This Note is written for those users who wish to get a deeper understanding of acoustic backscattering and its use in the LISST-ABS. It expands on the very simple page, How LISST-ABS Works.

What follows is a description of coherent and incoherent scattering, strength of scattering of sound by particles of single size, extension to multiple sizes, and how to interpret the logarithmic response of the LISST-ABS.

Scattering of Sound by Particles: Because the signals that are recorded by LISST-ABS are pressure of scattered sound, not mean square pressure, the following concepts are important to understand. The key idea is coherent vs incoherent scattering. In coherent scattering, the scattered sound pressure from individual particles would add (i.e. if they were in phase). In nature, however, the particles have random phases, so total sound scattering is expressed as incoherent scattering. In this case, the energy scattered by individual particles adds, not pressure. Consequently also, the scattered pressure sensed by the acoustic transducer is the square-root of the energy (or mean square pressure), not just the sum of scattering from individual grains.

Basics: Consider scattering by particles in the acoustic pulse. The scattering from each particle has some phase  and amplitude , so that the scattered wave from each particle can be written as a cos(ωt- φ), with amplitude a , frequency ω and phase φ. The phase φ is 2πR/λ, with R being rage from the transducer.

If we consider scattering by two particles, when these two scattered waves reach our receiver, the pressures add as:

P = a₁ cos(ωt- φ₁ ) + a₂ cos(ωt- φ₂ )                                  (1)

Coherent Scattering: If the two phases are equal,

P = a₁ cos(ωt- φ) + a₂ cos(ωt- φ), or

=  [a₁ + a₂] cos(ωt- φ)                                                           (2)

The intensity in this case of coherent scattering, being square of amplitude is:

I  =  1/2[a₁ + a₂]²  ,

[the factor 1/2 comes from average of the square of cos(ωt- φ) being 1/2.],  or

I  =  1/2[a₁² + a₂² + 2 a₁ . a₂]

It is easier if the two amplitudes are equal, then, the intensity is:

I  = 2 a².                                                                                 (3)

Eq.(3) shows that two particles add up to give 4 times the energy of the individual particles, 1/2a², because they are in phase.

[Note: this does not mean that two particles scattering coherently find new energy. Waves scattered by two particles in different directions will go in and out of phase, producing interference ‘fringes’, whose intensity varies from twice the average when phase difference is 0, to an intensity of 0 when phase difference is p!.]

Incoherent Scattering: For the case of scattering by real particles in nature, the two phases are not equal,

P = a₁ cos(ωt- φ₁ ) + a₂ cos(ωt- φ₂ )

I  = [ a₁ cos(ωt- φ₁ ) + a₂ cos(ωt- φ₂ )]²

which is

I  = 1/2[a₁² + a₂²] + 2 ʃa₁ . a₂ cos(ωt- φ­₁ ) . cos(ωt- φ₂ )

or,

I  = 1/2[a₁² + a₂²] + 2 ʃa₁ . a₂ [cos(ωt-( φ₁+ φ₂)) + cos(φ­₁ – φ­₂ )]

so that, when you take the time average, the first term in the cross-product cos(ωt -( φ₁+ φ₂)) goes to zero, and the second term remains:

I  = 1/2[a₁² + a₂² + 2 a₁ . a₂  cos(φ₁ – φ₂ )]                    (4)

For a suspension of particles with random phases f, the last terms just add with random signs and cancel out, leaving

I  = 1/2[a₁² + a₂²];

or, if the two amplitudes are equal to a,

I  = a²                                                                                     (5)

In other words, in the case of incoherent scattering, the total intensity is the sum of individual intensities. Recall, the coherent case gave us 2 times higher intensity, Eq.(3).

Strength of Acoustic Scattering Signal: Because the scattering from particles adds incoherently, the mean-square pressure (‘intensity’) from individual particles add, per Eq(5). Write this as:

P² = a₁² + a₂²+ a₃² + a₄²……..;

If all particles are of the same size, and there are N of them,

P² = N a²

Single Size Particle Scattering: The pressure (not intensity) scattered by a single particle is given by:

p  = f {m/Dρ}^1/2,

where we use lower case p for pressure, f is a ‘form factor’ (analogous to scattering efficiency in optics), D is grain diameter, ρ is grain density, and m for mass of a single particle, so that, for N number of particles, the intensity, which is the sum from Eq(5), becomes:

P² = N  f² {m/Dρ};

and combining Nm into mass concentration M, we have the simple result:

P² =  f² {M/Dρ};                                                                     (6)

Single Size Particles: If particles of a single size only are present, from Eq.(6), the root-mean-square(rms) pressure P is:

P =  f {M/Dρ}^1/2;                                                                   (7)

This implies that the pressure signal will increase as square-root of concentration. In a way, this is wonderful because it compresses the dynamic range of concentrations. An increase in concentration by a factor of 10⁶ will increase pressure by only 10³.

Particle Size Distributions:  If particles of different sizes are present, then incoherent scattering means that intensities from all different sizes add

P² = ∑ _i  f_i²{M_i/D_i ρ};                                                         (8)

This is weird. If we define a weight factor w_i

w_i  = f_i²/D_iρ;

then Eq. (8) becomes:

P² = ∑ _i w_i M_i                                                                       (9)

and

P_rms =  [∑ _i  w_i M_i ]^1/2                                                    (10)

In other words, when multiple size particles are present, the rms signal that our transducer sees is the square root of the weighted sum of Eq.(9).

This gives us a way to think about the relative importance of different sizes, represented by the weight factors w. The weight factors of Eq.(10) are shown below.

Fig. 1: Weight factors w_i for different size particles, an example PSD(red), and the contribution of all sizes to the total scattering (black). The total scattering will be the square-root of area under the black curve.

The Really Weird Part – Non-linear addition from multiple size classes:  Think of two experiments. In one, particles of single size are used. The signal strength will follow Eq.(7). The relative signals from different single-size particles will have the shape of the green curve.  When a distribution of sizes is present (red curve), the different size classes will be weighted in proportion to the green curve. Now, the small particles become even less significant!

In other words, if 2 different sizes are present, one large and one small, the pressures from the two won’t simply add. Intensities will. The sensed backscatter pressure will be the square-root of the sum of the two intensities. If a distribution of sizes is present (red curve) then the total intensity will be the sum of product of red and green curves. The weight factors are such that (green curve) the small particles make a smaller contribution to the total intensity. Now, since the stored rms pressure signal is square-root of the sum, it is hard for intuition to grasp this. Here’s an example that helps. Let there be two sizes of particles, size 1 and 2, such that w₁M₁ > w₂M₂. Then the total back-scattered pressure will be:

P_rms = [ w₁M₁ + w₂M₂]^1/2.

which can be rewritten as:

P_rms = (w₁M₁)^1/2 [ 1 + w₂M₂/ w₁M₁]^1/2                       (10a)

and since the second term is smaller than unity, further simplify as:

P_rms = (w₁M₁)^1/2  [ 1 + 1/2 w₂M₂/ w₁M₁ +…]                (10b)

That is, the weaker scatterers appear only half as important as their already lower scattering intensity!

LOGARITHMIC DATA

LISST-ABS reports analog/digital data proportional to the log of the acoustic backscattering signal. When two distinct particles are present in a suspension, the questions becomes: will the output be the sum of the output for the two components separately. The answer is no. Here’s why:

When the data are stored as log of signal strength, the output voltage will be, from Eq.(10a):

V = log{(w₁M₁)^1/2 [ 1 + w₂M₂/ w₁M₁]^1/2}                       (11)

which is,

V = 1/2 log(w₁M₁) + 1/2 log [ 1 + w₂M₂/ w₁M₁]                  (12)

In other words, the second component is modified; instead of being simply log(w₂M₂) the total is different. The two log outputs do not add (because, of course, that represents the log of the product of w₂M₂ and w₁M₁). In this regard, acoustics differ from optical sensors.  An adjustment of intuition is required!

Summary

The following points are worth remembering:

• Acoustic scattering is incoherent;
• The pressure at the transducer is the square-root of sum from all component sizes;
• The instrument senses and records the log of the rms pressure;
• The digital counts of data increase or decrease with log of concentration;
• The sensed signal strength for different sizes is not the sum of those for individual sizes. (see Eq.(12)).

Other Links: 

How LISST-ABS works

Sensitivity of Response of LISST-ABS and OBS to Size Distribution Changes.