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channel mixer math

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David Smithchannel mixer math
by on Feb 6, 2008 at 11:53:39 pm

I can't figure out how the math of the Channel Mixer works and it's really annoying me.

For example, let's say we have a .WAV with the following parameters:

L channel = -10 dB
R channel = -9 dB

If you go into the channel mixer and set the new left channel sliders to contain equal parts of the left and right channels (by putting 50 in each field), the new amplitude is 9.5dB. That makes sense because it's an average of the two channels.

But if we drop the left channel in our original example so that:

L channel = -20 dB
R channel = -9 dB

and apply the EXACT same settings, the new left channel becomes roughly...-12.8 dB. Huh? Shouldn't the new average be -14.5 dB?

Does anyone know how this thing works??

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Emmett AndrewsRe: channel mixer math
by on Feb 11, 2008 at 12:31:08 am

The channel mixer uses percentage inputs, not decibel inputs.

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