[Chris Hibner]

Originally Posted by Ether

Yeah, but I doubt that the counter in the FPGA is floating-point.

I want to know if the FPGA counter will not overflow during a match.




Does the above code actually write a new value to the FPGA counter itself? Or just zero the associated RAM variable in cRIO?


~

If WPI and NI did things properly, it really doesn't matter if the FPGA counter overflows or not.

Let's assume they did things correctly. One possible implementation is that the FPGA services the encoder interrupts and keeps track of the counter in a signed 16-bit register. Then the WPI interface samples that register at a given time interval. The WPI code would be like the following:

// variable declarations
s16 encoderCnt = 0;
s16 encoderCntPrev;
s16 encoderDiff;
double encoderOut;

// ...

// in the loop

encoderCntPrev = encoderCnt;
encoderCnt = getFPGAencoderCnt();
encoderDiff = encoderCnt - encoderCntPrev;

encoderOut = encoderOut + encoderDiff*inchesPerCount;

// etc.

(by the way, s16 is a typdef'd variable type for a signed 16 bit integer variable. The actual declaration will depend on your micro)


With the above code, it does not matter if the FPGA encoder counter overflows. If you don't believe me, try the following code:

s16 counter1, counter2, counter3;

counter1 = 32765;
counter2 = counter1 + 5;

cout << "counter1 = " << counter1 << "\n";
cout << "counter2 = " << counter2 << "\n";

counter3 = counter2 - counter1;

cout << "counter3 = " << counter3 << "\n";

If you don't like C/C++, you can do the same thing in LabVIEW. You'll notice that the two's compliment math handles the wraparound flawlessly and counter3 will equal 5. With the subtraction, you'll never know an overflow occurred - and it doesn't really matter.

This is actually a very slick way of handling angles to avoid having to do a lot of wrap-around math and checking. If you always scale your angles such that 360 degrees is equal to the overflow point, the two's compliment math automatically takes care of all of the wrapping for you.