24 bit/44.1?

Dseanm - It has more to do with the type of converter than number of bits. Dac in your Monarchy (PCM63 - now discontinued) is traditional DAC with laser trimmed resistor divider while most of 24/192 DACs (if not all) are Delta-Sigma. There is nothing wrong wit either approach - just different sound. Some people believe that Delta-Sigma is bad for audio and you can even find statement that Burr-Brown placed in the PCM63 datasheet saying that Delta-sigma are so noisy that they cannot even read lower three bits. Same company short time after made PCM1794 That has 6 highest bits of tradditional DAC and 18 lowest bits of Delta-Sigma. It is funny that they don't use word Delta-Sigma but "Advanced Segmented" instead. There is nothing wrong with Delta-Sigma and even SACD is byproduct of Delta-Sigma modulator before filtering. Same for DSD recording. Like everything else it is subjective and in your case you're not a fan of Delta-Sigma technology (or high oversampling, or digital filtering).

why would you think the 8 least significant bits would be zeros rather than the most significant?

Hi Bob,

The "least significant bit," as you may realize, corresponds to the smallest resolution increment, while the "most significant bit" corresponds to the largest.

For example, if the maximum possible value ("full scale") at the analog output is 2 volts, on the digital (SPDIF) output a logic "1" on the msb would indicate that the corresponding analog output is greater than 1 volt. The next most significant bit would have a weight of 1/2 that amount, so a 1 on the two most significant bits would indicate a value of greater than 1.5 volts. Etc. The least significant bit in a 24 bit word would have a weight of 2volts/2^24 (two volts divided by 2 to the 24th power), which is 0.000000119 volts.

So setting the 8 least significant bits to 0 would introduce very miniscule inaccuracy, while setting the msb's to 0 simply would not work.

For further confirmation of this, see the section in the middle of this page defining the time slots in the AES/EBU and SPDIF subframes:

http://en.wikipedia.org/wiki/AES/EBU

Regards,
-- Al

Yes, that looks right, Bob. Basically the 16 bit number is being left-shifted 8 places, which is equivalent to multiplying by 2^8 as you indicated.

The conversion from a 16-bit to a 24-bit representation is exact, but of course the additional 8 bits of resolution that a true 24-bit system would provide are lost.

Anything less than infinite resolution in a digital system can be thought of as a small noise component being added to the signal, and in fact is referred to as quantization noise. Which obviously is greater in the case of the left-shifted 16 bits than for a true 24-bit a/d.

Regards,
-- Al