Active Sonar Equation

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Active Sonar Equation
Adapting Passive Ideas
Passive Case:
LS/ N  SL  TL   NL  DI   DT
From our Sonar
TL  TL '  2TL
Target Strength describes the fraction
of energy reflecting back from the target
LS/ N  SL  2TL  TS   NL  DI   DT
Reverberation Noise
Reflections from non-target objects
is greater that noise.
Reverberation limited
RL  NL  DI
LS/ N  SL  2TL  TS  RL  DT
Active Sonar – Materials
• Typical piezoelectric materials
– Quartz
– PZT -Lead zirconate titanate
– Barium Titanate
Piezoelectricity
a
a+Da
Source Level for an Omnidirectional projector
SL  10 log
I 1 yd
I1 yd
I ref
p 2rms
Pwr


c Area at 1 yd
where Area at 1 yd  4 1 yd 


Pwr

2

4π
1
yd



  10 log Pwr c
SL  10 log
2
 p 2ref 
4 1 yd  p 2ref

c 

1000 kg/m  1500 m/s 
SL  10 log  Pwr   10 log
4 1 yd   0.9144 m
1x10
1 yd  
3
2
2
6
Pa 
2
 10 log  Pwr   171.5 dB
2
Electrical Efficiency
PwrElectric
Pwracoustic  PwrElectric  E
where E is the system efficiency thus:
SL  171.5 dB  10 log  PE  E 
Pwracoustic
SL  171.5 dB  10 log PE  10 log E
Efficiency may range from 20% to 70% for most sonar applications
Directional Arrays
DI T  10 log
I non directional
I directional
Principle of Reciprocity
b  ,  receiving  b  ,  Transmitting
SL  171.5 dB  10log PE 10log E  DIT
Transducer Sensitivity
How many dB for 1 volt input?
I
SV  10log  1V
 Iref

 p1V 

20log



p

 ref 
V2
PE 
R
SL  @1V   SV  171.5 10log R P  10log E  DI
Input impedance
Manufacturers typically advertise based on SV. To find SL, add 20logV.
Example
• Compute the source level for an circular piston
projector of diameter = 1 meter radiating 10 kW
acoustic power at a frequency of 15 kHz in water
SL  171.5 dB  10log PE 10log E  DIT
  1m  
 D 
DIT  10 log 
  29.94dB
  10 log 
  
 .1m 
2
Piston array:
2
SL  171.5 dB  10log104  29.94 dB  241.5 dB
Cavitation
Pressure Threshold
Power Threshold
pcav  patm  104 z
Pcav  S
pcav
(z in meters)
2
2 c
S = Tranducer Surface Area
SLcav  186  10log S  DI  20log 10  z 
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