Very small u control values in LQR initial response

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Alessandro Maria Laspina 2021-7-1

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https://ww2.mathworks.cn/matlabcentral/answers/869603-very-small-u-control-values-in-lqr-initial-response

评论： Paul 2021-7-5

I have a system described by xdot=Adx+Bdu and y=Cdx+Ddu, where dx and du are the changes in the states and control variables, based on the matrices A, B, C, and D which contain the linearized dynamics of the system. The lqr controller is set up the following way:

The A, B, C, and D matrices:

A =
   -0.0204    0.0994   -0.1976         0  -50.0000   12.0000         0    7.6460    0.6088
   -0.0994    0.0006    0.3023   50.0000         0  -12.0000   -7.5940   -1.7931    9.5517
    0.1927   -0.3085   -0.0359  -12.0000   12.0000         0    1.3239   -5.7965   -2.1517
   -0.0159   -0.0054   -0.0201         0         0         0         0         0         0
    0.2922    0.6400    2.7503         0         0         0         0         0         0
    0.0170    0.0215    0.0906         0         0         0         0         0         0
         0         0         0    1.0000    0.3724    1.2039    0.2035    0.4002         0
         0         0         0         0    0.9553   -0.2955   -0.1546         0         0
         0         0         0         0    0.4754    1.5369    0.2598    0.3135         0
B =
    0.0186   -0.0473    0.0202    0.0100
         0         0   -0.0316   -0.0817
    0.0903    0.0792    0.0843    0.0416
    0.0544   -1.1601    0.5681   -1.4700
   -0.0907   -1.9335    1.5596   -0.7709
   -1.6589    1.1871   -0.9469   -2.4499
         0         0         0         0
         0         0         0         0
         0         0         0         0
C =
   1.0e+03 *
   -0.0159   -0.0054   -0.0201         0         0         0         0         0         0
    0.2922    0.6400    2.7503         0         0         0         0         0         0
    0.0170    0.0215    0.0906         0         0         0         0         0         0
D =
   1.0e+03 *
    0.0544   -1.1601    0.5681   -1.4700
   -0.0907   -1.9335    1.5596   -0.7709
   -1.6589    1.1871   -0.9469   -2.4499

The lqr controller:

Q=1e-3*eye(9,9);
% Q=rand(10,10);
R=1e10*eye(4,4);
% N=eye(10,13)+1e-5;
% N=zeros(10,13);
% R=rand(13,13);
K=lqr(A,B,Q,R);
x0=ones(9,1);
sys=ss((A-B*K),B,C,D);
t=0:1/10:30;
[y,t,dx]=initial(sys,x0,t);

I then retrieve the values of du and apply it to u the following way:

u0 =
   -0.1014
    0.2692
   -0.6107
   -0.2415
   
for ii=1:length(t)
    du(:,ii)=pinv(D)*(y(ii,:)'-C*(dx(ii,:))');
end
u=u0+cumtrapz(t,du(1:4,:)')';

This then gives me the attached results. The controller returns to error, and the y values reach steady state of 0 after 15 or so seconds. The issue however Is that the control values seem incredibly small. In fact, the total change for the control values are in the order of 1e-14 units of control. I have tried also setting the R matrix to be larger. I understand that it may just be that "it is what it is", but I am wondering if I have defined the controls incorrectly or there is a output option for the initial function that gives the control?

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Alessandro Maria Laspina 2021-7-1

I've noticed I may have been calculating the control variables incorrectly.

For an lqr system, the u variables (or du in this case) can be found by:

x=-K*x (mentioned in this video: What Is Gain Scheduling? | Control Systems in Practice, Part 2 - YouTube)

I am not entirely sure, so a confirmation would be appreciated.

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Answer 1

Paul 2021-7-2

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/869603-very-small-u-control-values-in-lqr-initial-response#answer_738113

编辑：Paul 2021-7-2

在 MATLAB Online 中打开

For this dicussion, I'm going to just use x and u as the symbols for the state and control variables of the linearized model, instead of dx and du, but with the understand that x and u represent pertrubations to the state and control around which the system is linearized.

We start with the linear model:

xdot = A*x + B*u

y = C*x + D*u

A =[
   -0.0204    0.0994   -0.1976         0  -50.0000   12.0000         0    7.6460    0.6088
   -0.0994    0.0006    0.3023   50.0000         0  -12.0000   -7.5940   -1.7931    9.5517
    0.1927   -0.3085   -0.0359  -12.0000   12.0000         0    1.3239   -5.7965   -2.1517
   -0.0159   -0.0054   -0.0201         0         0         0         0         0         0
    0.2922    0.6400    2.7503         0         0         0         0         0         0
    0.0170    0.0215    0.0906         0         0         0         0         0         0
         0         0         0    1.0000    0.3724    1.2039    0.2035    0.4002         0
         0         0         0         0    0.9553   -0.2955   -0.1546         0         0
         0         0         0         0    0.4754    1.5369    0.2598    0.3135         0];
B =[
    0.0186   -0.0473    0.0202    0.0100
         0         0   -0.0316   -0.0817
    0.0903    0.0792    0.0843    0.0416
    0.0544   -1.1601    0.5681   -1.4700
   -0.0907   -1.9335    1.5596   -0.7709
   -1.6589    1.1871   -0.9469   -2.4499
         0         0         0         0
         0         0         0         0
         0         0         0         0];
C =   1.0e+03 * [
   -0.0159   -0.0054   -0.0201         0         0         0         0         0         0
    0.2922    0.6400    2.7503         0         0         0         0         0         0
    0.0170    0.0215    0.0906         0         0         0         0         0         0];
D =   1.0e+03 * [
    0.0544   -1.1601    0.5681   -1.4700
   -0.0907   -1.9335    1.5596   -0.7709
   -1.6589    1.1871   -0.9469   -2.4499];

Now we introduce the state feedback control

u = -K*x

Q=1e-3*eye(9,9);
R=1e10*eye(4,4);
K=lqr(A,B,Q,R);

and substitute:

xdot = A*x + B*(-K*x) = (A - B*K)*x

y = C*x + D*(-K*x) = (C - D*K)*x

So the resulting closed loop system is:

xdot = (A - B*K)*x

y = (C - D*K)*x

Note that the closed loop system no longer has a "B" or "D" matrix because the system is not being driven by an external input, like a reference command (which can be introduced if needed, but it's a different input than u, which is the control input that actually drives the plant). So the closed loop model in Matlab should be:

sys = ss(A-B*K,zeros(size(A,1)),C-D*K,0); % contrast with the code in the Question, especially the C-argument

Note that I just made the "B" and "D" matrices 0 because we have to put something in the ss model. Note also that ss automatically expands the scalar 0 in the D-argument to appropriate dimensions.

Now to simulate the IC response

x0=ones(9,1);
t=0:1/10:30;  % might want to consider a smaller dt?
[y,t,x]=initial(sys,x0,t);

To recover the control input that actually was driving the plant dynamics, the approach in your comment is correct:

u = (-K*x.').';

subplot(211);plot(t,y),grid

subplot(212);plot(t,u),grid

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Paul 2021-7-2

编辑：Paul 2021-7-5

Actually, the LQR controller objective is to drive the states to zero in a way that minimizes the LQR objective function, which essentially trades off the manner in which the states are regulated to zero with the control effort it takes to do so. Of course, if the states go to zero, then so will the control input, and so will the output.

Having said that, if the plant has D = 0 (your's does not), then you can select

Q = C.'*C % assuming C is such that Q is positive definite

in which case

x.' * Q * x = y.' * y

i.e., the LQR cost function is actually a specification on the output response.

If you want the output, y, to track some reference signal r, one approach (there may be others) would be to augment the plant model with integrators, and make each element of y an input to an integrator. Then design the LQR controller for the augmented plant. Then introduce the reference input r so that the input to each integrator becomes the error in y. effectively, the control scheme becomes:

u = -K1*x + K2/s*(r - y)

where K1 and K2 are partitions of the LQR feeback gain matrix K and assumes that full state vector x and the output vector y are measured for feedback. It's easier to explain in a picture than in words. It may be easier to understand for the case where the ouput, y, is one or more state variables, but that's not the case in the problem in the Question.

Alessandro Maria Laspina 2021-7-5

My states are perturbed states, so I would need some form of integral constraints on the on the control, i.e. my state-space is:

xdot=Adx+Bdu;

y=Cdx+Bdu;

I don't have any constraints on the states (dx) or the integral of the states (x), or the derivative (rates, which in this case they would be accelerations)

What are my integrator outputs, and why would they be 3?

I am simply trying to limit the values of the integral of my control du. How can I specify the integral of the control for an lqi controller?

Paul 2021-7-5

There are three additional integrators because the plant has three outputs and I assumed you wanted each to track its own reference signal.

As for the other questions, they've become somewhat far afield of Matlab itself. Again, suggest you undertake a search for good references on LQ control as it applies to the problem(s) that you are trying to solve.

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