Theory: Internal Model Design Questions #1215

Ok, I think that I might understand it now. Because it specifically states a "step reference input", this implies a constant value. Thus it has zero slope. Hence, x(dot) = 0. However, still unclear as to r(t).

Your reasoning for the first part is right: a step reference is constant, so its slope is zero, which is why xdot_r = 0.

The thing that ties it together is what equation (11.78) actually is. The reference is not just "a signal" here, it is modeled as the output of a small autonomous system (a signal generator) with its own state x_r:

xdot_r = A_r * x_r     (how the generator state evolves)
r       = d_r * x_r    (the reference is the output of that generator)

That pair is the general template. The "=>" in your tutorial means "specialize this to the step case", not "both hold at the same time for the same A_r". For a step you pick the generator whose free response is a constant, which is A_r = 0, so the first equation becomes xdot_r = A_r * x_r = 0 * x_r = 0. Same equation, with A_r = 0.

For r(t) it is the same move on the output equation. In general r = d_r * x_r, where d_r is just the output row that turns the generator state into the reference signal. For this scalar step model the generator has one state that holds the constant level, and the reference is just that state, so d_r = 1 and r = x_r. So x_r here is literally the constant step value being held, and r reads it out unchanged. rdot = 0 (eq. 11.80) then follows because r is a constant state.

The reason for setting it up this way is the internal model idea: you capture the class of reference you want to track by the eigenvalues (modes) of A_r, and the controller is built to contain a copy of A_r so the error goes to zero in steady state. A step is the mode at s = 0, so A_r = 0. For a ramp you would use A_r = [[0, 1], [0, 0]], and for a sinusoid at frequency w an A_r with eigenvalues +/- j*w. Same framework, just a bigger generator.