Step 7 counter card turn over problem

Bratt · Nov 5, 2009

I have a problem with a resolver connected to a ET200s counter module. I want to convert the value from the resolver to 0-->359 degrees and then start over. The axis can turn both backwards and forwards.

The following formula works the only problem is when the counter card hits the max value of +2147483647 its starts over and begins counting from -2147483648. The problem is that the value then reverses and begin counting backwards. The same goes when it hits min value -2147483648 then it starts over from +2147483647.

Anyone got any ideas?

Code:

      L     #Encoder.Feedback.Count_Val // Value from counter card
      L     L#1040                      // One turn on the resolver will give 1040 pulses
      /D    
      L     L#360    // Mod to get it to count from 0 to 359 and then start over
      MOD   
      T     #oVALUE  // Value 0 to 359

Peter Nachtwey · Nov 5, 2009

Bratt said:
The following formula works the only problem is when the counter card hits the max value of +2147483647 its starts over and begins counting from -2147483648.

Not if you shift it right once before using the counter. Now all the numbers are positive but you gave away half your resolution.
What kind of encoder returns 1040 counts per revolution?
Usually there are so many bits per turn and an upper turn counter. You should have something like 10 bits per revolution or 1024 counts, not 1040.

Anyone got any ideas?

Many

rdrast · Nov 5, 2009

Take advantage of signed integer math.

Every single scan:

Code:

Count_Delta = Current_Counter_Val - Last_Scan_Counter_Val

Last_Scan_Counter_Val = Current_Counter_Val

Encoder_Counts = Encoder_Counts + Count_Delta

Manglemender · Nov 5, 2009

If you are getting 1040 pulses per rev (odd number?) why not set the counter up in periodic counting mode rather than continuous count mode then angle becomes (count/1040)*360.

Nick

Jeebs · Nov 5, 2009

Code:

      L     #Encoder.Feedback.Count_Val // Value from counter card
      L     DW#16#7FFFFFFF     // Get rid of negative sign??
      AD
      L     L#1040      // One turn on the resolver will give 1040 pulses
      /D    
      L     L#360   // Mod to get it to count from 0 to 359 and then start over
      MOD   
      T     #oVALUE  // Value 0 to 359

But then again, would only work if the actual counts per revolution was 1024, which I thought was standard?
Would give you quite a few revs before it resets.

If the actual count is 1040 per 1°....well then you need to adjust your math a bit.

The part I added will get rid of the sign. (last time I checked it worked like that...)

Bratt · Nov 5, 2009

My mistake it was 1000 pulses per revolution

Bratt · Nov 5, 2009

Getting rid of the sign might work i just need to keep track of the boundaries and adjust the degree value when the boundaries get crossed. I'll give it a try

Bratt · Nov 5, 2009

Manglemender said:
If you are getting 1040 pulses per rev (odd number?) why not set the counter up in periodic counting mode rather than continuous count mode then angle becomes (count/1040)*360.

Nick

I can only adjust the max value in periodic counting mode when counting either negative or positive not both or am i missing something?

rdrast · Nov 5, 2009

Using a delta calculation (if this is a true, 32 bit signed counter) will also eliminate worrying about the sign.

Subtracting as I suggested will always give an appropriate value, and as a bonus, you never have to play with zeroing out a counter module, it's easy to pause, even easy to derive multiple virtual counters from.

To demonstrate with 8 bit signed math:

Code:

NEWCOUNT   OLDCOUNT   NEW-OLD (Delta)
25         15          10
127        126         1
-128       126         2

Bratt · Nov 5, 2009

rdrast said:
Using a delta calculation (if this is a true, 32 bit signed counter) will also eliminate worrying about the sign.
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Thank you!

I tried it seems to work for me. I'll try more tomorrow!
Seems like the best solution so far.

Step 7 counter card turn over problem

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