[Math] Will LQR act like MPC in reality

control theorylinear-controloptimal controloptimization

MPC is a predictive controller. Which means that MPC will analyse the best input values $u$ to get the shortest way from setpoint to reference point in trajectories $x$.

MPC is very well used in the industry. But my question is:

As I heard, LQR with saturation limits on $u$ is equal to MPC. Because LQR do the same math as MPC. The difference is that MPC have some limits in the input signal. I'm talking about the very basic MPC now.

That make me wonder what will be the difference between implementation of a controller with saturation and a controller with no saturation.

Image that we have a car and the car starting from 0 and the goal is 100.
The controller's mission is to speed up the car so the car can recive 100 at a few seconds, without over shoot.

So, let's assume that we are implementing a LQR controller inside the car and start the controller. The LQR gives full signal into the fuel injection module inside the car, but in reality, the LQR is implemented inside a computer and the computer's signal output is limited. Which results that the fuel injection model cannot give full power to the engine inside the car.

Question:

Due to the limits inside the car and the computer. The LQR controller will act like it has saturation in the input, and the results will be that LQR in reality will act like MPC in a simulation?

And this expand to a following question: If I want to simulate a process inside my computer, a MPC is better preferred that LQR, due to the built
in saturation/constraint limits in the MPC controller?

Best Answer

No, an LQR controller (or trivially saturated LQR controller) will not give the same control signal as an MPC controller. You can (and typically want to) tune he MPC controller though such that it coincides with the LQR feedback once the system enters the state where the LQR feedback is and remains unsaturated.

If LQR gave the same control as MPC we would never use MPC, as MPC is several orders of magnitudes more computationally expensive.

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