Transmission Loss
Review of Passive Sonar Equation
Terminology
• Signal to Noise
The ratio of received echo from target
to background noise produced by everything else.
• Detection Threshold (DT)
The measure of return signal required for an operator using
installed equipment to detect a target 50% of the time.
LS/N= LS - LN > DT
Terminology
• Source Level (SL)
– For ACTIVE sonar operations:
• The SONAR’s sonic transmission (transducer generated)
– For PASSIVE sonar operations:
• Noise generated by target
• Noise Level (NL = NLs  NLA)
– Self (NLs)
• Generated by own ship at the frequency of interest.
– Ambient (NLA)
• Shipping (Ocean Traffic), Wind and Weather - Sea State
(Hydrodynamic)
• Biologic and Seismic obtained from other methods
Terminology
• Directivity Index (DI)
– Receiver directional sensitivity.
– LN = NL - DI
• Transmission Loss (TL)
– Amount the Source Level is reduced due to
spreading and attenuation (absorption,
scattering).
Passive SONAR Equation
(Signal Radiated by the Target)
• SNR required for detection = DT
• To achieve detection > 50% of the time…
– SNR > DT
– LS – LN > DT
• LS = SL – TL (one way)
• LN = NL – DI
– Remember NL = NLs  NLa
• Therefore…
LS/N=SL - TL – (NL – DI) > DT
Passive Sonar Equation
LS/N=SL - TL – (NL – DI) > DT
The Passive Sonar Equation
LS/ N  SL  TL   NL  DI 
 IS 
SL  10log  
 I0 
 IN 
NL  10log  
 I0 
 IS 
TL  10log  
 IR 
DI  10log  d 
Making the Sonar Equations Useful
Passive Example
Known
Can Measure
Function of
Equipment
SL - TL - NL + DI > DT
ONLY UNKNOWN
Can Measure
Experimentally
Figure of Merit
• Often a detection threshold is established such that a trained
operator should be able to detect targets with that LS/N half of
the time he hears them. Called “Recognition Differential.”
(RD)
• Passive sonar equation is then solved for TL allowable at that
threshold. Called “Figure of Merit.” (FOM)
TLallowable = Figure of Merit = SL- LS/N Threshold - (NL-DI)
• Since TL logically depends on range, this could provide an
estimate of range at which a target is likely to be detected.
Called “Range of the Day.” (ROD)
• Any LS/N above the Recognition Differential is termed “Signal
Excess.” (SE) Signal Excess allows detection of targets
beyond the Range of the Day.
Range ???
• FOM helps to predict RANGE.
– The higher the FOM, the higher the signal loss that can
be suffered and, therefore, the greater the expected
detection range.
• Probability of Detection
– Passive
• If FOM > TL then > 50% prob det
• If FOM < TL then < 50% prob det
• Use Daily Transmission Loss (Prop Loss/FOM)
curve provided by Sonar Technicians
HW Example
• A submarine is conducting a passive barrier patrol against
a transiting enemy submarine. The friendly sub has a
directivity index of 15 dB and a detection threshold of 8
dB. The enemy sub has a source of 140 dB.
Environmental conditions are such that the transmission
loss is 60 dB and the equivalent isotropic noise level is 65
dB.
• What is the received signal level?
• What is the signal to noise ratio in dB?
• What is the figure of merit?
• Can the sub be detected? Why?
Prop Loss Curve
Max Range DP
Max Range BB
FOM = 70 dB
Prop Loss Curve
Max Range DP
Max Range CZ
FOM = 82 dB
Transmission Loss
• Sound energy in water
suffers two types of losses:
–Spreading
–Attenuation
Combination of these 2 losses:
TRANSMISSION LOSS (TL)
Spreading
• Spreading
–
–
–
–
Due to divergence
No loss of energy
Sound spread over wide area
Two types:
• Spherical
– Short Range: ro < 1000 m
TL  20 log r
• Cylindrical
– Long Range: ro> 1000 m
ro
r
TL  10 log  20 log
ro
1
Spherical component
Spherical Spreading
 IS 
TL  10log  
 IR 
P1  P2
I1 4r12  I 2 4r22
 r2 
I1 4r

  
I 2 4r
 r1 
2
2
2
1
2
 r2 
r
TL  20log    20log    20log r
1
 r1 
r1
r2
r3
Cylindrical Spreading
TL  10 log
I 1 yd 
I 1 yd 
I r 
 10 log
 10 log 0
I r 
I r0 
I r 
spherical cylindrical
r5
r4
Can be
approximated as
the sides of a
cylinder with a
surface area of
2r5H
H
r1
r2
r3
r4
r5
TL  20 log r0  10 log
r
r0
transition range
ro
Spherical to Cylindrical Transition Range
in a Mixed Layer
r0 
RH
8
H
H d
H  mixed layer thic kness
d  depth of the source
cn
R
 radius of curvature of sound ray
g cos n
Attenuation
• 2 Types
• Absorption
– Process of converting acoustic energy into heat.
• Viscosity
• Change in Molecular Structure
• Heat Conduction
– Increases with higher frequency.
• Scattering and Reverberation
– All components lumped into Transmission Loss Anomaly (A).
– Components:
• Volume: Marine life, bubbles, etc.
• Surface: Function of wind speed.
• Bottom Loss.
– Not a problem in deep water.
– Significant problem in shallow water; combined with refraction and
absorption into bottom.
Absorption
• Decrease in intensity, proportional to:
– Intensity
– Distance the wave travels
• Constant of Proportionality, a
dI  aIdr
I2
 a  r2  r1 
e
I1
Absorption Coefficient
I1
a r r
TL  10 log  10 log e  2 1 
I2
TL  a  r2  r1 10log e  4.343a  r2  r1 
TL    r2  r1 
  4.343a
TL    r2  r1  x10
Has units of dB/yard
3
 Has units of dB/kiloyard
Example
•
•
•
•
Spherical Spreading
Absorption coefficient,  = 2.5 dB/kyd
Find the TL from a source to 10,000 yards
Find the TL from 10,000 yards to 20,000 yards
 r2 
TL  20log      r2  r1  x103
 r1 
General Form of the Absorption
Coefficient
2
Af r f
 2 2
fr  f
fr = relaxation frequency. It is the reciprocal of the relaxation
time. This is the time for a pressure shifted equilibrium to
return to 1/e of the final position when pressure is released
f = frequency of the sound
When f << fr,
Af

fr
2
Estimating Absorption Coefficient
• Viscosity – Classical Absorption - Stokes
162 2

f
3
3c
3
  s   v
4
4
  2.75x10 f
2
Shear and volume viscosity
For seawater, dB/m, f in kHz
Chemical Equilibrium
Magnesium Sulfate:
3
2
4
MgSO 4  H 2 O  Mg  SO  H 2 O
2
40f

2
4100  f
f in kHz
Boric Acid:
B  OH 3   OH   B  OH 4


2
.1f

2
1 f
f in kHz
Scattering
• Scattering from inhomogeneities in seawater
  0.003dB / kyd
• Other scattering from other sources must be
independently estimated
All lumped together as Transmission Loss Anomaly
Attenuation Summary
Note that below 10000Hz,
attenuation coefficient is
extremely small and can be
neglected,


TL   r  103 dB
where

0.1 f 2
40 f 2
 4 2  dB
   0.003 


2
.
75

10
f 
2
2
kyd
1 f
4100  f


Transmission Loss Equations
TL = 10 log R + 30 +  R + A
Range  1000 meters
Transmission Loss Anomaly
Absorption
Cylindrical Spreading
TL = 20 log R +  R + A
Spherical Spreading
Absorption
TLA
Range < 1000 meters