Say I have a stable discrete State Space control system (feedback system) of the standard formulation:
x[k+1] = Ad*x[k] + Bd + u [k].
Why do the |discrete eigenvalues| approach the unit circle (z = 1) as sampling rate increases? I would have thought that due to a more stable system, the stability margin (eig distance from unit circle) would increase as sampling rate increases, and so
|discrete eig|-->0, ie more stable.
Background:
Ad and Bd are discrete matrices, discretized with a sampling rate Ts. Note: Since this is a feedback system, Ad includes the feedback mechanism, eg Ad = Ad_openLoop - Bd * Kd, where Kd could be eg. a discretized gain eg from Matlab's lqrd().
I build up the state space with varying loop rates to check for stability as a function of loop rate. I'd like to check stability by looking at the eigenvalues the system as the loop rate increases (Ts becomes smaller).
The eigenvalues of the continuous version of this system are stable. The eigenvalues of the discrete version of this system are also stable, but for smaller loop rates (larger Ts), the eigenvalues are closer to the origin. For larger loop rates approaching inf (Ts --> 0), the eigenvalues approach 1: |eig| --> 1
This seems counterintuitive. I would have thought faster rates would mean a more stable system, so larger eig margin, ie eig approaches origin (since |disc eig| >= 1 imply marginal stability or instability)
What am i missing conceptually?
The discretized system becomes trivial as the sample period $T \to 0.$ For now assume you have determined the controller and the loop is closed. Then the system takes the form,
$$\dot{x} = A\, x.$$
The exact discretization of this system with sample period $T$ is,
$$x(k) = \exp(T A)\, x(k-1).$$
It is straightforward to compute (use the Taylor series definition) that, as $T\to 0,$ we have that $\exp(T A) \to I$. The discrete time system matrix $\exp(T A)$ converges to a matrix with only 1 as the eigenvalues. This should not be surprising: if you decrease the sample period to zero, the "next sample" is necessarily equal to the current sample. This is what you are observing. If you instead computed the quantity
$$\frac{x(k) - x(k-1)}{T}$$
and see how that changes as $T\to 0$ you will end up finding it converges to $A x(k-1)$ which is what you actually expected. Fundamentally the issue here is that a discretized system matrix computes how you go from one state to the next while the continuous-time system matrix computes the rate of change at a given point.