BallBeam

BallBeam - Introduction The purpose of this project was to...

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Unformatted text preview: Introduction The purpose of this project was to gain experience in designing a ball and beam model using the methods of control design we have learnt in class, mainly frequency-domain design method, using bode plots, sisotool and simulink of Matlab to test the design. Also, we had to implement the design during a lab session, where we verify of the design works in a real world case and observe the performance of the ball and beam model. Basically, the ball and beam consists of a steel ball rolling on top of a long beam. The beam is mounted on the output shaft of an electric motor and so the beam can be tilted about its axis at the pivot point by applying an electrical control signal to the motor amplifier, that is a voltage that causes the beam to rotate with a rotation speed proportional to the voltage. The output signals are voltages representing the beam angle relative to the horizontal plane and the position of the ball. The control job is to automatically regulate the position of the ball on the beam by changing the angle of the beam. This is a difficult control task because the ball does not stay in one place on the beam but moves with an acceleration that is proportional to the tilt of the beam due to the force of gravity acting on it. In control theory, the system is open loop unstable because the system output (ball position) increases without limit for a fixed input (beam angle). Feedback control must be used to keep the ball in a desired position on the beam. EE 4580 Project 2 Ball and Beam Model Operation Principle The linear model is used where motion in only one plane is considered. The system is as follows: If the constant of friction is b, force due to friction = b(dx/dt) Component of force due to gravity acting on ball = Mg sin down the plane. Net force acting on ball : M(d 2 x/dt 2 ) + b(dx/dt) = -Mgsin ------- 1 For small angles, sin = Since sin = y/L, =y/L Also, we assume that there is no friction, therefore b= 0 Thus, from 1, we get G y (s) = X(s)/Y(s) = (g/L) / s 2 where g/L = 0.6223 The D.C motor servo is illustrated in the following diagram: 2 EE 4580 Project 2 Ball and Beam Model A m = 1.8408 and Tm = 1.0272. The position gain Kp and the velocity gain Kv have to be designed so that the closed loop system acts almost like a constant gain to the outer loop. The time constant of the inner loop must be small compared to the outer loop. 3 EE 4580 Project 2 Ball and Beam Model Design Design requirements: Steady state error, ess to a step input = 3% = 0.03 Percentage Overshoot, P.O = 10% = 0.1 Settling time, t s = 4 seconds Time to peak = 2 seconds From the design requirements, we get: Kv = 1/ess = 1/0.03 = 33.3 Outer Loop Design...
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BallBeam - Introduction The purpose of this project was to...

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