[DFTF] Closing the loop on Wheel Velocity...

[theprgramerdude]

Originally Posted by Ether

Or you could gear up the encoder a bit I suppose so it is spinning faster. Kind of kludgy but it might work since the encoder is virtually no load. Have to be careful not to exceed signal processing capability of Jag.

It has a 50 MHz Cortex M-3. Unless the PID algorithm onboard is rather complex and heavy on the cycles, how could anything from a digital encoder approach the Jag's limits?

[EricVanWyk]

Originally Posted by theprgramerdude

It has a 50 MHz Cortex M-3. Unless the PID algorithm onboard is rather complex and heavy on the cycles, how could anything from a digital encoder approach the Jag's limits?

The Stellaris microcontroller does quadrature decoding in dedicated hardware, so there is no additional processing burden for running the encoders faster. The upper limit of that circuitry is 1/4th the speed of the processor. For a part running at 50MHz*, that is roughly 750,000 RPM for a 250 count encoder. However, I doubt you'd get clean signals from the poor encoder at those speeds!

The Stellaris has more computational horsepower available than the entire IFI control system put together several times over. It is an amazing show of what Moore's law does to the industry.

* Edit: I can't remember what speed they actually run on the Jag - there simply is no need to run at the full 50 or 80 MHz. Suffice it to say that it is still overkill.

[Joe Johnson]

Originally Posted by EricVanWyk

The Stellaris microcontroller does quadrature decoding in dedicated hardware, so there is no additional processing burden for running the encoders faster. The upper limit of that circuitry is 1/4th the speed of the processor. <snip>

Do you know if the processor "does the right thing" in terms of speed calculations, by which I mean, does it get 4 readings per nominal encoder tick (rising edge channel A, rising edge channel B, falling edge channel A, falling edge channel B) AND create the time as rising edge to (the same) rising edge and falling edge to (again, the same) falling edge?

The reason I say this is that the encoder folks do a much better job at spacing their ticks than they do at getting exactly 50-50 ON-OFF duty cycles and spacing the phasing between the sensors A and B (remember their is only 1 encoder wheel, the 90 degree shift in the signal between the two phases is done by physically spacing the encoders N+/-(1/4) encoder pulses apart.

The reason I ask is that I want to get 4X the speed readings doing it this way provides over just a single speed reading per encoder tick which is the most obvious way to do things (see the above discussion about keeping high PID loop times at slow speeds -- without 1,000Hz sensor reading, it makes little sense to have a 1,000Hz update rate on the PID loop).

Joe J.