[tosehee]

Quote:

Originally Posted by jkeny /img/forum/go_quote.gif 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!

That topic is more suited for "Sound Science". I think it was talked like forever there with no agreement as with most of these topics.

 

[mmerrill99]

Sure, but I thought that so many here seemed to row in with what Dan had said to the extent that slim.a is posting it as an explanation for readers that I thought I must be missing something obvious & one simple post would stop me racking my brains to figure out what it is I'm missing. Quote:

Dan Lavry, from Lavry Engineering, corrected some misconceptions about longer digital cables being better and explains why a shorter cable is better than a longer cable.

Starting another thread would seem to be confusing as you guys are the ones that read Dan's posts & can explain it to a poor lost simpleton like me who as somebody here said, is travelling into the mist. Please save me from this fate!

 

[tosehee]

Quote:

Originally Posted by jkeny /img/forum/go_quote.gif Sure, but I thought that so many here seemed to row in with what Dan had said to the extent that slim.a is posting it as an explanation for readers that I thought I must be missing something obvious & one simple post would stop me racking my brains to figure out what it is I'm missing. Starting another thread would seem to be confusing as you guys are the ones that read Dan's posts & can explain it to a poor lost simpleton like me who as somebody here said, is travelling into the mist. Please save me from this fate!

Sound Science - Head-Fi: Covering Headphones, Earphones and Portable Audio

 

[mmerrill99]

Thanks Toshee,
I didn't know about that thread until now - & I see Lavry has posted there with the same posts - pity he didn't just reference it & I would have posted there instead of taking this thread off-topic. I really didn't know this had been discussed before even down to the long cable reference from DiyHi.org - Doh!

So I take it that nobody understands what Lavry is saying either, just believing it because of his industry name?

 

[Dan Lavry]

Quote:

Originally Posted by jkeny /img/forum/go_quote.gif 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...)

Regards
Dan Lavry

 

[punk_guy182]

Quote:

Originally Posted by jkeny /img/forum/go_quote.gif So I take it that nobody understands what Lavry is saying either, just believing it because of his industry name?

Just take it easy man! There are better causes worth fighting for.

 

[mmerrill99]

Dan,
Thank you so much for taking the time to post this - I know this took a lot of work, much appreciated. It is a lot clearer & obvious that you took time over phrasing it. I will digest it in time but one question, if you don't mind

- I have seen mentioned that reflections finally decay after 3-4 round-trips with best low-loss digital cables - what is your experience?

Edit: I meant to say, Dan ,that I took my posts over to the other thread before this reply above: http://www.head-fi.org/forums/6165724-post84.html

 

[Dan Lavry]

One can not say that final decay takes 3-4 round trips. It depends on the mismatch, on the cable length, and on the data rate. You can say that the longer the cable, the longer it takes to the reflections to settle. And of course, if the impedance matching is way off, the reflection amplitude is bigger, and it will take more reflections to get close to the desired level.

I guess there are a lot of things being "mentioned" in all sorts of places, that are not based on facts, and are often very wrong. This is sad. It makes one wonder why it is so.

Regards
Dan Lavry

 

[mmerrill99]

I'm not sure why you repeated the post in answer to my question - was it to prompt me to read it again?

I did that a number of times already & I have a question if you don't mind answering:
- your statement that you end up with the signal & it's reflections superimposed on it has me baffled because let's consider the signal as a wave that travels down the cable & takes a certain time. When it arrives at the receiver it is interpreted as either 1 or 0 depending on it's voltage level. At the same time a portion of this wave will be reflected (if there is an impedance mismatch). Now when this partial signal returns back to the receiver (200nsec round trip later) I presume it's not interpreted as a signal again (1.14V, 1.02V, etc)? How does this not happen?

- Now if this reflection makes enough round trips and dies out before the next signal is generated then we have no super imposing on the next signal (that's why my last question about round trips). I can understand shorter cables working in this case but it does depend on how quickly the reflection dies out!

 

[Dan Lavry]

Quote:

Originally Posted by jkeny /img/forum/go_quote.gif I'm not sure why you repeated the post in answer to my question - was it to prompt me to read it again? I did that a number of times already & I have a question if you don't mind answering: - your statement that you end up with the signal & it's reflections superimposed on it has me baffled because let's consider the signal as a wave that travels down the cable & takes a certain time. When it arrives at the receiver it is interpreted as either 1 or 0 depending on it's voltage level. At the same time a portion of this wave will be reflected (if there is an impedance mismatch). Now when this partial signal returns back to the receiver (200nsec round trip later) I presume it's not interpreted as a signal again (1.14V, 1.02V, etc)? How does this not happen? - Now if this reflection makes enough round trips and dies out before the next signal is generated then we have no super imposing on the next signal (that's why my last question about round trips). I can understand shorter cables working in this case but it does depend on how quickly the reflection dies out!

Did I repeat a post? If so it was by mistake.

The signal travels down the cable. Say it is 1V. But when it gets to the end it changes to 1.14 in our example. It is never seen as 1V by the receiver. So the receiver in our example does not have a the .5V threshold looking at 1V. It sees a 1.14V step.

The receiver is looking at the voltage across the termination resistor. Say the driver is sending a 1V into 100 Ohms cable. Initially the cable looks like a 100 Ohms resistive load, no matter what is at the end of the line, because the driver can not see ahead of time what is going to happen. So you are sending a current wave down the line, and the current wave is 1V/100 Ohm = 10mA.

But say the termination is 120 Ohms instead of 100 Ohms. When the 10mA current wave arrives at the termination resistor, the voltage becomes 10mA X 120 Ohms which is 1.2V (not 1 volt). That is the only way you can preserve the basic relationship between current, resistance and voltage - this is Ohms law, a very fundamental law.

When there are reflections, the change in voltage happens when the current waves arrives at the end and the beginning of the line. There is no "good time" to look at the signal before the reflections alter the signal. The alteration happens instantly when the signal hits the cable ends.

To that, add the impact of previous reflections that raise and/or lower the signal levels, and all that against a fixed threshold. So you end up with timing error in the detection, and that is jitter. As a rule the voltage alterations are not in sync with the signal. The signal transitions happen at a certain clock rate. The reflections alter the ave with timing that depends on the cable delay.

Regards
Dan Lavry

 

[mmerrill99]

Dan, thanks for the answer - I don't mind if we revert all this to the other thread to save these readers from boredom but if you continue here, I'll stay

Again, I thank you for the time you are putting into this.

Quote:

Originally Posted by Dan Lavry /img/forum/go_quote.gif Did I repeat a post? If so it was by mistake. The signal travels down the cable. Say it is 1V. But when it gets to the end it changes to 1.14 in our example. It is never seen as 1V by the receiver. So the receiver in our example does not have a the .5V threshold looking at 1V. It sees a 1.14V step. The receiver is looking at the voltage across the termination resistor. Say the driver is sending a 1V into 100 Ohms cable. Initially the cable looks like a 100 Ohms resistive load, no matter what is at the end of the line, because the driver can not see ahead of time what is going to happen. So you are sending a current wave down the line, and the current wave is 1V/100 Ohm = 10mA. But say the termination is 120 Ohms instead of 100 Ohms. When the 10mA current wave arrives at the termination resistor, the voltage becomes 10mA X 120 Ohms which is 1.2V (not 1 volt). That is the only way you can preserve the basic relationship between current, resistance and voltage - this is Ohms law, a very fundamental law.

OK, so this is how the signal is altered by the impedance mismatch & this signal is read by the receiver. Now what happens to the reflected current wave - it travels back to the transmitter, gets reflected & returns to the receiver again somewhat reduced - how is this not interpreted as another signal?

Quote:

When there are reflections, the change in voltage happens when the current waves arrives at the end and the beginning of the line. There is no "good time" to look at the signal before the reflections alter the signal. The alteration happens instantly when the signal hits the cable ends. To that, add the impact of previous reflections that raise and/or lower the signal levels, and all that against a fixed threshold. So you end up with timing error in the detection, and that is jitter. As a rule the voltage alterations are not in sync with the signal. The signal transitions happen at a certain clock rate. The reflections alter the ave with timing that depends on the cable delay.

Is this last sentence about cable delay, where the short cable comes into play? I'm not sure how it helps though - a short cable will just alter the wave in a different way to a long cable but it still alters the wave, no?

 

[mmerrill99]

Quote:

Originally Posted by Dan Lavry /img/forum/go_quote.gif When there are reflections, the change in voltage happens when the current waves arrives at the end and the beginning of the line. There is no "good time" to look at the signal before the reflections alter the signal. The alteration happens instantly when the signal hits the cable ends.

So according to this the cable length is immaterial as the original signal is instantly overlaid by any reflections

???

 

[slim.a]

Quote:

Originally Posted by jkeny /img/forum/go_quote.gif So according to this the cable length is immaterial as the original signal is instantly overlaid by any reflections ???

Jkeny, could please let go of this subject and let us move on to something else. Dan Lavry has given enough information. At this point either you are convinced or not.

Anyway, this thread has been taken way off track. There are already too many threads speaking mainly about jitter in digital cables.

From now on, I hope that everybody can respect the following :

- The intent of this thread is not discuss about theory of jitter or cables. It has been discussed enough and there are a ton of threads in head-fi and other forums that discuss those topics. There has been a lot of valuable information but enough has been said.

- We all know there are articles saying we can't hear jitter and differences between cables. Let assume that those posting in this thread believe otherwise. If people want to discuss about

- The intent of this thread is discuss about actual listening experiences using usb to spdif converters as transports and not about theory.
If people hear (or don't hear differences) between different usb to spdif converters and digital cables, they should be able to express/share their listening experience free from criticism in this thread.


My intent is not to sound harsh but to put back this thread in the right track.

 

[tosehee]

Quote:

Originally Posted by slim.a /img/forum/go_quote.gif Jkeny, could please let go of this subject and let us move on to something else. Dan Lavry has given enough information. At this point either you are convinced or not. My intent is not to sound harsh but to put back this thread in the right track.

Amen

 

[rosgr63]

slim.a I fully agree.
We got some very valuable info from Dan Lavry and Jkeny, and we are grateful for their time.
Back on track, when are the Mac drivers for the M2Tech hiFace going to be released?