Root Locus Analysis; Bode Plots and Gain Margin (20 marks): 22-Elec-A2 Systems and Control, Undated paper, Q2

Question 2: Root Locus Analysis; Bode Plots and Gain Margin (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. The same hydraulic loop as Question 1, i.e. open-loop transfer function $L(s)=K_p(s^2+10s+40)/[s(s+1)^2]$. Figure Q2.1 plots the frequency response with $K_p=1$; Figure Q2.2 is blank grid paper for the root-locus sketch.

Find. (1) verification of $K_{crit}$ and $\omega_{osc}$ from the Bode plot; (2) a complete root locus with asymptotes, centroid, break points and the imaginary-axis crossings; (3) confirmation of the stable gain range.

Check — what Figure Q2.1 actually shows. The caption calls it “the process $G(j\omega)$”, but the plotted curve is not $10/(s+1)^2$: that process is Type 0 with a flat 20 dB low-frequency asymptote and $0^\circ$ phase at DC. The printed curve starts near $+70$ dB falling at $-20$ dB/decade with a low-frequency phase of $-90^\circ$, and its phase dips below $-180^\circ$ only between roughly 1.5 and 4.3 rad/s. That is exactly the open-loop response $L(j\omega)=G_{PID}(j\omega)G(j\omega)$ evaluated at $K_p=1$ — the integrator supplies the $-20$ dB/decade slope and the $-90^\circ$ phase floor. All read-backs below use that (self-consistent) interpretation, and they reproduce Question 1’s Routh results to four figures.

[Figure not reproduced: Figure S2.1 - Redrawn Figure Q2.1: open-loop frequency response L(jw) = G_PID(jw)G(jw) at Kp = 1. The phase dips below -180 deg between the two marked crossings; 1/|L| there gives each critical gain. See the official exam paper.]

Approach. On a Bode plot marginal stability occurs where the phase is $-180^\circ$; the critical gain is the reciprocal of the magnitude there. The root locus is then constructed from the standard rules and must place its imaginary-axis crossings at the same frequencies.

Locate the phase crossovers. The phase of $L(j\omega)$ at $K_p=1$ passes through $-180^\circ$ twice, at$$\omega_1=1.456\ \text{rad/s}\qquad\text{and}\qquad\omega_2=4.345\ \text{rad/s},$$which are precisely the two $\omega_{osc}$ values from Question 1. Between them the phase dips to about $-190^\circ$; outside them it returns toward the $-90^\circ$ floor set by the integrator.
Apply the magnitude criterion at each crossover. Marginal stability needs $|K_pL_1(j\omega)|=1$ where $L_1$ is the response at unit gain, so $K_{crit}=1/|L_1(j\omega)|$:$$|L_1(j\omega_1)|=8.940\;(19.03\ \text{dB})\;\Rightarrow\;K_{crit,1}=\frac{1}{8.940}=0.1118,$$$$|L_1(j\omega_2)|=0.5592\;(-5.05\ \text{dB})\;\Rightarrow\;K_{crit,2}=\frac{1}{0.5592}=1.7882.$$Both agree with the Routh values to four significant figures, so $\boxed{K_{crit}=0.1118\text{ and }1.7882}$ is confirmed independently.
Set up the locus. The open-loop poles are $s=0$ and a double pole at $s=-1$; the open-loop zeros are the controller zeros $s=-5\pm j3.873$. With $n=3$ poles and $m=2$ zeros there is $n-m=1$ asymptote, at $180^\circ$.
Centroid. Using $\sigma_a=\dfrac{\sum p_i-\sum z_i}{n-m}$,$$\sigma_a=\frac{(0-1-1)-(-5-5)}{1}=+8.$$With a single $180^\circ$ asymptote the centroid is a degenerate anchor point: the escaping branch simply runs left along the real axis, so the numerical value of $\sigma_a$ carries no geometric meaning here beyond fixing the asymptote direction.
Real-axis segments. A real point lies on the locus when the number of real poles and zeros to its right is odd. There are no real zeros; counting the poles gives the locus on $-1\le s\le 0$ (one pole to the right) and on $s\le-1$ (three poles to the right). The whole negative real axis $(-\infty,0]$ therefore belongs to the locus.
Break points. Writing $K=-\dfrac{s(s+1)^2}{s^2+10s+40}$ and solving $dK/ds=0$ gives real stationary points at $s=-1$ (the double pole itself, $K=0$) and$$\boxed{s_b=-0.3527\ \text{with}\ K=0.00404}$$a break-away point: the branches leaving $s=0$ and $s=-1$ meet there and depart into the complex plane at very low gain.
Imaginary-axis crossings. These are the marginal-stability points already found, $s=\pm j1.456$ at $K_p=0.1118$ and $s=\pm j4.345$ at $K_p=1.7882$. The complex branches therefore leave the real axis at $K=0.004$, cross into the right half-plane at $K_p=0.1118$, cross back out at $K_p=1.7882$, and finally curve in to terminate on the zeros at $-5\pm j3.873$, while the third branch runs left to $-\infty$.

Figure S2.2 - Root locus of L(s) = Kp(s^2+10s+40)/(s(s+1)^2). Two jw-axis crossings (Kp = 0.1118 and 1.7882) bracket an UNSTABLE gain band.

Part (3): the locus construction confirms Question 1 exactly. The closed-loop poles sit in the left half-plane for small gain, migrate into the right half-plane over the band between the two crossings, and return for large gain, so the safe range is again$$\boxed{0\lt K_p\lt 0.1118\quad\text{or}\quad K_p\gt 1.7882}$$

Result	Value
Asymptote count / angle	$n-m=1$ at $180^\circ$
Centroid	$\sigma_a=+8$ (degenerate, single asymptote)
Real-axis locus	$(-\infty,\,0]$
Break-away point	$s=-0.3527$ at $K=0.00404$
Imaginary-axis crossings	$\pm j1.456$ at $K_p=0.1118$; $\pm j4.345$ at $K_p=1.7882$
Bode check of critical gains	$1/8.940=0.1118$; $1/0.5592=1.7882$
Stable range	$0\lt K_p\lt 0.1118$ or $K_p\gt 1.7882$

Topic: Root locus construction and the frequency-domain stability criteria for a conditionally stable loop.

Key relations: Magnitude criterion $|K_pL_1(j\omega_{pc})|=1$ at the phase crossover; asymptote angles $\theta_k=\dfrac{(2k+1)180^\circ}{n-m}$; centroid $\sigma_a=\dfrac{\sum p_i-\sum z_i}{n-m}$; break points from $dK/ds=0$.

Why this works: The root locus and the Bode plot are two views of the same characteristic equation $1+K_pL_1(s)=0$. A jump onto the imaginary axis in the $s$-plane is the same event as the $-180^\circ$ phase crossing in the frequency domain, so the two methods must return identical gains — which is exactly the cross-check part (1) asks for.

Common pitfall: Concluding instability from a single $-180^\circ$ crossing. When the phase dips below $-180^\circ$ and returns, there are two crossings and the gain margin has both a floor and a ceiling; the safe set is a union of intervals, not a single upper bound.

Source: Dorf & Bishop, Modern Control Systems, 13th ed., Ch. 7 (Root Locus), and Nise, Control Systems Engineering, 8th ed., Ch. 10 (Frequency Response Techniques).

Question 2 of 8: Root Locus Analysis; Bode Plots and Gain Margin (20 marks)

Question 2: Root Locus Analysis; Bode Plots and Gain Margin (20 marks)