Feedback and PID control

Task 10.1 · Open loop vs. P vs. PI

G(s) = 10 (0.1s + 1)s² + 10s + 1  H = 0.5   r(t) = 50·σ(t)

Final value theorem: for step inputs e_∞ = G_er(0)·r_ref + G_ed(0)·d_c. It only applies if the closed loop is asymptotically stable.

Open loop uses K_o = 1/A: perfect tracking only on paper, and the disturbance passes straight through. P control divides the error by 1 + kA. An integrator in K(s) makes G_er(0) = G_ed(0) = 0, so steps leave no error. A ramp needs a double integrator.

Task 10.2 · PID pole placement

G(s) = 1(1 + s)(1 + 0.5s)  K(s) = k(1 + 1T_Is + T_Ds)

Closed-loop characteristic polynomial: s³ + (1/a₁ + 1/a₂ + K_pkT_D/(a₁a₂)) s² + (K_pk + 1)/(a₁a₂) s + K_pk/(a₁a₂T_I). Match it to (s + βω_n)(s² + 2ζω_ns + ω_n²): three coefficients, three unknowns.

Pole placement fixes the poles only. The two PID zeros stay in the closed-loop numerator and add overshoot; compare with the dashed curve. Not every target is feasible with k > 0, T_I > 0, T_D ≥ 0: try a small ω_n.

Task 10.3 · PD for stabilization

G(s) = 1(s + 1)(s − 2)  unstable open-loop pole at +2

Closed-loop poles solve 1 + K(s)G(s) = 0. The root locus shows where they go as k grows from 0 to ∞: branches start at open-loop poles and end at open-loop zeros (or at infinity).

Task 10.4 · PID zeros and the root locus

G(s) = 1(s + 10)(s + 2)²

A PID controller adds one pole at 0 and two zeros. The zeros pull the branches; the asymptote centroid σ = (Σ poles − Σ zeros)/(n_p − n_z) decides whether the loop survives high gain.

Settling time: last time the response leaves the ±1 % band around its own final value. Overshoot is relative to that final value, so P control can "settle" while still missing the reference.