Stability by Root Locus, Bode and Routh–Hurwitz (compulsory): 22-Elec-A2 Systems and Control, May 2013, Q1

Question 1 — Stability by Root Locus, Bode and Routh–Hurwitz (compulsory) [7 + 7 + 6]

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. Unit-feedback loop, proportional gain $K_p$, open-loop transfer function $L(s)=K_p\,G(s)$ with $G(s)=\dfrac{100}{(s+2)(s+4)(s+10)}$. Expanding the denominator, $(s+2)(s+4)(s+10)=s^3+16s^2+68s+80$.

Find. Safe gain range, $K_{crit}$, $\omega_{osc}$, and the gain margin at $K_p=2$ — by three independent methods.

Figure Q1.1 — Unit-feedback loop, proportional control, $G(s)=100/[(s+2)(s+4)(s+10)]$.

Part A — Root locus and magnitude criterion [7]

Approach. Three open-loop poles ($-2,-4,-10$) and no finite zeros give three asymptotes; the branches that leave the real axis cross the imaginary axis at $\pm j\omega_{osc}$, and the magnitude criterion evaluated there gives $K_{crit}$.

Real-axis segments, asymptotes and centroid. The locus lies on the real axis to the left of an odd number of poles: on $[-4,-2]$ and on $(-\infty,-10]$. With $n-m=3$ branches to infinity, the asymptote angles are $\theta=\dfrac{(2k+1)180^\circ}{3}=60^\circ,180^\circ,300^\circ$, meeting the real axis at the centroid $$\sigma_a=\frac{(-2)+(-4)+(-10)}{3}=\boxed{-5.33}.$$
Break-away point. Setting $\dfrac{d}{ds}\big[(s+2)(s+4)(s+10)\big]=3s^2+32s+68=0$ gives $s=-2.93$ or $-7.74$; only $s=-2.93$ lies on a locus segment, so the break-away is at $s\approx-2.93$.
Imaginary-axis crossing (read from the sketch). The two branches leaving $-2.93$ sweep up to the asymptotes and cross the $j\omega$-axis. The crossing frequency read from the sketch is $\omega_{osc}\approx 8.2\ \text{rad/s}$ (confirmed analytically as $\sqrt{68}$ in Part C).
Critical gain from the magnitude criterion. On the locus, $K_p=\dfrac{\prod|s-p_i|}{100}$ evaluated at $s=j\sqrt{68}$: $$K_{crit}=\frac{|j\sqrt{68}+2|\,|j\sqrt{68}+4|\,|j\sqrt{68}+10|}{100}=\frac{\sqrt{72}\,\sqrt{84}\,\sqrt{168}}{100}=\boxed{10.08}.$$
Gain margin at $K_p=2$. $$\text{GM}=\frac{K_{crit}}{K_{op}}=\frac{10.08}{2}=5.04\ \text{V/V}=20\log_{10}(5.04)=\boxed{14.05\ \text{dB}}.$$

Figure Q1.2 — Root locus. Branches break away at $-2.93$ and cross $j\omega$ at $\pm 8.25\ \text{rad/s}$ (orange), where $K=K_{crit}=10.08$.

Part B — Bode plot of the open loop [7]

Approach. On the open-loop Bode plot the phase reaches $-180^\circ$ at the phase-crossover frequency $\omega_{pc}$; the gain margin is the amount by which the magnitude sits below 0 dB there, and $K_{crit}$ is the factor that lifts it to 0 dB.

Phase-crossover frequency. $\angle L(j\omega)=-\tan^{-1}\tfrac{\omega}{2}-\tan^{-1}\tfrac{\omega}{4}-\tan^{-1}\tfrac{\omega}{10}=-180^\circ$. Reading the sketch (and solving) gives $\omega_{pc}=\omega_{osc}\approx 8.25\ \text{rad/s}$.
Magnitude there, with $K_p=2$. $|G(j8.25)|=\dfrac{100}{\sqrt{72}\sqrt{84}\sqrt{168}}=0.0992$, so $|L|=2(0.0992)=0.198$. The gain margin is $$\text{GM}=\frac{1}{|L(j\omega_{pc})|}=\frac{1}{0.198}=5.04\ \text{V/V}=\boxed{14.05\ \text{dB}}.$$
Critical gain and oscillation frequency. $K_{crit}=1/|G(j\omega_{pc})|=1/0.0992=10.08$, and the corresponding $\omega_{osc}=8.25\ \text{rad/s}$ — identical to Part A.

Figure Q1.3 — Open-loop Bode plot of $K_pG(s)$ at $K_p=2$. Phase crosses $-180^\circ$ at $\omega\approx8.25$; the magnitude there is $14$ dB below 0 dB (the gain margin).

Part C — Routh–Hurwitz verification [6]

Approach. Build the Routh array of the closed-loop characteristic polynomial and set the $s^1$ row to zero for marginal stability; the auxiliary polynomial from the $s^2$ row gives $\omega_{osc}$.

Characteristic polynomial. $1+K_pG(s)=0\Rightarrow s^3+16s^2+68s+(80+100K_p)=0$.
Routh array. $$\begin{array}{c|cc}s^3&1&68\\ s^2&16&80+100K_p\\ s^1&\frac{16(68)-(80+100K_p)}{16}&0\\ s^0&80+100K_p&\end{array}$$ The $s^1$ entry is $\dfrac{1008-100K_p}{16}$.
Marginal stability. Set it to zero: $1008-100K_p=0\Rightarrow \boxed{K_{crit}=10.08}$.
Oscillation frequency. The auxiliary equation from the $s^2$ row at $K_{crit}$ is $16s^2+(80+1008)=0\Rightarrow s^2=-68\Rightarrow \boxed{\omega_{osc}=\sqrt{68}=8.25\ \text{rad/s}}$.
Safe range. All first-column entries positive requires $0\lt K_p\lt 10.08$.

All three methods agree: $K_{crit}=10.08$, $\omega_{osc}=8.25\ \text{rad/s}$, safe range $0\lt K_p\lt 10.08$, and gain margin $5.04$ V/V $=14.05$ dB at $K_p=2$.

Quantity (all three methods)	Result
Frequency of marginal oscillations	$\omega_{osc}=\sqrt{68}=8.25\ \text{rad/s}$
Critical gain	$K_{crit}=10.08$
Safe operating range	$0\lt K_p\lt 10.08$
Gain margin at $K_p=2$	$\boxed{\text{GM}=5.04\ \text{V/V}=14.05\ \text{dB}}$

Concept & Theory

Topic. Absolute stability of a unit-feedback loop by three equivalent tests.

Key relations. Magnitude criterion $K=\prod|s-p_i|/\prod|s-z_i|$ on the locus; gain margin $\text{GM}=1/|L(j\omega_{pc})|=K_{crit}/K_{op}$; Routh marginal condition (a zero first-column entry) with the auxiliary polynomial giving the jω-axis crossing.

Why this works. Root locus, Bode and Routh–Hurwitz are three views of the same characteristic equation. The point where a root-locus branch crosses $j\omega$, where the Bode phase hits $-180^\circ$, and where the Routh $s^1$ row vanishes are all the single condition "a closed-loop pole sits on the imaginary axis," so they must give the same $K_{crit}$ and $\omega_{osc}$.

Common pitfall. Reading the gain margin as $K_{crit}$ itself instead of the ratio $K_{crit}/K_{op}$, or forgetting that GM in dB is $20\log_{10}$ of the V/V value.

Source. Nise, Control Systems Engineering §6 (Routh–Hurwitz), §8 (root locus), §10 (gain/phase margins).

Question 1 of 8: Stability by Root Locus, Bode and Routh–Hurwitz (compulsory)

Question 1 — Stability by Root Locus, Bode and Routh–Hurwitz (compulsory) [7 + 7 + 6]

Part A — Root locus and magnitude criterion [7]

Part B — Bode plot of the open loop [7]

Part C — Routh–Hurwitz verification [6]