Creating 24bit CDs, Masterlink, only option??

Posted: Thu Aug 28, 2008 10:25 pm

Data discs don't have bit depth. The files on them could be 24-bit waves.

Can someone clear up:
Do 24bit converters provide more information about what's down in the low level of things, or do they provide 24bits of resolution over the same range of input volumes as 16bit?

To me it seems counterproductive to produce extra detail at a point where no-one can really tell, when the aim should be to provide greater resolution.

If that seems unclear, (it is 5am here), imagine a designer working on the finer points of an obscure part of a plan, when he should be working on refining the slightly sketchy (pun intended) overall layout?

Posted: Thu Aug 28, 2008 11:20 pm

As best as I can figure it, 16 bit gives 65535 points of resolution over 96db of range. Though I'm not sure what kind of db.

24 bit gives 16777216 points of resolution over 140db range.

Because db is logarithmic I can't "see" in my mind if that gives the same amount of accuracy but spread out over a larger field, or if it gives greater accuracy and a greater field.

Got nothing.

Posted: Fri Aug 29, 2008 11:58 am

Okay...here's the way to think of bit resolutions for audio (or anything for that matter.)

In binary, you've got 2 numbers to work with - 1 and 0. Just as in a base 10 numbering system, you think -Right to Left- (such as in base 10, the number 1,305 would be thought of as:
5*1 (to the power of 1) = 5
0*10 to the power of 1 = 00
3*100 to the power of 1 = 300
1*1000 to the power of 1

In binary, it's similar. (wikipedia does a good job of explaining it further.)

You must take into consdiration that:

000000001111111111111111

is exactly the same number as:
1111111111111111

That being said, a 16 bit number (representing a voltage) is identical to a 24 bit number if the leftmost bits in the 24 bit representation are all 0's. (A mild oversimplification given the different roles of certain bits).

Therefore, any voltage represented by a 24 bit figure up to the point represented above is identical to the 16 bit equivalent. As such, no extra points of resolution are provided to the bits that are identical in the rightmost 16 positions.

However, given that noise (digital and analog) inherits the lowest regions of the digital audio spectrum, the further down in the binary chain you can move it, the less objective and subjective impact it's going to have on the bits containing legitimate audio.

In addition, as IIRS points out, the further processing you do on this, the more you're going to muddle these lower bits. If they are below the audible range, you gain a perceived greater resolution.

You will not gain more points of resolution in the traditional sense.

On the other hand, since 0dBFS is a fixed voltage (or the constant), think of shifting the lower end of the audible spectrum down (which is where the increased dynamic range comes from). In other words, just because there are more numbers in a 24 bit string, doesn't mean it's louder than a 16 bit string. This makes the noise floor the variable (determined by the quantity of bits).

Just some thoughts.

J.

Posted: Fri Aug 29, 2008 12:28 pm

I can see now I used "points of resolution" in a whack context.

I want to be able to graph it, as in I need to know the lowest possible voltage and the higest possible voltage. The range.
And I need to know the accuracy at which the voltages are stored. This would give me the scale, or size of the increment between each possible step as it goes from lowest to highest.

Anyone know these things for 16bit and 24bit?

Posted: Fri Aug 29, 2008 1:04 pm

I'm not good at explaining things. I'll probably confusing things further, but here goes.

Bit depth and sample rate effect the audio differently. The bit depth represents the resolution of magnitude of the signal. The sample rate effects the frequency/time domain response.

Whether you are recording in 16 bit or 24 bit, you probably give yourself 6 to 12 dB of head room. Each bit represents a magnitude of two, and each 6 dB also represents a magnitude of 2. If you record with 12dB headroom, the left most 2 bits are your head room. Noise floor is about 102dB below the signal. 102dB / 6db per bit = 17 bits.

Code:

Bottom line, full scale in 16 bit is the same as full scale in 24 bit. The difference is in the least significant bits. 24 bit gives you:
1) a larger headroom without loosing quite signals.
2) Less error introduced when mixing or processing audio due to rounding.

Code:

voltage dB(full scale) dBv
1.23 0 +4 Full scale voltage
0.309 -12 -8 Signal level with 12 dB headroom
0.000002 -114 -110 noise floor

v = 1.23 * 10^(dB/20)

Posted: Fri Aug 29, 2008 1:46 pm

"Headroom" comes from analogue days, literally meaning how much room on the tapehead you had.

Is this a correct assumption? The wider the strip of tape the greater the dynamic range you could get from the quietest perceivable signal until tape saturation?

I'm getting lost in thinking about things in terms of analogue and digital...

Posted: Fri Aug 29, 2008 2:18 pm

OK. So 24bit just tacks on an extra 8bits for detail further down the volume ladder.

When something is AD'd the objective is to measure the amplitude from 0 (think of a sine wave graph). Use 1 bit to determine whether it's negative/positive (I believe audio is stored using signed values) and the other 23/15 are for the actual amplitude.

What I thought is that the available bits would be spread evenly across the volume.
0 in the graph scale (no input voltage) = a full complement of 0's.
1 in the graph scale (maximum input voltage) = a full complement of 1's.
0.5 in the graph scale would be 000000010000000 in 16bit or 00000000000100000000000 in 24bit mode.

However my understanding of the AD process is probably wrong.

Posted: Fri Aug 29, 2008 3:01 pm

Codemonkey wrote:

0 in the graph scale (no input voltage) = a full complement of 0's.
1 in the graph scale (maximum input voltage) = a full complement of 1's.
0.5 in the graph scale would be 000000010000000 in 16bit or 00000000000100000000000 in 24bit mode.

nice effort, but you are confusing some things.

The sign bit (if used) is the MSb or left most bit. Integers in binary are stored as the 2's complement. To convert to a positive integer you invert everything and add one. This is really beside the point however. 8 bit WAV files are stored as unsigned numbers. 16 and 24 bit are stored as two's compliment. Other file formats may be different.

Also for positive binary representation, four bits for simplicity, if 1111 was your full scale value = 1.0, then 1000 would be 0.5 not 0010 or 0100.

This thread has turned into a math forum.

each bit equals a multiple of 2.

least significant bit (lsb) = 1
next lsb = 2
next lsb = 4
msb = 8

so 1111 = 15. half of that is 7.5, round to 8
1000 = 8

for 4 bits signed integer:

0111 = 7
0110
0101
0100
0011
0010
0001 = 1
0000 = 0
1111 = -1
1110
1101
1100
1011
1010
1001
1000 = -8

for unsigned ints:
1111 = 15
1110
1101
1100
1011
1010
1001
1000 = 8
0111 = 7
0110
0101
0100
0011
0010
0001 = 1
0000 = 0

Code:

float integer binary
1.0 15 1111
0.93 14 1110
0.87 13 1101
0.80 12 1100
0.73 11 1011
0.67 10 1010
0.60 9 1001
0.53 8 1000
0.47 7 0111
0.40 6 0110
0.33 5 0101
0.27 4 0100
0.20 3 0011
0.13 2 0010
0.07 1 0001
0.00 0 0000

Posted: Fri Aug 29, 2008 9:29 pm

/fail

Should've remembered about 2's complement.
That has the nasty implication that you lose 1 number of resolution in the positive range...heheh.

Also should be shot for the binary fail. 00100 indeed. >_<

Posted: Sat Aug 30, 2008 11:31 am

Let me see if this is what you are looking for:
In 16-bit, below -60db there are 34 different possible levels (on both the positive and negative side of the graph) that can be represented by the way the bits are distributed across our voltage range.
The lowest values are as follows:
Value #0 = -96db
Value #1 = -90.3db
Value #2 = -84.28db
Value #3 = -80.76db
Value #4 = -78.26db

When you are working in 24-bit, you now have many more increments between these levels. Somewhere over 8,000 different possible voltage levels that can be represented under -60db. You can now record a level of -82.08 (sample value# 660) or -79.05db (sample value# 936) just as an example. The fact that having more bits has given smaller jumps in voltage that can be represented also allows us to now register those tiny voltages that were below our least significant bit in 16bit.
Another angle-- your number of bits gives you a certain number of different values possible. 65,536 possible values in 16bit, and 16,777,216 possible values in 24bit. In 16bit, each of those values represents a change of .00030518509476 volts. I don't know the exact number for the voltage steps in 24bit, but trust me it is a much smaller step. Very Happy

Posted: Sat Aug 30, 2008 11:35 am

OK so 24bit does actually provide greater detection of volume changes in the range that 16bit occupies.

At RO, we're all right!

Posted: Sat Aug 30, 2008 11:45 am

Codemonkey wrote:

OK so 24bit does actually provide greater detection of volume changes in the range that 16bit occupies.

Oddly enough, yes. Above, say, 60db either bitrate is so accurate reguarding levels that it don't make much difference anyway. But if you were to take some audio with a range of -60 to -96 in both 16bit and 24bit and crank them up to audible levels, the 24bit one would sound much better. Although, deep down I'm sure you already knew this.

Posted: Sat Aug 30, 2008 12:20 pm

Ah here we go, if I did my math right, the sample values in 24bit represent a change of .0000011920928955078125 volts.
Haha; more useless nerdy info. Wink

Posted: Sat Aug 30, 2008 2:44 pm

GeckoMusic wrote:

Bottom line, full scale in 16 bit is the same as full scale in 24 bit. The difference is in the least significant bits.

Thank you for stating that a little more eloquently and less confusing than my attempt above.

Lovin this discussion...

Posted: Sat Aug 30, 2008 2:47 pm

Reggie wrote:

Ah here we go, if I did my math right, the sample values in 24bit represent a change of .0000011920928955078125 volts.
Haha; more useless nerdy info. Wink

As a fellow nerd, how did you calculate that value?

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