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

rja


Yes there is a thing to watch for and that’s digital reflection (jitter?) this is said to be reduced when the interconnect is no shorter than 1.5mt, I think this would also apply to aes/ebu.
Can you link Al’s (rip) post on this please?

Cheers George
IMHO the digital reflection on SPDIF cables is primarily due to the impedance mismatch when using RCA connectors. If using 75 ohm bnc connectors this "reflection" should be very minimal and length should not be much of an effect. Length should also not be important for aes/ebu cable if the 110 ohm spec is respected.
Because you have heard it many times from audiophiles and cable vendors doesn't make it true.
mike, thanks for your useless input on this subject.
Please don’t post to my threads if you intend to be negative.

1. I’ve never heard it from a cable vendor.
2. There was at least one scientific paper pointed out by Al Marg concerning this issue.
His insights were frequently useful, always lucid. I believe he was an EE and audiophile so he knew his stuff.

oldears, I agree that bnc connections are probably better than rca because they are generally 75ohm although they come in other values as well.
Post removed 
@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. 
Except the 25nsec is not a set number, and neither is 5nsec/ft, both can be smaller or larger, not to mention there is hysteresis and settling time that pushes that transition values above/below 0V, which means that theoretical 12 and 15nsec could end up sitting on top of each other, which could happen anywhere from 1-2 meters system dependent. 1.5 meters may work, or it may be a worse case. There is simply not enough tolerance at 1.5m.


w.r.t fast edges, if the termination mismatch is not large, then fast edges will always be better. If the termination mismatch is large, you have the potential for again a worse case situation. It would be rarer, but still possible. The slow speed of TOSLINK couples with noise optical detectors and high gain. It is the worst of call conditions for jitter.


Of course, fortunately most system today are fairly jitter immune. Well $50 consumer DACs are relatively jitter immune. $5,000 audiophile DACs you are taking your chances.
So the answer is that 1.5 M is the best choice for AES digital cables if one doesn't know the input impedance of the receiving component?
@mesch   Input impedance of receiving component has nothing to do with it.  We assume 1.5m minimum for average system.  In some systems 0.5m might be enough, while others will require 2m.  The whole point was that longer cable might be better (counterintuitive).