Negative dB numbers

Jack Hammer

New member
I do not understand negative dB numbers. What does this really mean. Obviously, there is no negative sound, there is some level, sme amplitude, some volume, so that must correlate to positive something. So how come negative measurements are used or should I say negative designtions

I know this may seem like a dumb question but I am new to engineering. Sure, I have played an instrument for years and, yes, I have been in many, many studios but...it seems that it is taken for granted that you know certain things and possibly smoe people are afraid to ask obvious questions becuase they do not want ot sound dumb or inexperienced. God forbid your not the coolest engineer in town! But I want to know what the story is with these negative designations on the threshold contorl of a compressor for instance.

So, what is the story with negative dB threshold amonts.
 
Well... it's all a relative scale for different references....

If you're talking about digital Full-Scale, then 0db is where digital clipping occurs (which is to be avoided).... of course if 0dbFS is the top, then anything level less than that has to be a negative db value....

It all depends on what meter scale you're looking at, at any one time. For analog meters it can be different again....
 
The simple explanation is that 0db is the best quality signal. Anything above that is subject to distortion and anything below that is subject to lower signal to noise ratio.
 
If that didn't make sense figure that with digital, like BB said, 0db is the strongest possible signal. So if 0 is the loudest possible signal than you can only go down from there. I believe an increase of 6db is actually a 2x increase in strength. (I'm probably wrong on that number so please correct me anybody).

It would be too hard to measure from quiet on up to loud because how do you determine how quiet something is unless you compare it to something? It is easier to state a known maximum level, 0db, and go down from there.

Analog audio does have positive db values and the quality of equipment is often determined by how far it can go above 0db and still sound great.
 
what I still don't quite get is how come 24 bit audio or 8bit audio, or 64bit audio etc...all have the same maximum volume.

but, i am *much* closer to understanding the whole thing then i was before joining this board a year ago and having BB and Tex explain this same question to me...don't lose sleep over it anymore at least ;)
 
What may be causing the confusion is that the term dB is often misused to describe a number of different measurements.

By itself, dB is only an expression of a ratio, not an absolute measurement. As such, it is a way of comparing two sounds, but does not represent any absolute value by itself.

When dB is attached to a reference, then it actually has a specific value. The standard references differ, depending on application, and are represented by an additional letter added to dB, like dBV, dBW, dBm, dBu, etc.

If you can find it, there is an excellent discussion of dB in all it's forms in the December 2000 issue of Pro Audio Review in an article by Edward J. Foster. Not being very technical myself, I keep it handy for reference.
 
dB is relative. +6db is a higher level, and -6db is a lower level. 0db is the "reference" level, against which all other levels are measured.

Now which level is 0db is completely up to whoever makes the scale. As mentioned above, 0db on mixers is usually the "optimum" level.

0db on other stuff could be another level. For example, gear is often said to be -10db or +4db. This is information on what the optimum level of the inputs and outputs of the gear is. They use other reference levels, obviously.

Since levels are all relative like this, 8bit and 24bit audio does not really have the same maximum levels. Neither do they have different maximum levels. :-) It's just numbers, representing the availiable levels from 0 to maximum, with different amount of steps inbetween.

The "maximum level" doesn't mean anything more than "hey, go any higher and you get distortion". What this level is depends on other variables.

If you aren't more confused yet I can make you.
 
also a good rule of tracking is ever -3 db under -3 is 1 bit less of quality so -6 db under 0 your almost at 20bit!
 
Actually, a dB is 0ne tenth of a Bell, hence decibel. so 0dB is absolute quiet, but for audio aplications it is also a standard for any given piece of equipments clipping threshold.

The logrithmic relationship is to power. every multiple of six doubles spl. and to achive this power has to be multiplied by ten, hence at 1 watt my speaker puts out 89db spl, 95 at ten watts, 101 db at 100watts, 108 db at 1000w.....
 
dB

There's a mix of correct and slightly incorrect information above. Mostly correct, but ...

To be completely, totally, nitpickingly accurate, decibels are used to measure relative levels of power (if you don't believe me, look it up in a dictionary). So, say we measure the power of something (like a noise, say), and we want to say how "big" it is relative to something else. We can use decibels to describe the relationship: 10 * log10 (figure we're talking about / figure we're comparing it to) = decibels.

Decibels is a logarithmic scale. If a sound is 10 times more powerful than another sound, it's level is 10 dB greater. If it's 100 times more powerful, it's level is 20 dB greater. If it's 1000 times more powerful, it's level is 30 dB greater. If it's a tenth as powerful, its 10 dB smaller. If it's twice as powerful, it's 3.01 dB greater. If it's four times as powerful, it's 6.02 dB greater. If it 's half as powerful, its 3.01 dB less.

What does log10 mean? By definition log10 (A) equals the number that, when 10 is raised to it, is equal to A. In symbols:
log10 (A) = B means that 10^B = A.

So to put the definition of decibel another (perhaps more confusing way), the number of decibels is ten times the number that, when you raise 10 to it, equals the figure we're talking about divided by the figure we're comparing it to. Or, to say the same thing yet another way, if you raise 10 to the power dB/10, you'll get the figure we're talking about divided by the figure we're comparing it to. For example:

10 / 1 = 10^1 ... 1 x 10 = 10 ... 10 dB = 10 times the power.
100/1 = 10^2 ... 2 x 10 = 20 ... 20 dB = 100 times the power
1000/1 = 10^3 ... 3 x 10 = 30 ... 30 db = 1000 times the power

1 / 1 = 10^0 ... 0 x 10 = 0 ... 0 dB = the same amount of power

.1 / 1 = 10^-1 ... -1 x 10 = -10 ... -10 dB = 1/10th the power

2 / 1 = 10^.301 ... .301 x 10 = 3.01 ... 3.01 dB = 2 times the power.
4 / 1 = 10^.602 ... .602 x 10 = 6.02 ... 6.02 dB = 4 times the power
.5 / 1 = 10^-.301 ... -.301 x 10 = -3.01 ... -3.01 dB = half the power


Somewhat tricky part:

Voltage isn't a measure of power! Voltage is a measurement of electromotive force. Electromotive force (the "push" that makes an electrical current flow) transmits power (from the thing that creates the voltage to the thing that it runs through), but it depends on how much resistance there is in the thing that it runs through. For DC, power = voltage squared divided by resistance. Note that power varies with the square of voltage ... a doubling of voltage results in a quadrupling of power, for example.

Decibels are really used to compare power ... when we use the dB unit to compare two voltage levels, we're really talking about the power that those voltage levels will transmit in a given circuit.

So, by definition:

dB = 10 * log(Pm / P0) (P1=power we're measuring; P0=reference level, i.e. power that = 0 dB)

dB = 10 * log10 (Vm^2/R / V0^2/R)
Simplify the expression in parenthese slightly:
dB = 10 * log10 ( (Vm/V0)^2 )

Going back to the definition of "log10" for a second:

log10(A) = B means that 10^B = A. Thus, if
log10(A^2) = B, then
10^B = A^2, and
10^(B/2) = A.
So, if:
log10(A^2) = B, it follows that
10^(B/2) = A, and if that's true, reverse the definition and you see that
log10(A) = B / 2, which means that
2 * log10(A) = B

Put it all together:
log10(A^2) = 2 * log10(A)

Apply it to voltage:

10 * log10 ( (Vm/V0)^2 ) = 20 * log10 (Vm / V0)

for voltage:
dB = 20 * log10 (Vm / V0)

So, when we talk about voltage, the number of decibels is double what it is when we talk about power.

20 dB = 10 times the voltage
40 dB = 100 times the voltage
60 dB = 1000 times the voltage
-20 dB = 1/10th the voltage
6.02 dB = 2 times the voltage
12.04 dB = 4 times the voltage
-6.02 dB = 1/2 the voltage

Of course, none of this absolute -- it's just relative to some other figure (power, voltage) that we're comparing it to. To create an absolute measurement scale, we need to define a fixed level that we'll compare it to.

The two common references used to create an absolute scale for measuring voltage are:

dBV = .775V (1 mW power in 600 ohms)
dBu = 1V

+4 dBV = 1.2283 volts
-10 dBu = .3162 volts


People also have invented an absolute decibel scale to talk about sound pressure levels (SPL). SPL isn't power either -- like voltage it's the "push" (pressure) that transmits power. So the dB of a given SPL is also 20 * log10 (SPLm / SPL0). To create an absolute scale, people use a reference SPL that's equal to the threshold of hearing (i.e. the smallest sound you can year). So, when you go to a rock concert and the SPL is 100 dB, the SPL is 100,000 times the smallest SPL you can hear. What's perhaps more exciting is that it's 10,000,000,000 times more powerful.
 
uh.... uh... uh.... ! man!
it's no wonder everyone says that specs don't mean anything...

sjjohnston... that's a cool post...

what I wonder is ... 0db on my DAW computer ... is it the same as on somebody elses since you say it's relative measure... and what would be the implications if it's not... should I care???

Also I read when software says "internal 32 bit practiacaly no overload" or something like that... what's that about... I think I understand it as it has enough bits to digitize any signal withoud digital distortion .... right?


sjjohnston... dude, that was something... I guess some people have a neck for these things...
 
sjjohnston, in the definition of log in your post, are you saying that it's log base 10 of B? Or is that log base 10 of 10 (which happens to be 1). It's hard to tell because you can't make the 10 a subscript (I mean with your keyboard). And when you don't give it a base, it becomes 10. So is that log base 10 of 10 = 1, or log base 10 of B? Just trying to understand, it's a LITTLE confussing.
 
Fantastic_Mad said:
sjjohnston, in the definition of log in your post, are you saying that it's log base 10 of B? Or is that log base 10 of 10 (which happens to be 1). It's hard to tell because you can't make the 10 a subscript (I mean with your keyboard). And when you don't give it a base, it becomes 10. So is that log base 10 of 10 = 1, or log base 10 of B? Just trying to understand, it's a LITTLE confussing.

Just to clarify my typography (if not my substance), what I meant to do was always explicitly specify the base (though I can't make it a subscript) and always to enclose the argument in parentheses.

I slipped up one place and wrote "log( Pm / P0)" -- that really should read "log10( Pm / P0)."
 
Digital - "dBfs"

Fed said:
... what I wonder is ... 0db on my DAW computer ... is it the same as on somebody elses since you say it's relative measure... and what would be the implications if it's not... should I care??? ...

By convention, meters on digital recorders reference their decibel figures to "full scale." Sometimes this is abbreviated as "dBfs."

"Full scale" just means the limit of what the recorder can write -- it can't write a bigger positive (or smaller negative) value, because all of its bits are already "on." It's functionally the same as trying to write a number bigger than 999 with only three characters (you're trying to write a number bigger than 0xFFFF, on a 16-bit recorder).

The key thing to remember is that if you try to go over 0 dBfs, you'll get distortion, and it will become quite obvious and ugly if you try to go even a moderate way over.

Now, what voltage at the input jack on the recorder's A/D converter will produce 0 dBfs on the meter depends on the recorder -- indeed you can vary it on a lot of recorders, simply by turning an "input" level knob. Generally, you don't really need to worry about this, unless you're calibrating the recorder or you're having a problem making it work with some other gear.
 
Mickey, your math is way off, and the unit for power is wattage=current x voltage.

and the relationship to dB from wattage is logrithmic. And relitive volume doubles every 6 dB.
 
sjjohnston -

those are really good posts, thanks.

from reading that, here is my conclusion (i understand "headroom" and dynamic range etc.)..

but, if I track at 0 dB...and then put that CD in the car stereo, keeping the volume level the same.....

a 24 bit recording will be louder than a 20 bit recording will be louder than a 16 bit recording?

the way it has always been explained to me (and based on your math, and other things you said...I am thinking this is wrong)..

is that 24 bit, 20 bit, 16 bit..etc, will all playback at the exact same volume level...if tracked at 0db.

And the advantage of doing 24bit, is that I can record at lets say...(going by the 3db = 1bit thing above assuming that is accurage) -24dB, and still have the same resolution as a 16bit recording at 0db. I guess even by that, that answers my question...becuase when I get to -10db in 24bit, my volume will be louder...and when I get to 0db, it will be louder still.

My only false premise here might be that when you have more resolution in something, you have more volume (i.e, -7db vs. 0db). Don't see how I could be wrong on that part...which again, would mean that my original question, is 24bit louder than 16bit when tracked at 0db and when played back at the same setting..louder.

I guess I get it - but, let me know...thanks dude.
 
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