Using all of my bits...

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eclips1

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When recording sounds @ 24 bits when do you start loosing bits? Example: if I record a vocal track with peaks of -0.1 dbs, do I use more bits than if I were to record the same track with peaks of -4 dbs. I would really apreciate any responses on this. I get conflicting answers on this everywhere I search.

My ears are telling me some distortion is introduced when I'm trying to get a hot recording level and I go up to -.01 or even hit zero without blinking red in Vegas. But when I record with peaks of around -3 dbs, I can't tell the difference(besides one being louder), it accually sounds clearer. So this is how I record now, but am I using all my bits!?
 
You "loose" resolution as soon as you dont hit 0db. So, you should always try to have as high level as possible without distorting.

I know nothing about Vegas, so I can't say if you get digital distortion from bitclipping, or if the distortion comes from somewhere else. You could try to check the rest of thesignal path to see if the distortion comes from somewhere else.
 
At 24-bits of resolution...

...hitting maximum 0 is far less important than it was at 16-bits.... don't worry about taking the signal up so high... it's one of the advantages of going 24-bit...

Bruce
 
This is where I chime in with my stupid question....

Ok here goes, I've heard a lot about bits and know very little about what it actually does. I know and understand why its better to have a higher sampling rate but I know little about bit sizes. What is it actually doing when you increase or decrease bit size? See I told you, stupid question ;)
 
It's not "bit size," a bit is a bit -- a binary digit. It's "number of bits," how many bits are used to represent a sample. Just like regular numbers are made of decimal digits 0-9, binary numbers are made of bianry digits 0-1. Just like a 2-digit decimal number can have 100 different values from 00 to 99, a 2-digit binary number has 4 different values. OK, then, an 8-bit binary number has 2-to-the-8th possible values, which is 16; a 16-bit binary number has 2-to-the-16th possible values, which is 65,536... and a 24-bit number has 16,777,216 different values.

How does this apply to audio? In analog audio you constantly measure the voltage of a fluctuating waveform. In digital audio you take the voltage value of this fluctuating waveform and take measurements at incredibly brief instants (44,100 per second in CD-quality audio), and then express those voltages as binary number. Consider absolute silence to be 0 and the loudest sound to be the biggest number you have available. With 16 bits, then, the number of divisions between silence and the loudest sound is 65,536. But with 24 bits to use, the number of divisions is 16,777,216 -- much tinier.

It's a lot like when you use little dots of ink to represent letters or pictures on a printed page. If you have only 24 dots per square inch to use, things look pretty grainy. Use smaller dots and more of them, and it looks pretty darn smooth.
 
AlChuck left out a bit too. Not only do you have a smoother picture, but a BIGGER picture.

Because of sort of complicated stuff, 24 bit converters have a larger sound to noise ratio, which basically means that a larger maximum volume to lowest volume can be achieved.

On 16 bit converters, a 96dB s/n ratio is feasibly possible. Most every 24 bit converter has at least a 104dB s/n noise ratio. That is almost twice the s/n noise ratio that is possible because every 10dB in volume is in effect a doubling of the volume.

All of the above I just posted is drastically over simplified, and spec's from one A/D/A converter to the next can vary, sometimes significantly.

It is true that with 24 bit audio that you don't need to worry about achieving as hot levels during the A/D process. But, stay reasonable! One of the things that you want to achieve is a good RMS level. Think of RMS level as you "average" level of the recording, and if you are getting it around - 6 down to -12 on 24 bit, you will have exceptional sound quality. Unfortunately, most digital equipment HAS to show Peak volume, which means that little spikes of volume in the audio are going to show up on the meters, and give you a sense that you are in fact recording at very high volumes, when in fact, these near 0dB spikes may be problems in the recorded track, and making you lower your RMS level because of them.

Ahhhhhhhhhhhh, if this was all easy to do and understand, I could never charge someone for doing it....:)

Ed
 
Excellent thread ya'll, I apreciate the info.

Could someone explain the dithering process? I know what it does but if someone could explain how it does it, that would be cool. More specifically, how does it decide which bits to shed? or is it basically re-sampling my audio @ 16/44.1.
 
AlChuck said:
It's not "bit size," a bit is a bit -- a binary digit. It's "number of bits," how many bits are used to represent a sample.
Click to expand...
Bit size...bit number....you know what I mean,:D anyway thanx alot guys for the great info. I anxiously await your explainations on dithering.....
 
My momma always warned me that if I'd dithered I'd go blind...
 
No, you won't go blind, but you might start to grow hair on your mouse...
 
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