

−=N

n

n

N

ML k θθθθ

θ az

“Curse of dimensionality” Complexity is order QN for Q quantization

of angle.

40

ML Solution – 2 Sources

01

23

4

0

1

2

3

4-3

-2.5

-2

-1.5

-1

-0.5

0

x 104

θθ

log

p(z

k | θ

)

8/3,4/ ππθ =

41

Solutions for Multiple Source

Localization

Maximum-likelihood

Exponential complexity in number of sources QN

Alternating Maximization

Requires evaluation of projection matrices, multiple

iterations.

MUSIC (Multiple Source Identification and Classification)

Poor performance at low SNR.

Requires precise array calibration.

42

MUSIC Algorithm

IaazzR2

11

)()()()(1ˆ

v

N

n

H

nnn

k

l

H

z Pllk

σθθ +≈= ∑∑==

][][

.....2

121

H

N

H

sNsz

vMNN

UUUUR Λ=

==>>> + σλλλλλ

)()(

1)(

θθθ

aUUaH

NN

HJ =

NnJn ,...2,1),(maxarg == θθ θ

Compute sample correlation matrix

Perform the SVD to obtain

Compute the MUSIC spectrum

MUSIC direction-of-arrival estimates are

43

MUSIC Algorithm Solution

0 0.5 1 1.5 2 2.5 3 3.50

100

200

300

400

500

600

θ

Mu

sic

Sp

ectr

um

Angles =

σ2 = .1

σ2 = 1

σ2 = 10.

σ2 = 50.

θ = π/4, 3π/8

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