Patrick Dixon;153203 Wrote: 
> Please explain (using the laws of physics and engineering) how a 'jitter
> rejecting' DAC does actually, in reality, remove jitter!

It's explained quite clearly in the Lavry white paper, which was posted
here a while back.  I can dig it up if you can't find it.

The concept is simple - you just record the incoming stream of bits in
a buffer, and then play it out through the DAC using your own clock. 
In principle this completely eliminates tansport jitter because you're
not using the timing of the incoming bits as a clock.  The only jitter
here is in the DAC clock itself, but that is both independent of the
transport and extremely small.

The only problem is that the DAC clock will not be quite in sync with
the incoming bitstream, which means your buffer will either overflow or
underflow eventually.  The Lavry deals with this by checking the state
of its buffer periodically and adjusting the frequency of its own clock
to make sure it never fills or empties.  It only has to do this once
every few seconds or so, and the adjustment is tiny, so this will have
no effect on the frequency response.

So this completely removes transport jitter, unless it's so bad it's
causing bit errors - but in that case the problem isn't jitter, it's a
broken transport.  Does that answer the question?


-- 
opaqueice
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