I want to convert db into amplitude ratio.

I did some research but I only found online convertor, and I don't understand what's on Wikipedia very well...

Here is a little of backstory :
I am a programmer and I have to code an Equalizer; the problem is, I'm new to this DSP world.
I'm using a small documentation that an audio engineer in my company made a while ago and there are stuffs that I don't understand in it. (He's obviously not in the company anymore...)

In his documentation, he writes that the equalization is made step by step with an incrementation of ±0.1 dB.

So here are my questions :

  1. What's the formula to convert dB into a percentage ?
    On Wikipedia, there is 10 log10(x), but x is the power ratio, and I don't know how to get that either...

  2. Does processing step by step is the best/only solution ?
    Or can I calculate the ratio that I need each time (to limit the number of measures needed) ?

Thanks for you time.

  • $\begingroup$ Figures do not appear. Ad. 1. Percentage of what? Use the anti-log. Ad. 2. Simply pre-calculate the LUT of ratios and gains. $\endgroup$
    – jojek
    Apr 24, 2015 at 14:56
  • $\begingroup$ Am i the only one who can see the figures ? And yes, I might pre-calculate a LUT, but i needed to know how in the first place. $\endgroup$
    – Khaz42
    Apr 24, 2015 at 15:28

2 Answers 2


You will want to work with what Wikipedia calls "field quantities."

Your engineer is telling you to work in steps of .1dB, which would be an amplitude ratio of .1=20log10(x). To get x you do 10^(.1/20) = 1.01158

From x dB to ratio (amplitude): 10^(x/20)

From x ratio (amplitude) to dB: 20Log10(x)

If you can calculate the ratio, then you should be able to calculate the correction needed, round to the next .1dB step, and apply it in one step.


Normally deci Bell is a ratio (Power or energy).
Power have a ratio of $10 \log \tfrac{p_1}{P_0}$ here is $P_0$ the reverence value and $P_1$ the measured value.
Voltage and digital systems have normally $20 \log \tfrac{v_1}{v_0}$. where $V_0$ is the reference value of the voltage and $V_1$ is also measured value.
In a digital system the reference value is the maximum value. Let say this value is 1. a 16 bit system is the smallest value $\tfrac{1}{2^16 -1} = \tfrac{1}{65535}$ (A lot of software like Logic works with 24 Bit signal so they have more space before clipping).

So $20 \log \tfrac{v_1}{v_0} = 20 \log \tfrac{{}^1/_{65535}}{1} = -96.3295$ dB.

1) prefer no percentage but just values or deciBells

2) When you everything calculated in deciBells you can just add or subtract values from each other. In that case only on the input or output calculate to the original values


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.