@T Ansley posted:That graph is NOT the actual voltage, that is the digital signal. Read this, it clearly says there is no negative voltage. The digital signal rides above 0 volts. This is a good explanation.
Here's an excerpt:
"This is not an AC[15] signal, as the rail will have positive voltage or none at any point in time. What is happening is the current is changing direction as it moves from one rail to another. Measuring this with an oscilloscope will display a peak-to-peak signal, which many will claim supports their assertion that the signal is an AC waveform. Since this measurement is made with the track outputs floating, the trace indicates which connection was more positive than the other at a point in time.[16]
Mathematically, the signal would be expressed as X +jY + X − jY, where X is time and Y is the amplitude or voltage.
This method of transmitting data and power in the same signal results in a very robust signalling technique with a high signal to noise ratio while reducing the space charge around the rails and the electrostatic attraction of contaminants. A multifunction decoder can receive data regardless of the orientation of the locomotive, as either rail has data available."
Ted
Ted, thank you for trying to explain this. I clicked on your link and read it several times but I am still having trouble fully understanding it. I also read some other information on digital signals but they weren't a whole lot of help. I am not scientist nor an engineer but I would really like to understand how this works.
Okay, I get that the voltage never goes below 0 volts. Does the booster send the signal down one rail and use the other rail for a return or does the booster alternate sending the signal between rail A and rail B? If it does alternate the signal between rail A and rail B at what point in time does it switch rails? Can this be seen on the graph?
"The track voltage has phase, one rail is always the inverse of the other." What exactly does this statement mean?
Anther thing I found confusing was these two statements at the bottom of the page from your link:
"The 0V reference point is the chassis of the booster."
"The output of the booster is floating, as there is no fixed reference point such as a chassis ground for zero volts."
They seem to contradict one another.
During my research I found this Marklin forum where guys were talking about the same thing we are talking abut here.
https://www.marklin-users.net/...log-on-DCC-Mea-Cupla
Doesn't add much to the conversation but I thought it was interesting to know that we weren't the only ones discussing this.
