Balance root-mean-square in audio clips

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Tahariet Sharon 2017-11-24

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编辑： Nuchto 2017-11-27

I read in a paper that audio fragments were energy balanced "so that the root-mean-square power value was equal across clips". How does one balance the RMS power value across audio clips? Thanks.

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Tahariet Sharon 2017-11-25

编辑：Tahariet Sharon 2017-11-25

I tried your approach but it doesn't seem to maintain the ratio between left and right. For instance, the RMS= 0.1334 0.1265 (L and R), so the sqrt(rms(L)^2 + rms®^2) = 0.1838. When I plug this into the denominator to divide both L and R, the rms of this resulting quantity is 0.7256 0.0057.

David Goodmanson 2017-11-25

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Something must have gone wrong, because it does work

% make two channels with rms value .1334, .1265
a = rand(1,1000);
b = rand(1,1000);
a = .1334*a/rms(a);
b = .1265*b/rms(b);
rms_ab = [rms(a) rms(b)]
% normalize
anorm = a/sqrt(rms(a)^2 + rms(b)^2);
bnorm = b/sqrt(rms(a)^2 + rms(b)^2);
rmsnorm_ab = [rms(anorm) rms(bnorm)]
% same ratio for these two
rat = [rms(a)/rms(b)  rms(anorm)/rms(bnorm)]

I shall comment on sum of squares, but right now lunch is calling.

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Answer 1

David Goodmanson 2017-11-26

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Hi Tahariet,

You would like to normalize R and L so that the sum of R power and L power does not change from clip to clip. The ratio between R power and L power stays the same before and after normalization. Calling the channels a and b rather than R and L, then

anorm = a / sqrt(rms(a)^2 + rms(b)^2);
bnorm = b / sqrt(rms(a)^2 + rms(b)^2);

This works because, first, although people say it all the time (including in your question), the phrase "rms power" is a misnomer. rms is an amplitude.

In terms of voltage, assuming a signal with mean 0 for simplicity (and dropping the factor of 1/R since keeping it around does not change the conclusion), the average power is P = mean(V.^2). Root mean square = sqrt(mean(V.^2)) is linear in V, so it's an amplitude. And

P = rms^2.

You want to normalize so that Panorm + Pbnorm is the same from clip to clip, so

Panorm = Pa/(Pa+Pb)
Pbnorm = Pb/(Pa+Pb)

When you take the square root of both sides and use P = rms^2 everywhere, then you get the anorm and bnorm expressions.

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David Goodmanson 2017-11-27

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I see what you mean. Fortunately there is a remedy.

The normalization rescaled every clip so that its total power (Panorm + Pbnorm) = 1 as implied by the last two equations in the answer above. Taking the square root to get { the rms value that represents the total power }, then the rms value = 1 for normalized clips. The total power of every normalized clip is identical, which was the goal. But the rms value of 1 is too large and leads to clipping as you saw.

All that has to be done is to rescale all of the normalized clips by the same factor to bring them down to the right level.

For a given clip the rms value is

rms_clip = sqrt(rms(a)^2 + rms(b)^2).

In the example you showed,

rms_clip = sqrt(rms(a)^2+rms(b)^2) = .1838.

Presumably the rms value of all the other clips is in the same neighborhood. By surveying all the clips you can come up with an average value rms_clip_ave. One single number for the entire set of clips. Let's say it's somewhere around 0.2 (the normalized rms value is five times this, so no wonder there was clipping). Then the process is the same as before, only with the same constant factor in front of all of them:

anew = rms_clip_ave * (a / sqrt(rms(a)^2 + rms(b)^2));
bnew = rms_clip_ave * (b / sqrt(rms(a)^2 + rms(b)^2));

If there were still any clipping I would think it would not be by much, and you could multiply all the clips by a common factor not too much less than 1 to get rid of it.

p.s. I probably doesn't matter a whole lot if you find rms_clip_ave just by eyeballing all of the individual rms_clips, but it shouldn't be too surprising that for N clips the most formal definition is

rms_clip_ave = sqrt( (rms_clip_1^2 + rms_clip_2^2 + ... rms_clip_N^2) / N)

David Goodmanson 2017-11-27

Hi Nuchto, Well it does not quite make sense. In the expression for r there should be no inner sqrt function. One sqrt on the outside, as in the expression for rms_clip_ave, that's it. Not that the result is going to agree with .156, but at least it will be smaller. The expression for r is not a confidence builder so I hope that in the main calculations you will go back and make sure that there are no extra sqrts floating around. I hope your experiment turns out well.

Nuchto 2017-11-27