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Old 13th April 2007, 01:19 AM   #1
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Default Watts vs. decibels

I have oft wondered about this one.

I have been told that each increase of 3 dB requires 2X amplifier input.

I have also been told that a doubling of apparent volume is an increase of 10dB and requires and increase of 10X from the amplifier.

So translate that to my 96 dB PA system. In one instance it requires 1000 watts to achieve a 126 dB rating whereas with the other it takes 1024 watts.

It's not a big deal, I'd just like to know which is correct. I have a feeling the 10X is correct because it all scaled logirithmically but I'd like to know for sure.

Thanks.
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Old 13th April 2007, 01:45 AM   #2
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Cal,

innit? Apples and pears, that's the problem. You double the amp output using the same signal in one case. You need ten machines (or motorbikes, or whatever) with different signal outputs to achieve the same nuisance in the other. Add the fact that noise is logarithmic and you get as confused as I usually am.

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Old 13th April 2007, 01:49 AM   #3
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In short, 6db is doubling of gain.

For loudspeaker it's half, 3db.

The math is here: http://www.ac6v.com/db.htm

Edit: I suck at math. I take the 3 and 6db at faith. I struggled like an idiot with Log in high-school. Poobah, I think, said something about db should be called Log. Something such anyway. Suddenly things made sense, only years too late. Maybe I hadn't hated math if I had had a better teacher. But then, maybe I would.
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Old 13th April 2007, 01:54 AM   #4
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phn,

if I am not completely mistaken we have ~the same local time - another nightbird?

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Old 13th April 2007, 02:01 AM   #5
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I believe the +3dB->2x-power is an approximation of the other equation. Easier for most to comprehend.
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Old 13th April 2007, 02:02 AM   #6
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Indeed. I'm something of an insomnic. It's related to stress. I'm pretty sure of that. Not as bad as for my brother. He just returned to his job, different job, after a month. Doctor's order.
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Old 13th April 2007, 02:03 AM   #7
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Default Re: Watts vs. decibels

Cal,

Twice the power is +3dB

Twice the voltage (four times the power) is +6dB

Twice the loudness is (unfortunately) anywhere between +6dB (4 time the power) and plus +10dB (approximately 10 times the power). I think this depends on the frequency and loudness of the original signal. Check out "Equal Loudness Contours" or "Fletcher–Munson curves", or A-weighting vs B-weighting vs C-weighting.

On another forum, the question was posed "how much louder is the latest 600w amplifer than my 13w SET?" (close, anyway). The answer given was 16.5dB, or about 3 times as loud.

And, assume you listen to 2 1KHz tones, one at 30dB, and the other at 80dB. A difference of 50dB.
Now, listen to 2 40Hz tones at the same 'apparent loudness' as the 2 1KHz tones. A difference of 50Db, right? Wrong! You would only need to increase the 40Hz tone just over 25dB...
To hear the 40Hz tone at the same apparent loudness as the 30dB 1KHz would be 69dB; to hear the 40Hz tone at the same apparent loudness as the 1KHz tone at 80dB, would need to be 95dB.
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Old 13th April 2007, 02:23 AM   #8
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Default Re: Re: Watts vs. decibels

Quote:
Originally posted by bumblebee
I believe the +3dB->2x-power is an approximation of the other equation. Easier for most to comprehend.
Thanks bumblebee, I was wondering about that.

Quote:
Originally posted by Cloth Ears
Twice the loudness is
Thanks for that also CE but I was really wondering about the 10/10 ratio vs. the 2X=3dB not the "apparent loundness". Sorry I mislead you.
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Old 13th April 2007, 02:42 AM   #9
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Quote:
Originally posted by bumblebee
I believe the +3dB->2x-power is an approximation of the other equation. Easier for most to comprehend.
Uh. It's about +3.0103dB, or 10*log(2) for 2 times and +10dB for 10 times.
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Old 13th April 2007, 03:06 AM   #10
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Question officially answered. Thanks guys.
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