Sercos Interpolation

What does MDT stand for? In this case it is not magnetostrictive displacement transducer.

Also, does the master download integer counts or floating point position? Is this an option? There are implications either way. If counts are used then the feedback better be very fine. It the feedback is 32 bit floats then the resolution is dependent on the position. Floating point resolution changes as a funciton as follows: resolution=position/2^24.
 
MDT stands for Master Data Telegram. The master sends one of these to each slave on the SERCOS ring each SERCOS cycle. Each slave responds with and AT or Axis Telegram. These are not completely freely configurable but there is significant latitude.


I have only seen integers passed on SERCOS. They are typically formatted values with an implied decimal point whether they are 16 or 32 bit. I'm reasonably sure all the S parameters are this way. The P parameters are a bit more open so I can't say with any certainty if floats are supported their. The SERCOS standard may have something to say on this but, again, I'm not sure about that.

Keith
 
Peter Nachtwey said:
Also, does the master download integer counts or floating point position? Is this an option? There are implications either way. If counts are used then the feedback better be very fine. It the feedback is 32 bit floats then the resolution is dependent on the position. Floating point resolution changes as a funciton as follows: resolution=position/2^24.
Position is always a 32bit scaled DINT.
There are parameters to define if it is in inches, degrees, mm, or increments and where the assumed decimal place is.
 
This seems to becoming a weekly quiz...

Ok, at the end, can we make conclusion: what would make SERCOS better?
That was original question if I remember good...
 
Graphs showing Cubic Interpolation using 3 resolutions

ftp://ftp.deltacompsys.com/public/NG/Mathcad%20-%20Cubic%20Interpolation%20NG.pdf

I have some graphs showing a sinewave motion profile and the last part of a move to 2 meters. I used 3 resolutions. The first assumes positions downloaded are 64 bit floats. This should be close to infinite resolution for all practical purposes. The second set of graphs assumes the positioins are specified with a resolution of 1 nanometer. With this resolution the target position would be limited to -2 to +2 meters which isn't very far. The third set of graphs assumes the positions are speficied in microns. This allows for a much longer range of travel but the interpolation suffers.

Note that even when using 64 bit floats there is a bobble in the acceleration at the point where the motion segments join at 20 ms. This is due to the fact that the interpolation is occuring over two motion segments and that the interpolation routine really doesn't know that when it gets to 2 meters the velocity and acceleration should be 0. It would be better if the position, velocity and acceleration at the end points was speficied by the master so the slave does not have to interpolate.

Edit, Pandiani's solution was correct but not simplified.
 

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