Skip to main content

Creating 24bit CDs, Masterlink, only option?

Hi all,

I have a knowledge problem.

I have been recording in a Boss BR900CD and it records in 24 bit. I've done 2 tracks for a friend of mine and she is going to put them on her forthcoming debut album. Obviously I'm relatively new to recording.

She recorded most of the album at a commercial studio here in Ireland. The studio wants to mix the tracks and needs them as 24bit. I'm not sure how to give them a CD of 24bit tracks.

I've been trawling the forums and have read about the Alesis Masterlink. An experienced friend of mine also loves this product. I can get one, new from a local distributor for 500 euro.

I'm in the process of upgrading my gear to a Yamaha N12, Adam P11a pair, TL Audio 5052. I'm wondering:

1. How do I get a 24bit CD to the studio.
2. Would it be worth my while investing in a Masterlink in the long run as I hope to do a lot of tracking, mixing for local artists.

Sorry for the long post, all advice will be greatly appreciated,



BrianaW Sun, 08/24/2008 - 21:20

I don't really know anything about the machine you have, but I do see in the description that it has a USB port for syncing to a PC. You could transfer the WAV files that way, organize them, and burn them as a data disc. That's one way to do it, but sorting out the undo's, punch-in's, and time code/track placement issues is a completely different story. Hopefully there's a song maximize feature on the Boss that could help with this, or maybe even a software that came with it to make things easier. Just a suggestion, like I said, I don't know anything about that recorder. Someone else here may be able to give you a better solution. Maybe the Boss can burn direct data CD's?

VonRocK Sun, 08/24/2008 - 21:24

I quickly googled your product, found and downloaded your manual. On page 150, it explains how to do what you want.

The studio is going to want to mix your raw recorded tracks. That means, they don't want a mixed down stereo track on CD. They want the original files.

So, open your manual to page 150, and read how do burn the raw wav files on to CD-R. I did not read further, but I'm sure that you could transfer the wav files to your computer, and probably send them to the studio via the internet. Most studios will do this.

It would be best to give them the tracks without any effects or processing. Just the raw recording. Nothing else.

You could include a mixed down stereo CD to show them your mixing chops and vision. If you can, ask your friend if you can sit in on some of the mixing sessions.

anonymous Sun, 08/24/2008 - 21:27

"1. How do I get a 24bit cd to the studio."

Easy as pie, don't burn it as an audio cd burn it as a data cd. Just drag and drop the files onto a data cd.

Your best bet is to go and get a usb key, you can get a few gb for only a few dollars and use that as a portable data solution.

If you have larger data requirements get a portable hdd.

RemyRAD Sun, 08/24/2008 - 22:00

You know this is really one of those stupid things??!!

Standard audio CD's are only 16 bit. Nothing else is compatible as a Redbook standard CD than 16 bit. The request is idiotic. Greener is absolutely correct. If the studio needs a 24 bit CD it's not a studio. It's a beginner. You deliver on CD (under 700MB ROM data only) DVD, under 4.3GB, data only, hard disk drive (your choice of sizes) data only, commercial downloads servers such as Pando, real-time data transfers through AOL instant messenger & others, data only. Plus, you don't remix a CD. You remix from multitrack sources. So like I said earlier, beginners, clueless, recording school, beginners.

Sure. You can dither a 24-bit recording to 16 and a hide those additional eight bits in the noise floor. Big Deal. WTF ?? Like I said, beginners.

Get with the program.
Ms. Remy Ann David

anonymous Mon, 08/25/2008 - 03:36

Thanks guys. And thanks VonRock. Yeah the guys want the individual tracks in 24bit. I really don't understand why. I'm presuming the sound detail of 24 bit is better, fuller.

But to be honest the BR900cd isn't exactly an amazing machine but I learned a lot on it. So now I'm stepping up a gear. You can check out some of my tunes on if you like.

So do you think that Alesis Masterlink would be of any use considering the set up I mentioned in my post. I have a new MBP and was going to get an external hard drive 7200rpm.

VonRocK Mon, 08/25/2008 - 08:30

What method did you use? Did you transfer them to your PC first via USB, or just use the built in CD burner? You should be able to drag and drop them if connected via USB. Maybe you can get a card reader and read directly from the card.

Try burning an AIFF file, and see what that comes out as.

If the file is sitting there in 24 bit, you should be able to access it somehow. Experiment. Read the manual. Ask Roland.

anonymous Mon, 08/25/2008 - 09:34

Yeah I have a card reader but the br900 stores the tracks kinda weird. They're .bro files. So some kind of conversion is necessary I think. And the bro900, lo and behold converts them to 16bit.

I have a card reader and will try what you suggested guys.

The only fast solution I have come up with is using garageband to convert them to 24bit. Does that make any sense?

And thanks so much for the effort. I can't wait to get my Yamaha N12 with Cubase.

RemyRAD Mon, 08/25/2008 - 22:27

24 bits does not provide greater resolution. It provides greater dynamic range for inexperienced hands and incompetent postprocessing. They could & should take your 16-bit 44.1kHz material which they could up convert to 24-bit to suit their needs. The noise floor of your microphone preamps is greater than the noise floor of 16-bit digital processing. So none of this really makes any sense. Beginners, all of them. Clueless beginners. But if you have cut quality sounding tracks, that's all that should be necessary. Nobody can tell the difference between 16-bit & 24-bit when listing to playbacks. Inherent electronic signal to noise prevents any noise for from being lower than -105 DB generally speaking. It doesn't matter that 24-bit provides for 140 DB dynamic range when electronics can only deliver less than 110 DB and rock-and-roll typically isn't included in such specifications. Tell them to get off of their high horse and that you're delivering quality tracks. Let them play with it as they need to.

Silly silly silly boys
Ms. Remy Ann David

Thomas W. Bethel Tue, 08/26/2008 - 04:37

No matter what the bit depth or the sampling rate it still goes back to garbage in garbage out.

I get people who want to master at 196 kHZ and 24 or 32 bit. There stuff could well fit into 44.1/16 but they want this nirvana of audio perfection. Most times the song writing, the playing, the recording and the mixing is problematic at best and down right BAD at the worst so the quality of the audio DOES NOT REALLY MATTER ALL THAT MUCH.

Musicians and budding recording engineers would be better served by worrying about things like playing in tune and writing and recording GREAT music than chasing something that in the end really does not matter all that much to the overall "sound" of their recording.

196 kHZ and 32 bit is JUST a way for the manufactures to sell more equipment. DVD-A, SACD never caught on because basically people want to listen to their music when and where they chose and don't want to sit in a room with a calibrated multichannel monitoring system to listen to their IPOD with headphones. They are listening in their cars, in their bedrooms and out walking to classes.

If it were me I would keep it at 24 bit up until the final mastering and let the mastering engineer dither it down to 16 bits


IIRs Tue, 08/26/2008 - 08:32

RemyRAD wrote:
Sure. You can dither a 24-bit recording to 16 and a hide those additional eight bits in the noise floor. Big Deal. WTF ??

These posts of yours are embarassing Remy. Your bad advice to use 16 bit resolution is made even worse by the fact that your posts are otherwise very knowledgable and informative...

Your claim that 16 bit audio has plenty of dynamic range is misleading: the bottom end of that dynamic range is crunchy as hell.

Here's my standard dither test: I take a snare sample, run it through a long reverb impulse, then drop the gain by 40 or 50 dB (ie: only half way down the dynamic range that Remy claims for 16 bit audio). I then export a control version at 24 bit, another at 16 bit with no dither, and then further 16 bit files with the dither I wish to evaluate. The resulting files can then be normalised to hear the results at sensible listening levels:

1. [=""]24 bit file.[/]="http://platinumears…"]24 bit file.[/] This sounds just like the original version with a nice smooth reverb tail that decays to silence.

2. [[url=http://="http://platinumears…"]16 bit file with no dither.[/]="http://platinumears…"]16 bit file with no dither.[/] Note the horrible crunchy distortion, and the dead stop when the signal drops below the LSB.

3. [=""]16 bit file dithered with the free MDA dither plug.[/]="http://platinumears…"]16 bit file dithered with the free MDA dither plug.[/] Note that the reverb tail no longer sounds distorted, and that it continues for longer than the undithered version. But, the dither noise itself is quite obnoxious.

4. [[url=http://="http://platinumears…"]16 bit file dithered with the MegaBitMax Ultra algorithm included in the Izotope Ozone mastering plug.[/]="http://platinumears…"]16 bit file dithered with the MegaBitMax Ultra algorithm included in the Izotope Ozone mastering plug.[/] Note that the dither noise is now much less audible (despite actually having higher peak levels) because the noise has been pushed towards the ultra low and high frequencies to which we are less sensitive. (Encoding this file as an MP3 may have added some artifacts: hard to tell as I am currently using my built-in laptop speakers in a hotel lobby!)

Remember, all this is happening at -50dB. In other words, if you work at 16 bit resolution you are at best burying your low level detail in dither noise, and at worst turning it into a crunchy distorted mess. This will be compromising the subtle differences between left and right channels that provide good stereo imaging as well as the subtle details that provide the depth and space around a good acoustic recording.

Think back to your math lessons: you have to perform a series of calculations, and you need the final answer to 2 decimal places. If you round down to two decimal places after each calculation your errors accumulate each time, so the correct way to do it is to preserve as many decimal places as possible throughout each calculation and only round down once when you have the final answer.

I have no doubt that your 16 bit recordings sound good Remy. But I also have no doubt that they would sound better if you worked at a higher resolution up to the final stage, because this is a basic law of mathematics. Working at 16 bit means rounding down after each calculation and in these days of 24 bit converters and cheap data storage it is totally unneccesary.

IIRs Tue, 08/26/2008 - 10:43

If you record your snare at 16 bit resolution, while leaving 6dB headroom for unexpected peaks, you will end up with only 15 bits of information. If you then compress that audio by 6dB and add make-up gain you will end up with distorted rubbish filling the bottom 2 bits of your 16 bit file.

Every time you save a file at 16 bit resolution you are adding a layer of distortion or dither noise, which is cumulative, so even if your ears aren't sensitive enough to notice it after only one truncation, after 2 or 3 the sound will be noticeably colder and more brittle, with worse stereo imaging and less sense of depth and space.

If there were two types of analog tape available, both at the same price, but one had much better dynamic range and a lower noise floor, which would you choose?

anonymous Tue, 08/26/2008 - 10:56


I agree with your findings, but I don't think it applies to Remmy's real world microphone recording model that she is referring to. If the OP used effects heavily, that could be a different thing entirely.

With a sample, when you are listening to the tail of reverb on a snare you are listening to something that is almost entirely synthetic. In the real world, a -105dB noise floor, as Remmy mentioned, is just about the best case. In a musicians bedroom/home studio even a -96dB noise floor is good.

You could model this as a -96dB white noise with a -96dB artifact noise from the conversion, and -6dB of intended signal. (assuming the dynamic range of the song is 6dB)

Then the question is can you hear a -96dB signal with -96dB noise? Barely. And the real question: Will anyone notice it buried under the full tone signal? I doubt it.

The OP has recorded in 24bit, but can only export at 16bit. If the sound is normalize inside the BOSS recorder, and then exported he will have a dynamic range of 96dB.

In your mention on math lessons you are right, there will be rounding errors if processing audio at 16bits. This would be a problem if you were doing lots of processing at the 16 bit level. This is especially true of fader adjustments. However in the OP’s case there is minimal processing. The error is 1 / 2 bit at the 24 to 16 bit conversion inside the BOSS. In the DAW there is no loss from the 24 to 32 bit conversion, and there may be a 1 / 2 bit loss when converting 32 back to the target of 16bit.

So the biggest error is 1 a bit or -96dB. This is the same as the dynamic range of the target media.

IIRs Tue, 08/26/2008 - 11:10

If your equipment is limited to 16 bit resolution then you have no option but to make the best of it.

However, Remy has repeatedly advised people who use 24 bit converters and 32 bit or better audio processing that they should record and export 16 bit files. Whether the difference is audible in all cases or not is hardly relevant: the difference IS there as a mathematical inevitability, and multiple stages of truncation will accumulate those differences until eventually it is audible to anyone.

anonymous Tue, 08/26/2008 - 11:24

IIRs wrote: the difference IS there as a mathematical inevitability, and multiple stages of truncation will accumulate those differences until eventually it is audible to anyone.

I agree with you here. I thought you were referring to the OP's problem, and Remmy response to it, not a general trend that you saw.

Sorry for the confusion.


anonymous Tue, 08/26/2008 - 11:35

Wow there's a lot of info to take in.

Well in my situation I have not choice but to take the files from the recorder in 16bit form. The guys are just mixing the tracks. Im giving them a guide mix of course.

What I will do is give them both (16 and 24bit) even though the raw files are 16bit. I'm getting a Yamaha N12 with Cubase AI next week so I won't have this problem again. I'm a songwriter before an engineer BUT I'm determined to get a good grasp of as much technical Knowledge as I can to produce good sounding demos for artists and publishers.

If you want to hear some of my tracks recorded on the BR900CD go to

And I want to say thanks so much to all of you for your efforts and help. I'm shocked (in a good way) by the amount of feedback I got. It's very educational.

IIRs Tue, 08/26/2008 - 11:42

GeckoMusic wrote: [quote=IIRs]the difference IS there as a mathematical inevitability, and multiple stages of truncation will accumulate those differences until eventually it is audible to anyone.

I agree with you here. I thought you were referring to the OP's problem, and Remmy response to it, not a general trend that you saw.

Sorry for the confusion.


No worries. :-)

I probably should have posted the above in a different thread, but I'm sat in a hotel in Hamburg with nothing better to do and Remy's 16 bit proselytising finally annoyed me enough to dig out my dither test files...

anonymous Thu, 08/28/2008 - 12:13

Hey all. Just a quick add on question.

Do any of you own an Alesis Masterlink?
I'm asking because I work in a musical instrument store and the Alesis Distributor has offered me a unit for 500 euro. They're going for around 750 euro on websites in Europe.

I'm buying a T.L. Audio 5052 Ivory II stereo preamp/comp/eq for some tracking and basic mastering for my own demos. I'll be using a Yamaha N12, Macbook Pro and Cubase.

I was wondering would the results be better mixing down through the 5052 to the Alesis than mixing to Cubase? The 5052 has the option of digital out which would by pass the Masterlink converters. I've read they're not great. Good but not great!!

Codemonkey Thu, 08/28/2008 - 21:25

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?

anonymous Thu, 08/28/2008 - 22:20

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.

Cucco Fri, 08/29/2008 - 10:58'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:


is exactly the same number as:

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.


anonymous Fri, 08/29/2008 - 11:28

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?

anonymous Fri, 08/29/2008 - 12:04

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.

      In 24 bit:
| Signal Noise floor
| | |
24bit 000000000000000000000000
16bit 0000000000000000
clean audio lost due to bit 16 bit depth.

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.

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)

anonymous Fri, 08/29/2008 - 12:46

"Headroom" comes from analog 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 analog and digital...

Codemonkey Fri, 08/29/2008 - 13:18

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.

anonymous Fri, 08/29/2008 - 14:01

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
0001 = 1
0000 = 0
1111 = -1
1000 = -8

for unsigned ints:
1111 = 15
1000 = 8
0111 = 7
0001 = 1
0000 = 0

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

Reggie Sat, 08/30/2008 - 10:31

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. :D