Jack Ensor wrote: > Jon Elson wrote: > >> Jack Ensor wrote: >> >> >>> I noticed when jogging at my maximum rate of 90 ipm the 2 quadrature >>> signals coming at a rate of 1 Khz have approximately 50 micro seconds of >>> jitter. Is this excessive? How much would it contribute to tracking >>> error? My tracking error is insignifcant when homing but but huge when >>> running Axis.ngc. >>> >>> >> Would you please define this term "tracking error"? I do not >> know what it means. From context, I believe you mean to say the >> machine's position differs from the displayed position. Is that >> correct? >> >> > Yes, position displayed on the axis screen differs from what my dro says > except when I home they always agree. > >> How are you seeing the quadrature signals? On a "box" >> oscilloscope, or somehow with halscope? Unless halscope AND the >> software or hardware encoder input facility is sampling at a >> fast enough rate, you would miss some of the edges. >> > Yes, I understand the hal (storage scope) better now. When I used the > faster sample rate, I then got more resonable results > >> For >> instance, on my minimill, at 60 IPM, with 16 TPI leadscrews and >> 4:1 motor reduction, and with 500 CPR encoders producing 2000 >> counts/revolution, you get 128,000 counts per second. >> Therefore, counts are coming at a rate of one every 7.8 us. >> Obviously, my jitter must be less than yours. But, a scope >> would need to be sampling it at a rate of once a microsecond or >> better before you could even begin to discern jitter on the >> signal. If you are using an analog oscilloscope, then there is >> no sampling. But, without specifying the rate of encoder pulses >> when you see the 50 us jitter, it is hard to know what it means. >> If you had 50 us jitter when the count rate was one millisecond, >> it is not a big deal. If it was when the count rate was 50 us, >> it would be reducing the quadrature angle to zero, and would >> clearly cause errors. So, you have to compare the jitter to the >> count rate. >> > The rate as I originally stated was 1 Khz which translates to a pulse > period of .5 milliseconds low and .5 milliseconds high. So I suppose 50 > micreoseconds jitter isn't too bad then. (About 5%). > >> Ideally, there should be 90 degrees between the 4 >> states of the encoder's A and B signals. They never are, due to >> tiny errors in the manufacturing of the encoder's optics. The >> greater the error, the narrower some of the count states become, >> until they become so small the encoder counter's logic misses >> them. Then, the position will be off by multiples of 4 counts. >> >> When you say homing is OK, but axis is bad, is that all due to >> speed? >> >> > Slowing things down by a factor of ten makes no difference in position > error. It still jumps all over the place. > > Could you explain why the following speed calculation is in error? > I have a unipolar motor, driven in quadrature phase A, phase A not, > Phase B, and phase B not., where phase B lags phase A by 90 degrees. > Motor plate specifies 200 steps/rev > step down from motor to screw: 2.5 to 1 > Screw pitch: .2 in/rev > .2 in/rev x 1/2.5 rev/rev x1/200 rev/step = .0004 in/step. This is > correct because this is what I see the system do. > > For speed: > The max jog speed is set in emc to 90 ipm (1.5 in/sec). When jogging at > the max rate I measured a step frequency of 813 Hz on phase A. > Calculating the table speed: > 800 pulses/sec x 60 sec/min x .0004 in/step = 19.2 ipm > However just by looking at the table move, it is moving much faster than > that. > Is this because due to the nature of quadrature drive, the table > actually moves 4 times faster than the step rate?
Not quite, but close. It IS due to the nature of quadrature, but the table isn't moving faster than the step rate - the individual phase signals are changing 4 times slower than the step rate. When you see a 813 Hz square wave on one of the quadrature signals, that doesn't mean you are stepping at 813Hz. That means you are going through 813 complete quadrature cycles per second, but each cycle is four steps: A B 0 0 step 1 0 1 step 2 1 1 step 3 1 0 step 4 One cycle on phase A (or B) at 813 cycles per second means that you have gone four steps at 3252 steps per second. > This would put it more in the ball park of what I am seeing. Looks good to me. Regards, John Kasunich ------------------------------------------------------------------------- This SF.net email is sponsored by DB2 Express Download DB2 Express C - the FREE version of DB2 express and take control of your XML. No limits. Just data. Click to get it now. http://sourceforge.net/powerbar/db2/ _______________________________________________ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
