Feedback and Control Systems: Difference between revisions

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== Inverted Pendulum Project ==

== Inverted Penululm Project ==
[[Parameters]]
[[Parameters]]
% Double Pendulum Parameters (Tentative: There are two pendulums with different parameters. I'm not sure which these go to.)
% Run parameters
%f = input('Control Frequency (Hz) = ');
%crad = input('Pole Radius (1/s) = ');
%psi = input('Spreading Angle (deg) = ');
%eta = psi*pi/180;
%obshift = input('Observer Shift = ');
%Trun = input('Run Time (s) = ');
f=130;
crad=19;
psi=10;
eta=psi*pi/180;
obshift=2;
Trun=60;
kmax = round(f*Trun);
T = 1/f;
Maxpos = 0.25; % Max carriage travel +- 0.25 m
Maxangle = 0.175; % Max rod angle -- 10 deg
Maxvoltage = 20; % Max motor voltage, V
pstart = 0.005; % Carriage position starting limit, m
astart = 1*pi/180; % Angle starting limit, rad
g = 9.81; % m/s^2 Gravitational constant
% SYSTEM PARAMETERS
% Measured Mechanical Parameters
d1 = 0.323; % m Length of pendulum 1 (long)
d2 = 0.079; % m Length of pendulum 2 (short)
%mp1 = 0.0208; % kg Mass of pendulum 1
mp1 = 0.0318;
%mp2 = 0.0050; % kg Mass of pendulum 2
mp2 = 0.0085;
m = 0.3163; % kg Mass of carriage
rd = 0.0254/2; % m Drive pulley radius
md = 0.0375; % kg Mass of drive pulley (cylinder)
%mc1 = 0.0036; % kg Mass of clamp 1*
%mc2 = 0.0036; % kg Mass of clamp 2*
mc1 = 0.0085;
mc2 = mc1;
% *Clamp Dimensions
% Rectangular 0.0254 x 0.01143 m
% The pivot shaft is 0.00714 m from the end
% Motor Parameters (Data Sheet)
Im = 43e-7; % kg m^2/rad Rotor moment of inertia
R = 4.09; % ohms Resistance
kt = 0.0351; % Nm/A Torque constant
ke = 0.0351; % Vs/rad Back emf constant
% Derived Mechanical Parameters
% kg m^2/rad Moment of inertia, clamp 1
%Ic1 = mc1*(0.01143^2 + 0.0254^2)/12 + mc1*(0.0127-0.00714)^2;
Ic1 = mc1*(0.0098^2 + 0.0379^2)/12;
Ic2 = Ic1; % kg m^2/rad Moment of inertia, clamp 2
Id = md*(rd^2)/2; % kg m^2/rad Moment of inertia, drive pulley
Imd = Im + Id; % kg m^2/rad Moment of inertia, combined
J1 = Ic1 + mp1*(d1^2)/3; % Total moment of inertia, pendulum 1 (long)
J2 = Ic2 + mp2*(d2^2)/3; % Total moment of inertia, pendulum 2 (short)
Jd = Im + Id; % Total moment of inertia, motor drive
Mc = m + mc1 + mc2; % Total carriage mass
% Friction Test Data
% Carriage Slope = 19 deg; Terminal Velocity xdotss = 0.312 m/s; From
% twincarriage.m; formula b = m g sin(theta)/xdotss
% Pendulum 1 (long) Exponent a1 = 0.0756 1/s; From longfit.m
% Pendulum 2 (short) Exponent a2 = 0.2922 1/s; From shortfit.m
% formula b = 2 a J
%alpha = 19;
alpha = 12.2;
%xdotss = 0.312;
xdotss = 0.4852;
%a1 = 0.0756;
%a2 = 0.2922;
a1 = 0.0185;
a2 = 0.012;
% Ns/m Viscous friction of carriage system
b = (Mc + mp1 + mp2)*g*sin(alpha*pi/180)/xdotss;
b1 = 2*a1*J1; % Nms/rad Viscous friction of pendulum 1 (rotational)
b2 = 2*a2*J2; % Nms/rad Viscous friction of pendulum 2 (rotational)
scale = [rd*2*pi/4096 2*pi/4096 -0.05/250];
T = 1/f;

Revision as of 11:08, 23 February 2011

Inverted Pendulum Project

Parameters