1. r   Ball and Beam Ө = c*Input Voltage(u) Y = a*x System Equations State Space Model

2. Hardware Issues and Calibration Wiring Issues Sharp Sensor Output Issues Potentiometer Issues Mechanical Grip Issues

3. Final Calibration Voltage = (-1.0442e-06)*distance^4 + 0.00012262*distance^3 - 0.0040194*distance^2 -0.0091906*distance + 2.8626 y = (4.3862e-07)*x^3 – (3.8462e-07)*x^2 – (0.0099189)*x + 2.2135

4. Control Strategy Also employed the strategy of On-Off Controller

5. Modeling of DC Motor Position

6. Open Loop Transfer function of System:

7. Calculated Stall current = 1.5 Amps Motor Internal Resistance = R = Vs/Is(stall) = 24/1.5 = 16 Ohm Motor torque constant KT = Ts(stall)/(Is(stall)-Is(no-load)) = 54e-3/(1.5-0.12) = 0.0391304 Nm/Amp Motor Viscous Friction Constant = b = (KT*Is – Ts)/Rated Speed = 0.1014899e-6 Nms Electromotive force Constant: KE = (Vs – Is*Rm)/Rated Speed = 0.03836695 V/rad/sec Value of Constants:

8. Linearized Model of DC Motor (0.00037461*(z+0.99)*(z+5.843e-08)) -------------------------------------------------- (z*(z-1)*(z-0.9683)) Linearized Transfer Function: Sample time: 0.001 Analysis: There is a pole and zero very near to z = 0 that effectively cancel. -> Makes computations simpler. So, now our transfer function becomes: 0.00037461 (z+0.99) ------------------- (z-1) (z-0.9683)

9. State Variables Motor Position Motor Speed Armature Current Input: Armature Voltage Output: Rotational Position

10. DC Motor State Space Model

11. Step Response of Linearized Model of DC Motor

12. Dynamics of DC Motor: With command d2c(sys,'tustin'), we get Continuous-time zero/pole/gain model: -9.5146e-07 (s+3.981e05) (s-2000) --------------------------------- s (s+32.21) Analysis: Zero at 3.98 e05 contributes very little to the response of the plant. Step Closed Loop Response without Compensation

13. Gain Margin = 38.6 dB(at 254 rad/s); Phase Margin = 57.6 deg (at 20 rad/s) Gain crossover frequency = 3 Hz. So a sampling period of 0.001 seconds (frequency of 1000 Hz) is significantly faster than the dynamics of the plant.

14. Rise time: ≤ 0.05 sec Settling time: ≤ 0.25 sec Overshoot: ≤ 5% No steady state Error Bandwidth must be significantly larger than overall ball and beam system. Design Requirement for DC Motor Controller:

15. Closed Loop for DC Motor P=10; I = 2; D= 0.1; N = 100 20*(z-0.9501) ------------- (z-0.9) Discretized PID Controller:

16. Overall Closed Loop Response of DC Motor with Controller

17. Mass of the ball = m= 0.121 kg Radius of the ball = R = 0.0111m Gravitational acceleration = 9.8 m/s^2 Length of the beam (on left side of pivot)= 0.357m Ball's moment of inertia = J = (2/5)*M*(R^2) = 5.9634e-6 kg.m^2 Modeling of Ball and Beam Plant Value of Constants:

18. Overall Closed Loop Ball and Beam P=25; I = 1; D= 7; N = 100 725 z - 722.5 ------------- z - 0.9 Discretized PID Controller:

19. Rise time: 0.6613 sec Settling time = 9 sec Overshoot = 5% Overall Closed Loop Ball and Beam Step Response

20. Product Information Sheet R2-1, Rotary Ball & Beam Experiment, Quanser Inc., Markham, ON, Canada. R. Hirsch, Shandor Motion Systems, "Ball on Beam Instructional System," Shandor Motion Systems, 1999. K. C. Craig, J. A. de Marchi, "Mechatronic System Design at Rensselaer," in 1995 International Conference on Recent Advances in Mechatronics, Istanbul, Turkey, August 14-16. R. A. Pease, "What's All This Ball-On-Beam-Balancing Stuff, Anyhow," Electronic Design Analog Applications, p. 50-52, November 20, 1995. G. J. Kenwood, "Modem control of the classic ball and beam problem," B.S. thesis, Massachusetts Institute of Technology, Cambridge, MA, 1982. References