Ball and Beam Dynamics

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Ball and Beam Dynamics
Introduction This report does not omit any mathematical steps in order to reflect upon details of the mechanical operation of the process. 1. The Process The ball and beam model consists of a horizontal beam and a DC motor mechanically attached at the centre of the beam. The angle of the beam is controlled by the motor. The angle in turn influences the position of the ball. 2. Mathematical Model Composition The process model can be broken down into two separate models a. Model of the angle process with respect to the motor voltage H
φ

(s)

b. Model of the ball position with respect to the beam angle Hx(s) The total transfer function from the input voltage to the voltage that indicates the ball position is then H (s) = H ϕ (s) × H x (s) 3. Mathematical Model Derivation a. Model of beam angle vs. input voltage The relationship between the input voltage and the angle of the beam is defined by the DC motor transfer function. The DC motor, used for angle control application may be thought of as the ‘dirty integrator’ or the integrator with a filter action as shown in figure 1 on the next page.

Figure 1: General DC motor block diagram for angle control application

The K is the motor constant and the tau is the motor time constant. The actual model of the motor used for the project is shown in figure 2.

Figure 2: Actual block diagram of the DC motor used

The measured constants are summarized below:

K m = 229

rad

sec

V

τ = 0.4 sec K gear = 1 25 K var iable pot = 1.26 rad V
Therefore the DC motor transfer function becomes
θ( s)
V (s) = 11 .54 s (1 + 0.4 s )

b. Model of ball position vs. beam angle Consider the following sketch

Figure 3: Rolling ball free-body diagram

The inclination is considered the x-coordinate.

Let acceleration of the ball be denoted as

d 2 x •• =x dt2

The force due to translational motion is then

F tx= m x
The torque developed through ball rotation is determined by the force at the edge of the ball multiplied by the radius which can be further expressed as:
Tr = Frx R = J dwb d (v b / R ) d 2 ( x / R) J =J =J = x' ' dt dt R dt 2

••

where J = moment of inertia (for solid ball defined by J=2/5*mR2) Wb=angular velocity of the ball

Vb= speed of the ball along x axis The equation is arranged such that the final result is expressed solely in terms of position or its derivatives as well as variables associated with the ball. We now obtain the rotational force by dividing torque of the ball by its radius

Frx = Tr

R
2

=J

R2

x' '

substituting the moment of inertia into the equation we get

Frx = 2 / 5m R

x' ' = 2 m x' ' 5 R
2

In order to make the system independent of the mass of the ball we further express the above equations as

Frx + Ftx = mg sin α 2 mx' '+ mx' ' = mg sin α 5 2 x' '+ x' ' = g sin α 5
rearranging for x’’ gives

5 g sinα = x' ' 7
we utilize approximation sin α = α , since the angle of the beam will not exceed 20-30 degree inclination. This means that in radians, sine of the angle is approximately the angle itself, so the equation is further approximated as

5 gα = x' ' 7
taking Laplace transform of position with respect to angle (details omitted) gives

5 g X (s) H x ( s) = = 72 θ ( s) s

The constant in the numerator is a theoretical constant that neglects surface imperfections and friction. The measured constant is approximately 0.91 therefore
X ( s) 0.91 = 2 θ( s ) s

Now the overall transfer function of the system becomes
H (s) =

θ (s)
V (s)

×

X ( s) 10 .5 = 3 θ ( s) s ( 0.4 s + 1)

where, 10.5 is an approximated constant.

The block diagram of the overall system is

Figure 4: Entire system block diagram

The MATLAB provides easy conversion of the system into state space. >> num=[0 0 0 0 10.5]; >> den=[0.4 1 0 0 0]; >> [A, B, C, D]=tf2ss(num, den)
− 2.5  1 A =  0   0 1 0 B =  0   0 C = [0 0 D =0 0 0 1 0 0 0 0 1 0 0  0  0

0

26 .25 ]

The LQR control can be implemented by choosing Q and R values. The controllability is verified first as follows: >> rank(ctrb(A, B)) ans = 4

This indicates that the ball and beam system is completely state controllable. Next we select Q and R and calculate for controller gains in MATLAB. We get >> Q=[10 0 0 0; 0 10 0 0; 0 0 10 0; 0 0 0 10]; >> R=1; >> [K,S,E]=lqr(A,B,Q,R) K= 3.6048 10.5091 8.7444 3.1623

S= 3.6048 10.5091 8.7444 3.1623

10.5091 55.4111 50.2205 19.3050 8.7444 50.2205 72.5913 33.2327 3.1623 19.3050 33.2327 27.6524

E= -3.9559 -0.9059 -0.6215 + 0.7044i -0.6215 - 0.7044i We can see that the gain values are reasonable and therefore the actual system may perform well.

The simulation in simulink is done using the following block diagram:

Figure 5: Simulink state space model

The Scope measures the output while the Scope1 monitors the control effort. The snapshots are as shown in figures 6 and 7.

Figure 6: Output converging from IC

Figure 7: Control effort

The initial condition was set to 10cm from the centre at an angle of 5.7 degrees. It is can be seen that the system converges very slowly. The state variables are: - angular acceleration θ - angle of the beam θ - ball acceleration x - position of the ball x The position and the angle can be measured directly with sensors while the angular acceleration and the ball acceleration will have to be mathematically estimated. The control voltage V is then
V = K 1 θ + K 2θ + K 3 x + K 4 x
• •





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