Question 4 of 6: Identifying a non-minimum-phase plant from three frequency-response points

Question 4: Identifying a non-minimum-phase plant from three frequency-response points (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three measured points of the frequency response of a stable, first-order system whose transfer function may carry a numerator term.

Find. The transfer function $P(s)$, its unit-step response with the key features identified, and whether the loop closed around $C(s) = 1/s$ is stable.

Approach. The DC point fixes the gain and sign; the excess phase lag over the $-90^{\circ}$ a single lag can supply is the decisive clue and points to a right-half-plane zero; two phase readings then determine the two time constants, and the magnitudes provide an independent check. The step response and the closed-loop stability test follow directly.

1. Part (a) — read the DC gain and its sign. From the first row, $$|P(0)| = 10^{9.543/20} = 3.000, \qquad \angle P(0) = 0^{\circ} \;\Longrightarrow\; P(0) = +3 .$$ A zero-degree DC phase rules out any negative gain.
2. Recognise that a plain first-order lag cannot fit the data. The phase of $K/(1+\tau s)$ is $-\arctan(\tau\omega)$, which is bounded below by $-90^{\circ}$. The measurements reach $-139.4^{\circ}$, so an extra $49^{\circ}$ of lag must come from the numerator. A left-half-plane zero adds phase, so the only rational first-order structure that removes it is a zero in the right half-plane: $$P(s) = \frac{K\,(1 - \tau_z s)}{1 + \tau_p s}, \qquad K = 3,\;\; \tau_z,\tau_p \gt 0 .$$
3. Write the phase equation. For this structure both terms contribute lag: $$\angle P(j\omega) = -\arctan(\tau_z\omega) - \arctan(\tau_p\omega).$$ Substituting the two measured phases gives $$\arctan(\tau_z) + \arctan(\tau_p) = 108.4^{\circ}, \qquad \arctan(2\tau_z) + \arctan(2\tau_p) = 139.4^{\circ} .$$
4. Solve the pair. Trying the clean candidate $\tau_z = 1$, $\tau_p = 2$: $$45.00^{\circ} + 63.43^{\circ} = 108.43^{\circ}\;\checkmark, \qquad 63.43^{\circ} + 75.96^{\circ} = 139.40^{\circ}\;\checkmark,$$ both matching the recorded values to the quoted precision.
5. Confirm with the magnitudes, which were not used in the fit. $$|P(j\omega)| = \frac{3\sqrt{1+\omega^{2}}}{\sqrt{1+4\omega^{2}}} \;\Longrightarrow\; |P(j1)| = 1.897 = 5.563\ \text{dB},\quad |P(j2)| = 1.627 = 4.228\ \text{dB}.$$ Both agree with the table, so the identification is over-determined and correct: $$\boxed{P(s) = \frac{3(1-s)}{1+2s} = \frac{-1.5\,(s-1)}{s+0.5}} .$$ The plant is stable (pole at $-0.5$) but non-minimum phase (zero at $+1$).
6. Rule out the competing explanation. Excess lag can also come from a transport delay, $P = 3e^{-\theta s}/(1+\tau s)$. That model is excluded by the magnitudes: a delay is all-pass, so the gain would have to follow $3/\sqrt{1+\tau^{2}\omega^{2}}$, and the value needed to match 5.563 dB at 1 rad/s predicts 1.09 dB at 2 rad/s instead of the measured 4.228 dB. The right-half-plane zero is the only structure that fits gain and phase.
7. Part (b) — compute the unit-step response. With $Y(s) = P(s)/s$, $$Y(s) = \frac{3(1-s)}{s(1+2s)} = \frac{3}{s} - \frac{4.5}{s+0.5} \;\Longrightarrow\; \boxed{y(t) = 3 - 4.5\,e^{-t/2}} .$$
8. Identify the key features of that response. The bi-proper transfer function gives an immediate jump, and the right-half-plane zero makes it go the wrong way: $y(0^{+}) = P(\infty) = -1.5$, an initial undershoot equal to half the final value in the opposite direction. The response then rises monotonically, crossing zero at $$t_{0} = 2\ln\!\frac{4.5}{3} = 2\ln 1.5 = 0.811\ \text{s},$$ with time constant $\tau = 2$ s, final value $y(\infty) = 3$, and 2 % settling at $t_{s} = 2\ln 75 = 8.635$ s. There is no overshoot and no oscillation.
9. Part (c) — close the loop with the integral controller and test stability. With $C(s) = 1/s$ the loop gain is $L(s) = 3(1-s)/[s(1+2s)]$, so the characteristic equation is $$s(1+2s) + 3(1-s) = 2s^{2} - 2s + 3 = 0 .$$ The coefficients are $2$, $-2$, $3$: the sign change means the Routh first column changes sign twice, so there are two right-half-plane roots. Solving confirms $$s = 0.5 \pm j1.118 \;\Longrightarrow\; \boxed{\text{the closed loop is UNSTABLE}} .$$ The cause is structural rather than a matter of tuning: the integrator adds a further $90^{\circ}$ of lag at every frequency on top of a plant that already approaches $-180^{\circ}$, and the right-half-plane zero caps the bandwidth that could otherwise be traded for phase. A lead network, or simply a proportional gain in place of the integrator, would be needed.