2
\$\begingroup\$

Suppose I listen to some music on my headphones. Now change the amplitude (in volts, as measured e.g. with an oscilloscope) of the audio signal by a constant factor G.

Assuming that the amplifier in question is linear in the region we are talking about, how does this change the audio signal's power in dB, the loudness in dB(A) and the sound pressure level in dB?

\$\endgroup\$
  • \$\begingroup\$ Decibels are a power measurement, so the real question is how is power related to voltage? Think about it. \$\endgroup\$ – Dave Tweed Aug 25 '17 at 13:02
  • 2
    \$\begingroup\$ A general rule add 3 dB each time you double the power. If you double the voltage into a constant resistance (a rough approximation of a speaker) you end up getting 4 times the power (P=I*V and I=V/R). So as a first approximation you will get 6 dB more for a voltage doubling. However I'm not sure is dB(A) and sound pressure measurements have some other oddities that need to be taken into account. \$\endgroup\$ – Andrew Aug 25 '17 at 13:06
2
\$\begingroup\$

If the voltage is increased by G then the signal has increased by: -

\$20\cdot log_{10}(G)\$ decibels.

This would also mean the same decibel increase in SPL. However, for "A" weighting you should really take account of the actual SPL level before making comparisons. A quote from wiki says this: -

A-weighting is only really valid for relatively quiet sounds and for pure tones as it is based on the 40-phon Fletcher–Munson curves which represented an early determination of the equal-loudness contour for human hearing.

The reason is due to non-linearities in the ear and is typicfied by the above-mentioned fletcher Munson curve: -

enter image description here

But history has overtaken the FM curves and the equal loudness contours (shown in red above) are a better fit for what the ear does.

So, if your question was more aligned to understanding loudness as perceived by the ear, you must take into account the actual sound level and apply the more appropriate equal loudness curve.

EDIT to service extra question and comment

If a signal voltage rises by 6 dB then the power also increases by 6 dB and the sound pressure level also increases by 6 dB. It so happens that to increase the voltage by 6 dB you must double it and if the power increases by 6 dB it is quadrupled. This is because power is proportional to voltage squared hence a doubling of voltage (or pressure) results in four times the power.

So we often say a doubling of power is 3 dB and if it is doubled twice (i.e. made 4x bigger) it increases by 6 dB. Note that 6 dB is a slight approximation to the real figue i.e.: -

\$20\cdot log_{10}(2)\$ = more closely 6.0205 dB and

\$10\cdot log_{10}(4)\$ = the same 6.0205 dB

This is because \$log_{10}(x^2)\$ = \$2\cdot log_{10}(x)\$

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thanks for the good explanation of loudness curves. However, could you elaborate the first sentence in the answer? I assume, you mean that the increase in dB applies to the sound power, as explained by @Andrew in the comment below the question? Is that where the factor of 20 (instead of 10) comes from? \$\endgroup\$ – Fritz Aug 25 '17 at 13:27
  • \$\begingroup\$ PS: Edited question to explicitly reference "power" for clarification. \$\endgroup\$ – Fritz Aug 25 '17 at 13:29
  • \$\begingroup\$ I've also edited my answer - don't be surprised about how simple dB in power is the same as dB in voltage! \$\endgroup\$ – Andy aka Aug 25 '17 at 13:36
0
\$\begingroup\$

If we assume everything stays linear, not just the amplifier but also the headphones, and your ears, then the formula for a dB change based on voltage ratio G is \$20log_{10}(G)\$

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Neil, I have log ears. Are yours are linear? You should probably delete the ears bit. \$\endgroup\$ – Transistor Aug 25 '17 at 13:25
  • \$\begingroup\$ @Transistor Interesting comment, I've thought about that for a bit. My ears, eyes, spectrum analyser, power meter, all seem to be best interpretted in log terms, for dynamic range of simultaneous signals, and all have a noise floor, and a max signal level above which 'things go wrong'. For a single signal, in the middle portion of the range, I don't see a problem with talking about linearity, meaning in this case freedom from disotrtion (hence 'stays linear'), partciularly due to over-level. Neither of know how our ears work, so I'll let it stand, as a debating point. \$\endgroup\$ – Neil_UK Aug 25 '17 at 13:43

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.