>Maybe I misunderstand the program then. IF we're subtracting two signals and they're say .1db apart, then if you're really subtracting the whole signal, then the result should be - 0.1db - correct?
Nope. If you subtract signals, you don't subtract the dBs. You subtract the relative voltage values they represent, each of which would be 10^([dB_Value]/20). Subtracting dBs gives you another dB value, but that equates to effectively *dividing* the signal amplitudes (assuming they are same frequency and in phase...another complication, being glossed over here). If I wanted to divide one signal by another, I could subtract their related dB values. But I needed to subtract one signal from another, so I subtract their *voltage* values, not their dB values.
>Should matter not what the output level was before. If it was 90db and you subtract 89.9db - you get 0.1 db. Start with 70 and subtract 69.9 you get 0.1db. Am I missing something here or is it just time for a beer?
No, the misunderstanding is just about how dBs and logarithms work. (But a beer might be a good idea, still.)
Don't know if you've been around long enough to remember slide rules, but they basically were able to do multiplication by adding logarithms of numbers. When you add quantities expressed using dB, you are actually in effect multiplying the relative scalar quantities they represent.
As a crude example, say you limited discussion to powers of ten. You could represent the number 100 by counting the zeroes to the left of the decimal-- use "2" to represent the number 100. The number 10,000 would be represented as "4" because there are 4 zeroes. To find what 10x10,000 is, you can then just add the "2" to the "4", getting "6" -- 100x10,000=1 million, which is 1,000,000, which has "6" zeroes. That's essentially how dB and logarithms work.
A voltage that is 6dB higher than another has twice the voltage - it is the same as multiplying by 2. If you add 6dB more to it, you are really doubling the previous voltage again, so while the addition gets you (in dBs) 12dB,you've actually quadrupled the original voltage. Adding 6dB to a dB value doubles the voltage it represents (when you convert from dB back to volts). Decibel notation is a trick used by engineers who have to multiply and divide numbers all over the place and need to be able to do it in their heads when handling complicated systems. Multiplying or dividing in your head is difficult, adding or subtracting are relatively easy.
If you have a voltage that is 0.1dB higher than another you are comparing to (the original value), that really just means that it is about 1.012 times the other, if we talk in voltage ratios instead of in dBs. If I subtract the original relative VOLTAGE value (1 times the original) from the new relative VOLTAGE value (1.012 times the original), then the difference is 0.012, which back again in dB is -38dB, compared to the original voltage value of 1. The 1, in dB, is "0dB". If I just subtract 0dB from 0.1dB, that's just means not changing the 0.1dB (same thing as multiplying by 1 =0dB).
So, "0dB" doesn't mean "nothing" -- it means no change in level from what you compare to. BTW, There is no way to express "nothing" in dB -- it would have to be -infinity dBs.
(Sorry for hitting you with the math lecture, I just find that most audiophiles don't really "get" decibels, which makes discussions about signals like this difficult)
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People aren't generally assumed to be able to identify sound level changes of around a dB as being a change in volume. Meaning, they could not likely adjust by ear one volume, to match another, within a dB. But they might experience such a small volume difference as a change in clarity or sound character.
