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l3 - 6 6.1 Lab 3 Levitated Ball Outline The objective of...

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6 Lab 3: Levitated Ball 6.1 Outline The objective of this experiment is to design a control system to levitate a ball in a magnetic field. The system is unstable, with one pole in the right- half plane and two poles in the left-half plane. Root locus techniques will be used to design a stabilizing controller. Several physical variables can be measured allowing us to design a cascade control system. 6.1.1 Modeling and Linearization The experimental arrangement is illustrated conceptually in Figure 7. i x f mg Figure 7: Schematic diagram of the suspended ball The equation of motion of the ball is simply, m ¨ x = mg - f, (1) where m is the mass of the ball, g is the acceleration due to gravity and f is the magnetic force on the ball generated by the magnet. The magnetic force 36
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on the ball can be approximated by, f = a ( i - b ) ( x - c ) 2 , where a , b and c are constants which must be experimentally determined. Substituting this in Eq. 1 gives, ¨ y = g - a m ( i - b ) ( x - c ) 2 , (2) as the equation of motion for the system. Note that it is nonlinear. A FET drive circuit provides current for the coil. The circuit is illustrated in Figure 8. i V Coil FET 10K 1K 0.1 Ω +28 Vdc Figure 8: Magnet coil drive circuit The relationship between the input voltage, V , and the coil current, i , is modeled as, L d i dt + Ri = GV, (3) where L and R are the inductance and resistance of the coil respectively. G is an overall gain associated with the FET circuit. 37
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Equations 2 and 3 are our nonlinear, dynamic model of the ball levitation system. For design purposes a linearized transfer function between V and x is required. If a constant voltage is applied to V , a constant current will result. Denote the equilibrium voltage by V 0 and the equilibrium current by I 0 . Eq. 3, in steady state, yields the following relationship. I 0 = G R V 0 . Similarly, a fixed current, I 0 , will yield an equilibrium position for the ball. Denote this equilibrium as X 0 . By setting ¨ y = 0 in Eq. 2, we can determine the equilibrium relationship as, ( X 0 - c ) 2 = a ( I 0 - b ) mg . Measuring a sequence of I 0 , X 0 equilibrium pairs will allow us to determine a , b and c . We will now linearize about an equilibrium point, V 0 , I 0 and X 0 . Define incremental variables as follows, V = V 0 + δ v i = I 0 + δ i x = X 0 + δ x The ball position equation is now, ¨ X 0 + δ ¨ x = g - a m ( I 0 + δ i - b ) ( X 0 + δ x - c ) 2 . By performing an expansion in the δ i and δ x variables, and noting that ¨ X 0 = 0, we get, δ x g - a m ( I 0 - b ) ( X 0 - c ) 2 - a m ( X 0 - c ) 2 δ i + 2 a m ( I 0 - b ) ( X 0 - c ) 3 δ x. 38
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The equilibrium equation implies that, g - a m ( I 0 - b ) ( X 0 - c ) 2 = 0 , which leads to the linearized result, δ x - a m ( X 0 - c ) 2 δ i + 2 a m ( I 0 - b ) ( X 0 - c ) 3 δ x. - a ( I 0 - b ) δ i + 2 g ( X 0 - c ) δ x. At this point, we will drop the δ notation. Remember that the resulting linearized model refers only to small deviations from equilibrium. We will still have to take this equilibrium into account in setting up the experiment.
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