The numerical values given for the parameters correspond to the virtual experiment.

Electrical drive and wheel

System input

u (motor voltage)

Parameters

Rm = 1 Ω (motor resistance)

K = 0.1 N·m/A (motor torque/current constant)

N = 7 (reduction factor from motor to wheel)

Jw = 0.02 kg·m² (moment of inertia of the load)

R = 0.201 m (rails radius)

Auxiliary values

Mw (moment on the wheel produced by the motor)

theta (angle of the wheel)

thetaDer (derivative of theta)

thetaDDer (second derivative of theta)

F (tangential force applied by ball to wheel at contact point)

Equations

Mw = K N (u − K N thetaDer) / Rm

Jw thetaDDer = Mw - R F

Wheel and ball

Parameters

m = 0.1 kg (ball mass)

r = 27·10⁻³ m (rail contact radius of the ball)

Jb = 36·10⁻⁶ kg·m² (moment of inertia of the ball)

g = 9.81 m/s² (gravity acceleration constant)

Auxiliary values

thetaDDer (second derivative of theta)

phi (ball angle with respect to ground)

phiDer (derivative of phi)

phiDDer (second derivative of phi)

psi (ball position along rail as an angle from wheel center, 0 at top)

psiDer (derivative of psi)

psiDDer (second derivative of psi)

Equations

m (R + r) psiDDer = F + m g sin psi

Jb phiDDer = −F r

psi = theta + phi r / (R + r)

Eliminating phi and substituting F by its value in the first equation yields

Equation

m (R + r) psiDDer = Jb (R + r) (thetaDDer − psiDDer) / r² + m g sin psi

F = Jb (R + r) (thetaDDer − psiDDer) / r²

F doesn't appear anymore in the first equation which describes the effect of the wheel on the ball; but we still need it to take into account the effect of the ball on the wheel in the first subsystem.

State-space model

Combining these equations gives the following state-space model for the complete system. Because of the intricate influence between the wheel and the ball, to get explicit first-order differential equations, a 2x2 linear system must be solved; hence the auxilairy values a, b, c and d, coming from a 2x2 matrix, and what can be recognized as an inverse.

States

theta (angle of the wheel, 0 when rails are horizontal)

thetaDer (derivative of theta)

psi (ball position along rail as an angle from wheel center, 0 at bottom)

psiDer (derivative of p)

Auxiliary values (constant)

a = Jw + R Jb (R + r) / r²

b = −R Jb (R + r) / r²

c = Jb (R + r) / r²

d = m (R + r) + Jb (R + r) / r²

Equations (non-linear model)

theta' = thetaDer

thetaDer' = ((u − K N thetaDer) d K N / Rm − b m g sin psi) / (a d − b c)

psi' = psiDer

psiDer' = ((K N thetaDer − u) c K N / Rm + a m g sin psi) / (a d − b c)

These equations aren't linear with respect to states and input. It isn't a problem for numerical simulation (at least not a difficult one: we're using a Runge-Kutta 4th order integration algorithm with a fixed maximum time step), but many controller design methods, such as those in frequency domain, require linearity. Therefore we'll approximate sin psi ≈ psi, assuming psi is small.

This leads to the following linear model, with the same states.

Equations (linear model)

theta' = thetaDer

thetaDer' = ((u − K N thetaDer) d K N / Rm − b m g psi) / (a d − b c)

psi' = psiDer

psiDer' = ((K N thetaDer − u) c K N / Rm + a m g psi) / (a d − b c)

Decoupled model

In the state-space models above, the rotation of the wheel is perturbed by the ball. If we assume the ball effect is negligible, we start from the following equations:

Equations (decoupled system)

Jw thetaDDer = Mw

Mw = K N (u − K N thetaDer) / Rm

m (R + r) psiDDer = Jb (R + r) (thetaDDer − psiDDer) / r² + m g sin psi

Approximating sin psi ≈ psi, we eventually obtain the following decoupled linear state-space model:

Equations (decoupled linear model)

theta' = thetaDer

thetaDer' = K N u / (Rm Jw) − K² N² thetaDer / (Rm Jw)

psi' = psiDer

psiDer' = (Jb K N u / (Jw Rm) − Jb K² N² thetaDer / (Jw Rm) + m g r² psi / (R + r)) / (m r² + Jb)

We have here two subsystems connected in series: the first two equations for the d.c. motor and wheel with input u and output theta, and the last two equations for the ball with input theta and output psi. The controller must be robust enough to support all these approximations, in addition to other model structural assumptions, parameter uncertainties and external perturbations.

Two remarks on the ball subsystem:

• Its input is the second derivative of theta (angular acceleration). The wheel angle theta isn't required to simulate or control the ball position. This is also true for the non-linear model.
• The second derivative of psi depends on psi itself, not on its first derivative. This results in symmetric eigenvalues: the ball subsystem is unstable for any (positive) values of the parameters, which corresponds to our expectations. A controller can remedy that.

Controller structure

The system will be controlled with cascade controllers. The inner loop controls the wheel velocity thetaDer with a P controller whose control signal is the motor voltage u. Since the ball depends only on the wheel acceleration, a velocity steady-state error isn't harmful; no need for an integral term. We still allow for one, though.

Question: why? If you have difficulty in finding an answer, continue reading and experimenting first.

The outer loop controls the ball position; its control signal is the wheel velocity, the reference value passed to the inner controller. Since the ball position is actually controlled by the wheel acceleration, an integral term is required in the controller to reject perturbations, and no derivator term should be needed. An additional state is hence required.

The controller equations are given below.

Controller parameters

kp1 (proportional gain for the inner loop)

ki1 (integral gain for the inner loop)

kp2 (proportional gain for the outer loop)

ki2 (integral gain for the outer loop)

Controller state

psiInt (integrated psi)

Auxiliary value

thetaDerRef (reference value of the wheel velocity)

Equations (cascade P/PD controllers)

psiInt' = psi

thetaDerRef = −kp2 psi − ki2 psiInt

u = kp1 (thetaDerRef − thetaDer)

Type assist model and press the Enter button or hit the Return key repeatedly to see how. You can also modify commands, stop the assisted mode, start again, clear the output with clc or get help with doc.