Place command provided gain matrix which commands inputs to be too large

When using the place command on a MIMO system I recieve a gain matrix.

When testing this gain matrix output at an operating point where the output should be near zero, the output is far greater (30s).

[A,B,C,D] = linmod('LIN_F16Block_Modded', xeval, ueval);

[K,prec,message] = place(A,B,pp)

I checked that the poles of the closed look system were indeed where I had placed them with:

eig(A-B*K)

The inputs that I expect from:

u = -K*xeval

are an order of magnitude or more lower than the ones I recieve.

I considered this might be due to passing a large A matrix to the place command (18x18), so I reduced it to a 4x4 but the commands provided were still similarly an order of magnitude or more greater than expected.

Am I using the place command wrong? Is the place algorithm ill suited for MIMO systems?

[Paul]

What do you mean by:

"The inputs that I expect from:

u = -K*xeval

are an order of magnitude or more lower than the ones I recieve."

What inputs are you receiving from where?

[John Viljoen]

I have expected inputs, u, that I know ahead of time for this particular operating point. It should be a stable point basically, and therefore I expect the controller if poles are placed correctly to provide an input u which will keep it around this point.

The u values that the -K*xeval are providing are much larger than expected, and therefore leave the operating point in time simulation.

Apologies, and thank you!

[Paul]

To be clear on what the problem is ....

The model of the plant in LIN_F16Block_Modded is linearized around the operating point xeval, ueval. Is the operating point an equilibrium point, i.e., when evluated at xeval, ueval are all state derivatives xdot = 0?

The state feedback gain matrix is computed for the linearized model.

The state feedback control is then implemented in the model LIN_F16Block_Modded as u = -K*x?

Then the model is simulated. What are the initial states sent to in this simulation?

Finally, compare the control input u as computed in this simulation to -K * xeval?

Is this the workflow?

[John Viljoen]

• yes the operating point is an equilibrium point
• the state feedback gain matrix is computed for the linearised model
• The state feedback control is indeed implemented as u = -K*x for the model
• The initial states sent to the simulation are the states at the aforementioned equilibrium point
• Therefore the control input is already -K*xeval

That is the workflow!

I would post the whole code on here for brevity but its quite a lot of files so I am not sure if thats the way to go.

I cannot see where the error is right now and I believe I have distilled the problem into its fundamental parts here, many thanks for trying to help again!

[Paul]

Based on this detailed description, the control input at t = 0 should be -K*xeval. If that's not the case then it sounds like there is an error in setting the initial conditions on all of the states in the model, or in the formulation of u = -K*x (like incorrect ordering of the state variables), or both. Of course, as the simulation runs the control input u(t) will evolve in accordance with the closed loop dynamics.

[Mathieu NOE]

hello guys

are we sure that we are not comparing apples and oranges, I mean, when you say that u = -K*xeval , this applies in closed loop , so the closed loop equilibrium point is different from the open loop (another value for xeval , probably much larger, but don't take the open loop value of xeval for predicting what the control would be in closed loop when you do u = -K*xeval ; that's two different scenarios

[Paul]

@Mathieu NOE

I'm not sure which comment your responding to, but if you're responding to my comment that precedes yours, I don't see how your comment is any different than mine. I stated that u = -K*xeval is only true at t = 0 based on the OPs statement that the simulation is initialized at x = xeval. As I also stated, u(t) will then evolve in accordance with the closed loop dynamics (which may or may not lead to closed loop equilibrium point depending on the closed loop stability of the system). Are we not saying exactly the same thing?

[Mathieu NOE]

@Paul

yeap , you're right

sorry if I introduce some confusion in this post

[John Viljoen]

Apologies for the slow response:

I agree that the way the system will evolve will of course be different, however at T=0 the closed loop system should be attempting to keep the system at the position it is intialised at, resulting in zero change to the inputs to the plant from the controller. However this is not the case.

To give a more complete explanation if you have knowledge of aircraft: the aircraft is trimmed at an operating point, the trim conditions and trim inputs are found, the system is linearised at this aforementioned trim condition of states and inputs, the poles are placed using this linearised system, the gain matrix is implemented at the trim operating point and instantaneously provides incorrect inputs to the plant.

Things I have discovered on this adventure that may be of use:

• When all poles are 'placed' at exactly the same positions that they are already at, large gains are selected, when a zero gain matrix would do the same thing. This implies the algorithm for placing poles may be forcing certain poles to other poles desired positions, resulting in higher than expected gains and unexpected behaviour,
• I can get a rough controller functioning just by slapping a bunch of pid's together instead (what I have resorted to for now) (provides expected outputs)
• That the algorithm for placing poles is quite clever and I dont understand it well enough for me to use it in a critical use case currently haha.

Thanks again for you guys trying to solve my problem!

[Paul]

Two comments:

Regarding this statement: " ... at T=0 the closed loop system should be attempting to keep the system at the position it is intialised at ..." I'm not sure about this based on how this problem is described. The pair (xeval, ueval) defines a trim condition. But at T=0 in your closed loop simulation the initial state/input are (xeval/ -K*xeval), which is (likely) not a trim condition. Note that, in general, -K*xeval ~= ueval, so I don't think it is true that at T=0 there should be "zero change to the inputs." From initialization the closed loop dynamics wlll evolve and, if the closed loop is stable, come to steady state at a new trim condition (xfinal, -K*xfinal). Is that not what happens?

Regarding this statement: "When all poles are 'placed' at exactly the same positions that they are already at, ..." I assume this means that the requested closed loop poles are the same as the open loop poles. If so, then it's true that this could be achieved with K = 0. But I suspect that the system in question has more than one control input, so the selection of K is not, in general, unique for a desired set of closed loop poles. Some other restriction has to be applied and the doc page says the algorithm tries to pick K to meet some robustness goals. I don't know whether or not that algorithm typically resutls in "high gain" feedback. The high gain might also just be a consequece of the selection of the closed loop poles relative to the open loop poles.