ISEGORIA / MATH ENCYCLOPEDIA
Control theory: feedback, poles, and stability
Pole locations and step responses, PID feedback, and Bode and Nyquist diagrams.
Before you begin: Dynamical systems and complex analysis
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Poles decide the response
A second-order system has a pair of poles in the complex s-plane. Drag the upper pole (its mirror follows). Moving it left makes the response settle faster; moving it up makes it oscillate faster; crossing into the right half-plane makes it blow up.
Worked example. For \(\sigma<0\) the overshoot is \(e^{\pi\sigma/\omega}\), which depends only on the angle of the pole from the imaginary axis, and the envelope decays like \(e^{\sigma t}\).
Watch out. The steady-state gain is normalised to one. Real systems have zeros and more poles, which change the shape but not the rule that right-half-plane poles mean instability.
Which pole positions give the same overshoot?
Every pole on the same ray from the origin, since the overshoot depends only on \(\sigma/\omega\).
2. PID feedback
A sluggish third-order plant must follow a step in its set-point, and at t = 15 a load disturbance hits it. Tune the proportional, integral and derivative gains. The small plot shows the four closed-loop poles, so you can see why a tuning rings or goes unstable.
Worked example. Without the integral term a constant disturbance leaves a permanent error; any \(K_i>0\) drives it to zero, as long as the loop stays stable.
Watch out. The derivative acts on the measurement, not on the error, so a set-point step does not cause a derivative kick. Actuator limits are not modelled.
Why can too much gain destabilise a stable plant?
The plant lags by up to 270°; at the frequency where the lag reaches 180°, enough gain turns negative feedback into positive feedback.
3. Bode and Nyquist
The loop transfer function is traced two ways: as gain and phase against frequency, and as a single curve in the complex plane. Raise the gain or add delay; when the Nyquist curve wraps around −1 the closed loop is unstable, exactly when the Bode plot shows gain above 1 at a phase of −180°. In the formula, N counts clockwise encirclements of −1, P the open-loop poles in the right half-plane and Z the closed-loop ones: with P = 0, the closed loop is stable exactly when N = 0.
Worked example. With no delay the phase crosses \(-180^\circ\) at \(\omega=\sqrt3\), where \(|L|=K/8\): the gain margin is \(8/K\), so \(K=8\) is the edge of stability.
Watch out. This plant is itself stable, so the simple no-encirclement rule applies. An unstable plant needs the full Nyquist criterion with counted encirclements.
Why does delay spiral the Nyquist curve?
Delay adds phase \(-\omega\tau\) without changing gain, so as \(\omega\) grows the curve keeps winding while its radius shrinks.