Feedback and Control Systems: Difference between revisions
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== Inverted Pendulum Project == |
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#[[Parameters]] |
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#[http://www.ee.usyd.edu.au/tutorials_online/matlab/examples/pend/invpen.html Modeling the Inverted Pendulum] |
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#[http://engr.case.edu/merat_francis/eecs397/PLL.pdf Phase Locked Loops] |
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== |
== Octave Examples == |
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[[RLC Circuit of February 10, 2011]] |
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% Double Pendulum Parameters (Tentative: There are two pendulums with different parameters. I'm not sure which these go to.) |
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== Links == |
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% Run parameters |
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*[https://ia700708.us.archive.org/33/items/NetworkAnalysisFeedbackAmplifierDesign/Bode-NetworkAnalysisFeedbackAmplifierDesign.pdf H. W. Bode's "Network Analysis and Feedback Amplifier Design"] |
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%f = input('Control Frequency (Hz) = '); |
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*[http://www.cds.caltech.edu/~murray/books/AM05/pdf/am08-complete_30Aug11.pdf Feedback Systems (textbook) Astrom & Murray] |
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%crad = input('Pole Radius (1/s) = '); |
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*[http://www.eolss.net/sample-chapters/c18/e6-43-13-09.pdf Bernard Friedland, Reduced Order Observer] |
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%psi = input('Spreading Angle (deg) = '); |
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*[http://www.cds.caltech.edu/~murray/books/AM05/pdf/obc-kalman_22Dec09.pdf Optimization Based Control (Murray)] |
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%eta = psi*pi/180; |
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%obshift = input('Observer Shift = '); |
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%Trun = input('Run Time (s) = '); |
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f=130; |
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crad=19; |
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psi=10; |
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eta=psi*pi/180; |
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obshift=2; |
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Trun=60; |
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kmax = round(f*Trun); |
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T = 1/f; |
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Maxpos = 0.25; % Max carriage travel +- 0.25 m |
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Maxangle = 0.175; % Max rod angle -- 10 deg |
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Maxvoltage = 20; % Max motor voltage, V |
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pstart = 0.005; % Carriage position starting limit, m |
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astart = 1*pi/180; % Angle starting limit, rad |
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g = 9.81; % m/s^2 Gravitational constant |
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% SYSTEM PARAMETERS |
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% Measured Mechanical Parameters |
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d1 = 0.323; % m Length of pendulum 1 (long) |
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d2 = 0.079; % m Length of pendulum 2 (short) |
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%mp1 = 0.0208; % kg Mass of pendulum 1 |
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mp1 = 0.0318; |
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%mp2 = 0.0050; % kg Mass of pendulum 2 |
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mp2 = 0.0085; |
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m = 0.3163; % kg Mass of carriage |
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rd = 0.0254/2; % m Drive pulley radius |
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md = 0.0375; % kg Mass of drive pulley (cylinder) |
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%mc1 = 0.0036; % kg Mass of clamp 1* |
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%mc2 = 0.0036; % kg Mass of clamp 2* |
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mc1 = 0.0085; |
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mc2 = mc1; |
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% *Clamp Dimensions |
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% Rectangular 0.0254 x 0.01143 m |
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% The pivot shaft is 0.00714 m from the end |
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% Motor Parameters (Data Sheet) |
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Im = 43e-7; % kg m^2/rad Rotor moment of inertia |
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R = 4.09; % ohms Resistance |
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kt = 0.0351; % Nm/A Torque constant |
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ke = 0.0351; % Vs/rad Back emf constant |
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% Derived Mechanical Parameters |
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% kg m^2/rad Moment of inertia, clamp 1 |
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%Ic1 = mc1*(0.01143^2 + 0.0254^2)/12 + mc1*(0.0127-0.00714)^2; |
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Ic1 = mc1*(0.0098^2 + 0.0379^2)/12; |
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Ic2 = Ic1; % kg m^2/rad Moment of inertia, clamp 2 |
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Id = md*(rd^2)/2; % kg m^2/rad Moment of inertia, drive pulley |
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Imd = Im + Id; % kg m^2/rad Moment of inertia, combined |
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J1 = Ic1 + mp1*(d1^2)/3; % Total moment of inertia, pendulum 1 (long) |
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J2 = Ic2 + mp2*(d2^2)/3; % Total moment of inertia, pendulum 2 (short) |
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Jd = Im + Id; % Total moment of inertia, motor drive |
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Mc = m + mc1 + mc2; % Total carriage mass |
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% Friction Test Data |
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% Carriage Slope = 19 deg; Terminal Velocity xdotss = 0.312 m/s; From |
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% twincarriage.m; formula b = m g sin(theta)/xdotss |
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% Pendulum 1 (long) Exponent a1 = 0.0756 1/s; From longfit.m |
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% Pendulum 2 (short) Exponent a2 = 0.2922 1/s; From shortfit.m |
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% formula b = 2 a J |
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%alpha = 19; |
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alpha = 12.2; |
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%xdotss = 0.312; |
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xdotss = 0.4852; |
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%a1 = 0.0756; |
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%a2 = 0.2922; |
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a1 = 0.0185; |
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a2 = 0.012; |
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% Ns/m Viscous friction of carriage system |
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b = (Mc + mp1 + mp2)*g*sin(alpha*pi/180)/xdotss; |
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b1 = 2*a1*J1; % Nms/rad Viscous friction of pendulum 1 (rotational) |
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b2 = 2*a2*J2; % Nms/rad Viscous friction of pendulum 2 (rotational) |
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scale = [rd*2*pi/4096 2*pi/4096 -0.05/250]; |
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T = 1/f; |
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