NEED HELP TO BUY A SOUNDCARD 24bit 96k
- Thread starter charly666
- Start date
Big T... said:
Why not? Any pro/semi-pro card nowadays is going to be 24/96. Better headroom, better quality, more info to work with.
If he was just playing games then 24/96 would be overkill. But at the prices you can get them for these days, there's no need to go lower.
just curious... if i were making CD's... i'd be getting the best quality 16bit 44.1k Card i could find, not a lesser quality 24bit 96k card... not taking anything away from the audiophile, its a great card for the money... im just curious as to why he really wanted a 24bit 96k card thats all....
if the final recording is going to be on a cd, how is it relevent to record at 24 bit? its not going to improve the cd sound in anyway, if anything it'll prolly diminish it due to the extra stage of conversion... that's why i asked the initial question "may i ask why you need a 24bit 96k card?".
i agree with the world of difference part... recording and editing at 16bit 44.1k:
- saves almost half the hard drive space
- not to mention CPU and RAM usage
- and will save you time, and quality when recording a cd...
i dont like sending music thru unrequired conversion... just a personal thing is guess...
peace man... i need sleep...
i agree with the world of difference part... recording and editing at 16bit 44.1k:
- saves almost half the hard drive space
- not to mention CPU and RAM usage
- and will save you time, and quality when recording a cd...
i dont like sending music thru unrequired conversion... just a personal thing is guess...
peace man... i need sleep...
Big T... said:if the final recording is going to be on a cd, how is it relevent to record at 24 bit? its not going to improve the cd sound in anyway, if anything it'll prolly diminish it due to the extra stage of conversion... that's why i asked the initial question "may i ask why you need a 24bit 96k card?".
i agree with the world of difference part... recording and editing at 16bit 44.1k:
- saves almost half the hard drive space
- not to mention CPU and RAM usage
- and will save you time, and quality when recording a cd...
i dont like sending music thru unrequired conversion... just a personal thing is guess...
peace man... i need sleep...
I will say it again: More headroom. More room for error with levels. More information to work with, helpful when lots of plug-in processing is done.
Just because the end result is 16/44, doesn't mean to only use 16/44.
Would you buy a disc camera because you were only going to make 4x6 prints, or would you buy a 35mm camera because it is better and someday you might want to make 8x10's?
CD's may be the standard now, but they won't be forever. The Circuit City by me already a large selection of DVD Audio.
And yes, I can hear the difference between 24 and 16.
Slackmaster2K
Gone
BigT, the first part of the answer lies in processing.
If you were simply recording one stereo track and not processing it in any way, then recording at 16/44 or 24/96 wouldn't really make much difference.
But that's not what we're talking about. You're going to be doing a serious amount of "math" with the raw data files. The higher the resolution your raw tracks are at the better sound quality you end up with in the end. Consider a graphics artist doing artwork that will end up on the web. Just because the final distributable image will be rendered at 72dpi doesn't mean that the artist started at 72dpi. On the contrary, he/she was probably working with resolutions many times that.
The second part of the answer lies in "headroom."
Digital headroom is not the same as analog headroom, because in the digital realm nothing exists above 0db. However, higher the bit depth, the lower the levels you can record at without sacrificing quality. Remember that the decibel scale is logarithmic, which results in an interesting relationship between resolution and sound level.:
Every 6db of gain requires one more significant bit. Thus, in a 16bit system, -96db to -90db is represented by one bit; -96db to -84db is represented by 2 bits; and so on and so forth until you conclude that -96db to 0db is represented by 16bits, or 65536 (2^16) discrete values. Likewise in a 24bit system -144db to 0db is represented by 24bits, or 16777216 (2^24) values.
That's all pretty obvious, but now think about this. The range -96db to -90db in a 16bit system is represented by *one* bit, which yields two discrete values (1 or 0). Now, try to imagine what a waveform looks like when each sample can only consist of two voltage levels! But of course who cares, right? It's way down at -90db which is pretty hard to hear. Well, let's keep doing the math. The range -90db to -84db is represented by how many discrete values? 2^2 - 2^1 = 2. Again, two discrete values. The range -84db to -78db is represented by 2^3 - 2^2 = 4 values. The range -78db to -72db is represented by 2^4 - 2^3 = 8 values. Let's make a picture:
_____________________ -96db
0000 0000 0000 0000 = 0
0000 0000 0000 0001 = 1
_____________________ -90db
0000 0000 0000 0010 = 2
0000 0000 0000 0011 = 3
_____________________ -84db
0000 0000 0000 0100 = 4
0000 0000 0000 0101 = 5
0000 0000 0000 0110 = 6
0000 0000 0000 0111 = 7
_____________________ -78db
0000 0000 0000 1000 = 8
0000 0000 0000 1001 = 9
0000 0000 0000 1010 = 10
0000 0000 0000 1011 = 11
0000 0000 0000 1100 = 12
0000 0000 0000 1101 = 13
0000 0000 0000 1110 = 14
0000 0000 0000 1111 = 15
______________________ -72db
Continue on, and you soon realize that your prime tracking range of -6db to 0db is represented by 2^16 - 2^15 = 32768 discrete values. Can you now see why people using 16bit systems recommend trying to stay in the yellow? If you track at -12db to -6db, for instance you just halved your resolution, and your waveform is represented by just 16384 discrete levels.
Now a 24bit system looks like the following. Remember that 0db is our hard limit, thus when you increase your dynamic range by increasing the number of bits you use, that range is found at the bottom of the scale:
______________________________ -144db
0000 0000 0000 0000 0000 0000 = 0
0000 0000 0000 0000 0000 0001 = 1
______________________________ -138db
0000 0000 0000 0000 0000 0010 = 2
0000 0000 0000 0000 0000 0011 = 3
______________________________ -132db
0000 0000 0000 0000 0000 0100 = 4
0000 0000 0000 0000 0000 0101 = 5
0000 0000 0000 0000 0000 0110 = 6
0000 0000 0000 0000 0000 0111 = 7
Hmm, looks very similar to the 16bit illustration, except that we've sort of shifted everything way down to -144db. That's true, and at first it might not seem like a big deal, because who cares what happens at -144db? Nobody does. What's actually interesting is what happens up at the TOP!
In a 24bit system, the range 0db to -6db is represented by 2^24 - 2^23 = 8388608 discrete values. Continue on with the boring math, and you'll find that you don't even get into 16bit resolution until you hit -48db (ah, 6db * 8 extra bits = 48db!)!
That means that speaking purely from a resolution standpoint, tracking at -42db to -48db in a 24bit system is equivalent to tracking at -6db to 0db in a 16bit system! Of course our electronics aren't quite that good, and we still have converter self-noise (coming up next) to contend with, thus nobody would actually want to track that low. However, in a 24bit system you don't have to pound the yellow to maintain signal resolution like you do with 16bit....and if you do, you're simply rewarded with exceptionally higher resolution which can result in better math yeilding better end results.
Here's a fun example. Take any 16bit audio file and decrease the volume on it by 90db. This will effectively squash the sucker into just a couple bits. Apply the volume change, reopen the file (if necessary), and turn the volume back up by 90db. "ZZZZZZZZZZgarblegarbleZZZZZZZZZgarble" Well what did you expect? Do the same thing with a 24bit file and what do you end up with? Well, you basically get a 16bit version of your 24bit file...and that ain't so bad
The third part of the answer lies in noise.
Noise is an electronic problem that creeps into our digital world. No converter is going to be perfect, thus way down in lower sound level ranges you're going to have noise. Yick. In the extreme case, take something like an old Soundblaster 16. The noise floor (where the noise level sits) on those suckers was in the range of -40 to -30db! Extremely damaging! In the opposite extreme case, you might find that the noise floor in a very professional 16bit system is down in the -84db to -90db range, which is much more acceptable. But in a properly designed 24bit system, the noise floor can be way down in the -102db to -96db range, thus giving you the FULL 96db of dynamic range that was once promised by 16bit systems, and it can do so at much greater resolution!
Now I should put a disclaimer up. The most important thing about any converter is how it sounds. You want what you hear back out of your recording system to sound as close to what went into it as possible. Converters *damage* incoming sound in a very audible, unwanted way. A cheap card like a soundblaster damages the sound quite severely. A better card like an m-Audio Delta 1010 will still damage the sound in an audible way, but to a much less extent. The point I'm getting at is that it is possible to have 16bit converters that are less damaging than 24bit converters. In fact, I would suspect that a very pricey apogee 16bit converter would sound much better than the converters found in the $600 Delta 1010. Your DSP would sound worse on the 16bit data, but it would probably be worth it to preserve those sources that you've worked painstakenly to capture. The rule of thumb, though, is that 24bit converters really aren't that big of a deal to manufacturers anymore...thus on your better soundcards and converter boxes made today, you're just going to see 24bit converters...that's all there is to it.
Slackmaster 2000
If you were simply recording one stereo track and not processing it in any way, then recording at 16/44 or 24/96 wouldn't really make much difference.
But that's not what we're talking about. You're going to be doing a serious amount of "math" with the raw data files. The higher the resolution your raw tracks are at the better sound quality you end up with in the end. Consider a graphics artist doing artwork that will end up on the web. Just because the final distributable image will be rendered at 72dpi doesn't mean that the artist started at 72dpi. On the contrary, he/she was probably working with resolutions many times that.
The second part of the answer lies in "headroom."
Digital headroom is not the same as analog headroom, because in the digital realm nothing exists above 0db. However, higher the bit depth, the lower the levels you can record at without sacrificing quality. Remember that the decibel scale is logarithmic, which results in an interesting relationship between resolution and sound level.:
Every 6db of gain requires one more significant bit. Thus, in a 16bit system, -96db to -90db is represented by one bit; -96db to -84db is represented by 2 bits; and so on and so forth until you conclude that -96db to 0db is represented by 16bits, or 65536 (2^16) discrete values. Likewise in a 24bit system -144db to 0db is represented by 24bits, or 16777216 (2^24) values.
That's all pretty obvious, but now think about this. The range -96db to -90db in a 16bit system is represented by *one* bit, which yields two discrete values (1 or 0). Now, try to imagine what a waveform looks like when each sample can only consist of two voltage levels! But of course who cares, right? It's way down at -90db which is pretty hard to hear. Well, let's keep doing the math. The range -90db to -84db is represented by how many discrete values? 2^2 - 2^1 = 2. Again, two discrete values. The range -84db to -78db is represented by 2^3 - 2^2 = 4 values. The range -78db to -72db is represented by 2^4 - 2^3 = 8 values. Let's make a picture:
_____________________ -96db
0000 0000 0000 0000 = 0
0000 0000 0000 0001 = 1
_____________________ -90db
0000 0000 0000 0010 = 2
0000 0000 0000 0011 = 3
_____________________ -84db
0000 0000 0000 0100 = 4
0000 0000 0000 0101 = 5
0000 0000 0000 0110 = 6
0000 0000 0000 0111 = 7
_____________________ -78db
0000 0000 0000 1000 = 8
0000 0000 0000 1001 = 9
0000 0000 0000 1010 = 10
0000 0000 0000 1011 = 11
0000 0000 0000 1100 = 12
0000 0000 0000 1101 = 13
0000 0000 0000 1110 = 14
0000 0000 0000 1111 = 15
______________________ -72db
Continue on, and you soon realize that your prime tracking range of -6db to 0db is represented by 2^16 - 2^15 = 32768 discrete values. Can you now see why people using 16bit systems recommend trying to stay in the yellow? If you track at -12db to -6db, for instance you just halved your resolution, and your waveform is represented by just 16384 discrete levels.
Now a 24bit system looks like the following. Remember that 0db is our hard limit, thus when you increase your dynamic range by increasing the number of bits you use, that range is found at the bottom of the scale:
______________________________ -144db
0000 0000 0000 0000 0000 0000 = 0
0000 0000 0000 0000 0000 0001 = 1
______________________________ -138db
0000 0000 0000 0000 0000 0010 = 2
0000 0000 0000 0000 0000 0011 = 3
______________________________ -132db
0000 0000 0000 0000 0000 0100 = 4
0000 0000 0000 0000 0000 0101 = 5
0000 0000 0000 0000 0000 0110 = 6
0000 0000 0000 0000 0000 0111 = 7
Hmm, looks very similar to the 16bit illustration, except that we've sort of shifted everything way down to -144db. That's true, and at first it might not seem like a big deal, because who cares what happens at -144db? Nobody does. What's actually interesting is what happens up at the TOP!
In a 24bit system, the range 0db to -6db is represented by 2^24 - 2^23 = 8388608 discrete values. Continue on with the boring math, and you'll find that you don't even get into 16bit resolution until you hit -48db (ah, 6db * 8 extra bits = 48db!)!
That means that speaking purely from a resolution standpoint, tracking at -42db to -48db in a 24bit system is equivalent to tracking at -6db to 0db in a 16bit system! Of course our electronics aren't quite that good, and we still have converter self-noise (coming up next) to contend with, thus nobody would actually want to track that low. However, in a 24bit system you don't have to pound the yellow to maintain signal resolution like you do with 16bit....and if you do, you're simply rewarded with exceptionally higher resolution which can result in better math yeilding better end results.
Here's a fun example. Take any 16bit audio file and decrease the volume on it by 90db. This will effectively squash the sucker into just a couple bits. Apply the volume change, reopen the file (if necessary), and turn the volume back up by 90db. "ZZZZZZZZZZgarblegarbleZZZZZZZZZgarble" Well what did you expect? Do the same thing with a 24bit file and what do you end up with? Well, you basically get a 16bit version of your 24bit file...and that ain't so bad
The third part of the answer lies in noise.
Noise is an electronic problem that creeps into our digital world. No converter is going to be perfect, thus way down in lower sound level ranges you're going to have noise. Yick. In the extreme case, take something like an old Soundblaster 16. The noise floor (where the noise level sits) on those suckers was in the range of -40 to -30db! Extremely damaging! In the opposite extreme case, you might find that the noise floor in a very professional 16bit system is down in the -84db to -90db range, which is much more acceptable. But in a properly designed 24bit system, the noise floor can be way down in the -102db to -96db range, thus giving you the FULL 96db of dynamic range that was once promised by 16bit systems, and it can do so at much greater resolution!
Now I should put a disclaimer up. The most important thing about any converter is how it sounds. You want what you hear back out of your recording system to sound as close to what went into it as possible. Converters *damage* incoming sound in a very audible, unwanted way. A cheap card like a soundblaster damages the sound quite severely. A better card like an m-Audio Delta 1010 will still damage the sound in an audible way, but to a much less extent. The point I'm getting at is that it is possible to have 16bit converters that are less damaging than 24bit converters. In fact, I would suspect that a very pricey apogee 16bit converter would sound much better than the converters found in the $600 Delta 1010. Your DSP would sound worse on the 16bit data, but it would probably be worth it to preserve those sources that you've worked painstakenly to capture. The rule of thumb, though, is that 24bit converters really aren't that big of a deal to manufacturers anymore...thus on your better soundcards and converter boxes made today, you're just going to see 24bit converters...that's all there is to it.
Slackmaster 2000
jrosenstein
New member
best bit depth explanation ever
Jagular
Active member
Nice one Slack
I actually somewhat understood that. Bad part now is that you are making me want to get something with better converters than I was initially planning on. Except I can barely afford to get what I planned on (audiophile 2496, 1010lt, or 1010 if I can get a good deal used).
Knowledge sucks sometimes
I actually somewhat understood that. Bad part now is that you are making me want to get something with better converters than I was initially planning on. Except I can barely afford to get what I planned on (audiophile 2496, 1010lt, or 1010 if I can get a good deal used).
Knowledge sucks sometimes
James Argo
Fancy Rock N' Roll
...SEE !!! That's why he end up being Moderator. Because he's smart and cares !!! Well, I'm not that smart, but I bet if I knew all those knowledge, I'll be too lazy to write it down that long...
Way to go Slack... You're the man... 
Way to go Slack...
You're the man...