Question 2 of 5: Angle of departure, break-in point, asymptotes and the imaginary-axis crossing

Question 2: Angle of departure, break-in point, asymptotes and the imaginary-axis crossing (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 closely related second- and third-order loops, all inside unity feedback: $K(s+1)/(s^{2}+5s+17.33)$ for the departure angle, the same zero over the less-damped denominator $s^{2}+5s+10$ for the break-in, and $K/[(s+1)(s+2)(s+3)]$ for Part B.

Find. Part A: the departure angles from the complex pole pair, the break-in point on the real axis and the gain there. Part B: the asymptote intercept and angles, the gain range for stability, and the frequency at which the locus crosses the imaginary axis.

Approach. Part A(a) applies the angle condition in the immediate neighbourhood of a pole, which turns it into an equation for the departure direction. Part A(b) and (c) use the fact that a break-away or break-in point is a stationary point of $K(s)$ on the real axis, so it solves $dK/ds=0$; the gain there follows from the magnitude condition. Part B uses the asymptote formulas and the Routh array, and reads the imaginary-axis crossing off the auxiliary equation at the stability limit.

1. Part A (a) — locate the complex poles. Completing the square, $$s^{2}+5s+17.33=(s+2.5)^{2}+11.08 \;\Longrightarrow\; s=-2.500\pm3.329j ,$$ i.e. $\omega_{n}=4.163$ rad/s and $\zeta=0.6005$.
2. Part A (a) — apply the angle condition just off the pole. At a point infinitesimally displaced from the upper pole $p=-2.5+3.329j$ in the direction $\theta_{d}$, the angle condition $\sum\angle\text{zeros}-\sum\angle\text{poles}=-180^\circ$ becomes $$\theta_{z}-\theta_{\bar p}-\theta_{d}=-180^\circ ,$$ where $\theta_{z}$ is measured from the zero and $\theta_{\bar p}$ from the conjugate pole. Substituting $\theta_{z}=\arg(-1.5+3.329j)=114.25^\circ$ and $\theta_{\bar p}=\arg(6.657j)=90^\circ$ gives $$\boxed{\theta_{d}=114.25^\circ-90^\circ-180^\circ=-155.75^\circ \equiv204.25^\circ}$$ and by conjugate symmetry the lower pole departs at $+155.75^\circ$. Tracking the locus numerically for $K=4\times10^{-4}$ reproduces $-155.7^\circ$, confirming the direction.
3. Part A (b) — a break-in point is a stationary point of the gain. Solving the characteristic equation for the gain, $$K(s)=-\frac{s^{2}+5s+10}{s+1} ,$$ and setting $dK/ds=0$ gives $(2s+5)(s+1)-(s^{2}+5s+10)=s^{2}+2s-5=0$, whose roots are $s=-3.449$ and $s=+1.449$. Only the first lies on the real-axis part of the locus (the segment left of the zero at $-1$, which has one critical point to its right), so $$\boxed{s_{\text{break-in}}=-3.449}$$ and $+1.449$ is rejected because the count of real poles and zeros to its right is even.
4. Part A (c) — gain at the break-in point. The magnitude condition evaluated there gives $$K=-\frac{s^{2}+5s+10}{s+1}\bigg|_{s=-3.449} =-\frac{11.90-17.25+10}{-2.449}=\boxed{1.899} .$$ As a check, $s^{2}+5s+10+1.899(s+1)$ factors as $(s+3.449)^{2}$ — a genuine double root, which is what a break-in means.
5. Part B (a) — asymptote intercept and angles. With $n=3$ finite poles at $-1$, $-2$, $-3$ and $m=0$ finite zeros, $$\sigma_{a}=\frac{(-1-2-3)-0}{3-0}=\boxed{-2},\qquad \theta_{q}=\frac{(2q+1)180^\circ}{3}=\boxed{60^\circ,\ 180^\circ,\ -60^\circ} .$$ The centroid coincides with the middle pole here, which is a coincidence of the equally spaced pole set, not a general rule.
6. Part B (b) — Routh test on the closed-loop polynomial. $$1+\frac{K}{(s+1)(s+2)(s+3)}=0 \;\Longrightarrow\; s^{3}+6s^{2}+11s+(6+K)=0$$ The Routh array requires $6\times11>6+K$ and $6+K>0$, hence $$\boxed{-6<K<60}\qquad\text{i.e. }0<K<60\ \text{for positive gain.}$$ Spot checks confirm it: at $K=59$ all three roots are in the left half-plane, at $K=61$ two are not.
7. Part B (c) — the crossing from the auxiliary equation. At the limit $K=60$ the $s^{1}$ row of the array vanishes and the auxiliary polynomial is $6s^{2}+66=0$, so $$\boxed{s=\pm j\sqrt{11}=\pm j3.317\ \text{rad/s}} .$$ Equivalently $|(j\omega+1)(j\omega+2)(j\omega+3)|=60$ at $\omega=3.317$ rad/s with the phase exactly $-180^\circ$: the loop would sustain an undamped oscillation at 0.528 Hz.

Part A's two denominators differ only in their damping, and comparing them shows why the paper splits the question: with $\omega_{n}^{2}=17.33$ the pole pair is far enough off the real axis that the branches leave it steeply and never return, so there is no break-in at all — which is exactly why part (b) changes the denominator to $s^{2}+5s+10$ before asking for one. Part B's answers also interlock: the crossing frequency $\sqrt{11}$ and the limit gain 60 are the same fact read two ways, since $\omega_{pc}$ is where the phase reaches $-180^\circ$ and $K=60$ is where the magnitude reaches unity at that frequency.