I Sound power, amplitude, frequency, and decibels

[ngn]

TL;DR

If time-averaged power of a sinusoidal wave is proportional to the square of the amplitude of the wave and the square of the angular frequency of the wave, then why do decibels measurements only require an amplitude?

Hello,
It has been difficult to find a clear answer to this question. I've found some sources stating that the power of a sound wave depends upon both amplitude and frequency. I've found other sources stating that the power of a sound only depends on amplitude. I've found sources stating that power only depends on amplitude (and not frequency) for electromagnetic waves, but with mechanical waves (e.g., sound) both amplitude and frequency are important. And, I've found sources stating that amplitude and frequency only affect power with light waves but with sound waves, only amplitude matters.

If both amplitude and frequency are important for calculating the power of a sound wave, why is it that when it comes to calculating decibels, frequency does not matter? For example, if a 500 Hz and a 1000 Hz pure tone both have the same RMS normalized amplitude measurement (e.g., 0.50), then they both have the same decibel measurement (20*log(0.50) = -6 dB FS). Similarly, if a 500 Hz and 1000 Hz tone both have the same sound pressure (e.g., 10 Pa), then they both have the same dB SPL (20*log(10/.000020) = 114 dB SPL). The fact that you calculate decibels on amplitude-based measurements only suggests that frequency is not important in calculating power. Is this correct?

 

[renormalize]

I've found some sources stating that the power of a sound wave depends upon both amplitude and frequency.

Can you cite your source for this statement?

 

[ngn]

Can you cite your source for this statement?

Here is one source:
https://pressbooks.online.ucf.edu/osuniversityphysics/chapter/16-4-energy-and-power-of-a-wave/

There is a description of how the power of a mechanic wave is proportional to the amplitude and angular frequency. What amplitude measure is being considered here? I can understand this on some level because if the amplitude of displacement is considered, then yes, if a higher frequency wave and a lower frequency wave both have the same amplitude of particle displacement then the higher frequency wave will be more powerful because the particles travel greater distance in the same amount of time. What I don't understand is why that is not considered when calculating decibels from typical measures of amplitude.

One reason I was thinking is that displacement may not be a root-power quantity (i.e., we don't typically calculate decibels from displacement amplitude), and decibels are only calculated on root-power quantities like sound pressure, and sound pressure "controls" for frequency because it's based on particle velocity, which has frequency built into the measure. In other words, if a high frequency and low frequency sound have the same sound pressure, then the displacement of the higher frequency sound must be less than the displacement of the lower frequency sound (because their particle velocities are the same at equal pressure). Does this also apply to calculating decibels from normalized measures in a WAV file? Are those voltages? And do voltages have frequency built into the amplitude measure as well? A simple way to think about these issues will be helpful.

 

[tech99]

Maybe use an EM wave on an infinite transmission line as an analogy? In this case, the voltage and current waves are in-phase, and the power is just I*V. With sound, the pressure and velocity are roughly analogous to voltage and current, so to find power we can multiply them, or alternatively, if we know the properties of the medium, we can find the power from one of them. This is like finding the transmission line power from V^2/Z or I^2*Z, where Z is the characteristic impedance. For sound, power = velocity^2 * Z or pressure^2/Z.
When the frequency is altered, for the electrical case V and I do not alter, and for the sound case, pressure and velocity do not alter. But if you consider a transmission line, at low frequencies more charge is given to the conductors during each half cycle, because current flows for greater duration. So charge is dependent on both voltage and inversely with frequency. The same applies to sound, where displacement will be greater at lower frequencies, because the velocity lasts longer during each half cycle.

 

[Drakkith]

If both amplitude and frequency are important for calculating the power of a sound wave, why is it that when it comes to calculating decibels, frequency does not matter? For example, if a 500 Hz and a 1000 Hz pure tone both have the same RMS normalized amplitude measurement (e.g., 0.50), then they both have the same decibel measurement (20*log(0.50) = -6 dB FS). Similarly, if a 500 Hz and 1000 Hz tone both have the same sound pressure (e.g., 10 Pa), then they both have the same dB SPL (20*log(10/.000020) = 114 dB SPL). The fact that you calculate decibels on amplitude-based measurements only suggests that frequency is not important in calculating power. Is this correct?

Are both of those calculating power? Or just a decibel ratio of the amplitudes and sound pressures?

What I don't understand is why that is not considered when calculating decibels from typical measures of amplitude.

How is the power of a sound wave in decibels calculated from the amplitude of the wave?

 

[tech99]

If you have an electrical version of a sound, then it has come from a microphone. The microphone responds to either pressure or velocity, so the electrical power is proportional to the sound power (actually the power density of the sound in W/m2). Relative sound power is proportional to voltage squared, or 20 log V/Vref.