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Hey guys quick(ish) question:

I produce music and am having a discussion on the difference between two RMS values in a specific record (Nero-- The Thrill (Porter Robinson Remix)). In the verse, the RMS is -15dB, while in the chorus, the RMS value is -7dB. I would like to find out the value (in terms of a ratio) between these two.

The question is how many times louder is the chorus than the verse.

I have used the equation:

$$20log\Big(\frac{-7}{-15}\Big)$$

Then changed for syntax purposes changed it to:

$$20log\Big(\frac{7}{15}\Big)$$

So finally:

$$20log\Big(\frac{7}{15}\Big)=-6.62$$

I reckon That means that the dynamic range in the chorus has collapsed by 6.62dB, is that correct?

Maybe I'm over doing this, and I really should just set 7:15 up as a ratio (which means it was just over 3dB, or an increase of 2x in sound intensity).

Anyways, I'm really sorry for the word/number salad, as you can see I'm pretty out of me element here!

It should be noted that I posted this topic on the physics site and they said it would be better here. One of the results I got was:

The dB scale is logarithmic, so when you have two dB levels, their difference is their ratio. Going from -15dB to -7dB is an 8 dB step. It's that simple. --Floris

Is this correct? Do the above equations have any relevance to the problem?

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  • \$\begingroup\$ It IS that simple. Which is what makes the dB scale so useful. You might remember from high school math that multiplication, in the log domain, becomes addition (therefore division becomes subtraction). \$\endgroup\$ Apr 17 '15 at 10:04
  • \$\begingroup\$ So I could have put it in as 20log(8) and gotten 18dB? \$\endgroup\$
    – EGreen
    Apr 17 '15 at 10:27
  • \$\begingroup\$ No. The ratio is simply 8 dB. If you need that as a fraction or decimal ratio, you`re looking for the antilog operation : antilog(8/20), or 10^(8/20), or 10^(0.4) \$\endgroup\$ Apr 17 '15 at 10:45
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The difference is their ratio and this can be shown by converting from dB to an equivalent but arbitrary voltage.

  • -15dB converts to 177.8mV
  • -7dB converts to 446.7 mV

The ratio in these units is 2.512 and reconverting back to dB is 8dB

The math you did here: -

20log(-7/-15)

Is not applicable to the decibel ratio at all and is basically nonesense.

The conclusion here: -

I reckon That means that the dynamic range in the chorus has collapsed by 6.62dB, is that correct?

Is based on a meaningless premise i.e. the formula you used above.

Measuring dynamic range requires looking at the RMS and also looking at the peak value, and, in the case of a modern piece of digitized music, the peak will be somewhere close to 0dB.

BTW I also produce music!

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  • \$\begingroup\$ Those values were in RMS. Would it be better to convert the bytes to integers and then perform the 20log equation as peak values (highest to, lowest (set to 0))? \$\endgroup\$
    – EGreen
    Apr 17 '15 at 8:44
  • \$\begingroup\$ @EGreen6120 I have no idea what you are driving at? What bytes? \$\endgroup\$
    – Andy aka
    Apr 17 '15 at 8:46
  • \$\begingroup\$ Converting the samples in the 16-bit WAV (-32768 to +32767) and then comparing their peak values... But I'm guessing that too is improper/unrelated. Essentially the ratio is 8dB because dB is ITSELF a ratio. And since we only have RMS and not, say, an area, we can't compare anything else, such as power \$\endgroup\$
    – EGreen
    Apr 17 '15 at 8:57
  • \$\begingroup\$ The power ratio is directly related ie one signal being 8dB higher implies a power ratio of nearly 9 to 1 but, if you explained what you are trying to achieve I might be able to help you more. \$\endgroup\$
    – Andy aka
    Apr 17 '15 at 14:25
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Is this correct?

Yes it is.

Do the above equations have any relevance to the problem?

That's up to you.

I would like to find out the value (in terms of a ratio) between these two.

The key thing to remember is that you are free to do whatever you want. The question is what your goal is. Why do you want to calculate that "ratio"?

Dividing both dB values sure is a ratio and calculating the log of that value is certainly possible. As you found out, subtracting both dB values from one another also yields a value that expresses a ratio.

Maybe you want to add 42 to everything, maybe not. You could apply any mathematical operator/-ion to these numbers. Maybe you are observing a phenomenon empirically and it's in fact easier to describe it in terms of dividing both dB values instead of subtracting them.


The RMS is an indicator how "loud" a sound is. However, humans are not linear devices. As such, it depends on the frequency how loud a sound is (you think it is).

Take a look at this chart for clarification: the horizontal axis is the frequency in Hz enter image description here This is from here: http://en.wikipedia.org/wiki/Equal-loudness_contour

You stated that you are having a discussion, so clearly it is about what you perceive as loud. Given the chart above you can see that even if both parts of the song were let's say -5 dB, you could perceive one as louder than the other if it was made up of different frequencies.

If you want to find a measure for loudness, you have to take this into consideration.

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  • \$\begingroup\$ Sorry, the original question was how many times louder is the chorus than the verse. The answer should be 8dB I suppose, since we don't have anything else to compare with such as an area (for power). \$\endgroup\$
    – EGreen
    Apr 17 '15 at 8:48
  • \$\begingroup\$ thank you for clarifying this =), I added some information regarding the loudness of a sound to my answer. \$\endgroup\$ Apr 17 '15 at 9:18
  • \$\begingroup\$ Yes very good! So even though there's an 8dB difference, we won't quite be able to say how many times louder the chorus is because loudness is perceived (vs voltage or pressure which are exact quantities)? And to know those, I'd need a defined area and voltage from amp to speakers or something, correct? \$\endgroup\$
    – EGreen
    Apr 17 '15 at 9:33
  • \$\begingroup\$ @EGreen6120 you can derive a model of the speaker and try to figure out what pressure there is, yes. But I don't think you know all the required parameters of all the components. If you want to know the pressure, just measure it with a sound level meter. If you want to know how loud a person perceives the measured sound, you have to find the frequencies of the sound (FFT) and apply the chart above. To compare the results with those of another sound, repeat the measurements with that sound. \$\endgroup\$ Apr 17 '15 at 10:02

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