LQRCentralizedLTI
Updated:
Tags: control, documentation, lti
Sintax
[K_inf,P_inf] = LQRCentralizedLTI(A,B,Q,R,E)
[K_inf,P_inf] = LQRCentralizedLTI(A,B,Q,R,E,opts)
Description
Consider a generic LTI system of the form
\[\mathbf{x}(k+1)=\mathbf{A}\mathbf{x}(k)+\mathbf{B}\mathbf{u}(k)\;\]where $\mathbf{x}(k)\in\mathbb{R}^{n}$ is the state vector, $\mathbf{u}(k)\in \mathbb{R}^{m}$ is the input vector, and $\mathbf{A}$ and $\mathbf{B}$ are constant matrices of appropriate dimensions.
Consider a standard LQR regulator
\[\mathbf{u}(k) = -\mathbf{K}(k)\mathbf{x}(k)\:,\]where $\mathbf{K}(k)\in\mathbb{R}^{m\times n}$ is the regulator gain. It is often very useful to find a steady-state constant gain $\mathbf{K}_{\infty}$ instead. Consider a regulator cost function
\[J(k) = \sum_{\tau=k}^{\infty}\left(\mathbf{x}^T(\tau)\mathbf{Q}\mathbf{x}(\tau)+\mathbf{u}^T(\tau)\mathbf{R}\mathbf{u}(\tau)\right)\:,\]where $\mathbf{Q} \succeq \mathbf{0}\in\mathbb{R}^{n\times n}$ and $\mathbf{R} \succ \mathbf{0}\in\mathbb{R}^{m\times m}$ are the weighing matrices of the state vector and control action, respectively. It can be shown that there is a matrix $\mathbf{P}(k) \succ \mathbf{0}\in\mathbb{R}^{n\times n}$ such that
\[J(k) = \mathbf{x}^T(k)\mathbf{P}(k)\mathbf{x}(k)\:.\]The commands
[K_inf,P_inf] = LQRCentralizedLTI(A,B,Q,R,E)
[K_inf,P_inf] = LQRCentralizedLTI(A,B,Q,R,E,opts)compute the well-known steady-state centralized gain that solves the optimization problem
Input arguments
Required
A: matrix $\mathbf{A}$ of the dynamics of the LTI systemB: matrix $\mathbf{B}$ of the dynamics of the LTI systemQ: state weighting matrix, $\mathbf{Q} \succeq \mathbf{0}$R: control action weighting matrix, $\mathbf{R} \succeq \mathbf{0}$
Optional
opts: struct of optional arguments (assumes default value for each parameter which is not assigned by the user)epsl: minimum relative improvement on the objective function of the optimization problem (default:opts.epsl = 1e-5)
maxIt: maximum number of iterations until convergence (default:opts.maxIt = 1000)verbose: display algorithm status messages (default:opts.verbose = false)
Output Arguments
K_inf: steady-state regulator gain matrix $\mathbf{K}_{\infty}$P_inf: steady-state matrix $\mathbf{P}_{\infty}$
Examples
Synthetic system
To open this example execute the following command in the MATLAB command window
open LQRCentralizedLTITutorial
Use synthetic system matrices $\mathbf{A}$, $\mathbf{B}$, $\mathbf{Q}$, $\mathbf{R}$, and $\mathbf{E}$
n = 5;
m = 3;
A = [0.6085 0.0188 0.9615 0.6161 0.0494;
0.3959 0.6575 0.7305 0.5338 0.4611;
0.1743 0.9322 0.3419 0.2229 0.1466;
0.5204 0.7850 0.2257 0.4315 0.8538;
0.8603 0.8842 0.1132 0.7550 0.3850];
B = [0.3843 0.7494 0.4509;
0.1446 0.5369 0.0092;
0.6133 0.6413 0.6322;
0.3401 0.4020 0.3725;
0.7084 0.4744 0.0031];
Q = [0.8057 0.9316 0.9227 0.7569 0.6049;
0.9316 2.6200 1.2024 1.4863 1.6318;
0.9227 1.2024 1.1229 0.9413 0.6800;
0.7569 1.4863 0.9413 1.3068 1.2506;
0.6049 1.6318 0.6800 1.2506 1.5626];
R = [1.0134 0.5867 0.9654;
0.5867 0.4666 0.4427;
0.9654 0.4427 1.5383];
E = [1 1 1 0 1;
0 0 1 0 0;
0 0 1 1 1];
Synthesize the centralized regulator gain (with some optional parameters)
opts.verbose = true;
[Kinf,Pinf] = LQRCentralizedLTI(A,B,Q,R,opts);
Kinf
----------------------------------------------------------------------------------
Computing centralized LQR gain with: epsl = 1e-05 | maxIt = 1000.
Convergence reached with: epsl = 1e-05 | maxIt = 1000.
A total of 2 iterations were run.
----------------------------------------------------------------------------------
Kinf =
0.1064 -0.0372 -0.7317 -0.0232 0.1165
0.9813 1.3256 1.3547 1.0714 0.7619
-0.3304 -0.2918 0.1180 -0.2828 -0.2849