I was reading the source code in PID_v1.cpp in conjunction with associated blog posts on Brett's site, and I spotted something interesting.
On line 71-72 of PID_v1.cpp, we have the following:
/*Add Proportional on Measurement, if P_ON_M is specified*/
if(!pOnE) outputSum-= kp * dInput;
To my understanding, kp * dInput is the P term when P_ON_M is specified.
However, if we had P_ON_E specified instead, the P term is added to the output instead of being accumulated into outputSum (lines 77-83):
/*Add Proportional on Error, if P_ON_E is specified*/
double output;
if(pOnE) output = kp * error;
else output = 0;
/*Compute Rest of PID Output*/
output += outputSum - kd * dInput;
In this case, I think kp * error is the P term when we have P_on_E specified.
Why is the P term treated differently when it is proportional on measurement? Why is it being accumulated into outputSum, which seems to be relevant only to the I term?
I am not sure if this is correct and something eclipsed my understanding, or if this is a mistake that should be fixed (if so, I can drop a pull request to fix it). If someone could enlighten me on this question, it would be much appreciated. Thanks!
I was reading the source code in
PID_v1.cppin conjunction with associated blog posts on Brett's site, and I spotted something interesting.On line 71-72 of
PID_v1.cpp, we have the following:To my understanding,
kp * dInputis the P term whenP_ON_Mis specified.However, if we had
P_ON_Especified instead, the P term is added to theoutputinstead of being accumulated intooutputSum(lines 77-83):In this case, I think
kp * erroris the P term when we haveP_on_Especified.Why is the P term treated differently when it is proportional on measurement? Why is it being accumulated into
outputSum, which seems to be relevant only to the I term?I am not sure if this is correct and something eclipsed my understanding, or if this is a mistake that should be fixed (if so, I can drop a pull request to fix it). If someone could enlighten me on this question, it would be much appreciated. Thanks!