L03 & L04

Question 1

In which frequency region should hydrophones operate?

Answer 1

Below their resonance frequency, eg. TC 4014 less than 400kHz because of the
Flat frequency response in that region.

Question 2

How does is an underwater sound measuring system exemplary constructed and why?
Why is the bandwidth of a signal important to determine when measuring uw sound signals?

Answer 2

1. A hydrophone measures incoming pressure and converts ot into voltage with a negative gain.
2. A preamplifier compensates the negative gain partially .
3. A BPF cuts unwanted frequency from the received signal with a negative Gain (-70-0dB)
4. A preamplifier once more compensates the loss partially with 0-50dB.
5. AD converter makes the analog signal ready for digital signal processing.

• the bandwidth is important to determine for the identification or filtering of signals from different objects like marine mammals or animals

Question 3

What is the power spectral density used for? Why do we need the bandwidth for that?
What is the spectral Level?

Answer 3

To identify dominant frequencies in the received signal.
The properties of sound source can be determined that way.
It’s calculated by the Normalized Fourier Transform.

The spectral level is an measure of power of a singal at a certain frequency.
It helps identifying different features measured in the water.

Question 4

What is a good power spectral density of wind to take into account for the construction of measuring systems uw?

Answer 4

The area between
50dB_re1mikroPa^2/Hz at
10Hz
and
15dB_re1mikroPa/Hz at
100kHz
is suitable, because there we have
sea state 0 considering a
low wind, wave and splash condition as
lowest noise level measurable under water.

Question 5

Does Ship traffic have a higher impact on noise measurable under water than the sea state?

Answer 5

Yes, with heavy ship traffic we can measure more than 90dBre1mikroPa^2/Hz.
In sea State 7 we can have up to appr. 75dBre1mikroPa^2/Hz.
(Under stormy conditions)

Question 6

What is the transmission loss?
How is the Transmission Loss determined?
How is it determined in a spherical wave?

Answer 6

TL is the decay of sound intensity I(r) compared to I(ro) at reference distance:

TL = 10 log_10 ( I(ro)/I(r) )
[20 log_10 ( p_ampl(r_0) / p_ampl(r) )]

It is always extrapolated from far field to reference distance r_0 =1m (too much turbulence there)

Spherical: r»1m
I(r) = Phi_0^2*w^2 / (2cr^2)
-> decreases with r^2!

TL: 20log_10(r/1m)

Question 7

How is the sound power of a spherical wave determined?

Answer 7

P = 4pi* r^2 * I(r)
I(r) = p_rms^2 / rho_0*c

Question 8

What is the difference of the sound pressure level and the source level SL @ distance r?
Is it defined by the amplitudes?

Answer 8

The SPL takes the squared measured power and reference power, the
SL doesn’t:

SL = 20* log_10 ( p(ro)/p(r) ) (p_tilde)

No, it is defined by the rms values = p_amp*(1/sqrt(2))

Question 9

Which two principles are important concerning reflection and transmission at interfaces?
What condition must hold for that?

Answer 9

1. The continuity of pressure, when there is no net force:
   (p_i + p_r) exp(jwt) = p_t * exp(jwt)

k_1 = w/c_1 and k_2 varies with different c_2

2. Continuity of Particle velocity v (fluids remain in contact)

v_i + v_r = v_t

The boundary of the interface must not move for that condition.
(x=0)

Question 10

@ normal incidence:
How is the Pressure Reflection Coefficient calculated?
Which cases are there for the pressure reflection coefficient?

Answer 10

R = p_Amp,r / p_Amp,i
= ((Z2/Z1) - 1) / ((Z2/Z1) + 1)

1. Z2/Z1 -> inf. -> R = 1
   rigid boundary,
   p_xo = 2 * p_Amp,i
2. Z2/Z1 = 1 -> R = 0
   Media acoustically same, no reflection
   eg used for shielded sonar
3. Z2/Z1 -> 0 -> R = -1
   Pressure release boundary!
   p_Amp,xo = 0 (particles move,
   eg. Water -> Air)

Question 11

Multiple interfaces:
How is the Transmission index simplified to calculate?
x=L=Thickness of the wall
For which frequencies does it hold?

Answer 11

1. Z1 = Z3
2. Z2/Z1 * sin(k2L)&raquo_space; 2
3. k2L«1 (low freq. approx.)

eg. Lambda = 1m (330Hz), then L = 1m is too thin to approximate this

-> Intensity transmission coefficient:
T_i = 2/(k_2*L) * Z1/Z2
-> for low frequencies!!!

Question 12

How is the wave bended at oblique incidence?

Answer 12

After snells law the
-> sine of the incidence angle by c1 has to be equal to the
-> sine of the transmission angle by c2

-> at the same frequency the speed of sound changes @interface and the
-> angle is adapted by an adapted k vector

Question 13

When does total reflection occur?

Answer 13

Observable when a wave travels from a medium with higher speed into a medium with lower speed (water -> air)
At oblique incidence @interface:
Beyond the critical angle the wave is not transmitted but completely reflected.

Critical angle: Theta_c

Sin(Theta_c) = c1/c2

Question 14

What does Berger’s mass law tell about the sound reduction index?

Answer 14

The Sound reduction at the interface between two media is higher when increasing the mass per unit area:

Red = 20 log_10 ( wm_pua / 2rhoc )

-> the higher the frequency the less mass is necessary for a spec. Reduction

Question 15

How does the Sound Reduction index vary with the applied frequency?
What is the critical frequency?

Answer 15

R is increased by 20dB per Dec if f <fc

R is 0 if f=fc (acoustically transparent media)

R is increased by 60dB per Dec if f>fc

The critical frequency is
f at which the second media are
transparent for the sound.

Question 16

What’s special at sound transmission through elastic media (solids)?

Answer 16

They store energy due to their ability to be compressed.

Important is Hook’s law:
stress = young’s modulus * strain

Here we use the elasticity module (N/m^2) and the poisson ratio (dimensionless), which determines the spatial expansion perpendicular to the compressing force.