使用 Python 代码实现更快的正交解码器循环 [英] Faster quadrature decoder loops with Python code

问题描述

我正在使用 BeagleBone Black 并使用 Adafruit 的 IO Python 库.编写了一个简单的正交解码函数，当电机以大约 1800 RPM 运行时，它工作得非常好.但是当电机以更高的速度运行时，代码开始丢失一些中断，编码器计数开始累积错误.你们对我如何使代码更高效或者是否有可以以更高频率循环中断的函数有什么建议吗.

I'm working with a BeagleBone Black and using Adafruit's IO Python library. Wrote a simple quadrature decoding function and it works perfectly fine when the motor runs at about 1800 RPM. But when the motor runs at higher speeds, the code starts missing some of the interrupts and the encoder counts start to accumulate errors. Do you guys have any suggestions as to how I can make the code more efficient or if there are functions which can cycle the interrupts at a higher frequency.

谢谢，凯尔

代码如下:

# Define encoder count function
def encodercount(term):
global counts       
global Encoder_A
global Encoder_A_old
global Encoder_B
global Encoder_B_old
global error


Encoder_A = GPIO.input('P8_7')  # stores the value of the encoders at time of interrupt
Encoder_B = GPIO.input('P8_8')

if Encoder_A == Encoder_A_old and Encoder_B == Encoder_B_old:
# this will be an error
    error += 1
    print 'Error count is %s' %error

elif (Encoder_A == 1 and Encoder_B_old == 0) or (Encoder_A == 0 and Encoder_B_old == 1):
# this will be clockwise rotation
    counts += 1
    print 'Encoder count is %s' %counts
    print 'AB is %s %s' % (Encoder_A, Encoder_B)

elif (Encoder_A == 1 and Encoder_B_old == 1) or (Encoder_A == 0 and Encoder_B_old == 0):
# this will be counter-clockwise rotation
    counts -= 1
    print 'Encoder count is %s' %counts
    print 'AB is %s %s' % (Encoder_A, Encoder_B)

else:
#this will be an error as well
    error += 1
    print 'Error count is %s' %error

Encoder_A_old = Encoder_A     # store the current encoder values as old values to be used as comparison in the next loop
Encoder_B_old = Encoder_B       

# Initialize the interrupts - these trigger on the both the rising and falling 
GPIO.add_event_detect('P8_7', GPIO.BOTH, callback = encodercount)   # Encoder A
GPIO.add_event_detect('P8_8', GPIO.BOTH, callback = encodercount)   # Encoder B

# This is the part of the code which runs normally in the background
while True:
    time.sleep(1)

推荐答案

让代码更高效...

def encodercount(term):
global counts       
global Encoder_A
global Encoder_A_old
global Encoder_B
global Encoder_B_old
global error

Encoder_A,Encoder_B = GPIO.input('P8_7'),GPIO.input('P8_8')

if ((Encoder_A,Encoder_B_old) == (1,0)) or ((Encoder_A,Encoder_B_old) == (0,1)):
# this will be clockwise rotation
    counts += 1
    print 'Encoder count is %s
AB is %s %s' % (counts, Encoder_A, Encoder_B)

elif ((Encoder_A,Encoder_B_old) == (1,1)) or ((Encoder_A,Encoder_B_old) == (0,0)):
# this will be counter-clockwise rotation
    counts -= 1
    print 'Encoder count is %s
AB is %s %s' % (counts, Encoder_A, Encoder_B)

else:
#this will be an error
    error += 1
    print 'Error count is %s' %error

Encoder_A_old,Encoder_B_old = Encoder_A,Encoder_B

# Initialize the interrupts - these trigger on the both the rising and falling 
GPIO.add_event_detect('P8_7', GPIO.BOTH, callback = encodercount)   # Encoder A
GPIO.add_event_detect('P8_8', GPIO.BOTH, callback = encodercount)   # Encoder B

# This is the part of the code which runs normally in the background
while True:
    time.sleep(1)

最大的好处将来自print的单一调用.打印到 stdout 通常很慢，这会限制程序的性能.您应该考虑每 20 次打印一次或稍微减少打印一次.

The greatest benefit will come from the single call of print. Printing to the stdout is slow in general and this will limit the performance of your program. You should consider to print out only every 20th time or somewhat less often.

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