24 bit/44.1?


The digi out on my B&K phono 10 is supposedly ( according to the manual) a 24bit/44.1 signal? Does this make sense ? BTW - whatever it is -sounds great thru my Becnchmark DAC-1.
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Showing 3 responses by almarg

The SPDIF protocol provides locations in each subframe for 24 bits. They might simply (and misleadingly) be basing their statement on that, and setting the 8 least signficant bits to 0, or they might really be using some or all of those bits. There's no way to tell without more information than is provided in the manual (which I took a look at, btw).

Regards,
-- Al
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