The figure on the right shows a simplified version of a magnetic levitation system that can be built in a laboratory, wh

[answerhappygod]

The Figure On The Right Shows A Simplified Version Of A Magnetic Levitation System That Can Be Built In A Laboratory Wh 1 (335.29 KiB) Viewed 31 times

The Figure On The Right Shows A Simplified Version Of A Magnetic Levitation System That Can Be Built In A Laboratory Wh 2 (2.61 KiB) Viewed 31 times

QUESTION BELOW

The Figure On The Right Shows A Simplified Version Of A Magnetic Levitation System That Can Be Built In A Laboratory Wh 3 (93.7 KiB) Viewed 31 times

The figure on the right shows a simplified version of a magnetic levitation system that can be built in a laboratory, where a ball of magnetic material is suspended by means of an electromagnet whose current is controlled by feedback from the optically measured ball position. To quote the excellent reference [1], "this system has the basic ingredients of systems constructed to levitate mass, used in gyroscopes, accelerometers, and fast trains”. Several such examples are given in many technical references, some of which I have listed and are posted on Blackboard In any case, the equation of motion of the ball is mj = -kỷ + mg +Fy,i) (1) where m is the mass of the ball, y 20 is the vertical (downward) position of the ball measured from a reference point y = 0 when the ball is next to the coil), k is a viscous friction coefficient, g is the gravitational acceleration constant, and Fly,i) is the force generated by the electromagnet, i 2 0 being the coil current. The force F(y,1) is related to the energy E(y,i) = L(y)i2/2 (where L(y) is the position-dependent coil inductance) stored in the coil by F(1,1) = E(7.1) Lola cy 20(1 + y/a)2 where Lo, a are positive constants. This completes the system model if one controls the position by manipulating the coil current directly. The equations for controlling the ball position by varying the coil voltage v include the additional equation v= Ri + d(Li), dt Ri- Li di (3), but this makes the problem harder (you can try this a(1+y/a) (1+ y/a) dt if you want!), and so we will stick to designing the current and not the voltage. Loi j+
Gr(s) = y(5) s 1 3.132 2 5 +0.015+196.18 SS
(3) For your linearized plant G, (s), design a linear compensator G (5) to regulate the value of the output to y=r=0.05m. A very important observation is that your linear compensator deals with the deviations from the equilibrium values !! In other words, you are not designing the current i, but the current i,(t)=i-1. Do not use integral control! Perform simulations in simulink with your compensator design to verify that the specifications have been satisfied. Only your compensator should be linear, the plant should be the nonlinear, in other words, you must design for a linear system, but evaluate your controller's performance on the nonlinear system. Plots should indicate that the responses of the various signals are satisfactory.