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## How to decrease AAC (lav) audio volume by 40%?

Started by thoste, July 24, 2022, 09:38:06 AM

#### thoste

I have a *.mp4 video with an audio too loud.

So I want to keep the video encoding (H.2645) but decrease the volume level by 40%.

How can I achieve this?

I selected "AAC (lav)" from audio drop down and clicked on "Filters".
And then?

I tried Gain Mode = Manual (db)
Gain Value = -7

but that not a percentage value and requires a lot of trial and error encoding runs

No percentage mode possible?

#### dosdan

#1
Quote from: thoste on July 24, 2022, 09:38:06 AMNo percentage mode possible?

Be aware that the brain perceives volume changes logarithmically. An rotary analogue volume control is a potentiometer (variable resistor). You can have a linear-law pot or a log-law pot. If you were to drop to "50%" (half-way in the rotation) using a linear pot, you would notice the difference (-6dB) but it would not be that much, whereas with a log pot at 50% rotation it's -20dB which is perceived as half the volume of 100% rotation (0dB).

-20dB is 1/10th the voltage or 1/100th the power (since for the same load resistance, the power-increase/loss is the square of the voltage-increase/loss).

Standard audio stuff is specified in dBs rather than percentages. People use dBs rather than fractions or percentages as changes in volume/gain can be calculated with simple additions or subtractions rather than multiplications and divisions And most audio level meters show dBs.

Foe example:
-6dB is a 1/2x voltage or 1/4x power drop.  (+6dB is 2x voltage or 4x power boost)
-12dB is a 1/4x voltage or 1/16x power drop.
-18db is a 1/8x voltage or 1/64x power drop.

-3db is 0.707x voltage or 1/2x power drop.

So:
-9dB would be -6db -3dB which is 0.353x voltage or 1/8x power.

Much easier to comprehend volume/gain/attenuation changes in dBs than in a linear percentage drop in voltage which for -9dB is 35%.

Dan.

#### thoste

#2
Quote from: dosdan on July 24, 2022, 10:30:15 AMMuch easier to comprehend volume/gain/attenuation changes in dBs than in a linear percentage drop in voltage which for -9dB is 35%.

Hmm, I doubt that this is the case.
It is easier to think in percentage than in logarithmic scales.

You explained how to estimate db changes. I know this roughly before, but thank you.

But is there a way to apply changes in percent anyway?