22-Elec-A2 Systems and Control · May 2018

Question 6 of 8: Root Locus of a Plant with a Right-Half-Plane Pole

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, 16-Elec-A2 Systems and Control, May 2018 — 3 hours, closed book (approved Casio or Sharp calculator plus one double-sided handwritten 8.5 × 11″ formula sheet). A short Laplace-transform table is printed on page 2, and the $\zeta$–overshoot, $\zeta$–resonant-peak and phase-margin design charts plus the second-order model, controllability/observability and state-space transfer-function definitions are on page 3. Questions 1 and 2 are compulsory; five questions constitute a complete paper, so a candidate chooses three of Q3–Q8. Each question carries 20 marks, for 100 in total. All eight questions are worked below so the set is a complete study resource.

Reference texts. N. S. Nise, Control Systems Engineering (7th ed., Wiley) — block-diagram and signal-flow reduction with Mason’s rule (Ch. 5), steady-state error and the static error constants (Ch. 7), Routh–Hurwitz (Ch. 6), root locus (Ch. 8), cascade compensator design (Ch. 9), frequency response and gain/phase margins (Ch. 10), frequency-domain lead design (Ch. 11), state space and controllability/observability (Ch. 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — second-order correlations, dominant-poles modelling and pole placement by state feedback. Every block diagram, root locus, pole–zero map, Bode plot, closed-loop magnitude curve and step response below is redrawn as an inline figure.

Reading the supplied design charts. Page 3 gives $PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}$ for percent overshoot, $M_r/K_{dc} = 1/\left(2\zeta\sqrt{1-\zeta^{2}}\right)$ for the closed-loop resonant peak, and the phase-margin correlation $\Phi_m \approx 100\zeta$ (equivalently $\zeta \approx 0.01\,\Phi_m$). Those three relations, together with $T_{settle(\pm 2\%)} \approx 4/(\zeta\omega_n)$ and $T_{rise(0-100\%)} = (\pi - \cos^{-1}\zeta)/\omega_d$, carry most of the numerical work in Questions 2, 5, 7 and 8.

Question 6: Root Locus of a Plant with a Right-Half-Plane Pole (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. Unity-feedback loop under proportional control $K_p$, with $G(s) = 1/[(s-5)(s^{2}+6s+40)]$. The open-loop poles are $s = +5$ and $s = -3 \pm j5.568$ (the roots of $s^{2}+6s+40$); there are no finite zeros.

Find. The critical gain(s) and oscillation frequency(ies); a fully annotated root locus; and a verdict on whether a $\zeta = 0.707$ second-order approximation is attainable.

Approach. Form the closed-loop characteristic polynomial and run a Routh array in $K_p$ — a right-half-plane open-loop pole means the loop is unstable at low gain, so expect a bounded stable window with two critical gains. Then assemble the locus from the standard construction rules and finally test the achievable damping along the complex branch.

Characteristic polynomial. Expanding the open-loop denominator, $$(s-5)(s^{2}+6s+40) = s^{3}+s^{2}+10s-200$$ so the closed-loop characteristic equation $1 + K_pG(s) = 0$ becomes $$\Delta(s) = s^{3} + s^{2} + 10s + (K_p - 200) = 0$$ Note the negative constant term at $K_p = 0$, consistent with the unstable open-loop pole at $+5$.
Routh array and the stable window. $$\begin{array}{c|cc} s^{3} & 1 & 10 \\ s^{2} & 1 & K_p-200 \\ s^{1} & 210-K_p & 0 \\ s^{0} & K_p-200 & \end{array}$$ The $s^{0}$ row requires $K_p \gt 200$ and the $s^{1}$ row requires $K_p \lt 210$, so $$\boxed{\;200 \lt K_p \lt 210\;}$$ The loop is stable only inside a 10-unit window — it is unstable both below and above it. This is the characteristic signature of an open-loop-unstable plant: enough gain is needed to pull the right-half-plane pole across the axis, but too much then drives the complex pair across in the other direction.
The two critical gains and their frequencies. At the lower edge the $s^{0}$ element vanishes, which places a root at the origin: $$K_{crit,1} = 200 \;\Longrightarrow\; \Delta = s(s^{2}+s+10) \;\Longrightarrow\; \boxed{\;\omega_{osc,1} = 0\ \text{rad/s}\;}$$ At the upper edge the $s^{1}$ element vanishes and the auxiliary equation from the $s^{2}$ row gives a genuine oscillation: $$K_{crit,2} = 210 \;\Longrightarrow\; s^{2} + (210-200) = 0 \;\Longrightarrow\; \boxed{\;\omega_{osc,2} = \sqrt{10} = 3.162\ \text{rad/s}\;}$$ Factoring confirms it: at $K_p = 210$, $\Delta = (s^{2}+10)(s+1)$, so the roots are $\pm j3.162$ and $-1$. The lower crossing at $K_p = 200$ is a real-axis crossing, not an oscillation, which is why its frequency is zero.
Root-locus construction rules. With $n = 3$ poles and $m = 0$ finite zeros, all three branches run to infinity along asymptotes: $$\sigma_a = \frac{\sum \text{poles} - \sum \text{zeros}}{n-m} = \frac{5 + (-3) + (-3)}{3} = \boxed{\;-\tfrac{1}{3}\;}, \qquad \theta_a = \frac{(2k+1)180^{\circ}}{3} = \pm 60^{\circ},\ 180^{\circ}$$ The real-axis locus consists of points with an odd number of real poles and zeros to their right; the only real pole is at $+5$, so the entire segment $(-\infty,\,5]$ lies on the locus.
Break points and departure angles. Writing $K_p = -(s-5)(s^{2}+6s+40)$ and differentiating, $$\frac{dK_p}{ds} = 0 \;\Longrightarrow\; 3s^{2}+2s+10 = 0$$ whose discriminant is $4 - 120 = -116 \lt 0$. There are therefore no real break-away or break-in points. This is confirmed directly: along the real-axis locus $K_p(\sigma) = (5-\sigma)(\sigma^{2}+6\sigma+40)$ increases monotonically as $\sigma$ decreases (the quadratic factor has no real roots and stays positive, with a minimum of 31 at $\sigma=-3$), so the single real branch simply travels left without ever meeting another. The angle of departure from the upper complex pole $-3+j5.568$ follows from the angle criterion: $$\theta_d = 180^{\circ} - \left[\angle(p - 5) + \angle(p - \bar{p})\right] = 180^{\circ} - \left[145.2^{\circ} + 90^{\circ}\right] = \boxed{\;-55.2^{\circ}\;}$$ with $+55.2^{\circ}$ from the conjugate.
Assemble the locus. The branch starting at $+5$ moves left along the real axis, reaches the origin at $K_p = 200$ and continues to $-\infty$ along the $180^{\circ}$ asymptote. The complex pair leaves $-3 \pm j5.568$ at $\mp 55.2^{\circ}$, curves right, crosses the imaginary axis at $\pm j3.162$ when $K_p = 210$, and then follows the $\pm 60^{\circ}$ asymptotes into the right half plane. All three branches are accounted for without any break point, which is consistent with the discriminant test.
Is $\zeta = 0.707$ attainable? No — on two independent counts. First, the damping ratio of the complex pair is largest at $K_p = 0$, where the pair sits at the open-loop poles: $$\zeta_{max} = \frac{3}{\sqrt{3^{2}+5.568^{2}}} = \frac{3}{6.325} = 0.474$$ and it decreases monotonically as the gain rises towards the imaginary-axis crossing, reaching zero at $K_p = 210$. The $\zeta = 0.707$ ray (at $45^{\circ}$ to the negative real axis) therefore never intersects the locus at any gain, let alone inside the stable window. Second, even if it did, no dominant-pair approximation is legitimate here: at a representative stable gain of $K_p = 205$ the roots are $-0.244 \pm j3.113$ and $-0.513$, a separation ratio of only $$\frac{0.513}{0.244} = 2.1 \;\lt\; 5$$ so the real pole is as close to the imaginary axis as the complex pair and contributes comparably to the transient. $$\boxed{\;\text{A } \zeta = 0.707 \text{ second-order model is not possible}\;}$$

Question 6 root locus (this replaces the blank grid of Figure Q6.1). Crosses mark the open-loop poles $+5$ and $-3 \pm j5.568$. The real branch crosses the origin at $K_p = 200$; the complex pair departs at $\mp 55.2^{\circ}$ and crosses the imaginary axis at $\pm j3.162$ when $K_p = 210$. Asymptote centroid $-1/3$, asymptote angles $\pm 60^{\circ}$ and $180^{\circ}$; there are no real break points.

Quantity	Result
Characteristic polynomial	$s^{3}+s^{2}+10s+(K_p-200)$
Stable gain range	$200 \lt K_p \lt 210$
$K_{crit,1}$ / $\omega_{osc,1}$	$200$ / $0\ \text{rad/s}$ (real-axis crossing)
$K_{crit,2}$ / $\omega_{osc,2}$	$210$ / $3.162\ \text{rad/s}$
Asymptote centroid / angles	$-1/3$ / $\pm60^{\circ}$, $180^{\circ}$
Break-away / break-in points	none (discriminant $-116 \lt 0$)
Angle of departure	$\mp 55.2^{\circ}$
Maximum $\zeta$ on the complex branch	$0.474$ (at $K_p=0$)
$\zeta = 0.707$ model possible?	No

Topic. Root-locus construction and Routh analysis for a plant with an open-loop right-half-plane pole, and the limits of dominant-pole approximation.

Key relations. Characteristic equation $1+K_pG(s)=0$; asymptote centroid $\sigma_a = (\sum p_i - \sum z_i)/(n-m)$ and angles $(2k+1)180^{\circ}/(n-m)$; break points from $dK_p/ds = 0$; angle of departure $\theta_d = 180^{\circ} - \sum\angle(\text{other poles}) + \sum\angle(\text{zeros})$; $\zeta = \cos\theta$ for a pole at angle $\theta$ from the negative real axis; and the rule of thumb that a pole must be at least five times further left than the dominant pair to be neglected.

Why this works. Every point on the locus satisfies the angle condition, and the construction rules are simply that condition evaluated in limiting cases — far from the poles (asymptotes), on the real axis (the odd-count rule), and immediately adjacent to a complex pole (departure angle). The Routh array supplies the exact gains at which branches cross the imaginary axis, which anchors the sketch quantitatively. The unusual feature of this plant is that stability requires enough gain as well as not too much: at low gain the branch emerging from the pole at $+5$ has not yet been pulled into the left half plane, so the loop is unstable, giving the bounded window $200 \lt K_p \lt 210$. That same geometry explains the answer to part 3, since the complex branch bends towards the imaginary axis as gain rises and its damping can only fall.

Common pitfall. Reporting a single critical gain. Two Routh rows can vanish for different $K_p$, and here they do — missing the lower one at $K_p = 200$ loses half the answer and hides the fact that the loop is unstable at low gain. A related error is quoting a non-zero $\omega_{osc}$ at $K_p = 200$: that crossing is a root passing through the origin, so the oscillation frequency is zero. Finally, students often solve $dK_p/ds = 0$, find the complex roots $-0.333 \pm j1.795$, and plot them as break points; complex solutions of that equation are only genuine break points when the corresponding gain is real and positive, which it is not here.

Source. Nise, Control Systems Engineering 7th ed., Ch. 8 (root-locus sketching rules, break-away points, angles of departure, imaginary-axis crossings via Routh) and §4.8 (validity of the dominant-pole approximation); Ogata 5th ed., Ch. 6.

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