Protools 9 LE now has automatic latency compensation. I am able to trigger SD2 within a protools session with the hardware buffer set down to 64 samples. The sample rate for the session is 88.2. Am I correct to understand that the buffer of 64 samples causes latency ONLY in hearing ( monitoring ) the drum performance as it is being recorded?
After recording ( the midi performance ) will SD2 play back perfectly in time with the midi notes, or is there still 64 samples of delay to my ears? What about when the drum tracks are bounced? Will the wave files be right in line with the original midi notes, or do I need to nudge them back in time 64 samples? If this is even an issue, does protools 9 take care of it now? I might not be understanding the basic principals. The goal is obviously to situate the drum files in exact alignment with the drummer’s performance. A detailed explanation would be so appreciated!
The latency is the result of the Buffer Size, which the computer needs to be able to interpret the incoming signal and decide how to deal with it, so to speak.
Now, 64 samples Buffer Size @ 88,2kHz would (at least in theory) mean 0,72 ms latency (in live practice it would be at least times 2). This you cannot hear. Most ordinary musicians do not notice a latency of 10-12 ms. Drummers are generally more sensitive, of course.
Pro Tools 9 (and most other DAWs) takes any buffer sizes into consideration when recording and aligns audio and MIDI correctly. When playing back and latency inducing plugins are present, the correct amount of samples are compensating each track’s delay, making use of the Automatic Delay Compensation.
IOW, don’t worry and don’t nudge.
Best Regards,
John
John Rammelt - Toontrack
Technical Advisor
Thanks so much for the info. I have a related question, that when answered, should put to rest my latency concerns, and my incomplete understanding of the issues, although much better now. “In theory”… at a sample rate of 44.1 with hardware buffer size of 64, is the latency identical to a sample rate of 88.2 with buffer size 128 samples? Makes sense in my mind, but trying to confirm the math. Also…what is the formula “in theory” for calculating ms of latency based on buffer size? How did you come up with the numbers in your reply. Thanks, and that should do it for me!
Buffer Size / Sample Rate ≈ Latency (times 2 if it’s a I/O round trip and, depending on hardware involved, external delay)
128 / 44,1 ≈ 2,9 ms
64 / 44,1 ≈ 1,45 ms
128 / 88,2 ≈ 1,45 ms
At least this is how I calculate, someone better informed may correct me, should I be wrong.
John Rammelt - Toontrack
Technical Advisor
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