Sonar Equation: The Wave Equation - University of Washington

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Sonar Equation: The Wave Equation The Great Wave Hokusai LO: Recognize physical principles associated with terms in sonar equation.

Transcript of Sonar Equation: The Wave Equation - University of Washington

Sonar Equation: The Wave Equation

The Great Wave Hokusai

LO: Recognize physical principles associated with terms in sonar equation.

…the Punchline If density too high to resolve individual organisms, then:

E[energy from volume] = n * energy from an individual

energy from individual = σbs= backscattering cross section

energy from volume = sv = volume backscatter coefficient

E[sv] = n σbs So… - measure energy from volume - assume, measure, or model energy from representative individual - can calculate density of individuals

Definitions

Wave Number: k = 2π/λ (units rad/m)

Angular Frequency: ω = 2π/Τ=2πf (units rad/s)

Pressure (p): force/area (units Pascals = N/m2, 1µPa = 10-6 Pa)

p(r) = p(ro) ro/r

Intensity (I): power/area (units W/m2)

I(r) = I(ro) (ro/r)2 I ∝ p2

Hugen’s Principle

Every point in a wave field acts as a point source

Interference among sources (e.g. ripples in a pond)

+ = + =

Constructive Destructive

Propagating Waves

1 10 x Amp

D1(x) D2(x)

D2(x) = D1(x-10) = D1(x - 2 ms-1* 5 s)

D(x) = D1(x-ct)

Since c = λ/T

D(x) = D1(x-λ/T* t) = D1(x-λft)

speed 2 ms-1

Make non-dimensional: divide by λ

D(x/λ) = D1(x/λ – ft)

Non-dimensional Wave Equation

Make non-dimensional: divide by λ

D(x/λ) = D1(x/λ – ft)

Recall 2π/λ = k; 2πf = ω

D(k,x) = D1(kx - ωt) +x direction

D(k,x) = D1(kx + ωt) -x direction

Harmonic Motion

x m

k Force = -kx F = ma Therefore: ma = -kx

kxdt

xdm −=2

2

2

2

0dt

xdmkx −−=

Divide by –m:

mkx

dtxd

+= 2

2

0

Harmonic Motion

x made up of 2 parts:

= t

mkAx cos1

= t

mkBx sin2

So:

+

= t

mkBt

mkAx sincos

mk

=ωsince

( ) ( )tBtAx ωω sincos +=

( )αω −= tCx cosA,B,C complex constants

22 BAC +=

= −

AB1tanαSinusoidal form with magnitude and phase

k Units k = N/kg = (kg/s2)/kg = s-2 Square root = s-1

= rad/s = ω

Alternate Solution for Harmonic Motion

x made up of 2 parts:

So:

e tiDx ω=1 e tiEx ω−=2

ee titi EDx ωω −+=

Wave Equation for Acoustics

222

22 1

ctpp ∗=∇

δδ For plane waves in 1 direction:

2

22

xδδ

=∇

222

2

2

2 1ct

px

p∗=

δδ

δδ

LaPlacian operator

Plane Wave Incident Solution

Forward Traveling Planar Wave:

p(x,t)= C cos(kx-ωt), where C is reference pressure (po)

( ) ( ) ( )ee tkxitkxi BAtxp ωω +− +=,forward reverse

Alternate form: ( )e tkrioo

rrpp ω−= r = x = range

Planar Incident Pressure

( ))( tkri

t

oortkri

ooinc errperpp t

t

ωω

−−

==

Spherical Scattering

rrr δδ

δδ 2

2

22 +=∇LaPlacian

Wave Equation 2

2

22

2 12tp

crp

rrp

δδ

δδ

δδ

=+ r is distance from source center

Solution ( ) ( )ee tkriootkri

rrp

rAp ωω −− ==

amplitude frequency

Directivity: D

=

= θθ

θθ sin

2sinc

sin

sin2

sin)( kL

Lk

kL

D

sinc=(sin(x)/x)

Transducer Directivity: D

Di = 10log(D) Directivity Index Di

Di = 10 log(4πa/λ) where a = active transducer area

Di = 10log(2.5/sin(β1/2) sin(β2/2))

where β’s = beam width at –3dB points

from transducer

from beam angles

where: Io = radiated intensity at acoustic axis = mean intensity over all directions

D = Io/

Transducer Beam Pattern

* sound pattern is independent of intensity (analogous to flashlight with low batteries)

Shading used to form main lobe through constructive and destructive interference

Don’t Forget Transmission Losses

Spherical Spreading Δp α 1/r

ΔI α 1/r2

Absorption p2 = p1 10- αr/20

α = 1/ro 10 log10(p12/p2

2)

I2 = I1 10- αr/10 α = 1/ro 10 log10(I1/I2)

I ∝ p2

ro = unit distance

Putting Components Together

Pressure p(r) = p(ro) ro/r 10- α(r-ro/20)

Intensity I(r) = I(ro) (ro/r)2 10- α(r-ro/10)

Sonar Equation for Pressure

p(r) = p(ro) ro/r 10- α(r-ro/20)

Divide by reference pressure po

p(r)/po = p(ro) /po ro/r 10- α(r-ro/20)

Take 20 log10 of both sides:

20 log10(p(r)/po) = 20 log10 (p(ro) /po) + 20 log10(ro/r) - α(r-ro)

Sonar Equation for Pressure

20 log10(p(r)/po) = 20 log10 (p(ro) /po) - 20 log10(ro/r) - α(r-ro) SPL

Sound Pressure Level

Echo Level

= SL

Source Level

- TL TL -

Transmission Losses

(one way)

Scattering from an Object

Ii

Incident sound on target

Ir

Intensity of reflected sound at 1 meter

ρ1c1

ρ2c2

=

i

r

IITS 10log10

Point or Spherical Scattering

( ) fr

eerrpp

s

ikrtkri

t

ooscat

st ××= −ω f = complex scattering amplitude

(a.k.a. complex scattering length)

reflectivity, target size, orientation, geometry of transmit & receive

Scattered Field

amplitude frequency

incident field reflected field

Acoustic Impedance Scattering is caused by an Acoustic Impedance mismatch

What is acoustic impedance (Z)? Z = ρ c

Reflection (1 direction) Scattering (all directions)

small objects (e.g. piling) LARGE objects (e.g. breakwater)

pr pi

pi

R = pr/pi

g = ρ2/ρ1 h = c2/c1

density sound speed

g = ρ2/ρ1 h = c2/c1

11

1

1

11

22

11

22

1122

1122

+−

=+

−=

+−

=ghgh

cccc

ccccR

ρρρρ

ρρρρ

density contrast sound speed contrast

Echoes: Acoustic Impedance Mismatch

Reflection

proportion of sound reflected at an interface

11

R,,

,,, +

−=

sbfbsbfb

sbfbsbfbsbfb hg

hg

Incident

R fb,sb

R w,fb T T Rw,fb fb,w fb,sb

fb

sb

Reflection Coefficient

R = reflected T = transmit w = water fb = fish body sb = swim bladder

Reflectivity Examples

Lead in water g = 11.35 h =1.49 R =14/16 ≅ 1

Air – water interface g = 0.0012 h =0.22 R = 0-1/0+1= -1

Artic krill (Euphasia superba) g = 1.0357 h = 1.0279 R = 0.06/2.06 ≅ 0.03

Atlantic cod (Gadus morhua) Body Swimbladder g = 1.039 g = 0.0012 h =1.054 h = 0.232 R = 0.045 R = - 0.99

Perfect reflector Pressure release surface

Backscatter Definition f = complex scattering amplitude (a.k.a. complex scattering length)

- if f is big then target scatters lots of sound - backscatter is most common geometry for a transducer

σbs – backscattering cross-sectional area (units m2)

TS – target strength (units dB re 1 µPa)

2*bsbsbsbs fff =×=σ

( ) ( )bsbsfTS σ102

10 log10log10 ==

* complex conjugate

Some Final Definitions

EL = echo level ( )

= 2

2

10 1log10

Papr

µ

SL = source level ( )

= 22

22

10 )1(1log10

mParp ii

µ

TS = target strength

= 210 )1(

log10m

bsσ

TL = transmission loss

= 2

2target

10 )1(log10

mr

assumes no absorption

Two Way Scattering Equation

( ) bssource

ooscat rrrpp σ×××= 22

target

22 11

EL = SL – TLto target + TLto source + TS Rearrange:

TS = EL – SL + 2 TL (anything missing?)

Sonar Equation Schematic

Transmit

Receive

Active vs Passive Systems SPL = SL - TL

Passive energy received = energy transmitted * fraction not absorbed * fraction not spread

= 10

2

10orr

oor r

rIIα

source transmission loss

Active SPLrec = SL - 2TL + TS

+ noise (Σ thermal, electrical, anthropogenic)

one-way system

two-way system

Active vs Passive Applications Passive

- determine how far away we can hear animals - determine how far away animals can hear us (and corresponding intensity level) - determine range of animal communication - census populations - monitor behavior

Active - determine how much power needed to detect animal - determine range of predator detecting prey - census and size populations - map distributions of animals

Dolphin Sound Reception

Norris 1968

Additional Structures:

Pathway of sound to the cochlea is via the lower jaw (cf. Bullock et al. 1968; McCormick et al. 1970)