If I double sound on my PC, how would the sound intensity increase?

[Lotto]

TL;DR Summary

If I double sound output on my PC, how would the sound intensity increase?

If I have on my PC let's say sound on 50% and I increase it on 100%, how would the sound intensity change? Would it be twice bigger? But I think that function of sound intensity level dependent on sound percets is a linear function, isn't it? How to explain it? Why is it the linear function?

 

[Drakkith]

I seriously doubt the increase is linear since human hearing isn't linear, but more closely approximates a logarithmic function. A sound that is twice as 'loud' as another in terms of perception is possibly 5x or 10x higher in terms of sound pressure intensity. In addition, hardware and software often have different filters and effects that affect the output in non-linear ways. AND, on top of that, frequency response of the ear isn't the same, so some frequencies need higher or lower intensities to have the same loudness, which your pc may or may not take into account in its volume control.

 

[Lotto]

I seriously doubt the increase is linear since human hearing isn't linear, but more closely approximates a logarithmic function. A sound that is twice as 'loud' as another in terms of perception is possibly 5x or 10x higher in terms of sound pressure intensity. In addition, hardware and software often have different filters and effects that affect the output in non-linear ways. AND, on top of that, frequency response of the ear isn't the same, so some frequencies need higher or lower intensities to have the same loudness, which your pc may or may not take into account in its volume control.

And how to write a function of sound intensity level dependent on sound percents? I have no idea how to do it.

 

[Lotto]

I seriously doubt the increase is linear since human hearing isn't linear, but more closely approximates a logarithmic function. A sound that is twice as 'loud' as another in terms of perception is possibly 5x or 10x higher in terms of sound pressure intensity. In addition, hardware and software often have different filters and effects that affect the output in non-linear ways. AND, on top of that, frequency response of the ear isn't the same, so some frequencies need higher or lower intensities to have the same loudness, which your pc may or may not take into account in its volume control.

And I made a measurement and according to the datas, it seems to be a linear function

 

[Drakkith]

And I made a measurement and according to the datas, it seems to be a linear function

What did you measure? What were the results?

And how to write a function of sound intensity level dependent on sound percents? I have no idea how to do it.

I suppose you would make volume percent as the x-axis, and whatever output you were measuring as the y-axis of a graph. That would give you x,y pairs you could turn into a function of some sort.

 

[Lotto]

What did you measure? What were the results?I suppose you would make volume percent as the x-axis, and whatever output you were measuring as the y-axis of a graph. That would give you x,y pairs you could turn into a function of some sort.

6,25% - 30,0 dB
12,5% - 40,0 dB
25% - 53,2 dB
50% - 65,0 dB
100% -73,2 dB

To me it seems as a linear function (the datas are not 100% correct).

 

[Drakkith]

It is certainly not linear, as decibels are a logarithm scale. Going from 30 to 60 db is more than a 1,000 fold increase in power and about a 300x increase in amplitude, but less than an 8x increase in your percentage. Besides, plotting your data points gives me what looks like a logarithmic function.

 

[Mister T]

If I have on my PC let's say sound on 50% and I increase it on 100%, how would the sound intensity change?

According to your measurements the loudness went up by 8.2 dB. Using the formula $\beta=10 \log(I_2/I_1)$, and $\beta$ equal to 8.2 dB, I get a ratio of intensities of about 6.6. So, to answer your question, the intensity increases by a factor of about 6.6.

And how to write a function of sound intensity level dependent on sound percents? I have no idea how to do it.

You don't need that function to answer your question.

 

[f95toli]

And how to write a function of sound intensity level dependent on sound percents? I have no idea how to do it.

Depends on what you mean. This can in principle be done if you were playing say white noise through the speaker by measuring using a calibrated sound pressure meter. However, if you mean perceived (=what we hear) sound intensity then that is nearly impossible since it depends on the frequency content of what you are playing (which in turn also depends on which speakers you are using etc). There are weighted dB scales that try to take this into account, but I don't believe they work very well for say music.

You can find plots (in 2D) of this in any good book about acoustics.

 

[Mister T]

Let's say you have a volume control knob that's marked in percents. Turn the knob through half of its maximum rotation and you have 50%. Turn the knob all the way to its maximum rotation and you have 100%. The same argument can apply to slider bars. To determine the relationship between the percent reading and the sound intensity you need to take measurements. Unless its provided in the technical specs.

 

[DaveC426913]

6,25% - 30,0 dB
12,5% - 40,0 dB
25% - 53,2 dB
50% - 65,0 dB
100% -73,2 dB

To me it seems as a linear function (the datas are not 100% correct).

You have no idea what those percentages really mean. There is no reason to assume they correlate directly with any particular corresponding output. The software control input goes through a lot of software processing and then through a lot of hardware processing before it gets to your speaker.

It is quite likely that some OS developer has written a function that anticipates the non-linearity of the input and produces some more subjectively-useful level of output.