Originally Posted by jkeny
Can I respectfully ask someone to explain Dan's short cable operation in simple terms (as was given in the two quotes I gave about long cables - that the reflection arrives back after the SPDIF transition has taken place & thus doesn't causing distortion at the point of transition)? I really want to understand what a lot of others seem to accept as correct but I fail to grasp it. Please help me in my quest for clarity!
OK, jkeny, I will try to explain, and it will take some of my dear time. In return, you have to take the time and make an effort to understand what I say. It is a bit complicated to explain, and may require some concentration.
Ideally, a digital signal has 2 very distinct levels. For example let’s stick to 0V and 1V, and call those levels 0 and 1. The signal changes between the levels, and it does take some time for the change (transition) to occur. The time it takes to go from 0 to 1 is called rise time. The time it takes to go from 1 to 0 is called fall time. The rise and fall time are (by definition) the time it takes to go between 10% and 90% of the voltage levels. In our example, 10% is 0.1V and 90% is 0.9V. For simplicity sake, you can imagine a straight line (with a slope) connecting the 10% and the 90% "points". This line has a slope to it. Say a transition from 0V to 1V got to 0.1V at some time, and 10 nsec LATER it got to .9V. Then your rise time is 10 nsec.
Now, you want to detect (with a receiver circuit) the digital signal. You want to know if the signal is 0 or 1, and WHEN the transition happened. The timing is important, and the detection must be accurate (time wise), or else you have jitter.
So you can place a circuit (such as a comparator) that will detect when the signal is UNDER OR OVER 0.5V. You call that 0.5V the threshold level. If the input signal is at 1V, your receiver reads it as 1, and if the input 0V it is read as 0. But what do you do if the signal is at say, 0.4V? Well it is still under the 0.5V threshold, so it is read as a 0. And so on...
So in fact you are "crossing 2 lines". One of the lines is a horizontal constant level at 0.5V. The other line is has a slop to it (it is the rise time of a transition). If you were to move the threshold from say 0.5V to 0.6V, the lines cross at a different point in time. The signal has to rise from 0 to 0.6V to detect a 1, and that takes longer than for a 0.5V threshold. That is pretty easy to grasp, but that is not our issue.
But our issue is very similar to moving the threshold. Say you keep the threshold at 0.5V but ADD 0.1V to the input signal. Now, instead of having a 0V and 1V signal, you have 0.1V and 1.1V signal. But the threshold did not move, it is still at 0.5V. So now it takes less time to detect a 1. The signal starts off with the 0 (low level) at .1V thus only .4V away from the transition threshold; to the detection will happen earlier. You can see that moving the threshold down has the same effect as moving the signal up.
Note that there is noting wrong with moving either the threshold or the signal by some constant amount for all the transition detections. That would only make the detection earlier or later, and no one cares if the music is played at sat 10nsec later. So where is the problem?
When you have reflections, the signal is moving up and down all the time, not by a constant amount. The reflections DO IMPACT THE SIGNAL LEVEL DURING THE TRANSITION DETECTION AND AT ALL OTHER TIMES. The statement that the reflection HAPPENS LATER, after the transition, is WRONG! When you send a signal down a cable, and there are reflections, you do not end up with a 2 level signal, you end up with a signal that is the sum of the original 2 level with the reflections ADDED to it.
The pattern of the levels of the reflections is complex. It depends on:
1. The digital signal pattern itself.
2. The amount of load end reflection (Reflection coefficient) at the cable end
3. The amount of driver end reflection (Reflection coefficient) at the cable input.
And the timing of the pattern of voltage level changes depends on the cable delay, which has nothing to do with the data rate down the cable. So when you have reflections you end up with a mess.
Here is an example. Say the cable is 75 Ohms, the load is 100 Ohms (instead of 75) and the line delay is 100nsec. Send a single transition of 0-1V down the cable. 100nsec later, at the receiver end you have 1.14V (instead of 1V), and that will LAST FOR the next 200nsec. During that 200nsec, a reflection is going backwards to the driver, which (is typically just a few ohms impedance) and that will invert the reflection and send it back to the load. So at 300nsec the load side sees 1.02V, which LASTS FOR another 200nsec.... It gets closer to the desired 1V but still not where it should be...
So for a single transition at the input, the output looks like 1.14V for 200nec, then 1.02V for 200 nsec and so on…eventually, with enough reflections it gets to be near 1V.
It takes a number of back and forth to make the signal settle near enough to the desired 1V. And that is just the beginning of the mess. Say while the reflections of the first transition are bouncing back and forth, the driver sends another transition. Now the second transition cause a new set of reflections, and those are ADDED to the reflections of the first transition... Now, if the time between transitions is very long, you could say that for all practical purposes, the reflections of the first transition decayed way down. For example, if one is using a word clock for a 44.1Khz clock, the time between transitions is around 11.3.usec (2 transitions per cycle), there is enough time to fit a lot of reflections for say 100nsec (0.1usec) cable delay. You could have 113 reflections (around 56 round trips). Even at .5 reflection coefficient (50% of the signal is reflected each round trip), you end up with 56 multiplications of .5 (.5 X .5 X ..... .5), and that is .5^56 = .000000000000000014 disturbance before the next transition.
But such is not the case here for SPDIF or AES. We are not taking about a 44.1KHz word clock. We are talking about signal which is around 5.65Mhz (for 44.1KHz sample rate). So you have an input transition every 89nsec. Now take a 100nsec cable, and this time you have a new input transition before the first reflection is over, so the signal levels at the load are way off, and that impacts the detection time...
At 2 meters cable (9nsec), and you have around 20 back and forth reflections between transitions, for 44.1KHz SPDIF. That is not bad, but with 1 foot (1.5nsec cable) you have 118 back and forth reflections, so the decay before the next transition is MUCH BETTER then that of the 2 meter.
If you cut the cable length by a factor of 2, the number N of round trips reflections (for any transition) doubles, and the end result is the reflection coefficient to the POWER N. Say the reflection coefficient .5 for example (you can choose any other example number between -1 and 1). Say you have 2 reflections then you have an error of .5^2 = .25. If you have 4 reflections (half the cable length) the error is .0625 which is exactly 4 times better
The improvement (when reflections do happen) is EXPONENTIAL. Each halving of the cable length yields 4 times improvement. You see why shorter is not just better, it is a lot BETTER.
And of course, that is ON TOP of what I already explained in the previous posts - when the cable is short with respect to the rise time, you do not have a reflection issue to begin with...
So in this post I explained that even if the rise time were a theoretical 0nsec, which is much shorter then say any (non zero length) cable delay, the shorter cable offers a lot better reflections decay, because with a short cables you have more reflection round trips between transitions, making the reflection impact on the signal levels much lower. The improvement is EXPONENTIAL.
The statement that with longer cable the reflection happens later is WRONG. The reflections do not go away because the cable is longer. In fact it, they last longer, impacting the time when the digital signal crosses the threshold. The reflections ALTER the signal, from a 2 state signal (such as 0V and 1V) to multiple state signal, and the alteration is rather messy, so the timing detection is messy. The alteration is due to reflections ,and those take LONGER to die out when the cable is longer.
That did take a long time to write. I am pretty sure that some folks will not understand it, I hope others will. If someone has difficulties with my explanation, and are technically minded, I suggest going to the literature (or one can look at it with a variable rise time pulse generator, a coax cable and a fasts scope...)