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
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