Feedback and Control Systems
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Inverted Penululm Project
[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;