Root Locus Analysis and Gain Selection, Second Order Model (20 marks): 22-Elec-A2 Systems and Control, Undated paper, Q6

Question 6: Root Locus Analysis and Gain Selection, Second Order Model (20 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. A unity-feedback loop stabilised by a proportional controller $K_p$, with process

Quantity	Symbol	Value
Process	$G(s)$	$\dfrac{(s+3)(s+6)}{s^2(s+2)}$
Open-loop poles	$p_i$	$0,\ 0,\ -2$
Open-loop zeros	$z_i$	$-3,\ -6$
Target overshoot	$PO$	$\approx5\%$

Find. (1) the complete root locus with all relevant coordinates; (2) $K_{op}$ for $PO\approx5\%$ plus estimates of $T_{s(\pm5\%)}$, $T_{r(0-100\%)}$ and $e_{ss(step\%)}$; (3) comment on model-versus-actual response.

Figure S6.1 - Figure Q6.1: proportional control of the Question 6 process.

Approach. Build the locus from the standard rules, test for an imaginary-axis crossing with the Routh array, then locate the gain at which the dominant pair reaches the required damping ratio.

Count branches and asymptotes. With $n=3$ poles and $m=2$ zeros there is $n-m=1$ asymptote, at $180^\circ$, and the centroid is$$\sigma_a=\frac{(0+0-2)-(-3-6)}{1}=+7,$$again a degenerate anchor for the single real asymptote.
Real-axis segments. Counting real poles and zeros to the right of a test point: on $(-2,0)$ there are two (the double pole at the origin) — even, so no locus; on $(-3,-2)$ there are three — locus; on $(-6,-3)$ there are four — no locus; and for $s\lt-6$ there are five — locus. Hence the real-axis locus is$$[-3,\,-2]\ \cup\ (-\infty,\,-6].$$
Departure from the double pole at the origin. Two branches leave $s=0$ at $\pm90^\circ$ (a double pole splits perpendicular to the real axis), curve into the left half-plane and eventually return to the real axis far out.
Break points. With $K=-\dfrac{s^2(s+2)}{(s+3)(s+6)}$, solving $dK/ds=0$ gives real roots at $s=0$ ($K=0$), $s=-1.547$ and $s=-3.630$ (both with $K\lt0$, so they are not on the positive-gain locus), and$$\boxed{s=-12.82\ \text{with}\ K=26.55}$$a break-in point where the two complex branches rejoin the real axis; one then moves right to the zero at $-6$ and the other left to $-\infty$.
Test for an imaginary-axis crossing. The characteristic equation is $s^2(s+2)+K(s+3)(s+6)=0$, i.e. $s^3+(2+K)s^2+9Ks+18K=0$. The Routh $s^1$ element is$$c_1=\frac{(2+K)(9K)-18K}{2+K}=\frac{K(18+9K-18)}{2+K}=\frac{9K^2}{2+K},$$which is positive for every $K\gt0$. The first column never changes sign, so$$\boxed{\text{there is no }K_{crit}\text{ and no }\omega_{osc}\text{: the loop is stable for all }K_p\gt0.}$$This is why the question says “if applicable”. Geometrically the same conclusion follows from the asymptote structure: with only one asymptote, at $180^\circ$, no branch has anywhere to go but leftward along the real axis.
Find $K_{op}$ for $PO\approx5\%$ (part 2). Inverting the overshoot relation, $PO=5\%$ needs$$\zeta=\frac{-\ln(0.05)}{\sqrt{\pi^2+\ln^2(0.05)}}=0.6901.$$Sweeping $K$ and tracking the complex pair, the $\zeta=0.6901$ ray meets the locus at$$\boxed{K_{op}=13.70}$$where the closed-loop poles are $s=-6.427\pm j6.740$ and $s=-2.843$, so $\omega_n=|{-6.427+j6.740}|=9.313$ rad/s.
Estimate the specifications from the dominant pair.$$T_{s(\pm5\%)}=\frac{3}{\zeta\omega_n}=\frac{3}{6.427}=0.467\ \text{s},\qquad T_{r(0-100\%)}=\frac{\pi-\arccos\zeta}{\omega_n\sqrt{1-\zeta^2}}=0.346\ \text{s}.$$For the error, the open loop has a double pole at the origin, so the system is Type 2: the position constant $K_{pos}=\lim_{s\to0}K_pG(s)$ is infinite and$$\boxed{e_{ss(step\%)}=0\%}$$A Type-2 loop also tracks a ramp with zero error; only a parabolic input leaves a finite offset.

Figure S6.2 - Root locus of Kp(s+3)(s+6)/(s^2(s+2)). The locus never reaches the jw axis: the loop is stable for every Kp > 0.

Result	Value
Asymptote	1 branch at $180^\circ$, centroid $\sigma_a=+7$
Real-axis locus	$[-3,-2]\cup(-\infty,-6]$
Departure angles at $s=0$ (double)	$\pm90^\circ$
Break-in point	$s=-12.82$ at $K=26.55$
Imaginary-axis crossing	none — stable for all $K_p\gt0$
Design gain	$K_{op}=13.70$
Closed-loop poles	$-6.427\pm j6.740$, $-2.843$
Model $T_{s(\pm5\%)}$ / $T_{r}$	0.467 s / 0.346 s
Steady-state step error	0% (Type 2)

Part 3 — expected versus actual response. The differences here are unusually large and are worth quantifying. Simulating the true closed loop at $K_{op}=13.70$ gives $PO=22.5\%$, $T_{s(\pm5\%)}=0.447$ s and $T_{r(0-100\%)}=0.113$ s, against the model predictions of $5\%$, $0.467$ s and $0.346$ s. Two effects explain the gap. First, dominance is genuinely marginal: the third closed-loop pole sits at $-2.843$, which is closer to the imaginary axis than the pair’s real part of $-6.427$, so the usual five-times separation test fails outright. What rescues the second-order picture is that the open-loop zero at $-3$ is retained in the closed loop and very nearly cancels that pole (separation only $0.157$), leaving a near-dipole whose residue is small. Any claim of dominance here must cite that near-cancellation explicitly. Second, and dominating the overshoot error, the closed-loop transfer function keeps both open-loop zeros, at $-3$ and $-6$. The zero at $-6$ is close to the pair ($\omega_n=9.3$) and differentiates the response, adding a large early transient: this is what lifts the overshoot from $5\%$ to $22.5\%$ and cuts the rise time to a third of the predicted value. The settling time, which depends mainly on the envelope decay $e^{-\zeta\omega_n t}$ rather than on peak shape, is the one estimate that survives well (0.447 s versus 0.467 s predicted). The practical lesson is that the standard $\zeta$-from-$PO$ design rule assumes a pole-only second-order model, and on a plant whose zeros migrate into the closed loop it will systematically under-predict overshoot; the gain should be trimmed downward, or the specification checked by simulation, before committing to $K_{op}$.

Topic: Root-locus gain selection, system Type, and the limits of dominant-pole estimates.

Key relations: Routh $s^1$ row decides the existence of a $j\omega$ crossing; $\zeta=-\ln(PO/100)/\sqrt{\pi^2+\ln^2(PO/100)}$; $T_{s(\pm5\%)}=3/(\zeta\omega_n)$; Type $N$ = number of open-loop poles at the origin, and a Type-2 loop has $e_{ss}=0$ for step and ramp inputs.

Why this works: Closed-loop zeros equal the open-loop zeros for a unity-feedback loop with the controller in the forward path, so the zeros of $G(s)$ appear in $G_{cl}(s)$ and shape the transient even though they do not affect stability.

Common pitfall: Reporting a $K_{crit}$ that does not exist. When the Routh $s^1$ element is positive for all $K\gt0$, the correct answer is to say plainly that no crossing exists and support it with both the Routh row and the asymptote geometry. The second pitfall is quoting the dominant-pole overshoot without checking the closed-loop zeros.

Source: Nise, Control Systems Engineering, 8th ed., Ch. 8 (Root Locus Techniques) and Sec. 4.8 (effect of zeros); Dorf & Bishop, Modern Control Systems, 13th ed., Ch. 7.

Question 6 of 8: Root Locus Analysis and Gain Selection, Second Order Model (20 marks)

Question 6: Root Locus Analysis and Gain Selection, Second Order Model (20 marks)