Signal integrity question

[ggennai]

Hi everybody, I´m new in SI world. I´m reading Dr Bogatin´s book "Signal and Power Integrity simplified" and I found this figure.
Could somebody explain why the far end of the line 1 is high if the transition y from 0V to 1V ?

 

[danadakk]

Is it a 1/4 wave stub response ?

Regards, Dana.

 

[FvM]

The example is incomplete, no specification of source and load impedance.

 

[danadakk]

Isn't one line shorted and the other open ?

Regards, Dana.

 

[FvM]

Isn't one line shorted and the other open

Despite of unclarities with the test setup, the lines must terminated at one end, otherwise you get multiple reflections.

I get similar waveforms with no source termination and matched differential + common mode load termination.

Could somebody explain why the far end of the line 1 is high if the transition y from 0V to 1V ?

I admit that the behavior of coupled lines isn't obvious at first sight, but the answer to this particular question seems self-evident. You think about the expectable steady state response after all transients died away.

 

[danadakk]

Is this a problem with unterminated lines ? Poster has to clarify this.


Regards, Dana.

 

[vfone]

This is the text from the mentioned book affiliated to the posted figure 11-21:

Any voltage can be applied to the front end of a
differential pair, such as a microstrip differential pair.
If we were to launch the voltage pattern of a 0-v to 1-v
signal in line 1 and a 0-v constant signal in line 2, we
would find that as we moved down the lines with the
signal, the actual signal on the lines would change.
There would be far-end cross talk between line 1 and
line 2. Noise would be generated on line 2, and as the
noise built, the signal on line 1 would decrease.
Figure 11-21 shows the evolution of the voltages on the
two lines as the signals propagate. The voltage pattern
launched into the differential pair changes as it
propagates down the line. In general, any arbitrary
voltage pattern we launch into a pair of transmission
lines will change as it propagates down the line.

Is no information about the length and terminations of the differential pair lines (should be).
Later in the book the author speak about cases when the differential pair lines are terminated, which confirm that the lines in figure 11-21 are not terminated.

 

[FvM]

Later in the book the author speak about cases when the differential pair lines are terminated, which confirm that the lines in figure 11-21 are not terminated.

Don't agree. Without termination, the waveform would show multiple reflections. You can try in a simulation. I rather guess that the author has "abstracted" from termination problem to simplify the discusion at this point.

Instead of terminating the line, you could extend it to the right so that any reflection happens outside the observed time scale. The picture neither suggests a termination nor a continuation, in so far it's either incomplete or incorrect.

In general, any arbitrary
voltage pattern we launch into a pair of transmission
lines will change as it propagates down the line.

Specifically, it changes for asymmetrical patterns due to different propagation speed of odd and even waves. The differential voltage waveform V1-V2 does not change.

--- Updated Jul 5, 2020 ---

Despite of my objections, thanks for clarifying the context of figure 11-21. To absorb the asymmetrical wave pattern at the far end completely, you need a combination of differential and common mode termination.

 

[vfone]

When I mentioned that the lines in figure 11-21 are not terminated, I mean that the author of the book didn't terminate the lines and is no information in this regards.
My oppinion is that the lines should have termination information.
Anyway, there are many ambiguities in this book, many of them having a lack of setup explanations...

 

[ggennai]

Hi I think that the termination does not care cause in the image the slope did not arrives to the far end yet.
I think the author shows secesive images of the traveling of a slope in a long line. This is the active line. In adition he shows the crosstalk in the quiet line.
My question : Is the slope low to high as is written in the image text ? or there is a mistake and the image is of a hight to low slope ?

 

[FvM]

I agree with your conclusion about far end termination outside the picture's scope. The waveform is plausible and can be reproduced in a simulation, see post #5.

The cause for the step in V1 and the negative pulse in V2 is different propagation speed of odd and even wave. Your simulation setup must include this property to get a realistic waveform.

 

[ggennai]

Well, I found the mistake. It was in the x axis. It is not "Position along the line". It is simply time.
In the image my simulation with LTSpice.

 

[missmoradi]

thanks for good information

https://electrobenyamin.ir