Controlling a Difficult Systems


#1

Last summer we hired a college student to design a difficult to control system. The goal was to design the system so the natural frequency and damping factor is very low. He succeeded. The natural frequency turn out to be about 6.5 Hz and the damping factor about 0.1 to 0.15. The main problem is is the damping factor. Motion controllers with only a PID will not be able to control the system in any meaningful way because the gains would need to be too low. The RMC controller has a second derivative gain that compensates for the low damping factor. This help use place the closed loop poles on the negative real axis in the s-plane so the response to any disturbance will not over shoot.

The video shows how I first attempt to use the auto tuner ( gain optimizer ) using only the PID gains. You can see that the control is poor. Then I use the same data to calculate the controller gains for the PID2, a PID with second derivative gain. You can see this is successful.

The question that should be asked is why does the second derivative gain make such a difference?
The answer can be found by looking at the characteristic equation of a hydraulic cylinder and load

\small CE\left(s\right)={ s }^{ 4 }+\left(K2\cdot K\cdot { \omega }_{ n }^{ 2 }+2\cdot\zeta\cdot{ \omega }_{ n } \right)\cdot{ s }^{ 3 }+\left( { Kd\cdot K\cdot { \omega }_{ n }^{ 2 }+\omega }_{ n }^{ 2 }\right)\cdot { s }^{ 2 }+Kp\cdot K\cdot{ \omega }_{ n }^{ 2 }\cdot s+Ki\cdot K\cdot{ \omega }_{ n }^{ 2 }

You can see that every power of s has a controller gain. Notice that the second derivative gain is in the s^3 term. Notice also that the damping factor is also in the s^3 term. The second derivative gain can be increased to compensate for the damping factor decreases. If there is no second derivative gain then K2=0 so there is no way to compensate for low damping factors. In fact, a PID controller will only work if the damping factor so high that the second derivative gain is not needed.