cd --> wav --> cd --> ?

P

PopD

New member
Here my situation:

I am running my stuff off a laptop with a cd-rom/dvd drive running that to an external burner. . .aka --> buffer underun HELL!

So, this morning I used 'Audio Catalyst' to rip the cd tracks off the source cd and copied them to the hard drive as 44.1kHz / 16-bit / stereo wav files. Then I used adaptec to burn these wav files to the destination cd-r in the external burner. also, worked like a charm, no skips, no buffer underruns (yet) , no prollems.

My question is: am I losing any audio quality this way, by converting cdda --> wav --> cdda? I listened to both and I cannot discern any difference, but this doesn't mean that there is no difference. I guess a test I could run would be to rip the first track off the original cd and the first track off the cd-r and then diff the two.

Anyone know. . .?
 
My guess would be the answer is no, you're not losing audio quality. Since it's the same sample rate and bit quality, in theory you shouldn't lose anything in the transfer.
 
The answer is NO

The answer is NO. When ripping from a CD into a WAV file, no quality is lost. However, when ripping from a CD into an MP3 file, and then burning that MP3 to a CD-R, THEN you lose quality. The reason being? MP3's are compressed forms of WAV files. The inaudible noises that the human ear cannot pick up are deleted and that is why MP3's are 1/10 the size of a WAV. The final result? Even if you ARE burning MP3's onto a CD, the loss of quality is so unnoticeable that it really doesn't matter in the long run.
 
-thanks for the helpage

-I have a few more questions:

-so when I copy cd to cd, the cddb information is retained in the burned copy, i.e. winamp tells me whose cd it is and what the track names are. When I go cd --> wav --> cd, I assume that this info is lost. (haven't tested it yet, cuz I only have winamp at work) Where exactly is this info stored on the cd? TOC?

-when a wav is encoded to mp3, is this inaudible info compressed (as in winzip) or is it completely stripped out forever? I am under the impression that wav(a) gets encoded to mp3(a) which can be converted to wav(b) and in coder talk: wav(a) ! = wav(b). What is the actual difference between mp3(a) and wav(b)? The original wav information from wav(a) is not restored to wav(b) so is it basicallly like just changing the file extension or is it more complicated?:cool:
 
Here's the answer to your questions:

-so when I copy cd to cd, the cddb information is retained in the burned copy, i.e. winamp tells me whose cd it is and what the track names are. When I go cd --> wav --> cd, I assume that this info is lost. (haven't tested it yet, cuz I only have winamp at work) Where exactly is this info stored on the cd? TOC?

Yes, the info is lost, simply decoding CD to WAV will not retain any of the information that was on the CD. All you get is the music.

-when a wav is encoded to mp3, is this inaudible info compressed (as in winzip) or is it completely stripped out forever? I am under the impression that wav(a) gets encoded to mp3(a) which can be converted to wav(b) and in coder talk: wav(a) ! = wav(b). What is the actual difference between mp3(a) and wav(b)? The original wav information from wav(a) is not restored to wav(b) so is it basicallly like just changing the file extension or is it more complicated?

No, it is not compressed, it is completely erased. This explains why MP3's are so much smaller than WAV files and why they can never fully equal CD quality. When you've decoded a WAV from an MP3, that WAV will not be the same quality as the WAV that was ripped from the CD. (Something I don't understand is why when you decode an MP3, why the WAV file becomes so big again.) But anyways, like I said before, the difference is basically unnoticeable.
 
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