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    Linear Systems and Equilibrium Point

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    Consider the ping pong ball levitator shown in the attachment. Assume that a nozzle makes a cone of air with velocity V_air(h) = u(t) / (h + alpha)^2.

    u(t) is the opening of an air supply valve, h is the height above the nozzle and alpha is a constant. The drag force on the ball is f_up = c_d * A(v_air - v_ball)^2. "A" is the ball cross sectional area and c__d is the drag coefficient. The model of the ball dynamics is then,

    dv_ball / dt = (1/m)[(c_d)(A)(u(t)/(h+alpha)^2 - v_ball)^2 - mg]

    The output of interest is the total energy of the ball, E = [(mv_ball)^2]/ 2 + mgh_ball.

    Find the linearisation of the system around the equilibrium point with v_ball = 0, h_ball = 0.1. Take m = 0.03kg, g = 9.8Nm/s^2, A = 1E-3m^2, c_d = 0.4, alpha = 0.05m.

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    https://brainmass.com/engineering/electrical-engineering/linear-systems-equilibrium-point-37544

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    Solution uses information about the equilibrium point and linearized system matrices. All calculations are formatted in the attached Word document.

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