2. digital
  3. 1311630964

how can a cheap cd drive equal a fine transport?

Once a "perfect" file is on a HD, I grasp why playback may be better than reading data from a drive in real-time. But when ripping a cd, the digital data stream is read by a cd drive, i.e, a flimsy, cheap transport. The best transports produce a data stream with less error, or jitter. Large amounts of error correction are audible, so presumably, the less error correction applied, the better. So at the point where the cd is read by a drive, before applying error correction, before it even reaches a HD (or the prior optimal solution, a Genesis Digital Lens), how can a cheap computer drive produce a data stream comparable to a good transport? How can programs which try it 64 times, or whatever, produce a better result? Aren't they just using error correction (or checksum algorthms to determine which attempt got the best result, out of many error-laden reads) compensate for high initial error rates? Are fine transports almost pointless, now?
lloydc

Showing 1 response by shadorne

shadorne's avatar
shadorne
7,477 posts
 
Assuming a decent CD that can be easily read, then the transport can make a difference if the DAC operates as a slave to the transport clock. Only an asynchronous DAC (where there is no connection between clocks) will be totally immune from the type of transport used.

Only if the transport influences the DAC clock in some way can it possibly make a difference. So if a different transport makes a difference then it actually says more about the DAC jitter immunity than the transport.
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