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.
2.
3.
4.
5.
“. . . ship to stop” −⇒ reduce self-noise
“. . . tube in water” −⇒ transducer
“. . . to your ear” −⇒ receiver
“. . . will hear ships” −⇒ detection
“. . . at a great distance” −⇒ low attenuation
1687 Issac Newton: first theoretical prediction of sound speed
1
ρπ
� 2
c
ρp
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
2
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
3
1 INTRODUCTION TO SONAR
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
4
1 INTRODUCTION TO SONAR
Reverberation vs. Noise-limited Range
• Active Sonar
• Range independent noise vs. range dependent reverberation
• Define Echo Level: EL = SL − 2TL + TS
Echo Level
Level
(dB)
Reverberation
Level
Noise Level
reverberation
limited range
Range (time)
Echo Level
Level
(dB)
Reverberation
Level
Noise Level
noise
limited range
Range (time)
5
1 INTRODUCTION TO SONAR
Definition of sound intensity
dB = decibel (after Alexander Graham Bell)
For acoustics:
dB = 10 log10
I
Iref
6
2 ARRAYS
2
Arrays
Summary of array formulas
Source Level
• SL = 10 log I I = 10 log pp2
ref
2
ref
• SL = 171 + 10 log P
• SL = 171 + 10 log P + DI
(general)
(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( 2�L )
• DI = 10 log(( ��D )2) = 20 log( ��D )
4�L L
• DI = 10 log( �x2 y )
(line array)
(disc array)
(rect. array)
3-dB Beamwidth α3dB
• α3dB = ± 25.3�
L deg.
• α3dB = ± 29.5�
D deg.
25.3�
,
±
• α3dB = ± 25.3�
Lx
Ly deg.
(line array)
(disc array)
(rect. array)
7
2 ARRAYS
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
0
30
theta (degrees)
f = 12 kHz D = 0.5 m beamwidth = +−7.325 deg
60
90
−30
60
90
60
90
−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/� equal
to 2, 4, and 8. Note that the beampattern gets narrower as the diameter is
increased.
8
2 ARRAYS
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 α) ⎤
�
b(α) =
1
2 kL sin α
�
whereas for the disc array, the beam pattern is
⎣2
2J1 ( 12 kD sin α) ⎤
�
b(α) =
1
2 kD sin α
�
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.
9
2 ARRAYS
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, 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:
l2 = r2 + z 2 − 2rz cos �.
If we factor out r 2 from the left hand side, and substitute sin λ for cos �, we get:
2
l =r
2
2z
z2
1−
sin λ + 2
r
r
⎝
�
and take the square root of each side we get:
2z
z2
l =r 1−
sin λ + 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 λ + 2 )
2
r
r
⎝
�
10
2 ARRAYS
Finally, making the assuming that z << r, we can drop the term
z2
r2
to get
l�
= r − z sin λ
Field calculation
For an element of length dz at position z, the amplitude at the field position (r, λ) is:
A 1 −i(kl−λt)
e
dz
Ll
We obtain the total pressure at the field point (r, λ) due to the line array by integrating:
dp =
p=
but l �
= r − z sin λ, so we can write:
A � L/2 1 −i(kl−λt)
dz
e
L −L/2 l
A −i(kr−λt) � L/2
1
p= e
eikz 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
1
and move it outside the integral:
r
p=
1
r−z sin �
with
A −i(kr−λt) � L/2 ikz sin �
e
dz
e
rL
−L/2
Next, we evaluate the integral:
A −i(kr−λt) eikz sin �
p=
e
rL
ik sin λ
�L/2
�
1
⎝
Next move the term
1
L
−L/2
1
⎣
A −i(kr−λt) � e 2 ikL sin � − e− 2 ikL sin � ⎤
e
p=
rL
ik sin λ
into the square brackets:
�
⎣
1
1
A
e 2 ikL sin � − e− 2 ikL sin � ⎤
p = e−i(kr−λt) �
ikL sin λ
r
and, using the fact that sin(x) =
eix −e−ix
,
2i
we can write:
A −i(kr−λt) sin( 12 kL sin λ)
p= 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( 12 kL sin λ)
b(λ) =
1
kL sin λ
2
⎝
�2
11
2 ARRAYS
Steered Line Array
Recall importance of phase:
• Spatial phase: kz sin λ = 2�� z sin λ
• Temporal phase: πt = 2T� ; T = f1
z
L/2
r
θ
θ0
A/L
x
-L/2
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 ) iλt
e dz
e
r
We can write
zk sin λ0 = πz
(1)
sin λ0
c
z sin λ0
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 λ0 = πT0 (z) ; T0 (z) =
eiz(k sin �−k sin �0 ) eiλt = eikz sin � e−iλ(t+T0 (z))
integrating Equation 1 yields:
sin( k2L [sin λ − sin λ0 ])
A
p = e−i(kr−λt)
r
( k2L [sin λ − sin λ0 ])
⎝
�
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)
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 α − sin α0 ]) ⎤
�
b(α) =
( kL
2 [sin α − sin α0 ])
�
for α0 = 0 and α0 = 20 degrees.
90
120
13
2 ARRAYS
Example 1: Acoustic Bathymetry
�
�
2θ3dB
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
τ = 2d tan α3dB
=d ·
Depth resolution, λ
TF =
2d
c
(earliest arrival time)
TL =
2r
c
(latest arrival time)
λ = (TL − TF ) · c/2
= d( cos 1λ3dB − 1)
For d = 2 km
τ=
λ =
=d·
14
2 ARRAYS
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)
3 PROPAGATION PART I: SPREADING AND ABSORPTION
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
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
216
800000
900000
264
1000000 (1 MHz)
315
1140
2000000
2520
3000000
4440
4000000
6920
5000000
9940
6000000
7000000
13520
17640
8000000
9000000
22320
10000000 (10 MHz)
27540
15
3 PROPAGATION PART I: SPREADING AND ABSORPTION
Absorption of sound in sea water (from 13.851 class notes).
16
3 PROPAGATION PART I: SPREADING AND ABSORPTION
17
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).
� � 3.0 × 10
−3
44f 2
0.1f 2
+ 2.75 × 10−4 f 2
+
+
2
2
1+f
4100 + f
18
3 PROPAGATION PART I: SPREADING AND ABSORPTION
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)
xi+1 = xi −
f (xi )
f � (xi )
To solve for range given TL, we have:
f (xi ) = T L − 20 log xi − 10−3 �xi
f � (xi ) = −
�=
i xi
0
1
2
3
4
20
− 10−3 �
r
TL =
20 log xi
0.001�xi f (xi )
x0 =
f � (xi )
f (xi )/f � (xi ) xi+1 = xi −
f (xi )
f � (xi )
3 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 =
=
�=
DI =
SL =
TL = 20 log r + � ≈ r ≈ 10−3 =
�=
Rt =
r=
8680
�
=
(w/o absorption)
r=
(with absorption)
19
20
3 PROPAGATION PART I: SPREADING AND ABSORPTION
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, � = 0.003 dB/km).
f (xi ) = T L − 20 log xi − 10−3 �xi =
f � (xi ) = −
�=
i xi
20
− 10−3 � =
r
x0 =
TL =
20 log xi
0 500,000 114
1 459,500 13.2
2 445,000 112.96
3
4
0.001�xi f (xi )
f � (xi )
1.5
1.38
1.34
−3.7 × 10−5
−4.05 × 10−5
-1.5
-0.6
-0.3
f (xi )/f � (xi ) xi+1 = xi −
40540
14805
459500
444694
f (xi )
f � (xi )
3 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 =
=
�=
DI =
TL = 20 log r + � ≈ r ≈ 10−3 =
�=
Rt =
8680
�
=
r=
(w/o absorption)
r=
(w/o spreading)
r=
f (xi ) = T L − 20 log xi − 10−3 �xi =
(with absorption)
21
22
3 PROPAGATION PART I: SPREADING AND ABSORPTION
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, � = 30 dB/km).
f � (xi ) = −
�=
20
− 10−3 � =
r
x0 =
TL =
i xi
20 log xi
0.001�xi f (xi )
f � (xi )
0
1
2
3
4
74
69.6
69.35
150
90.9
88.05
-0.034
-0.037
-0.368
5000
3030
2935
2925
-67
-3.5
-0.4
f (xi )/f � (xi ) xi+1 = xi −
1970
95
10.9
3030
2935
2925
f (xi )
f � (xi )
23
4 PROPAGATION PART II: REFRACTION
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
= 4.6 m/sec/C�
ρT
ρc
= 1.34 m/sec/ppt
ρS
ρc
= 0.016 m/sec/m
ρD
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
24
4 PROPAGATION PART II: REFRACTION
Sound across an interface
p
r
θ1
k1 = ω/c1
p
i
θ1
��
��
��
��
��
��
��
��
��
��
��
��
ρ , c1��
1
��
��
��
y
p
t
θ2
k2 = ω/c2
x
ρ , c2
2
1
p1 = pi + pr = Ie−i(k1 x cos λ1 +k1 y sin λ1 ) + Re−i(−k1 x cos λ1 +k1 y sin λ1 )
p2 = pt = T e−i(k2 x cos λ2 +k2 y sin λ2 )
At x = 0, we require that p1 = p2 (continuity of pressure)
(I + R)e−ik1 y sin λ1 = T e−ik2 y sin λ2
Match the phase to get Snell’s law:
sin α1
sin α2
=
c1
c2
25
4 PROPAGATION PART II: REFRACTION
If c2 > c1 , then α2 > α1
p
t
y
θ2
t
x
θ1
c2
p
i
c1
If c2 < c1 , then α2 < α1
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
26
4 PROPAGATION PART II: REFRACTION
c
z
x
z
Linear sound
speed profile�
Radius of curvature R is constant
(arc of a circle)
Goal: prove that the radius of curvature R is constant for a linear sound
speed gradient.
Use the following:
1. Snell’s law: sinc λ = constant = δ or c = sinα λ
(δ is the horizontal slowness or ray parameter)
2. radius of curvature: R =
3. gradient: g =
dS
dλ
dc
dz
4. dz = dS cos α
dx
dS
θ
dz�
First use Equation 4 in Equation 2:
dS
dz
1
=
·
dα
dα cos α
dz dc
1
R=
·
dc dα cos α
R=
Then use Equation 3 for
dz
dc
:
R=
1
dc
·
g cos α dα
27
4 PROPAGATION PART II: REFRACTION
but from Equation 1 we can write:
dc
cos α
=
dα
δ
and so we can write:
1
gδ
Hence, for linear sound speed gradient, radius of curvature is constant.
R=
=� ray paths are arcs of a circle
c
z
x�
z�
For gradient g > 0, upward refraction occurs.
c
z�
x�
z
For gradient g < 0, downward refraction occurs.
28
4 PROPAGATION PART II: REFRACTION
Other forms for R
Since
sin λ
c�
= δ = constant for a ray, we can choose any known value.
For example, choose
sin λ
c�
at the turning point z = zt .
x
θ
z�
t
θ t = π/2
z�
At α =
�
sin α(zt )
1
=�
=
=δ
c(zt )
2
c(zt )
Then:
R=
c(zt )
g
But
c(zt ) = c(z0 ) + g(zt − z0 )
So we can write:
R=
for z0 any given depth.
c(z0 )
+ (zt − z0 )
g
29
4 PROPAGATION PART II: REFRACTION
Example 1: Arctic propagation
1440 m/s
Xc
c
x
g=0.016
θ
4000 m
1504 m/s z
z
θ t = π/2
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 ?
R−z
�
AB = 2 2Rz − z 2
R
R
A
S = 2R cos−1 (
B
z
S
Distance Xc is the chord:
�
Xc = 2 2Rz − z 2 =
Arc length S is the distance traveled by the ray:
S = 2R cos
−1
⎦
R−z
=
R
⎛
R−z
)
R
30
4 PROPAGATION PART II: REFRACTION
What is the launch angle α(z0 ) for this ray?
1
sin(α(z0 ))
=
c(z0 )
c(zt )
α(z0 ) = sin
⎨
�
−1 ⎩ c(z0 ) ⎡
c(zt )
=
Example 2: Bi-linear duct
1490 m/s
DU
1540 m/s
c
DL
x
θ
g=−0.05
z=1000m
za
g=0.015
z
zt
z t =4333 m
z=5000m
z
1550 m/s
sea bottom
Upper:
RU =
Lower:
c(zt )
=
g
�
DU = 2 2RU za − za2 =
c(zt )
=
g
c(za )
zt = RL −
+ za =
g
Rl =
�
DL = 2 2RL (zt − za ) − (zt − za )2 =
31
5 REFLECTION AND TARGET STRENGTH
5
Reflection and target strength
Interface reflection
1
y
p
p
i
r
θ1
θ1
ρ c
1
�
�
1
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
ρ c
2
�
x
2
θ2
p
t
pi = Ieiπte−i(k1x sin λ1−k1 y cos λ1)
pr = Reiπte−i(k1x sin λ1+k1y cos λ1)
pt = T eiπte−i(k2x sin λ2−k2y cos λ2)
32
5 REFLECTION AND TARGET STRENGTH
Boundary condition 1: continuity of pressure:
@ y=0
p 1 = pi + pr = pt = p2
Boundary condition 2: continuity of normal particle velocity:
@ y=0
Momentum equation
v 1 = vi + vr = vt = v2
1 φp
φv
=− ·
φt
π φy
φv
= iτv (since v ≡ eiπt )
φt
i φp
=� v =
τπ φy
but
Continuity of pressure:
pi |y=0 + pr |y=0 = pt |y=0
(I + R)e−ik1x sin λ1 = T e−ik2x sin λ2
sin λ
(I +
−iπx c 1
1
R)e
sin λ
=
−iπx c 2
2
Te
And so using Snell’s law, we can write:
I +R=T
Continuity of normal velocity:
vi =
i
cos α1
· ik1 cos α1pi = −
pi
τπ1
π1 c 1
(2)
5 REFLECTION AND TARGET STRENGTH
33
cos α1
pr
π1 c 1
cos α2
vt = −
pt
π2 c 2
vi |y=0 + vr |y=0 = vt |y=0
vr =
π2c2 cos(α1)(I − R) = π1c1 cos(α2)T
(3)
5 REFLECTION AND TARGET STRENGTH
We can solve Equations 1 and 2 to get the reflection and
transmission coeficients R and T :
R=
R π2c2 cos α1 − π1c1 cos α2
=
I
π2c2 cos α1 + π1c1 cos α2
T
2π2c2 cos α1
=
I
π2c2 cos α1 + π1c1 cos α2
Recall that given α1, we can compute α2 with Snell’s law:
T =
sin α1 sin α2
=
c2
c1
Special case: α = 0 (normal incidence)
R=
Z2 − Z 1
π2 c 2 − π 1 c 1
�
Z2 + Z 1
π2 c 2 + π 1 c 1
T =
2π2c2
2Z2
�
Z2 + Z 1
π2 c 2 + π 1 c 1
where Z = πc is defined as the acoustic impedance.
34
35
5 REFLECTION AND TARGET STRENGTH
Example 1: source in water
p
y
t
θ2
Air
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
θ1
i
�
x
Water
p
�
θ1
p
r
πa = 1.2 kg/m3, ca = 340 m/sec =� Za = 408 kg/m2 s
πw = 1000 kg/m3 , cw = 1500 m/sec =� Zw = 1.5 × 106 kg/m2s
408 cos α1 − 1.5 × 106 cos α2
R=
� −1
408 cos α1 + 1.5 × 106 cos α2
T = 1 + R � 0 (no sound in air)
36
5 REFLECTION AND TARGET STRENGTH
Example 2: source in air
y
p
p
i
r
θ1
θ1
Air
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
x
Water
θ2
p
t
1.5 × 106 cos α1 − 408 cos α2
R=
� +1
1.5 × 106 cos α1 + 408 cos α2
T = 1 + R � 2 (double sound in water!)
Does this satisfy your intuition?
Consider intensity:
p2a
p2a
Ia =
=
πaca 408
p2w
4p2a
Iw =
=
πw c w πw c w
4p2a
p2a
=
=
1.5 × 106 3.75 × 105
5 REFLECTION AND TARGET STRENGTH
=� Iw �
Does this satisfy your intuition?
1
Ia
1000
37
38
5 REFLECTION AND TARGET STRENGTH
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 =�
πc
Pinc = �r02Iinc
Assume Pscat = Pinc (omnidirectional scattering)�
Iscat
Pscat �r02Iinc
=� 2�=�
4�r2�
4�r
For r = rref = 1 meter:
r02
TS = 10 log
4
If r0 = 2 meters, then TS = 0 dB
(Assuming r0 >> �)
6 DESIGN PROBLEM: TRACKING NEUTRALLY BUOYANT FLOATS
6
39
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
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 =
40
41
7 REVERBERATION
7
Reverberation
Surface reverberation
Plan View
Side View
*
�
����
�
�
����
�
dφ ���
I0
r
Irev
*
Iinc
dΑ
Iscat
dΑ = (1/2)cτ rdφ
dΑ
Irev =
�
I · b(α, δ) ·
A 0
1 Iscat 1 ≤
· · b (α, δ)dA
·
r2 Iinc r2
I
Define the ratio Iscat as ss
inc
I0 �
Irev = 4 ss A b(α, δ)b≤ (α, δ)dA
r
From figure:
cθ
rdδ
2
consider only beampatterns of the form:
dA =
⎞
⎠
b(0, δ) = b≤(0, δ) = ⎧
Irev =
1 , |δ| ⇒ �/2
0 , |δ| > �/2
I0 cθ
ss r�
r4 2
42
7 REVERBERATION
Define A as the insonified area
reverberation level RLs:
cα
2 r�.
Take logs to define the
RLs = SL − 40 log r + Ss + 10 log A
(No absorption)
43
7 REVERBERATION
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:
S + 10 log A ⎡
SE = SL − 2TL − (NL − DIr ) + ⎩ s
− 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 =
44
7 REVERBERATION
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
2
Drum is finite cylinder =� TS = 10 log aL
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 r4 =
EL |drum - EL|boulder =
45
7 REVERBERATION
Side scan sonar analysis
Slant range correction:
Rs
� ���
�����
���
Hf
Target
Rh
�
By Pythagorus: Rh = Rs2 − Hf2
Object height:
Rs
Target
Hf
Ht
Rh
By similar triangles: Ht =
Ls H f
Rs + L s
Ls
46
7 REVERBERATION
*
2θ
3dB−V
RSmin
H
RSmax
θ1
θ2
R
Hmin
RHmax
Side view of sidescan geometry.
Tow direction
R Hmax
R Hmin
Xmin
D
Xmin
2θ3dB−H
Plan view of sidescan geometry.
Xmax
47
7 REVERBERATION
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 ∼ T f
• 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
48
8 NOISE
8
Noise
Noise levels in the deep ocean (from Urick).�
8 NOISE
Sources of noise in the ocean (from Urick).
49
50
8 NOISE
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
f < 1 Hz
Band I:
• 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
ρ ≡ eiπt =� v =
dρ
= iτρ
dt
Noise power due to microseisms
p = πcv = iτπcρ = i2�f πcρ
|p| = 2�f πcρ = 1.4P a =� NL = 123 dB re 1 µPa
51
8 NOISE
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)
52
8 NOISE
500 Hz < f < 50 KHz
Band IV: surface waves
• 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
k
g�
c2p =
2�
if � � 2000 km, then cp � 1500 m/sec =� radiate noise!
cp =
Band V: Thermal noise
NL = −15 + 20 log f
50 KHz < f
53
8 NOISE
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
54
9 DESIGN PRINCIPLES
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
(line array)
• α3dB = ± 25.3�
L deg.
(disc array)
• α3dB = ± 29.5�
D deg.
25.3�
(rec. array)
• α3dB = ± 25.3�
Lx , ± Ly deg.
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.
θ
Δr = cτ/2
Δrh
5. Characteristics of the transducer determine minimum pulse
length.
55
10 SIDESCAN SONAR
• For a narrowband transducer, you must have at least 10 to 15
cycles of the carrier frequency.
10
Sidescan sonar
*
2θ
3dB−V
H
RSmin
RSmax
θ1
θ2
R
Hmin
RHmax
56
10 SIDESCAN SONAR
Side view of sidescan geometry.
Tow direction
R Hmax
R Hmin
Xmin
D
Xmin
2θ3dB−H
Plan view of sidescan geometry.
Xmax
57
10 SIDESCAN SONAR
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 ∼ T f
• 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
58
11 SUMMARY OF IMPORTANT FORMULAE
11
Summary of important formulae
Source Level
• SL = 171 + 10 log P + DI
Directivity Index
• DI = 10 log( 2�L )
• DI = 20 log( ��D )
• DI = 10 log( 4�L�x2 Ly )
(line array)
(disc array)
(rectangular array)
3-dB Beamwidth α3dB
• α3dB = ± 25.3�
L deg.
29.5�
• α3dB = ± D deg.
25.3�
• α3dB = ± 25.3�
Lx , ± Ly deg.
(line array)
(disc array)
(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 >> �):
2
• TS = 10 log r40
Target strength of a cylinder: (at normal incidence, a = radius, L =
length)
2
• TS = 10 log aL
2�
Rule-of-thumb for picking frequency given maximum range
• � × r × 0.001 = 10 dB [for r in meters]
Range resolution
• �r =
cπ
2
11 SUMMARY OF IMPORTANT FORMULAE
Absorption of sound in sea water (from Prof. Dyer’s 13.851 class notes).
59
11 SUMMARY OF IMPORTANT FORMULAE
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
47.4
200000
63.1
300000
83.1
400000
108
500000
139
600000
174
700000
800000
216
900000
264
1000000 (1 MHz)
315
2000000
1140
3000000
2520
4000000
4440
5000000
6920
6000000
9940
7000000
13520
8000000
17640
9000000
22320
10000000 (10 MHz)
27540
Noise levels in the deep ocean (from Urick).
60