Does the 1.5m 75ohm rule apply to 110ohm?


I’ve read many times that 1.5m is the best length for SPDIF.
Al Marg’s (RIP) excellent explanations come to mind.
Does this rule apply to XLR 110ohm digital cables as well?

Please don’t turn this thread into a ’bits are bits’ carnival.
If you have nothing but negative opinions please don’t post.
128x128rja
@georgehifi, search the digital cable thread for almarg’s comments on the spdif question. The thread lasted a week or more at the time, 2yrs ago maybe. ??

1.5m rule is based on transition and propagation times. Reflection from RCA connector might be stronger, but there will be reflection on any impedance boundary, unless DAC’s digital input impedance is perfectly match to cable’s characteristic impedance (that can be a little different than 110 ohm). Transition time is unknown (only estimated) as well as propagation time (that depends on dielectric). 1.5m is good place to start (based on 25ns transition and 5ns/m propagation). Very short cables (less than 1ft) might also work, but are not very practical.
What happened to the post with the links? It appear to have been deleted?


I did check one of the linked articles. Conceptually it was okay, but the math was questionable. They considered the rise time to determine a mid-point, but did not consider at 0 time (start of rise), there is no signal to reflect and what you need to consider is both the amplitude of the initial arrival plus the amplitude of the reflection which is signal and mismatch dependent. With poor matching, that 1.5m could be longer (or shorter) and there will be one length that will be approximately the worst, and that is probably near 1.5m though it could be anywhere between 1-2 meters. 


One of the articles talks about speeding up the edges, but the best thing to do is just to use proper impedance matching. 
sugabooger

Reflection starts from the beginning of transition (first knee).  Assuming 5ns/m it will take about 7.5ns to travel to the end of the 1.5m cable, reflect and another 7.5ns to come back - total of 15ns.  It will miss 12.5ns midpoint of 25ns transition and that is the whole point.  Reflections  often create staircases, that in presence of noise induce jitter.  There can be many more transitions in the cable going back and forth creating multiple step staircase patterns (Bergeron diagrams), but the first reflection in this case is the worst one.  Following reflections will affect next edge only and will be greatly reduced since next edge comes in more than 1us.                                               
As for speeding the edges - it should reduce noise induced jitter since noise with vertical crossing of the threshold point will affect  exact moment of level recognition less.  Slow transitions are the weakness of Toslink, making it susceptible to system noise, in spite of being immune to ambient electrical noise.  Some transports might have transitions as fast as 5ns, but you will really have to match characteristic impedance well.