I don't fully understand what the benefit is to having the last 2 samples in integral calculation. Isn't the area from t-1 to t-2 already represented in previous integral? Why not just do Ts((e(t)+e(t-1))/2)+previous integral? Wouldn't that also give you the integral? I know this video is quite old but I hope that you might still read the comments and are willing to explain this a bit more.
@Pickering-Control5 ай бұрын
Hey, yes you’re right. You could just take the area from the last integrated error, and sum that up with the previous integrals.
@eduardoc125211 ай бұрын
Great video! I have a question regarding the proportional (kp), integral (ki), and derivative (kd) gains in 17:30. Are these gains the same as those used in the continuous-time domain?