Ultrasound Attenuation: Why Power Ratio?

In summary, the conversation discusses the relationship between ultrasound attenuation and the log of the power ratio. The coefficient of attenuation, α, is a parameter describing the medium rather than the attenuation of a given beam of ultrasound. The ratio of powers is commonly used to describe the attenuation of a given beam, but the use of logarithms and decibels is preferred due to the ear being a logarithmic detector. The use of power rather than voltage in this ratio can be explained by the fact that the effects of the wave depend on its intensity, which is power per unit area, rather than field quantities.
  • #1
Karlos
6
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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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  • #2
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.
[itex]P=P_o e^{- \alpha x} [/itex]
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.
 
  • #3
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.
 
  • #4
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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Related to Ultrasound Attenuation: Why Power Ratio?

1. What is ultrasound attenuation?

Ultrasound attenuation refers to the decrease in intensity of an ultrasound wave as it travels through a medium, such as tissue. This attenuation can be caused by a variety of factors, including absorption, scattering, and reflection.

2. Why is understanding ultrasound attenuation important?

Understanding ultrasound attenuation is important because it affects the quality and accuracy of ultrasound imaging. The amount of attenuation can provide information about the tissue properties, such as density and composition, and can help identify abnormalities or diseases.

3. What is power ratio in ultrasound attenuation?

Power ratio in ultrasound attenuation refers to the comparison of the intensity of the ultrasound wave after it passes through a medium to the intensity of the wave before it enters the medium. It is often represented in decibels (dB) and can be used to measure the amount of attenuation that occurs.

4. How does power ratio affect ultrasound imaging?

The power ratio in ultrasound attenuation directly affects the quality of ultrasound imaging. A higher power ratio means less attenuation and a stronger ultrasound signal, resulting in clearer and more detailed images. A lower power ratio indicates higher attenuation and can result in weaker and less detailed images.

5. What are some factors that can affect power ratio in ultrasound attenuation?

There are several factors that can affect the power ratio in ultrasound attenuation, including the frequency of the ultrasound wave, the distance the wave travels through the medium, and the properties of the medium itself (such as tissue density and composition). Additionally, the type of ultrasound transducer and the settings used can also impact the power ratio.

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