1 Introduction to Sonardspace.mit.edu/bitstream/handle/1721.1/35715/2-011Fall... · 2019-09-12 · 1 Introduction to Sonar History of Sonar “If you cause your ship to stop, and

1 Introduction to Sonar

History of Sonar

“If you cause your ship to stop, and place the head of a long tube in the water and place the outer extremity to your ear you will hear ships at a great distance from you”

— Leonardo da Vinci, 1490.

1. “. . . ship to stop” −⇒ reduce self-noise 2. “. . . tube in water” −⇒ transducer 3. “. . . to your ear” −⇒ receiver 4. “. . . will hear ships” −⇒ detection 5. “. . . at a great distance” −⇒ low attenuation

1687 Issac Newton: first theoretical prediction of sound speed

ρπ 1 2ρp

� c

1847 Culladon and Sturm:

• first accurate measurement of speed of sound in water • light flash/underwater bell

1900 Submarine Signal Company: first commerical application.

• range from ship to lighthouse • simulataneous sounding of underwater bell and foghorn

1914 Fessendon: first active sonar system (detect iceberg 2 miles)

World War I: experiments.

• passive sonar operational equipment • active sonar experiments

1 INTRODUCTION TO SONAR 2

Passive Sonar Equation

NL

DI

TLDT

range absorption SL

SL − TL − (NL − DI) = DT

Active Sonar Equation

SL RL

Iceberg* TL TS

NL DI

DT

SL − 2TL + TS − ( NL − DI

) = DT RL


* SL

TL

NL

USADT

DI

Hawaii

Tomography

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 •


Reverberation vs. Noise-limited Range

Active Sonar •

• Range independent noise vs. range dependent reverberation

• Define Echo Level: EL = SL − 2TL + TS

Level (dB)

Echo Level

Reverberation Level

Noise Level

reverberation Range (time) limited range

Echo Level

Level (dB)

Level Reverberation

Noise Level

noise Range (time) limited range


Definition of sound intensity

dB = decibel (after Alexander Graham Bell)

For acoustics:

I dB = 10 log10 Iref

�

2 ARRAYS 6

2 Arrays

Summary of array formulas

Source Level I 2 • SL = 10 log Iref

= 10 log p2 (general)

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)

L • DI = 10 log(2 � ) (line array)

D D • DI = 10 log((�� )

2) = 20 log(� ) (disc array)

• DI = 10 log(4�LxLy ) (rect. array)

�2

3-dB Beamwidth α3dB

• α3dB = 25.3� deg. (line array) L±29.5�• α3dB = D deg. (disc array) ±25.3� 25.3� • α3dB = ± Lx

, ± Ly deg. (rect. array)

90

2 ARRAYS 7

f = 12 kHz D = 0.25 m beamwidth = +−14.65 deg S

ourc

e le

vel n

orm

aliz

ed to

on−

axis

res

pons

e

−20

−40

−60

−80 −90 −60 −30 0 30 60

theta (degrees) f = 12 kHz D = 0.5 m beamwidth = +−7.325 deg

−20

−40

−60

−80 −90 −60 −30 0 30 60

theta (degrees) f = 12 kHz D = 1 m beamwidth = +−3.663 deg

−20

−40

−60

−80 −90 −60 −30 0 30 60

theta (degrees)

This figure shows the beam pattern for a circular transducer for D/� equal to 2, 4, and 8. Note that the beampattern gets narrower as the diameter is increased.

90

90

�

�

0

2 ARRAYS 8

comparison of 2*J1(x)/x and sinc(x) for f=12kHz and D=0.5m

−10

−20

−30

−40

−50

−60

−70

−80

Sou

rce

leve

l nor

mal

ized

to o

n−ax

is r

espo

nse

sinc(x)

2*J1(x)/x

−90 −60 −30 0 30 60 theta (degrees)

This figure compares the response of a line array and a circular disc transducer. For the line array, the beam pattern is:

sin(1 kL sin α) ⎣2

� ⎤b(α) = 2 1 kL sin α2

whereas for the disc array, the beam pattern is

2J1(1 kD sin α)

⎣2

� ⎤b(α) = 2 1 kD sin α2

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.

90

��

2 ARRAYS 9

Line Array

��

��

� �

�

L/2

dz

A/L

−L/2

r

l

α θ

z

x

Problem geometry

Our goal is to compute the acoustic field at the point (r, λ) 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 λ. From the law of cosines, we can write:

2 2l2 = r + z − 2rz cos �.

If we factor out r2 from the left hand side, and substitute sin λ for cos �, we get: ⎝

2z z2 �

2l2 = r 1 − sin λ + 2r r

and take the square root of each side we get: ⎝ � 1

2z z2 2

l = r 1 − sin λ + 2r r

We can simplify the square root making use of the fact:

p(p − 1)(p − 2)(1 + x)p = 1 + px +

p(p − 1)+ +

2! 3! · · ·

1and keeping only the first term for p = 2 :

1 1 (1 + x) 2 = 1 + x�

2

Applying this to the above expression yields: ⎝

1 2z z2 �

= r 1 + sin λ + )l �2(−

r r2

2 ARRAYS 10

2zFinally, making the assuming that z << r, we can drop the term r

to get 2

= r − z sin λl �

Field calculation

For an element of length dz at position z, the amplitude at the field position (r, λ) is:

A1 dp = e −i(kl−λt)dz

Ll

We obtain the total pressure at the field point (r, λ) due to the line array by integrating: � L/2A 1

p = e −i(kl−λt)dz L −L/2 l

but l �= r − z sin λ, so we can write: � L/2A 1−i(kr−λt)p = e e ikz sin � dz

L −L/2 r − z sin λ

Since we are assuming we are in the far field, r >> z sin λ, so we can replace r−z

1sin � with

1 and move it outside the integral: r

� L/2A −i(kr−λt)p = e e ikz sin �dz rL −L/2

Next, we evaluate the integral:

A ⎝

ikz sin � �L/2

−i(kr−λt) ep = e

rL ik sin λ −L/2

� ⎣ 1 ikL sin �

2A −i(kr−λt) e2 ikL sin � − e− 1

p = e � ⎤ rL ik sin λ

Next move the term 1 into the square brackets: L

� ⎣ 1 ikL sin �

2A −i(kr−λt) e2 ikL sin � − e− 1

p = e � ⎤ r ikL sin λ

and, using the fact that sin(x) = eix−e−ix , we can write:

2i ⎝ �

A sin( 1 kL sin λ)−i(kr−λt) 2p = e 1 r kL sin λ

2

which is the pressure at (r, λ) due to the line array. The square of the term in brackets is defined as the beam pattern b(λ) of the array:

⎝ sin( 1 kL sin λ)

�2

b(λ) = 2 1 kL sin λ2

2 ARRAYS 11

Steered Line Array

Recall importance of phase:

Spatial phase: kz sin λ = 2 �� z sin λ•

Temporal phase: πt = 2T� ; T =

f 1 •

L/2

A/L

-L/2

r

θ

z

x

θ 0

k sin λ = vertical wavenumber k sin λ0 = vertical wavenumber reference

To make a steered line array, we apply a linear phase shift −zk sin λ0 to the excitation of the array:

dp = A/L iz(k sin �−k sin �0)e iλtdze (1)

r

We can write sin λ0

zk sin λ0 = πz c

z sin λ0zk sin λ0 = πT0(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.

iz(k sin �−k sin �0)e iλt ikz sin � −iλ(t+T0(z))e = e e

integrating Equation 1 yields:

⎝ �

LA sin( k 2 [sin λ − sin λ0])−i(kr−λt)p = e Lr ( k [sin λ − sin λ0])2

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 λ = λ0 instead of λ = 0.

�

�

12 2 ARRAYS

steered and unsteered line array (theta_0 = 20 deg)

−80

−70

−60

−50

−40

−30

−20

−10

0

Sou

rce

leve

l nor

mal

ized

to o

n−ax

is r

espo

nse

unsteered beam

steered beam

−90 −60 −30 0 30 60 90 120theta (degrees)

This plot shows the steered line array beam pattern

sin( kL 2

⎣2

b(α) = [sin α − sin α0])

⎤ (kL

2 [sin α − sin α0])

for α0 = 0 and α0 = 20 degrees.

��

• •

• •

• •

2 ARRAYS 13

Example 1: Acoustic Bathymetry

� �

2θ 3dB

D = 0.5 m

Given: Compute:

f = 12 kHz � =

Baffled disc transducer DI =

D = 0.5 m SL =

• Acoustic power P = 2.4 W • α3dB =

Spatial resolution, τ Depth resolution, λ

τ = 2d tan α3dB TF = 2d (earliest arrival time) c

= d TL = 2cr (latest arrival time) ·

λ = (TL − TF ) · c/2

1 − 1)= d(cos λ3dB

= d · For d = 2 km

τ =

λ =

�

��

2 ARRAYS 14

Example 2: SeaBeam Swath Bathymetry

��

2 degrees

90 degrees

Transmit: 5 meter unsteered line array (along ship axis)

� �

��

steered

No returns No returns

returns received

Receive array: 5 meter line array (athwartships)

only from +− 2 degrees

��

2 degrees

Net beam (plan view)

Ship’sShip’s TrackTrack

One beam, 2 by 2 degrees 100 beams, 2 by 2 degrees

(without steering) (with steering)

3 PROPAGATION PART I: SPREADING AND ABSORPTION 15

3 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 50000 15.9

108 139 174 216 264 315 1140 2520 4440 6920 9940 13520 17640 22320

40000 11.8

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 900 0.056 600000 1000 (1kHz) 0.063 700000 2000 0.12 800000 3000 0.18 900000 4000 0.26 1000000 (1 MHz) 5000 0.35 2000000 6000 0.46 3000000 7000 0.59 4000000 8000 0.73 5000000 9000 0.90 6000000 10000 (10 kHz) 1.08 7000000 20000 3.78 8000000 30000 7.55 9000000

frequency [Hz] alpha [dB/km]

10000000 (10 MHz) 27540


Absorption of sound in sea water (from 13.851 class notes).


Absorption of sound in sea water

Relaxation mechanism

(conversion of acoustic energy to heat)

Four mechanisms

shear viscosity (θ � 10−12 sec) •

structural viscosity (θ � 10−12 sec) •

magnesium sulfate — MgSO4 (θ � 10−6 sec) [1.35 ppt] •

boric acid (θ � 10−4 sec) [4.6 ppm]•

Relaxation time, θ

• if σθ << 1, then little loss. • if σθ � 1 or greater, then generating heat (driving the fluid too fast).

The attenuation coefficient � depends on temperature, salinity, pressure and pH. The following formula for � in dB/km applies at T=4 � C, S=35 ppt, pH = 8.0, and depth = 1000 m. (Urick page 108).

0.1f 2 44f 2

� � 3.0 × 10−3 + + + 2.75 × 10−4f 2

1 + f 2 4100 + f 2


Solving for range given transition loss TL

The equation 20 log r + 10−3�r = 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)

f (xi) xi+1 = xi −

f �(xi)

To solve for range given TL, we have:

f (xi) = T L − 20 log xi − 10−3�xi

20 f �(xi) = − − 10−3�

r

� = TL = x0 =

i xi 20 log xi 0.001�xi f (xi) f �(xi) f (xi)/f �(xi) xi+1 = xi − f (xi) f �(xi)

0 1 2 3 4

�

� =


L = 2 km

Example 1: whale tracking

Passive sonar equation:

Given:

• f0 = 250 HzDT = 15 dB• P = 1 Watt (omni) •

NL = 70 dB • line array: L = 2 km •

Question: How far away can we hear the whale?

TL = =

� =

DI =

SL =

TL = 20 log r + � ≈ r ≈ 10−3 =

Rt = 8680 =

r = (w/o absorption) r = (with absorption)


TL vs range for alpha=0.003 dB/km, f=250 Hz 140

120

100

TL

(dB

)

80

60

40 100 1000 10000 100000 1000000

range (m)

Figure 1: TL vs. range for whale tracking example (f=250 Hz, � = 0.003 dB/km).

f (xi) = T L − 20 log xi − 10−3�xi =

20f �(xi) = − − 10−3� =

r

� = TL = x0 =


0 500,000 114 1.5 -1.5 −3.7 × 10−5 40540 459500 1 459,500 13.2 1.38 -0.6 −4.05 × 10−5 14805 444694 2 445,000 112.96 1.34 -0.3 3 4

�

� =

� =


L = 1 m

Example 2: dolphin tracking

Passive sonar equation:

Given:

• f0 = 125 kHzDT = 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 = =

DI =

TL = 20 log r + � ≈ r ≈ 10−3 =

Rt = 8680 =

r = (w/o absorption) r = (with absorption)

r = (w/o spreading)

f (xi) = T L − 20 log xi − 10−3�xi =


TL vs range for alpha=30 dB/km, f=125 kHz 400

350

TL

(dB

) 300

250

200

150

100

50

0 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

range (m)

Figure 2: TL vs. range for dolphin tracking example (f=125 kHz, � = 30 dB/km).

20 f �(xi) = − − 10−3� =

r

� = TL = x0 =


0 5000 74 150 -67 -0.034 1970 3030 1 3030 69.6 90.9 -3.5 -0.037 95 2935 2 2935 69.35 88.05 -0.4 -0.368 10.9 2925 3 2925 4

4 PROPAGATION PART II: REFRACTION 23

4 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: ρc ρc ρc

= 4.6 m/sec/C� = 1.34 m/sec/ppt = 0.016 m/sec/m ρT ρS ρD

For example:

• �T = 25� =� �c = 115 m/sec

• �S = 5 ppt =� �c = 6.5 m/sec

• �D = 6000 m =� �c = 96 m/sec

sound speed c

D S T c

z

depth

24 4 PROPAGATION PART II: REFRACTION

Sound across an interface

�� y

x

θ2

θ1

θ1

k = ω/c1 1 k = ω/c2 2

ρ c1 1 cρ2 2p

p

i

r

, ,

pt

1

p1 = pi + pr = Ie−i(k1x cos λ1+k1y sin λ1) + Re−i(−k1x cos λ1+k1y sin λ1)

p2 = pt = Te−i(k2x cos λ2+k2y sin λ2)

At x = 0, we require that p1 = p2 (continuity of pressure)

(I + R)e−ik1y sin λ1 = Te−ik2y sin λ2

Match the phase to get Snell’s law:

sin α1 sin α2 =

c1 c2


If c2 > c1, then α2 > α1

y

xθ1

pi

c1

c2

pt

θ2 t

If c2 < c1, then α2 < α1

y

x

θ2

θ1

pi

c2

c1

pt

Sound bends towards region with low velocity

Sound bends away from region with high velocity


c x

Linear sound

(arc of a Radius of curvature

circle) R is constant

z z

speed profile�

Goal: prove that the radius of curvature R is constant for a linear sound speed gradient.

Use the following: sin λ1. Snell’s law: sin λ = constant = δ or c =

c α

(δ is the horizontal slowness or ray parameter)

2. radius of curvature: R = dS dλ

dc3. gradient: g = dz

4. dz = dS cos αdx

θ dS

dz�

First use Equation 4 in Equation 2:

dS dz 1 R = =

dα dα ·cos α

dzdc 1 R =

dcdα·cos α

Then use Equation 3 for dz :dc

1 dc R =

g cos α·dα


but from Equation 1 we can write:

dc cos α =

dα δ

and so we can write: 1

R = gδ

Hence, for linear sound speed gradient, radius of curvature is constant.

=� ray paths are arcs of a circle

c x�

z z�

For gradient g > 0, upward refraction occurs.

c x�

zz�

For gradient g < 0, downward refraction occurs.


Other forms for R

Since sin λ = δ = constant for a ray, we can choose any known value. c�

For example, choose sin λ at the turning point z = zt.c�

z� t

z�

x

θ 2t

θ

= π/

� sin α(zt) 1 At α = =

2 �

c(zt)=

c(zt)= δ

Then: c(zt)

R = g

But c(zt) = c(z0) + g(zt − z0)

So we can write: c(z0)

R = + (zt − z0) g

for z0 any given depth.

x


Example 1: Arctic propagation

Xc1440 m/s c

g=0.016

4000 m 1504 m/s zz θt = π/2

θ

At zt = 4000 m, αt = �/2, c(zt) = 1504 m/s.

Then radius of curvature R = c(zt) = g

What is range to first turning point, Xc?

A

R

S

z

R−z

R

B

2AB = 2 �

2Rz − z

S = 2R cos−1( R − z

)R

Distance Xc is the chord:

Xc = 2 �

2Rz − z2 =

Arc length S is the distance traveled by the ray:

S = 2R cos−1 ⎦ R − z

⎛

= R

�

�


What is the launch angle α(z0) for this ray?

sin(α(z0)) 1=

c(z0) c(zt) ⎨ �

⎩ ⎡ α(z0) = sin−1 c(z0)=

c(zt)

Example 2: Bi-linear duct

1490 m/s 1540 m/s DLDUc

g=−0.05

z=1000m za

g=0.015

z

zt zt =4333 m

z=5000m1550 m/s z sea bottom

θ

Upper: c(zt)

RU = = g

DU = 2 2RU za − za 2 =

Lower: c(zt)

Rl = = g

c(za) zt = RL − + za =

g

DL = 2 2RL(zt − za) − (zt − za)2 =

x

�

5 REFLECTION AND TARGET STRENGTH 31

5 Reflection and target strength

Interface reflection

1

�

x

y p

tp

r pi

θ θ

θ2

11

� � � � � � � � � � � � � � � � � � � ρ c 1 1

ρ c 2 2

pi = Ie iπt e−i(k1x sin λ1−k1y cos λ1)

pr = Re iπt e−i(k1x sin λ1+k1y cos λ1)

pt = Te iπt e−i(k2x sin λ2−k2y cos λ2)

32 5 REFLECTION AND TARGET STRENGTH

Boundary condition 1: continuity of pressure:

@ y = 0 p1 = pi + pr = pt = p2

Boundary condition 2: continuity of normal particle velocity:

@ y = 0 v1 = vi + vr = vt = v2

Momentum equation φv 1 φp

= φt

− π · φy

φv iπt)but = iτv (since v ≡ eφt

i φp =� v =

τπ φy

Continuity of pressure:

pi y=0 + pr y=0 = pt |y=0| |(I + R)e−ik1x sin λ1 = Te−ik2x sin λ2

−iπx sin λ2 c1 c2(I + R)e

−iπx sin λ1 = Te

And so using Snell’s law, we can write:

I + R = T (2)

Continuity of normal velocity:

i cos α1 vi = ik1 cos α1pi = pi

τπ1 · −

π1c1


cos α1 vr =

π1c1 pr

cos α2 vt = −

π2c2 pt

| |vi y=0 + vr y=0 = vt |y=0

π2c2 cos(α1)(I − R) = π1c1 cos(α2)T (3)


We can solve Equations 1 and 2 to get the reflection and transmission coeficients R and T :

R π2c2 cos α1 − π1c1 cos α2 =R =

I π2c2 cos α1 + π1c1 cos α2

T 2π2c2 cos α1 = =T

I π2c2 cos α1 + π1c1 cos α2

Recall that given α1, we can compute α2 with Snell’s law:

sin α1 sin α2 =

c1 c2

Special case: α = 0 (normal incidence)

π2c2 − π1c1 Z2 − Z1 R = π2c2 + π1c1

� Z2 + Z1

2π2c2 2Z2 =T

π2c2 + π1c1 �

Z2 + Z1

where Z = πc is defined as the acoustic impedance.

�


Example 1: source in water

�

x

y tp

θ1

� � � � � � � � � � � � � � � � � � �Air

Water θ1

θ2

r pp

i

πa = 1.2 kg/m3, ca = 340 m/sec =� Za = 408 kg/m2s

πw = 1000 kg/m3, cw = 1500 m/sec =� Zw = 1.5 × 106 kg/m2s

R = 408 cos α1 − 1.5 × 106 cos α2

408 cos α1 + 1.5 × 106 cos α2 � −1

T = 1 + R � 0 (no sound in air)

�


Example 2: source in air

�

x

y p

tp

r pi

θ θ

θ2

11

� � � � � � � � � � � � � � � � � � �Air

Water

R = 1.5 × 106 cos α1 − 408 cos α2

1.5 × 106 cos α1 + 408 cos α2 � +1

T = 1 + R � 2 (double sound in water!)

Does this satisfy your intuition?

Consider intensity:

2 2p aIa = a = p

πaca 408

p2 4p2 24p2 pa a aIw = w = = = πwcw πwcw 1.5 × 106 3.75 × 105


1 =� Iw Ia�

1000

Does this satisfy your intuition?

=�


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

IscatTS = 10 log r=rrefIinc

|

r

r 0

2�

Ip

inc =�

I

Pπc

inc = �r02Iinc

Assume Pscat = Pinc (omnidirectional scattering)�

scat =�Pscat �r0

2Iinc

4�r2� 4�r2�

For r = rref = 1 meter:

TS = 10 log r0

2

(Assuming r0 >> �)4

If r0 = 2 meters, then TS = 0 dB

6 DESIGN PROBLEM: TRACKING NEUTRALLY BUOYANT FLOATS 39

6 Design problem: tracking neutrally buoyant floats

Sonar design problem

2r0 = 25 cm

Require:

• Track to ± 3∗ bearing

• Range error: λ ± 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)

6 DESIGN PROBLEM: TRACKING NEUTRALLY BUOYANT FLOATS 40

receiver DI: DIR =

pulse length: θ =

array length: L =

source level: SL =

noise level: NL =

transmission loss: TL =

wavelength: � =

source DI: DIT =

time-of-flight: T =

ping interval: Tp =

frequency: f =

target strength: TS =

range resolution: λ =

acoustic power: P =

average acoustic power: P =

�

7 REVERBERATION 41

7 Reverberation

Surface reverberation

��

* * dφ

dΑ

dΑ

r

dΑ = (1/2)cτ rdφ

I0

I rev

Iscat

Iinc

Side View Plan View

� 1 Irev =

A I0 · b(α, δ) · 1

Iscat

2 b≤(α, δ)dA

r2 ·

Iinc ·r

·

Iscat as ssDefine the ratio Iinc

I0Irev =

4 ss b(α, δ)b≤(α, δ)dA

Ar

From figure: cθ

dA = rdδ 2

consider only beampatterns of the form: ⎞

b(0, δ) = b≤(0, δ) = ⎠ 1 , |δ| ⇒ �/2 ⎧ 0 , δ > �/2| |

I0 cθ Irev =

r4 ss r�

2

7 REVERBERATION 42

Define A as the insonified area cα r�. Take logs to define the 2 reverberation level RLs:

RLs = SL − 40 log r + Ss + 10 log A (No absorption)

� =

7 REVERBERATION 43

Example: 100 kHz side-scan sonar

• f = 100 kHz (� = 30 db/km)

• α3dB−H = 0.5∗

• α3dB−V = 30∗

• SL = 201 dB re 1 µ Pa @ 1 meter (rectangular array)

DT = 15 dB•

NL = 35 dB •

θ = 0.1 ×10−3 sec•

Sonar equation: ⎨ �

SE = SL − 2TL − (NL − DIr) +⎩ Ss + 10 log A

⎡ − DT TS

How much signal excess for rock bottom vs clay bottom at 300 meter range?

•

• Lx =

• Ly =

• DIR =

2 TL =•

• 10 log A =

Rock: Ss = -20 dB =� SE =

Clay: Ss = -40 dB =� SE =

7 REVERBERATION 44

Compare the echo level of a steel drum to the reverberation level of rocks and clay at 300 m.

EL = SL - 2TL + TS •

• RL = SL - 2TL + SS + 10 log A

Drum is finite cylinder =� TS = 10 log aL2 = 2�

a a = 0.45 m

L = 1.5 m

L

EL =

=RL rocks |RL clay =|

What if drum is in a boulder field with boulders of diameter 2 meters? (assume spherical boulders)

2 TS|boulder = 10 log r = 4

=EL drum - EL|boulder |

�

7 REVERBERATION 45

Side scan sonar analysis

Slant range correction:

��

Rs

H f Target

Rh

By Pythagorus: Rh = R

Object height:

2H− f2 s

Rs

TargetH f ��

�� Ls

H t

Rh

LsHfBy similar triangles: Ht =

Rs + Ls

7 REVERBERATION 46

*

RSmaxRSmin

R

2θ3dB−V

θ1

θ2

H

Hmin

Hmax R

Side view of sidescan geometry.

Tow direction

D Xmin

Xmin

Xmax2θ3dB−H

RHmax

RHmin

Plan view of sidescan geometry.

7 REVERBERATION 47

Summary of variable definitions for side-scan sonar.�

• λ1 is the angle from the vertical to the near edge of the fan beam in the vertical plane.

• λ2 is the angle from the vertical to the far edge of the fan beam in the vertical plane.

λ2 = λ1 + 2 � λ3dB-V

• RSmax is the maximum slant range.

RSmax = H/ cos(λ2)

• RSmin is the minimum slant range.

RSmin = H/ cos(λ1)

• RHmax is the maximum horizontal range.

RHmax = H tan(λ2)

• RHmin is the minimum horizontal range.

RHmin = H tan(λ1)

• Xmax is the along-track resolution at maximum range.

Xmax = 2RHmax � tan λ3dB-H

• Xmin is the along-track resolution at minimum range

Xmin = 2RHmin � tan λ3dB-H

• Tf is the time-of-flight, which is the time required for the sonar ping to travel to maximum range and back.

Tf = 2 � RSmax/c

• Tp is the pulse repetition interval, which is the time between pings. To avoid overlap of echos from one ping to the next:

Tp ∼ Tf

• v is the velocity of the towfish.

• D is the distance traveled by the sonar towfish during one ping cycle. To avoid gaps in the sonar coverage:

D = v � Tp � Xmin

8 NOISE 48

8 Noise

Noise levels in the deep ocean (from Urick).�

8 NOISE 49

Sources of noise in the ocean (from Urick).

| | �

8 NOISE 50

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 = πgH � 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 � 1 7 Hz

On land, displacements are

ρ � 10−6meters

Assume harmonic motion dρ

ρ ≡ e iπt =� v = = iτρ dt

Noise power due to microseisms

p = πcv = iτπcρ = i2�fπcρ

p = 2�fπcρ = 1.4P a = NL = 123 dB re 1 µPa

8 NOISE 51

Band II: Oceanic turbulence 1 Hz < f < 20 Hz

Possible mechanisms:

• 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)

�

8 NOISE 52

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

τ = kg

τ cp =

k g�2 c =p 2�

if � � 2000 km, then cp � 1500 m/sec =� radiate noise!

Band V: Thermal noise 50 KHz < f

NL = −15 + 20 log f

8 NOISE 53

Directionality of noise

Vertical

• Low frequency

– distant shipping dominates

– low attenuation at horizontal θ

NL

−90

90

0

• High frequency

– sea surface noise

– local wind speed dominates

– high attenuation at horizontal

0

−90

90

θ

NL

Horizontal

• Low frequency: highest in direction of shipping centers

• High frequency: omnidriectional

9 DESIGN PRINCIPLES 54

9 Design Principles

Summary of important design principles

1. Required maximum range determines maximum frequency

Dyer’s rule : � × r × 0.001 = 10 dB [for r in meters]

2. Required angular resolution determines array size

• α3dB = 25.3� deg. (line array) L±29.5� • α3dB = D deg. (disc array) ±25.3� 25.3� • α3dB = ± Lx

, ± Ly deg. (rec. array)

3. Required range resolution determines maximum pulse length cθ

�r = 2

4. In a side-scan sonar, the horizontal (across-track) range resolution �rh is determined by the beam geometry:

�rh = �r sin α

where α is the angle between a ray drawn at the maximum range and the vertical.

Δr = cτ/2θ

Δrh

5. Characteristics of the transducer determine minimum pulse length.

10 SIDESCAN SONAR 55

• For a narrowband transducer, you must have at least 10 to 15 cycles of the carrier frequency.

10 Sidescan sonar

*

RSmaxRSmin

R

2θ3dB−V

θ1

θ2

H

Hmin

Hmax R


Side view of sidescan geometry.

Tow direction

D Xmin

Xmin

Xmax2θ3dB−H

RHmax

RHmin

Plan view of sidescan geometry.


Summary of variable definitions for side-scan sonar.

• λ1 is the angle from the vertical to the near edge of the fan beam in the vertical plane.

• λ2 is the angle from the vertical to the far edge of the fan beam in the vertical plane.

λ2 = λ1 + 2 � λ3dB-V

• RSmax is the maximum slant range.

= H/ cos(λ2)RSmax

• RSmin is the minimum slant range.

= H/ cos(λ1)RSmin

• RHmax is the maximum horizontal range.

= H tan(λ2)RHmax

• RHmin is the minimum horizontal range.

= H tan(λ1)RHmin

• Xmax is the along-track resolution at maximum range.

Xmax = 2RHmax � tan λ3dB-H

• Xmin is the along-track resolution at minimum range

Xmin = 2RHmin � tan λ3dB-H

• Tf is the time-of-flight, which is the time required for the sonar ping to travel to maximum range and back.

Tf = 2 � RSmax/c

• Tp is the pulse repetition interval, which is the time between pings. To avoid overlap of echos from one ping to the next:

Tp ∼ Tf

• v is the velocity of the towfish.

• D is the distance traveled by the sonar towfish during one ping cycle. To avoid gaps in the sonar coverage:

D = v � Tp � Xmin

11 SUMMARY OF IMPORTANT FORMULAE 58

11 Summary of important formulae

Source Level

• SL = 171 + 10 log P + DI

Directivity Index LDI = 10 log( 2 � ) (line array) • DDI = 20 log( �� ) (disc array) •

DI = 10 log( 4�LxLy ) (rectangular array) �2•

3-dB Beamwidth α3dB

• α3dB = 25.3� deg. (line array) L±

29.5�• α3dB = D deg. (disc array) ±

25.3� 25.3� • α3dB = ±Lx

, ±Ly

deg. (rectangular array)

Transmission loss, spherical spreading and absorption:

TL = 20 log r + 10−3�r •

Transmission loss, cylindrical spreading and absorption:

TL = 10 log r + 10−3�r •

Target strength of a sphere (r0 = radius, assumes r0 >> �): 20• TS = 10 log r4

Target strength of a cylinder: (at normal incidence, a = radius, L = length)

TS = 10 log aL2

2�•

Rule-of-thumb for picking frequency given maximum range

• � × r × 0.001 = 10 dB [for r in meters]

Range resolution

�r = cπ 2•


Absorption of sound in sea water (from Prof. Dyer’s 13.851 class notes).


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

Noise levels in the deep ocean (from Urick).

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