Principles Of Sonar

Question 1

What is a Decibel (dB)?

Answer 1

1/10th of a bel (B)
these are two signals that have a power ratio of 10^(1/10)

Question 2

Why use Decibels (dB) in acoustics?

Answer 2

Used due to the very wide dynamic range of many systems
Sound levels can range over scales from 1 to 10^16

Question 3

Conversion to dB

Answer 3

Level in dB = 20log10(P_measured / P_reference)

Example for an underwater sound of 200μPa
= 20log10(200/1)
= 46 dB re 1μPa

Question 4

Why do we use logs?

Answer 4

They are good for dealing with really big numbers or really small numbers
Its for large dynamic ranges

Question 5

Sound propagation - Calculating Received level

Answer 5

Received level = Source Amplitude x Loss Factor

Question 6

What is the Sound level?

Answer 6

Its a theoretical value that appears to radiate from an ‘acoustic centre’ of a source when it is observed at a distance from the source in the far field independent of the environment.
Reference to 1m
i.e. dB re 1μPa-m
Not the same as received level

Question 7

Propagation equations in dB

Answer 7

RL = SL - TL
Received level = Source level - Transmission level

Question 8

What is Transmission loss?

Answer 8

AKA propagation loss
It is modelled in most sound fields
It depends on the physics of the environment as the sound energy travels through it

Question 9

How does sound radiate

Answer 9

Spherical spreading

Energy spreads over a larger and larger area

Square Wave law

Energy measured on the surface is lower the further away from the source you are

Question 10

What is Spherical Spreading

Answer 10

Intensity (I) α 1 / 4πR^2

In dB 20log10(R)

R is range in m from the source

Question 11

How can radiating sound be affected by boundaries?

Answer 11

Boundaries will be different medias
E.g. Surface or the seabed

–> Refraction - sound bends travelling through mediums
–> Reflection - sound strikes medium boundary in a way that it bounces back
–> Combination of both - some sound is refracted, some is reflected

Question 12

What is the free field?

Answer 12

The area of water where sound propagates without encountering boundaries or obstacles.
E.g not near the surface or seabed

Question 13

What is the surface and seabed like as a boundary?

Answer 13

Surface
–>It is very good at reflecting
–>Creates a large boundary as there is a large pressure differential between water and air.

Seabed
–> It might get some reflection and refraction

Question 14

What is the law of refraction in underwater acoustics?

Answer 14

Snell’s law

n1 / n2 = sin(θ1) / sin(θ2) = ρ1 x c1 / ρ2 x c2

where ρ x c is acoustic impedance
ρ is density in kgm^-3
c is sound velocity in ms^-1

Question 15

What is Cylindrical spreading?

Answer 15

When sound spreads from a source but is restricted between two boundaries. E.g. between the surface and seabed.

Similar to how fibre optics works

It has less geometrical losses than spherical spreading as energy is trapped between boundaries

Question 16

How do you calculate Transmission loss (TL) only considering geometric spreading?

Answer 16

TL = Nlog10(R) in dB

For Spherical
N = 20
TL = 20log10(R) in dB
Intensity (I ) α 1 / 4πR^2

For Cylindrical
N = 10
TL = 10log10(R) in dB
Intensity (I) α 1/2πRD

Question 17

What is Absorption / Attenuation (α) ?

Answer 17

Another loss effect

Absorption is different for given medias and different frequencies

Sound is absorbed by medium

It relates to a mediums molecules and elasticity

For absorption in salt water –> use the attenuation coefficients graphs to find the α value

Question 18

How do you calculate Transmission loss (TL) considering geometric spreading and Attenuation?

Answer 18

TL =Nlog10(R) + α(R) in dB

N is dependant on type of spreading
α is attenuation coefficient
R is range from source

Question 19

What are the limitations of geometric models?

Answer 19

Which N value do you use if there is interaction with both surface and bottom? The N value makes a huge difference

The range changes with the environment. E.g. Bathymetry –> Deeper, shallower, slopes, ground not flat

Is the signal broadband? Multiple frequencies involved? Which α value?

Question 20

What is the critical angle and total internal reflection?

Answer 20

Critical angle
When the angle of incidence of sound strikes a boundary and undergoes total internal reflection. It is the minimum angle of incidence required for total internal reflection

Total internal reflection
Complete reflection of a wave at the boundary between two mediums
Prisms work like this

Question 21

What is geometric dispersion?

Answer 21

Sound separates into components due to varying wave velocities or frequencies.

Question 22

What are thermoclimbs and how do they affect sound?

Answer 22

Thermoclimbs are the distinct layers/ zones in the ocean. Created from temperature changes

They can be treated as different mediums so there are multiple boundaries.
Sound rays will therefore bend dues to sound velocity and density with depth.

Question 23

Names results caused by refraction

Answer 23

Shadow zones
–> Waves refract toward the normal if objects are in the way, there are areas where its harder/ less waves reach that area
–> some areas the sound can’t reach at all which are shadow zones

Convergence zones
–> where multiple waves bend towards one point
–> from ‘constructive interference’ - most common paths taken
–> easier for energy to reach these areas
–> higher net energy in these areas

Question 24

What is the Lloyds mirror effect

Answer 24

Sound waves reflect off the surface and submerged objects. They interfere creating distinct interference patterns and spatial variations in received sound field.

Question 25

Monopole Vs Dipole sources

Answer 25

Monopole source
–> a sound source that radiates waves in all directions equally
–> similar to omnidirectional point source
–> spherical wavefronts propagate outward from source
–> E.g speaker

Dipole source
–> a sound source consisting of 2 sources even if one is virtual
–> the sources oscillate in opposite directions creating a dipole sound field
–> sound propagates primarily in perpendicular direction
–> E.g. fish

Question 26

What is the low frequency cut off?

Answer 26

Mode stripping
–> when lower frequencies cannot propagate well dues to the channel being too shallow

Low frequency cut off
–> the point where signals will not propagate well

Question 27

What is the Passive Sonar Equation?

Answer 27

SE = SL - TL - NL - DT

where
SE - Signal excess
amount of signal you have left. Ideally +ve, -ve means not enough signal, 0 means just about enough

SL - Source level
in dB re 1µPa-m (-m part makes it source level)

TL - Transmission loss
the one way propagation loss from the target in dB

NL - Noise level
in dB - e.g. ambient noise, ship noise
high noise, signal excess is lower

DT - Detection level
detection threshold in dB

Sometimes DI (Directivity Index) is included for directionality
- e.g. sound is lower when someone is talking in front of you compared to behind

Values are all in dB

Question 28

What is the Wenz curves ?

Answer 28

Aka Wenz spectrum

A graphical representation of the power spectral density of underwater ambient noise as a function of frequency

Tells you about the distribution and intensity of noise at different frequencies underwater.

They came from experiments looking at how noisy the ocean is

Question 29

What is the active sonar equation?

Answer 29

SE = SL + TS - 2TL - NL - DT

where
SE - Signal excess

SL - Source level in dB re 1µPa-m

TS - Target strength

2TL - 2x Transmission loss
accounts for the 2 journeys made, there and back

NL - Noise Level
ambient noise, ship noise, reverberation power summed

DT - Detection Threshold

All in dB

Question 30

What is Target Strength?

Answer 30

TS = 10log(Iscat / Iinc)

where
Iscat
- scattered intensity referenced to a distance of 1m from target centre
Iinc
- intensity of the transmitted signal incident on the target

if 0dB then everything sent out comes back
can be greater than 0dB

e.g. with divers their air cavities just look like balloons from the lung capacity on an echo

For a large rigid sphere, of radius a metres:

TS = 10log(a^2 / 4)
where a 2m radius sphere has 0dB TS, larger spheres have positive target strengths

Generally:

TS = 10log(σs / 4π)
where σs is the effective scattering cross section
e.g might not see an individual fish but you will seen many fish densely packed together in a shoal which is an effective target

Question 31

Near field Vs Far Field Vs Free Field

Answer 31

Near field
–> try to avoid
–> region close to source
–> field dominated by complex interactions
–> significant variations in pressure and particle velocity
–> individual parts can be detected
–> highly spatial and very wave dependant

Far field
–> source should only be calculated from here
–> sounds appear to come from same virtual infinitely small point
–> region father away from source
–> more stable
–> characterized by spherical wavefronts and decreasing intensity with distance following the inverse square law.

Free field
–> point where measured radiated energy only travels a ‘direct path’ from source to reciever
–> free of any multi paths
–> no reflections
–> free to radiate spherically

Question 32

What is the acoustic centre?

Answer 32

The reference point where waves are considered to originate from
Its a virtual point / theoretical value observed from the far field
usually referenced to 1m

Question 33

What is the Equal energy hypothesis?

Answer 33

all frequencies of sound have equal energy

loud sound –> short time
lower sound –> longer time

Both do same damage (Hearing damage)–> area under curve is the same