Introduction to Sonor

1 INTRODUCTION TO SONAR 1 1 Introduction to Sonar Passive Sonar Equation NL DI TL DT range absorption SL SL − T L − (N L − DI) = DT Active Sonar Equation RL SL Iceberg * TL TS NL DI DT SL − 2T L + T S − ( N L − DI ) = DT RL 1 INTRODUCTION TO SONAR 2 Tomography NL SL DT * DI TL Hawaii SL − T L − (N L − 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 USA 1 INTRODUCTION TO SONAR 3 Summary of array formulas Source Level • SL = 10 log I I = 10 log pp2 ref 2 (general) ref • SL = 171 + 10 log P • SL = 171 + 10 log P + DI (omni) (directional) Directivity Index • DI = 10 log( IID ) O (general) ID = directional intensity (measured at center of beam) IO = omnidirectional intensity (same source power radiated equally in all directions) • DI = 10 log( 2λL ) πD 2 • DI = 10 log(( πD λ ) ) = 20 log( λ ) 4πL L • DI = 10 log( λx2 y ) (line array) (disc array) (rect. array) 3-dB Beamwidth θ3dB • θ3dB = ± 25.3λ L deg. • θ3dB = ± 29.5λ D deg. 25.3λ • θ3dB = ± 25.3λ Lx , ± Ly deg. (line array) (disc array) (rect. array) 1 INTRODUCTION TO SONAR 4 f = 12 kHz D = 0.25 m beamwidth = +−14.65 deg −20 Source level normalized to on−axis response −40 −60 −80 −90 −60 −30 60 90 −30 60 90 60 90 0 30 theta (degrees) f = 12 kHz D = 0.5 m beamwidth = +−7.325 deg −20 −40 −60 −80 −90 −60 0 30 theta (degrees) f = 12 kHz D = 1 m beamwidth = +−3.663 deg −20 −40 −60 −80 −90 −60 −30 0 theta (degrees) 30 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) Source level normalized to on−axis response −10 2*J1(x)/x −20 −30 −40 −50 −60 −70 −80 −90 −60 −30 0 theta (degrees) 30 60 90 This figure compares the response of a line array and a circular disc transducer. For the line array, the beam pattern is: ⎡ ⎤2 sin( 12 kL sin θ) ⎦ ⎣ b(θ) = 1 2 kL sin θ whereas for the disc array, the beam pattern is ⎡ ⎤2 2J1 ( 12 kD sin θ) ⎦ ⎣ b(θ) = 1 2 kD 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
 dz
r  α θ A/L x −L/2 Problem geometry Our goal is to compute the acoustic field at the point (r, θ) in the far 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 + z 2 − 2rz cos α. If we factor out r 2 from the left hand side, and substitute sin θ for cos α, we get:  2 l =r 2 2z z2 1− sin θ + 2 r r  and take the square root of each side we get:  2z z2 l =r 1− sin θ + 2 r r 1 2 We can simplify the square root making use of the fact: (1 + x)p = 1 + px + p(p − 1) p(p − 1)(p − 2) + + ··· 2! 3! and keeping only the first term for p = 12 : 1 1 (1 + x) 2 ∼ =1+ x 2 Applying this to the above expression yields:   1 2z z2 l∼ = r 1 + (− sin θ + 2 ) 2 r r 1 INTRODUCTION TO SONAR 7 Finally, making the assuming that z << r, we can drop the term z2 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 −i(kl−ωt) e dz Ll We obtain the total pressure at the field point (r, θ) due to the line array by integrating: dp = p= A  L/2 1 −i(kl−ωt) dz e L −L/2 l but l ∼ = r − z sin θ, so we can write: A −i(kr−ωt)  L/2 1 p= e eikz sin θ dz L −L/2 r − z sin θ Since we are assuming we are in the far field, r >> z sin θ, so we can replace 1 and move it outside the integral: r p= 1 r−z sin θ with A −i(kr−ωt)  L/2 ikz sin θ e dz e rL −L/2 Next, we evaluate the integral:  A −i(kr−ωt) eikz sin θ p= e rL ik sin θ ⎡ L/2 −L/2 1 1 ⎤ A −i(kr−ωt) ⎣ e 2 ikL sin θ − e− 2 ikL sin θ ⎦ p= e rL ik sin θ Next move the term 1 L into the square brackets: ⎡ 1 ⎤ 1 A e 2 ikL sin θ − e− 2 ikL sin θ ⎦ p = e−i(kr−ωt) ⎣ r ikL sin θ and, using the fact that sin(x) = eix −e−ix , 2i we can write:  A −i(kr−ωt) sin( 12 kL sin θ) p= e 1 r kL sin θ 2  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:  sin( 12 kL sin θ) b(θ) = 1 kL sin θ 2 2 1 INTRODUCTION TO SONAR 8 Steered Line Array Recall importance of phase: • Spatial phase: kz sin θ = 2λπ z sin θ • Temporal phase: ωt = 2Tπ ; T = f1 z L/2 r θ θ0 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: dp = A/L iz(k sin θ−k sin θ0 ) iωt e e dz r We can write zk sin θ0 = ωz (1) sin θ0 c z sin θ0 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. zk sin θ0 = ωT0 (z) ; T0 (z) = eiz(k sin θ−k sin θ0 ) eiωt = eikz sin θ e−iω(t+T0 (z)) integrating Equation 1 yields:  [sin θ − sin θ0 ]) sin( kL A 2 p = e−i(kr−ωt) kL r ( 2 [sin θ − sin θ0 ])  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 Source level normalized to on−axis response −10 −20 −30 −40 −50 −60 unsteered beam −70 steered beam −80 −90 −60 −30 0 30 theta (degrees) 60 This plot shows the steered line array beam pattern ⎡ ⎤2 sin( kL 2 [sin θ − sin θ0 ]) ⎦ ⎣ b(θ) = ( kL 2 [sin θ − sin θ0 ]) for θ0 = 0 and θ0 = 20 degrees. 90 120 1 INTRODUCTION TO SONAR 10 Example 1: Acoustic Bathymetry 2θ3dB D = 0.5 m Given: Compute: • f = 12 kHz • λ= • Baffled disc transducer • DI = • D = 0.5 m • SL = • Acoustic power P = 2.4 W • θ3dB = Spatial resolution,   = 2d tan θ3dB =d · Depth resolution, δ TF = 2d c (earliest arrival time) TL = 2r c (latest arrival time) δ = (TL − TF ) · c/2 = d( cos 1θ3dB − 1) =d· For d = 2 km = 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 meter steered line array (athwartships)                 returns received only from +− 2 degrees 2 degrees No returns No returns Net beam (plan view) Ship’s Track Ship’s Track One beam, 2 by 2 degrees (without steering) 100 beams, 2 by 2 degrees (with steering) 2 PROPAGATION PART I: SPREADING AND ABSORPTION 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] 1 0.003 10 0.003 100 0.004 200 0.007 300 0.012 400 0.018 500 0.026 600 0.033 700 0.041 800 0.048 900 0.056 1000 (1kHz) 0.063 2000 0.12 3000 0.18 4000 0.26 5000 0.35 6000 0.46 7000 0.59 8000 0.73 9000 0.90 10000 (10 kHz) 1.08 20000 3.78 30000 7.55 40000 11.8 frequency [Hz] alpha [dB/km] 50000 15.9 60000 19.8 70000 23.2 80000 26.2 90000 28.9 100000 (100 kHz) 31.2 200000 47.4 300000 63.1 400000 83.1 500000 108 600000 139 700000 174 800000 216 264 900000 315 1000000 (1 MHz) 1140 2000000 2520 3000000 4440 4000000 6920 5000000 9940 6000000 13520 7000000 17640 8000000 22320 9000000 10000000 (10 MHz) 27540 12 2 PROPAGATION PART I: SPREADING AND ABSORPTION 13 10-2 103 10 1 MgSO4 Shear Viscosity 10-1 102 Boron 10-3 S = 35 % T = 4 oC P = 300 ATM 10-5 10-2 rl, Range in km (from α rl = 10 dB) α, Absorption in dB/km Structural 104 106 1 102 f, Frequency, kHz 104 Absorption of Sound in Sea Water Figure by MIT OCW. Absorption of sound in sea water. 2 PROPAGATION PART I: SPREADING AND ABSORPTION 14 Absorption of sound in sea water Relaxation mechanism (conversion of acoustic energy to heat) Four mechanisms • shear viscosity (τ ≈ 10−12 sec) • structural viscosity (τ ≈ 10−12 sec) • magnesium sulfate — MgSO4 (τ ≈ 10−6 sec) [1.35 ppt] • boric acid (τ ≈ 10−4 sec) [4.6 ppm] Relaxation time, τ • if ωτ << 1, then little loss. • if ωτ ≈ 1 or greater, then generating heat (driving the fluid too fast). The attenuation coefficient α depends on temperature, salinity, pressure and pH. The following formula for α in dB/km applies at T=4◦ C, S=35 ppt, pH = 8.0, and depth = 1000 m. (Urick page 108). α ≈ 3.0 × 10 −3 0.1f 2 44f 2 + + + 2.75 × 10−4 f 2 2 2 1+f 4100 + f 2 PROPAGATION PART I: SPREADING AND ABSORPTION 15 Solving for range given transition loss TL The equation 20 log r + 10−3 αr = T L cannot be solved analytically. If absorption and spreading losses are comparable in magnitude, you have three options: • In a ”back-of-the-envelope” sonar design, one can obtain an initial estimate for the range by first ignoring absorption, then plugging in numbers with absorption until you get an answer that is “close enough”. • For a more systematic procedure, one can do Newton-Raphson iteration (either by hand or with a little computer program), using the range without absorption as the initial guess. • Another good strategy (and a good way to check your results) is to make a plot of TL vs. range with a computer program (e.g., Matlab). Newton-Raphson method: (Numerical recipes in C, page 362) xi+1 = xi − f (xi ) f  (xi ) To solve for range given TL, we have: f (xi ) = T L − 20 log xi − 10−3 αxi f  (xi ) = − TL = α= i xi 0 1 2 3 4 20 − 10−3 α r 20 log xi 0.001αxi f (xi ) x0 = f  (xi ) f (xi )/f  (xi ) xi+1 = xi − f (xi ) f
(xi ) 2 PROPAGATION PART I: SPREADING AND ABSORPTION Example 1: whale tracking L = 2 km Passive sonar equation: Given: • f0 = 250 Hz • DT = 15 dB • P = 1 Watt (omni) • NL = 70 dB • line array: L = 2 km Question: How far away can we hear the whale? = TL = λ= DI = SL = TL = 20 log r + α ∗ r ∗ 10−3 = α= Rt = r= 8680 α = (w/o absorption) r= (with absorption) 16 2 PROPAGATION PART I: SPREADING AND ABSORPTION 17 TL vs range for alpha=0.003 dB/km, f=250 Hz 140 120 TL (dB) 100 80 60 40 100 1000 10000 100000 range (m) 1000000 Figure 1: TL vs. range for whale tracking example (f=250 Hz, α = 0.003 dB/km). f (xi ) = T L − 20 log xi − 10−3 αxi = f  (xi ) = − α= i xi 20 − 10−3 α = r TL = 20 log xi 0 500,000 114 1 459,500 13.2 2 445,000 112.96 3 4 x0 = 0.001αxi f (xi ) f  (xi ) f (xi )/f  (xi ) xi+1 = xi − 1.5 1.38 1.34 −3.7 × 10−5 −4.05 × 10−5 40540 14805 -1.5 -0.6 -0.3 459500 444694 f (xi ) f
(xi ) 2 PROPAGATION PART I: SPREADING AND ABSORPTION Example 2: dolphin tracking L = 1 m Passive sonar equation: Given: • f0 = 125 kHz • DT = 15 dB • SL 220 dB re 1 µ Pa at 1 meter • NL = 70 dB • line array: L = 1 m Question: How far away can we hear (detect) the dolphin? = TL = λ= DI = TL = 20 log r + α ∗ r ∗ 10−3 = α= Rt = 8680 α = r= (w/o absorption) r= (w/o spreading) r= f (xi ) = T L − 20 log xi − 10−3 αxi = (with absorption) 18 2 PROPAGATION PART I: SPREADING AND ABSORPTION 19 TL vs range for alpha=30 dB/km, f=125 kHz 400 350 300 TL (dB) 250 200 150 100 50 0 0 1000 2000 3000 4000 5000 6000 range (m) 7000 8000 9000 10000 Figure 2: TL vs. range for dolphin tracking example (f=125 kHz, α = 30 dB/km). f  (xi ) = − α= 20 − 10−3 α = r x0 = TL = i xi 20 log xi 0.001αxi f (xi ) f  (xi ) f (xi )/f  (xi ) xi+1 = xi − 0 1 2 3 4 74 69.6 69.35 150 90.9 88.05 -0.034 -0.037 -0.368 1970 95 10.9 5000 3030 2935 2925 -67 -3.5 -0.4 3030 2935 2925 f (xi ) f
(xi ) 3 PROPAGATION PART II: REFRACTION 3 20 Propagation Part II: refraction In general, the sound speed c is determined by a complex relationship between salinity, temperature, and pressure: c = f (S, T, D) Medwin’s formula is a useful approximation for c in seawater: c = 1449.2 + 4.6T − 0.055T 2 + 0.00029T 3 +(1.34 − 0.010T )(S − 35) + 0.016D where S is the salinity in parts per thousand (ppt), T is the temperature in degrees Celsius, and D is the depth in meters. (See Ogilvie, Appendix A.) Partial derivatives: ∂c = 4.6 m/sec/C◦ ∂T ∂c = 1.34 m/sec/ppt ∂S ∂c = 0.016 m/sec/m ∂D For example: • ∆T = 25◦ =⇒ ∆c = 115 m/sec • ∆S = 5 ppt =⇒ ∆c = 6.5 m/sec • ∆D = 6000 m =⇒ ∆c = 96 m/sec c sound speed D depth z S T c 3 PROPAGATION PART II: REFRACTION 21 Sound across an interface p y r p θ1 k1 = ω/c1 t θ2 k2 = ω/c2 θ1 ρ , c1 1 p x ρ , c2 2 i 1 p1 = pi + pr = Ie−i(k1 x cos θ1 +k1 y sin θ1 ) + Re−i(−k1 x cos θ1 +k1 y sin θ1 ) p2 = pt = T e−i(k2 x cos θ2 +k2 y sin θ2 ) At x = 0, we require that p1 = p2 (continuity of pressure) (I + R)e−ik1 y sin θ1 = T e−ik2 y sin θ2 Match the phase to get Snell’s law: sin θ2 sin θ1 = c1 c2 3 PROPAGATION PART II: REFRACTION 22 If c2 > c1 , then θ2 > θ1 p t y θ2 t x θ1 c2 p i c1 If c2 < c1 , then θ2 < θ1 y p t θ2 x θ1 p i c1 c2 Sound bends towards region with low velocity Sound bends away from region with high velocity 4 REFLECTION AND TARGET STRENGTH 4 23 Reflection and target strength Interface reflection 1 y p p i r θ1 θ1 ρ c 1 1 x ρ c 2 2 θ2 p t pi = Ieiωte−i(k1x sin θ1−k1y cos θ1) pr = Reiωte−i(k1x sin θ1+k1y cos θ1) pt = T eiωte−i(k2x sin θ2−k2y cos θ2) 4 REFLECTION AND TARGET STRENGTH 24 Boundary condition 1: continuity of pressure: @ y=0 p 1 = pi + pr = pt = p2 Boundary condition 2: continuity of normal particle velocity: @ y=0 v 1 = vi + vr = vt = v2 Momentum equation ∂v 1 ∂p =− · ∂t ρ ∂y ∂v = iωv (since v ∝ eiωt) ∂t i ∂p =⇒ v = ωρ ∂y but Continuity of pressure: pi |y=0 + pr |y=0 = pt |y=0 (I + R)e−ik1x sin θ1 = T e−ik2x sin θ2 (I + sin θ −iωx c 1 1 R)e = sin θ −iωx c 2 2 Te And so using Snell’s law, we can write: I +R=T Continuity of normal velocity: vi = i cos θ1 pi · ik1 cos θ1pi = − ωρ1 ρ1 c 1 (2) 4 REFLECTION AND TARGET STRENGTH 25 cos θ1 pr ρ1 c 1 cos θ2 vt = − pt ρ2 c 2 vi |y=0 + vr |y=0 = vt |y=0 vr = ρ2c2 cos(θ1)(I − R) = ρ1c1 cos(θ2)T (3) 4 REFLECTION AND TARGET STRENGTH We can solve Equations 1 and 2 to get the reflection and transmission coeficients R and T : R= R ρ2c2 cos θ1 − ρ1c1 cos θ2 = I ρ2c2 cos θ1 + ρ1c1 cos θ2 T 2ρ2c2 cos θ1 = I ρ2c2 cos θ1 + ρ1c1 cos θ2 Recall that given θ1, we can compute θ2 with Snell’s law: T = sin θ1 sin θ2 = c2 c1 Special case: θ = 0 (normal incidence) R= Z2 − Z 1 ρ2 c 2 − ρ 1 c 1 ≡ Z2 + Z 1 ρ2 c 2 + ρ 1 c 1 T = 2ρ2c2 2Z2 ≡ Z2 + Z 1 ρ2 c 2 + ρ 1 c 1 where Z = ρc is defined as the acoustic impedance. 26 4 REFLECTION AND TARGET STRENGTH 27 Example 1: source in water p y t θ2 Air x Water θ1 p i θ1 p r ρa = 1.2 kg/m3, ca = 340 m/sec =⇒ Za = 408 kg/m2s ρw = 1000 kg/m3, cw = 1500 m/sec =⇒ Zw = 1.5 × 106 kg/m2s 408 cos θ1 − 1.5 × 106 cos θ2 R= ≈ −1 408 cos θ1 + 1.5 × 106 cos θ2 T = 1 + R ≈ 0 (no sound in air) 4 REFLECTION AND TARGET STRENGTH 28 Example 2: source in air y p p i r θ1 θ1 Air x Water θ2 p t 1.5 × 106 cos θ1 − 408 cos θ2 R= ≈ +1 1.5 × 106 cos θ1 + 408 cos θ2 T = 1 + R ≈ 2 (double sound in water!) Does this satisfy your intuition? Consider intensity: p2a p2a = Ia = ρaca 408 p2w 4p2a 4p2a p2a Iw = = = = ρw cw ρw cw 1.5 × 106 3.75 × 105 4 REFLECTION AND TARGET STRENGTH =⇒ Iw ≈ Does this satisfy your intuition? 1 Ia 1000 29 4 REFLECTION AND TARGET STRENGTH 30 Target Strength Assumptions: • large targets (relative to wavelength) • plane wave source – no angular variation in beam at target – curvature of wavefront is zero Example: rigid or soft sphere TS = 10 log Iscat | Iinc r=rref r r 0 p2 Iinc = ρc Pinc = πr02Iinc Assume Pscat = Pinc (omnidirectional scattering) Iscat Pscat πr02Iinc = = 4πr2 4πr2 For r = rref = 1 meter: r02 TS = 10 log 4 If r0 = 2 meters, then TS = 0 dB (Assuming r0 >> λ) 5 DESIGN PROBLEM: TRACKING NEUTRALLY BUOYANT FLOATS 5 31 Design problem: tracking neutrally buoyant floats Sonar design problem 2r0 = 25 cm Require: • Track to ± 3◦ bearing • Range error: δ ± 10 meters • Maximum range: R = 10 km • Active sonar with DT = 15 dB • sonar and float at sound channel axis • baffled line array (source and receiver) • Noise from sea surface waves (design for Sea State 6) 5 DESIGN PROBLEM: TRACKING NEUTRALLY BUOYANT FLOATS receiver DI: DIR = pulse length: τ = array length: L= source level: SL = noise level: NL = transmission loss: TL = wavelength: λ= source DI: DIT = time-of-flight: T = ping interval: Tp = frequency: f = target strength: TS = range resolution: δ= acoustic power: P= average acoustic power: P = 32 6 NOISE 6 33 Noise Figure removed for copyright reasons. Figure 7.5, "Average deep-water ambient-noise spectra." Source: Urick, Robert J. Principles of Underwater Sound. Los Altos, CA: Peninsula Publishing, 1983. 6 NOISE 34 Figure removed for copyright reasons. Graph of ocean noise, by frequency. Source: Urick, Robert J. Ambient Noise in the Sea. Los Altos, CA: Peninsula Publishing, 1986. 6 NOISE 35 Five bands of noise: I. f < 1 Hz: hydrostatic, seismic II. 1 < f < 20 Hz: oceanic turbulence III. 20 < f < 500 Hz: shipping IV. 500 Hz < f < 50 KHz: surface waves V. 50 kHz < f : thermal noise Band I: f < 1 Hz • Tides f ≈ 2 cycles/day p = ρgH ≈ 104 · H Pa noise level: NL = 200dB re 1 µPa − 20 log H example: 1 meter tide =⇒ NL = 200 dB re 1 µPa • microseisms f ≈ 17 Hz On land, displacements are η ≈ 10−6meters Assume harmonic motion η ∝ eiωt =⇒ v = dη = iωη dt Noise power due to microseisms p = ρcv = iωρcη = i2πf ρcη |p| = 2πf ρcη = 1.4P a =⇒ NL = 123 dB re 1 µPa Band II: Oceanic turbulence Possible mechanisms: 1 Hz < f < 20 Hz 6 NOISE 36 • hydrophone self-noise (spurious) • internal waves • upwelling Band III: Shipping 20 Hz < f < 500 Hz Example: ≈ 1100 ships in the North Atlantic assume 25 Watts acoustic power each SL = 171 + 10 log P = 215 dB re 1 µPa at 1 meter Mechanisms • Internal machinery noise (strong) • Propeller cavitation (strong) • Turbulence from wake (weak) 6 NOISE 37 Band IV: surface waves 500 Hz < f < 50 KHz • Observations show NL is a function of local wind speed (sea state) • Possible mechanisms: – breaking waves (only at high sea state) – wind flow noise (turbulence) – cavitation (100-1000 Hz) – long period waves  ω = kg ω cp = k gλ c2p = 2π if λ ≈ 2000 km, then cp ≈ 1500 m/sec =⇒ radiate noise! Band V: Thermal noise NL = −15 + 20 log f 50 KHz < f 6 NOISE 38 Directionality of noise Vertical • Low frequency – distant shipping dominates – low attenuation at horizontal θ 90 0 NL −90 • High frequency – sea surface noise – local wind speed dominates – high attenuation at horizontal θ 90 0 NL −90 Horizontal • Low frequency: highest in direction of shipping centers • High frequency: omnidriectional