ezkcdude;148736 Wrote: 
> Let's illustrate this graphically, because it can be confusing. Imagine
> this is the 16-bit signal:
> 
> (LSB)1000 0100 0010 0001(MSB) 
> 
> MSB=most significant bit
> LSB=least significant bit
> 
> Now, if we pad that 16-bit signal with 8 bits from the LSB side, we
> get:
> 
> (LSB)0000 0000 1000 0100 0010 0001(MSB)
> 
> O.k., now let's attenuate the signal by a factor of two (just for the
> sake of simplicity):
> 
> (LSB)0000 0001 0000 1000 0100 0010(MSB)
> 
> Just for good measure, let's do it again (factor of two attenuation):
> 
> (LSB)0000 0010 0001 0000 1000 0100(MSB)
> 
> So far, so good. Now, what will happen to that "1" on the left (LSB)
> side of the signal if we keep attenuating like this a few more times
> (well, 7X to be exact)? Right, information from the original 16-bit
> signal will be lost. However, that is theoretical. What is more of a
> problem in the real world is how many times we can reduce the volume
> until that bit is lost in the noise floor of the DAC. Are there truly 8
> bits to play around with? The answer should be obvious. Also, it should
> be clear that if the signal was sent to a 16-bit DAC, that bit would
> have been lost on the first attenuation step, because we left-shifted
> the data.


Whilst not disagreeing with your example, I presume you do agree that a
16 bit DAC will simply ignore any value in the 8 LSB's? - and that the
16th bit is lost in the noise floor (probably).


-- 
Phil Leigh
------------------------------------------------------------------------
Phil Leigh's Profile: http://forums.slimdevices.com/member.php?userid=85
View this thread: http://forums.slimdevices.com/showthread.php?t=27832

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