So, with my scenario, where the MT was recieving 24bits, what do you think the MT did with the extra 8bits of info? Is the MT smart enough to discard the least significant 8 bits?
In danlynch's scenario, his ADC delivered 16 bits of data to the MT and the MT stored the 16 bits of data in a 24 bit format by simply padding the least significant 8 bits with zeroes. In other words, the MT stored all the data delivered by the ADC, and truncating the least significant 8 bits results in a 16-bit file identical to the file the MT would have stored had it saved the data in a 16 bit format.
In your scenario, the ADC delivered a full 24 bits of data from the V3, but the MT only stored 16 bits of that data by truncating the least significant 8 bits. The MT does not include a mechanism for dithering from 24 to 16 bits. In other words, the MT stored only 2/3 of the data (16 bits) provided by the ADC, and "threw away" 1/3 (8 bits).
That said, none of the portable 24 bit gear tapers use achieves full 24 bit resolution, including the V3. One bit = 6 dB of dynamic range. So the full dynamic range of 24 bits is 144 dB (24 * 6) and the full dynamic range of 16 bits is 96 dB (16 * 6). The V3 has a dynamic range of 110 dB, which is only about 18-1/3 bits of resolution (110 / 6). The rest of the resolution (i.e. the additional 5-2/3 bits, or 24 - 18-2/3) is comprised of data that falls below the V3s noise floor. And that's why many people don't hear a difference: because only 2-1/3 bits, instead of the full 8, of the truncated 8-bits contains meaningful audio data. At least, that's how I understand it (I'd rather not count the number of times I've been wrong).
Topic: MT Set @ 16 Bit with V3 @ 24 bit - Whoops? (Read 9605 times)