Ultrasound Attenuation: Why Power Ratio?

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    Attenuation Ultrasound
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Discussion Overview

The discussion revolves around the concept of ultrasound attenuation, specifically questioning why attenuation is expressed as a function of the logarithm of the power ratio rather than field quantities like voltage. Participants explore the physical basis for this relationship and the implications of using power versus voltage in calculations.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant asks why attenuation is a function of the log of the power ratio, seeking the physical basis for this relationship.
  • Another participant argues that attenuation is not a function of the power ratio but can be measured by it, emphasizing that attenuation depends on frequency and medium properties.
  • A participant clarifies that the logarithmic expression is often used to describe attenuation in decibels, noting that the ear detects sound logarithmically.
  • There is a discussion about the relationship between power and voltage, with one participant providing a mathematical expression that relates power to voltage and questioning why power is used instead of voltage directly.
  • Another participant suggests that the definition of using power is based on its usefulness and meaningfulness, indicating that wave effects depend on intensity, which is power per unit area.

Areas of Agreement / Disagreement

Participants express differing views on the nature of attenuation and the rationale behind using power ratios versus voltage. There is no consensus on the definitions or the underlying reasons for these choices.

Contextual Notes

Participants mention the need for logarithmic expressions in relation to decibels and the distinction between power and field quantities, but do not resolve the implications of these definitions or their applications in practice.

Who May Find This Useful

This discussion may be of interest to those studying ultrasound physics, wave propagation, or anyone looking to understand the principles of attenuation and its mathematical representation.

Karlos
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Hello, I have a simple physics question.

It is about ultrasound attenuation.

Why is the attenuation a function of the log of the POWER ratio? i.e. why must it be the ratio of power quantities rather than field quantities? what is the physical basis of this relationship?

thank you!
 
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Attenuation is not a function of the power ratio but can be measured by this ratio. Attenuation is a function of the frequency and properties of the medium.

I am not sure if you are taking about decibel scale or the coefficient of attenuation. They both may involve logarithms.
The effect of attenuation on amplitude (or power) of the wave can be usually described by an exponential decay as a function of distance.
P=P_o e^{- \alpha x}
where α is the coefficient of attenuation. If you want to express α as a function of the power, you will need to use a logarithm.
But α is a parameter describing the medium rather than the attenuation of a given beam of ultrasound.

To describe how much a given beam was attenuated in a given experiment you can use the ratio of the powers. Like .1 or 10%.
But more often people prefer to take the logarithm of the ratio and multiply it by 10 and give the answer in decibels. One reason is that the ear is a logarithmic detector rather a linear one.
 
Thank you for your reply nasu.

My question was more about the quantities used to compute the ratio itself.

If you are comparing an output voltage to an input voltage for a give value of x (distance) and a given medium you could write:

##10log\frac{P}{P_0} = 10log\frac{V^2}{V^2_0}=20log\frac{V}{V_0}##

Where P is the power and V is the voltage.

I want to understand why it is the POWER and not the voltage directly that you must input. If using the voltage directly you must either square it, or multiply the logarithm by 2.
 
OK, then you are asking why something is defined the way is defined.
And the answer can be only that it is useful or meaningful. It's not that it has to be this way.
You can think that the effects of the wave (sound or EM) depend on the wave's intensity, which is power per unit area.
Sounds with the same intensity may have different field quantities (pressure, displacement, etc).
 
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