4
$\begingroup$

I dont understand the way we measure loudness. I know that it is in some way related to the magnitude of pressure variations. please help me figure out where my understanding fails. Let $p(t)$ be the pressure at a point at time $t$. The sound pressure level $L_P$, measured in $Db$, is defined to be \begin{equation}L_p=20\log_{10}(\dfrac{P}{P_0})\end{equation} where P is the RMS pressure, itself defined as \begin{equation}P=\sqrt{\dfrac{1}{T_2-T_1}\int^{T_2}_{T_1}p(t)^2dt}\end{equation}

The function of RMS pressure seems to be to provide an average pressure. Obviously it is not possible to merely use the mean pressure of a wave form because the contributions to the sum above and below the equilibrium pressure will largely cancel. But why not define the average to be \begin{equation}Q=\sqrt{\dfrac{1}{T_2-T_1}\int^{T_2}_{T_1}|p(t)|dt}\end{equation} This way all the contributions are positive, but it is a true average rather than being off by some fraction (due to the fact that the sum of the squares is not equal to the square of the sum). In trying to answer this question independently I came across this question, which is skeletally the same as mine but to do with voltages instead of pressure. In the answers to that question it is pointed out that the RMS of a varying AC voltage is the constant voltage a DC source must sustain to provide the same average power, a fact which itself is due to the relation that voltage is proportional to power squared.

Similarly, the intensity of a wave is proportional to the square of the pressure. The same justification (given in an answer to the question I linked) implies that a wave with constant pressure equal to the RMS pressure of a given wave would have the same average intensity (over some time interval).

In order for this to be a justification for $P$ (or its logarithm) measuring loudness, intensity must have some effect on loudness. If I'm correct so far, my question is now: what has intensity to do with percieved loudness? I understand the percieved loudness of a sound is a complex issue, affected by frequency response etc. But I imagine that the volume knob on a stereo controls a parameter which is loosely proportional to percieved loudness. Is that parameter $L_P?$ If so, why not $L_{Q}$?

$\endgroup$
3
$\begingroup$

The intensity depends on the average of the pressure squared so it is that average which must be used. Taking the squared bit out of the log bracket leaves you, as you have noted, with the rms pressure which is not the same as the mean of the magnitude of the pressure. The same is true in ac theory where the rms value is used because the power depends on the current squared.

In the end it boils down to the measurement of energy transfer per second and because that measure is proportional to the pressure squared it is that quantity that is averaged giving greater weight in the averaging process to the larger pressure variations.

To get the loudness you must factor in the sensitivity of the ear (response to different frequencies) to your $L_p$ which is a measure of the intensity of a sound wave relative to some chosen standard intensity. The process of factoring in the sensitivity of the ear is a complex process which is usually present as graphs of the type shown here. This is not of course the response of your ears but a "typical" ear for a person of a certain age.
So if you increase the intensity of a sound that is within the audible range of frequencies then the sound will become louder but that will not be so for frequencies outside that range.

$\endgroup$
  • $\begingroup$ Thanks for the answer. If it boils down the energy tranfer per second why isn't intensity used in place of sound pressure level? $P$ is the square root of a quantity proportional to intensity - why measure volume in this roundabout way instead of simply calculating intensity? $\endgroup$ – Lammey Feb 20 '16 at 0:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.