Question 4 of 5: Bode diagram with a stability limit, and a zero-degree root locus

Question 4: Bode diagram with a stability limit, and a zero-degree root locus (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.

Find. Part 1: the magnitude and phase Bode plots of \(G(j\omega)\), and the largest gain \(K\) that keeps the unity-feedback loop stable. Part 2: a valid sketch of the root locus for negative gain, with the real-axis segments, asymptote count and endpoints justified.

Approach. Part 1 is built from asymptotes — a \(-20\) dB/decade slope from the integrator, steepening by 20 dB/decade at each corner — after which the gain margin is read at the \(-180^\circ\) crossing and cross-checked against Routh–Hurwitz. Part 2 uses the zero-degree versions of the locus rules, in which the real-axis test counts an even number of real poles and zeros to the right.

1. Part 1(a) — put \(G\) in Bode form and locate the corners. Writing the plant with unity constant terms in each factor, $$G(s) = \frac{24}{s(s+2)(s+6)} = \frac{2}{s\left(1+\frac{s}{2}\right)\left(1+\frac{s}{6}\right)} .$$ So the low-frequency behaviour is \(2/s\): a \(-20\) dB/decade line passing through \(20\log_{10}2 = 6.02\) dB at \(\omega=1\) rad/s and 0 dB at \(\omega = 2\) rad/s. The slope steepens to \(-40\) dB/decade at the first corner \(\omega=2\) and to \(-60\) dB/decade at the second corner \(\omega=6\). The phase starts at \(-90^\circ\) from the integrator and falls a further \(90^\circ\) through each corner, approaching \(-270^\circ\) at high frequency.
2. Part 1(a) (continued) — the two crossover frequencies. The phase reaches \(-180^\circ\) when the two arctangents add to \(90^\circ\), which happens exactly when the product of their arguments is one: $$\arctan\frac{\omega}{2} + \arctan\frac{\omega}{6} = 90^\circ \;\Longleftrightarrow\; \frac{\omega}{2}\cdot\frac{\omega}{6} = 1 \;\Longrightarrow\; \omega_{pc} = \sqrt{2\times6} = \sqrt{12} = 3.464 \text{ rad/s} .$$ That closed form is worth remembering: for \(K/[s(s+a)(s+b)]\) the phase crossover is always \(\sqrt{ab}\), independent of the gain. The gain crossover has no closed form and follows from \(|G(j\omega)|=1\), which a bisection on \(24 = \omega\sqrt{\omega^2+4}\,\sqrt{\omega^2+36}\) solves as \(\omega_{gc} = 1.536\) rad/s.
3. Part 1(a) (continued) — read the two margins. At the phase crossover, $$|G(j\omega_{pc})| = \frac{24}{\omega_{pc}\sqrt{\omega_{pc}^2+4}\sqrt{\omega_{pc}^2+36}} = \frac{24}{ab(a+b)} = \frac{24}{2\cdot6\cdot8} = 0.25 ,$$ so $$\boxed{\mathrm{GM} = \frac{1}{0.25} = 4 = 12.04\text{ dB at }\omega_{pc}=3.464\text{ rad/s}}$$ and at the gain crossover the phase is \(-90^\circ - \arctan(1.536/2) - \arctan(1.536/6) = -141.90^\circ\), giving \(\mathrm{PM} = 180^\circ - 141.90^\circ = 38.10^\circ\). Both margins are positive, so the loop as given (\(K=1\)) is stable with modest damping.

4. Part 1(b) — the maximum gain, from the gain margin. Multiplying the plant by \(K\) lifts the magnitude curve by \(20\log_{10}K\) dB without touching the phase, so the loop reaches the verge of instability when the lift equals the gain margin: $$\boxed{K_{\max} = \mathrm{GM} = \frac{ab(a+b)}{24} = \frac{2\cdot 6\cdot 8}{24} = 4}$$ At \(K = 4\) the closed-loop poles sit on the imaginary axis at \(\pm j\omega_{pc} = \pm j3.464\) rad/s, and the loop oscillates at that frequency.
5. Part 1(b) (continued) — confirm with Routh–Hurwitz. The characteristic polynomial is $$s(s+2)(s+6) + 24K = s^3 + 8s^2 + 12s + 24K .$$ The Routh array’s \(s^1\) entry is \((8\times12 - 24K)/8\), which stays positive while \(96 > 24K\), i.e. \(K < 4\); the \(s^0\) entry \(24K\) is positive for any \(K>0\). So the range is \(0 < K < 4\), agreeing exactly with the frequency-domain answer, and at \(K = 4\) the auxiliary polynomial \(8s^2 + 96 = 0\) gives \(s = \pm j\sqrt{12}\) — the same oscillation frequency. ✓
6. Part 2 — establish the pole–zero set and the branch count. Factoring, $$G(s) = \frac{K(s^2+5)(s^2-3)}{(s^2+6)(s^2-4)} = \frac{K(s-j\sqrt5)(s+j\sqrt5)(s-\sqrt3)(s+\sqrt3)} {(s-j\sqrt6)(s+j\sqrt6)(s-2)(s+2)} .$$ There are four finite poles (\(\pm j2.449\), \(\pm2\)) and four finite zeros (\(\pm j2.236\), \(\pm 1.732\)), so \(n_p = n_z = 4\): the locus has four branches, and $$\boxed{n_p - n_z = 0 \text{ — there are no asymptotes}}$$ Every branch therefore terminates on a finite zero. Note also that the plant is open-loop unstable, with a pole at \(s=+2\), and has a right-half-plane zero at \(s=+\sqrt3\).
7. Part 2 (continued) — apply the zero-degree real-axis rule. For \(K<0\) the angle condition becomes \(\sum\theta_z - \sum\theta_p = k\cdot 360^\circ\), so a point on the real axis belongs to the locus when the number of real poles and real zeros strictly to its right is even (the opposite of the familiar \(K>0\) rule). Ordering the real critical points \(-2\) (pole), \(-\sqrt3\) (zero), \(+\sqrt3\) (zero), \(+2\) (pole): $$\boxed{\text{segments: } (-\infty,-2],\quad [-\sqrt3,\;\sqrt3],\quad [2,\;\infty)}$$ The two gaps \((-2,-\sqrt3)\) and \((\sqrt3,2)\) each have an odd count to the right and are not on the negative-gain locus — they belong to the \(K>0\) locus instead.
8. Part 2 (continued) — the imaginary axis is also part of the locus. Putting \(s = j\omega\), $$\frac{G(j\omega)}{K} = \frac{(5-\omega^2)(\omega^2+3)}{(6-\omega^2)(\omega^2+4)} ,$$ which is real for every \(\omega\) — so the whole imaginary axis satisfies the angle condition for real gain, with the sign deciding which half belongs to which locus. The ratio is positive for \(\omega<\sqrt5\) and for \(\omega>\sqrt6\), and negative in between; since \(1+KG=0\) needs \(KG=-1\), the segments \(|\omega|\le\sqrt5\) and \(|\omega|\ge\sqrt6\) belong to the \(K<0\) locus. Hence the branches leave the poles at \(\pm j\sqrt6\) moving outward along the imaginary axis, and separate branches arrive at the zeros \(\pm j\sqrt5\) from the origin side. The origin itself is a closed-loop root at \(K = -1/[G(0)/K] = -1/0.625 = -1.6\).
9. Part 2 (continued) — the branch that passes through infinity. Even though \(n_p = n_z\), one branch does reach infinity: the characteristic polynomial \((s^2+6)(s^2-4) + K(s^2+5)(s^2-3)\) has \(s^4\) coefficient \(1+K\), which vanishes at \(K = -1\). As \(K\) falls from \(0\) to \(-1\), the two real poles at \(\pm2\) run outward along the real axis to \(\pm\infty\); past \(K=-1\) the leading coefficient changes sign and two real roots re-enter from \(\mp\infty\), travelling inward until they finish on the real zeros \(\pm\sqrt3\) as \(K\to-\infty\). That is exactly why both \([2,\infty)\) and \((-\infty,-2]\) are locus segments in a system with no asymptotes.