# The finite step response of a system pole at the origin is unbounded, how is this critically stable?

This is a question one of my students brought up in class and I thought it was worth posting.

When considering the stability of a linear time-invariant system it is often useful to look at the pole positions on the pole-zero map. Usually one of the first things that is taught in an undergraduate systems engineering course is that poles located on the LHS s-plane are stable, while poles located on the RHS s-plane are unstable. Furthermore, poles located on the imaginary axis are considered "critically stable".

An example of such critically stable systems would be: $$G_1 = \frac{Y_1}{U_1} = \frac{1}{s^2+\omega_n^2}$$ $$G_2 = \frac{Y_2}{U_2} = \frac{1}{s}$$ The exact value of $\omega_n$ is not important for this question. The poles of $G_1$ and $G_2$ are plotted on the pole-zero map below.

So lets take a look at the behaviour of these critically stable poles. If subjected to a finite step input, $G_1$ will produce a bounded oscillation with constant amplitude:

If subjected to a finite step input, $G_2$ will produce an unbounded response (linear response approaching infinity as time approaches infinity):

I thought that only unstable systems had unbounded finite step responses.

If the finite step response of a LTI system with a pole at the origin is unbounded, why do we say that the system is critically stable?

The main issue here is a misunderstanding of how stability is defined. Stability is not defined by the step response, but is instead defined by the system response to initial conditions (or an impulse response). From Katsuhiko Ogata Modern Control Engineering, 5th Ed.:

A linear time-invariant control system is stable if the output eventually comes back to its equilibrium state when the system is subjected to an initial condition. A linear time-invariant control system is critically stable if oscillations of the output continue forever. It is unstable if the output diverges without bound from its equilibrium state when the system is subjected to an initial condition.

[Emphasis mine].

## From a purely physical perspective

Another way of phrasing this is: in the presence of zero input and non-zero initial conditions, does the energy of the system increase, decrease or remain constant over time? If the energy decreases over time, the system is stable. If the energy increases over time, the system is unstable. If the energy of the system remains constant over time, the system is critically stable.

Let us analyze the previous systems from a purely physical perspective and not worry too much about the systems engineering side of things. An example of a critically stable system that is equivalent to $G_1$ would be a spring-mass system with no damping: $$m\ddot{y}_1(t) + ky_1(t) = u_1(t)$$ Where $y_1$ is the position of the spring, $u_1$ is an external force applied to the spring, $k$ is the spring constant, and $m$ is the mass. If you pull a spring away from its resting position or if it has some initial velocity (or both), and there is no friction or any kind of energy dissipation, and there is no external force performing work on the system ($u_1=0$) then it is clear that the system will oscillate freely forever. As it oscillates, it transfers energy between kinetic and potential forms (energy stored in the mass vs. energy stored in the spring). No energy is being created or dissipated, therefore critically stable.

Now lets look at the $G_2$ case. This system could be represented as a mass sliding on a frictionless surface: $$m\dot{y}_2(t)=u_2(t)$$ Where $y_2$ is the velocity of the mass, $m$ is the mass, and $u_2$ is an external force applied to the mass. If you apply some initial velocity to the mass, and there is no friction or any kind of energy dissipation, and there is no external force performing work on the system, then it will remain at that velocity forever. The energy in the system is all kinetic, and is stored in the mass. No energy is being created or dissipated, therefore critically stable.

## From a systems engineering perspective

You can represent non-zero initial conditions using the impulse response of the system. So for the purposes of systems engineering, if the impulse response of the system is bounded, then the system is at least critically stable. If the impulse response of the system approaches zero, then the system is stable. If the impulse response of the system is unbounded, then it is unstable.

Example showing how the impulse response is equivalent to non-zero IC's: $$m\dot{y}(t) = u(t)$$ Taking the Laplace transform: $$m(sY(s) - y(0)) = U(s)$$ $$Y(s) = \frac{U(s)/m + y(0)}{s}$$ For zero input and non-zero initial conditions: $$Y(s) = \frac{y(0)}{s}$$ If $y(0) = 1$ then: $$Y(s) = \frac{1}{s}$$ Similarly, for zero initial conditions and a impulse scaled to $m$ (i.e. $u(t) = m\delta(t)$, where $\delta(t)$ is the dirac delta function): $$Y(s) = \frac{1}{s}$$