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Estimating UnderwaterEstimating Underwater Acoustic PropagationAcoustic Propagation

Ethem Mutlu SEthem Mutlu SöözerzerResearch EngineerResearch Engineer

MIT Sea Grant College ProgramMIT Sea Grant College Program

OutlineOutline

�� What is decibel?What is decibel?�� Transducers and hydrophonesTransducers and hydrophones�� Underwater acoustic propagationUnderwater acoustic propagation�� Ray tracingRay tracing�� Delay and signal strength calculationsDelay and signal strength calculations�� Channel impulse responseChannel impulse response

�� Estimating the range of a sourceEstimating the range of a source�� Estimating the direction of a sourceEstimating the direction of a source

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DeciDeci--belbel (dB)(dB)

�� A decibel (dB) is a unit for measuring theA decibel (dB) is a unit for measuring the relative strength of a signal in logarithmic scalerelative strength of a signal in logarithmic scale

�� P(dBP(dB) = 10 log10(P/Pr)) = 10 log10(P/Pr) = 20 log10(V/Vr)= 20 log10(V/Vr)

�� V(dBV(dB) = 10 log10(V/Vr)) = 10 log10(V/Vr) = 10 (log10(V)= 10 (log10(V) –– log10(Vr))log10(Vr))

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Transducer and HydrophoneTransducer and Hydrophone SpecificationsSpecifications

�� Open Circuit Receiving Response (OCRR)Open Circuit Receiving Response (OCRR)�� Transmitting Voltage Response (TVR)Transmitting Voltage Response (TVR)�� Directionality PatternDirectionality Pattern�� PrePre--AmplifierAmplifier

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Transmitting Voltage ResponseTransmitting Voltage Response (TVR)(TVR)

�� output SIL generated per 1 Voutput SIL generated per 1 V of input Voltage at 1m rangeof input Voltage at 1m range as a function of frequencyas a function of frequency

�� dB redB re µµPaPa / V/ V

V = 200 V @@ fcfc=22 kHz=22 kHz

SIL (µµPaPa) = V (V) x TVR (µµPaPa / V)/ V)

SIL = 10 log10(V) + TVR(fc)

ITC 1001 Transducer VTR = 26 + 144 = 170 dB re µµPaPa

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110

2 10

120

130

140

150

18Frequency in kHz

dB re

µPa

/V @

1m

26 34

Transmitting Voltage Response

Figure by MIT OCW.

Open Circuit Receiving ResponseOpen Circuit Receiving Response (OCRR)(OCRR)

�� output voltage (V) generatedoutput voltage (V) generated by the transducer perby the transducer per µµPaPa ofof sound pressure as a functionsound pressure as a function of frequencyof frequency

�� dB re 1V /dB re 1V / µµPaPa

SIL = 190 dB re µµPaPa @@ fcfc=22 kHz=22 kHz

V = SIL (V/µµPaPa) x OCRR (µµPaPa))

VdB = SIL + OCRR(fc)

ITC 1001 Transducer OCRR = 190 + (-190) = 0 dB re 1V

VdB re 1V = 10 log10 ( V / 1 )

V = 10( VdB / 10 ) = 1 V

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102

-220

-210

-200

-190

-180

18 26 34Frequency in kHz

dB re

1V

/µPa

Open Circuit Receiving Response

Figure by MIT OCW.

Directionality PatternDirectionality Pattern

�� ITC 1001 spherical transducerITC 1001 spherical transducer �� ITC 2010ITC 2010 toroidaltoroidal transdtransducerucer

�� Uniform resUniform respponse oonse ovver aller all �� More gain over the sidesMore gain over the sides angles ( 0 to 2angles ( 0 to 2ππ) on both) on both (horizontal plane) than the(horizontal plane) than the horizontal and vertical planehorizontal and vertical plane over the top and bottomover the top and bottom

(vertical plane)(vertical plane)

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Directivity Pattern at 2.0 kHz

240

10dB/div

270

300

120

90

60

210 180 150

330 0 30

Directivity Pattern at 18.0 kHz

240

10dB/div

270

300

120

90

60

210 180 150

330 0 30

Figures by MIT OCW.

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PrePre--AmplifierAmplifier

Amplifies the signals generated at the receiving element Amplifies the signals generated at the receiving element (hydrophone or transducer)(hydrophone or transducer)Gain defined in dBGain defined in dB

mV

mV

amplifier

V

pre-amplifier

Vin = 1 mV, Gain = 20 dB

Vout(dB) = Vin(dB) + Gain(dB)

Vin(dB) = 10 log10(Vin) = 10 log(10e-3) = -30 dB

Vout(dB) = -30 + 20 = 0 dBVout = 10(-10/10) = 0.1 Volt

Shallow Water PropagationShallow Water Propagation

�� Assumptions:Assumptions:�� CoConstant sound speed (c = 1500nstant sound speed (c = 1500 m/sm/s))�� Surface and bottom are smoothSurface and bottom are smooth

r=100m

d=20m

h=80m

source destination

θ2 θ2

θ1 θ1

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Length of propagation pathsLength of propagation paths Direct path => d0 = 100mDirect path => d0 = 100mSurface reflection => d1 = 2d/cos(Surface reflection => d1 = 2d/cos(θθ1) = 107.7 m1) = 107.7 m

θθ1 = atan(r/2d)1 = atan(r/2d)Bottom reflection => d2 = 2h/cos(q2) = 188.7 mBottom reflection => d2 = 2h/cos(q2) = 188.7 m

θθ2 = atan(r/2h)2 = atan(r/2h)SBS reflection => d3 = 2(2d/cos(SBS reflection => d3 = 2(2d/cos(θθ3)+ h/cos(3)+ h/cos(θθ3)) = 260 m3)) = 260 mBSB reflection => d4 = (2d/cos(BSB reflection => d4 = (2d/cos(θθ4)+ 2(h/cos(4)+ 2(h/cos(θθ4))) = 399.5 m4))) = 399.5 m

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r=100m

d=20m

h=80m

source destination

θ2 θ2

θ1 θ1

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Time of arrival to the receiverTime of arrival to the receiver

�� ττii == ddii/c/c�� ττ0 = 66.70 = 66.7 msecmsec�� ττ1 = 71.81 = 71.8 msecmsec�� ττ2 = 125.82 = 125.8 msecmsec�� ττ3 = 173.33 = 173.3 msecmsec�� ττ4 = 266.34 = 266.3 msecmsec

Transmission lossTransmission loss �� fcfc=22kHz=22kHz�� T = 15T = 15 ˚̊C (fmC (fm==100100 kcycleskcycles/sec, A=6e/sec, A=6e--4, B=2.4e4, B=2.4e--7)7)�� a = (A fm f2)/(f2+fm2)+Bf2 dB/ma = (A fm f2)/(f2+fm2)+Bf2 dB/m�� Surface reflection loss (Surface reflection loss (RLsRLs) = 1 dB) = 1 dB�� Bottom reflection loss (Bottom reflection loss (RLbRLb) = 3 dB) = 3 dB

TL = TLs + TLa + RLs + RLb TLi = 20log(di) + adi + RLs + RLb

TL0 = 40.3 dB TL1 = 42.0 dB TL2 = 49.1 dB TL3 = 54.0 dB TL4 = 60.2 dB

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Impulse ResponseImpulse Response

�� output of a system to anoutput of a system to an excitation with a unit impulseexcitation with a unit impulse

⎩ ⎨ ⎧ =

=)( otherwise n

n 0

11δ

r=100m

d=20m

h=80m

source destination

θ1 θ1

θ2 θ2

Transmitted and Received SILTransmitted and Received SIL

�� ITC1001,ITC1001, fcfc=22kHz, V=100 V=22kHz, V=100 V�� SILsSILs = 144 + 10 log(100) = 170 dB re= 144 + 10 log(100) = 170 dB re mPamPa at 1mat 1m

�� SILriSILri == SILsSILs –– TLiTLi�� SILr0 = 129.7 dBSILr0 = 129.7 dB�� SILr1 = 128.0 dBSILr1 = 128.0 dB�� SILr2 = 120.9 dBSILr2 = 120.9 dB�� SILr3 = 116.0 dBSILr3 = 116.0 dB�� SILr4 = 109.8 dBSILr4 = 109.8 dB

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Received VoltageReceived Voltage

�� OCRR =OCRR = --162 dB re V /162 dB re V / µµPAPA�� PrePre--amplifier gain, G = 40 dBamplifier gain, G = 40 dB�� VdBVdB == SILriSILri --162 + 40162 + 40�� V = 10(VdB/10)V = 10(VdB/10)

�� V0 = 5.9 VV0 = 5.9 V�� V1 = 4.0 VV1 = 4.0 V�� V2 = 0.8 VV2 = 0.8 V�� V3 = 0.2 VV3 = 0.2 V�� V4 = 0.1 VV4 = 0.1 V

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Results for 1000mResults for 1000m

Determining the Range of a SourceDetermining the Range of a Source

�� Tracker sends a pulse,Tracker sends a pulse, p(tp(t) = A sin(2) = A sin(2ππffc1c1t), 0<t<Tst), 0<t<Ts�� Target replies, p1(t) = A sin(2Target replies, p1(t) = A sin(2ππffc2c2(t(t--ττpp--ττtt))))�� Tracker receives, p2(t) = A sin(2Tracker receives, p2(t) = A sin(2ππffc2c2(t(t--ττpp--ττtt--ττpp))))�� How can we measureHow can we measure ττpp++ττtt++ττ ??pp

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CorrelationCorrelation

�� Definition:Definition:

�� Autocorrelation:Autocorrelation:

( ) ∫ ∞

∞−

−= λ)dta(t)b(tλR b)(a,

( ) ∫ ∞

∞−

−= λ)dta(t)a(tλR a)(a,

autocorrelation

Estimating the RangeEstimating the Range

�� CorrelateCorrelate p(tp(t) with) with p(tp(t--ττpp--ττtt--ττpp))�� Find the peak of the correlation,Find the peak of the correlation, λλ�� λλ = 2= 2ττpp--ττtt

�� ττpp is the propagation delayis the propagation delay�� Range is, d=cRange is, d=c ττpp = 1500= 1500 ττpp

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Correlation ResultsCorrelation Results

100m, delay estimate is 66.7 msec 100m, delay estimate is 666.7 msec

Determining the Direction of theDetermining the Direction of the TargetTarget

θθ r1-r4

d

Quadrant 1

Quadrant 2Quadrant 3

Quadrant 4

Η1

Η2Η3

Η4

�� Four hydrophonesFour hydrophones�� Measure delay atMeasure delay at

each hydrophoneeach hydrophone�� Compare delay pairsCompare delay pairs

((ττ11,, ττ22), (), (ττ22,, ττ33),), ((ττ33,, ττ44), (), (ττ44,, ττ11) to find) to find which quadrantwhich quadrant

�� Estimate the angleEstimate the angleθ = sign(r1-r4)acos( |r1-r4| / d)

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