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1 Introduction to Sonar 1 1 INTRODUCTION TO SONAR 1 1 Introduction to Sonar Passive Sonar Equation NL DI TL DT range absorption SL SL − TL− (NL− DI)=DT Active Sonar Equation SL RL Iceberg * TL TS NL DI DT NL− DI SL − TL TS − DT 2 + ( RL )= 1 INTRODUCTION TO SONAR 2 Tomography NL SL DT DI USA * TL Hawaii SL − TL− (NL− DI)=DT Parameter definitions: • SL = Source Level • TL = Transmission Loss • NL = Noise Level • DI = Directivity Index • RL = Reverberation Level • TS = Target Strength • DT = Detection Threshold 1 INTRODUCTION TO SONAR 3 Summary of array formulas Source Level I p2 • SL = 10 log I = 10 log 2 (general) ref pref • SL = 171 + 10 log P (omni) • SL = 171 + 10 log P + DI (directional) Directivity Index • DI = 10 log(ID ) (general) IO ID = directional intensity (measured at center of beam) IO = omnidirectional intensity (same source power radiated equally in all directions) 2L • DI = 10 log( λ ) (line array) πD 2 πD • DI = 10 log(( λ ) ) = 20 log( λ ) (disc array) • 4πLxLy DI = 10 log( λ2 ) (rect. array) 3-dB Beamwidth θ3dB 25.3λ • θ3dB = ± L deg. (line array) • ±29.5λ θ3dB = D deg. (disc array) 25.3λ 25.3λ • dB ± ± θ3 = Lx , Ly deg. (rect. array) 1 INTRODUCTION TO SONAR 4 f = 12 kHz D = 0.25 m beamwidth = +−14.65 deg −20 −40 −60 −80 −90 −60 −30 0 30 60 90 theta (degrees) f = 12 kHz D = 0.5 m beamwidth = +−7.325 deg −20 −40 −60 −80 −90 −60 −30 0 30 60 90 theta (degrees) f = 12 kHz D = 1 m beamwidth = +−3.663 deg Source level normalized to on−axis response −20 −40 −60 −80 −90 −60 −30 0 30 60 90 theta (degrees) This figure shows the beam pattern for a circular transducer for D/λ equal to 2, 4, and 8. Note that the beampattern gets narrower as the diameter is increased. 1 INTRODUCTION TO SONAR 5 comparison of 2*J1(x)/x and sinc(x) for f=12kHz and D=0.5m 0 sinc(x) −10 2*J1(x)/x −20 −30 −40 −50 −60 Source level normalized to on−axis response −70 −80 −90 −60 −30 0 30 60 90 theta (degrees) This figure compares the response of a line array and a circular disc transducer. For the line array, the beam pattern is: ⎡ ⎤ 1kL θ 2 ⎣sin( 2 sin ) ⎦ b(θ)= 1 2kL sin θ whereas for the disc array, the beam pattern is ⎡ ⎤ J 1kD θ 2 ⎣2 1(2 sin ) ⎦ b(θ)= 1 2kD sin θ where J1(x) is the Bessel function of the first kind. For the line array, the height of the first side-lobe is 13 dB less than the peak of the main lobe. For the disc, the height of the first side-lobe is 17 dB less than the peak of the main lobe. 1 INTRODUCTION TO SONAR 6 Line Array z L/2 l ¡ ¡ ¢ r dz ¢ α θ A/L x −L/2 Problem geometry Our goal is to compute the acoustic field at the point (r, θ)inthefar field of a uniform line array of intensity A/L. First, let’s find an expression for l in terms of r and θ.From the law of cosines, we can write: l2 = r2 + z2 − 2rz cos α. If we factor out r2 from the left hand side, and substitute sin θ for cos α, we get: 2zz2 l2 = r2 1 − sin θ + rr2 and take the square root of each side we get: 1 2zz2 2 l = r 1 − sin θ + rr2 We can simplify the square root making use of the fact: p(p − 1) p(p − 1)(p − 2) (1 + x)p =1+px + + + ··· 2! 3! p 1 and keeping only the first term for = 2 : 1 ∼ 1 (1 + x) 2 = 1+ x 2 Applying this to the above expression yields: 12zz2 l ∼= r 1+ (− sin θ + ) 2 rr2 1 INTRODUCTION TO SONAR 7 z<<r z2 Finally, making the assuming that , we can drop the term r2 to get l ∼= r − z sin θ Field calculation For an element of length dz at position z, the amplitude at the field position (r, θ)is: A 1 dp = e−i(kl−ωt)dz Ll We obtain the total pressure at the field point (r, θ) due to the line array by integrating: A L/2 1 p = e−i(kl−ωt)dz L −L/2 l but l ∼= r − z sin θ,sowecanwrite: A L/2 1 p = e−i(kr−ωt) eikz sin θdz L −L/2 r − z sin θ r>>z θ 1 Since we are assuming we are in the far field, sin , so we can replace r−z sin θ with 1 r and move it outside the integral: A L/2 p = e−i(kr−ωt) eikz sin θdz rL −L/2 Next, we evaluate the integral: A eikz sin θ L/2 p = e−i(kr−ωt) rL ik θ sin −L/2 ⎡⎤ 1 ikL sin θ − 1 ikL sin θ A e 2 − e 2 p = e−i(kr−ωt) ⎣⎦ rL ik sin θ 1 Next move the term L into the square brackets: ⎡⎤ 1 ikL sin θ − 1 ikL sin θ A e 2 − e 2 p = e−i(kr−ωt) ⎣⎦ rikLsin θ x eix−e−ix and, using the fact that sin( )=2i ,wecanwrite: A sin( 1 kL sin θ) p = e−i(kr−ωt) 2 r 1 kL θ 2 sin which is the pressure at (r, θ) due to the line array. The square of the term in brackets is defined as the beam pattern b(θ) of the array: 2 sin( 1 kL sin θ) b(θ)= 2 1 kL θ 2 sin 1 INTRODUCTION TO SONAR 8 Steered Line Array Recall importance of phase: • kz θ 2π z θ Spatial phase: sin = λ sin • ωt 2π T 1 Temporal phase: = T ; = f z L/2 r θ θ 0 A/L x -L/2 k sin θ = vertical wavenumber k sin θ0 = vertical wavenumber reference To make a steered line array, we apply a linear phase shift −zk sin θ0 to the excitation of the array: A/L dp = eiz(k sin θ−k sin θ0)eiωtdz (1) r We can write sin θ0 zk sin θ0 = ωz c z sin θ0 zk sin θ0 = ωT0(z); T0(z)= c The phase term is equivalent to a time delay T0(z) that varies with position along the line array. We can re-write the phase term as follows. eiz(k sin θ−k sin θ0)eiωt= ee ikz sin θ −iω(t+T0(z)) integrating Equation 1 yields: kL A sin( [sin θ − sin θ0]) p = e−i(kr−ωt) 2 r kL θ − θ ( 2 [sin sin 0]) The resulting beam pattern is a shifted version of the beampattern of the unsteered line array. The center of the main lobe of the response occurs at θ = θ0 instead of θ =0. 1 INTRODUCTION TO SONAR 9 steered and unsteered line array (theta_0 = 20 deg) 0 −10 −20 −30 −40 −50 −60 Source level normalized to on−axis response unsteered beam −70 steered beam −80 −90 −60 −30 0 30 60 90 120 theta (degrees) This plot shows the steered line array beam pattern ⎡ ⎤ kL θ − θ 2 ⎣⎦sin( 2 [sin sin 0]) b(θ)= kL ( 2 [sin θ − sin θ0]) for θ0 =0andθ0 = 20 degrees. 1 INTRODUCTION TO SONAR 10 Example 1: Acoustic Bathymetry 2θ 3dB D = 0.5 m Given: Compute: • f =12kHz • λ = • Baffled disc transducer • DI = • D =0.5m • SL = • Acoustic power P =2.4W • θ3dB = Spatial resolution, Depth resolution, δ d θ T 2d =2 tan 3dB F = c (earliest arrival time) 2r = d · TL = c (latest arrival time) δ =(TL − TF ) · c/2 = d( 1 − 1) cos θ3dB = d · For d =2km = δ = 1 INTRODUCTION TO SONAR 11 Example 2: SeaBeam Swath Bathymetry Transmit: 5 meter unsteered line array (along ship axis) ¡ 2 degrees 90 degrees Receive array: 5 metersteered line array (athwartships) ¢£ ¢ ¦ ¦ ¦ ¦ ¢ ¢ ¢ ¦ ¦ ¦ ¦ ¢ ¢ ¢ returns received only from +− 2 degrees 2 degrees No returns No returns Net beam (plan view) Ship’s Ship’s Track Track One beam, 2 by 2 degrees 100 beams, 2 by 2 degrees (without steering) (with steering) 2 PROPAGATION PART I: SPREADING AND ABSORPTION 12 2 Propagation Part I: spreading and absorption Table of values for absorption coefficent (alpha) 13.00 Fall, 1999 Acoustics: Table of attentuation coefficients frequency [Hz] alpha [dB/km] frequency [Hz] alpha [dB/km] 1 0.003 50000 15.9 10 0.003 60000 19.8 100 0.004 70000 23.2 200 0.007 80000 26.2 300 0.012 90000 28.9 400 0.018 100000 (100 kHz) 31.2 500 0.026 200000 47.4 600 0.033 300000 63.1 700 0.041 400000 83.1 800 0.048 500000 108 900 0.056 600000 139 1000 (1kHz) 0.063 700000 174 2000 0.12 800000 216 3000 0.18 900000 264 4000 0.26 1000000 (1 MHz) 315 5000 0.35 2000000 1140 6000 0.46 3000000 2520 7000 0.59 4000000 4440 8000 0.73 5000000 6920 9000 0.90 6000000 9940 10000 (10 kHz) 1.08 7000000 13520 20000 3.78 8000000 17640 30000 7.55 9000000 22320 40000 11.8 10000000 (10 MHz) 27540 2 PROPAGATION PART I: SPREADING AND ABSORPTION 13 103 10-2 Structural 10 1 = 10 dB) l r MgSO4 α Shear Viscosity 10-1 102 , Absorption in dB/km , α , Range in km (from Boron l r -3 10 104 S = 35 % T = 4oC P = 300 ATM 10-5 106 10-2 1 102 104 f, Frequency, kHz Absorption of Sound in Sea Water Figure by MIT OCW.
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