Recirculation Gain and the Limits of a Linear Stability Certificate
A two-tank thermal loop is linearised about a nominal operating point. The deviation state x = [T1, T2, T3]^T in kelvin satisfies dx/dt = A x, where A = [[0, 1, 0], [0, 0, 1], [-k, -3, -2]] and k > 0 is the recirculation-valve gain in h^-1. At k = 4 the measured settling time is 6.8 min and the response is monotone. At k = 5.8 a student reports a growing 0.9 rad/min oscillation that later appears to persist at constant amplitude. The linear model neglects actuator saturation at ±2 K and a thermocouple lag of 20 s. The design team wants to certify the loop for 0 < k ≤ 5.5 using only this linear model and asks whether the k = 5.8 observation contradicts the model. The characteristic polynomial is s^3 + 2s^2 + 3s + k.
1. Determine the exact range of k for asymptotic stability. What does the linear model predict at k = 6, and how does that compare with the reported 0.9 rad/min oscillation?
Read the worked answer
Form the Routh array for s^3 + 2s^2 + 3s + k:
- s^3: 1, 3
- s^2: 2, k
- s^1: (6 - k)/2, 0
- s^0: k
For asymptotic stability every first-column entry must be positive: k > 0 and 6 - k > 0, so 0 < k < 6. At k = 6 the s^1 row vanishes. The auxiliary equation from the s^2 row is 2s^2 + 6 = 0, giving s = ±j√3, i.e. ω = 1.732 rad/min, period 3.63 min. The reported 0.9 rad/min corresponds to a period of about 6.98 min, roughly twice as long. More importantly, k = 5.8 is inside the stable interval, so the linear model predicts a decaying oscillation, not a constant-amplitude one. The observation therefore does not match the linear prediction; a nonlinear mechanism such as actuator saturation or valve stiction is a better explanation than a simple eigenvalue crossing.
2. The team proposes to certify 0 < k ≤ 5.5. Quantify the stability margin at k = 5.5 and explain why the 20 s thermocouple lag could invalidate that certification. Suggest one conservative action.
Read the worked answer
At k = 5.5 the s^1 entry is (6 - 5.5)/2 = 0.25. All first-column entries are positive, so the loop is stable, but the margin is small: the real part of the critical root pair moves toward zero as k approaches 6. The 20 s lag is about 0.33 min. Near a 1.7 rad/min oscillation, the added phase lag is roughly ωτ = 1.7 × 0.33 ≈ 0.56 rad, or 32 degrees. That phase lag can turn an apparently safe gain into an oscillatory one, especially with a small Routh margin. A purely proportional certificate based on the undelayed polynomial is therefore not conservative. A defensible action is to reduce the certified upper limit, for example to k ≤ 5.0, or to include a first-order sensor lag in the characteristic equation and redo the Routh test. Simply averaging the measured settling time would not address the phase margin.
3. If actuator saturation is causing the constant-amplitude oscillation at k = 5.8, outline how you would estimate the limit-cycle frequency without simulating the full nonlinear system. State what measurement would confirm your estimate.
Read the worked answer
Use a describing-function argument. Replace the saturation element by its equivalent gain N(A), which depends on the assumed oscillation amplitude A. The limit cycle is predicted when the linear loop with gain k·N(A) has a pair of imaginary-axis roots, i.e. when the characteristic polynomial s^3 + 2s^2 + 3s + k·N(A) has roots at ±jω. Substitute s = jω and separate real and imaginary parts: from the imaginary part, ω^2 = 3, so ω = √3 = 1.732 rad/min; from the real part, k·N(A) = 2ω^2 = 6. Solve for A from N(A) = 6/k. At k = 5.8 this gives N(A) ≈ 1.034, which is close to the saturation limit and implies a small-amplitude limit cycle. The predicted frequency is 1.732 rad/min, not 0.9 rad/min. A confirming measurement would be the actual period of the sustained oscillation, together with the amplitude at the valve output; if the period is near 3.63 min and the valve is clipping, the saturation hypothesis is supported.