[python-control-discuss] SF.net SVN: python-control:[22] trunk/examples/pvtol-lqr.py

[<mur...@us...>]

Revision: 22
          http://python-control.svn.sourceforge.net/python-control/?rev=22&view=rev
Author:   murrayrm
Date:     2010-06-06 21:24:07 +0000 (Sun, 06 Jun 2010)

Log Message:
-----------
preliminary LQR example (just MATLAB version for now)

Added Paths:
-----------
    trunk/examples/pvtol-lqr.py

Added: trunk/examples/pvtol-lqr.py
===================================================================
--- trunk/examples/pvtol-lqr.py	                        (rev 0)
+++ trunk/examples/pvtol-lqr.py	2010-06-06 21:24:07 UTC (rev 22)
@@ -0,0 +1,192 @@
+% pvtol_lqr.m - LQR design for vectored thrust aircraft
+% RMM, 14 Jan 03
+
+aminit;
+
+%%
+%% System dynamics
+%%
+%% These are the dynamics for the PVTOL system, written in state space
+%% form.
+%%
+
+pvtol_params;			% System parameters
+
+% System matrices (entire plant: 2 input, 2 output)
+xe = [0 0 0 0 0 0]; ue = [0 m*g];
+[A, B, C, D] = pvtol_linearize(xe, ue);
+
+%%
+%% Construct inputs and outputs corresponding to steps in xy position
+%%
+%% The vectors xd and yd correspond to the states that are the desired
+%% equilibrium states for the system.  The matrices Cx and Cy are the 
+%% corresponding outputs.
+%%
+%% The way these vectors are used is to compute the closed loop system
+%% dynamics as
+%%
+%%	xdot = Ax + B u		=>	xdot = (A-BK)x + K xd
+%%         u = -K(x - xd)		   y = Cx
+%%
+%% The closed loop dynamics can be simulated using the "step" command, 
+%% with K*xd as the input vector (assumes that the "input" is unit size,
+%% so that xd corresponds to the desired steady state.
+%%
+
+xd = [1; 0; 0; 0; 0; 0];  Cx = [1 0 0 0 0 0];
+yd = [0; 1; 0; 0; 0; 0];  Cy = [0 1 0 0 0 0];
+
+%%
+%% LQR design
+%%
+
+% Start with a diagonal weighting
+Qx1 = diag([1, 1, 1, 1, 1, 1]);
+Qu1a = diag([1, 1]);
+K1a = lqr(A, B, Qx1, Qu1a);
+
+% Close the loop: xdot = Ax + B K (x-xd)
+H1a = ss(A-B*K1a, B*K1a*[xd, yd], [Cx; Cy], 0);
+[Y, T] = step(H1a, 10);
+
+figure(1); clf; subplot(321);
+  plot(T, Y(:,1, 1), '-', T, Y(:,2, 2), '--', ...
+    'Linewidth', AM_data_linewidth); hold on;
+  plot([0 10], [1 1], 'k-', 'Linewidth', AM_ref_linewidth); hold on;
+
+  amaxis([0, 10, -0.1, 1.4]); 
+  xlabel('time');   ylabel('position');
+
+  lgh = legend('x', 'y', 'Location', 'southeast');
+  legend(lgh, 'boxoff');
+  amprint('pvtol-lqrstep1.eps');
+
+% Look at different input weightings
+Qu1a = diag([1, 1]); K1a = lqr(A, B, Qx1, Qu1a);
+H1ax = ss(A-B*K1a,B(:,1)*K1a(1,:)*xd,Cx,0);
+
+Qu1b = 40^2*diag([1, 1]); K1b = lqr(A, B, Qx1, Qu1b);
+H1bx = ss(A-B*K1b,B(:,1)*K1b(1,:)*xd,Cx,0);
+
+Qu1c = 200^2*diag([1, 1]); K1c = lqr(A, B, Qx1, Qu1c);
+H1cx = ss(A-B*K1c,B(:,1)*K1c(1,:)*xd,Cx,0);
+
+[Y1, T1] = step(H1ax, 10);
+[Y2, T2] = step(H1bx, 10);
+[Y3, T3] = step(H1cx, 10);
+
+figure(2); clf; subplot(321);
+  plot(T1, Y1, 'b-',  'Linewidth', AM_data_linewidth); hold on;
+  plot(T2, Y2, 'b-', 'Linewidth', AM_data_linewidth); hold on;
+  plot(T3, Y3, 'b-', 'Linewidth', AM_data_linewidth); hold on;
+  plot([0 10], [1 1], 'k-', 'Linewidth', AM_ref_linewidth); hold on;
+
+  amaxis([0, 10, -0.1, 1.4]); 
+  xlabel('time');   ylabel('position');
+
+  arcarrow([1.3 0.8], [5 0.45], -6);
+  text(5.3, 0.4, 'rho');
+
+  amprint('pvtol-lqrstep2.eps');
+
+% Output weighting - change Qx to use outputs
+Qx2 = [Cx; Cy]' * [Cx; Cy];
+Qu2 = 0.1 * diag([1, 1]);
+K2 = lqr(A, B, Qx2, Qu2);
+
+H2x = ss(A-B*K2,B(:,1)*K2(1,:)*xd,Cx,0);
+H2y = ss(A-B*K2,B(:,2)*K2(2,:)*yd,Cy,0);
+
+figure(3); step(H2x, H2y, 10);
+legend('x', 'y');
+
+%%
+%% Physically motivated weighting
+%%
+%% Shoot for 1 cm error in x, 10 cm error in y.  Try to keep the angle
+%% less than 5 degrees in making the adjustments.  Penalize side forces
+%% due to loss in efficiency.
+%%
+
+Qx3 = diag([100, 10, 2*pi/5, 0, 0, 0]);
+Qu3 = 0.1 * diag([1, 10]);
+K3 = lqr(A, B, Qx3, Qu3);
+
+H3x = ss(A-B*K3,B(:,1)*K3(1,:)*xd,Cx,0);
+H3y = ss(A-B*K3,B(:,2)*K3(2,:)*yd,Cy,0);
+figure(4);  clf; subplot(221);
+step(H3x, H3y, 10);
+legend('x', 'y');
+
+%%
+%% Velocity control
+%%
+%% In this example, we modify the system so that we control the
+%% velocity of the system in the x direction.  We ignore the 
+%% dynamics in the vertical (y) direction.  These dynamics demonstrate
+%% the role of the feedforward system since the equilibrium point 
+%% corresponding to vd neq 0 requires a nonzero input.
+%%
+%% For this example, we use a control law u = -K(x-xd) + ud and convert 
+%% this to the form u = -K x + N r, where r is the reference input and
+%% N is computed as described in class.
+%%
+
+% Extract system dynamics: theta, xdot, thdot
+Av = A([3 4 6], [3 4 6]);
+Bv = B([3 4 6], 1);
+Cv = [0 1 0];				% choose vx as output
+Dv = 0;	
+
+% Design the feedback term using LQR
+Qxv = diag([2*pi/5, 10, 0]);
+Quv = 0.1;
+Kv = lqr(Av, Bv, Qxv, Quv);
+
+% Design the feedforward term by solve for eq pt in terms of reference r
+T = [Av Bv; Cv Dv];			% system matrix
+Nxu = T \ [0; 0; 0; 1];			% compute [Nx; Nu]
+Nx = Nxu(1:3); Nu = Nxu(4);		% extract Nx and Nu
+N = Nu + Kv*Nx;				% compute feedforward term
+
+%%
+%% Design #1: no feedforward input, ud
+%%
+
+Nv1 = [0; 1; 0];
+Hv1 = ss(Av-Bv*Kv, Bv*Kv*Nx, Cv, 0);
+step(Hv1, 10);
+
+%%
+%% Design #2: compute feedforward gain corresponding to equilibrium point
+%%
+
+Hv2 = ss(Av-Bv*Kv, Bv*N, Cv, 0);
+step(Hv2, 10);
+
+%%
+%% Design #3: integral action
+%%
+%% Add a new state to the system that is given by xidot = v - vd.  We
+%% construct the control law by computing an LQR gain for the augmented
+%% system.
+%%
+
+Ai = [Av, [0; 0; 0]; [Cv, 0]];
+Bi = [Bv; 0];
+Ci = [Cv, 0];
+Di = Dv;
+
+% Design the feedback term, including weight on integrator error
+Qxi = diag([2*pi/5, 10, 0, 10]);
+Qui = 0.1;
+Ki = lqr(Ai, Bi, Qxi, Qui);
+
+% Desired state (augmented)
+xid = [0; 1; 0; 0];
+
+% Construct the closed loop system (including integrator)
+Hi = ss(Ai-Bi*Ki,Bi*Ki*xid - [0; 0; 0; Ci*xid],Ci,0);
+step(Hi, 10);
+


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