Response and stability

Approaching a target in one simulation does not establish stability under every condition. Even two stable responses can differ greatly in speed and overshoot. This chapter builds on the states and exponential responses in Dynamic models.

Free and forced responses

A linear response separates into a contribution from the initial state and one from the input. For constant heating,

$$\theta (t)={\theta }_{\infty }+({\theta }_0-{\theta }_{\infty })e^{-t/\tau }.$$

The second term captures decay of the initial deviation. Changing initial temperature and changing the reference are different experiments. Record initial conditions, inputs, disturbances, observation time and actuator limits.

For proportional thermal control without delay or saturation,

$$\dot{x}=-\frac{k+P_{\max }{K}_p}{C_{\mathrm{th}}}x.$$

With positive parameters and $K_p=0.04$, the closed-loop time constant is $2000/(20+40)=33.33$ s. The response approaches equilibrium monotonically. Increasing positive gain does not create another state. Real high-gain oscillations require investigation of omitted filters, actuator dynamics, integral action or delay.

Two states allow oscillation

A mass–spring–damper obeys $m\ddot{q}+b\dot{q}+k_{s}q=F$. Position and velocity are separate states: the mass can cross equilibrium while still moving. Damping dissipates energy. A normalized second-order model is

$$\ddot{y}+2\zeta {\omega }_n\dot{y}+{\omega }_n^{2}y={\omega }_n^{2}r,$$

with ${\omega }_n=\sqrt{k_s/m}$ in rad/s and dimensionless $\zeta =b/(2\sqrt{m{k}_s})$. It is underdamped for $0<\zeta <1$, critically damped at 1, and overdamped above 1.

For a unit step, zero initial state and $0<\zeta <1$, let ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}$. Then

$$y(t)=1-e^{-\zeta {\omega }_{n}t}\left[\cos ({\omega }_{d}t)+\frac{\zeta }{\sqrt{1-{\zeta }^2}}\sin ({\omega }_{d}t)\right].$$

The sinusoid oscillates within a decaying envelope. Overshoot is $M_p=e^{-\pi \zeta /\sqrt{1-{\zeta }^2}}$. For ${\omega }_n=1,\zeta =0.5$, the first peak occurs at $t_p=\pi /{\omega }_d\approx 3.63$ s and overshoot is 16.3%.

Compare output and velocity while changing damping. Velocity is zero at an output peak, but usually nonzero when output crosses the target. Critical damping avoids oscillation in this standard model; it is not a universal fastest-response prescription.

Which stability claim?

With $\zeta >0$, the standard model's modes decay. At $\zeta =0$, free oscillations remain bounded but do not decay; a bounded resonant sinusoid can produce an unbounded response. Bounded free motion therefore does not prove BIBO stability. Negative damping amplifies deviations.

An energy function also gives evidence: $V=\frac{1}{2}m{v}^2+\frac{1}{2}{k}_s{q}^2$ has $\dot{V}=-b{v}^2\le 0$ in the unforced mechanical system. Energy cannot increase. Asymptotic convergence requires the additional observation that the only trajectory able to remain in $\dot{V}=0$ is the origin; this is the entry to invariance methods.

Response metrics need definitions

Specify rise-time thresholds, the settling tolerance, and the overshoot normalization. The estimate $t_s\approx 4/(\zeta {\omega }_n)$ is an engineering approximation for a 2% band in underdamped second-order systems. A finite plot only identifies the last observed entry into a band, not a guarantee about the future. A zero target needs an absolute tolerance. Report stability, error, speed and control effort separately.

Check your understanding

Does constant-amplitude free oscillation establish asymptotic or BIBO stability?

At fixed $\zeta =0.5$, what happens when ${\omega }_n$ doubles from 1 to 2?

Further reading

The Caltech control course connects modeling, state space, phase portraits and stability. Next we introduce the controller's own dynamic states.