opaqueice;194637 Wrote: 
> Sorry, I thought my statement was suffciently clear, but apparently not.
> The good approximation is that the voltage is the same everywhere along
> the wire *at a given time*.  Obviously it changes with time, but it
> doesn't depend on position along the wire.

This is one of those (rare) phenomena that can be easily quantified.
The formula for the wavelength of a signal propagating along a linear
conductor is v=fl, where v is the wave velocity, f is the frequency,
and l is the wavelength.

In a coaxial cable, v is about 80% the speed of light, or 240M m/s. The
raw data rate (fundamental) on an S/PDIF cable for 44.1kHz sample rate
is around 5.6 Mbit/s, which gives a wavelength of 43 metres. So, a 1
metre cable is short compared to the wavelength, but not totally
insignificant.

That's not the whole story, though, because it's not the data rate that
matters, but the edge speed. Fast edges are desirable in this case
because the faster the edge, the more accurately its position can be
determined - and the lower the jitter. Data edges can easily contain
harmonic components well over 100MHz, which have a wavelength
comparable with the length of the cable. Transmission line effects
aren't just significant - they completely describe the propagation of
the signal.

The Squeezebox does have a very clean digital output, with rise and
fall times in the order of 10ns or so. Depending on the length and
impedance of the cable, and how it is terminated, the receiver may see
a nice, clean, monotonic edge, or it may see a waveform which is quite
different. S-shaped edges which hover around the 1<>0 threshold are
surprisingly common, as are 'edges' which cross the threshold one way,
then go back across it the other way, before finally crossing it again
on the way up to the final voltage. All these cause problems recovering
a low jitter clock.


-- 
AndyC_772
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