Ultrasound Attenuation: Why Power Ratio?

[Karlos]

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!

 

[nasu]

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.

 

[Karlos]

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.

 

[nasu]

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).